Problems Ted Eisenberg, Section Editor *********************************************************

This section of the Journal offers readers an opportunity to exchange interesting mathematical problems and solutions. Please send them to Ted Eisenberg, Department of Mathematics, Ben-Gurion University, Beer-Sheva, Israel or fax to: 972-86-477-648. Questions concerning proposals and/or solutions can be sent e-mail to . Solutions to previously stated problems can be seen at .

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Solutions to the problems stated in this issue should be posted before March 15, 2018

• 5475: Proposed by Kenneth Korbin, New York, NY √ a + b = 14 ab − 48,  √ Given positive integers a, b, c and d such that b + c = 14√bc − 48, c + d = 14 cd − 48, with a < b < c < d. Express the values of b, c, and d in terms of a.

• 5476: Proposed by Ed Gray, Highland Beach, FL

Find all triangles with integer area and perimeter that are numerically equal.

• 5477: Proposed by Daniel Sitaru, “Theodor Costescu” National Economic College, Drobeta Turnu-Sevrin, Meredinti, Romania

Compute: √ √ √ ! 1 − 1 + x2 3 1 + x2 · ... · n 1 + x2 L = lim ln n + lim . n→∞ x→0 x2

• 5478: Proposed by D. M. Btinetu-Giurgiu, “Matei Basarab” National Collge, Bucharest, Romania and Neculai Stanciu, “George Emil Palade” Secondary School, Buzu, Romania

Compute: Z π/2  π  π  cos2 x sin x sin2 cos x + cos x sin2 sin x dx. 0 2 2

• 5479: Proposed by Jos´eLuis D´ıaz-Barrero, Barcelona Tech, Barcelona, Spain

Let f : [0, 1] → < be a continuous convex function. Prove that

2 Z 1/3 3 Z 2/3 5 Z 8/15 f(t)dt + f(t)dt ≥ f(t)dt. 5 0 10 0 8 0

1 • 5480: Proposed by Ovidiu Furdui, Technical University of Cluj-Napoca, Cluj-Napoca, Romania

Let n ≥ 1 be a nonnegative integer. Prove that in C[0, 2π]

span{1, sin x, sin(2x),..., sin(nx)} = span{1, sin x, sin2 x, . . . , sinn x}

if and only if n = 1.

k X We mention that span{v1, v2, . . . , vk} = ajvj, aj ∈ <, j = 1, . . . , k, denotes the set of j=1 all linear combinations with v1, v2, . . . , vk.

Solutions

• 5457: Proposed by Kenneth Korbin, New York, NY

12 Given angle A with sin A = . A circle with radius 1 and a circle with radius x are each 13 tangent to both sides of the angle. The circles are also tangent to each other. Find x.

Solution by Anthony J. Bevelacqua, University of North Dakota, Grand Forks, ND

The angle bisector passes through the centers C and D of the two circles, and the radii from the centers to the points of tangency P and Q of the circles with a side of the angle make right angles 6 CPO and 6 DQO. Thus we have a pair of similar right triangles as follows.

D

C

O P Q

Here 6 DOQ = A/2 and |CD| = 1 + x. Suppose x > 1. Then |CP | = 1 and |DQ| = x. We have

|CP | 1 sin(A/2) = = |CO| |CO|

2 so |CO| sin(A/2) = 1. And |DQ| |DQ| x sin(A/2) = = = |DO| |DC| + |CO| 1 + x + |CO| so sin(A/2) + x sin(A/2) + |CO| sin(A/2) = x. Thus 1 + sin(A/2) x = . 1 − sin(A/2) p Now sin A = 12/13 so cos A =  1 − sin2 A = 5/13, and thus r 1 − cos(A) √ √ sin(A/2) = = 2 13/13 or 3 13/13. 2 Therefore √ √ 13 + 2 13 13 + 3 13 x = √ ≈ 3.491 or x = √ ≈ 10.908. 13 − 2 13 13 − 3 13 If x < 1 then scale the plane by 1/x and appeal to the last paragraph. This gives two more values of x: √ √ 13 − 2 13 13 − 3 13 x = √ ≈ 0.286 or x = √ ≈ 0.092. 13 + 2 13 13 + 3 13

Editor0s Comment : David Stone and John Hawkins of Georgia Southern University, Statesboro, GA generalized the problem. First, they proved the following lemma:

Let A be an angle, 0 < A < π. Suppose a circle C1 of radius r = 1 is inscribed in A and a larger circle C2 of radius R = x is also inscribed in A, with C2 tangent to C1. Then 1 + sin α 1 x = , α = A. 1 − sin α 2

They proved this lemma by showing that it held when angle A is acute and also obtuse. Then they magnified the entire figure by a factor of r, so that the smaller circle C1 has a radius of r and the larger circle C2 has a radius of R = rx, and this allowed them to generalize the lemma: Let A be an angle, 0 < A < π. Suppose that two circles, circle C1 of radius r and C2 of radius R are also inscribed in A, with C2 tangent to C1. Then 1 + sin α 1 R = r, α = A. 1 − sin α 2

They went on to say that “with the result stated in this way, we see the co-dependency between r and R − if we know one we know the other.” Applying the lemma they went on to solve the problem. In conclusion they stated the following: We can apply this result in several interesting ways. For example, as a Corollary, if 1 A = 60◦, then sin α = sin 30◦ = . Let circle C of radius 1 be inscribed in A. Then we 2 0 have a larger inscribed circle C1 tangent to C0 which has radius 1 + sin α 1 + 1/2 R = · 1 = = 3. 1 1 − sin α 1 − 1/2

3 And in continuing on in this manner we have a larger inscribed circle C2 tangent to C1 which has radius 1 + sin α R = · 3 = 32. 2 1 − sin α There is an infinite sequence of expanding inscribed pairwise tangent circles having radii n Rn = 3 , n ≥ 0.

Not to be outdone, we have a smaller inscribed circle C−1, tangent to C0 which has 1 − sin α 1 radius R = · 1 = −1 1 + sin α 3 Continuing, there is an infinite sequence of shrinking inscribed pairwise tangent circles of 1 radii R = , n ≥ 0. −n 3n We could carry out this construction for any angle A, (but the numbers won’t work out so nicely.) In summary they stated: Given values for R and r, we can solve the Corollary equation for α and then find A. That is, given tangent circles of radii r and R, with r < R, we R − r  can compute the angle A which will “circumscribe” the circles: A = 2 sin−1 . R + r

Also solved by Arkady Alt, San Jose, CA; Charles Burnette, Academia Sinica, Taipei, Taiwan; Bruno Salgueiro Fanego, Viveiro, Spain; Ed Gray, Highland Beach, FL; Paul M. Harms, North Newton, KS; David A. Huckaby, Angelo State University, San Angelo TX; Kee-Wai, Lau, Hong Kong, China; David E. Manes, Oneonta, NY; Charles McCracken, Dayton, OH; Vijaya Prasad, Nalluri, India; Trey Smith, Angelo State University, San Angelo, TX; Ioannis D. Sfikas, National and Kapodistrian University of Athens, Greece; Albert Stadler, Herrliberg, Switzerland; David Stone and John Hawkins, Georgia Southern University, Statesboro, GA, and the proposer.

• 5458: Proposed by Michal Kremzer, Gliwice, Silesia, Poland

Find two pairs of integers (a, b) from the set {1, 2, 3, 4, 5, 6, 7, 8, 9} such that for all positive integers n, the number c = 537aaa b . . . b 18403 | {z } is composite, where there are 2n numbers b between a and 1 in the string above.

Solution 1 by Bruno Salgueiro Fanego, Viveiro, Spain

Note that c = 18403 + b · 105 · 1...1 +a · 102n+5 · 111 + 537 · 102n+8 |{z} 2n times = 18403 + b00000 · 1...1 +102n+5 · 537aaa. |{z} 2n times

Thus, if (a, b) ∈ {(2, 7), (9, 7)}, since 18403 ≡7 0, 700000 ≡7 0, 537222 ≡7 0, and 537999 ≡7 0, where ≡7 denotes congruence modulo 7, then 2n+5 c ≡7 0 + 0 · 1 1...1 +10 · 0 ≡7 0, |{z} 2n times

4 so c is divisible by 7 and, hence composite.

Solution 2 by Ed Gray, Highland Beach, FL

The two pairs which guarantee that c = 537aaa bbbbbb . . . bb 18403 is always composite | {z } 2n times are: a = 2, b = 7 and a = 9, b = 7. We will show that with these integers, c is always divisible by 7.

A test for divisibility by 7 is as follows: double the last digit and subtract it from the remaining truncated number. If the result is divisible by 7, then so was the original number. As a simple example, consider the number 826. Double the last digit which gives 12, and subtract it from the leading truncated number, which is 82. Then 82 − 12 = 70, which is divisible by 7, so 826 is divisible by 7.

Now consider our number. It’s last digit is 3, and we double it to get 6. Subtracting 6 from the “truncated” number, we have 537aaa bbbbbb . . . bb 1834. | {z } 2n times We note that 1834 is divisible by 7; that if we let b = 7, every b will be divisible by 7. It remains to find 2 values for a such that 537aaa is divisible by 7. If a = 2, we have the number 537222 = 7 · 76746, and if a = 9, we have the number 537999 = 7 · 76857. This concludes the proof.

Solution 3 by David E. Manes, Oneonta, NY

Two pairs of integers (a, b) that satisfy the problem are (2, 7) and (9, b) where b is any nonnegative integer. For the pair (2, 7), the integer c is always divisible by the prime 7 and for the pair (9, b), c is always divisible by 11. n n−1 Given: N is a positive integer and N = an10 + an−110 + ··· + a110 + a0. Then

N ≡ (100a2 + 10a1 + a0) − (100a5 + 10a4 + a3) + (100a8 + 10a7 + a6) − · · · (mod 7).

For this case, N is divisible by 7 if and only if N ≡ 0 (mod 7). Moreover,

n n−1 N ≡ (−1) an + (−1) an−1 + · · · − a1 + a0 (mod 11) and N is divisible by 11 if and only if N ≡ 0 (mod 11). Let n be a positive integer and define Cn = 537aaab . . . b18403 where the number of digits b is 2n. If a = 2 and b = 7, then C1 = 5372227718403, C2 = 537222777718403 and C3 = 53722277777718403. Therefore, modulo 7,

C1 ≡ 403 − 718 + 227 − 372 + 5 ≡ 4 − 4 + 3 − 1 + 5 ≡ 0 (mod 7),

C2 ≡ 403 − 718 + 777 − 222 + 537 ≡ 4 − 4 + 0 − 5 + 5 ≡ 0 (mod 7),

C3 ≡ 403 − 718 + 777 − 277 + 722 − 53 ≡ 4 − 4 + 0 − 4 + 1 − 4 ≡ 0 (mod 7).

Thus, C1,C2 and C3 are all divisible by 7 and hence, each one is composite. Furthermore, C3n+1 ≡ C1 (mod 7), C3n+2 ≡ C2 (mod 7) and C3n ≡ C3 (mod 7) for all positive integers n. Hence, if a = 2 and b = 7, then Cn is always composite since all of these integers are divisible by 7.

5 If a = 9 and b is any nonnegative integer, then the number of digits in Cn is always odd and

Cn ≡ 5 − 3 + 7 − a + a − a + b − b + ··· + b − b + 1 − 8 + 4 − 0 + 3 ≡ 9 − a (mod 11).

Therefore, for all positive integers n, the prime 11 is a divisor of Cn if and only if a = 9 and the value of b is superfluous. Hence, Cn is always composite.

Solution 4 Anthony J. Bevelacqu, University of North Dakota, Grand Forks, ND

We have 2n z }| { c = 18403 + 105 · b · (1 ··· 1) + 105+2n · (a · 111 + 103 · 537). Note that c > 18403 = 7 · 11 · 239. 2n z }| { Since 10 ≡ −1 mod 11 we have (1 ··· 1) ≡ 0 mod 11 for any n and so c ≡ −(a + 2) mod 11. Thus c will be divisible by 11 when a = 9 for any choice of the digit b and for any non-negative n. Now if b = 7 we have c ≡ 105+2n(6a + 2) mod 7. Thus c will be divisible by 7 when a = 2 for any number of digits b = 7. Therefore c will be composite when (a, b) = (9, b) for any choice of the digit b and when (a, b) = (2, 7).

Editor0s comments : Most of the other solvers of this problem noticed that an even number of b digits forces the number formed by them alone, to be divisible by 11. Hence, they found the value a = 9 makes the number 537aaa18403 divisible by 11, and so (9, any digit) solves the problem. The solutions listed above pick up another ordered pair. But then The Honor Students at Ashland University in Ashland, Ohio upped the ante by finding additional ordered pairs to (9, any other digit). Using MAPLE they found 6 pairs of values for (a, b) that satisfy the problem. They checked these values for all positive integers n ≤ 25. Letting c = 537aaa b . . . b 18403 they found | {z } that:

(a, b) c is divisible by (2, 7) 7    (4, 1) 29    (4, 5) 13    (6, 5) 17    (6, 9) 59  (7, 5) 89 David Stone and John Hawkins of Georgia Southern University, Statesboro, GA found all of the solutions in the above table and an additional one (4, 8), which is divisible by 13. They also found that if b = 0 were allowed, then (2, 0) is divisible by 7, for all n ≥ 0. With respect to the pair (4, 8) they stated that it seemingly has a unique property. If cn = 537aaab....b18403 as defined in the problem, then no single prime divides all cn, but 7 divides all c3k, 3 divides all c3k+1, and and 13 divides all c3k+2.

Also solved by Brian D. Beasley, Presbyterian College, Clinton, SC; Pat Costello, Eastern Kentucky University, Richmond, KY; Kee-Wai Lau, Hong

6 Kong, China; Zachary Morgan, student at Eastern Kentucky University, Richmond, KY; Nathan Russell, Eastern Kentucky University, Richmond, KY; Albert Stadler, Herrliberg, Switzerland; David Stone and John Hawkins, Georgia Southern University, Statesboro, GA, and the proposer.

• 5459: Proposed by Arsalan Wares, Valdosta State University, Valdosta, GA

Triangle ABC is an arbitrary acute triangle. Points X,Y , and Z are midpoints of three sides of 4ABC. Line segments XD and XE are perpendiculars drawn from point X to two of the sides of 4ABC. Line segments YF and YG are perpendiculars drawn from point Y to two of the sides of 4ABC. Line segments ZJ and ZH are perpendiculars drawn from point Z to two of the sides of 4ABC. Moreover, P = ZJ ∩ FY,Q = ZH ∩ DX, and R = YG ∩ XE. Three of the triangles, and three of the quadrilaterals in the figure are shaded. If the sum of the areas of the three shaded triangles is 5, find the sum of the areas of the three shaded quadrilaterals.

7 Solution 1 by David A. Huckaby, Angelo State University, San Angelo, TX

Let a be the area of triangle ABC. Since Y and Z are the midpoints of AC and AB, respectively, 4AY Z is similar to 4ACB with a scale factor of 2, so that the area of 1 1 4AY Z is 4 a. Similarly, the areas of 4BXZ and 4CXY are each 4 a, and therefore the 1 area of 4XYZ is also 4 a.

1 The area of rectangle GY ZH is 2 a, since it has the same height and base as 4XYZ. 1 Similarly, the areas of rectangles FYXD and EXZJ are each 2 a.

Consider the sum of the areas of these three rectangles:

area of three rectangles = area of six outer white triangles +2(area of three pink triangles) + 3(area of 4XYZ),

that is,

1 1 3( a) = area of six outer white triangles + 2(5) + 3( a), 2 4

3 so that the sum of the areas of the six outer white triangles is 4 a − 10.

Now consider the sum of the areas of triangles AY Z, BXZ, and CXY :

area of triangles AY Z, BXZ, and CXY = area of six outer white triangles +area of three pink triangles + area of three blue quadrilaterals,

that is,

1 3  3( a) = a − 10 + 5 + area of three blue quadrilaterals, 4 4

so that the sum of the areas of the three blue quadrilaterals is 5.

8 Solution 2 by Andrea Fanchini, Cant´u,Italy

We use barycentric coordinates and the usual Conway’s notations with reference to the triangle ABC. Then we have X(0 : 1 : 1),Y (1 : 0 : 1),Z(1 : 1 : 0). • Coordinates of points D, E, F, G, H, J. Line segments XD and XE perpendiculars drawn from point X to two of the sides of 4ABC are

2 2 XAB∞⊥ :(c + SA)x − SBy + SBz = 0, XAC∞⊥ :(b + SA)x + SC y − SC z = 0 therefore the points D,E have coordinates

2 2 D = XAB∞⊥ ∩ AB = (SB : c + SA : 0),E = XAC∞⊥ ∩ AC = (SC : 0 : b + SA) then cyclically

2 2 2 2 G = (0 : SC : a +SB),J = (b +SC : 0 : SA),F = (c +SB : SA : 0),H = (0 : a +SC : SB)

• Coordinates of point P, Q, R. Coordinates of point P are

2 2 P = ZAC∞⊥ ∩ Y AB∞⊥ = (2S − a SA : SASC : SASB) then cyclically

2 2 2 2 Q = (SBSC : 2S − b SB : SASB),R = (SBSC : SASC : 2S − c SC ) • Areas of the three shaded triangles. Areas of the three shaded triangles are S S S S S S [PZY ] = B C [ABC], [QZX] = A C [ABC], [RXY ] = A B [ABC] 4S2 4S2 4S2 If the sum of the areas of the three shaded triangles is 5, we have

[ABC] [PZY ] + [QZX] + [RXY ] = , ⇒ [ABC] = 20 4 • Areas of the three shaded quadrilaterals. Area of the quadrilateral [AF P J] is given from [AF J] + [PFJ] so

S2 S2 S S S2 (S2 + S S ) [AF J] = A [ABC], [PFJ] = A B C [ABC], ⇒ [AF P J] = A B C [ABC] 4b2c2 4b2c2S2 4b2c2S2 then cyclically

S2 (S2 + S S ) S2 (S2 + S S ) [BDQH] = B A C [ABC], [CERG] = C A B [ABC] 4a2c2S2 4a2b2S2 therefore a2S2 (S2 + S S ) + b2S2 (S2 + S S ) + c2S2 (S2 + S S ) [AF P J]+[BDQH]+[CERG] = A B C B A C C A B [ABC] 4a2b2c2S2 but [ABC] = 20 then

S2(a2S2 + b2S2 + c2S2 ) + S S S (a2S + b2S + c2S ) [AF P J]+[BDQH]+[CERG] = 5 A B C A B C A B C a2b2c2S2

9 2 2 2 2 now a SA + b SB + c SC = 2S then a2S2 + b2S2 + c2S2 + 2S S S [AF P J] + [BDQH] + [CERG] = 5 A B C A B C a2b2c2 2 2 2 2 2 2 2 2 2 finally a SA + b SB + c SC + 2SASBSC = a b c so we have that also [AF P J] + [BDQH] + [CERG] = 5.

Solution 3 by Nikos Kalapodis, Patras, Greece

We denote with [S] the area of shape S. Let XL, YM and ZN be the heights of triangle XYZ and K its orthocenter. Then the quadrilaterals PZKY , QXKZ and RYKX are parallelograms. It follows that [PZY ] = [KZY ], [QZX] = [KZX], and [RXY ] = [KXY ]. Therefore [PZY ] + [QZX] + [RXY ] = [KZY ] + [KZX] + [KXY ] = [XYZ] (1). Furthermore since the triangles AZY , BXZ, CYX and XYZ are congruent with orthocenters P , Q, R and K respectively, it easily follows that [AF P J] = [XNKM], [BHQD] = [Y LKN] and [CERG] = [ZMKL]. Therefore [AF P J] + [BHQD] + [CERG] = [XNKM] + [Y LKN] + [ZMKL] = [XYZ] (2). From (1) and (2) we get that [AF P Z] + [BHQD] + [CERG] = [PZY ] + [QZX] + [RXY ] = 5.

Solution 4 by Angel´ Plaza, University of Las Palmas de Gran Canaria, Spain

Triangles ∆AZY , ∆ZBX and ∆YXC are equal. Also the sum of the areas of the saded triangles is equal to the area of for example ∆AZY . Also the sum of the areas of the

10 three shaded quadrilaterals is equal to the area of one of the triangles, for example ∆AZY . Therefore, the requested sum is equal to 5.

Also solved by Bruno Salgueiro Fanego, Viveiro, Spain; Ed Gray, Highland Beach, FL; Kee-Wai, Lau, Hong Kong, China; David E. Manes, Oneonta, NY; Sachit Misra, Nelhi, India; Neculai Stanciu,“George Emil Palade” School, Buzau,˘ Romania and Titu Zvonaru, Comanesti,˘ Romania; David Stone and John Hawkins, Georgia Southern University, Statesboro, GA and the proposer.

• 5460: Proposed by Angle´ Plaza,Universidad de Las Palmas de Gran Canaria, Spain

If a, b > 0 and x, y > 0 then prove that

a3 b3 a2 + b2 + ≥ . ax + by bx + ay x + y

Solution 1 by Dionne Bailey, Elsie Campbell, and Charles Diminnie, Angelo State University, San Angelo,TX

Since a, b, x, y > 0, we have

a3 (bx + ay)(x + y) + b3 (ax + by)(x + y) − a2 + b2
(ax + by)(bx + ay) = a2 (bx + ay)[a (x + y) − (ax + by)] + b2 (ax + by)[b (x + y) − (bx + ay)] = a2 (a − b) y (bx + ay) + b2 (b − a) y (ax + by) = (a − b) y a2 (bx + ay) − b2 (ax + by) = (a − b) y ab (a − b) x + a3 − b3
y = (a − b)2 y abx + a2 + ab + b2
y ≥ 0, (1)

with equality if and only if a = b.

Since a, b, x, y > 0, we need only to divide (1) by the positive quantity (ax + by)(bx + ay)(x + y) and to re-arrange terms to obtain the desired inequality. Further, equality is attained if and only if a = b.

Solution 2 by Henry Ricardo, Westchester Area Math Circle, NY

Using the Engel form of the Cauchy-Schwarz inequality (or Bergstr¨om’sinequality) and the AGM inequality, we see that

a3 b3 a4 b4 + = + ax + by bx + ay a2x + aby b2x + aby (a2 + b2)2 ≥ (a2 + b2)x + 2aby (a2 + b2)2 a2 + b2 ≥ = . (a2 + b2)x + (a2 + b2)y x + y

Solution 3 by Anna Valkova Tomova, Varna, Bulgaria

11 We move the expression to the left of the right side of the inequality. Now we have to prove that the new left part is non-negative. Again we will use the capabilities of the mathematical site to examine the transformed look of this new left-hand side of the inequality.

a3 b3 a2 + b2 + − ax + by bx + ay x + b

y(a4y + a3bx − a3by − 2a2b2x + ab3x − ab2y + b4y) = . (x + y)(ay + bx)(ax + by)

y(a − b)2(a2y + abx + aby + b2y) = . (x + y)ay + bx)(ax + by)

Since the numbers involved in the expression are conditionally positive, we have proved the inequality because the equivalent expression is positive too.

Conclusion: The application of information technology enhances the quality of education in mathematics in all of its stages of study. Of course, it should be checked at every stage so as not to allow ridiculous errors. In this sense, “E-Mathematics” does not replace the classic, it continues development with new, more efficient vehicles.

Editor0s Comments : Brian Bradie of Christopher Newport University in Newport News VA stated that this problem is a generalization of two inequalities that appeared in Problem B-1201 in the February 2017 issue of the Fibonacci Quarterly:

a3 b3 a2 + b2 + ≥ aFn + bFn+1 bFn + aFn+1 Fn+2

a3 b3 a2 + b2 + ≥ aLn + bLn+1 bLn + aLn+1 Ln+2

Three other generalizations of this problem were made by D.M.Batinetu-Giurgiu˘ of the “Matei Basarab” National College in Bucharest, Romania.

1. A generalization with “two variables:” am+2 bm+2 a2 + b2 If m ≥ 0 and a, b, x, y > 0, then + ≥ . (ax + by)m (bx + ay)m (x + y)m Proof: am+2 bm+2 a2m+2 b2m+2 + = + (ax + by)m (bx + ay)m (a2x + aby)m (b2x + aby)m

J.Radon AM≥GM (a2)m+1 (b2)m+1 z}|{ (a2 + b2)m+1 z}|{ (a2 + b2)m+1 = + ≥ ≥ (a2x + aby)m (b2x + aby)m (a2 + b2)x + 2aby)m ((a2 + b2)x + (a2 + b2)y)m

12 (a2 + b2)m+1 a2 + b2 = = . Q.E.D. (a2 + b2)m(x + y)m (x + y)m

Corollary 1. If m = 1, then we obtain the problem 5460.

2. A generalization with “three variables:”

If m ≥ 0 and a, b, c, x, y, z > 0, then am+2 bm+2 cm+2 a2 + b2 + c2 + + ≥ . (ax + by + cz)m (bx + cy + az)m (cx + ay + bz)m (x + y + z)m Proof: am+2 bm+2 cm+2 + + (ax + by + cz)m (bx + cy + az)m .(cx + ay + bz)m

(a2)m+1 (b2)m+1 (c2)m+1 = + + (a2x + aby + acz)m (b2x + bcy + abz)m (c2x + acy + bcz)m

J.Radon z}|{ (a2 + b2 + c2)m+1 ≥ ((a2 + b2 + c2)x + (ab + bc + ca)y + (bc + ca + ab)z)m

(a2 + b2 + c2)m+1 a2 + b2 + c2 ≥ = , Q.E.D. (a2 + b2 + c2)m(x + y + z)m (x + y + z)m

In the last inequality we are utilizing the fact that a2 + b2 + c2 ≥ ab + bc + ca where a, b, c > 0.

3. A generalization with “n variables:”

If t, x, y, ak > 0, n ∈ N, n ≥ 2, n ∈ {1, 2, . . . , n} such that n n X 2 X t ak ≥ akak+1, an+1 = a1, then t=1 k=1

n 3 n X ak 1 X 2 ≥ ak. xak + yak+1 x + ty k=1 k=1

Proof:

n !2 X 2 Bergstrom ak n 3 n 2 2 X ak X (ak) z}|{ k=1 = 2 ≥ n xak + yak+1 xa + ya a k=1 k=1 k k k+1 X 2 (xak + yakak+1) k=1

13 n !2 n !2 X X a2 a2 k k n 1 X k=1 = k=1 = a2, Q.E.D. n n n n x + ty k X 2 X X 2 X 2 k=1 x ak + y akak+1 x ak + ty ak k=1 k=1 k=1 k=1

Also solved by Arkady Alt (3 solutions), San Jose, CA; Bruno Salgueiro Fanego Viveiro, Spain; D.M.Batinetu-Giurgiu˘ of the “Matei Basarab” National College in Bucharest, Romania; D.M.Batinetu-Giurgiu˘ of the “Matei Basarab” National College in Bucharest, Romania with Neculai Stanciu, “George Emil Palade” School, Buzau,˘ Romania; Brian Bradie, Christopher Newport University, Newport News, VA; Ed Gray, Highland Beach, FL; Paul M. Harms, North Newton, KS; Kee-Wai, Lau, Hong Kong, China; Nikos Kalapodis, Patras, Greece; Moti Levy, Rehovot, Israel; David E. Manes, Oneonta, NY; Paolo Perfetti, Department of Mathematics, Tor Vergata University, Rome, Italy; Ioannis D. Sfikas, National and Kapodistrian University of Athens, Greece; Albert Stadler, Herrliberg, Switzerland; Neculai Stanciu, “George Emil Palade” School in Buzau, Romanina with Titu Zvonaru of Comanesti,˘ Romania; and the proposer.

• 5461: Proposed by Jos´eLuis D´ıaz-Barrero, Barcelona Tech, Barcelona, Spain

Compute the following sum: ∞ X cos (2n − 1) 2 . n=1 (2n − 1)

Solution 1 by Brian Bradie, Christopher Newport University, Newport, VA

π Consider the function f(x) = − x on the interval [0, π]. Because 2 2 Z π π  − x dx = 0 π 0 2 and, for positive integer n, ( 4 2 Z π π  , n odd − x cos nx dx = n2π π 0 2 0, n even

it follows that the Fourier cosine series for f is

∞ 4 X cos(2n − 1)x . π (2n − 1)2 n=1 The function f is continuous at x = 1, so

∞ 4 X cos(2n − 1) f(1) = ; π (2n − 1)2 n=1

14 therefore, ∞ X cos(2n − 1) π π π  = f(1) = − 1 . (2n − 1)2 4 4 2 n=1

Solution 2 by Ed Gray, Highland Beach, FL

Many of these infinite series can be solved by finding a function whose Fourier series expansion results in the given series. The series at hand reprsents an even function so suitable candidates are functions like f(x) = x2, f(x) = |x|, etc. A perusal of some functions reveals that the function f(x) = |x|, π < x < π, seems just was we need.

The expression is: ∞ π 4 X cos(kx) 1. f(x) = − . 2 π k2 k≥1,odd Since the sum involves odd terms only, we let k = 2n − 1. Further, we eliminate x by letting x = 1.(Since an even function, x = −1 would do just as well.) In either case, f(1) = f(−1) = |1| = 1 and equation (1) becomes: ∞ π 4 X cos(2n − 1) 2. 1 = − , or 2 π (2n − 1)2 n=1 ∞ 4 X cos(2n − 1) π 3. = − 1. π (2n − 1)2 2 n=1 π Multiplying by . 4 ∞ X cos(2n − 1) π2 π 4. = − . (2n − 1)2 8 4 n=1

Solution 3 by Paolo Perfetti, Department of Mathematics, Tor Vergata University, Rome, Italy

∞ ∞ X cos(2n − 1) 1 X Z 1 Z 1 = (ei(2n−1) + e−i(2n−1)) t2n−2dt u2n−2du (2n − 1)2 2 n=1 n=1 0 0

∞ ∞ e−i Z 1 dt Z 1 du X ei Z 1 dt Z 1 du X = (tuei)2k + (tue−i)2k 2 t2 u2 2 t2 u2 0 0 n=1 0 0 n=1

e−i Z 1 Z 1 e2i ei Z 1 Z 1 e−2i = dt du 2 2i + dt du 2 −2i . 2 0 0 1 − (tu) e 2 0 0 1 − (tu) e

The change x = tu, y = u yields

ei Z 1 dy Z y 1 e−i Z 1 dy Z y 1 dx 2 2i + dx 2 −2i 2 0 y 0 1 − x e 2 0 y 0 1 − x e

15 i Z 1 Z y   −i Z 1 Z y   e dy 1 i e dy 1 1 = dx i + 11 + xe + dx −i + −i 4 0 y 0 1 − xe 4 0 y 0 1 − xe 1 + xe

1 Z 1 dy h i 0 1 Z 1 dy h i y i i = Ln(1 − xe + Ln(1 + xe 4 0 y y 4 0 y 0

1 Z 1 dy h i 0 1 Z 1 dy h i y −i −i = Ln(1 − xe + Ln(1 + xe 4 0 y y 4 0 y 0

−1 Z 1 Ln(1 − yei) 1 Z 1 Ln(1 + yei) −1 Z 1 Ln(1 − ye−i) 1 Z 1 Ln(1 + ye−i) = dy + dy + dy + dy 4 0 y 4 0 y 4 0 y 4 0 y

i i −i −i −1 Z e Ln(1 − y) 1 Z −e Ln(1 − y) −1 Z e Ln(1 − y) 1 Z −e = dy + dy + dy + Ln(1 − y)ydy 4 0 y 4 0 y 4 0 y 4 0

1 1 1 1 = Li (ei) − Li (−ei) + Li (e−i) − d Li (−e−i). 4 2 4 2 4 2 4 2

The relation

1 π2 (Ln(−z))2 Li ( ) + Li (z) = − − 2 z 2 6 2 gives

1  π2 (Ln(−ei))2  1  π2 (Ln(ei))2  1 i2 π2 − 2π − − − − − = − (i(π − 1))2 + = . 4 6 2 4 6 2 8 8 8

Solution 4 by Ioannis D. Sfikas, National and Kapodistrian University of Athens, Greece

A given 2π-periodic function f can be represented as by the convergent series

∞ a0 X f(x) = + [a cos(nx) + b sin(nx)]. 2 n n n=1

The convergence of the series means that the sequence (sn(x)) of partial sums, defined by

n a0 X s (x) = + [a cos(kx) + b sin(kx)], n 2 n n k=1 converges at a given point x to f(x), sn(x) → f(x). Consider f to be a 2π-periodic function defined by f(x) = |x| for x ∈ [−π, π]. Note that f is even. Since the product of Z n even functions with the odd function is odd, it follows that f(x) sin(nx)dx = 0. −n

Hence bn = 0 for all n ≥ 1. To compute an note that the product of an even function with an even function is even, so that 1 Z n 2 Z π 2 Z π an = f(x) cos(nx)dx = f(x) cos(nx)dx = x cos(nx)dx. n −n n 0 n 0

16 2 Z π 4 If n = 0 then a0 = xdx = π. Hence an = 0 if n is even and an = − 2 when n is n 0 n π odd, and hence ∞ π 4 X cos(2n − 1)x f(x) ∼ − . 2 π (2n − 1)2 n=1

1 Since the series P converges, the M−Weierstarss test implies that the above series n2 converges. Furthermore, f is continuous at every point and it is smooth except at points mπ, with m odd. Hence the Fourier series of f converges to f at every point. In particular, ∞ π 4 X cos(2n − 1)x |x| = − , 2 π (2n − 1)2 n=1 for x ∈ [−π, π]. Substituting x = 0, we find that

∞ X 1 n2 = , (2n − 1)2 8 n=1 and substituting x = 1, we find that

∞ X cos(2n − 1) π π  = − 1 ≈ 0.448302. (2n − 1)2 4 2 n=1

Also solved by Bruno Salgueiro Fanego Viveiro, Spain; Kee-Wai Lau, Hong Kong, China; Moti Levy, Rehovot, Israel; Henry Ricardo, Westchester Area Math Circle, NY; Albert Stadler, Herrliberg, Switzerland; Anna V. Tomova, Varna, Bulgaria, and the proposer.

• 5462: Proposed by Ovidiu Furdui, Technical University of Cluj-Napoca, Cluj-Napoca, Romania

Let n ≥ 1 be an integer. Calculate

π Z 2 cos x  n dx. 0 1 + psin(2x)

Solution 1 by Moti Levy, Rehovot, Israel

First simplification by setting t = 4x − π, √ Z π Z π t π
Z π t 2 cos x 1 cos 4 + 4 2 cos 4 In :=  √ n dx =  q n dt =  q n dt 0 1 + sin 2x 4 −π t 4 0 t 1 + cos 2 1 + cos 2 √ 1− cos t √ 2 Further simplification by the change of variable w = t : 1+ cos 2 √ s s t 1 − w 2 t 2 1 − w 2 t 1 − w 4 cos = , cos = 1 + , sin = 1 − 2 1 + w 4 2 1 + w 2 1 + w

17 1 Z 1  1 √  n−2 √ In = n (1 + w) − w dw. 2 0 w By the binomial theorem,

n−2 X n − 2 (1 + w)n−2 = wk, n ≥ 2 k k=0 after interchanging integration and summation,

n−2   Z 1 1 X n − 2  k− 1 k+ 1  I = w 2 − w 2 dw n 2n k k=0 0 n−2 ! 1 X n − 2 1 1 = − 2n k k + 1 k + 3 k=0 2 2 n−2 n−2
1 X = k , n ≥ 2. 2n k + 1
k + 3
k=0 2 2 For n = 1,

1 Z 1 1 − w 1 Z 1 1 − x2 √ I1 = dw = 2 dx 2 0 1 + w w 0 1 + x Z 1  2  π = 2 − 1 dx = − 1. 0 1 + x 2

Solution 2 by Albert Stadler, Herrliberg, Switzerland π The transformation x → − x yields π 2 π Z 2 cos x Z 2 sin x  n dx =  n dx. 0 1 + psin(2x) 0 1 + psin(2x)

Therefore

π π π p Z 2 cos x 1 Z 2 cos x + sin x 1 Z 2 (cos x + sin x)2  n dx =  n dx =  n dx 0 1 + psin(2x) 2 0 1 + psin(2x) 2 0 1 + psin(2x)

π p π p π √ 1 Z 2 (1 + sin(2x) Z 4 (1 + sin(2x) 1 Z 2 1 + sin x =  n dx =  n dx =  √ n dx 2 0 1 + psin(2x) 0 1 + psin(2x) 2 0 1 + sin x

√ y= sin x Z 1 p 2 Z 1 z}|{ 1 + y y 1 y = n · p dy = n · p dy 0 (1 + y) 1 − y4 0 (1 + y) 1 − y2

2z y= 2 1+z Z 1 1 2z 1 + z2 2(1 − z2) Z 1 z(1 + z2)n−2 Z ∞ z(1 + z2)n−2 z}|{= · · · dz = 4 dz = dz, 2z 1 + z2 1 − z2 (1 + z2)2 (1 + z)2n (1 + z)2n 0 1 + (1+z2)n 0 0 Z 1 z(1 + z2)n−2 Z ∞ z(1 + z2)n−2 where we have used that n dz = n dz, 0 (1 + z) 0 (1 + z)

18 as follows by performing the change of variables z → 1/z.

