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Problems 5475-5522 Ssma Mathematical Magazine 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 <[email protected]>. Solutions to previously stated problems can be seen at <http://www.ssma.org/publications>. ||||||||||||||||||||{ Solutions to the problems stated in this issue should be posted before March 15, 2018 • 5475: Proposed by Kenneth Korbin, New York, NY p 8a + b = 14 ab − 48; < p Given positive integers a; b; c and d such that b + c = 14pbc − 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: p p p ! 1 − 1 + x2 3 1 + x2 · ::: · n 1 + x2 L = lim ln n + lim : n!1 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] spanf1; sin x; sin(2x);:::; sin(nx)g = spanf1; sin x; sin2 x; : : : ; sinn xg if and only if n = 1. k X We mention that spanfv1; v2; : : : ; vkg = ajvj; aj 2 <; 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 jCDj = 1 + x. Suppose x > 1. Then jCP j = 1 and jDQj = x. We have jCP j 1 sin(A=2) = = jCOj jCOj 2 so jCOj sin(A=2) = 1. And jDQj jDQj x sin(A=2) = = = jDOj jDCj + jCOj 1 + x + jCOj so sin(A=2) + x sin(A=2) + jCOj 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) p p sin(A=2) = = 2 13=13 or 3 13=13: 2 Therefore p p 13 + 2 13 13 + 3 13 x = p ≈ 3:491 or x = p ≈ 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: p p 13 − 2 13 13 − 3 13 x = p ≈ 0:286 or x = p ≈ 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 f1; 2; 3; 4; 5; 6; 7; 8; 9g 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 f(2; 7); (9; 7)g, 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.
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