π Z 2 cos x Obviously, for n = 0, dx = 1.  0 0 1 + p(2x)

For n = 1 we have π ∞ ∞ Z 2 cos x Z z Z  1 1  π  dx = 2 2 2 dz = 2 − 2 dz = − 1. 0 1 + p(2x) 0 (1 + z) (1 + z ) 0 1 + z (1 + z) 2

For n ≥ 2 we use the Binomial Theorem to expand the integrand. Z ∞ z(1 + z2)n−2 Z ∞ z(1 + 2z + z2 − 2z)n−2 2 2n dz = 2 2n dz 0 (1 + z) 0 (1 + z) n−2 n−2 X n − 2 Z ∞ zj+1 X n − 2 (j + 1)!2 =2 (−2)j dz = 2 (−2)j j (1 + z)2j+4 j (2j + 3)! j=0 0 j=0 n−2 X (j + 1)! = −(n − 2)! (−2)j+1(j + 1) , (n − 2 − j)!(2j + 3)! j=0 where we have used that Z ∞ zj+1 j + 1 Z ∞ zj (j + 1)j Z ∞ z(j−1) dz = dz = dz = 2j+4 2j+3 (2j+2) 0 (1 + z) 2j + 3 0 (1 + z) (2j + 3)(2j + 2) 0 (1 + z) (j + 1)j Z ∞ 1 (j + 1)!2 ... = dz = , applying repeated (j+3) (2j + 3)(2j + 2) ··· (j + 3) 0 (1 + z) (2j + 3)! integration by parts. So the integral evaluates to a rational number for all natural numbers n except for n = 1.

Solution 3 by Kee-Wai Lau, Hong Kong, China

Denote the integral of the problem by In. We show that

 π − 2  , for n = 1  2   n−1 n−1
In = X (1) n k − 2n−1  2k + 1  k=0  , for n ≥ 2.  (n − 1)2n−1

π Z 2 sin x π Let Jn =  n dx. By substituting x = − y into In, we see that 0 1 + psin(2x) 2 2 In = Jn. Since 1 + sin(2x) = (cos x + sin x) , so √ Z π p Z π In + Jn 1 2 1 + sin(2x) 1 1 + sin x In = =  n dx =  √ n dx. 2 2 0 1 + psin(2x) 4 0 1 + sin x

By putting x = π − y we see that √ √ Z π 1 + sin x Z π/2 1 + sin y  √ n dx = √
n dy. π 1 + sin y 2 1 + sin x 0

19 √ Z π/2 1 1 + sin x 2 Hence, In =  √ n dx. By putting sin x = cos θ so that 2 0 1 + sin x Z π/2 cos θdθ cos xdx = −2 sin θ cos θdθ and so In = n . We have 0 (1 + cos θ)

Z π/2 Z π/2 Z /2 cos θdθ π dθ π 1 2 θ π − 2 I1 = = − = − sec d = . 0 1 + cos θ 2 0 1 + cos θ 2 2 0 2 2

For n ≥ 2, integrating by parts, we have

Z π/2 cos θdθ Z π/2 d(sin θ) Z π/2 sin2 θdθ Z π/2 (1 − cos θ)dθ In = n = n = 1−n n+1 = 1−n n . 0 (1 + cos θ) 0 (1 + cos θ) 0 (1 + cos θ) 0 (1 + cos θ)

Z π/2 Z π/4 dθ n 2n Hence (1 − n)In = 1 − n n = 1 − n−1 sec θdθ. 0 (1 + cos θ) 2 0

By putting t = tan θ, we obtain

n − 1 π/4 1 1 n−1   ! n−1 Z Z n−1 Z X n − 1 X k sec2n θθ = 1 + t2
dt + t2k dt = . . k 2k + 1 0 0 0 k=0 k=0

Thus (1) holds and this completes the solution.

Also solved by Arkady Alt, San Jose, CA; Ed Gray, Highland Beach, FL; Paolo Perfetti, Department of Mathematics, Tor Vergata University, Rome, Italy, and the proposer.

20 Problems Ted Eisenberg, Section Editor *********************************************************

This section of the Journal offers readers an opportunity to exchange interesting mathematical problems and solutions. Please send them to Ted Eisenberg, Department of Mathematics, Ben-Gurion University, Beer-Sheva, Israel or fax to: 972-86-477-648. Questions concerning proposals and/or solutions can be sent e-mail to . Solutions to previously stated problems can be seen at .

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Solutions to the problems stated in this issue should be posted before April 15, 2018

5481: Proposed by Kenneth Korbin, New York, NY A triangle with integer area has integer length sides (3, x, x + 1). Find five possible values of x with x > 4.

5482: Proposed by Daniel Sitaru, “Theodor Costescu” National Economic College, Drobeta Turnu-Severin, Mehedinti, Romania Prove that if n is a natural number then (tan5◦)n (tan4◦)n (tan3◦)n 3 + + ≥ . (tan4◦)n + (tan3◦)n (tan3◦)n + (tan2◦)n (tan2◦)n + (tan1◦)n 2

5483: Proposed by D.M. Ba˘tinetu-Giurgiu, “Matei Basarab” National College, Bucharest and Neculai Stanciu, “George Emil Palade” School Buza˘u, Romania  π  If a, b > 0, and x ∈ 0, then show that 2 sin x 2ab tan x 6ab (i)(a + b) · + · ≥ . x a + b x a + b √ (ii) a · tan x + b · sin x > 2x ab.

5484: Proposed by Mohsen Soltanifar, Dalla Lana School of Public Health, University of Toronto, Canada

Let X1,X2 be two continuous positive valued random variables on the real line with corresponding mean, median, and mode x1, x˜1, xˆ1 and x2, x˜2, xˆ2 respectively. Assume for their associated CDFs, (Cummulative Distribution Functions) we have

FX1 (t) ≤ FX2 (t)(t > 0). Prove or give a counter example:

1 (i) x2 ≤ x1, (ii)x ˜2 ≤ x˜1, (iii)x ˆ2 ≤ xˆ1.

5485: Proposed by Jos´eLuis D´ıaz-Barrero, Barcelona Tech, Barcelona, Spain Let x, y, z be three positive real numbers. Show that Y X (2x + 3y + z + 1) (4x + 2y + 1)−3 ≥ 3. cyclic cyclic

5486: Proposed by Ovidiu Furdui, Technical University of Cluj-Napoca, Cluj-Napoca, Romania

Let (xn)n≥0 be the sequence defined by x0 = 0, x1 = 1, x2 = 1 and ∞ X xn x = x + x + x + n, ∀n ≥ 0. Prove that the series converges and find n+3 n+2 n+1 n 2n n=1 its sum.

Solutions

5463: Proposed by Kenneth Korbin, New York, NY Let N be a positive integer. Find triangular numbers x and y such that x2 + 14xy + y2 = 72N 2 − 12N − 1
2.

Solution 1 by Dionne Bailey, Elsie Campbell, and Charles Diminnie, Angelo State University, San Angelo, TX n (n + 1) The nth triangular number T is given by T = . To simplify matters, we will n n 2 assume that x ≤ y. Then, by trial and error, we found the following solutions for the first four values of N.

N 72N 2 − 12N − 1 x y 1 59 T4 = 10 T6 = 21 2 263 T10 = 55 T12 = 78 . 3 611 T16 = 136 T18 = 171 4 1103 T22 = 253 T24 = 300

This leads to the conjecture that one solution consists of

(6N − 2) (6N − 1) x = T = = (3N − 1) (6N − 1) = 18N 2 − 9N + 1 (1) 6N−2 2 and 6N (6N + 1) y = T = = 3N (6N + 1) = 18N 2 + 3N. (2) 6N 2

2 After some algebraic simplification, we obtain 2 x2 + 14xy + y2 = 18N 2 − 9N + 1
+ 14 18N 2 − 9N + 1
18N 2 + 3N
2 + 18N 2 + 3N
= 5184N 4 − 1728N 3 + 24N + 1 2 = 72N 2 − 12N − 1
and hence, (1) and (2) provide a solution for each N ≥ 1.

Solution 2 by Albert Stadler, Herrliberg, Switzerland (aN + b)(aN + b − 1) (cN + d)(cN + d − 1) We put x = , y = . 2 2

(aN + b)2(aN + b − 1)2 (aN + b)(aN + b − 1)(cN + d)2(cN + d − 1)2 (cN + d)2(cN + d − 1)2 +14 + 4 4 4 = (72N 2 − 12N − 1)2. By comparing the coefficients of N 4,N 3,N 2,N and the statement of the problem we find the solutions (6N − 1)(6N − 2) (6N + 1)6N  (x, y) = , and 2 2 (6N + 1)6N (6N − 1)(6N − 2) (x, y) = , . 2 2

Also solved by Brian D. Beasley, Presbyterian College, Clinton, SC; Anthony Bevelacqua, University of North Dakota, Grand Forks, ND; Jeremiah Bartz, University of North Dakota, Grand Forks, ND; Bruno Salgueiro Fanego, Viveiro, Spain; Ed Gray, Highland Beach, FL; Kee-Wai Lau, Hong Kong, China; David E. Manes, Oneonta, NY; Ioannis D. Sfikas, National and Kapodistrian University of Athens, Greece; Titu Zvonaru, Comanesti˘ and Neculai Stanciu,“George Emil Palade” School Buzau,˘ Romania; David Stone and John Hawkins, Georgia Southern University, Statesboro, GA, and the proposer.

5464: Proposed by Ed Gray, Highland Beach, FL Let ABC be an equilateral triangle with side length s that is colored white on the front side and black on the back side. Its orientation is such that vertex A is at lower left, B is its apex, and C is at lower right. We take the paper at B and fold it straight down along the bisector of angle B, thus exposing part of the back side which is black. We continue to fold until the black part becomes 1/2 of the existing figure, the other half being white. The problem is to determine the position of the fold, the distance defined by x (as a function of s) which is the distance from B to the fold.

Solution 1 by David E. Manes, Oneonta, NY √ 3 √ If x = (2 − 2)s, then the area of the resulting black triangle equals the sum of the 2 areas of the two resulting white triangles.

3 √ Introduce the coordinates A(−s/2, 0),B(0, 3s/2) and√C(s/2, 0). Then triangle ABC is an equilateral triangle with side length s and altitude 3s/2). In view of the coordinates x and y, let t denote the distance from vertex B to the fold. The equation of the line L √  s containing the points B and C is y = − 3 x − . Note that for a given value of t, the √ 2 √ √ 3 3s 3s value of y is given by y = s − t. For example, let t = . Then y = and 2 4 4 vertex B has been moved to the origin, thus creating three equilateral triangles two of s which are white. Substituting the above value of y in the equation for L yields x = so 4 √ ! s 3 that the point P , s is a base vertex for the black triangle and an apex for one of 4 4 the white triangles with side PC. By symmetry, the side length for the black triangle is √ ! √ s s 3 s2 3 2 = so that its area A is given by A = = s2. However, the 4 2 B B 4 2 16 v u √ !2 us2 3s s side length PC = t + = , hence the sum of the areas A of the two 4 4 2 W √ ! √ 3 s2 3 white triangles is A = 2 = s2. Since A > A , it follows that the W 4 2 8 W B √ √ 3s 3 value of t is greater than . For this reason, let t = s + k, where k is a real √4 4 3 number such that 0 < k < s. Then 4 √ √ √ ! √ 3 3 3 3 y = s − t = − s + k = s − k. 2 2 4 4 √ s 3 Substituting this value of y in the equation for L, one obtains x = + k. Hence, the 4 3 √ √ ! s 3 3 point P = + k, s − k is a base vertex for the black triangle and an apex for 4 3 4 the white triangle with side PC. Therefore, the side length of the black triangle is √ ! √ √ √ !2 s 3 s 2 3 3 s 2 3 2 + k = + k so that its area is A = + k . Moreover, for 4 3 2 3 B 4 2 3 the white triangle v u √ !2 √ !2 √ u 3 s 3 s 2 3 PC = t k − + s − k = − k. 3 4 4 2 3

√ √ !2 3 s 2 3 Therefore, the sum of the areas of the two white triangles is A = − k . W 2 2 3 Setting AW = AB, we get √ √ ! √ √ ! 3 s2 2 3 4 3 s2 2 3 4 − sk + k2 = + sk + k2 . 2 4 3 3 4 4 3 3 This equation simplifies to the following quadratic equation in k: 4 √ s2 k2 − 2 3sk + = 0 3 4 4 √ √ 3 3 √  3 with roots k =  2 s. The positive square root of 2 yields a value of k > s 2 2 √ 4 3 3 √  and so is inadmissable. Therefore, k = − 2 s, whence 2 2 √ √ 3 3 √ t = + k = (2 − 2)s. 4 2 Observe that using this value of t, the area of the black triangle as well as the sum of √ 3 √  the areas of the two white triangles is 3 − 2 s2. 2 Solution 2 by Kee-Wai Lau, Hong Kong, China When B reaches the midpoint of AC, the black part is only 1/3 of the existing figure, which is a trapezium. So we need to push B downwards further. The black part is then x2 an equilateral triangle with base 2x tan 30◦, height x and hence area √ . The distance √ 3 3s between the fold and AC equals − x. The white part now consists of two congruent 2 √ ! 3s 2x equilateral triangles of lengths − x sec 30◦ = s − √ . Since the area of the 2 3 √ 3  2x 2 x2 white part equals the area of the black part, we have s − √ = √ . Solving, we √ √ 2 3 3 3 2 − 2
s obtain x = . 2 Editor0s Comment: David Stone and John Hawkins both Georgia Southern University in Statesboro, GA generalized the statement as follows: “The problem asked for the configuration in which the black triangle covered half of the final figure. We could just as well determine when the black triangle covers any given portion of the final figure; say one fourth or nine tenths.” 1 1 They did this by looking at two cases: 1) 0 < λ < and 2) < λ < 1, where the Black 3 3 1 Area=λTotal Area. The constant comes from when the vertex of the Black Triangle 3 lies on the base of the White Triangle. Letting x be the length of the height of the black triangle (measured from its vertex to the fold line) they found that for the first case, where the vertex√ of the Black Triangle lies in the interior of the White Triangle that: r λ 3 x = · s and in the second case, where the vertex of the Black Triangle lies in 1 + λ 2 2λ − p2λ(1 − λ) the exterior of the White Triangle that x = · h, where h is the altitude 3λ − 1 1 of the given White Triangle. When λ = we obtain the statement of the problem and 2 using their formula reaffirms the above answers. 2a2 In concluding their comment they noted, “Another nice example: with λ = , 2a2 + 1  1  which is very close to 1, we find that x = 1 − h. That is, in a precisely 2a + 1 measurable way, we must fold almost all the way down to get a figure which is almost all black.”

5 Also solved by Brian D. Beasley, Presbyterian College, Clinton, SC; Bruno Salgueiro Fanego (two solutions), Viveiro, Spain; David A. Huckaby, Angelo State University San Angelo, TX; and the proposer.

5465: Proposed by Arsalan Wares, Valdosta State University, Valdosta, GA Quadrilateral ABCD is a rectangle with diagonal AC. Points P,R,T,Q and S are on sides AB and DC and they are connected as shown. Three of the triangles inside the rectangle are shaded pink, and three are shaded blue. Which is larger, the sum of the areas of the pink triangles or the sum of the areas of the blue triangles?

Solution by David A. Huckaby, Angelo State University, San Angelo, TX

Let p be the sum of the areas of the pink triangles, b the sum of the areas of the blue triangles, and w the sum of the areas of the three white polygons below the diagonal.

p + w = the sum of the areas of triangles DPQ, QRS, and STC 1 1 1 = (AD · DQ) + (AD · QS) + (AD · SC) 2 2 2 1 = [AD · (DQ + QS + SC)] 2 1 = (AD · DC) 2 = the area of triangle ADC = b + w

So p = b, that is, the sum of the areas of the pink triangles is equal to the sum of the areas of the blue triangles.

6 Also solved by Michael N. Fried, Ben-Gurion University of the Negev, Beer-Sheva, Israel; Ed Gray, Highland Beach, FL; Kee-Wai Lau, Hong Kong, China; Albert Stadler, Herrliberg, Switzerland; David Stone and John Hawkins, Georgia Southern University, Statesboro, GA, and the proposer.

5466: Proposed by D.M. Ba˘tinetu−Giurgiu, “Matei Basarab” National College, Bucharest, Romania and Neculai Stanciu, “Geroge Emil Palade” School, Buza˘u, Romania Let f : (0, +∞) → (0, +∞) be a continuous function. Evaluate

2 (n√+1) Z n+1 (n+1)! x lim f dx. n→∞ n2 n√ n n!

Solution 1 by Moti Levy, Rehovot, Israel

The mean value theorem of the integral calculus states: Let f (x) be continuous function, then

Z b f (x) dx = (b − a) f (ξ) , a ≤ ξ ≤ b. a Therefore,

2 (n√+1) 2 2 !   2 2 Z n+1 (n+1)! x (n + 1) n ξ n (n + 1) f dx = − √n f , √n ≤ ξ ≤ . n2 n+1p n+1p n√ n (n + 1)! n! n n! (n + 1)! n! Taking limits of both sides,

2 (n√+1) 2 2 !   Z n+1 (n+1)! x (n + 1) n ξ lim f dx = lim − √n lim f . n→∞ n2 n→∞ n+1p n→∞ n√ n (n + 1)! n! n n! Since f (x) is continuous then

 ξ   ξ   n  lim f = f lim = f lim √ n→∞ n n→∞ n n→∞ n n! Using Stirling’s asymptotic formula, we have √ n n n! ∼ . (1) e By (1), n n2 √ ∼ e, √ ∼ e · n, n n! n n! which implies that  n  f lim √ = f (e) n→∞ n n! and that (n + 1)2 n2 − √ ∼ e. n+1p(n + 1)! n n!

7 We conclude that 2 (n√+1) Z n+1 (n+1)! x lim f dx = ef (e) . n→∞ n2 n√ n n!

Solution 2 by Bruno Salgueiro Fanego,Viveiro, Spain Let’s proceed as in http://www./oei.es/historico/oim/revistaoim/numero 53/261 Bruno.pdf:

Let n ∈ N; since f is continuous on (xn, xn+1), by the mean value theorem of integral Z xn+1   x ξn calculus, we have that f dx = f (xn+1 − xn)for some ξn ∈ (xn, xn+1). xn n n x ξ x Since n < n < n+1 , n n n

(n + 1)n+1 r 2n r n n  n xn n n n n (n + 1)! (n + 1) (n + 1)n! n + 1 lim = lim = lim = lim n = lim = lim = e n→∞ n n→∞ nnn! n→∞ n! n→∞ n n→∞ (n + 1)!nn n→∞ n n! x x n + 1 x n + 1 and lim n+1 = lim n+1 · = lim n+1 · lim = e · 1 = e, by the n→∞ n n→∞ n + 1 n n→∞ n + 1 n→∞ n ξ sandwich rule we obtain that lim n = e, and, hence, n→∞ n ξ   ξ  lim f n = lim n = f(e). n→∞ n n→∞ n Moreover, from Stolz’ rule,

(xn+1 − xn) xn lim (xn+1 − xn) = lim = lim = e. n→∞ n→∞ (n + 1) − n n→∞ n

So, the required limit is equal to

Z xn+1   x ξn lim f dx = lim f lim (xn+1 − xn) = ef(e). n→∞ n→∞ n→∞ xn n n

Solution 3 by Ioannis D. Sfikas, National and Kapodistrian University of Athens, Greece This particular problem is similar to Problem 121, which was proposed by D.M. B˘atinetu-Giurgiu(“Matei Basarab” National College, Bucharest, Romania) and Neculai Stanciu (“George Emil Palade” School, Buz˘au,Romania) to the Math Problems Journal, Volume 5, Issue 2 (2015), pp. 420-421. We’ll use the following lemma.

Lemma: Let f :[a, b] → < be continuous and (xn)n, (yn)n two convergent sequences of [a, b] that have the same limit c, then

Z yn f(t)dt = f(c)(yn − xn) + O(yn − xn). xn Proof: Let  > 0 , then there exists δ > 0 such that |f(t) − f(c)| < , whenever |x − c| < δ. Since xn, yn → c, the there is an n0 ∈ N such that xn, yn ∈ (C − δ, C + δ),

8 whenever n > n0. Therefore,

Z yn Z yn

f(t)dt − f(c)(yn − xn) ≤ |f(t) − f(c)| dt ≤  |yn − xn| . xn xn Note that the given integral equals

(n+1)2 n+1√ Z n (n+1)! In = n f(t)dt, n n√ n! x this comes directly from the substitution t = . Let x , y be the lower, upper bound of n n n n the last integral respectively then xn, yn → e, since √ → e, and thus n n! (n + 1)2 n + 1 (n + 1) = −→ e, as n −→ ∞. Note that by Stolz’ theorem n n+1p(n + 1)! n n+1p(n + 1)!

(n + 1)2 n2 n(yn − xn) = − √ −→ e, as n −→ ∞. n+1p(n + 1)! n n!

By the lemma we have

In = n [f(c)(yn − xn) + O(yn − xn)] = ef(e) + O(1), which proves that the limit equals ef(e).

Also solved by Arkady Alt, San Jose, CA; Kee-Wai Lau, Hong Kong, China; Soumitra Mandal, Scottish Church College, Chandan -Nagar, West Bengal, India; Paolo Perfetti, Department of Mathematics, Tor Vergata University, Rome, Italy; Albert Stadler, Herrliberg, Switzerland; Anna V. Tomova, Varna, Bulgaria, and the proposers.

5467: Proposed by Jos´eLuis D´ıaz-Barrero, Barcelona Tech, Barcelona, Spain In an arbitrary triangle 4ABC, let a, b, c denote the lengths of the sides, R its circumradius, and let ha, hb, hc respectively, denote the lengths of the corresponding altitudes. Prove the inequality

a2 + bc b2 + ca c2 + ab 3abc r 1 + + ≥ 3 , b + c c + a a + b 2R ha · hb · hc and give the conditions under which equality holds.

Solution 1 by Paolo Perfetti, Department of Mathematics, Tor Vergata University, Rome, Italy

We know that ha = (bc)/(2R) and cyclic so the inequality actually is

1 2 2 2 3 a + bc b + ca c + ab 3abc  8R  3 1 + + ≥ = 3(abc) 3 . b + c c + a a + b 2R (abc)2 We prove the stronger one

9 a2 + bc b2 + ca c2 + ab + + ≥ a + b + c, b + c c + a a + b that is

a2 + bc  b2 + ca  c2 + ab  − a + − b + − c ≥ 0, b + c c + a a + b or

(a − b)(a − c) (b − c)(b − a) (c − a)(c − b) + + ≥ 0. b + c a + c a + b We can suppose a ≥ b ≥ c by symmetry so we come to

(a − b)(a − c) (a − c)(b − c) (a − b)(b − c) + ≥ . b + c a + b a + c This is implied by

(a − b)(a − c) (a − b)(b − c) (a − b)(b − c) + ≥ , b + c a + b a + c | {z } a−c≥a−b or

a − c b − c b − c + ≥ . b + c a + b a + c This is in turn implied by

a − c b − c b − c + ≥ a + c a + b a + c | {z } a≥b and this evidently holds true by a − c ≥ b − c ≥ 0. The equality case is a = b = c.

Solution 2 by Dionne Bailey, Elsie Campbell, and Charles Diminnie, Angelo State University, San Angelo, TX We will prove the following slight improvement:

a2 + bc b2 + ca c2 + ab + + ≥ a + b + c b + c c + a a + b √ ≥ 3 3 abc 3abc r 1 = 3 , (1) 2R ha · hb · hc with equality if and only if a = b = c. To begin, we note that

a2 − b2
2 + b2 − c2
2 + c2 − a2
2 a4 + b4 + c4 − a2b2 − b2c2 − c2a2 = 2 ≥ 0, (2)

10 with equality if and only if a2 = b2 = c2. Since a, b, c > 0, it follows that equality is attained in (2) if and only if a = b = c. Next, we use (2) to obtain

a2 + bc b2 + ca c2 + ab + + b + c c + a a + b a2 − b2
+ b2 + bc
b2 − c2
+ c2 + ca
c2 − a2
+ a2 + ab
= + + b + c c + a a + b a2 − b2 b2 − c2 c2 − a2 = + + + a + b + c b + c c + a a + b a2 − c2
+ c2 − b2
b2 − c2 c2 − a2 = + + + a + b + c b + c c + a a + b  1 1   1 1  = a2 − c2
− + b2 − c2
− + a + b + c b + c a + b c + a b + c a − c b − a = a2 − c2
+ b2 − c2
+ a + b + c (a + b)(b + c) (b + c)(c + a) a2 − c2
2 + b2 − c2
b2 − a2
= + a + b + c (a + b)(b + c)(c + a) a4 + b4 + c4 − a2b2 − b2c2 − c2a2 = + a + b + c (a + b)(b + c)(c + a) ≥ a + b + c, (3) with equality if and only if a = b = c. Also, the Arithmetic - Geometric Mean Inequality implies that √ a + b + c ≥ 3 3 abc, (4) with equality if and only if a = b = c. For the final step, let K = area (4ABC). Then, 1 1 1 K = ah = bh = ch 2 a 2 b 2 c and hence, 2K 2K 2K h = , h = , and h = . a a b b c c abc Since R = , we have 4K r r 3abc 3 1 12KR 3 abc = 3 2R ha · hb · hc 2R 8K

6K √ = 3 abc 2K

√ = 3 3 abc. (5)

11 If we combine (3), (4), and (5), statement (1) follows and equality is attained throughout if and only if a = b = c.

Solution 3 by Arkady Alt, San Jose, CA

Let F = [ABC] (area) and let s be its semi-perimeter. 2F 2F 2F Since h = , h = , h = and abc = 4RF then a a b b c c

r r √ 3 1 3 abc 1 3 = 3 = abc and hahbhc 8F 2F

3abc r 1 √ 3 = 3 3 abc. 2R hahbhc Thus, original inequality becomes a2 + bc b2 + ca c2 + ab √ (1) + + ≥ 3 3 abc. b + c c + a a + b 4a2 Since ≥ 4a − b − c ⇐⇒ (2a − b − c)2 ≥ 0 we have b + c

X a2 + bc X a2 X bc X 4a − b − c X bc = + ≥ + b + c b + c b + c 4 b + c cyc cyc cyc cyc cyc

r a + b + c X bc X b + c bc  X b + c bc = + = + ≥ 2 · 2 b + c 4 b + c 4 b + c cyc cyc cyc

√ q√ √ √ X 3 √ 3 = bc ≥ 3 bc · ca · ab = 3 abc. cyc

Solution 4 by Nicusor Zlota,“Traian Vuia” Technical College, Focsani, Romania, and Corneliu-Manescu Avram, Ploiesti, Romania Assume that a ≥ b ≥ c.

a2 + bc b2 + ca c2 + ab First, we will prove that + + ≥ a + b + c ⇐⇒ b + c c + a a + b a2 + bc b2 + ca c2 + ab − a + − b + − c ≥ 0 ⇐⇒ b + c c + a a + b (a − b)(a − c) (b − c)(b − a) (c − a)(c − b) + + ≥ 0 ⇐⇒ b + c c + a a + b a − c b − c  b − a c − a a − c b − c  (a − b) − + (b − a) − + (c − a) − ≥ 0 b + c c + a c + a a + b b + c c + a a + b b + c c + a (a − b)2 + (b − c)2 + (c − a)2 ≥ 0. (b + c)(c + a) (a + b)(c + a) (a + b)(b + c)

12 Then, it suffices to prove that r r √ √ 3abc 3 1 3abc 3 abc 3abc 1 3 3 a = b = −c ≥ = 3 = abc = abc, 2R hahbhc 2R 8S 2R 2S which is the AM-GM inequality. Equality holds for a = b = c. Also solved by Hatef I. Arshagi, Guilford Technical Community College, Jamestown, NC; Bruno Salgueiro Fanego, Viveiro, Spain; Ed Gray, Highland Beach, FL; Kee-Wai Lau, Hong Kong, China; Moti Levy Rehovot, Israel; Ioannis D. Sfikas, National and Kapodistrian University of Athens, Greece; Albert Stadler, Herrliberg, Switzerland; Neculai Stanciu, “George Emil Palade” School, Buzau,˘ Romania and Titu Zvonaru, Comanesti,˘ Romania; David Stone and John Hawkins, Georgia Southern University, Statesboro,GA, and the proposer. Editor0s note: Hatef I. Arshagi’s solution was dedicated to the memory of Mrs. Alieh Ataee.

5468: Proposed by Ovidiu Furdui and Alina Sˆinta˘ma˘rian, both at the Technical University of Cluj-Napoca, Cluj-Napoca, Romania Find all differentiable functions f : < → < with f(0) = 1 such that f 0(x) = f 2(−x)f(x), for all x ∈ <.

Solution 1 by Moti Levy, Rehovot, Israel Let us differentiate both sides of the given differential equation,

00 0 0 f (x) = −2f(−x)f (−x) f(x) + f 2(−x)f (x). (1)

The following two equation are direct consequence of the original equation.

0 f (−x) = f 2(x)f(−x), (2) f 0 (x) f 2(−x) = . (3) f(x)

After substitution of (2) and (3) in (1), we get differential equation (4) with initial conditions at x = 0,

 0 2 f (x) 00 0 0 f (x) + 2f 2(x)f (x) − = 0, f (0) = f (0) = 1. (4) f(x)

By the substitution f (x) = pg (x), √ f = g,

0 1 0 f = √ g , 2 g

00 1 00 1  0 2 f = √ g − √ g , 2 g 4 g
3

13 we arrive at the equivalent differential equation

00 0 1  0 2 0 g + 2gg − g = 0, g (0) = 1, g (0) = 2. (5) g

0 dg 00 d2g dz Now we want to lower the order of (5) by the substitution g = dx = z, g = dx2 = z dg , dz 1 z + 2gz − z2 = 0, dg g or dz g − z = −2g2. (6) dg The solution of (6) is z = cg − 2g2. The initial conditions on g dictate that c = 4, thus we obtain the following differential equation for g, dg = 4g (x) − 2g2 (x) . dx or dx 1 = . dg 4g − 2g2 After integration over g, we get 1 g x = ln + c 4 2 − g or e4x g (x) = 2k . ke4x + 1 Again, the initial condition g (0) = 1 dictates k = 1,

2e4x g (x) = . e4x + 1 We conclude that the √ e2x f (x) = 2√ . e4x + 1

Solution 2 by Ioannis D. Sfikas, National and Kapodistrian University of Athens, Greece The defining relation

f 0(x) = f 2(−x)f(x), (1) implies that f is continuously differentiable. Setting −x in the relation (1), one gets for every = x, f 0(−x) = f 2(x)f(−x), Multiplying (1) by f(x) yields for every x

0 0 f (x)f(x) = f 2(−x)f 2(x) = f 2(x)f(−x) f(−x) = f (−x)f(−x),

14 Z x Z x that is x → f 0(x)f(x) is even. Therefore, f 0(t)f(t)dt = 2 f 0(t)f(t)dt, and since −x 0 f 2 an antiderivative of f 0f is , this implies that for every x, f 2(x) + f 2(−x) = 2. 2 Replacing f 2(−x) in the defining relation one get for every x

f 0(x) − 2f(x) − f 2(x).

This non-linear differential equation seems to have only one solution, namely

√ 2e2x x → √ (2) e4x + 1 Conversely, it is easily checked that (2) is indeed a solution to the equation.

Solution 3 by Kee-Wai Lau, Hong Kong, China

We show that √ 2e2x f(x) = √ . (1) 1 + e4x From the given equation , we obtain f(x)f 0(x) = f 2(−x)f 2(x), so that

f(−x)f 0(−x) = f(x)f 0(x) (2)

Integrating (2) with respect to x, and making use of the fact that f(0) = 1, we obtain

f 2(−x) = 2 − f 2(x). (3)

Substituting (3) into the given equation, we obtain f 0(x) = 2 − f 2(x)
f(x) or d(f(x)) = dx. Integrating boh sides we obtain (1 − f 2(x))f(x)

ln (f(x)) ln 2 − f 2(x)
− = x + C, 2 4 where C is a constant. Since f(0) = 1, so C = 0. Now (1) follows easily by simple algebra.

Editor0s comment: Anna Tomova of Varna Bulgaria expressed her solution in ex terms of a hyperbolic function; f(x) = √ , f(0) = 1. cosh 2x Also solved by Albert Stadler, Herrliberg, Switzerland; Anna V. Tomova, Varna, Bulgaria, and the proposers.

15 Problems Ted Eisenberg, Section Editor *********************************************************

This section of the Journal offers readers an opportunity to exchange interesting mathematical problems and solutions. Please send them to Ted Eisenberg, Department of Mathematics, Ben-Gurion University, Beer-Sheva, Israel or fax to: 972-86-477-648. Questions concerning proposals and/or solutions can be sent e-mail to . Solutions to previously stated problems can be seen at .

————————————————————–

Solutions to the problems stated in this issue should be posted before May 15, 2018

5487: Proposed by Kenneth Korbin, New York, NY p √ (x + 1)4 b + b − b Given that 2 = a with x = p √ . Find positive integers a and b. x(x − 1) b − b − b

5488: Proposed by Daniel Sitaru, “Theodor Costescu” National Economic College, Drobeta, Turnu-Severin, Mehedinti, Romania Let a, and b be complex numbers. Solve the following equation:

x3 − 3ax2 + 3(a2 − b2)x − a3 + 3ab2 − 2b3 = 0.

5489: Proposed by D.M. Ba˘tinetu-Giurgiu, Bucharest, Romania, and Neculai Stanciu, “George Emil Palade” School Buza˘u, Romania Z a If a > 0, compute x2 − ax + a2
arctan(ex − 1)dx. 0

5490: Proposed by Moshe Stupel, “Shaanan” Academic College of Education and Gordon Academic College of Education, and Avi Sigler, “Shaanan” Academic College of Education, Haifa, Israel Triangle ABC whose side lengths are a, b, and c lies in plane P . The segmentA1A, BB1,CC1 satisfy:

A1A ⊥ P,B1B ⊥ P,C1C ⊥ P,

where A1A = a, B1B = b and C1C = c, as shown in the figure. Prove that 4A1B1C1 is acute -angled.

1 5491: Proposed by Roger Izard,Dallas, TX Let O be the orthocenter of isosceles triangle ABC, AB = AC. Let OC meet the line segment AB at point F. If m = FO, prove that c4 ≥ m4 + 11m2c2.

5492: Proposed by Jos´eLuis D´ıaz-Barrero, Barcelona Tech, Barcelona, Spain Let a, b, c, d be four positive numbers such that ab + ac + ad + bc + bd + cd = 6. Prove that

r abc r bcd r cda r dab r2 + + + ≤ 2 . a + b + c + 3d b + c + d + 3a c + d + a + 3b d + a + b + 3c 3

Solutions

5469: Proposed by Kenneth Korbin, New York, NY Let x and y be positive integers that satisfy the equation 3x2 = 7y2 + 17. Find a pair of larger integers that satisfy this equation expressed in terms of x and y.

Solution 1 by Bruno Salugueiro Fanego, Viveiro, Spain

2 It suffices to find a pair of the type (ax + by, cx + dy) where a, b, c, and d are positive integers and 3(ax + by)2 − 7(cx + dy)2 = 3x2 − 7y2. Since (ax + by)2 − 7(cx + dy)2 = (3a2 − 7c2)x2 + (3b2 − 7d2)y2 + 2(3ab − 7cd)xy, it is sufficient that a, b, c, and d satisfy the relations:  2 2  3a − 7b = 3 3b2 − 7d2 = −7  3ab − 7cd = 0. The pair (a, c) = (55, 36) of positive integers verifies 3a2 − 7b2 = 3, and if it assumed that d = a = 55 then it only remains to find a positive integer b such that 3 · b − 7 · 36 = 0 and 3b2 − 7 · 552 = −7. Since b = 84 satisfies 3b2 − 7 · 552 = −7, the pair of larger integers (55x + 84y, 36x + 55y) solves the problem.

Solution 2 by Ed Gray, Highland Beach, FL Clearly, by inspection, the equation is satisfied by x = 8, y = 5. Let the larger integers which satisfy the equation be x + k = 8 + k and y + m = 5 + m then we have:

1. 3(8 + k)2 = 7(5 + m)2 + 17, expanding 2. 3(64 + 16k + k2) = 7(25 + 10m + m2) + 17 3. 192 + 48k + 3k2 = 175 + 70m + 7m2 + 17 4. 48k + 3k2 = 70m + 7m2. Let (k, m) = r. So, 5. k = ra, m = rb, (a, b) = 1, and substituting into 4) 6. 48ra + 3r2a2 = 70rb + 7r2b2 7. 48a + 3ra2 = 70b + 70rb2 8. 3a(16 + ra) = 7b(10 + rb). Suppose 9. 3a = 10 + rb and that 10. 7b = 16 + ra. Multiply step 9 by a and step10 by b 11. 3a2 = 10a + arb 12. 7b2 = 16b + arb 13. 3a2 − 7b2 = 10a − 16b 14. 3a2 − 10a = 7b2 − 16b, and by inspection b = 4, a = 6 15. 3a2 − 10a = 3(6)2 − (10)(6) = 108a2 − 60 = 48 16. 7b2 − 16b = 7(4)2 − (16)(4) = 112 − 64 = 48. From step 10 17. 7b = 16 + ra or (7)(4) = 28 = 16 + 6r, and r = 2. From step 5 18. k = ra = (2)(16) = 12, m = rb = (2)(4) = 8.

Hence the larger integers are x + k = x + 12 = 20 and y + m = y + 8 = 13. As a check, substituting (20, 13) into 3x2 = 7y2 + 17, gives equality.

Solution 3 by David E. Manes, Oneonta, NY We will show that if (x, y) is a solution of 3x2 = 7y2 + 17, then (55x + 84y, 36x + 55y) is another solution of this equation. Furthermore, all positive integer solutions of this equation are given by

3 √ ! √ ! 5 21 √ 5 21 √ x = 4 + (55 + 12 21)n + 4 − (55 − 12 21)n, n 6 6 √ ! √ ! 4 21 5 √ −4 21 5 √ y = + (55 + 12 21)n + + (55 − 12 21)n, n 7 2 7 2 where n ≥ 1 and  = 1. Re-write the equation as (1) 3x2 − 7y2 − 17 = 0 and note that the two least positive solutions (x, y) are (8, 5) and (20, 13). Construct the recurrent sequences

xn+1 = αxn + βyn

yn+1 = γxn + δyn, where α, β, γ and δ are unknowns and assume that (xn+1, yn+1) is a solution of (1). 2 2 Then 3(αxn + βyn) − 7(γxn + δyn) − 17 = 0 which expands to 2 2 2 2 2 2 (3α − 7γ )xn + (3β − 7δ )yn + (6αβ − 14γδ)xnyn − 17 = 0. Comparing this equation with (1), we get (2) 3α2 − 7γ2 = 3, (3) 3β2 − 7δ2 = −7, (4) 3αβ = 7γδ. Squaring equation (4), we get (3α2)(3β2) = 49γ2δ2. Using (2) and (3), one obtains (3 + 7γ2)(−7 + 7δ2) = 49γ2δ2 or 3δ2 − 7γ2 = 3. Subtracting this equation from (2) results in α = δ. Substituting this value in (4), we 7 get 3(δ)β = 7γδ or β =  γ. Finally, substituting this value in (3) along with δ = α 3 yields the equation 3α2 − 7γ2 = 3 which is equation (2). The smallest positive integer 7 solution of this equation is α = 55, γ = 36 so that β = γ = 84. Since we want positive 3 solutions only, we let 7 x = αx + γy = 55x + 84y n+1 n 3 n n n yn+1 = γxn + αyn = 36xn + 55yn. 55 84 Define the 2 × 2 matrix A over the reals as follows: A = . By construction, if 36 55 x x 55x + 84y is a positive integer solution vector of equation (1), then A = is y y 36x + 55y x  8 x  20 also a solution vector with larger integers. Let 0 = , and 1 = . Then y0 5 y1 13 all positive integer solutions of equation (1) are given by

 0     00    xn n 8 xn n 20 0 = A and 00 = A yn 5 yn 13  8  20 for n ≥ 1. Noting that A = , we can shorten the above description to −5 13

x   8  n = An , yn 5

4 where n ≥ 1 and  = 1. To get a closed form expression for the positive integer solutions of equation 3x2 − 7y2 − 17 = 0, we note that A has two distinct eigenvalues √ √ −→ √ λ1 = 55 + 12 21 and λ2 = 55 − 12 21 with corresponding eigenvectors v1 = ( 21, 3) −→ √ and v2 = (− 21, 3). Therefore, A is diagonalizable; that is, there exists a diagonal λ 0  matrix D = 1 , where λ , λ are the eigenvalues of A and an invertible matrix 0 λ 1 2 √ √ 2  21 − 21 P = consisting of the two eigenvectors in columns such that 3 3 P −1AP = D. Therefore, A = PDP −1 so that for each positive integer n, An = PDnP −1. Hence, x   8   8  n = An = PDnP −1
, yn 5 5 which computationally yields √ ! √ ! 5 21 √ 5 21 √ x = 4 + (55 + 12 21)n + 4 − (55 − 12 21)n, n 6 6 √ ! √ ! 4 21 5 √ −4 21 5 √ y = + (55 + 12 21)n + + (55 − 12 21)n, n 7 2 7 2 where n ≥ 1 and  = 1. 2 2 Some examples of solutions (xn, yn) to the equation 3x = 7y − 17 with  = 1 are (x1, y1) = (860, 563), (x2, y2) = (94592, 61925), (x3, y3) = 10 404 260, 6 811 187) and (x4, y4) = (1 144 374 008, 749 168 645). Examples of solutions with  = −1 are (x1, y1) = (20, 13), (x2, y2) = (2192, 1435), (x3, y3) = (241100, 157837) and (x4, y4) = (26 518 808, 17 360 635).

Solution 4 by Jeremiah Bartz, University of North Dakota, Grand Forks, ND Put z = 3x and rewrite 3x2 = 7y2 + 17 as z2 − 21y2 = 51. This is a Pell-like equation with two classes of solutions with initial solutions (z, y) of (24, 5) and (60, 13). The√ fundamental solution of the corresponding Pell equation z2 − 21y2 = 1 is 55 + 12 21. From Pell theory,√ solutions√ can be generated√ via zi+1 + yi+1 21 = (zi + yi 21)(55 + 12 21). This is equivalent to

zi+1 = 55zi + 252yi

yi+1 = 12zi + 55yi or back substituting

xi+1 = 55xi + 84yi

yi+1 = 36xi + 55yi. 2 2 Thus (xi+1, yi+1) is a larger pair of positive integer solutions to 3x = 7y + 17 than the positive integer solutions (xi, yi). The initial solutions are listed below and separated by class. i xi yi xi yi 1 8 5 20 13 2 860 563 2192 1435 3 94592 61925 241100 157837 4 10404260 6811187 26518808 17360635 5 1144374008 749168645 2916827780 1909512013

5 Editor0s notes: Kenneth Korbin, proposer of problem 5469 stated the following:

Given the sequence t = (··· , 33, 7, 2, 3, 13, ··· ) with 5tN = tN−1 + tN+1. Let a and b be a a + b a − b pair of consecutive terms in this sequence with both odd. If x = , y = then 2 2 33 + 7 33 − 7 3x2 = 7y2 + 17. Example : x = , y = . 2 2 Brian D. Beasley of Presbyterian College in Clinton, SC noted that using 2 2 2 2 Brahmagupta’s identity (see [1]), which notes that if x1 − Ny1 = k1 and x2 − Ny2 = k2, then 2 2 (x1x2 + Ny1y2) − N(x1y2 + x2y1) = k1k2. 2 7
2 2 7
2 17 Since 55 − 3 36 = 1, if x and y are positive integers with x − 3 y = 3 , then 7 2 7 2 17 (55x + 3 · 36y) − 3 (55y + 36x) = 3 . Hence the solution (x, y) produces the larger solution (55x + 84y, 55y + 36x).

Reference: [1] https://en.wikipedia.org/wiki/Brahmagupta%27s identity David Stone and John Hawkins, both at Georgia Southern University in Statesboro, GA approached the problem by looking at the graph of the given statement as a hyperbola, with the question asking us to prove that this graph contains infinitely many lattice points with both coordinates being integers. They did this and then listed a few of the lattice points, two of them being (94592, 61925) and (241100, 157837). They went on to state the following: r3 The asymptotes of the given hyperbola are y =  x ≈ 0.6546536707x. Our lattice 7 y points demonstrate the closeness of the curve to the asymptote. We compute : x

6129 ≈ 0.6546536705 94592

157837 ≈ 0.6546536707. 241100 Very close! We’d need a piece of graph paper the size of a football field to see that these points are on the hyperbola but not on the asymptote.

Also solved by Brian D. Beasley, Presbyterian College, Clinton, SC; Anthony Bevelacqua, University of North Dakota, Grand Forks, ND; Kee-Wai Lau, Hong Kong, China; Ioannis D. Sfikas, National and Kapodistrian University of Athens, Athens, Greece; Trey Smith, Angelo State University, San Angelo, TX; Albert Stadler, Herrliberg, Switzerland; David Stone and John Hawkins, Georgia Southern University, Statesboro, GA; Anna V. Tomova, Varna, Bulgaria, and the proposer.

5470: Proposed by Moshe Stupel, “Shaanan” Academic College of Education and Gordon Academic College of Education, and Avi Sigler, “Shaanan” Academic College of Education, Haifa, Israel

6 Prove that there are an infinite number of Heronian triangles (triangles whose sides and area are natural numbers), whose side lengths are three consecutive natural numbers.

Solution 1 by Kenneth Korbin, New York, NY Given a Heronian Triangle with consecutive integer length sides (b − 1, b, b + 1). Then, a larger such triangle has sides (b2 − 3, b2 − 2, b2 − 1), and another such triangle has sides  √ √ √  −1 + 2b + 3b2 − 12, 2b + 3b2 − 12, 1 + 2b + 3b2 − 12 .

Therefore there are infinitely many such triangle. Examples, (3, 4, 5), (13, 14, 15), etc.

Solution 2 by Albert Stadler, Herrliberg, Switzerland 3n Let (n − 1, n, n + 1) be the sides of the triangle. Then the semiperimeter is s = . The 2 area is given by Heron’s formula as: n ps(s − n + 1)(s − n)(s − n − 1) = p3(n + 2)(n − 2). (1) 4 Clearly n must be even if (1) represents an integer. We must show that there are infinitely many pairs of integers (m, n) such that 3(2n + 2)(2n − 2) = 4m2 or equivalently m2 − 3n2 = −3. We see that m must be divisible by 3 and we need therefore to find pairs of integers (m, n) such that n2 − 3m2 = 1. This is Pell’s equation √ √ k whose general solution is given by n − 3m = 2 − 3
, where k is an integer. We √ k √ k conclude that (1) is an integer if and one if n = 2 − 3
+ 2 + 3
, k = 0, 1, 2, etc.

Solution 3 by Ioannis D. Sfikas, National and Kapodistrian University of Athens, Athens, Greece A triangle whose sides and area are rational numbers is called a rational triangle. If the rational triangle is right-angled, it is called a right-angled rational triangle or a rational Pythagorean triangle or a numerical right triangle. If the sides of a rational triangle is of integer length, it is called an integer triangle. If further these sides have no common factor greater than unity, the triangle called primitive integer triangle. If the integer triangle is right-angled, it is called a Pythagorean triangle. A Heronian triangle (named after Heron of Alexandria) is an integer triangle with the additional property that its area is also an integer. A Heronian triangle is called a primitive Heronian triangle if sides have no common factor greater than unity. In the 7th century, the Indian mathematician Brahmagupta studied the special case of triangles with consecutive integer sides. Assume that the consecutive sides of a Brahmagupta triangle are d − 1, d, d + 1 , where 3d d > 4 is a positive integer. The semiperimeter is s = , and by Heron’s formula the 2 area A is

v u " 2 # du d A = t3 − 1 . (1) 2 2

But A must also be an integer, then the base d of the triangle must be even and the altitude h of the triangle must be an integer multiple of 3 since 16A2 = 3d2(d2 − 4). dh Since A = , this equation reduces to 2 7 4h2 = 3(d2 − 4). (2)

If d were odd, then the factors on the right side of (2) would all be odd, a contradiction. Thus d is even and we may write d = 2y (y is a positive integer). The area of the triangle is then A = hy , and it follows that h is a rational number. But

h2 = 3(y2 − 4). (3) is an integer and h itself has to be an integer and hence a multiple of 3. If h = 3x, we reduce (3) to the Pell equation y2 − 3x2 = 1. The Pell equation has an infinity of integer solutions. If (x, y) = (U, T ), where T > 0, U > 0 the solution with least positive x, all the solutions are given by √  √ n xy 3 =  T + U 3 , where n is an arbitrary integer (Mordell, 1969, p. 53). [1] Mordell, L.J. (1969). Diophantine equations. London Academic Press, Inc.

Solution 4 by Paul M. Harms, North Newton, KS

Let n be a natural number and let n − 1 and n + 1 be the sides of the triangle. To be a triangle, n must be at least 3. If s is the semi-perimeter, the area of this triangle is r 3n n  n  n p ps(s − (n − 1))(s − n)(s − (n + 1)) = − 1 + 1 = 3 (n2 − 4). 2 2 2 22

For area to be a natural number one requirement is that 3 n2 − 4
be the square of a natural number. Since 3 is a factor inside the square root we need 3 n2 − 4
= (3t)2 for some natural number t. The last equation is equivalent to n2 − 3t2 = 4. Letting n = 2x and t = 2y where x and y are natural numbers, the equation becomes x2 − 3y3 = 1 which is a Pell equation. One solution is x = x1 = 2 and y = y1 = 1. There exist an infinite number of solutions of natural numbers found by equating coefficients of the √ √
k equation xk + yk 3 = 2 + 3 for k = 1, 2, 3, 4, etc.

For example, when k = 3, x3 = 26 and y3 = 15. In this case, the n for the triangle given above is n = 2x3 = 52. We n = 2xk we need to make sure that the area is a natural number. Replacing n by 2xk in the area formula, we obtain x q q k 4(3) x2 − 1
= x 3 x2 − 1
= x (3y ) . 2 k k k k k Thus the area is the product of natural numbers so it is a natural number.

Editor 0s comments: Bruno Salgueiro Fanego of Viveiro, Spain mentioned in his solution that: “Discussions of the more general problem of finding all the Heronian triangles whose side lengths are in arithmetic progression can be found, for example, in the articles Heron Triangles with Sides in Arithmetic Progression and Heronian Triangles with Sides in Arithmetic Progression: An Inradius Perspective by J. A. MacDougall and by Herb

8 Bailey and William Gosnell, respectively.” http://www.jstor.org/stable/10.4169/math.mag.85.4.290
; Mathematics Magazine Vol. 85, No. 4 (October 2012), pp. 290-294. This generalization was also mentioned in the solution submitted by David Stone and John Hawkins both of Georgia Southern University in Statesboro, GA. They showed in their solution that all primitive Heronian triangles with sides in arithmetic progression had to have one as the difference in the side lengths. Having a common difference greater than 1 produced a Heronian Triangle, but not a primitive one. Brian D. Beasley of Presbyterian College in Clinton, SC also stated that this problem is well-known, as noted in [1] (below), and that the sequence {ni} (giving the infinitely many Heronian triangles with side lengths (ni − 1, ni, ni + 1), where {ni} is defined by

n1 = 4, n2 = 14, and ni+2 = 4ni+1 − ni for i ≥ 1 may also be given in the closed form √ √ i i ni = (2 + 3) + (2 − 3) .

Reference: [1] H. W. Gould, A Triangle with Integral Sides and Area, The Fibonacci Quarterly, Vol. 11(1) 1973, 27-39.

Also solved by Brian D. Beasley, Presbyterian College, Clinton, SC; Bruno Salgueiro Fanego of Viveiro, Spain; Ed Gray, Highland Beach, FL; Kee-Wai Lau, Hong Kong, China; David E. Manes, Oneonta, NY; Trey Smith, Angelo State University, San Angelo, TX; David Stone and John Hawkins, Georgia Southern University, Statesboro, GA, and the proposers.

5471: Proposed by Arkady Alt, San Jose, CA For natural numbers p and n where n ≥ 3 prove that

1 1 n np > (n + p) (n+1)(n+2)(n+3)···(n+p) .

Solution 1 by Moti Levy, Rehovot, Israel

1 The function f (x) = x x is strictly monotone decreasing for x ≥ 3 > e, since 0 1 x 1 f (x) = x x2 (1 − ln x) < 0, for x > e. Hence n + p > n implies

1 1 n n > (n + p) (n+p) .

It follows that 1 1 1  1  np−1   np−1 n n > (n + p) (n+p) , or 1 1 1 1   np−1 p−1 n n = n np > (n + p) (n+p)n . To complete the solution, we note that

1 1 (n + p) np−1(n+p) > (n + p) (n+1)(n+2)···(n+p) .

9 Solution 2 by Kee-Wai Lau, Hong Kong, China We prove the equivalent inequality

ln n ln(n + p) > , (1) np (n + 1)(m + 2) ··· (n + p) by induction on p. ln x 1 − ln x For x ≥ 3 let f(x) = . Since f 0(x) = < 0, so f(x) is strictly decreasing. x x2 Hence f(n) > f(n + 1) and so (1) is true for p = 1. Suppose that (1) is true for p = k ≥ 1. By the induction assumption, we have

ln n 1 ln n ln(n + k) = > = nk+1 n nk n(n + 1)(n + 2) ··· (n + k)

ln(n + k + 1) k ln(n + k) = + + (n + 1)(n + 2) ··· (n + k)(n + k + 1) n(n + 1)(n + 2) ··· (n + k)2

1 + (f(n + k) − f(n + k + 1)) (n + 1)(n + 2) ··· (n + k)

ln(n + k + 1) > . (n + 1)(n + 2) ··· )n + k)(n + k + 1)

Thus (1) is true for p = k + 1 as well and hence true for all positive integers p.

Also solved by Ed Gray, Highland Beach, FL; Paolo Perfetti, Department of Mathematics, Tor Vergata University, Rome, Italy; Ioannis D. Sfikas, National and Kapodistrian University of Athens, Athens, Greece; Albert Stadler, Herrliberg, Switzerland, and the proposer.

5472: Proposed by Francisco Perdomo and Angel´ Plaza, both at Universidad Las Palmas de Gran Canaria, Spain Let α, β, and γ be the three angles in a non-right triangle. Prove that

1 + sin2 α 1 + sin2 β 1 + sin2 γ 1 + sin α sin β 1 + sin β sin γ 1 + sin γ sin α + + ≥ + + cos2 α cos2 β cos2 γ 1 − sin α sin β 1 − sin β sin γ 1 − sin γ sin α.

Solution 1 by Albert Stadler, Herrliberg, Switzerland We prove more generally that

1 + a2 1 + b2 1 + c2 1 + ab 1 + bc 1 + ca + + ≥ + + , if 0 ≤ a, b, c < 1. 1 − a2 1 − b2 1 − c2 1 − ab 1 − bc 1 − ca

The special case follows by putting a = sin α, b = sin β, c = sin γ, with α + β + γ = π.

10 Indeed, 1 1 + x2 1 1 + y2 1 + xy (x − y)2(1 + xy) · + · − = ≥ 0. 2 1 − x2 2 1 − y2 1 − xy (1 − x2)(1 − y2)(1 − xy) So 1 + a2 1 + b2 1 + c2 1 + ab 1 + bc 1 + ca 1 1 + a2 1 1 + b2 1 1 + ab + + − − − = · + · − · + 1 − a2 1 − b2 1 − c2 1 − ab 1 − bc 1 − ca 2 1 − a2 2 1 − b2 2 1 − ab 1 1 + b2 1 1 + c2 1 1 + bc 1 1 + c2 1 1 + a2 1 1 + ca · + · − · + · + · − · ≥ 0. 2 1 − b2 2 1 − c2 2 1 − bc 2 1 − c2 2 1 − a2 2 1 − ca

Solution 2 by Moti Levy, Rehovot, Israel Let a = sin α, b = sin β and c = sin γ. Then the inequality becomes 1 + a2 1 + b2 1 + c2 1 + ab 1 + bc 1 + ca + + ≥ + + , 1 − a2 1 − b2 1 − c2 1 − ab 1 − bc 1 − ca 2x 1 + x2 and since 1 + = , then it is equivalent to the following inequality, 1 − x2 1 − x2 a2 b2 c2 ab bc ca + + ≥ + + , 0 ≤ a, b, c, < 1. 1 − a2 1 − b2 1 − c2 1 − ab 1 − bc 1 − ca The function x2 f (x) := (1) 1 − x2

0 2x is monotone increasing and convex in 0 ≤ x < 1, since f (x) = ≥ 0, and (1 − x2)2 2 00 3x + 1 f (x) = 2 > 0 for 0 ≤ x < 1. (1 − x2)3 By definition (1) the right hand side is ab ca bc √  √ √  + + = f ab + f ca
+ f bc , 1 − ab 1 − ca 1 − bc and the left hand side is a2 b2 c2 + + = f (a) + f (b) + f (c) . 1 − a2 1 − b2 1 − c2 Without loss of generality, we may assume that a ≥ b ≥ c. Then the vector (a, b, c) a + b c + a b + c majorizes the vector , , . 2 2 2 By the Majorizing Inequality, a + b c + a b + c f (a) + f (b) + f (c) ≥ f + f + f . (2) 2 2 2 a + b √ c + a √ b + c √ By the AM-GM inequality ≥ ab, ≥ ca and ≥ bc. The function 2 2 2 f (x) is monotone increasing, hence a + b c + a b + c √  √ √  f + f + f ≥ f ab + f ca
+ f bc . (3) 2 2 2

11 Inequalities (2) and (3) imply the required result. In order to make this solution self-contained, the definition of majorizing and the Majorizing Inequality are explained here. The explanations are excerpted from a nice short article by Murray S. Klamkin (1921-2004) who was one of the greatest problems composer. M. S. Klamkin, On a “Problem of the Month”, Crux Mathematicorum, Volume 28, Number 2, page 86, 2002. “If A and B are vectors (a1, a2, . . . , an) , (b1, b2, . . . , bn) where a1 ≥ a2 ≥ ... ≥ an, b1 ≥ b2 ≥ ... ≥ bn, and a1 ≥ b1, a1 + a2 ≥ b1 + b2, a1 + a2 + ··· + an−1 ≥ b1 + b2 + ··· + bn−1, a1 + a2 + ··· + an = b1 + b2 + ··· + bn, we say that A majorizes B and write it as A B. Then, if F is a convex function,

F (a1) + F (a2) + ··· + F (an) ≥ F (b1) + F (b2) + ··· + F (bn) .”

Also solved by Arkady Alt, San Jose, CA; Hatef I. Arshagi, Guilford Technical Community College, Jamestown, NC; Soumava Chakraborty, Kolkata, India; Pedro Acosta De Leon, Massachusetts Institute of Technology Cambridge, MA; Bruno Salgueiro Fanego, Viveiro, Spain. Ed Gray, Highland Beach, FL; Kee-Wai Lau, Hong Kong, China; Ioannis D. Sfikas, National and Kapodistrian University of Athens, Athens, Greece, and the proposers.

5473: Proposed by Jos´eLuis D´ıaz-Barrero, Barcelona Tech, Barcelona, Spain

Let x1, ··· , xn be positive real numbers. Prove that for n ≥ 2, the following inequality holds:

n ! n ! n X sin xk X cos xk 1 X 1 ≤ 1/2 1/2 2 xk. k=1 ((n − 1)xk + xk+1) k=1 ((n − 1)xk + xk+1) k=1 (Here the subscripts are taken modulo n.)

Solution 1 by Moti Levy, Rehovot, Israel The following three facts will be used in this solution: 1) n ! n ! n !2 X X 1 X a sin x a cos x ≤ a . (4) k k k k 2 k k=1 k=1 k=1 This can be shown by expanding the left hand side and using the facts that 1 sin xk cos xk ≤ 2 and sin xj cos xk + cos xj sin xk ≤ 1. 2) n √ !2 n X ak X ak ≤ . (5) n n k=1 k=1

This is implied from M 1 ≤ M1 where Mk are power means. 2 3) 1 1  1 1 ≤ + , p, q ≥ 0 and p + q = 1. (6) px + qy 2 x y This can be shown by Jensen’s inequality.

12 Now let 1 ak := 1 . ((n − 1) xk + xk+1) 2 Then n ! n ! X sin xk X cos xk LHS := 1 1 k=1 ((n − 1) xk + xk+1) 2 k=1 ((n − 1) xk + xk+1) 2 n ! n ! n !2 X X 1 X = a sin x a cos x ≤ a . k k k k 2 k k=1 k=1 k=1 By (5),

n !2 n n 1 X n X n X 1 LHS ≤ ak ≤ ak = 2 2 2 (n − 1) xk + xk+1 k=1 k=1 k=1 n 1 X 1 = . 2 n−1 x + 1 x k=1 n k n k+1 n−1 1 Set p = n and q = n , then by (6) n n n 1 X 1 1 X  1 1  1 X 1 n−1 1 ≤ + = . 2 x + x 4 xk xk+1 2 xk k=1 n k n k+1 k=1 k=1

Solution to 2 by Kee-Wai Lau, Hong Kong, China Since 2ab ≤ a2 + b2 for any real numbers a and b, so by the Cauchy-Schwarz inequality, we have

n ! n ! X sin xk X cos xk 2 1/2 1/2 k=1 ((n − 1)xk + xk+1) k=1 ((n − 1)xk + xk+1)

n !2 n !2 X sin xk X cos xk ≤ + 1/2 1/2 k=1 ((n − 1)xk + xk+1) k=1 ((n − 1)xk + xk+1)

n 2 n 2 ! X sin xk X cos xk ≤ n + (n − 1)xk + xk+1 (n − 1)xk + xk+1 k=1 k=1

n X 1 = n . (n − 1)xk + xk+1 k=1

1 Applying Jensen’s inequality to the convex function for x > 0, we have x ! n − 1 1 1 n2 + ≥ n = . (n−1)x +x xk xk+1 k k+1 (n − 1)xk + xk+1 n 13 n n n ! n X 1 1 X n − 1 X 1 X 1 It follows that n ≤ + = . (n − 1)xk + xk+1 n xk xk+1 x k=1 k=1 k=1 k=1

Thus the inequality of the problem holds.

Solution 3 by Arkady Alt , San Jose, CA

s s n sin x n 1 n By Cauchy Inequality P k ≤ P · P sin2 x 1/2 k k=1 ((n − 1) xk + xk+1) k=1 (n − 1) xk + xk+1 k=1 s s n cos x n 1 n and P k ≤ P · P cos2 x . 1/2 k k=1 ((n − 1) xk + xk+1) k=1 (n − 1) xk + xk+1 k=1

Also by AM-GM inequality s n s n  n n  n P 2 P 2 1 P 2 P 2 1 P 2 2
n sin xk · cos xk ≤ sin xk + cos xk = sin xk + cos xk = . k=1 k=1 2 k=1 k=1 2 k=1 2

! ! n sin x n cos x n n 1 Thus, P k P k ≤ P 1/2 1/2 k=1 ((n − 1) xk + xk+1) k=1 ((n − 1) xk + xk+1) 2 k=1 (n − 1) xk + xk+1 and it remains to prove the inequality n n 1 1 n 1 n 1 1 n 1 P ≤ P ⇐⇒ P ≤ P . 2 k=1 (n − 1) xk + xk+1 2 k=1 xk k=1 (n − 1) xk + xk+1 n k=1 xk

By the Cauchy Inequality     n − 1 1 2 1 1 n − 1 1 ((n − 1) xk + xk+1) + ≥ n ⇐⇒ ≤ 2 + xk xk+1 (n − 1) xk + xk+1 n xk xk+1 n n   n P 1 1 P n − 1 1 1 P 1 then ≤ 2 + = . k=1 (n − 1) xk + xk+1 n k=1 xk xk+1 n k=1 xk

Also solved by Ed Gray, Highland Beach, FL; Ioannis D. Sfikas, National and Kapodistrian University of Athens, Athens, Greece; Albert Stadler, Herrliberg, Switzerland, and the proposer.

5474: Proposed by Ovidiu Furdui, Technical University of Cluj-Napoca, Cluj-Napoca, Romania

Let a, b ∈ <, b 6= 0. Calculate  a b n 1 −  n2 n    lim   . n→∞  b a  1 + . n n2

Solution 1 by Bruno Salgueiro Fanego, Viveiro, Spain

14  a b  1 − √  n2 n  b2n2 + a2 Let M =  . The eigenvalues of M are 1 ∓ .   2  b a  n 1 + n n2

Since they are distinct, M is diagonalizable. It can be obtained by following the √  b2n2 + a2  1 − 0  n2    −1   diagonalization of M : M = PDP , where D =   is the  √   b2n2 + a2  0 1 + n2 diagonal matrix whose principal diagonal are its eigenvalues and √ √ −a − b2n2 + a2 −a + b2n2 + a2   bn bn    P =   is the matrix whose columns are the     1 1 respective eigenvectors which form a basis of R2. Hence,

M n = P · Dn · P −1

√ n √ √  b2n2+a2    −a− b2n2+a2 −a+ b2n2+a2  1 − n2 0 bn bn   =   ·   ·   √ n 1 1 b2n2+a2 0 1 + n2 √ 2 2 2  √ −bn −a√+ b n +a  2 b2n2+a2 2 b2n2+a2 m m  ·  , that is M n = 11 12 where √  2 2 2  m21 m22 √ bn a+√ b n +a 2 b2n2+a2 2 b2n2+a2

2  q n  q n  q n  q n an b
2 a
2 b
2 a
2 1 b
2 a
2 b
2 a
2 m11 = m22 = √ 1 − + 2 − 1 + + 2 + 1 − + 2 + 1 + + 2 2 b2n6 + a2n4 n n n n 2 n n n n and 3  q n  q n bn b
2 a
2 b
2 a
2 m12 = m21 = √ 1 + + 2 − 1 − + 2 . 2 b2n6 + a2n4 n n n n

Thus, as n → ∞,

r r r r 2  2 2  2 ! 2  2 2  2 ! an2 −n ( b ) + a n ( b ) + a 1 −n ( b ) + a n ( b ) + a m = m ∼ e n n2 − e n n2 + e n n2 + e n n2 11 22 2|b|n3 2 ∼ a   1   ∼ 1   e−|b| − e|b| + e−|b| + e|b| 0 + e−|b| + e|b| = cosh |b| n → ∞ 2|b|n 2 n → ∞ 2 and as n → ∞

15 r r 2  2 2  2 ! bn3 n ( b ) + a −n ( b ) )+ a m = m ∼ e n n2 − e n n2 12 21 2|b|n3

b   b ∼ e|b| − e−|b| = sinh |b|, so, 2|b|n |b|

 b  cosh |b| |b| sinh |b| lim M n =   . n→∞  b  |b| sinh |b| cosh |b|

Solution 2 by Brian Bradie, Christopher Newport University, Newport News, VA

Let  a b  1 −  n2 n  A =  b a  . 1 + n n2 The eigenvalues of A are

1 p 1 p λ = 1 + a2 + n2b2 and λ = 1 − a2 + n2b2. + n2 − n2 Because b 6= 0, these two eigenvalues are distinct, which implies that A is diagonalizable. An eigenvector associated with λ+ is

b ! 1 p , (a + a2 + n2b2) n and an eigenvector associated with λ− is

b ! 1 p . (a − a2 + n2b2) n Thus, if we set b b ! P = 1 p 1 p (a + a2 + n2b2) (a − a2 + n2b2) n n and  1 p  1 + a2 + n2b2 0  n2  D =  1 p  , 0 1 − a2 + n2b2 n2 then A = PDP −1 and An = PDnP −1. Now,

 b b  1  −|b| −b  lim P = and lim P −1 = − . n→∞ |b| −|b| n→∞ 2b|b| −|b| b

16 Moreover,   1 p n  1 + a2 + n2b2 0 n2 Dn =   ,   1 p n   0 1 − a2 + n2b2  n2 so  e|b| 0  lim Dn = . n→∞ 0 e−|b| Thus,

1  b b   e|b| 0   −|b| −b  lim An = − n→∞ 2b|b| |b| −|b| 0 e−|b| −|b| b b ! cosh |b| |b| sinh |b| = b |b| sinh |b| cosh |b|  cosh b sinh b  = . sinh b cosh b

Remark: This problem is very similar to Problem 1113 from the November 2017 issue of The College Mathematics Journal.

Editor0s comment : David Stone and John Hawkins both at Georgia Southern Uni- versity in Statesboro, GA accompanied their solution with the following comment: “At a first, we accidentally used 1 − in the (2, 2) spot of the matrix and the limit was the same. n2 Perhaps there is a lot of flexibility about this term (since the limit does not depend upon a).”

Also solved by Ulich Abel, Technische Hochschule Mittelhessen, Germany; Hatef I. Arshagi, Guilford Technical Community College, Jamestown, NC; Kee-Wai Lau, Hong Kong, China; Moti Levy, Rehovot, Israel; Ioannis D. Sfikas, National and Kapodistrian University of Athens, Greece; Francisco Perdomo and Angel´ Plaza, Universidad de Las Palmas de Gran Canaria, Spain; Albert Stadler, Herrliberg, Switzerland; David Stone and John Hawkins, Georgia Southern University, States- boro, GA; Anna V. Tomova, Varna, Bulgaria, and the proposer.

Late Solutions

A late solution was received from Paul M. Harms of Newton, KS to problem 5467.

17 Problems Ted Eisenberg, Section Editor *********************************************************

This section of the Journal offers readers an opportunity to exchange interesting mathematical problems and solutions. Please send them to Ted Eisenberg, Department of Mathematics, Ben-Gurion University, Beer-Sheva, Israel or fax to: 972-86-477-648. Questions concerning proposals and/or solutions can be sent e-mail to . Solutions to previously stated problems can be seen at .

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Solutions to the problems stated in this issue should be posted before June 15, 2018

5493: Proposed by Kenneth Korbin, New York, NY Convex quadrilateral ABCD is inscribed in a circle with diameter AC = 729. Sides AB and CD each have positive integer length. Find the perimeter if BD=715.

5494: Proposed by Moshe Stupel, “Shaanan” Academic College of Education and Gordon Academic College of Education, and Avi Sigler, “Shaanan” Academic College of Education, Haifa, Israel If a ≥ b ≥ c ≥ d are the lengths of four segments from which an infinite number of convex quadrilaterals can be constructed, calculate the maximal product of the lengths of the diagonals in such quadrilaterals.

5495: Proposed by D.M. Ba˘tinetu-Giurgiu, “Matei Basarab” National College, Bucharest and Neculai Stanciu, “George Emil Palade” School Buza˘u, Romania √ √ 3 pn Let {xn}n≥1, x1 = 1, xn = 1 · 3!! · 5!! · ... (2n − 1)!!. Find:  (n + 1)2 n2  lim √ − √ . n→∞ n+1 xn+1 n xn

5496: Daniel Sitaru, “Theodor Costescu” National Economic College, Drobeta Turnu-Severin, Mehedinti, Romania Let a, b, c be real numbers such that 0 < a < b < c. Prove that: √ ab X   X  b  ea−b + eb−a ≥ 2a − 2c + 3 + . a cyclic cyclic

5497: Proposed by Jos´eLuis D´ıaz-Barrero, Barcelona Tech, Barcelona, Spain

1 n−1 Y kπ  For all integers n ≥ 2, show that 2 sin is an integer and determine it. n k=1

5498: Proposed by Ovidiu Furdui and Alina Sˆintam˘ a˘rian, both at the Technical University of Cluj-Napoca, Cluj-Napoca, Romania Prove that ∞ X {n!e} Z 1 ex − 1 = dx n x n=1 0 where {a} denotes the fractional part of a.

Solutions

5475: Proposed by Kenneth Korbin, New York, NY √ a + b = 14 ab − 48,  √ Given positive integers a, b, c and d such that b + c = 14√bc − 48, with a < b < c < d. c + d = 14 cd − 48, Express the values of b, c, and d in terms of a.

Solution 1 by David E. Manes, Oneonta, NY

If a = 1 < b = 97 < c = 18817 < d = 3650401, then it is easily verified that these positive integers√ satisfy the system of equations. For the first equation a + b = 14 ab − 48, square both sides, simplify and write the equation as a quadratic in b. Then one obtains b2 − (194a)b + (a2 + 9408) = 0 with roots b = 97a  56p3(a2 − 1). Note that if a = 1, then b = 97. For the second equation, by symmetry, c = 97b  56p3(b2 − 1). If b = 97, then c = 972  56p3(972 − 1) = 18817 or 1. Therefore, c = 18817 since b < c. Writing c in terms of a, we first note that  p 2 b2 = 97a  56 3(a2 − 1) p = 18817a2  10864a 3(a2 − 1) − 9408. Therefore, p c = 97b  56 3(b2 − 1) r  p   p  = 97 97a  56 3(a2 − 1)  56 3 18817a2  10864a 3(a2 − 1) − 9409 .

If a = 1, then c = 18817 using the positive radicals. At this point, let α = 56p3(a2 − 1) r  and β = 56 3 18817a2  10864ap3(a2 − 1) − 9409. With these substitutions, the equation for c reads c = 972a  97α  β and if a = 1, then α = 0 and β = 9408. For the last equation, d = 97c  56p3(c2 − 1) again by symmetry. If c = 18817, then d = 3650401 or 97 and c < d implies d 6= 97. Expressing d in terms of a, observe that 2 c2 = 972a  97α  β
= 974a2 + 972α2 + β2  2 · 973aα  2 · 972aβ  2 · 97αβ.

2 Therefore, p d = 97c  56 3(c2 − 1) p = 973a  972α  97β  56 3 (974a2 + 972α + β2  2 · 973aα  2 · 972aβ  194αβ − 1).

Finally, note that if a = 1, then α = 0, β = 9408 and d = 3650401 for the positive signs. This completes the solution.

Solution 2 by Ioannis D. Sfikas, National and Kapodistrian University of Athens, Greece √ Squaring the equation x + y = 14 xy − 48 with x, y positive integers and x < y, yields the quadratic equation of y = y(x).

y2 − 194xy + x2 + 9408 = 0 (1)

The discriminant ∆ is ∆ = 37632(x2 − 1). Since x and y are positive integers, hence p 2 p 2 x ≥ 1 and y1 = 97x + 56 3(x − 1) or y2 = 97x − 56 3(x − 1).

Nevertheless, there is x < y2 which holds for 1 ≤ x < 7. Applying (1) to the given system there is p 2 p 2 p 2 b2 = 97a − 56 3(a − 1), c2 = 97b2 − 56 3(b2 − 1) and d2 = 97c2 − 56 3(c2 − 1), with 1 ≤ a < 7.

(1) For a = 1, then b2 = 97, c2 = 1 a contradiction. (2) For a = 2, then b2 = 26, c2 = 2, a contradiction. (3) For a = 3, 4, 5, or 6, b2 is not an integer. So, the solution (a, b2, c2, d2) is rejected.

Furthermore, there is x < y1, which holds for x ≥ 1. We may generalize the problem: n √ given positive integers {xi}i=1 such that xi + xi+1 = 14 xixi+1 − 48 with xi < xi+1, then we have the recursive relation q 2 xi+1 = 97xi + 56 3(xi − 1).

Again, applying (1) to the given system there are the following recursive relations: q q p 2 2 2 b1 = 97a+56 3(a − 1), c1 = 97b1+56 3(b1 − 1) and d1 = 97c1 +56 3(c1 − 1), with a ≥ 1.

So, we may list some solutions:

a b1 c1 d1 1 97 18817 3650401 2 362 70226 13623482 7 1351 262087 50843527 26 5042 978122 189750626 97 18817 3650401 708158977 We can assume that equation (1) is a Diophantine equation. Then, possible solutions are (x, y) = (1, 97), (2, 26), (92, 362), (26, 5042), (97, 18817). Equation (1) reduces to a Diophantine equation of the Pell type. We may write x2 − 194xy + y2 in the matrix

3  1 −97  x,  form x, y
. This matrix has eigen vectors (1, 1), which leads −97 1 y us to consider u1 = x + y and v1 = y − x. Then, equation (1) becomes 2 2 6 2 49v1 − 48u1 + 9408 = 0. Since 9408=2 · 3 · 7 , this impies that u1 = 7u and v1 = 12v and so, u2 − 3v2 = 4, a Pell equation. The Pell equation has an infinity of integer solutions in general and the Pell equation implies √  √ n  √ n 3 h √ n  √ ni u = 2 − 3 + 2 + 3 , v = 2 + 3 − 2 − 3 , 3 with n ∈ N, or, u − v 7u − 12v 1 h √   √ n  √   √ ni x = 1 1 = = 7 + 4 3 2 − 3 + 7 − 4 3 2 + 3 . 2 2 2 So, we may list some solutions:

n x y c1 d1 1 2 362 70226 13623482 2 1 97 18817 3650401 3 2 362 70226 13623482 4 7 1351 262087 50843527 5 26 5042 978122 189750626 6 97 18817 3650401 708158977 7 362 70226 13623482 2642885282 8 1351 262087 50843527 9863382151 9 5042 978122 189750626 36810643322 10 18817 3650401 708158977 137379191137

Solution 3 by Kee-Wai Lau, Hong Kong, China We show that either (a, b, c, d) = (2, 26, 5042, 978122) or √ √ (2 + 3)k + (2 − 3)k p p , 97a + 56 3(a2 − 1), 18817a + 10864 3(a2 − 1), 2  p 3650401a + 2107560 3(a2 − 1) , for nonnegative integers k, there are k solutions. Squaring both sides of the given equations, we obtain respectively,

b2 − 194ab + a2 + 9408 = 0, (1)

c2 − 194bc + b2 + 9408 = 0, (2)

d2 − 194cd + c2 + 9408 = 0. (3)

By subtracting (1) from (2), we obtain (c − a)(c + a − 194b) = 0. Since c > a, so c = 194b − a. Similarly by subtracting (2) from (3), we obtain d = 194c − b = 37635b − 194a. From (1), we obtain b = 97a  56p3(a2 − 1). Since b is an integer, so 3(a2 − 1) is a square, which leads to the Pell Equation

4 √ √ (2 + 3)k + (2 − 3)k a2 − 3t2 = 1. It is well known that a must be of the form . 2 We first suppose that: b = 97a − 56p3(a2 − 1), with a > 1. Since b > a, we deduce with some algebra that a < 7. This gives a = 2 and hence b = 26, c = 5042, d = 978122. We next suppose that b = 97a + 56p3(a2 − 1), where a ≥ 1. We the obtain the solutions stated at the beginning and this completes the solution.

Solution 4 by Kenneth Korbin (the proposer) NewYork, NY

Sequence x = (1, 2, 7, 26, 97, 362, 1351, . . . , xN ,... ) with xN+1 = 4xN − xN−1

xN + xN+4 = 14xN+2 p = 14 (xN )(xN+4) − 48

1 + 97 = 14(7) = 14p(1)(97) − 48 2 + 362 = 14(26) = 14p(2)(362) − 48 7 + 1351 = 14(97) = 14p(7)(1351) − 48 etc.

If a = xN , then b = xN+4, c = xN+8, d = xN+12, Sequence

y = (xN , xN+4, xN+8, xN+12, xN+16, ··· )

y = (a, b, c, d, xN+16, xN+20, ··· ) c = 194b − a ac − b2 = 9408 b2 + 9408 Therefore, c = a b2 + 9408 194b − a = = c a √ Therefore, b = 97a + 56 3a2 − 3 c = 194b − a √ Therefore, c = 18817a + 210864 3a2 − 3 d = 194c − b √ Therefore, = 36590401a + 2107560 3a2 − 3

Editor0s Comment : As with previous problems, David Stone and John Hawkins of Georgia Southern University in Statesboro, GA attached comments about the problem and their solution that I believe are both informative and instructive to the readership. They are listed below. Comment 1: The points (a, b), (b, c) and (c, d) all lie on the hyperbola x2 − 194xy + y2 = −9408.

5 This is actually the hyperbola 48x2 − 49y2 = 4704, after a 45-degree rotation. √ 2 Our hyperbola lies in the first quadrant; it asymptotes are y = 4 3 + 7
x and 1 y = √ 2 x, and all of the points (a, b), (b, c), and (c, d) lie very close to the steep 4 3 + 7
asymptote whose equation is “almost” y = 194x. p 2 Comment 2: Our function f should probably be denoted f+(x) = 97x + 56 3 (x − 1). (See solutions 1, 2 or 3 for the motivation of its derivation.) The companion function p 2 f−(x) = 97x − 56 3 (x − 1) actually serves as the inverse of f with suitable restrictions −1 on the domain: f+ : [1, ∞) → [97, ∞) is a bijection and (f+) = f− : [97, ∞) → [1, ∞). Without the ordering conditions on a, b, c and d, we could use f+ and f− randomly to generate solutions based upon an appropriate value for a.

Also solved by Ed Gray, Highland Beach, FL; David Stone and John Hawkins, Georgia Southern University, Statesboro, GA.

5476: Proposed by Ed Gray, Highland Beach, FL Find all triangles with integer area and perimeter that are numerically equal.

Solution 1 by Henry Ricardo, Westchester Area Math Circle, NY If we assume that any triangle satisfying our condition must have integer sides, then this is an old problem, that of finding all equable Heronian triangles [1], [2].

The solution is that there exist only five triangles with numerically equal area and perimeter—those with sides (6, 8, 10), (5, 12, 13), (6, 25, 29), (7, 15, 20), (9, 10, 17). (The first two are the only right triangles.)

Prielipp [3] has proved that a triangle has equal area and perimeter if and only if it can be circumscribed about a circle of radius 2. Kilmer [4] uses Prielipp’s result to generate triangles of equal area and perimeter. Markowitz [5] shows that there exists an infinite number of right triangles having rational side lengths for which the area equals the perimeter. Bates [6] has shown that a right triangle ∆ABC with 6 C = 900 has numerically equal area and perimeter if and only if a + b − c = 4.

References [1] “Heronian Triangle,” Wikipedia article. [2] “Equable Shape,” Wikipedia article. [3] “Area = Perimeter,” Robert Prielipp, Math.Teacher 78 (February 1985), 90; 127. [4] “Triangles of Equal Area and Perimeter and Inscribed Circles,” Jean Kilmer, Math.Teacher 71 (January 1988), 65-70

[5] “Area = Perimeter,” Lee Markowitz, Math.Teacher 74 (March 1981), 222-223. [6] “Serendipity on the Area of a Triangle,” Madelaine Bates, Math.Teacher 72 (April 1979), 273-275.

Solution 2 by Ioannis D. Sfikas, National and Kapodistrian University of Athens, Greece

6 A triangle whose sides and area are rational numbers is called a rational triangle. If the rational triangle is right-angled, it is called a right-angled rational triangle or a numerical right triangle. If the sides of a rational triangle is of integer length, it is called an integer triangle. If further these sides have no common factor greater than unity, the triangle called primitive integer triangle. A Heronian triangle (named after Heron of Alexandria) is an integer triangle with the additional property that its area is also an integer [1]. In 1904, W.A. Whitworth and D. Biddle proved that there are only five Heronian triangles with integer sides whose area equals the perimeter, namely (9,10,17), (7,15,20), (6,25,29), (6,8,10) and (5,12,13) [2]. These Heronian triangles are called perfect triangles. In 1955, R.R. Phelps suggested the problem of finding all Heronian triangles whose integer valued sides add up to twice its area [3]. N. J. Fine solved the last problem and found that there is only one such triangle, namely (3,4,5) [4]. Subbarao [5] proved that the number of integer triangles, whose integer valued-sides add up to λ times their area A is finite, namely P = λA, with λ > 0 and P is√ perimeter. Furthermore, he showed that the number of integer triangles for λ > 8 is zero, while√ he didn’t determine the particular values of sides of integer triangles for λ ≤ 8 [specifically, he mentioned other articles about perfect triangles and the right-angled triangle (3,4,5)]. A similar problem concerns the search of the number of Heronian triangles whose integer 1 valued-sides add up to times their area A, namely A = nP , with n ∈ N. Markov [6], n [7] found an algorithm for the listing of all Heronian triangles with the property A = nP . In 1985, Goehl [8] solved that particular problem in the special case of right triangles. In addition, the (3,4,5) right-angled triangle is the integer-sided triangle for which the A ratio is a rational number less than 1, it actually has the smallest such ratio [9]. P In the usual notation, we have from the hypothesis and the classical area formula of Heron 2s = m ps(s − a)(s − b)(s − c). (1) With x = s − a, y = s − b, z = s − c, we have s = x + y + z and (1) becomes

4(x + y + z) = xyz. (2)

Suppose (without loss of generality) x ≤ y ≤ z. Then 3 < x implies xyz ≥ 16z and x + y + z ≤ 3z so that 4(x + y + z)) ≤ 12z < 16z ≤ xyz, and (2) would be impossible. Hence, we need try only x = 1, 2, 3. (a) For x = 1, (2) becomes (y − 4)(z − 4) = 20. For y ≥ 9 , then y − 4 ≥ 5 and (y − 4)(z − 4) ≥ 25, a contradiction. Furthermore, for y ≤ 4 and y = 7, a contradiction. So, for y = 5, 6, 8, we have the following three perfect triangles: T1 = (6, 25, 29),T2 = (7, 15, 20),T3 = (9, 10, 17). (b) For x = 2, (2) becomes (y − 2)(z − 2) = 8. For y ≥ 5, then y − 2 ≥ 3 and (y − 2)(z − 2) ≥ 9, a contradiction. Furthermore, for y = 1, 2, a contradiction. So, for y = 3, 4 we have the following two perfect triangles: T4 = (5, 12, 13) and T5 = (6, 8, 10). Notice that each of the pairs (T1,T5) and (T3,T5) have a common side. These pairs can be placed along their common sides to form a large triangle in each case [10]. In particular, the perfect triangle T4 and T5 are right-angled triangles [11]. (c) For x = 3, (2) becomes (3y − 4)(3z − 4) = 52. Then, y 6= 3, since 3y − 4 = 5 is not a factor of 52. Furthermore, y ≤ 3, since 4 ≤ y ≤ z implies 8 ≤ 3y − 4, whence

7 (3y − 4)(3 − 4) ≥ 64, a contradiction. References [1] Carmichael, Robert D. (1915). Diophantine analysis. New York: John Wiley and Sons. [2] Dickson, Leonard Eugene (2005). History of the theory of numbers, Volume II: Diophantine Analysis. Dover Publications. [3] Bankoff, Leon; Olds, C. D.; Phelps, R. R.; Lehner, Joseph and Linis, Viktors (1955). Problems for solution: E1166-E1170. The American Mathematical Monthly, 62 (5): 364-365. [4] Phelps, R. R. and Fine, N. J. (1956). E1168. The American Mathematical Monthly, 63 (1): 43-44. [5] Subbarao, M. V. (1971). Perfect triangles. The American Mathematical Monthly, 78 (4): 384-385. [6] Markov, Lubomir P. (2006). Pythagorean triples and the problem A = mP for triangles. Mathematics Magazine, 79 (2): 114-121. [7] Markov, Lubomir (2007). Heronian triangles whose areas are integer multiples of their perimeters. Forum Geometricorum, 7: 129-135. [8] Goehl Jr., John F. (1985). Area = k(perimeter). The Mathematics Teacher, 78 (5): 330-332. [9] Dolan, Stan (2016). Less than equable Heronian triangles. The Mathematical Gazette, 100 (549): 482-489. [10] Rabinowitz, Stanley (1992). Index to Mathematical Problems 1980-1984 (Indexes to mathematical problems). Mathpro Press. [11] Markowitz, Lee (1981). Area = Perimeter. The Mathematics Teacher, 74 (3): 222-223.

Comment submitted by Dionne Bailey, Elsie Campbell, Charles Diminnie, and Trey Smith, Angelo State University, San Angelo, TX There are only five examples of triangles with integer sides for which the area and perimeter are equal and integer-valued. These are the triangles for which (a, b, c) = (5, 12, 13), (6, 8, 10), (6, 25, 29), (7, 15, 20), and (9, 10, 17). These can be found by using an algorithm described in Reference [1]. However, it’s possible to find an infinite number of examples of triangles where at least one side is irrational and yet the area and perimeter are equal and integer-valued. For example,√ if n is an integer and n ≥ 4, let √ a = 2n − n2 − 12, b = 2n, and c = 2n + n2 − 12. When n = 4, (a, b, c) = (6, 8, 10). It is easily shown that when n > 4, n2 − 12 cannot be a perfect square and hence, a and c are irrational. We note also that for all n ≥ 4, a < b < c and  √  a + b − c = 2 n − n2 − 12 > 0. Consequently, we have a < b < c and a + b > c, which guarantees that there is a non-degenerate triangle with sides a, b, and c. For this triangle, the perimeter P = 6n and the semi-perimeter s = 3n. Then, Heron’s Formula

8 for the area A yields

A = ps (s − a)(s − b)(s − c) r  p   p  = (3n) n + n2 − 12 (n) n − n2 − 12 p = 3n2 [n2 − (n2 − 12)] √ = 36n2 = 6n = P.

References: [1] T. Leong, D. Bailey, E. Campbell, C. Diminnie, and P. Swets, Another Approach to Solving A = mP for Triangles, Mathematics Magazine 80, pp. 363 - 368, 2007.

Editor0s comment : Some readers asked if the side lengths had to be integers, and from the history of the problem we can see that that was the intent originally. But as mentioned in the comment by Bailey, Campbell, Diminnie, and Smith, the side lengths need not be integers and this is the territory where the solution of Stone and Hawkins took them. Daivd Stone and John Hawkins of Georgia Southern University, seem to have rediscovered a version of the result cited by Bailey, et.al., that the side lengths need not be rational. In their solution they stated and proved the following algorithm: For any integer P ≥ 21, there are infinitely many triangles with A = P . All such triangles are given by the following prescription; Let P ≥ 21 be an integer. Choose b such that b > 4 and 2b3 − P b2 + 16P ≤ 0. 16P Compute z = b2 − . P − 2b P − b 1√ P − b 1√ Let a = − z and c= + z. 2 2 2 2 Then a, b, c form a triangle with area and perimeter P . (After verifying the above algorithm they presented the following examples.) Example 1: The first example has two irrational sides, but still has A = P = integer. Let P = 26. Then we must choose b such that 2b3 − 26b2 + 432 ≤ 0. That is, 5.146 < b < 11.399. √ 112 √ 4 35 Let b = 8, then z = , so z = . 5 5 Thus the other two√ sides of our√ triangle are√ 26 − 8 1 4 35 2 35 45 − 2 35 a = − = 9 − = ≈ 6.6336 2 2 5 5 5 and √ 45 + 2 35 c = ≈ 11.3664 5 Example 2: This example demonstrates that the minimum value for P, 21 is actually achieved. Let P = 21. Then we must choose b such that 2b3 − 21b2 + 336 ≤ 0. That is, 6.405 < b < 7.562.

9 1 13 1 15 Let b = 7. Then z = 1 so a = 7 − = and = 7 + = . 2 2 2 2 13 14 15 Here we have the triangle , , with rational, non-integer sides with 2 2 2 A = P = 21. Example 3: Note that the previous example used b = P/3. This is always a valid value for b. In this case, we have the triangle P 1√ P P 1√  − P 2 − 432, , + P 2 − 432 3 6 3 3 6 which has A = P . This triangle has rational sides only for P = 21, 24, 31, 39, 56 and 109.

Also solved by Kee-Wai Lau, Hong Kong, China David E. Manes, Oneonta, NY; Albert Stadler, Herrliberg, Switzerland, and the proposer

5477: Proposed by Daniel Sitaru, “Theodor Costescu” National Economic College, Drobeta Turnu-Sevrin, Meredinti, Romania Compute: √ √ √ ! 1 − 1 + x2 3 1 + x2 · ... · n 1 + x2 L = lim ln n + lim . n→∞ x→0 x2

Solution 1 by Ed Gray, Highland Beach, FL We rewrite the expression as: h 2 1 + 1 + 1 +...+ 1 i 1 − (1 + x ) 2 3 n n 1. lim . x→0 x2 k=n X 1 2. Let N = , i.e., the harmonic series −1 k k=2 1 − 1 + x2)N  3. Now consider lim . x→0 x2 We expand (1 + x2)N by the Binomial Theorem: N(N − 1) 4. (1 + x2)N = 1 + Nx2 + x4 + ... 2! Then   N(N − 1)  1 − 1 + Nx2 + x4 + ... 2 5. lim , or x→0 x2  −N(N − 1)  −Nx2 + x4 + ... 2 −Nx2 6. lim = = −N. x→0 x2 x2 The original limit becomes

k=n ! X 1 7. lim (ln(n) − N) = lim ln(n) − = lim (ln(n) + 1 − Harmonic series). n→∞ n→∞ k n→∞ k=2 The Euler-Mascheroni Constant is defined as γ = lim T he Harmonic series − ln(n). n→∞ Therefore our expression in step 7 equals 1 − γ.

10 Solution 2 by Bruno Salgueiro Fanego, Viveiro, Spain Since

1 + 1 +... 1 H −1 1 − 1 + x2
2 3 + n 1 − 1 + x2
n 0  = lim = lim = Indet. x→0 x2 x→0 x2 0 L0Hospital

Hn−2 0 − (Hn − 1) (1 + x) 2x 2 Hn−2 lim = (1 − Hn) lim 1 + x ) = 1 − Hn, x→0 2x x→0

L = lim (ln n + 1 − Hn) = 1 − lim (Hn − ln n) = 1 − γ, n→∞ n→∞ where Hn is the n-th harmonic number and γ is the Euler-Mascheroni constant.

Solution 3 by Paolo Perfetti, Department of Mathematics, Tor Vergata University, Rome, Italy

p p3 pn 2 1 + 1 + 1 2 H −1 1 + x2 1 + x2 ··· 1 + x2 = (1 + x ) 2 3 n = (1 + x ) n We have  1 − (1 + x2)Hn−1  lim ln n + lim n→∞ x→0 x2 Now 1 − (1 + x2)Hn−1 lim = −Hn + 1 x→0 x2 thus L = lim (ln − ln n − γ + o(1) + 1) = −γ + 1 n→∞

Solution 4 by Julio Cesar Mohnsam and Mateus Gomes Lucas, both from IFSUL, Campus Pelots-RS, Brazil, and Ricardo Capiberibe Nunes of E.E. Amlio de Caravalho Bas, Campo Grande-MS, Brazil

 2 Hn−1    1 − (1 + x ) 2 Hn−2 L = lim ln n + lim = lim ln n + lim(1 − Hn)(1 + x ) . n→∞ x→0 x2 n→∞ x→0 because,

2 Hn−1 0 2 Hn−1 0 [1 − (1 + x ) ] 0 [1 − (1 + x ) ] 2 Hn−2 lim = lim = lim(−Hn+1)(1+x ) = −Hn+1 x→0 (x2) x→0 (x2)0 x→0

Therefore:

L = lim (ln n − Hn + 1) = lim (ln n − Hn) + 1 = 1 + lim (ln n − Hn) = 1 − γ n→∞ n→∞ n→∞

1 1 Note: γ is Euler-Mascheroni constant and Hn = 1 + 2 + ··· + n . Also solved by Yen Tung Chung, Taichung, Taiwan; Serban George Florin, Romania; Kee-Wai Lau, Hong Kong, China; Moti Levy, Rehovot, Israel;

11 Angel Plaza, University of Las Palmas de Granada Canaria Spain; Ravi Prakash, New Delhi, India; Henry Ricardo, Westchester Area Math Circle, NY; Ioannis D. Sfikas, National and Kapodistrian University of Athens, Greece; Shivam Sharma, New Delhi, India; Albert Stadler, Herrliberg, Switzerland, and the proposer.

5478: Proposed by D. M. Btinetu-Giurgiu, “Matei Basarab” National Collge, Bucharest, Romania and Neculai Stanciu, “George Emil Palade” Secondary School, Buzu, Romania Compute: Z π/2  π  π  cos2 x sin x sin2 cos x + cos x sin2 sin x dx. 0 2 2

Solution 1 by Karl Havlak, Angelo State University, San Angelo, TX √ Let u = cos x. Then du = − sin xdx and sin(x) = 1 − u2. We may rewrite the given integral as Z 1  π  u3 π p  u2 sin2 u + √ sin2 1 − u2 du. 2 0 2 1 − u 2 √ Considering the second term in the integrand, we let v = 1 − u2 so that u dv = −√ du and u2 = 1 − v2. We may write the integral above as 1 − u2

Z 1 π  Z 1 π  u2 sin2 u du + (1 − v2) sin2 v dv. 0 2 0 2 Z 1 2 π  1 This reduces to sin v dv, which can be easily shown to be 2 . 0 2 Solution 2 by Moti Levy, Rehovot, Israel)

π Z 2  π  π  I := cos2 x
sin x sin2 cos x + cos x sin2 sin x dx 0 2 2 π π Z 2 π  Z 2 π  = cos2 x sin x sin2 cos x dx + cos2 x cos x sin2 sin x dx 0 2 0 2 Change of variables, u = cos x in the first integral and v = sin x in the second integral gives

Z 1 π  Z 1 π  I = u2 sin2 u du + 1 − u2
sin2 u du 0 2 0 2 Z 1 π  1 = sin2 u du = . 0 2 2

Solution 3 by Bruno Salgueiro Fanego, Viveiro, Spain

Z π/2  π  π  I = cos2 x sin x sin2 cos x + cos x sin2 sin x dx 0 2 2 12 Z π/2 π  Z π/2 π  = 1 − sin2 x
sin x sin2 cos x + cos3 x sin2 sin x dx 0 2 0 2

= I1 − I2 + I3 where

Z π/2 Z π/2 2 π  3 2 π  I1 = sin x sin cos x ,I2 = sin x sin cos x dx and 0 2 0 2

Z π/2 3 2 π  I3 = cos x sin sin x dx. 0 2

Since

Z π/2 1 − cos(π cos x) Z π/2 sin x sin x(cos π cos x) I1 = sin x dx = − dx 0 2 0 2 2

 cos x sin(π cos x)x=π/2 = − − 2 2π x=0 cos(π/2) sin(π cos(π/2))  cos 0 sin(π cos 0) = − − − − 2 2π 2 2π

1 1 = 0 − 0 + + 0 = . 2 2

π With the substitution t = − x, one obtains that 2 Z π/2 Z 0 3 2 π  3 π  2 π π  I2 = sin x sin cos x dx = sin − t sin cos − t (−dt) 0 2 π/2 2 2 2

Z π/2 3 2 π  = cos t sin sin t dt = I3. 0 2

1 The value of the given integral is therefore I = I = . 1 2 Also solved by Yen Tung Chung, Taichung, Taiwan; Ed Gray, Highland Beach, FL; Kee-Wai Lau, Hong Kong, China; Paolo Perfetti, Department of Mathematics, Tor Vergata, Rome, Italy; Angel´ Plaza, University of Las Palmas de Gran Canaria, Spain; Ravi Prakash, New Delhi, India; Ioannis D. Sfikas, National and Kapodistrian University of Athens, Greece; Shivam Sharma, New Delhi, India; Albert Stadler, Herrliberg, Switzerland, and the proposers.

13 5479: Proposed by Jos´eLuis D´ıaz-Barrero, Barcelona Tech, Barcelona, Spain Let f : [0, 1] → < be a continuous convex function. Prove that

2 Z 1/3 3 Z 2/3 5 Z 8/15 f(t)dt + f(t)dt ≥ f(t)dt. 5 0 10 0 8 0

Solution 1 by Dionne Bailey, Elsie Campbell, Charles Diminnie, and Trey Smith, Angelo State University, San Angelo, TX If c ∈ (0, 1), then for all t ∈ [0, 1], ct ∈ [0, c] ⊂ [0, 1] and hence, f (ct) is continuous on [0, 1]. Further, by making the substitution u = ct, we get

Z 1 1 Z c 1 Z c f (ct) dt = f (u) du = f (t) dt 0 c 0 c 0 and thus, Z c Z 1 f (t) dt = c f (ct) dt. (1) 0 0 By (1),

1 2 2 Z 3 3 Z 3 f (t) dt + f (t) dt 5 0 10 0 2 1 Z 1 1   3  2 Z 1 2  = f t dt + f t dt 5 3 0 3 10 3 0 3 1 Z 1 2 1  3 2  = f t + f t dt. (2) 3 0 5 3 5 3

1 2 Since f (t) is convex on [0, 1] and 3 t, 3 t ∈ [0, 1] for all t ∈ [0, 1], Jensen’s Theorem implies that

2 1  3 2  2 1  3 2  f t + f t ≥ f t + t 5 3 5 3 5 3 5 3  8  = f t (3) 15 for all t ∈ [0, 1]. By combining (1), (2), and (3), we obtain

1 2 1 2 Z 3 3 Z 3 1 Z 2 1  3 2  f (t) dt + f (t) dt = f t + f t dt 5 0 10 0 3 0 5 3 5 3 1 Z 1  8  ≥ f t dt 3 0 15 8 1 15 Z 15 = f (t) dt 3 8 0 8 5 Z 15 = f (t) dt. 8 0

Solution 2 by Michael Brozinsky, Central Islip, NY

14 We have by the Mean Value Theorem for Integrals that there exists constants C on  1  2  8  1 2 0, , D on 0, , E on 0, , S on , such that 3 3 15 3 3 Z 1/3 1 Z 2/3 2 Z 8/15 8 f(t)dt = f(C), f(t)dt = f(D), f(t)dt = f(E) and so 0 3 0 3 0 15 1 Z 8/15 Z 8/15 Z 1/3 8 1 f(S) = f(t)dt = f(t)dt − f(t)dt = f(E) − f(C) and 5 1/3 0 0 15 3 2 Z 2/3 Z 2/3 Z 8/15 2 8 f(T ) = f(t)dt = f(t)dt − f(t)dt = f(D) − f(E)dt. 15 8/15 0 0 3 15 3 5 The first of these last two equations gives f(E) = f(S) + f(C) and so the second 8 8 1 3 1 then gives f(D) = f(T ) + f(S) + f(C). 5 10 2 2 Z 1/3 3 Z 12/3 5 Z 8/15 The desired inequality f(t)dt + f(t)dt ≥ f(t)dt can be written 5 10 8 as 0 0 0 2 3 2 5 8 f(C) + · · f(D) ≥ · · f(E), or equivalently as 15 10 3 8 15 2 1 1 3 1  1 3 5  f(C) + f(T ) + f(S) + f(C) ≥ f(S) + f(C) which is equivalent to 15 5 5 10 2 3 8 8  2 1 5  1 1 3  + − f(C) + f(T ) ≥ − f(S) and then to 15 10 24 25 8 50 1 1 13 f(C) + f(T ) ≥ f(S) and finally to 40 25 200 5 8 f(C) + f(T ) ≥ f(S) which is true by the convexity since C < S < T . 13 13

Note: A function is convex on [a, b] means that for all 0 ≤ λ ≤ 1 whenever a ≤ X < Z ≤ b we have f(X) + λ (f(Z) − f(X)) ≤ f(X + λ(Z − X)) which can be cast as (1 − λ)f(X) + λf(Z) ≥ f (X + λ(Z − X)) and so in the above taking X = C,Z = T , S − C and λ = we have X + λ(Z − X) = S. T − C Solution 3 by Henry Ricardo, Westchester Area Math Circle, NY If f is convex on [0, 1], then for all x, y ∈ [0, 1] and for all λ ∈ [0, 1], we have λf(x) + (1 − λ)f(y) ≥ f(λx + (1 − λ)y). (1) Setting λ = 2/5, x = t/3, and y = 2t/3 in (1), 0 ≤ t ≤ 3/2, we get 2 3 f(t/3) + f(2t/3) ≥ f(8t/15). 5 5 Integrating from t = 0 to t = 1 yields 2 Z 1 3 Z 1 Z 1 f(t/3)dt + f(2t/3)dt ≥ f(8t/15)dt. (2) 5 0 5 0 0 15 Setting u = t/3 in the first integral of (2), we have

2 Z 1 6 Z 1/3 f(t/3)dt = f(u)du. 5 0 5 0 Similarly, setting u = 2t/3 in the second integral, we get

3 Z 1 9 Z 2/3 f(2t/3)dt = f(u)du. 5 0 10 0 Finally, setting u = 8t/15, we find that

Z 1 15 Z 8/15 f(8t/15)dt = f(u)du. 0 8 0 Substituting these into (2) and dividing by 3, we obtain

2 Z 1/3 3 Z 2/3 5 Z 8/15 f(u)du + f(u)du ≥ f(u)du. 5 0 10 0 8 0

Solution 4 by Angel´ Plaza, University of Las Palmas de Gran Canaria, Spain Z a Z αa By changing variables it follows that α f(t) dt = f(s/α) ds. Therefore, the 0 0 left-hand side of the proposed inequality, say LHS, is

2 Z 1/3 3 Z 1/3 LHS = f(t) dt + f(2t) dt 5 0 5 0 Z 1/3 2 6  ≥ f t + t dt 0 5 5 Z 1/3 8  = f t dt 0 5 2 3 2 6  where we have used that f is convex, so f(t) + f(2t) ≥ f t + t . Since the 5 5 5 5 5 Z 8/15 Z 1/3 8  right-hand side is f(t) dt = f t dt, the conclusion follows. 8 0 0 5 Also solved by Bruno Salgueiro Fanego, Viveiro, Spain; Ed Gray, Highland Beach, FL; Kee-Wai Lau, Hong Kong, China; Moti Levy, Rehovot, Israel; Paolo Perfetti, Department of Mathematics, Tor Vergata University, Rome, Italy; Albert Sadler, Herrliberg, Switzerland; Ioannis D. Sfikas, National and Kapodistrian University of Athens, Greece, and the proposer.

5480: Proposed by Ovidiu Furdui, Technical University of Cluj-Napoca, Cluj-Napoca, Romania Let n ≥ 1 be a nonnegative integer. Prove that in C[0, 2π]

span{1, sin x, sin(2x),..., sin(nx)} = span{1, sin x, sin2 x, . . . , sinn x} if and only if n = 1.

16 k X We mention that span{v1, v2, . . . , vk} = ajvj, aj ∈ <, j = 1, . . . , k, denotes the set of j=1 all linear combinations with v1, v2, . . . , vk.

Solution 1 by Moti Levy, Rehovot, Israel

If n = 1 then the spans are trivially equal. Let n > 1. Suppose that sin2 x can be expressed as a linear combination of the functions {1, sinx, sin(2x), ..., sin(nx)} ,

n 2 X sin x = a0 + ak sin (kx) . (1) k=1

By setting x = 0, we have a0 = 0. The following definite integral vanish for integer k.

Z 2π 1 Z 2π sin2x · sin(kx)dx = (1 − cos (2x)) · sin(kx)dx (2) 0 2 0 1 Z 2π 1 Z 2π = sin(kx)dx − cos (2x) · sin(kx)dx = 0. 2 0 2 0

Now we multiply both sides of (1) by sin2 x and integrate from 0 to 2π,

Z 2π n Z 2π 4 X 2 sin (x) dx = ak sin x sin (kx) dx. (3) 0 k=1 0

3 The right hand side of (3) is equal to 4 π but the left hand side is equal to zero, by (2). This contradiction leads to the conclusion that sin2 x cannot be expressed as a linear combination of the functions {1, sinx, sin(2x), ..., sin(nx)} , hence the spans are not equal for n > 1.

Solution 2 by David Stone and John Hawkins, Georgia Southern University, Statesboro, GA For n = 1 it is certainly true that span{1, sin x} = span{1, sin x}. For n > 1,

span{1, sin x, sin(2x), sin(3x),..., sin(nx)}= 6 span{1, sin x, sin2 x, sin3 x, . . . , sinn x} because sin2 x cannot be written as a linear combination of 1, sin x, sin(2x), sin(3x),..., sin(nx). To see this, suppose that 2 sin x = c0 · (1) + c1 sin(x) + c2 sin(2x) + c3 sin(3x) + ..., +cn sin(nx). Then this equation must hold for all values of x in C [0, 2π].

Letting x = 0, shows that c0 = 0. π Letting x = gives us 1 = 0 + c · 1 + c · 0 + c · (−1) + c · 0 + c · 1 + ... 2 1 2 3 4 5 or (1) 1 = c1 − c3 + c5 − c7 + c9 + ....

17 3π Letting x = gives us 1 = 0 + c · (−1) + c · 0 + c · 1 + c · 0 + c · (−1) + ... 2 1 2 3 4 5 or (2) 1 = −c1 + c3 − c5 + c7 − c9 + .... The final term in each summation depends upon the of parity of n, but the terms on the right hand sides match up in any case. So, adding (1) + (2) gives us 2 = 0, which is certainly a contradiction. Thus, sin2 x cannot be written in terms of 1, sin x, sin(2x), sin(3x),..., sin(nx).

Also solved by Kee-Wai Lau, Hong Kong, China; Ioannis D. Sfikas, National and Kapodistrian University of Athens, Greece, and the proposer

Mea Culpa

The names Dionne Bailey, Elsie Campbell, and Charles Diminnie all at Angelo State University in San Angelo, TX were inadvertently omitted from the list of those who had solved problem 5470. Paolo Perfetti of the Mathematics Department at Tor Vergata University in Rome, Italy should be credited for having solved 5472.

18 Problems Ted Eisenberg, Section Editor *********************************************************

This section of the Journal offers readers an opportunity to exchange interesting mathematical problems and solutions. Please send them to Ted Eisenberg, Department of Mathematics, Ben-Gurion University, Beer-Sheva, Israel or fax to: 972-86-477-648. Questions concerning proposals and/or solutions can be sent e-mail to . Solutions to previously stated problems can be seen at .

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Solutions to the problems stated in this issue should be posted before October 15, 2018

5499: Proposed by Kenneth Korbin, New York, NY Given a triangle with sides (21, 23, 40). The sum of these digits is 2 + 1 + 2 + 3 + 4 + 0 = 12. Find primitive pythagorean triples in which the sum of the digits is 12 or less.

5500: Proposed by Moshe Stupel, “Shaanan” Academic College of Education and Gordon Academic College of Education, and Avi Sigler, “Shaanan” Academic College of Education, Haifa, Israel 3 Without the use of a calculator, show that: 8 sin 20◦ · sin 40◦ · sin 60◦ · sin 80◦ = . 2

5501: Proposed by D.M. Ba˘tinetu-Giurgiu, Bucharest, Romania, Neculai Stanciu, “George Emil Palade” School Buza˘u, Romania and Titu Zvonaru, Coma˘nesti, Romania Determine all real numbers a, b, x, y that simultaneously satisfy the following relations: (1) ax + by = 5     2 2 (2) ax + by = 9

(3) ax3 + by3 = 17    (4) ax4 + by4 = 33.

5502: Proposed by Daniel Sitaru, “Theodor Costescu” National Economic College, Drobeta Turnu-Severin, Mehedinti, Romania Prove that if a, b, c > 0 and a + b + c = e then e e e 2 2 2 eac · eba · ecb > ee · abe · bce · cae .  1 n Here, e = lim 1 + 1 n→∞ n 5503: Proposed by Jos´eLuis D´ıaz-Barrero, Barcelona Tech, Barcelona, Spain

Let a1, a2, . . . , an be positive real numbers with n ≥ 2. Prove that m m m m (a1 a2 + a2 a3 + ··· + an a1) 1 m m m m+1 ≤ , (a1 + a2 + ··· an ) n where m is a positive integer.

5504: Proposed by Ovidiu Furdui and Alina Sˆınt˘am˘arianboth at the Technical University of Cluj-Napoca, Cluj-Napoca, Romania Let n ≥ 0 be an integer. Calculate Z 1 xn  1 dx, 0 x where bxc denotes the integer part of x.

Solutions

5481: Proposed by Kenneth Korbin, New York, NY A triangle with integer area has integer length sides (3, x, x + 1). Find five possible values of x with x > 4.

Solution 1 by Dionne Bailey, Elsie Campbell, Charles Diminnie, and Trey Smith, Angelo State University, San Angelo, TX For our approach, we will need to find positive integer solutions for the equation m2 − 8k2 = 9. (1) One way to do so is to first solve the Pell Equation X2 − 8Y 2 = 1 (2) and then set m = 3X and k = 3Y . Following the usual process for solving (2), we note that the solution with the smallest X value is X = 3, Y = 1. Then, all solutions (Xn,Yn) of (2) can be found by setting √  √ n Xn + Yn 8 = 3 + 8 for all n ≥ 1. Then, as described above, we get solutions for (1) by setting mn = 3Xn and kn = 3Yn. The first six solutions for (1) and (2) are listed in the following table:

n Xn Yn mn kn 1 3 1 9 3 2 17 6 51 18 3 99 35 297 105 (3) 4 577 204 1, 731 612 5 3, 363 1, 189 10, 089 3, 567 6 19, 601 6, 930 58, 803 20, 790

2 For the problem at hand, the semiperimeter s of our triangle is

3 + x + (x + 1) s = = x + 2 2 and Heron’s Formula for the area A yields

A = ps (s − 3) (s − x)(s − x − 1) = p(x + 2) (x − 1) (2) (1) p = 2 (x2 + x − 2).

For A to be a positive integer, we must find a positive integer k for which

A2 = 2 x2 + x − 2
= 4k2 or x2 + x − 2 − 2k2 = 0. (4) By the Quadratic Formula, the positive solution of (4) is

−1 + p1 + 8 (k2 + 1) x = √ 2 −1 + 8k2 + 9 = . 2 For x to be a positive integer, we will need

8k2 + 9 = m2 or m2 − 8k2 = 9 for some odd positive integer m. However, table (3) gives us six solutions to use. In each case, √ −1 + 8k2 + 9 m − 1 x = = and A = 2k. 2 2

The solution m1 = 9 and k1 = 3 yields x = 4, which is ruled out in the statement of the problem. The other five entries in the table provide five plausible values of x for which A is a positive integer. These values are listed in our final table:

n mn kn x A 2 51 18 25 36 3 297 105 148 210 4 1, 731 612 865 1, 224 5 10, 089 3, 567 5, 044 7, 134 6 58, 803 20, 790 29, 401 41, 580

Solution 2 by Bruno Salgueiro Fanego, Viveiro, Spain

3 Let p and S be the semi perimeter and the area of such a triangle respectively. Then 2p = 3 + x + x + 1 = 2x + 4 and, by Heron’s formula √ S = pp(p − 3)(p − x)(p − (x + 1)) = 2x2 + 2x − 4 must be an integer. It can be easily verified that for each of the five values of x ∈ {25, 148, 865, 5044, 29401} one obtains triangles that have areas of 36, 210, 1224, 7134, 41580, respectively. x  25 More generally, if 1 = , then the recurrence given by S1 36 x  3 2  x  1 x  n+1 = n + for any integer n ≥ 1, gives a pair n+1 Sn+1 4 3 Sn 2 Sn+1 where xn+1 is the length of a triangle with integer length sides (3, xn+1, xn+1 + 1) and Sn+1 is the integer area of that triangle.

Solution 3 by Julio Cesar Mohnsam and Luiz Lemos Junior, both at IFSUL Campus Pelotas-RS, Brazil 3 + x + x + 1 Let p be the semi-perimeter p = = x + 2 2 The area by Heron is given by: A = pp(p − 3)(p − x)(p − x − 1) = p(x + 2)(x − 1)(2)(1) Then (x + 2)(x − 1)(2) must be a square, that is, 2x2 + 2x − 4 = y2, follow that:

2x2 + 2x − y2 − 4 = 0 (1) Multiplying (1) by 8 we have:

16x2 + 16x − 8y2 − 32 = 0 (2) Adding 4 on both sides of (2) we have:

(4x + 2)2 − 8y2 − 36 = 0 (3) Now make X = 4x + 2 and Y = y, we have:

X2 − 8Y 2 − 36 = 0 (4) (4) is a diophantine equation of the form ax2 − by2 + c = 0 in the case of c = −1 we have the particular case of the equation of Pell x2 − Dy2 = 1. If (a, b)|c, the equation has a solution. Let’s solve (4) using the method Florentin Smarandache [1]. We consider the equation

aX2 − bY 2 + c = 0 (5) and

aα2 − bβ2 = a (6) We set the matrix A from (6) as follows:

 α b β  A = 0 a 0 β0 α0 where (α0, β0) are initial solutions of (6).

4 Now let (X0,Y0) are initial solutions of (5), then the general solutions of (5) are given by the following recurrence relation:  X   X  n = An 0 Yn Y0 Thus, by solving (4) we have to first assemble matrix A from X2 − 8Y 2 = 1, note that this Pell equation has initial solution (3, 1), so we have:  3 8  A = 1 3 But (4) has initial solution (6, 0). So we have to:

 X   3 8   6   18  1 = = Y1 1 3 0 6

Like 18 = X1 = 4x1 + 2 → x1 = 4 but we have to find x > 4. Thus we calculate A2, such that:

 X   3 8 2  6   17 48   6   102  2 = = = Y2 1 3 0 6 17 0 36

As 102 = X2 = 4x2 + 2 → x2 = 25 and the lengths of the first triangle are (3, 25, 26). To find the other values of x we will diagonalize the matrix A. We know that An = PDnP −1.

 λ 0  A = PDP −1 = P 1 P −1 0 λ2 √ √ Eigenvalues λ1 = 3 + 2√ 3 and λ2 = 3 − 2 3√  2 3   −2 3  Eigenvectors ~v = and ~v = 1 1 2 1 Therefore x = {25, 148, 565, 5044, 29401} [1] Smarandache F. “Un metodo de resolucion de la ecuacion diofantica. Gazeta Matematica, Serie 2, Vol. 1, Nr. 2, 1988. Madrid. p. 151-157.

Solution 4 by Brian D. Beasley, Presbyterian College, Clinton, SC Given such a triangle, its semiperimeter is s = (3 + x + x + 1)/2 = x + 2. Then by Heron’s formula, its area is ∆ = p((x + 2)(x − 1)(2)(1) = p2(x2 + x − 2), so we seek integers ∆ and x with x > 4 such that ∆2 = 2x2 + 2x − 4. This equation in turn is equivalent to ∆2 2x + 12 2∆2 + 9 = (2x + 1)2, or 2 + 1 = . 3 3 We let a = (2x + 1)/3 and b = ∆/3 in order to solve the Pellian equation 2b2 + 1 = a2. This equation has infinitely many integer solutions for a and b, which we may describe with the sequences

a0 = 1, a1 = 3, and an+2 = 6an+1 − an for n ≥ 0; b0 = 0, b1 = 2, and bn+2 = 6bn+1 − bn for n ≥ 0.

5 Thus there are infinitely many integers x which satisfy the requirements of the problem, given by the terms xn (with n ≥ 2) of the sequence

x0 = 1, x1 = 4, and xn+2 = 6xn+1 − xn + 2 for n ≥ 0. In particular, the next five values of x after 4 are 25, 148, 865, 5044, and 29401.

√ Addenda.√(i) We may also describe the above sequences by letting γ = 3 + 2 2 and n n δ = 3 − 2 2. Then an = (γ + δ )/2 for each n ≥ 0, which implies that n n xn = (3γ + 3δ − 2)/4.

(ii) We further note that the ratios an/bn for n ≥ 1 occur as every√ other term in the sequence of converging to the continued fraction representation of 2.

Comments by Editor : Ioannis D. Sfikas of Athens Greece started his solution off with some nomenclature and bit of history about the problem. “A triangle whose sides and area are rational numbers is called a rational triangle. If the rational triangle is right-angled, it is called a right-angled rational triangle or a rational Pythagorean triangle or a numerical right triangle. If the sides of a rational triangle is of integer length, it is called an integer triangle. If further these sides have no common factor greater than unity, the triangle is called a primitive integer triangle. If the integer triangle is right-angled, it is called a Pythagorean triangle. A Heronian triangle (named after Heron of Alexandria) is an integer triangle with the additional property that its area is also an integer. A Heronian triangle is called primitive Heronian triangle if sides have no common factor greater than unity. In the 7th century, the Indian mathematician Brahmagupta studied the special case of triangles with consecutive integer sides.”

Kenneth Korbin, the proposer of this problem√ stated that triangles sides with lengths (3, x, x + 1) with x ≥ 4 have an area of 2x2 + 2x − 4, and are associated with the sequences of (25, 148, 865, 5044, 29401, . . . , xN ,...) that satisfies the recursion of xN+1 = 6xN − xN−1 + 2. David Stone and John Hawkins of Southern Georgia University asked in their solution, why the values of x ≥ 4? Why not x > 0? They then stated that: If

x = 1, the triangle (3, 1, 2) is degenerate;√ x = 2, the triangle (3, 2, 3) has area √8 and is not Heronian; x = 3, the triangle (3, 3, 4) has area 20 and is not Heronian; and x = 4, the right triangle (3, 5, 5) is too easy.

They then asked: What about those Heronian triangles of the form (3, x, x + 2), where x is an integer. Applying Heron’s Formula they obtained that 16A2 = 20x2 + 40x − 25 and stated that there are no integer solutions to this for x ≥ 1 because the left-and side is even while the right hand side is odd. They then looked at triangles of the form (3, x, x + 3) and stated that the only triangle of this form is degenerate. Moreover, no triangle of the form (3, x, x + d) can exist for d > 3. They continued on with the following: “Thus the problem poser selected the one form that does admit solutions. Still to be

6 investigated: finding the Heronian triangles of the form (4, x, x + 1), and those of the form (4, x, x + 2), etc.” Their solution ended with the statement: “There are no Heronian triangles of the form (n, x, x + d) for positive integers n and d having opposite parity.”

Also solved by Hatef I. Arshagi, Guilford Technical Community College, Jamestown, NC; Jeremiah Bartz, University of North Dakota, Grand Forks, ND; Anthony Bevelacqua University of North Dakota, Grand Forks, ND; Ed Gray, Highland Beach, FL; Paul M. Harms, North Newton, KS; Kee-Wai Lau, Hong Kong, China; Carl Libis, Columbia Southern University, Orange Beach, AL; David E. Manes, Oneonta, NY; Ioannis D. Sfikas, National and Kapodistrian University of Athens, Greece; Albert Stadler, Herrliberg, Switzerland; Neculai Stanciu,“George Emil Palade” School, Buzau˘ and Tito Zvonaru, Comanesti,˘ Romania; David Stone and John Hawkins of Southern Georgia University, Statesboro, GA, and the proposer.

5482: Proposed by Daniel Sitaru, “Theodor Costescu” National Economic College, Drobeta Turnu-Severin, Mehedinti, Romania Prove that if n is a natural number then (tan5◦)n (tan4◦)n (tan3◦)n 3 + + ≥ . (tan4◦)n + (tan3◦)n (tan3◦)n + (tan2◦)n (tan2◦)n + (tan1◦)n 2

Solutions 1 and 2 by Henry Ricardo, Westchester Area Math Circle, NY Solution 1. Since, for a fixed natural number n, (tan x)n is an increasing positive function for x ∈ [0, 90◦), we have (tan 5◦)n (tan 5◦)n 1 ≥ = , (tan 4◦)n + (tan 3◦)n (tan 5◦)n + (tan 5◦)n 2 (tan 4◦)n (tan 4◦)n 1 ≥ = , (tan 3◦)n + (tan 2◦)n (tan 4◦)n + (tan 4◦)n 2 (tan 3◦)n (tan 3◦)n 1 ≥ = , (tan 2◦)n + (tan 1◦)n (tan 3◦)n + (tan 3◦)n 2 so that adding these inequalities gives us the desired result. Equality holds if and only if n = 0 (assuming that 0 is considered a natural number).

Solution 2. Since, for a fixed natural number n, (tan x)n is an increasing positive function for x ∈ [0, 90◦), we have (tan 3◦)n (tan 3◦)n ≥ , (tan 2◦)n + (tan 1◦)n (tan 4◦)n + (tan 5◦)n and (tan 4◦)n (tan 4◦)n ≥ , (tan 3◦)n + (tan 2◦)n (tan 3◦)n + (tan 5◦)n

7 so that X (tan 3◦)n (tan 5◦)n (tan 4◦)n (tan 3◦)n ≥ + + . (tan 2◦)n + (tan 1◦)n (tan 4◦)n + (tan 3◦)n (tan 3◦)n + (tan 5◦)n (tan 4◦)n + (tan 5◦)n cyclic

Setting a = (tan 3◦)n, b = (tan 4◦)n, and c = (tan 5◦)n, we see that the right-hand side of the last inequality has the form a b c + + , b + c c + a a + b for a, b, c > 0, which is greater than or equal to 3/2 by Nesbitt’s inequality. Equality holds if and only if n = 0 (assuming that 0 is considered a natural number).

Solution 3 by Ed Gray, Highland Beach, FL First we retrieve the required values: 1. tan 1◦ = .017455065 2. tan 2◦ = .034920769 3. tan 3◦ = .052407779 4. tan 4◦ = .069926812 5. tan 5◦ = .087488664

We rewrite the problem’s equation as:

1 1 1 3 + + ≥ tan 4◦ tan 3◦ tan 3◦ tan 2◦ tan 2◦ tan 1◦ 2 + + + tan 5◦ tan 5◦ tan 4◦ tan 4◦ tan 3◦ tan 3◦ Substituting the values from steps 1-5 and performing the indicated divisions we define:

1 1 1 f(n) = + + . (.799267114)n + (.599023652)n (.794551256)n + (.499433116)n (.66632797)n + (.333062483)n We note that f(n) is an increasing function of n since the denominators clearly decrease as n increases. 3 Finally we note that f(1) = .715158838 + 1.248899272 + 1.000609919 = 2.964668029 > . 2 Then the equality holds for all n since f(n) is an increasing function.

Solution 4 by David Stone and John Hawkins, Georgia Southern University, Statesboro, GA cx Lemma: For fixed positive reals a, b, c with a < c, b < c let f(x) = for x ≥ 0. bx + ax 1 Then f(x) ≥ , for x ≥ 0, with equality holding only for x = 0. 2 Proof: We calculate the derivative:

(bx + ax) cx ln c − cx (ax ln a + bx ln b) f 0(x) = (bx + ax)2

8 (bx + ax) ln c − (ax ln a + bx ln b) = cx (bx + ax)2

bx (ln c − ln b) + ax (ln c − ln a) = cx . (bx + ax)2

The ln function is increasing, so ln c > ln b and ln c > ln a; thus we see that the derivative 1 is positive. Hence the function f is increasing, so = f(0) ≤ f(x) for x ≥ 0. Because 2 1 the derivative is strictly positive, the function f actually grows: so f(x) > for x > 0. 2 To verify the inequality of the problem, we note that the tangent function is increasing, so in each summand the tangent term in the numerator is larger that each tangent term in the denominator. Hence we can apply the lemma to each of the three summands, 3 forcing the sum ≥ . Note that equality holds if and only if n = 0. 2 Comment: We can apply the lemma to obtain some ugly inequalities which are clearly true: 3n 4n 5n (n + 2)n n + + + ··· + ≥ , and 1n + 2n 2n + 3n 3n + 4n nn + (n + 1)n 2

[(n + 2)!]n 1 ≥ . [n!]n + [(n + 1)!]n 2

Also solved by Arkady Alt, San Jose, CA (two solutions); Paul M. Harms, North Newton, KS; Kee-Wai Lau, Hong Kong, China; Paolo Perfetti, Department of Mathematics, University of Tor Vergata, Rome, Italy; Angel Plaza, University of Las Palmas de Gran Canaria, Spain; Ioannis D. Sfikas, National and Kapodistrian University of Athens, Greece; Albert Stadler, Herrliberg, Switzerland, and the proposers

5483: Proposed by D.M. Ba˘tinetu-Giurgiu, “Matei Basarab” National College, Bucharest and Neculai Stanciu, “George Emil Palade” School Buza˘u, Romania  π  If a, b > 0, and x ∈ 0, then show that 2 sin x 2ab tan x 6ab (i)(a + b) · + · ≥ . x a + b x a + b √ (ii) a · tan x + b · sin x > 2x ab.

Solution 1 by Paolo Perfetti, Department of Mathematics, Tor Vergata University, Rome, Italy Proof of (i).

9 The AHM yields 4 2 a + b ≥ 1 1 ⇐⇒ (a + b) ≥ 4ab a + b and then sin x 2ab tan x 4ab sin x 2ab tan x (a + b) · + · ≥ · + · x a + b x a + b x a + b x Thus we prove 4ab 2ab 6ab · sin x + · tan x − x ≥ 0 a + b a + b a + b This is equivalent to . f(x) = 4 sin x + 2 tan x − 6x ≥ 0 2 f 0(x) = 4 cos x + − 6 cos2 x 4 sin x  4  f 00(x) = −4 sin x + = 4 sin x − 1 > 0 cos3 x cos3 x via cos x ∈ (0, 1) for 0 < x < π/2. Since f 0(0) = f(0) = 0 we get f(x) ≥ 0.

Proof of (ii). Let √ f(x) = a · tan x + b · sin x − 2x ab, f(0) = 0 r a √ a √ ab cos x √ f 0(x) = + b cos x − 2 ab ≥ + b cos x − 2 ab ≥ 2 − 2 ab = 0 cos2 x cos x cos x and this concludes the proof.

Solution 2 by Arkady Alt, San Jose, CA (i) First we will prove inequality tan x 2 sin x tan x + 2 sin x > 3x ⇐⇒ + > 3, x ∈ (0, π/2) . x x 1 Let h (x) := tan x + 2 sin x − 3x, x ∈ (0, π/2) . Since h0 (x) = + 2 cos x − 3 = cos2 x (2 cos x + 1) (1 − cos x)2 > 0, x ∈ (0, π/2) then h (x) > h (0) = 0. cos2 x sin x 2ab tan x 2ab  2 sin x Hence, (a + b) + · > (a + b) sin x + · 3 − = x a + b x a + b x sin x  4ab  6ab sin x (a − b)2 6ab 6ab a + b − + = · + ≥ . x a + b a + b x a + b a + b a + b √ a √ (ii) Let h (x) := a tan x + b sin x− 2x ab. Since h0 (x) = + b cos x− 2 ab ≥ √ cos2 x r a √ √ 1 − cos x √ 2 2 · b cos x− 2 ab = 2 ab · > 0 then h (x) > h (0) = 0 ⇐⇒ cos x √ cos x a tan x + b sin x > 2x ab.

Solution 3 by Kee-Wai Lau, Hong Kong, China

10  π  x3 x3 It is well known that for x ∈ 0, , we have sin x − and tan x ≥ x + . Since 2 6 3 4ab + (a − b)2 4ab a + b = ≥ , so the left side of (i) is greater than or equal to a + b a + b 2ab 2 sin x + tan x 2ab   x2   x2  6ab ≥ 2 1 − + 1 + = , a + b x a + b 6 3 a + b as required.

 π  x2 x4 It is also well known that for x ∈ 0, , cos x ≤ 1 − + , so that 2 2 24

 x3 2  x2 x4  x4(12 − x2) sin2 x = x2 cos x ≥ x − − x2 1 − + = ≥ 0. 6 2 24 72

Hence, r √ sin2 x √ a · tan x + b · sin x ≥ 2p(a tan x)(b sin x) = 2 abx ≥ 2 abx x2 cos x and (ii) holds.

Also solved by Ioannis D. Sfikas, National and Kapodistrian University of Athens, Greece; Albert Stadler, Herrliberg, Switzerland; David Stone and John Hawkins, Southern Georgia University, Statesboro, GA, and the proposers

5484: Proposed by Mohsen Soltanifar, Dalla Lana School of Public Health, University of Toronto, Canada

Let X1,X2 be two continuous positive valued random variables on the real line with corresponding mean, median, and mode x1, xe1, xb1 and x2, xe2, xb2 respectively. Assume for their associated CDFs, (Cumulative Distribution Functions) we have

FX1 (t) ≤ FX2 (t)(t > 0).

Prove or give a counter example:

(i) x2 ≤ x1, (ii) xf2 ≤ xf1, (iii) xb2 ≤ xb1.

Solution 1 by Albert Stadler, Herrliberg, Switzerland Z ∞ Z ∞

(i) We have x2 = E(X2) = (1 − FX 2(t))dt ≤ (1 − FX1 (t))dt = E (X1) = x1. 0 0     1 (ii) By definition, FX Xf1 = FX Xf2 = . The functions t → FX (t) and t → FX (t) 1 2 2 1 2   are monotonically increasing. Therefore F X ≤ F (X ) implies x ≤ x . X2 f2 X1 f1 f2 f1 (iii) We construct a counter example as a follows:

Let p1(x) = 0, if x ≤ 1/3 or x ≥ 1, p1(x) = 9x − 3, if 1/3 ≤ x ≤ 2/3, and p1(x) = 9 − 9x , if 2/3 ≤ x ≤ 1.

11 Let p2(x) = 36x, if 0 ≤ x ≤ 1/6, p2(x) = 12 − 36x, if 1/6 ≤ x ≤ 1/3, p2(x) = 0, if x ≤ 0 or x ≥ 1/3. Z 1 Z 1 p1(x) and p2(x) are probability density functions since p1(x)dx = p2(x)dx = 1. 0 0 Z x Z x Obviously, F (x) = p (t)dt ≤ p (t)dt = F (x), however x = 3, x = 6. X1 1 2 X2 f1 f2 0 0 Solution 2 by Kee-Wai Lau, Hong Kong, China We answer (i) and (ii) in the affirmative and (iii) in the negative.

Let X be a continuous random variable on the real line with CDF FX (t) and mean x. It Z ∞ Z 0 is known that x = (1 − FX (t))dt − FX (t)dt, provided at least one of the two 0 −∞ Z 0 Z 0 integrals is finite. Since X1,X2 are positive, so FX1 (t)dt = FX2 (t)dt = 0 and −∞ −∞ (i) follows from the fact that Z ∞ Z ∞

x2 − x1 = ((1 − FX2 (t)) − (1 − FX1 (t))) dt = (FX1 (t) − FX2 (t)) dt ≤ 0. 0 0

Next we consider the medians, assuming that xe is the least number a satisfying 1 F (a) = . Suppose, on the contrary, that x > x . Since F (t) is a non-decreasing X 2 f2 f1 X x + x function and x < f1 f2 < x , we have 1 2 f2 1 x + x  x + x  1 F (x ) = > F f1 f2 ≥ F f1 f2 ≥ F (x ) = , X2 f2 2 X2 2 X1 2 X1 f1 2 which is false. This proves (ii).

We now show that (iii) does not necessarily hold. Define the probability density functions fX1 (t) and fX2 (t) of X1 and X2 as follows:  0 x ≤ 0,     0 x ≤ 0,  x   0 ≤ x ≤ 4,   16 2x fX (t) = and fX (t) = 0 ≤ x ≤ 5, 1 2 25 8 − x   4 ≤ x ≤ 8,   16   0 x ≥ 5.   0 x ≥ 8 Then

12  0 x ≤ 0,     0 x ≤ 0,  x2   0 ≤ x ≤ 4,     32 x2 FX1 (t) = and FX2 (t) = 0 ≤ x ≤ 5,  2 25 −x + 16x − 32   4 ≤ x ≤ 8,   32   1 x ≥ 5.   1 x ≥ 8 It is easy to check that F (t) ≤ F (t) for t > 0, but x = 4 < 5 = x . X1 X 2 f1 f2 This completes the solution.

Also solved by the proposer.

5485: Proposed by Jos´eLuis D´ıaz-Barrero, Barcelona Tech, Barcelona, Spain Let x, y, z be three positive real numbers. Show that Y X (2x + 3y + z + 1) (4x + 2y + 1)−3 ≥ 3. cyclic cyclic

Solution 1 by Neculai Stanciu, “George Emil Palade” School, Bazau˘ Romania and Tito Zvonaru, Comanesti,˘ Romania We denote 4x + 2y + 1 = a, 4y + 2z + 1 = b, and 4z + 2x + 1 = c. We must prove that

(a + b)(b + c)(c + a)  1 1 1  + + ≥ 3 (∗) 8 a3 b3 c3 By the AM-GM inequality we have that √ √ √ (a + b)(b + c)(c + a) 2 ab · 2 bc · 2 ca 8abc ≥ = = abc, (1) 8 8 8 1 1 1 r 1 1 1 3 + + ≥ 3 · 3 · · = , (2) a3 b3 c3 a3 b3 c3 abc By (1) and (2) we obtain

(a + b)(b + c)(c + a)  1 1 1  3 + + ≥ abc · = 3. I.e. (∗) 8 a3 b3 c3 abc

Solution 2 by Nikos Kalapodis, Patras, Greece By the AM-GM inequality we have

Y X −3 Y 3 (2x + 3y + z + 1) (4x + 2y + 1) ≥ (2x + 3y + z + 1) Y . cyclic cyclic cyclic (4x + 2y + 1) cyclic Y Y So, it suffices to prove that (2x + 3y + z + 1) ≥ (4x + 2y + 1). cyclic cyclic

13 After expanding the inequality reduces to 2(x3 +y3 +z3)+x2 +y2 +z2 +3(xy2 +yz2 +zx2) ≥ 3(x2y+y2z+z2x)+xy+yz+zx+6xyz.

Since x2 + y2 + z2 ≥ xy + yz + zx, it remains to prove that 2(x3 + y3 + z3) + 3(xy2 + yz2 + zx2) ≥ 3(x2y + y2z + z2x) + 6xyz. This follows again by using the AM-GM inequality properly: 2(x3 + y3 + z3) + 3(xy2 + yz2 + zx2) = 2(x3 + xy2) + 2(y3 + yz2) + 2(z3 + zx2) + (xy2 + yz2 + zx2) ≥ 4x2y + 4y2z + 4z2x + (xy2 + yz2 + zx2) = 3(x2y + y2z + z2x) + (x2y + y2z + z2x + xy2 + yz2 + zx2) ≥ 3(x2y + y2z + z2x) + 6xyz.

Also solved by Arkady Alt, San Jose, CA; Kee-Wai Lau, Hong Kong, China; Paolo Perfetti, Department of Mathematics, Tor Vergata University, Rome, Italy; Kevin Soto Palacios, Huarmey, Per´u;Ioannis D. Sfikas, National and Kapodistrian University of Athens, Greece; Albert Stadler, Herrliberg, Switzerland, and the proposer.

5486: Proposed by Ovidiu Furdui, Technical University of Cluj-Napoca, Cluj-Napoca, Romania

Let (xn)n≥0 be the sequence defined by x0 = 0, x1 = 1, x2 = 1 and ∞ X xn x = x + x + x + n, ∀n ≥ 0. Prove that the series converges and find n+3 n+2 n+1 n 2n n=1 its sum.

Solution 1 by Angel´ Plaza, University of Las Palmas de Gran Canaria, Spain. The recurrence sequence may be unmasked by generating functions. Let F (z) be the ∞ X n n+3 associated generating function. That is, F (z) = xnz . Multiplying by z the n=0 recurrence relation defining (xn) and taking into account the initial values it is obtained that z4 F (z) − z + z2
= z (F (z) − z) + z2F (z) + z3F (z) + (1 − z)2 z(1 − z)2 + x4 from where F (z) = . (z − 1)2(1 − z − z2 − z3) ! 1 q3 √ 2 Since F (z) converges for |z| < 17 + 3 33 − − 1 ∼ 0.5436..., then p3 √ 3 17 + 3 33 ∞ X xn = F (1/2) = 6. 2n n=1 Solution 2 by Paolo Perfetti, Department of Mathematics, Tor Vergata University, Rome, Italy Answer: 6.

14 Clearly xn increases and xn ≥ 1.

p p p−3 X xn x1 x2 X xn 3 X xn+3 = + + = + = 2n 2 4 2n 4 2n+3 n=1 n=3 n=0   p−3 p−3 3 X xn+2 xn+1 xn  X n = +  + +  + = 4  2n+3 2n+3 2n+3  2n+3 n=0 | {z } | {z } | {z } n=0 I1 I2 I3 p−1 p−2 p−3 p−3 3 X xn X xn X xn X n = + + + + = 4 2n+1 2n+2 2n+3 2n+3 n=2 n=1 n=1 n=0 | {z } | {z } | {z } I1 I2 I3 p p 3 1 X xn xp X xn xp−1 xp = + − + − + − − + 4 4 2n+1 2p+1 2n+2 2p+1 2p+2 n=1 n=1 | {z } | {z } I1 I2 p p−3 X xn xp−2 xp−1 xp X n + − − − + 2n+3 2p+1 2p+2 2p+3 2n+3 n=1 n=0 | {z } | {z } I3 →1/4 as p→∞

It follows p 1 X xn 1 h xp xp−1 xp xp−2 xp−1 i 1 = − + + + + + 8 2n 2 2p+1 2p+1 2p+2 2p+1 2p+2 4 n=1 Now we prove the

k Lemma xk/2 → 0. Proof of the Lemma k First step: the sequence xk/2 in monotonic not increasing. x x + x + x + k x x + x + k x k+3 = k+2 k+1 k ≤ k+2 ⇐⇒ k+1 k ≤ k+2 2k+3 2k+3 2k+2 2k+3 2k+3 that is x(k−1)+2 + x(k−1)+1 + (k − 1) + 1 ≤ x(k−1)+3 and this is implied by

x(k−1)+2 + x(k−1)+1 + (k − 1) + 1 ≤ x(k−1)+2 + x(k−1)+1 + xk−1 + (k − 1) = x(k−1)+3

k via xk−1 ≥ 1. The monotonicity of the sequence means that the limit L of xk/2 does exist and moreover 0 ≤ L < +∞. If L = 0 the proof is concluded yielding

p ∞ 1 X xn 3 X xn lim = ⇐⇒ = 6 p→∞ 8 2n 4 2n n=1 n=1 L 6= 0 is impossible as shown by the following argument. We employ the Cesaro–Stolz theorem that states: x x − x lim k = lim k+1 k k→∞ 2k k→∞ 2k+1 − 2k 15 provided that the second limit does exist. We write x − x x + x + k 1 x 1 x k k+3 k+2 = k+1 k = k+1 + k + 2k+3 − 2k+2 2k+2 2 2k+1 4 2k 2k+2 x The existence of the limit L = lim k would imply k→∞ 2k 1 1 L = L + L =⇒ L = 0 2 4

Solution 3 by Arkady Alt, San Jose, CA

For any sequence (xn)n≥0 let T (xn) := xn+3 − xn+2 − xn+1 − xn, n ∈ N ∪ {0} . Obvious that such defined operator T (we will call it Tribonacci Operator) is linear.  n n + 3 n + 2 n + 1 n Since T − = − + + + = n then denoting 2 2 2 2 2 n u := x + , n ∈ N ∪ {0} n n 2 n we obtain x = u − , n ∈ N ∪ {0} where T (u ) = 0 and, n n 2 n 1 3 2 u = 0, u = 1 + = , u = 1 + = 2. 0 1 2 2 2 2 Let (tn)n≥0 be the sequence defined by t0 = 0, t1 = 1, t2 = 1 and T (tn) = 0, n ∈ N ∪ {0} . (Tribonacci Sequence). We have t3 = 2, t4 = 4, t5 = 7, t6 = 13, t7 = 24, t8 = 44, ... 0 1 1 Since det 1 1 2 6= 0 then for any sequence (xn)n≥0 there is triple (c1, c2, c3) of real 1 2 4 numbers such that xn = c2tn + c2tn+1 + c3tn+2, that is sequences (tn)n≥0 , (tn+1)n≥0 , (tn+2)n≥0 form a basis of 3-dimesion space ker T := {(xn)n≥0 | T (xn) = 0, n ∈ N ∪ {0}} . We will find representation un as linear combination of tn, tn+1, tn+2, namely, un = c1tn + c2tn+1 + c3tn+2, n ∈ N ∪ {0} . We 3 have u = c t +c t +c t ⇐⇒ c +c = 0, u = c t +c t +c t ⇐⇒ c +c +2c = , 0 1 0 2 1 3 2 2 3 1 1 1 2 2 3 3 1 2 3 2 u2 = c1t2 + c2t3 + c3t4 ⇐⇒ c1 + 2c2 + 4c3 = 2. From this system of equations we obtain 3 1 1 c = −c , c − c = , c − 2c = 2.Hence, c = 1, c = − , c = and since 3 2 1 2 2 1 2 1 2 2 3 2 t t t t n 2t − t + t − n u = t − n+1 + n+2 we obtain x = t − n+1 + n+2 − = n n+1 n+2 . n n 2 2 n n 2 2 2 2 ∞ ∞ P n−1 P n−1 1 Since radius of convergence of seies nx is 1 and nx = 2 n=1 n=1 (1 − x) ∞ ∞ P n 1 P n 1 1 then n = n−1 = 2 = 2 and, therefore, for convergency of n=1 2 2 n=1 2 2 (1 − 1/2) ∞ ∞ P xn P tn n suffice to prove convergency of series n . n=1 2 n=1 2 n n We can prove that using another basis of ker T which form sequences (α )n≥0, (β )n≥0, n (γ )n≥0 where α, β, γ are roots of characteristic equation x3 − x2 − x − 1 = 0. 4u + 1 Substitution x = in equation x3 − x2 − x − 1 = 0 give us equivalent equation 3 19 4u3 − 3u = 8 16 1  1 19 which we solve using substitution u := t + . Then equation 4u3 − 3u = 2 t 8 1  13 1  1 19 1 19 becomes 4 t + − 3 · t + = ⇐⇒ + t3 = .Denoting z := t3 2 t 2 t 8 t3 4 we obtain √ √ √ √ 1 19 19 − 3 33 19 + 3 33 19 − 3 33 19 + 3 33 + z = ⇐⇒ z = , ⇐⇒ t3 = , . z 4 √ √ 8 8 8 8 19 − 3 33 19 + 3 33 1  1 Since · = 1 and u = t + then suffices to 8 √ 8 2 t 19 + 3 33 find t3 = . 8 p3 √ 19 + 3 33 2kπ We have t = r (cos ϕ + i sin ϕ) , where r = and ϕ = , k = 1, 2, 3. 2 3 p3 √ 19 + 3 33 2π 2π that is t = ωk, k = 1, 2, 3 and ω = cos + i sin , ω3 = 1. k 2 3 3 p3 √ p3 √ Thus, denoting θ := 19 + 3 33, θ∗ := 19 − 3 33 we obtain 1 + θ + θ∗ 1 + ωθ + ω2θ∗ α = , β = , 3 3 1 + ω2θ + ωθ∗ γ = , the three roots of the equation x3 − x2 − x − 1 = 0. 3 1 + θ + θ∗ We will prove that α = < 2. 3 First note that by Power Mean–Arithmetic Mean inequality r √ √ p3 √ p3 √ 3 19 + 3 33 + 19 − 3 33 √ √ p := 19 + 3 33 + 19 − 3 33 < 2 = 2 3 19 < 2 3 27 = 6. 2 p3 √ p3 √ √ Since 19 + 3 33 · 19 − 3 33 = 3 192 − 9 · 33 = 4 then p3 √ p3 √ p3 = 38 + 3 19 + 3 33 · 19 − 3 33 · p = 38+ 12p < 38+ 12 · 6 = 110 < 125 = 53. 1 + θ + θ∗ Hence, α < 2. Also, we obtain |β| , |γ| ≤ < 2. 3 ∞ ∞  n ∞ P αn P β P γ n Since series , , are convergent and tn is linear combination n=1 2 n=1 2 n=1 2 of ∞ n n n P tn (α )n≥0, (β )n≥0, (γ )n≥0 then series n convergent as well. n=1 2 ∞ P xn Now we ready to find sum of series n . n=1 2 n n P tn P n Let sn := n and s (x) = tn+1x . Note also that function k=1 2 k=0 1 generates 1 − x − x2 − x3 ∞ 1 P n Tribonacci numbers. Indeed, let 2 3 = anx . Then 1 − x − x − x n=0 ∞ P n 2 3
anx · 1 − x − x − x = 1 n=0 and since ∞ ∞ ∞ ∞ P n 2 3
P n P n+1 ∞ n+2 P n+3 anx · 1 − x − x − x = anx − anx −n=0 anx − anx = n=0 n=0 n=0 n=0 ∞ 2 P n+3 a0 + (a1 − a0) x + (a2 − a1 − a0) x + (an+3 − an+2 − an+1 − an) x then n=3 a0 = 1, a1 − a0 = a2 − a1 − a0 = 0 implies a1 = 1, a2 = 2 and

17 an+3 − an+2 − an+1 − an = 0, n ∈ N ∪ {0} . Thus, an = tn+1, n ∈ N ∪ {0} and, therefore, n P n 1 tn+1x = s (x) = 2 3 . In, k=0 1 − x − x − x ∞ ∞   P tn 1 P tn 1 1 particular, n = n−1− = s = n=1 2 2 n=1 2 2 2 1 1 · = 4. 2 1 12 13 1 − − − 2 2 2 ∞ ∞ ∞ ∞ ∞ P xn P tn P tn+1 P tn+2 P n Then, n = n − n+1 + 2 n+2 − n+1 = n=1 2 n=1 2 n=1 2 n=1 2 n=1 2 ∞ ∞ ∞ P tn P tn ∞ tn 1 P n n − n + 2n=3 n − n = n=1 2 n=2 2 2 2 n=1 2 ∞ ∞     t1 P tn 1 1 P tn t1 t2 1 1 1 1 + 2 n − · 2 = − + 2 n − 2 1 + 2 = − + 2 · 4 − 2 + = 6. 2 n=3 2 2 2 n=1 2 2 2 2 2 4 Solution 4 by Brian D. Beasley, Presbyterian College, Clinton, SC We show that the given series converges by first using induction to prove that n xn < 1.95 for each positive integer n. Note that this claim holds for n ∈ {1, 2, 3}. n Given a positive integer k, if xn < 1.95 for n ∈ {k, k + 1, k + 2}, then k+2 k+1 k k xk+3 < 1.95 + 1.95 + 1.95 + k = 1.95 (6.7525) + k. Thus it suffices to show that 1.95k(6.7525) + k ≤ 1.95k+3, or equivalently k ≤ 1.95k(0.662375). This latter inequality holds for each positive integer k (using a n separate induction argument). Hence xn < 1.95 for n ≥ 1, so for any positive integer m, m ∞ ∞ n X xn X xn X 1.95 0.975 < < = = 39. 2n 2n 2n 1 − 0.975 n=1 n=1 n=1 Since its sequence of partial sums is increasing and bounded above, the given series converges.

∞ X xn Next, we let = L. Then 2n n=1 ∞   ∞ 1 1 X xn+2 + xn+1 + xn + n 3 1 1 1 1 X n L = + + = + L − + L + L + . 2 4 2n+3 4 2 2 4 8 2n+3 n=0 n=0 ∞ X n Since = 2, we conclude L = 7 L + 1 + 1 (2) and hence L = 6. 2n 8 2 8 n=0 Also solved by Kee-Wai Lau, Hong Kong, China; Ioannis D. Sfikas, National and Kapodistrian University of Athens, Greece; Albert Stadler (two solutions), Herrliberg, Switzerland; David Stone and John Hawkins, Southern Georgia University, Statesboro, GA, and the proposer.

Mea − Culpa

18 Arkady Alt of San Jose, CA should have been credited with having solved 5477, and 5478. Dionne Bailey, Elsie Campbell, Charles Diminnie, and Trey Smith, all of Angelo State University in San Angelo, TX should have been credited with having solved 5475. Paul M. Harms, of North Newton, KS should have been credited for having solved 5476. Anna Valkova Tomova of Varna, Bulgaria should have been credited with having solved 5475 and 5477. Mea Culpa.

19 Problems Ted Eisenberg, Section Editor *********************************************************

This section of the Journal offers readers an opportunity to exchange interesting mathematical problems and solutions. Please send them to Ted Eisenberg, Department of Mathematics, Ben-Gurion University, Beer-Sheva, Israel or fax to: 972-86-477-648. Questions concerning proposals and/or solutions can be sent e-mail to . Solutions to previously stated problems can be seen at .

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Solutions to the problems stated in this issue should be posted before December 15, 2018

5505: Proposed by Kenneth Korbin, New York, NY Given a Primitive Pythagorean Triple (a, b, c) with b2 > 3a2. Express in terms of a and b the sides of a Heronian Triangle with area ab(b2 − 3a2). (A Heronian Triangle is a triangle with each side length and area an integer.)

5506: Proposed by Daniel Sitaru, “Theodor Costescu” National Economic College, Drobeta Turnu-Severin, Mehedinti, Romania " # 1 5 100  25 −5 100 Find Ω = det + . 5 25 −5 1

5507: Proposed by David Benko, University of South Alabama, Mobile, AL A car is driving forward on the real axis starting from the origin. Its position at time 0 ≤ t is s(t). Its speed is a decreasing function: v(t), 0 ≤ t. Given that the drive has a finite path (that is lim s < ∞), that v(2t)/v(t) has a real limit c as t → ∞, find all t→∞ possible values of c.

5508: Proposed by Pedro Pantoja, Natal RN, Brazil Let a, b, c be positive real numbers such that a + b + c = 1. Find the minimum value of a b c f(a, b, c) = + + . 3ab + 2b 3bc + 2c 3ca + 2a

1 5509: Proposed by Jos´eLuis D´ıaz-Barrero, Barcelona Tech, Barcelona, Spain Let x, y, z be positive real numbers that add up to one and such that x y z π 0 < , , < . Prove that y z x 2 s r y  r  z  x 3√ x cos + y cos + z cos < 5. z x y 5

5510: Proposed by Ovidiu Furdui and Alina Sˆınt˘am˘arianboth at the Technical University of Cluj-Napoca, Cluj-Napoca, Romania

Calculate ∞ X [4n (ζ(2n) − 1) − 1] , n=1 where ζ denotes the Riemann zeta function.

Solutions

5487: Proposed by Kenneth Korbin, New York, NY p √ (x + 1)4 b + b − b Given that 2 = a with x = p √ . Find positive integers a and b. x(x − 1) b − b − b Solution 1 by David E. Manes, Oneonta, NY p √ p √ b + b − b 2b 2 b − b If x = p √ , then x + 1 = p √ and x − 1 = p √ . Moreover, b − b − b b − b − b √ b − b − b 4 4 16b 2 4(b − b) (x + 1) = p √ and (x − 1) = p √ . Therefore, (b − b − b)4 (b − b − b)2 16b4 4 √ √ (x + 1) 4 √(b− b− b) a = 2 = √ √ x(x − 1) (b+ b− b)(4(b− b) √ √ (b− b− b)3 16b4 = √ p √ p √ 4(b − b)(b + b − b)(b − b − b) 4b4 = √ √ . b3 − b2 − b2 b + 2b b − b √ Note that the two√ terms with√ b have opposite signs and cancel off if b = 2. Let b = 2. Then b3 − b2 − b2 b + 2b b − b = 2 and a = 26/2 = 32. Hence, b = 2 and a = 32 is the unique solution.

Solution 2 by Anthony J. Bevelacqua, University of North Dakota, Great Falls, ND

2 p √ b + c 2b For notational convenience set c = b − b. We have x = so x + 1 = and b − c b − c 2c x − 1 = . Thus a is b − c (x + 1)4  2b 4 b − c b − c2 = · · x(x − 1)2 b − c b + c 2c 4b4 = . (b2 − c2)c2 and so a(b2 − c2)c2 = 4b4. Now √ √ (b2 − c2)c2 = (b2 − b + b)(b − b) √ = (b3 − b2 − b) + (2b − b2) b and so √ a((b2 − b − 1) + (2 − b) b) = 4b3. √ Thus (2 − b) b is a rational number. Therefore either b = 2 or b = d2 for some positive integer d. In the first case our last displayed equation yields a · 1 = 4 · 23 and so a = 32. Thus a = 32 and b = 2 is a solution to our problem. In the second case we have √ (b2 − b − 1) + (2 − b) b = d4 − d3 − d2 + 2d − 1.

Call this n. We have an = 4b3. Since a and b are positive so is n. Since d and n are relatively prime we see that n must be a divisor of 4. If n = 1 we have

d4 − d3 − d2 + 2d − 1 = 1 and so d4 − d3 − d2 + 2d − 2 = 0.

By the rational root theorem the only possible positive integer d would be 1 and 2, but neither of these are roots. Similarly n = 2 gives d4 − d3 − d2 + 2d − 3 = 0 and n = 4 gives d4 − d3 − d2 + 2d − 5 = 0, but, again, neither of these have positive integer roots. Thus the only solution to our problem is a = 32 and b = 2.

Solution 3 by Brian D. Beasley, Presbyterian College, Clinton, SC p √ Let c = b − b − b. Then x + 1 = 2b/c and x − 1 = 2(b − c)/c, so (x + 1)4 16b4 c3 4b4 √ √ a = 2 = 4 · p √ = . x(x − 1) c 4(b − c)2(b + b − b) (b2 − b + b)(b − b) √ √ This in turn yields a = 4b4/(b3 − b2 b − b2 + 2b b − b). Since a is a positive integer, we must have either b = n2 for some positive integer n or −b2 + 2b = 0. If b = n2, then 4(n3 + n2 − 3n + 2) a = 4n2 + 4n + 8 + ; n4 − n3 − n2 + 2n − 1 the fraction in this latter expression is not an integer for 1 ≤ n ≤ 5 and is strictly between 0 and 1 for n > 5, so a is not a positive integer. Thus −b2 + 2b = 0, so b = 2 and hence a = 32.

Also solved by Michel Bataille, Rouen, France; Ed Gray, Highland Beach, FL; Khanh Hung Vu (Student), Tran Nghia High School, Ho Chi Minh,

3 Vietnam; Kee-Wai Lau, Hong Kong, China; Ioannis D. Sfikas, National and Kapodistrain University of Athens, Greece; David Stone and John Hawkins, Georgia Southern University, Statesboro, GA, and the proposer.

5488: Proposed by Daniel Sitaru, “Theodor Costescu” National Economic College, Drobeta, Turnu-Severin, Mehedinti, Romania Let a, and b be complex numbers. Solve the following equation:

x3 − 3ax2 + 3(a2 − b2)x − a3 + 3ab2 − 2b3 = 0.

Solution 1 by Dionne Bailey, Elsie Campbell, Charles Diminnie, and Trey Smith, Angelo State University, San Angelo, TX

To begin, we note that

x3 − 3ax2 + 3 a2 − b2
x − a3 + 3ab2 − 2b3 can be re-written as

x3 − 3ax2 + 3a2x − a3
− 3b2x + 3ab2 − 2b3 or (x − a)3 − 3b2 (x − a) − 2b3. Hence, if we substitute y = x − a, the given equation becomes

y3 − 3b2y − 2b3 = 0. (1)

Next, the left side of equation (1) can be re-grouped to obtain

y3 − 3b2y − 2b3 = y3 + b3
− 3b2 (y + b) = (y + b) y2 − by + b2
− 3b2 = (y + b) y2 − by − 2b2
= (y + b)2 (y − 2b) .

Therefore, the solutions of (1) are y = 2b and y = −b (double solution). Finally, since y = x − a, the solutions of the original equation are x = a + 2b and x = a − b (double solution).

Solution 2 by Michel Bataille, Rouen, France Let p(x) denote the polynomial on the left-hand side. Then, a short calculation gives

p(X + a) = X3 − 3b2X − 2b3 = (X + b)2(X − 2b) which has 2b as a simple root and −b as a double one. It immediately follows that the solution of the given equation are a − b, a − b, a + 2b.

Solution 3 by Paul M. Harms, North Newton, KS

4 The equation can be written as (x − a)3 − 3ab2(x − a) − 2b3 = 0. If b = 0, the solution is x = a. If b is not zero, let x − a = yb. Then the equation become b3(y3 − 3y − 2) = 0. We have y3 − 3y − 2 = (y − 2)(y + 1)2 = 0. The y solutions are 2, −1 and −1. The solutions of the equation in the problem are x = a + 2b and x = a − b as a double root.

Solution 4 by G. C. Greubel, Newport News,VA 0 = x3 − 3 a x2 + 3(a2 − b2) x − (a3 − 3ab2 + 2b3) = x3 − 3ax2 + (a − b)(3a + 3b) x − ((a2 − 2ab + b2)(a + 2b) = x3 − (2(a − b) + (a + 2b)) x2 + (a − b)((a − b) + 2(a + 2b)) x − (a − b)2(a + 2b) = (x2 − 2(a − b) x + (a − b)2)(x − (a + 2b)) = (x − (a − b))2 (x − (a + 2b)). From this factorization the solutions of the cubic equation are x ∈ {a − b, a − b, a + 2b}.

Editor0s comment: David Stone and John Hawkins made an instructive comment in their solution that merits being repeated. They wrote: “We confess - we did not immediately recognize the factorization. We originally used Cardano’s Formula to find the solutions. However, there is a line of heuristic reasoning which would lead to the solution. If we consider a = b, the equation become x3 − 3ax2 = 0, which has x = 0 as a double root. Hence, the difference a − b could be significant. Trying x = a − b (via synthetic division) then proves to be productive.”

Also solved by Brian D. Beasley, Presbyterian College, Clinton, SC; Anthony J. Bevelacqua, University of North Dakota, Great Falls, ND; Bruno Salgueiro Fanego, Viveiro, Spain; Ed Gray, Highland Beach, FL; Kee-Wai Lau, Hong Kong, China; David E. Manes, Oneonta, NY; Paolo Perfetti, Department of Mathematics, Tor Vergata University, Rome, Italy; Angel´ Plaza, University of Las Palmas de Gran Canaria, Spain; Ioannis D. Sfikas, National and Kapodistrain University of Athens, Greece; Neculai Stanciu “George Emil Palade” School, Buzau,˘ Romania and Titu Zvonaru, Comanesti,˘ Romania (two solutions); David Stone and John Hawkins, Georgia Southern University, Statesboro, GA, and the proposer.

5489: Proposed by D.M. Ba˘tinetu-Giurgiu, Bucharest, Romania, and Neculai Stanciu, “George Emil Palade” School Buza˘u, Romania Z a If a > 0, compute x2 − ax + a2
arctan(ex − 1)dx. 0 Solution by Soumitra Mandal, Chandar Nagore, India Let x = a − y ⇒ dx = −dy, when x = 0, y = a; when x = a; y = 0. Z a Ω = (x2 − xa + a2) tan−1(ex − 1)dx 0 5 Z 0 = − {(a − y)2 − a(a − y) + a2} tan−1(ea−y − 1)dy a Z a = (y2 − ay + a2) tan−1(ea−y − 1)dy, therefore, 0 Z a 2Ω = (x2 − ax + a2){tan−1(ex − 1) + tan−1(ea−x − 1)}dx 0 Z a x a−x 2 2 −1 e − 1 + e − 1 = (x − xa + a ) tan x a−x dx 0 1 − (e − 1)(e − 1)

x=a ! Z a π x3 x2 5πa3 2 2 −1 2 = (x − ax + a ) tan (1)dx = − a + a x = . 0 4 3 2 24 x=0 5πa3 Therefore, Ω = . 48

Also solved by Ioannis D. Sfikas, National and Kapodistrain University of Athens, Greece, and the proposers.

5490: Proposed by Moshe Stupel, “Shaanan” Academic College of Education and Gordon Academic College of Education, and Avi Sigler, “Shaanan” Academic College of Education, Haifa, Israel Triangle ABC whose side lengths are a, b, and c lies in plane P . The segmentA1A, BB1,CC1 satisfy:

A1A ⊥ P,B1B ⊥ P,C1C ⊥ P, where A1A = a, B1B = b and C1C = c, as shown in the figure. Prove that 4A1B1C1 is acute -angled.

6 Solution 1 by Michel Bataille, Rouen, France −→ −→ −→ −→ We shall use the dot product, recalling that U · V has the same sign as cos(6 (U, V )). We calculate −−−→ −−−→ −−→ −−→ −−→ −−→ −→ −−→ A1B1 · A1C1 = (A1A + AB + BB1) · (A1A + AC + CC1) −−→ −→ = a2 + 0 − ac + 0 + AB · AC + 0 − ab + 0 + bc 1 −−→ −→ = (a2 + b2 + c2 − 2ac − 2ab + 2bc) (since 2AB · AC = b2 + c2 − a2) 2 1 = (b + c − a)2. 2 −−−→ −−−→ Thus, A1B1 · A1C1 > 0 and so 6 B1A1C1 is acute. −−−→ −−−→ 1 2 Similarly, we obtain B1C1 · B1A1 = (c + a − b) > 0 and −−−→ −−−→ 2 1 2 6 6 C1A1 · C1B1 = 2 (a + b − c) > 0 and therefore C1B1A1 and A1C1B1 are acute as well. Solution 2 by Muhammad Alhafi, Al Basel High School, Aleppo, Syria

2 2 2 We will prove that B1C1 < B1A1 + A1C1 . 2 2 2 If we draw a line through C1 parallel to BC we will see that a + (b − c) = B1C1 . In the same manner we have:

7 2 2 2 2 2 2 A1B1 = c + (a − b) , A1C1 = b + (a − c) . So the inequality is equivalent to: a2 + (b − c)2 < c2 + (a − b)2 + b2 + (a − c)2

⇐⇒ 2ab + 2ac < a2 + b2 + c2 + 2ab

⇐⇒ 2a(b + c) < a2 + (b + c)2, which follows from the AM−GM inequality.

2 2 2 Following this line of reasoning we can prove: B1A1 < B1C1 + A1C1 and that 2 2 2 A1C1 < B1A1 + B1C1 . Hence, 4A1B1C1 is acute.

Solution 3 by Michael N. Fried, Ben-Gurion University, Beer Sheva, Israel Suppose we are given an arbitrary triangle such as ABC with sides BC = a, AC = b, and AB = c. Let the lines AA0, BB0, CC0 with lengths a, b, and c, respectively, be drawn perpendicular to the plane of ABC (see figure 1). Then the triangle A0B0C0 with sides B0C0 = a0, A0C0 = b0, and A0B0 = c0 is acute.

Let us consider first the special case when ABC is an isosceles triangle. First, it is obvious that if ABC is isosceles then also A0B0C0 will be isosceles. Moreover, if BC is the base and the angle at A is already acute then the angle at A0 will also be acute since a = a0 and c0 = b0 > b = c so that the angle at A0 will be less than the angle at A. So we need only consider the case when A is obtuse. In that case, also a > b = c. It makes life easier to consider A0B0C0 with respect to the plane UVW drawn through C0 (or B0) and parallel to ABC so that also UVW =∼ ABC. In that case, VW coincides with B0C0 and UA0 = a − c (or a − b) (see figure 2).

8 With that out of the way, we need to show that if α is the apex angle at A0 then α < 90◦, or, by the law of cosines, that 2c02 cos α = 2c02 − a2 > 0. Or since c02 = c2 + (a − c)2:

2c2 + 2(a − c)2 − a2 > 0

Or, opening parentheses and rearranging:

4c2 − a(4c − a) > 0

Note that by the triangle inequality, 2c − a > 0 so that certainly 4c − a > 0. By the arithmetic/geometric mean inequality, then, we have (keeping in mind that a 6= 4c − a since otherwise 2c = a which is impossible):

a + (4c − a)2 4c2 = > a(4c − a) 2

So, indeed, 4c2 − a(4c − a) > 0 and α < 90◦. Now, let us consider the case in which ABC is not isosceles. Let us assume that a > b > c. As before, consider A0B0C0 with respect to the plane UVW drawn through C0 and parallel to ABC. Then we have WB0 = b − c and UA0 = a − c (see figure 3).

9 We have then: a02 = a2 + (b − c)2 b02 = b2 + (a − c)2 c02 = c2 + (a − b)2 Observe that as a > b > c, also a0 > b0 > c0, for consider a02 − b02:

a02 − b02 = a2 + (b − c)2 − b2 − (a − c)2 = (a − b)2c > 0 so that a02 > b02. Similarly, we can show that b02 > c02. Since a0 is thus the longest side of A0B0C0, the angle at A0, which we call α0, is the largest angle. Therefore, it suffices to show that α0 < 90◦. Again, by the law of cosines this means we must show:

2b0c0 cos α0 = b02 + c02 − a02 > 0

Substituting the expressions above for a0, b0, and c0, we have to show:

b2 + (a − c)2 + c2 + (a − b)2 − a2 − (b − c)2 > 0

After some algebra, the expression on the left-hand side can be rewritten as follows:

c2 − (a − b)(2c − (a − b))

Notice that a − b > 0 since we are assuming that a is the longest side of ABC. Also since by the triangle inequality we have c − (a − b) = b + c − a > 0 , it is certainly true that 2c − (a − b) > 0. Therefore, again by the arithmetic/geometric-mean inequality, we have:

(a − b) + (2c − (a − b)2 c2 = > (a − b)(2c − (a − b)) 2

10 So, indeed, b02 + c02 − a02 = c2 − (a − b)(2c − (a − b)) > 0 From which we have α0 < 90◦.

Also solved by Yagub N. Aliyev, Problem Solving Group of ADA University, Baku Azerbeaijan; Kee-Wai Lau, Hong Kong, China; David E. Manes, Oneonta, NY; Ioannis D. Sfikas, National and Kapodistrain University of Athens, Greece; Titu Zvonaru, Com˘anesti,Romania and Neculai Stanciu, “George Emil Palade” School, Buz˘au,Romania; David Stone and John Hawkins, Georgia Southern University, Statesboro, GA, and the proposers.

5491: Proposed by Roger Izard,Dallas, TX Let O be the orthocenter of isosceles triangle ABC, AB = AC. Let OC meet the line segment AB at point F. If m = FO, prove that c4 ≥ m4 + 11m2c2.

Solution 1 by Ed Gray, Highland Beach, FL We assume that c is one of the two equal legs. We re-write the inequality by dividing by c4, so: m4 m2 1) 1 ≥ + 11 . We attempt to prove the inequality by finding the maximum c c m value of . We shall use the following notation: vertex A is the apex (top) with angle c 2t. We note that 2t < 90, otherwise O = A, or O is external to the triangle. Vertex B is at lower left, and has value 90 − t. Vertex C is at lower right, also having a value of 90 − t. Let P be the mid-point of BC, y = BF, c − y = AF, m = OF , and the base, s BC = s, so that BP = PC = . We note that 4F AC is a right triangle, so 2 6 ACF = 90 − 2t. Since 6 ACB = 90 − t, by subtraction, 2) 6 FCB = t. From 4AOF , m 3) tan(t) = . From 4FCB, c − y y 4) sin(t) = , or y = s · sin(t). From 4ABP , s s s 5) sin(t) = , or c = . Substituting (4) and (5) into (3), 2c 2 sin(t) s 6) m = tan(t) . Dividing (6) by (5), (2 sin(t)) − s · sin(t) m sin(t) s t 7) = · − s · sin(t) · 2 sin , or c cos(t) 2 sin(t) s m sin(t) − 2 sin3(t) 8) = c cos(t) d m cos(t) cos(t) − 6 cos(t) sin2(t)
− sin(t) − 2 sin3(t)
(− sin(t)) 9) = . Simplifying, dt c cos2(t) 10) 8 sin4(t) − 6 sin2(t) + 1 = 0. This is a quadratic equation in sin2(t) with roots: 11) 16 sin2(t) = 6  p(36 − 32), or 1 1 12) sin2(t) = , or sin2(t) = . The former is impossible, since t = 45, and 2t = 90, 2 4

11 1 which would put O = A. Therefore, sin(t) = , and t = 30, 2t = 60, and we have an 2 c equilateral triangle. Then c = s, y = , and from (3) 2 m 13) tan(30) = c , and 2 √ m 1 3 m2 3 1 m4 1 14) = tan(30) = , = = , = , so c 2 6 c 36 12 c 144 m4 m2 1 11 15) + 11 = + < 1, and the conjecture is proved. Q.E.D. c c 144 12

Solution 2 by Albert Stadler, Herrliberg, Switzerland π The angle α at the vertex A is ≤ , because OC meets the line segment AB. Clearly 2 α α OF m AF = AC cos α and OF = AF tan = AC cos α tan . Furthermore = . 2 2 AC c Therefore we need to prove that α α α cos4 α tan4 + 11 cos2 α tan2 ≤ 1, for 0 ≤ α ≤ . (1) 2 2 2 We note that

  2 α 1 − tan 2 α  2 α  α 2 α α 2 1 − x y = cos α tan = 2 cos − 1 tan =  α − 1 tan = tan α = x 2 , 2 2 2 1 + tan2 2 2 1 + tan2 1 + x 2 2 α α h αi where we have put x = tan . Clearly the function x = tan maps the interval 0, 2 2 2 to the interval [0, 1]. We claim that √ 1 − x2 q√ 5 − 1 max x = 5 − 2 . 0≤x≤1 1 + x2 2 Indeed, √ √ √ 2
 p   p  d 1 − x2 1 − 4x2 − x4 − x + 2x + 5 x − 5 − 2 x + 5 − 2 x = = , dx 1 + x2 (1 + x2)2 (1 + x2)2 1 − x2 p√ so the maximum of x in the interval [0, 1] is assumed at 5 − 2 and equals 1 + x2 √ √ p√ 3 − 5 p√ 5 − 1 5 − 2√ = 5 − 2 . 5 − 1 2 Therefore √ !4 √ !4 α α p√ 5 − 1 p√ 5 − 1 cos4 α tan4 + 11 cos2 α tan2 ≤ 5 − 2 + 11 5 − 2 = 1, 2 2 2 2 and (1) is proven.

Solution 3 by Kee-Wai Lau, Hong Kong, China Without loss of generality, let b = c = 1. Let AB = AC and AO is perpendicular to BC π so AO bisects 6 BAC. Let 6 BAC = 2θ, where 0 < θ ≤ . 4 12 By considering triangles AOF and ACF , we obtain respectively m = AF tan θ and AF = cos 2θ, so that m = tan θ cos 2θ. Let t = tan θ, so that 0 < t ≤ 1. Then t(1 − t2) dm 1 − 4t2 − t4 p√ m = . We have = , which vanishes when t = 5 − 2, at 1 + t2 dt (1 + t2)2 r √ 5 5 − 11 which m attains its maximum value of . Hence 2 √ √ 123 − 55 5 55 5 − 121 m4 + 11m2 ≤ + = 1, 2 2 and this completes the solution.

Also solved by Ioannis D. Sfikas, National and Kapodistrain University of Athens, Greece, David Stone and John Hawkins, Georgia Southern University, Statesboro, GA, and the proposer.

5492: Proposed by Jos´eLuis D´ıaz-Barrero, Barcelona Tech, Barcelona, Spain Let a, b, c, d be four positive numbers such that ab + ac + ad + bc + bd + cd = 6. Prove that

r abc r bcd r cda r dab r2 + + + ≤ 2 . a + b + c + 3d b + c + d + 3a c + d + a + 3b d + a + b + 3c 3

Solution 1 by Kee-Wai Lau, Hong Kong, China By the inequality of Cauchy-Schwarz, the left side of the inequality of the problem does r abc bcd cda dab not exceed 2 + + + . a + b + c + 3d b + c + d + 3a c + d + a + 3b d + a + b + 3c 6 − ab − bc − ca From the given relation, we have d = , so that a + b + c abc 2abc(a + b + c) abc(a + b + c) = ≤ . a + b + c + 3d (a − b)2 + (b − c)2 + (c − a)2 + 36 18

Similarly,

bcd bcd(b + c + d) ≤ b + c + d + 3a 18 cda cda(c + d + a) ≤ c + d + a + 3b) 18 (dab dab(d + a + b) ≤ . d + a + b + 3c) 18

Hence the inequality of the problem will follow from

abc(a + b + c) + bcd(b + c − d) + cda(c + d + a) + dab(d + a + b) ≤ 12. (1)

Now it can be checked readily that the left side of (1) equals

13 2(ab + ac + ad + bc + bd + cd)2 − (a − b)2(c − d)2 − (b − c)2(d − a)2 − (c − a)2(b − d)2 , 6 (ab + ac + ad + bc + bd + cd)2 which does not exceed = 12. 3 This completes the solution.

Solution 2 by Ed Gray, Highland Beach, FL 1) Let n = a + b + c + d. Then: 2) n2 = a2 + b2 + c2 + d2 + 2ab + 2ac + 2ad + 2bc + 2bd + 2cd = a2 + b2 + c2 + d2 + 12 3) Suppose that a = b = c = d = a. Then (2) becomes: 4) (4a)2 = 4a2 = 12, and a = 1. The left side of the inequality becomes: 5) 4p1/6) = 2p4/6 = 2p2/3, and we see that the inequality becomes an equality. We need show that the expression is a maximum when a = b = c = d. We do this by leaving a = b = 1, c = .99, d = 1.01 so that the constant n = a + b + c + d is maintained. Substituting the new values into the left side, √ 6) p.99/6.02 + p.9999/6 + .9999 + p1.01/5.98 = 7) .405526605 + .408227878 + .407549194 + .410969976 = 1.632273698 < 1.632993162 = 2p2/3. Hence the function is a maximum for a = b = c = d, and the inequality is proven.

Solution 3 by Neculai Stanciu “George Emil Palade” School, Buzau,˘ Romania and Titu Zvonaru, Comanesti,˘ Romania With the Cauchy-Buniakovski-Schwarz inequality we have abc bcd cda dab + + + a + b + c + 3d b + c + d + 3a c + d + a + 3b d + a + b + 3c

 abc bcd cda dab  ≤ 4 + + + . a + b + c + 3d b + c + d + 3a c + d + a + 3b d + a + b + 3c With the AM−HM inequality we have

abc abc 1  abc abc abc  = ≤ + + a + b + c + 3d a + d + b + d + c + d 9 a + d b + d c + d

abc bcd cda dab + + + ≤ a + b + c + 3d b + c + d + 3a c + d + a + 3b d + a + b + 3c

1  abc bcd abc acd abc abd bcd acd bcd abd abd acd  ≤ + + + + + + + + + + + = 9 a + d a + d b + d b + d c + d c + d a + b a + b a + c a + c b + c b + c

1 2 = (bc + ac + ab + cd + bdf + ad) = . 9 3 14 Hence, by the inequalities from above we obtain the desired inequality!

Solution 4 by Marian Urs˘arescu,National College “Roman-Voda,” Roman, Romania

Cauchy’s Inequality implies

r !2 X abc X abc 4 ≥ ⇒ a + b + c + 3d a + b + c + 3d

r r X abc X abc ≤ 2 ⇒ a + b + c + 3d a + b + c + 3d

r s X abc X abc ≤ 2 (1) a + b + c + 3d (a + d) + (b + d) + (c + d)

 1 1 1 1 1 1 1 1 But, (x + y + z) + + ≥ 9 ⇒ ≤ + + , which implies x y z x + y + z 9 x y z

1 1 1 1 1  ≤ + + (2) (a + d) + (b + d) + (c + d) 9 a + d b + d c + d

From (1) and(2) we obtain,

r r X abc 2 X  1 1 1  ≤ abc + + . (3) a + b + c + 3d 3 a + d b + d c + d

But

X  1 1 1  abc + + = a + d b + d c + d

abc abc abc bcd bcd bcd + + + + + + a + d b + d c + d a + b a + c a + d

cda cda cda dab dab dab + + + + + + = b + a b + c b + d c + a c + b c + d

bc(a + d) ac(b + d) ab(c + d) bc(a + b) 4d(a + c) ad(4 + d) = + + + + + = a + d b + c c + d a + b a + c 4 + d

15 ab + ac + ad + bc + 4d + cd = 6. (4)

Equations (3) and (4) implies that r r X abc 2√ 2 ≤ 6 = 2 a + b + c + 3d 3 3

Solutions 5 and 6 by Paolo Perfetti, Department of Mathematics, Tor Vergatta University, Rome, Italy √ First Proof The first step uses the concavity of the function x yielding r s r X abc X abc 2 ≤ 2 ≤ 2 a + b + c + 3d a + b + c + 3d 3 cyc cyc that is X abc 2 ≤ a + b + c + 3d 3 cyc Cauchy–Schwarz reversed yields 1 1 1 9 + + ≥ a + d b + d c + d a + b + c + 3d so it suffices to prove

1 abc abc abc bcd bcd bcd + + + + + + 9 a + d b + d c + d d + a b + a c + a ! cda cda cda dab dab dab 2 ab + bc + ca + ad + bd + cd + + + + + + ≤ a + b c + b d + b a + c b + c d + c 3 6

We can rewrite it as abc bcd abc cda abc dab bcd cda bcd dab + + + + + + + + + + a + d d + a b + d d + b c + d d + c b + a a + b c + a a + c cda dab + + ≤ ab + bc + ca + ad + bd + cd c + b c + b

This is actually an equality since abc bcd + = bc a + d d + a and so on for the other five cases. This concludes the proof. √ Proof 6 (Computer assisted) The first step uses the concavity of the function x yielding r s r X abc X abc 2 ≤ 2 ≤ 2 a + b + c + 3d a + b + c + 3d 3 cyc cyc

16 that is X abc 2 ≤ a + b + c + 3d 3 cyc First case d = 0. The inequality is abc 2 ≤ (1) a + b + c 3 We know that √ 2 3pabc(a + b + c) = 3abc(a + b + c) ≤ (ab + bc + ca)2 ⇐⇒ abc(a + b + c) ≤ (ab)2 + (bc)2 + (ca)2 and this holds true by the AGM (ab)2 + (ac)2 ≥ a2bc and cyclic. Based on this we can write √ 6 = ab + bc + ca ≥ 3pabc(a + b + c) ⇐⇒ abc(a + b + c) ≤ 12 which inserted in (1) gives 12 1 3 ≤ 2 ⇐⇒ (a + b + c)2 ≥ 18 a + b + c a + b + c This follows easily by (a + b + c)2 ≥ 3(ab + bc + ca) = 18 Second case d = 1 which is allowed by the homogeneity of the inequality after writing

X abc 2 ab + bc + cd + da + ac + bd ≤ a + b + c + 3d 3 6 cyc

For d = 1 the above inequality becomes abc bc ca ab 2 + + + ≤ (2) a + b + c + 3 b + c + 1 + 3a c + 1 + a + 3b 1 + a + b + 3c 3 This is a algebraic symmetric inequality in three variables and we employ the so called “UVW” theory. Thus we change variables

a + b + c = 3u, ab + bc + ca = 3v2, abc = w3

By expanding (2) we get X A(a, b, c) 8ab + 3a + 16a2 + 26a3 + 16a4 − 150a2b2 + 8a4bc + 36a2b2c + cyc  +42a3bc + 36a2bc − 150a2b2 + 26a3b3 + 3a5 + X  +A(a, b, c) −11a3b2c − 11a3b + 3a5b + 16a4b2 − 11a3b2 + 8a4b + sym . +A(a, b, c)(−150a2b2c2 + 42abc) = A(a, b, c)B(a, b, c)

A(a, b, c) = −9(a + b + c + 3)(b + c + 1 + 3a)(c + 1 + a + 3b)(1 + a + b + 3c)

17 Now we prove the Lemma that: The polynomial B(a, b, c) is a concave parabola in the variable w3.

Proof of the Lemma We concentrate on the terms of order six, the only terms containing w6. X X (8a4bc + 26a3b3) + (−11a3b2c + 3a5b + 16a4b2) − 150(abc)2 (3) cyc sym and once introduced the new variables (u, v, w), we are interested in those terms containing w6. We have, X X a4bc = abc a3 = w3(3w3 + 27u3 − 27uv2) cyc cyc X X a3b3 = 27v6 − 27uv2w3 + 3w6, a3b2c = w3(9uv2 − 3w3), cyc sym Moreover, X X X X X X X X a5b = a a5 − a6, a5 = a3 a2 − 2 a3b2 sym cyc cyc cyc cyc cyc cyc sym X X Since a2 + b2 + c2 = 9u2 − 6v2, in a5b only a6 contains w6 and precisely sym sym X a6 = 729u6 − 1458u4v2 + 729u2v4 + 162u3w3 − 54v6 − 108uv2w3 + 3w6 cyc X X X X a4b2 = a2 a4 − a6 sym cyc cyc cyc The coefficient of the term w6 of (3) is

24 + 26 · 3 + 11 · 3 − 3 · 3 − 16 · 3 − 150 = −72 and so the Lemma has been proved.

Since A(a, b, c) < 0, −B(a, b, c) is a convex parabola whose maximum is attained at one or both the extreme points of variations of w3. The “UVW” theory states that once fixed the values of u and v, the minimum value of w occurs when abc = 0 = w3 or when b = c (or cyclic) while the maximum value occurs when b = c (or cyclic). So we need to study two cases.

First case. c = 0.

ab + bc + cd + da + ac + bd = 6 ⇐⇒ a = (6 − b)/(1 + b) Inequality (2) becomes (5b2 − 16b + 14 − ≤ 0 3(7 + b + b2) which evidently holds true.

Second case. c = b.

18 ab + bc + cd + da + ac + bd = 6 ⇐⇒ a = (6 − b2 − 2b)/(1 + 2b) Inequality (2) becomes

(b2 − 7)(7b4 − 18b3 − 27b2 − 64b − 114)(b − 1)2 − ≤ 0 (4) 3(4b + 7b2 + 7)(−2b + b2 + 19)(3 + b2 + 2b) √ Clearly a ≥ 0 so b ≤ 7 − 1 and then b2 − 7 ≤ 0. Moreover √ 7b4 − 18b3 − 27b2 ≤ 0 ⇐⇒ b ≤ (9 + 270)/7 and thus 7b4 − 18b3 − 27b2 − 64b − 114 ≤ 0. The conclusion is that (4) holds true and this completes the proof.

Also solved by Michel Bataille, Rouen, France; Ioannis D. Sfikas, National and Kapodistrain University of Athens, Greece, and the proposer.

Mea Culpa

Brian D. Beasley of Presbyterian College in Clinton, SC should have been credited with having solved problem 5510.

19 Problems Ted Eisenberg, Section Editor *********************************************************

This section of the Journal offers readers an opportunity to exchange interesting mathematical problems and solutions. Please send them to Ted Eisenberg, Department of Mathematics, Ben-Gurion University, Beer-Sheva, Israel or fax to: 972-86-477-648. Questions concerning proposals and/or solutions can be sent e-mail to . Solutions to previously stated problems can be seen at .

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Solutions to the problems stated in this issue should be posted before January 15, 2019

5511: Proposed by Kenneth Korbin, New York, NY √ √ A trapezoid with perimeter 58 + 14 11 is inscribed in a circle√ with diameter 17 + 7 11. Find its dimensions if each of its sides is of the form a + b 11 where a and b are positive integers.

5512: Proposed by Angel´ Plaza, Universidad de Las Palmas de Gran Canaria. Spain n n 2 If ak > 0, (k = 1, 2, . . . , n) then n − n ≥ . X 1 X 1 n + 1 1 + a ak k=1 k k k=1

5513: Proposed by Michael Brozinsky, Central Islip, NY In an n × n × n cube partitioned into n3 congruent cubes by n − 1 equally spaced planes parallel to each pair of parallel faces, there are 20 times as many non-cubic rectangular parallelepipeds that could be formed as were cubic parallelepipeds. What is n?

5514: Proposed by D. M. Batinetu-Giurgiu, “Matei Basarab” National College, Bucharest, Romania and Neculai Stanciu, “George Emil Palade” School, Buza˘u, Romania ! !  π  √ b · n+1p(2n + 1)!! If a ∈ 0, and b = arcsin a, then calculate lim n n! sin − a . 2 n→∞ pn (2n − 1)!!

1 5515: Proposed by Jos´eLuis D´ıaz-Barrero, Barcelona Tech, Barcelona, Spain Let n be a positive integer. Prove that  2 n s 2 1 X 1 n − 1 + ≥ 1. 2n  n2 k − 1  k=1

5516: Proposed by Ovidiu Furdui and Alina Sˆınt˘am˘arianboth at the Technical University of Cluj-Napoca, Cluj-Napoca, Romania

∞ X  1 1 1  Calculate n ζ(3) − 1 − − · · · − − . 23 n3 2n2 n=1

Solutions

5493: Proposed by Kenneth Korbin, New York, NY Convex quadrilateral ABCD is inscribed in a circle with diameter AC = 729. Sides AB and CD each have positive integer length. Find the perimeter if BD=715.

Solution by Bruno Salgueiro Fanego, Viveiro, Spain Let a = AB, b = BC, c = CD and d = DA. Since AC is a diameter of the circumcircle π of ABCD, 6 CBA = = 6 ADC and hence the Pythagorean theorem can be applied on 2 4ABC and 4ACD: a2 + b2 = 7292 = c2 + d2 . Since ABCD is cyclic, by Ptolemy’s theorem ac + bd = 729 × 715. Thus, (7292 − a2)(7292 − c2) = b2d2 = (729 · 715 − ac)2, that is, the point with positive integer coordinates (a, c) lies on the ellipse whose equation is 729x2 − 1430xy + 729y2 − 14737464 = 0. From this it follows that (a, c) ∈ {(279, 405), (405, 279), (715, 729), (729, 715)} and since a < 729 and c < 729,

(a, c) ∈ {(279, 405), (405, 279)}, so the lengths of the sides of ABCD are √ √ √ √ (a, b, c, d) ∈ {(279, 180 14, 405, 162 14), (279, 162 14, 405, 180 14)} √ and hence, the perimeter of ABCD is a + b + c + d = 342(2 + 14).

Editor0s comment : Ioannis D. Sfikas’ solution to this problem started off with comments about its related history. “In Euclidean geometry, Ptolemy’s inequality relates the six distances determined by four points in the plane or in a higher-dimensional space. It states for any four points A, B, C, D the following inequality holds: AB · CD + BC · DA ≥ AC · BD.

2 As a special case, Ptolemy’s theorem states that the inequality becomes an equality exactly when the four points lie in cyclic order on a circle. The inequality does not generalize from Euclidean spaces to arbitrary metric spaces. The spaces where it remains valid are called the Ptolemaic spaces; they include the inner product spaces, Hadamard spaces, and shortest path distances on Ptolemaic graphs. In other words, Ptolemy’s theorem is a relation between the four sides and two diagonals of a cyclic quadrilateral (a quadrilateral whose vertices lie on a common circle). The theorem is named after the Greek astronomer and mathematician Ptolemy. Ptolemy used the theorem as an aid to creating his table of chords, a trigonometric table that he applied to astronomy.” Ioannis then went on to solve the problem in the above manner.

Also solved by the Brian D. Beasley, Presbyterian College, Clinton, SC; Cartesian Gains Student Problem Solving Group, Mountain Lakes High School, Mountain Lakes, NJ; Ed Gray, Highland Beach, FL; Paul M. Harms, North Newton, KS; Kee-Wai Lau, Hong Kong, China; Ioannis D. Sfikas, National and Kapodistrian University of Athens, Greece; Albert Stadler, Herrliberg, Switzerland; David Stone and John Hawkins, Georgia Southern University, Statesboro, GA, and the proposer.

5494: Proposed by Moshe Stupel, “Shaanan” Academic College of Education and Gordon Academic College of Education, and Avi Sigler, “Shaanan” Academic College of Education, Haifa, Israel If a ≥ b ≥ c ≥ d are the lengths of four segments from which an infinite number of convex quadrilaterals can be constructed, calculate the maximal product of the lengths of the diagonals in such quadrilaterals.

Solution 1 by Ioannis D. Sfikas, National and Kapodistrian University of Athens, Greece A classical result of Claudius Ptolemy of Alexandria (circa 85AD−165AD), known as Ptolemy’s theorem, states that for a cyclic quadrilateral with side lengths a, b, c, d (in that order) and diagonals of lengths p and q, the product of the lengths of the diagonals equals the sum of the products of the lengths of the opposite sides, pq = ac + bd. For a general convex quadrilateral, we have Ptolemy’s inequality: Theorem 1. For a convex quadrilateral with sides of length a, b, c, d (in that order) and diagonals of length p and q, we have pq ≤ ac + bd. For the above problem, we have to order α, β, and γ, where α = ab + cd, β = ac + bd and γ = ad + bc. Then, we have: (i) If we have α ≥ β, that means ab + cd ≥ ac + bd, or a(b − c) + d(c − b) ≥ 0, which holds. (ii) If we have α ≥ γ, that means ab + cd ≥ ad + bc, or b(a − c) + d(c − a) ≥ 0, or (b − d)(a − c) ≥ 0, which holds. So, if a ≥ b ≥ c ≥ d are the lengths of four segments, from which an infinite number of convex quadrilaterals can be constructed, then the maximal product of the lengths of the diagonals in such quadrilaterals is ab + cd. [1] Alsina, Claudi and Nelsen, Roger B. (2009). When less is more: visualizing basic inequalities, p. 82. Mathematical Association of America.

3 Solution 2 by Kee-Wai Lau, Hong Kong, China We show that the maximal product of the length of the diagonals in such quadrilaterals equals ab + cd. Let WXYZ be a convex quadrilateral such that WX = w, XY = x, YZ = y, ZW = z. By a result of C.A. Bretschneider, the product of the lengths of the diagonals equals pw2y2 + x2z2 − 2wxyz cos(6 XWZ + 6 XYZ), which does not exceed pw2y2 + x2z2 + 2wxyz = wy + xz. Note that

ab + cd = ac + bd + (b − c)(a − d) ≥ ac + bd and ab + cd = ad + bc + (a − c)(b − d) ≥ ad + bc. Hence in order to obtain the maximum product of the diagonals, we put w = a, x = c, y = b, and z = d. It is easy to check that when WXYZ is a cyclic quadrilateral, we have s s (ab + cd)(bc + ad) (ab + ca)(ac + bd) WY = and XZ = , (ac + bd) (bc + ad) so that the maximum product ab + cd is attained. Hence our claimed maximum.

Solution 3 by Albert Stadler, Herrliberg, Switzerland The German mathematician Carl Anton Bretschneider derived in 1842 the following generalization of Ptolemy’s theorem, regarding the product pq of the diagonals in a convex quadrilateral

p2q2 = a2c2 + b2d2 − 2abcd cos(A + C) (1)

This relation can be considered to be a law of cosines for a quadrilateral. Since 4cos(A + C) ≥ −1, it also gives a proof of Ptolemy’s inequality. We note that the product p2q2 in (1) is maximal if cos(A + C) = −1, i.e., if A + C = 180◦ which implies that the product pq is maximal if the quadrilateral is a cyclic quadrilateral. In that case we get pq = ac + bd. It remains to determine for which permutation of the sides the term ac + bd is maximal. There are three possibilities, namely ab + cd, ac + bd, ad + bc. Of these three expressions ab + cd is maximal, since ab + cd − (ac + bd) = (a − d)(b − c) ≥ 0, and ab + cd − (ad + bc) = (a − c)(b − d) ≥ 0. References [1] Titu Andreescu & Dorian Andrica, Complex Numbers from A to . . . Z, Birkh¨auser, 2006, pp. 207-209.

Solution 4 by David Stone and John Hawkins, Georgia Southern University, Statesboro, GA The maximal product of the lengths of the diagonals is ab + cd. This maximum is achieved when (AB, BC, CD, DA) = (a, c, b, d) is a cyclic quadrilateral.

4 By considering the vertices as hinges, Thomas proves [1] that any convex quadrilateral can be deformed into a cyclic quadrilateral (having the same side lengths). In any convex quadrilateral, Ptolemy’s Inequality tells us that the product of the diagonals is less than or equal to the sum of the products of the lengths of opposite sides. In a cyclic quadrilateral, Ptolemy’s Theorem, tells us that the product of the diagonals equals the sum of the products of the lengths of opposite sides. Given our four appropriate segments, a, b, c, d, there are six ways to arrange them in a convex quadrilateral. By symmetry, only three of these are distinct. We show these three possibilities with corresponding bound on the product of the diagonals, AC · BD: (AB, BC, CD, DA) = (a, b, c, d); AC · BD ≤ ac + bd (AB, BC, CD, DA) = (a, b, d, c); AC · BD ≤ ad + bc (AB, BC, CD, DA) = (a, c, b, d); AC · BD ≤ ab + cd. The third case gives the largest possible value (because we’ve placed the two largest sides opposite one another). Algebraically, ac + bd ≤ ab + cd ⇐⇒ 0 ≤ (a − d)(b − c) which is true by the given ordering, and ad + bc ≤ ab + cd ⇐⇒ 0 ≤ (a − c)(b − d) which is also true by the given ordering. Summarizing: when the four segments are arranged in a quadrilateral ABCD, the product of the diagonals is ≤ AB · CD + BC · AD; the largest possible value for AB · CD + BC · AD is ab + cd, which is achieved when a, b and c, d are opposite sides of a cyclic quadrilateral. Reference: 1. Peter, Thomas, Maximizing the Area of a Quadrilateral, The College Mathematics Journal, Vol. 34, No. 4 (September 2003), pp. 315-316.

Also solved by Kenneth Korbin, New York, NY; David E. Manes, Oneonta, NY, and the proposers.

5495: Proposed by D.M. Ba˘tinetu-Giurgiu, “Matei Basarab” National College, Bucharest and Neculai Stanciu, “George Emil Palade” School Buza˘u, Romania √ √ 3 pn Let {xn}n≥1, x1 = 1, xn = 1 · 3!! · 5!! · ... (2n − 1)!!. Find:  (n + 1)2 n2  lim √ − √ . n→∞ n+1 xn+1 n xn

Solution 1 by Moti Levy, Rehovot, Israel

(2n)! (2n − 1)!! = . (1) 2nn!

5 Using Stirling’s asymptotic formula, we have nn n! ∼ . (2) en Applying (2) to (1) yields 1 ! (2n)2n n 2n pn (2n − 1)!! ∼ = (3) e2n2nn! e

Now we use (3) to approximate xn,

n n n n n 2 n n Y 2 n! n 2 n x ∼ 2ke = ∼ e = , n en en e2n k=1 or √ 2n n x ∼ . n e2 Hence, (n + 1)2 n2 e2 e2 e2 √ − √ ∼ (n + 1) − n = , n+1 xn+1 n xn 2 2 2 ! (n + 1)2 n2 e2 and we conclude that lim √ − √ = =∼ 3.694 5. n→∞ n+1 xn+1 n xn 2

Solution 2 by Kee-Wai Lau, Hong Kong, China e2 We show that the limit of the problem equals . 2 We need the following knows results for positive integers n.

 1  1  ln(n!) = n + ln n − n + A + O , (1) 2 n n X 1  1  = ln n + B + O , (2) k n k=1  1  1  1  ln 1 + = + O . (3) n n n2

where A and B are constants. By (1) we have

ln 2  1  ln ((2k)!) − ln(k!) = ln(k!) + (2 ln 2 − 1)k + + O . (4) 2 k

Next we show that (1 + ln 2) ln n = n ln n + (ln 2 − 2)n + + O(1) (5) n 2 6 In fact by (4) we have

n n X ln((2k − 1)!!) X ln((2k)!) − ln(k!) − (ln 2)k ln x = . = n k k k=1 k=1

n X  ln 2  1  = ln k + ln 2 − 1 + + O 2k k2 k=1

n ln 2 X 1 = ln(n!) + (ln 2 − 1)n + + O(1), 2 n k=1 and (5) follows from (1) and (2). ln x Let f(n) = 2 ln n − n . By (5), we obtain n ln n f(n) = ln n + 2 − 2 + O . (6) n

We next show that 1 ln n f(n + 1) − f(n) = + O . (7) n n2 In fact ln x ln x  f(n + 1) − f(n) = 2(ln(n + 1) − ln n) − n+1 − n n + 1 n

 1  ln x ln(2n + 2)! − ln(n + 1)! − (ln 2)(n + 1) = 2 ln 1 + + n − , n n(n + 1) (n + 1)2 and (7) follows readily from (3),(5) and (4). By the mean value theorem, we have

ef(n+1) − ef(n) = (f(n + 1) − f(n)) et, (8) where t is a number lying between f(n) and f(n + 1). By (6), both ef(n+1) and ef(n) e2n  ln n equal 1 + O . Hence, by (7) and (8), 2 n

(n + 1)2 n2 e2  ln n √ − √ = ef(n+1) − ef(n) = 1 + O , n+1 xn+1 n xn 2 n and our claim for the limit follows.

Solution 3 by Bruno Salgueiro Fanego, Viveiro, Spain

We will use the lemma from Solution 3 to Problem 5398 that appeared in this Column (see Nov. 2016 issue) that stated: “If the positive sequence (pn) is such that pn+1 √ √ p n lim = p > 0, then lim ( n+1 pn+1 − pn) = .” n→∞ npn n→∞ e 7 n2n We let {pn}n≥1, pn = . Then xn

(n + 1)2n+2 p x n+1p(2n + 1)!! lim n+1 = lim n n→∞ np n→∞ n2n n n xn

(n + 1)2n+2 (n + 1)2n(n + 1)2 = lim = lim n→∞ n2n+1 n+1p(2n + 1)!! n→∞ n2nn n+1p(2n + 1)!!

s n + 1n2 n+1p(n + 1)2(n+1) n2n = lim lim = e2 lim n n→∞ n n→∞ n+1pnn+1(2n + 1)!! n→∞ (n − 1)n(2n − 1)!!

(n + 1)2(n+1) nn+1(2n + 1)!! (n + 1)2n(n + 1)2(n − 1)n = e2 lim = e2 lim root crition n→∞ n2n n→∞ nnn2n(2n + 1) (n − 1)n(2n − 1)!!

(n + 1)2n(n + 1)2(n − 1)n = e2 lim n→∞ n2nn(2n + 1)nn

(n + 1)2n (n + 1)2 (n − 1)n = e2 lim lim lim n→∞ n2n n→∞ n2n+1 n→∞ nn

n + 1n2 1  1 n 1 = e2 lim lim 1 − = e2e2 e−1 n→∞ n 2 n→∞ n 2

e3 = =: p > 0, which implies by the lemma mentioned above, 2 √ √ p e2 n that the required limit is lim ( n+1 pn+1 − pn) = = . n→∞ e 2

Editor0s comment : In addition to the above solution Bruno Salgueiro Fanego stated that a more general form of the problem was published by the authors0 of 5495 in the journal La Gaceta de la Real Sociedad Matem´atica Espana˜ola vol. 17 (3), 2014, pp. 523-524. (available at http://gaceta.rsme.es/abrir.phd?id=1218.) Therein they showed:

If {an}n≥1 is a sequence of real positive numbers such that lim (an+1 − an) = a 6= 0, n→∞ then   (n + 1)2 (n)2 e lim − = .  q pn n  n→∞ n+1 n+1 Q ca Q k=1 f(ak) k=1 f(ak)

pn Letting an = n and f(n) = (2n − 1)!! gives the desired result.

8 Solution 4 by Michel Bataille, Rouen, France

1/n n2  n2n  e2 Let un = √n = . We show that lim (un+1 − un) = . xn xn n→∞ 2 (2n)! To this end, we first recall that (2n − 1)!! = and the following asymptotic 2n(n!) expansion as n → ∞: ln(n) √ 1 ln(n!) = n ln(n) − n + + ln( 2π) + + o(1/n), 2 12n from which we readily deduce (2n)! ln(2) 1 ln = ln((2n)!) − ln(n!) = n ln(n) + n(2 ln(2) − 1) + − + o(1/n). n! 2 24n Now, we have

n  1/k! ln(xn) 1 Y 1 (2k)! sn ln(u ) = 2 ln(n)− = 2 ln(n)− ln · = 2 ln(n)+ln(2)− (1) n n n 2 k! n k=1

n P 1  (2k)!  where sn = k · ln k! . k=1 Consider the sequence {yn}n≥2 defined by 1 + ln(2) y = s − n ln(n) − (2 ln(2) − 2)n − · ln(n). n n 2 For n → ∞, we calculate 1 + ln(2)  1  y − y = s − s − n ln(n) + (n − 1) ln(n − 1) − (2 ln(2) − 2) + · ln 1 − n n−1 n n−1 2 n 1  (2n)!  1  1 − ln(2)  1  = ln + n ln 1 − − ln(n) − ln 1 − + (2 − ln(2)) n n! n 2 n ln(2) 1  1 1 1  = 2 ln(2) − 1 + − + o(1/n2) + n − − − + o(1/n3) 2n 24n2 n 2n2 3n3 1 − ln(2)  1 1  − − − + o(1/n2) + 2 − 2 ln(2) 2 n 2n2 a = + o(1/n2) n2 1+2 ln(2) P∞ where we set a = − 8 . Thus, the series n=2(yn − yn−1) is convergent. Let S Pn denotes its sum. Then, we may write k=2(yk − yk−1) = S + o(1) and so yn = b + o(1) as n → ∞ (where b = S + y1). It follows that 1 + ln(2) s = n ln(n) + (2 ln(2) − 2)n + · ln(n) + b + o(1) n 2 as n → ∞. From (1), we now obtain 1 + ln(2) ln(n) b ln(u ) = ln(n) + 2 − ln(2) − · − + o(1/n). n 2 n n 9 ln(n)+2−ln(2) o(1) First, we deduce that ln(un) = ln(n) + 2 − ln(2) + o(1), hence un = e · e e2 and so un ∼ n · 2 . Second, the calculation of ln(un+1) − ln(un) easily leads to  1  1 ln(u ) − ln(u ) = ln 1 + + o(1/n) = + o(1/n). n+1 n n n

(Note that ln(n) ln(n+1) 1 −1

1  ln(n)  n − n+1 = n (ln n) 1 − (1 + 1/n) + o(1) = n − n + o(1) = o(1/n) as n → ∞.) Since u    n+1 ln(un+1)−ln(un) un+1 − un = un − 1 = un e − 1 un we finally arrive at

e2  1  e2 u − u ∼ u (ln(u ) − ln(u )) ∼ n · · ∼ n+1 n n n+1 n 2 n 2 and the result follows.

Also solved by Paolo Perfetti, Department of Mathematics, Tor Vergata University, Rome, Italy; Ioannis D. Sfikas, National and Kapodistrian University of Athens, Greece; Albert Stadler, Herrliberg, Switzerland; Marian Ursarescu - Romania, and the proposers.

5496: Daniel Sitaru, “Theodor Costescu” National Economic College, Drobeta Turnu-Severin, Mehedinti, Romania Let a, b, c be real numbers such that 0 < a < b < c. Prove that: √ ab X   X  b  ea−b + eb−a ≥ 2a − 2c + 3 + . a cyclic cyclic

Solution 1 by Henry Ricardo, Westchester Area Math Circle, NY For x > 0 we apply the known inequality ex > x + 1 to x = a − b, b − c, and a − c to get

ea−b > a − b + 1, eb−c > b − c + 1, ea−c > a − c + 1, respectively. Adding these inequalities yields

ea−b + eb−c + ea−c > 2a − 2c + 3. (1)

For x > y, we see that √ √ √ ex−y > (x/y) xy ⇐⇒ x − y > xy ln(x/y) ⇐⇒ xy < (x − y)/(ln x − ln y), which is the left-hand member of the logarithmic mean inequality. Thus we have, since 0 < a < b < c, √ √ √ √  b  ab c bc  c  ac a ac eb−a > , ec−b > , ec−a > > . (2) a b a c

10 Adding (1) and (2) , we find that √ ab X   X  b  ea−b + eb−a > 2a − 2c + 3 + . a cyclic cyclic

Solution 2 by Albert Stadler, Herrliberg, Switzerland We will prove the slightly stronger inequality √ ab X   X  b  ea−b + eb−a ≥ a − c + 3 + . a cyclic cyclic

We will use the inequalities ex ≥ 1 + x, x real, (1) √  y  xy 1 ≥ , 0 ≤ y ≤ x, (2) x √  y  xy ey−x ≥ , y ≥ x, (3) x (1) and (2) are clear, while (3) is equivalent to each of the following lines: √  y  y − x ≥ xy log , x r y rx  y  − ≥ log , x y x 1 Z x  1 1 x − − log x = 1 + 2 − dt ≥ 0, x ≥ 1 which holds true. x 1 t t Thus

√ √ ab bc X    b  b ea−b + eb−a ≥ 1 + a − b + + 1 + b − c + + 1 + c − a + aa−c a c cyclic

√ √  b  ab c bc = 3 + + + ea−c a b √ √  b  ab c bc ≥ 3 + + + 1 + a − c a b √ √ √  b  ab c bc a bc ≥ 3 + + + + a − c. a b c

Also solved by Ed Gray, Highland Beach, FL; Kee-Wai Lau, Hong Kong, China, and the proposer.

5497: Proposed by Jos´eLuis D´ıaz-Barrero, Barcelona Tech, Barcelona, Spain

11 n−1 Y kπ  For all integers n ≥ 2, show that 2 sin is an integer and determine it. n k=1 Solution 1 by Kee-Wai Lau, Hong Kong, China

It is proved as formula 12 on p. 227 Chapter XII of [1] that

n−1 Y  kπ  sin nθ = sin θ 2 sin θ + . n k=1

n−1 sin nθ Y  kπ  Since lim = n, so 2 sin θ + = n. θ→∞ sin θ n k=1 Reference: 1. D.V. Durell and A. Robinson: Advanced Trigonometry, Dover Publication, Inc., New York 2003.

Solution 2 by David E. Manes, Oneonta, NY Subtracting the complex equation e−ix = cos(−x) + i sin(−x) = cos x − i sin x from the equation eix = cos x + i sin x, one obtains the formula 1 2 sin x = eix − e−ix
. i Therefore,

n−1 n−1 Y kπ  Y 1   2 sin = eiπk/n − e−iπk/n n i k=1 k=1 n−1 n−1 ! n−1 ! 1 Y Y   = eiπk/n 1 − e−2iπk/n . i k=1 k=1 Note that

n−1  n−1 X X n−1  iπk/n (iπ/n) k Y   eiπk/n = e k=1 = e k=1 = e(iπ/n)((n−1)(n)/2) k=1  n−1 = e(iπ/2)(n−1) = eiπ/2 = (cos(π/2) + i sin(π/2))n−1 = in−1.

Hence, n−1 n−1 Y kπ  Y   2 sin = 1 − e−2iπk/n = f(1), n k=1 k=1 n−1 Y   where f(x) = x − e−2iπk/n . The zeros of the polynomial f(x) are the non-trivial k=1 nth roots of unity so that xn − 1 f(x) = = 1 + x + x2 + ··· + xn−1. x − 1 12 Therefore, f(1) = n. Hence, if n ≥ 0, then

n−1 Y kπ  2 sin = n. n k=1

Solution 3 by Anthony J. Bevelacqua, University of North Dakota, Grand Forks, ND Let ζ = eπi/n. Then ζ2 = e2πi/n is a primitive n-th root of unity. So ζ2, . . . , ζ2(n−1) are the roots of xn − 1 = xn−1 + ··· + 1 x − 1 and therefore n−1 Y xn−1 + ··· + 1 = (x − ζ2k). k=1 Let x = 1 to find

n−1 Y n = (1 − ζ2k) k=1 n−1 Y = −ζk(ζk − ζ−k). k=1

Since ζk = ekπi/n = cos(kπ/n) + i sin(kπ/n) we have ζk − ζ−k = 2i sin(kπ/n). Thus

n−1 Y kπ  n = −2iζk sin . n k=1 Finally, since each sin(kπ/n) > 0 and each | − 2iζk| = 2 for k = 1, . . . , n − 1 we have

n−1 Y kπ  n = 2 sin n k=1 by taking the absolute value of the last expression.

Editor0s comment : Paul M. Harms of North Newton KS mentioned in his solution to 5497 that Wikipedia’s “List of Trigonometric Identities” includes n−1 Y kπ  n sin = , and from this the value of n immediately follows. n 2n−1 k=1 Also solved by Michel Bataille, Rouen, France; Bruno Salgueiro Fanego (three solutions), Viveiro, Spain; Ed Gray, Highland Beach, FL; Paul M. Harms, North Newton KS; Moti Levy, Rehovot, Israel; Paolo Perfetti, Department of Mathematics, Tor Vergata University, Rome, Italy; Angle´ Plaza, University of Las Palmas de Gran Canaria, Spain; Henry Ricardo, Westchester Area Math Circle, NY; Ioannis D. Sfikas, National and Kapodistrian University of Athens, Greece; Albert Stadler, Herrliberg, Switzerland; Marian Urs˘arescu,Romania, and the proposer.

13 5498: Proposed by Ovidiu Furdui and Alina Sˆintam˘ a˘rian, both at the Technical University of Cluj-Napoca, Cluj-Napoca, Romania Prove that ∞ X {n!e} Z 1 ex − 1 = dx n x n=1 0 where {a} denotes the fractional part of a.

Solution 1 by Pedro H. O. Pantoja, Natal/RN, Brazil By Taylor’s formula, 1 1 1 eα e = 1 + + + ··· + + , α ∈ (0, 1), 1! 2! n! (n + 1)! and this implies that

 1 1 1  eα n!e = n! + + ··· + + , α ∈ (0, 1). 1! 2! n! n + 1

Therefore,  1 1 1  {n!e} = n!e − bn!ec = n! e − 1 − − − · · · − ⇒ 1! 2! n! ∞  1 1  X 1 {n!e} = n! + + ··· = . (n + 1)! (n + 2)! (n + 1)(n + 2) ··· (n + k) k=1 We have,

∞ ∞ ∞ X {n!e} X X 1 = n n(n + 1)(n + 2) ··· (n + k) n=1 n=1 k=1

∞ ∞ X X 1 Z 1 = (1 − x)kxn−1dx k! n=1 k=1 0

∞ ∞ Z 1 X (1 − x)k X = xn−1dx k! 0 k=1 n=1

Z 1 1 = (e1−x − 1) · dx 0 1 − x

Z 1 e1−x − 1 = dx 0 1 − x Z 1 ey − 1 = dy, 0 y where in the last integral, we used the substitution y = 1 − x.

14 Solution 2 by Paolo Perfetti, Department of Mathematics, Tor Vergata University, Rome, Italy Both the sides of the equality are equal to ∞ X 1 k · k! k=1  1 1 1 1   1 1  {n!e} = n! 2 + + + ... + + ... = + + ... 2 3! m! (m + 1)! n + 1 (n + 1)(n + 2) Since ∞ ∞ X 1 X < 2−k = 1 (n + 1)(n + 2) ... (n + k) k=1 k=1 it follows that ∞  1 1  X 1 + + ... = n + 1 (n + 1)(n + 2) (n + 1)(n + 2) ... (n + k) k=1 and ∞ ∞ ∞ X {n!e} X X 1 = = n n(n + 1)(n + 2) ... (n + k) n=1 n=1 k=1 ∞ ∞ ∞ X X 1 X 1 = = (1) n(n + 1)(n + 2) ... (n + k) k · k! k=1 n=1 k=1 and finally ∞ ∞ ∞ Z 1 ex − 1 Z 1 X xk−1 X Z 1 xk−1 X 1 dx = dx = dx = x k! k! k · k! 0 0 k=1 k=1 0 k=1

For proving (1) let’s write an = 1/(n(n + 1) ··· (n + k)).

an+1 n = ⇐⇒ an+1(n + 1) − nan = −an+1k an n + k + 1 Telescoping

N N X X an+1(n + 1) − nan = aN+1(N + 1) −a1 = −k an+1 n=1 | {z } n=1 →0 and ∞ X 1 + k 1 a = = k k(k + 1)! k · k! k=1

Solution 3 by Michel Bataille, Rouen, France ex−1 P∞ xn−1 From x = n=1 n! for x ∈ (0, 1] and ∞ Z 1 n−1 ∞ X x X 1 dx = < ∞, n! n · (n!) n=1 0 n=1 15 we deduce that ∞ ! ∞ ∞ Z 1 ex − 1 Z 1 X xn−1 X Z 1 xn−1  X 1 dx = dx = dx = . (1) x n! n! n · (n!) 0 0 n=1 n=1 0 n=1 On the other hand, for n ≥ 1 we have

∞ ∞ X 1 X 1 (n!)e = (n!) = a + j! n (n + 1) ··· (n + k) j=0 k=1 n
n
n
where an = n! + (n − 1)! 1 + (n − 2)! 2 + ··· + 1! n−1 + 1 is a positive integer and ∞ ∞ X 1 X 1 1 0 < < = ≤ 1. (n + 1) ··· (n + k) (n + 1)k n k=1 k=1 It follows that ∞ X 1 {n!e} = (n + 1) ··· (n + k) k=1 and so ∞ ∞ ∞ X {n!e} X X 1 = n n(n + 1) ··· (n + k) n=1 n=1 k=1 ∞ ∞ X X 1 = n(n + 1) ··· (n + k) k=1 n=1 ∞ ∞ ! X 1 X  1 1  = − k n(n + 1) ··· (n + k − 1) (n + 1) ··· (n + k) k=1 n=1 ∞ X 1 1 = · . k 1 · 2 ··· k k=1 ∞ ∞ P {n!e} P 1 Finally we obtain n = k·(k!) , and comparing with (1) gives the required result. n=1 k=1 Also solved by Ed Gray, Highland Beach, FL; Kee-Wai Lau, Hong Kong, China; Moti Levy, Rehovot, Israel; Ioannis D. Sfikas, National and Kapodistrian University of Athens, Greece; Albert Stadler, Herrliberg, Switzerland, and the proposers.

Mea Culpa

Brian D. Beasley of Presbyterian College in Clinton, SC should have been credited with having solved 5482, and Albert Stadler of Herrliberg, Switzerland should have been credited with having solved 5488. Titu Zvonaru of Comanesti,˘ Romania noted that proof number 2 (of the 6 shown) for problem 5492 is incomplete. The question asked us to prove that a certain inequality

16 held that was subject to a constraint on the variables. The author of the solution found values of the variables that produced equality, and then by taking other values in small epsilon neighborhoods around this point that produced equality, showed that the resulting values of the expression were smaller than the value that gave equality. Up to here, everything is fine. But it was then concluded that the point giving equality was a local maximum. The method used was very similar to the one that is often used in obtaining saddle and extrema points vis-a-vis Lagrange Multipliers. Admittedly there is some hand-waving in using this approach, and this is what Titu noticed. The approach used in this problem can tell us when the inequality goes awry, but it cannot be used to prove with absolute certainty that the inequality holds. For that, derivative tests within the theory of Lagrange Multipliers, must be used.

17 Problems Ted Eisenberg, Section Editor *********************************************************

This section of the Journal offers readers an opportunity to exchange interesting mathematical problems and solutions. Please send them to Ted Eisenberg, Department of Mathematics, Ben-Gurion University, Beer-Sheva, Israel or fax to: 972-86-477-648. Questions concerning proposals and/or solutions can be sent e-mail to . Solutions to previously stated problems can be seen at .

————————————————————–

Solutions to the problems stated in this issue should be posted before February 15, 2019

5517: Proposed by Kenneth Korbin, New York, NY Find positive integers (a, b, c) such that  a   b   c  arccos = arccos + arccos with a < b < c. 1331 1331 1331

5518: Proposed by Roger Izard, Dallas, TX Let triangle P QR be equilateral and let it intersect another triangle ABC at points U, U 0, W, W 0,V,V 0 such that WU 0,UV 0,VW 0 are equal in length, and triangles AU 0W, BV 0U, CW 0V are equal in area (see Figure 1). Show that triangle ABC must then also be equilateral

1 5519: Proposed by Titu Zvonaru, Coma˘nesti, Romania Let a, b, c be positive real numbers. Prove that a2 b2 c2 2abc 11 + + + ≥ . b2 c2 a2 a3 + b3 + c3 3

5520: Proposed by Raquel Le´on(student) and Angel Plaza, University of Las Palmas de Gran Canaria, Spain Let n be a positive integer. Prove that

2n X 2n + k2n(−1)k 1 = 0. k k 2k k + 1 k=0

5521: Proposed by Jos´eLuis D´ıaz-Barrero, Barcelona Tech, Barcelona, Spain

Let a > 0 be a real number. If f is an odd non-constant real function having second derivative in the interval [−a, a] and f 0(−a) = f 0(a) = 0, then prove that there exists a point c ∈ (−a, a) such that 1 |f(a)| f 00(c) ≥ 2 a2

5522: Proposed by Ovidiu Furdui and Cornel V˘alean from Technical University of Cluj-Napoca, Cluj-Napoca, Romania and Timi¸s,Romania, respectively

Calculate Z 1 Z 1 log(1 − x) − log(1 − y) dxdy. 0 0 x − y

Solutions

5499: Proposed by Kenneth Korbin, New York, NY Given a triangle with sides (21, 23, 40). The sum of these digits is 2 + 1 + 2 + 3 + 4 + 0 = 12. Find primitive pythagorean triples in which the sum of the digits is 12 or less.

Solution 1 by Anthony J. Bevelacqua, University of North Dakota, Grand Forks, ND

The two best-known primitive pythagorean triples (3, 4, 5) and (5, 12, 13) have this property. We will give two infinite families of such triples. Recall that a = s2 − t2, b = 2st, c = s2 + t2 is a primitive pythagorean triple for any s > t ≥ 1 with gcd(s, t) = 1 and s and t of opposite parity.

2 1. For any n ≥ 1 let s = 10n + 1 and t = 10n. Then

a = s2 − t2 = 2 · 10n + 1 b = 2st = 2 · 102n + 2 · 10n c = s2 + t2 = 2 · 102n + 2 · 10n + 1

is a primitive pythagorean triple. The sum of the digits in (a, b, c) is

2 + 1 + 2 + 2 + 2 + 2 + 1 = 12.

2. For any n ≥ 1 let s = 102n + 1 and t = 10n. Then

a = s2 − t2 = 104n + 102n + 1 b = 2st = 2 · 103n + 2 · 10n c = s2 + t2 = 104n + 3 · 102n + 1

is a primitive pythagorean triple. The sum of the digits in (a, b, c) is

1 + 1 + 1 + 2 + 2 + 1 + 3 + 1 = 12.

Solution 2 by Dionne Bailey, Elsie Campbell, Charles Diminnie, and Trey Smith, Angelo State University, San Angelo, TX

To begin, we note the well-known result that (a, b, c) is a primitive Pythagorean triple, with a2 + b2 = c2 and a odd if and only if a = m2 − n2, b = 2mn, and c = m2 + n2 for some positive integers m and n such that m > n, gcd (m, n) = 1, and m and n have opposite parity. We can provide an infinite family of primitive Pythagorean triples for which the sum of the digits is 12 by choosing m = 10k + 1 and n = 10k with k ≥ 0. Then,

2  k  2k k ak = 10 + 1 − 10 = 2 × 10 + 1,

 k   k 2k k bk = 2 10 + 1 10 = 2 × 10 + 2 × 10 , and 2  k  2k 2k k ck = 10 + 1 + 10 = 2 × 10 + 2 × 10 + 1 for k ≥ 0. As noted above, (ak.bk, ck) is a primitive Pythagorean triple for each k ≥ 0. Further, in each case, the sum of the non-zero digits for ak, bk, and ck is (2 + 1) + (2 + 2) + (2 + 2 + 1) = 12. In particular, when k = 0, we have (a0, b0, c0) = (3, 4, 5), the best known primitive Pythagorean triple. Another example is (a, b, c) = (5, 12, 13). However, we haven’t been able to generalize this in a manner similar to that shown above. Also, we haven’t found any other examples of primitive Pythagorean triples (a, b, c) for which the sum of the digits of a, b, and c is 12 or less.

Editor0s comment : David Stone and John Hawkins, both of Georgia Southern University in Stateboro, GA stated that a computer search revealed no triples with total digit sum < 12. In each triple with total digit sum 12, the 12 was achieved as

3 3+4+5. They also found no triples with total digit sum 13 or 14. They went on to find the above mentioned infinite class with digit sum15 and ended their solution with the comment: “We do not know whether there are other triples with total digit sum 12. Note that x + y + z = 2ab + b2 − a2
+ b2 + a2
= 2b(ab). For any integer w we know that w = Digitsum(w) mod 3. Thus Digitsum (x)+ Digitsum (y)+ Digitsum (z) = x + y + z = 2b(a + b) mod 3. So if the total digit sum is 12, then b = 0 mod 3 or a + b = 0 mod 3. That is, there are restrictions on the generators a and b.”

Also solved by Ed Gray, Highland Beach, FL; David E. Manes, Oneonta, NY; Ioannis D. Sfikas, National and Kapodistrian University of Athens, Greece; Albert Stadler, Herrliberg, Switzerland, and the proposer.

5500: Proposed by Moshe Stupel, “Shaanan” Academic College of Education and Gordon Academic College of Education, and Avi Sigler, “Shaanan” Academic College of Education, Haifa, Israel 3 Without the use of a calculator, show that: 8 sin 20◦ · sin 40◦ · sin 60◦ · sin 80◦ = . 2 Solution 1 by David A. Huckaby, Angelo State University, San Angelo, TX

8 sin 20◦ · sin 40◦ · sin 60◦ · sin 80◦ = 8 sin(30◦ − 10◦) · sin(30◦ + 10◦) · sin 60◦ · sin(90◦ − 10◦) = 8[sin 30◦ · cos 10◦ − cos 30◦ · sin 10◦] · [sin 30◦ · cos 10◦ + cos 30◦ · sin 10◦] √ 3 · · [sin 90◦ · cos 10◦ − cos 90◦ · sin 10◦] 2 " √ # " √ # √ 1 3 1 3 3 = 8 · cos 10◦ − · sin 10◦ · · cos 10◦ + · sin 10◦ · · [cos 10◦] 2 2 2 2 2 √ h √ i h √ i = 3 cos 10◦ · cos 10◦ − 3 sin 10◦ · cos 10◦ + 3 sin 10◦ √ = 3 cos 10◦ · cos2 10◦ − 3 sin2 10◦ √ = 3 cos3 10◦ − 3 sin2 10◦ · cos 10◦ √ = 3 cos 30◦ √ √ 3 = 3 · 2 3 = 2

Solution 2 by Cartesian Gains Student Problem Solving Group, Mountain Lakes High School, Mountain Lakes, NJ

eiθ − e−iθ We use the well known formula: sin θ = . 2i Converting to radians, we rewrite the left side of the equation as

 1 4         8 · · eiπ/9 − e−iπ/9 · ei2π/9 − e−i2π/9 · ei3π/9 − e−i3π/9 · ei4π/9 − e−i4π/9 . 2i

4 Expanding gives: 1   · ei(10π/9) + e−i(10π/9) − ei(8π/9) − e−i(8π/9) − ei(6π/9) − e−i(6π/9) + 2 (1) 2

We use the fact that on the unit circle ei(10π/9) represents the same complex number as e−i(8π/9). Similarly, ei(8π/9) = e−i(10π/9). These terms cancel out in our equation. Additionally,

   6π 6π −6π −6π  − ei(6π/9) + e−i(6π/9) = − cos + i sin + cos + i sin = −2 cos(6π/9) = 1. 9 9 9 9

1 3 Therefore, equation (1) reduces to: (1 + 2) = . 2 2 Editor0s comments: Albert Stadler of Herrliberg, Switzerland and several other solvers, noticed that this problem is a special case of problem 5497, which asked us to find a closed form of n−1 Y kπ  2 sin . n k=1 He showed that n−1 Y kπ  xn − 1 2 sin = lim 1 = n. n x x − 1 k=1 kπ  n − k)π  By symmetry sin = sin , so if n is odd then n n

n−1 Y2 kπ  √ 2 sin = n. n k=1 So problem is 5500 is the special case with n = 9.

Yagub Alyiev of ADA University in Baku, Azerbaijan, mentor to the two students listed below from his university who solved the problem, sent two web addresses wherein animated solutions can be found. See: https://www.youtube.com/watch?v=Tc58b2AGFf4 (and) https://www.youtube.com/watch?v=zAiXPhPvWpct=187s.

Also solved by Arkady Alt; San Jose, CA; Michel Bataille, Rouen, France; Brian D. Beasley (two solutions), Presbyterian College, Clinton, SC; Anthony J. Bevelacqua, University of North Dakota, Grand Forks, ND; Scott H. Brown, Montgomery, AL; Michael Brozinsky, Central Islip, NY; Elsie Campbell, Dionne T. Bailey, Charles Diminnie, and Trey Smith, Angelo State University, San Angelo, TX; : Michael C. Faleski, Delta College, University Center, MI; Bruno Salgueiro Fango, Viveiro, Spain; Ed Gray, Highland Beach, FL; Vagif Hamzayev(student), ADA University, Baku, Azerbaijan; Paul M. Harms, North Newton, KS; Kee-Wai Lau, Hong Kong, China; Carl Libis, Columbia Southern University, Orange Beach, AL;

5 David E. Manes, Oneonta, NY; Kamal Mustafayev (student), ADA University, Baku, Azerbaijan; Pedro H.O. Pantoja, Natal/RN, Brazil; Paolo Perfetti, Department of Mathematics, Tor Vergata University, Rome, Italy; Ioannis D. Sfikas (two solutions), National and Kapodistrian University of Athens, Greece; Digby Smith, Mount Royal University, Calgary, Canada; Albert Stadler of Herrliberg, Switzerland; Neculai Stanciu, “George Emil Palade” School Buz˘au,Romania and Titu Zvonaru, Com˘anesti,Romania; David Stone and John Hawkins of Georgia Southern University, Statesboro, GA; Daniel Vacaru,˘ Pitesti, Romania, and the proposers.

5501: Proposed by D.M. Ba˘tinetu-Giurgiu, Bucharest, Romania, Neculai Stanciu, “George Emil Palade” School Buza˘u, Romania and Titu Zvonaru, Coma˘nesti, Romania Determine all real numbers a, b, x, y that simultaneously satisfy the following relations: (1) ax + by = 5     2 2 (2) ax + by = 9

(3) ax3 + by3 = 17    (4) ax4 + by4 = 33.

Solution 1 by Stanley Rabinowitz, Chelmsford, MA (1) ax + by = 5. (2) ax2 + by2 = 9. (3) ax3 + by3 = 17. (4) ax4 + by4 = 33. (5) ax = 5 − by, by (1). (6) (5 − by)xy + by3 = 9y, by (2) and (5). (7) ax2 = 9 − by2, by (2). (8) (9 − by2)x + by3 = 17, by (3) and (7). Subtracting (8) from (6) gives (9) 5xy − 9x = 9y − 17. (10) ax3 = 17by3, by (3). (11) (17 − by3)x + by4 = 33, by (4) and (10). Multiplying (8) by y and subtracting (11) yields (12) 9xy − 17x = 17y − 33. Subtracting 5 times (12) from 9 times (9) and dividing the result by 4 gives (13) x = −y + 3. Substituting this value of x into (9) and simplifying yields: −5(y − 1)(y − 2) = 0. Therefore, y = 1 or 2. Suppose y = 1. Then, x = 2, by (13). Thus, 2a + b = 5 and 4a + b = 9, by (1) and (2). Hence, a = 2 and b = 1. That is, (x, y, a, b) = (2, 1, 2, 1). Similarly, if y = 2, then (x, y, a, b) = (1, 2, 1, 2).

6 Note that this result holds in any commutative ring with unity, which has no zero divisors and 5 6= 0.

Solution 2 by David E. Manes, Oneonta, NY Writing 5 = 22 + 1, 9 = 23 + 1, 17 = 24 + 1 and 33 = 25 + 1, one notes that two of the solutions (a, b, x, y) for the system of equations are (1, 2, 1, 2) and (2, 1, 2, 1). We will show that these are the only solutions. Let A be the augmented 4 × 3 matrix for the system of equations where a and b are regarded as the unknowns and the powers of x and y are regarded as the coefficients. Then

 x y 5     2 2  x y 9    A =   .   x3 y3 17     x4 y4 33

Row-reducing A, we find that it is row-equivalent to the matrix R given by

 5y−9  1 0 x(y−x)     0 1 9−5x   y(y−x)    R =   .   0 0 17 − 9y − 9x + 5xy      0 0 33 − 9(x2 + y2 + xy) + 5xy(y + x)

If x = 1 and y = 2, then the two expressions 17 − 9y − 9x + 5xy and 33 − 9(x2 + y2 + xy) + 5xy(y + x) both equal 0. Therefore, a = 1 and b = 2 since 5y−9 9−5x x(y−x) = 1 and y(y−x) = 2 when x = 1 and y = 2. If x = 2 and y = 1, then 17 − 9x − 9y + 5xy = 33 − 9(x2 + y2 + xy) + 5xy(y + x) = 0 so that a = 2 and b = 1. Working with residues modulo 3, one finds that the equation 17 − 9y − 9x + 5xy ≡ 0 (mod 3) if and only if x ≡ 1 (mod 3) and y ≡ 2 (mod 3) or x ≡ 2 (mod 3) and y ≡ 1 (mod 3). Furthermore, these residues have to be least residues since otherwise, the residues can be made to satisfy the first equation in the system, but not the second.

Also solved by Arkady Alt; San Jose, CA; Hatef Arshagi, Guilford Technical Community College, Jamestown, NC; Michel Bataille, Rouen, France; Brian D. Beasley, Presbyterian College, Clinton, SC; Anthony Bevelacqua, University of North Dakota, Grand Forks, ND; Cartesian Gains Student Problem Solving Group, Mountain Lakes High School, Mountain Lakes, NJ; Dionne Bailey, Elsie Campbell, Charles Diminnie, and Trey Smith, Angelo State University, San Angelo, TX; Ed Gray, Highland Beach,FL; Paul M. Harms, North Newton, KS; Kee-Wai Lau, Hong Kong, China; Perfetti Paolo, Department of Mathematics, Tor Vergata University, Rome, Italy; Ioannis D. Sfikas (two solutions), National and Kapodistrian University of Athens, Greece; Digby Smith, Mount Royal University, Calgary, Canada; Albert Stadler of Herrliberg, Switzerland; David Stone and John Hawkins,

7 Georgia Southern University, Statesboro, GA; Daniel Vacaru,˘ Pitesti, Romania, and the proposers.

5502: Proposed by Daniel Sitaru, “Theodor Costescu” National Economic College, Drobeta Turnu-Severin, Mehedinti, Romania Prove that if a, b, c > 0 and a + b + c = e then

e e e 2 2 2 eac · eba · ecb > ee · abe · bce · cae .  1 n Here, e = lim 1 + n→∞ n Solution 1 by Paolo Perfetti, Department of Mathematics, Tor Vergata University, Rome, Italy The inequality is equivalent to

ace + bae + cbe > e + be2 ln a + ce2 ln b + ae2 ln c that is a(ce − e2 ln c) + b(ae − e2 ln a) + c(be − e2 ln b) > e Let f(x) = xe − e2 ln x. e2 f 00(x) = e(e − 1)xe−2 + > 0 x2 Thus by Jensens’s inequality e X a a + b + c a + b + c e (ce − e2 ln c) ≥ e − a2 ln = e e e e cyc

Solution 2 by Moti Levy, Rehovot, Israel The function ln x is monotone increasing, then by applying log function on both sides of the inequality, we get

ace + bae + cbe > e + be2 ln a + ce2 ln b + ae2 ln c, (1) or a b c b c a  ce + ae + be > 1 + e2 ln a + ln b + ln c . (2) e e e e e e The function ln x is concave, hence ab + bc + ca b c a ln ≥ ln a + ln b + ln c. (3) e e e e Thus we get for the right hand side of inequality (2) : b c a  1 − e2 + e2 ln (ab + bc + ca) ≥ 1 + e2 ln a + ln b + ln c . (4) e e e The function xe is convex, hence we get for the left hand side of inequality (2): a b c ab + bc + cae ce + ae + be ≥ . (5) e e e e

8 By (4) and (5), to finish the solution, we have to show that ab + bc + cae > 1 − e2 + e2 ln (ab + bc + ca) . (6) e Let us denote x := (ab + bc + ca)e . (7) e2 Since ab + bc + ca ≤ 3 , then e2 e 0 < x ≤ . (8) 3 Setting (7) in (6), we need to show that

x e2 e > 1 − e2 + e ln x, for 0 < x ≤ , ee 3 or that e2 e f (x) := x − e1+e ln x + ee e2 − 1
> 0, for 0 < x ≤ . (9) 3 e 0 e1+e  e2  One can easily check that f (x) = 1 − x < 0 for 0 < x ≤ 3 . Hence, f (x) is e  e2  monotone decreasing function for 0 < x ≤ 3 . Moreover, limx→0 f (x) = +∞ and e e e  e2    e2  1+e   e2   e 2
∼ f 3 = 3 − e ln 3 + e e − 1 = 7.478 9 > 0. These and the e 1+e e 2
 e2  monotonicity of f (x) imply that x − e ln x + e e − 1 > 0, for 0 < x ≤ 3 .

Solution 3 by Kee-Wai Lau, Hong Kong, China For 0 < x < 1, let f(x) be the convex function xe − e2 ln x. By taking logarithms, we see that the inequality of the problem is equivalent to

af(c) + bf(a) + cf(b) > e. (1) a b c Let γ = , γ = and γ = . By Jensen’s inequality, the left side of (1) 1 e 2 e 3 e ab + bc + ca is greater than or equal to ef(γ c + γ a + γ b) = ef . 1 2 3 e e(xe − e) Since f 0(x) = < 0 and x 2(a + b + c)2 − (a − b)2 − (b − c)2 − (c − a)2 e3 ab + bc + ca = ≤ , so 6 3 ab + bc + ca e f ≥ f = 1.49 ··· > 1. e 3 Thus (1) holds and this completes the solution.

Solution 4 by Michel Bataille, Rouen, France Taking logarithms and arranging, we see that the inequality is equivalent to a b c b c a  · ce + · ae + · be > 1 + e2 · ln a + · ln b + · ln c . e e e e e e

9 Since the functions x 7→ xe and x 7→ ln x are respectively convex and concave on (0, ∞), Jensen’s inequality yields

a b c ab + bc + cae · ce + · ae + · be ≥ e e e e and b c a ab + bc + ca · ln a + · ln b + · ln c ≤ ln . e e e e Therefore, it is sufficient to prove that

U e − e2 ln U − 1 > 0 (1)

ab+bc+ca where U = e . 2 2 2 2 2 e Since e = (a + b + c) = a + b + c + 2(ab + bc + ca) ≥ 3(ab + bc + ca), we have U ≤ 3 , hence U ∈ (0, 1). e 2 0 e(xe−e) Now, let f(x) = x − e ln x − 1. The function f satisfies f(1) = 0 and f (x) = x . It follows that f is strictly decreasing on the interval (0, 1] and so f(U) > f(1), which is the desired inequality (1).

Also solved by Ed Gray, Highland Beach, FL; Ioannis D. Sfikas, National and Kapodistrian University of Athens, Greece; Albert Stadler of Herrliberg, Switzerland, and the proposer.

5503: Proposed by Jos´eLuis D´ıaz-Barrero, Barcelona Tech, Barcelona, Spain

Let a1, a2, . . . , an be positive real numbers with n ≥ 2. Prove that

m m m m (a1 a2 + a2 a3 + ··· + an a1) 1 m m m m+1 ≤ , (a1 + a2 + ··· an ) n where m is a positive integer.

Solution 1 by Michel Bataille, Rouen, France

Let Im(a1, a2, . . . , an) denote the proposed inequality. First we suppose that I1(a1, a2, . . . , an) holds for all a1, . . . , an > 0, in other words that

2 n(a1a2 + a2a3 + ··· + ana1) ≤ (a1 + a2 + ··· + an) (1) for all positive a1, . . . , an and we show that if m is an integer with m ≥ 2, then Im(a1, . . . , an) also holds for all positive a1, . . . , an. m m Let m be an integer with m ≥ 2 and let a1, . . . , an > 0. Applying (1) with a1 , . . . , an m m 2 m m m m instead of a1, . . . , an, respectively, we obtain (a1 + ··· + an ) ≥ n(a1 a2 + ··· + an a1 ) so that

m m m+1 m m 2 m m m−1 m m m m m m m−1 (a1 +···+an ) = (a1 +···+an ) (a1 +···+an ) ≥ n(a1 a2 +···+an a1 )(a1 +···+an ) .

But from Holder’s inequality, we have

m m m m m m m−1 m m m m (a1 a2 + ··· + an a1 )(a1 + ··· + an ) ≥ (a1 a2 + a2 a3 + ··· an a1)

10 m m m+1 m m m m and it follows that (a1 + ··· + an ) ≥ n(a1 a2 + a2 a3 + ··· an a1) , which is the desired inequality Im(a1, . . . , an). Now, we show that (1) holds for all positive a1, . . . , an if and only if n ≤ 4. Suppose that n ≥ 5. If (1) holds for all a1, a2, . . . , an > 0, then in particular it holds if we take a1 = a2 = ··· = an−2 = ε and an−1 = an = 1 where ε is an arbitrary positive number. This provides the inequality n((n − 3)ε2 + 2ε + 1) ≤ ((n − 2)ε + 2)2. Letting ε → 0+, we obtain n ≤ 4, a contradiction. Thus, we must have n ≤ 4. Conversely, if n = 2 (resp. n = 3, resp, n = 4), it is easily checked that (1) is equivalent 2 2 2 2 to (a1 − a2) ≥ 0 (resp. (a1 − a2) + (a2 − a3) + (a3 − a1) ≥ 0, resp. 2 (a1 − a2 + a3 − a4) ≥ 0) and so I1(a1, . . . , an) holds for all a1, . . . , an > 0 when n = 2, 3 or 4. In conclusion, Im(a1, a2, . . . , an) holds for any positive integer m and all positive real numbers a1, . . . , an if and only if n ≤ 4.

Solution 2 by Albert Stadler, Herrliberg, Switzerland The statement is wrong in general. It is true for n = 2 and for any real m ≥ 1, since by H¨older’sinequality, m m m m m m m−1 m−1 m m m  m m 1 m m m−1  (x y + xy ) = x y (x + y ) ≤ x y (1 + 1 ) m (x + y ) m = 1 = xmym2(xm + ym)m−1 ≤ (xm + ym)m+1, 2 m m 1 m m 2 and the last inequality is equivalent to x y ≤ 4 (x + y ) , which is clearly true. The statement is true as well for n = 3 and any real m ≥ 1, since by H¨older’sinequality, xm−1 ym−1 xm−1 zm−1 m (xmy + ymz + zmx)m = xmymzm + + + ≤ z x z y

1 !m   m m m m 1 1 1 m m m m−1 ≤ x y z + + (x + y + z ) m = zm xm ym 1 = (xmym + ymzm + zmxm)(xm + ym + zm)m−1 ≤ (xm + ym + zm)m+1, 3 1 2 and the last inequality is equivalent to ab + bc + ca ≤ 3 (a + b + c) , with a = xm, b = ym, c = zm, which is clearly true, since it is equivalent to ab + bc + ca ≤ a2 + b2 + c2 (which is true because of Cauchy-Schwarz). The problem statement is not true in general, We construct counterexamples as follows:

Let a = 1 for 1 ≤ i ≤ k, ai = 0 for k + 1 ≤ i ≤ n. Then m m m m+1 m+1 m+1 (a1 a2 + a2 a)3 + ··· + an a1 = k − 1 and a1 + a2 + ··· + an = k. The stated inequality then reads as 1 (k − 1)m ≤ km+1, (2) n which fails for an infinity of triples (k, m, n). For instance (2) is wrong for (n − 2, 1, n), if n ≥ 5, it is wrong for (n − 3, 2, n), if n ≥ 8 and it is wrong for (n − 4, 3, n) if n ≥ 12.

Purists may argue that these are not real counter-examples, since ai = 0, for k + 1 ≤ i ≤ n, so that not all ai are strictly positive. However we may replace 0 by  > 0 and make  sufficiently small to reach the same conclusion.

11 Also solved by Arkady Alt; San Jose, CA; Ed Gray, Highland Beach, FL; Perfetti Paolo, Department of Mathematics, Tor Vergata University Rome, Italy; Angel´ Plaza, University of Las Palmas de Gran Canaria, Spain; Ioannis D. Sfikas (two solutions), National and Kapodistrain University of Athens, Greece, and the proposer.

5504: Proposed by Ovidiu Furdui and Alina Sˆınt˘am˘arian bothat the Technical University of Cluj-Napoca, Cluj-Napoca, Romania Let n ≥ 0 be an integer. Calculate

Z 1 xn  1 dx, 0 x where bxc denotes the integer part of x.

Solution 1 by Albert Stadler, Herrliberg, Switzerland We have

Z 1 n K Z 1 K Z 1 K   x X k n X k n X k 1 1  1 dx = lim k x dx = lim k x dx = lim n+1 − n+1 0 K→∞ 1 K→∞ 1 K→∞ n + 1 k (k + 1) x k=1 k+1 k=1 k+1 k=1

K K 1 X  1 k + 1 − 1  1 X  1 1 1  = lim − = lim − + n + 1 K→∞ kn (k + 1)n+1 n + 1 K→∞ kn (k + 1)n (k + 1)n+1 k=1 k=1

K ! 1 1 X 1 ζ(n + 1) = lim 1 − + = . n + 1 K→∞ (K + 1)n (k + 1)n+1 n + 1 k=1

Solution 2 by Stanley Rabinowitz, Chelmsford, MA

 1  We start by breaking the interval (0, 1) up into subintervals over which the function x is constant.

1 n ∞ 1 n Z x X Z k x  1  dx =  1  dx 0 1 x k=1 k+1 x ∞ 1 n X Z k x = dx 1 k k=1 k+1 ∞ 1 X  xn+1  k = k(n + 1) 1 k=1 k+1   n+1  ∞ 1
n+1 1 1 X k+1 =  k −  n + 1  k k  k=1 ∞ 1 X  1 1  = − n + 1 kn+2 k(k + 1)n+1 k=1 12 ∞ 1 X = [A − B] n + 1 k=1 where ∞ ∞ X 1 X 1  1  A = = ζ(n + 2) and B = kn+2 k (k + 1)n+1 k=1 k=1 ∞ X 1 and where ζ(n) = is the Riemann zeta function. kn k=1 Now note that by the formula for the sum of a geometric progression,

n+1 X 1 1 1  1  = − . (k + 1)i k(k + 1) k (k + 1)n+1 i=2 So ∞ X 1  1  B = k (k + 1)n+1 k=1 ∞ " n+1 # X 1 X 1 = − k(k + 1) (k + 1)i k=1 i=2 ∞ ∞ "n+1 # X  1  X X 1 = − k(k + 1) (k + 1)i k=1 k=1 i=2 ∞ n+1 " ∞ # X  1 1  X X 1 = − − k k + 1 (k + 1)i k=1 i=2 k=1 n+1 X = 1 − (ζ(i) − 1) i=2 n+1 X = n + 1 − ζ(i) i=2

Hence Z 1 xn 1  1  dx = [A − B] 0 x n + 1 " n+1 !# 1 X = ζ(n + 2) − n + 1 − ζ(i) n + 1 i=2 "n+2 # 1 X = ζ(i) − (n + 1) n + 1 i=2 "n+2 # 1 X = ζ(i) − 1. n + 1 i=2

Solution 3 by Angel´ Plaza, University of Las Palmas de Gran Canaria, Spain

13 1 1  1   1  For x ∈ (0, 1] if < x ≤ , then k ≤ < k + 1. That is for x, = k. k + 1 k x x Therefore,

∞ Z 1 xn X Z 1/k xn = dx  1  k 0 x k=1 1/k+1

∞ X 1  1 1  = − k(n + 1) kn+1 (k + 1)n+1 k=1

∞ 1 X  1 1  = − n + 1 kn+2 k(k + 1)n+1 k=1

∞ 1 X  1 1 1 1  = + + ··· + − , n + 1 kn+2 (k + 1)n+1 k + 1 k k=1 from where n+2 X ζ(j) 1 n Z x j=2  dx = −1 + . 0 1 n + 1 x

Solution 4 by Moti Levy, Rehovot, Israel

1 The first step is to substitute y = x and then to split the integration range into intervals [k, k + 1] , k ≥ 1.

∞ Z 1 xn Z ∞ y−n−2 X Z k+1 y−n−2 dx = dy = dy  1  byc k 0 x 1 k=1 k ∞ 1 X  1 1  = − (n + 1) kn+2 n+1 k=1 k (k + 1) ∞ ! 1 X 1 = ζ (n + 2) − . (n + 1) n+1 k=1 k (k + 1)

P∞ 1 Let Sn := k=1 k(k+1)n .

∞ X  1 1  S − S = − n+1 n n+1 k (k + 1)n k=1 k (k + 1) ∞ X 1  1 1  = − (k + 1)n k (k + 1) k k=1 ∞ X 1 = − n+1 = 1 − ζ (n + 1) . (10) k=1 (k + 1)

14 It is easy to see that ∞ X 1 S := = 1. (11) 1 k (k + 1) k=1 It follows from (10) and (11) that

n−1 X Sn = n − ζ (k + 1) . k=1

Z 1 xn 1  1 dx = (ζ (n + 2) − Sn+1) 0 x n + 1 n ! 1 X = ζ (n + 2) − (n + 1) + ζ (k + 1) n + 1 k=1 n+2 ! 1 X = ζ (k) − 1. n + 1 k=2

Solution 5 by David Stone and John Hawkins, Georgia Southern University, Statesboro, GA We shall show that the value of the integral is (the average of the“first” n + 1 values of ζ(2) + ζ(3) + ... + ζ(n + 2) the Riemann zeta function) minus 1; that is, − 1. n + 1 We shall need the following result, which seems interesting in its own right. Lemma 1: For m ≥ 1, 1 + x + x(x + 1) + x(x + 1)2 + x(x + 1)3 + ... + x(x + 1)m−1 = (x + 1)m. Proof by induction. The identity is clearly true for m = 1. Upon the induction hypothesis, 1 + x + x(x + 1) + x(x + 1)2 + x(x + 1)3 + ... + x(x + 1)m−1 = x(x + 1)m

= (x + 1)m + x(x + 1)m

= (x + 1)m(1 + x)

= (x + 1)m+1.

This leads to the following result about a partial fractions decomposition. Lemma 2: 1 1 1 1 1 1 1 = − − − − · · · − − . k(k + 1)m k (k + 1)m (k + 1)m−1 (k + 1)m−2 (k + 1)2 k + 1 Proof: After clearing fractions, we see that this identity is equivalent to 1 = (k + 1)n − k − k(k + 1) − k(k + 1)2 − ... − k(k + 1)n−2 − k(k + 1)n−1, which is true by Lemma 1. Now we are in position to calculate the given integral. Note that  1  1 1 1 = k ⇐⇒ k ≤ < k + 1 ⇐⇒ x ≤ . x x k + 1 k

15 Thus

1 n 1 n " n+1 n+1# Z k x Z k x 1 1  1   1  1 1  1 1  1 dx = dx = − = − . 1   1 k k n + 1 k k + 1 k n + 1 kn+1 (k + 1)n+1 k+1 x k+1

Therefore,

1 n ∞ 1 n ∞   Z x X Z k x X 1 1 1 1  1 dx =  1 dx = n+1 − n+1 0 1 k n + 1 k (k + 1) x k=1 k+1 x k=1

∞ 1 X  1 1  = − n + 1 kn+2 k(k + 1)n+1 k=1

∞ 1 X  1  1 1 1 1 1 1  = − − − − − · · · − − n + 1 kn+2 k (k + 1)n+1 (k + 1)n (k + 1)n−1 (k + 1)2 k + 1 k=1

∞ 1 X  1 1 1 1 1 1 1  = − + + + + ··· + + by Lemma 2 n + 1 kn+2 k (k + 1)n+1 (k + 1)n (k + 1)n−1 (k + 1)2 k + 1 k=1

( ∞ ∞ ∞ ∞ ∞ ) 1 X 1 X 1 X 1 X 1 X  1 1  = + + + ··· + − − . n kn+2 (k + 1)n+1 (k + 1)n (k + 1)2 k k + 1 k=1 k=1 k=1 k=1 k=1

The final sum in this expression telescopes, and its sum is 1. Each of the other sums is a shifted version of the zeta function:

∞ X 1 1 1 1  1  1 1 1 = + + + ··· = −1 + + + + + ··· = −1 = ζ(m). (k + 1)m 2m 3m 4m 1m 2m 3m 4m i=1 Therefore,

( ∞ ∞ ∞ ∞ ∞ ) Z 1 xn 1 X 1 X 1 X 1 X 1 X  1 1  dx = + + + ··· + − −  1  n + 1 kn+2 (k + 1)n+1 (k + 1)n (k + 1)2 k k + 1 0 x k=1 k=1 k=1 k=1 k=1

1 = {[−1 + ζ(n + 2)] + [−1 + ζ(n + 1)] + [−1 + ζ(n)] + [−1 + ζ(n − 1)] + ··· + [−1 + ζ(2)] − 1} n + 1

1 = {ζ(2) + ζ(3) + ··· + ζ(n) + ζ(n + 1) + ζ(n + 2) − n · 1 − 1 } n + 1

1 = {ζ(2) + ζ(3) + ··· + ζ(n) + ζ(n + 1) + ζ(n + 2)} − 1. n + 1 16 There are n + 1 terms inside the braces, so we have our promised result:

Z 1 xn  1 dx = 0 x (the average of the first n + 1 values of the Riemann zeta function) minus 1.

Comment: There are other variants of this answer, because the values of the zeta function for even n can be expressed in terms of the Bernoulli numbers. That would not make the answer any nicer though.

Also solved by Michel Bataille, Rouen, France; Ed Gray, Highland Beach, FL; Kee-Wai Lau, Hong Kong, China; Ioannis D. Sfikas (two solutions), National and Kapodistrian University of Athens, Greece; Perfetti Paolo, Department of Mathematics, Tor Vergata University Rome, Italy; Daniel Vacaru,˘ Pitesti, Romania, and the proposers.

Mea Culpa Ioannis D. Sfikas of National and Kapodistrian in University of Athens, Greece should have been credited with having solved 5495 and 5496. Carl Libis of Columbia Southern University in Orange Beach, AL should have been credited with having solved 5497.

Correction: Problem 5514 in the November 2018 issue of this column should have been stated as: ! ! √ b · n+1p(2n + 1)!! If a ∈ (0, 1) and b = arcsin a, then calculate lim n n! sin − a . n→∞ pn (2n − 1)!!

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