Junior Problems J289. Let a Be a Real Number Such That 0 - DocsLib

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Junior problems

J289. Let a be a real number such that 0 ≤ a < 1. Prove that

1 1 a 1 + + 1 = . 1 − a 1 − a

Proposed by Arkady Alt, San Jose, California, USA

Solution by Ángel Plaza, University of Las Palmas de Gran Canaria, Spain Since 0 ≤ a < 1, then 0 < 1 − a ≤ 1. 1 1 1 k − 1 k 1 If k ∈ such that < 1 − a ≤ , then k ≤ < k + 1, and ≤ a < , so = k. N 1 + k k 1 − a k k + 1 1 − a On the other hand, for the left-hand side of the proposed identity we have

1 a 1 + + 1 = ba (1 + k)c + 1 1 − a = k − 1 + 1 = k.

Also solved by Archisman Gupta, RKMV, Agartala, Tripura, India; Joshua Benabou, Manhasset High School, NY, USA; Daniel Lasaosa, Pamplona, Navarra, Spain; Arber Igrishta, Eqrem Qabej, Vushtrri, Ko- sovo; Mathematical Group Galaktika shqiptare, Albania; Paolo Perfetti, Università degli studi di Tor Vergata Roma, Italy; Alessandro Ventullo, Milan, Italy; Polyahedra, Polk State College, FL, USA; AN-anduud Pro- blem Solving Group, Ulaanbaatar, Mongolia; Ioan Viorel Codreanu, Satulung, Maramures, Romania; Viet Quoc Hoang, University of Auckland, New Zealand.

Mathematical Reflections 1 (2014) 1 J290. Let a, b, c be nonnegative real numbers such that a + b + c = 1. Prove that

p3 p3 p3 13a3 + 14b3 + 13b3 + 14c3 + 13c3 + 14a3 ≥ 3.

Proposed by Titu Andreescu, University of Texas at Dallas, USA

Solution by Shohruh Ibragimov, Lyceum Nr.2 under the SamIES, Samarkand, Uzbekistan From Hölder’s inequality we easily obtain (13a3 + 14b3)(13 + 14)(13 + 14) ≥ (13a + 14b)3 =>

p3 13a + 14b 13a3 + 14b3 ≥ (1) 9 Similarly, we have

p3 13b + 14c p3 13c + 14a 13b3 + 14c3 ≥ , 13c3 + 14a3 ≥ (2) 9 9 From (1) and (2) we get

p3 p3 p3 27(a + b + c) 13a3 + 14b3 + 13b3 + 14c3 + 13c3 + 14a3 ≥ = 3 9

Also solved by Bodhisattwa Bhowmik, RKMV, Agartala, Tripura, India; Daniel Lasaosa, Pamplona, Navarra, Spain; Nicuşor Zlota, Traian Vuia Technical College, Focşani, Romania; Arkady Alt, San Jose, California, USA; An Zhen-ping, Xianyang Normal University, China; Viet Quoc Hoang, University of Auc- kland, New Zealand; Ioan Viorel Codreanu, Satulung, Maramures, Romania; AN-anduud Problem Solving Group, Ulaanbaatar, Mongolia; Polyahedra, Polk State College, FL, USA; Paolo Perfetti, Università degli studi di Tor Vergata Roma, Italy; Ángel Plaza, University of Las Palmas de Gran Canaria, Spain; Sayak Mukherjee, Kolkata, India; Sayan Das, Indian Statistical Institute, Kolkata, India; Neculai Stanciu, Buzău,

Romania and Titu Zvonaru, Comănes, ti, Romania; Alessandro Ventullo, Milan, Italy; Georgios Batzolis, Mandoulides High School, Thessaloniki, Greece.

Mathematical Reflections 1 (2014) 2 J291. Let ABC be a triangle such that ∠BCA = 2∠ABC and let P be a point in its interior such that PA = AC and PB = PC. Evaluate the ratio of areas of triangles P AB and P AC.

Proposed by Panagiote Ligouras, Noci, Italy

Solution by Polyahedra, Polk State College, USA

C

P

D B A ∑

E

Let Σ be the circle with center A and radius AC. Suppose that the bisector of ∠ACB intersects AB at D and Σ at E. Then PD is the perpendicular bisector of BC. Since ∠AEC = ∠ACE = ∠BCE, EA k BC. Thus ∠BAE = ∠ABC, so PD is the perpendicular bisector of AE as well. Hence 4AP E is equilateral. Therefore, 1 ◦ ◦ ◦ ◦ ◦ ∠PCE = 2 ∠P AE = 30 , ∠AP C = 30 +B, and ∠AP B = 60 +∠EPB = 60 +∠AP C = 90 +B. Finally, let [·] denote area, then

[P AB] sin AP B cos B 2 = ∠ = = √ . [P AC] sin ∠AP C sin (30◦ + B) 1 + 3 tan B

Also solved by Andrea Fanchini, Cantú, Italy; Daniel Lasaosa, Pamplona, Navarra, Spain; Arkady Alt, San Jose, California, USA.

Mathematical Reflections 1 (2014) 3 J292. Find the least real number k such that for every positive real numbers x, y, z, the following inequality holds: Y (2xy + yz + zx) ≤ k(x + y + z)6. cyc

Proposed by Dorin Andrica, Babes-Bolyai University, Romania

Solution by Albert Stadler, Herrliberg, Switzerland 64 We claim that k = . 729 Let x = y = z = 1 ⇒ Q (2xy + yz + zx) cyc 64 = ≤ k (x + y + z)6 729 By the Cauchy-Schwarz inequality , xy + yz + zx ≤ x2 + y2 + z2. So 3(xy + yz + zx) ≤ 2xy + 2yz + 2zx + x2 + y2 + z2 = (x + y + z)2. By AM-GM:

P(2xy + yz + zx)3 4 P xy 3 2 3 Y cyc cyc 4(x + y + z) 64 (2xy + yz + zx) ≤ = ≤ = (x + y + z)6,  3   3  9 729 cyc

64 Which proves that k ≤ and the conclusion follows. 729 Also solved by Daniel Lasaosa, Pamplona, Navarra, Spain; Alessandro Ventullo, Milan, Italy; Titu Zvona-

ru, Comănes, ti, Romania and Neculai Stanciu, Buzău, Romania; Shohruh Ibragimov, Lyceum Nr.2 under the SamIES, Samarkand, Uzbekistan; Sayan Das, Indian Statistical Institute, Kolkata, India; Sayak Mukherjee, Kolkata, India; Paolo Perfetti, Università degli studi di Tor Vergata Roma, Italy; Sun Mengyue Lansheng, Fudan Middle School, Shanghai, China; Arkady Alt, San Jose, California, USA; Georgios Batzolis, Mandou- lides High School, Thessaloniki, Greece; Ioan Viorel Codreanu, Satulung, Maramures, Romania; AN-anduud Problem Solving Group, Ulaanbaatar, Mongolia; Polyahedra, Polk State College, FL, USA; Jan Jurka, Brno, Czech Republic.

Mathematical Reflections 1 (2014) 4 J293. Find all positive integers x, y, z such that

2 x + y2 + z2 − 8xyz = 1.

Proposed by Aaron Doman, University of California, Berkeley , USA

Solution by Alessandro Ventullo, Milan, Italy We rewrite the equation as

x2 + 2x(y2 + z2 − 4yz) + (y2 + z2)2 − 1 = 0.

Since x must be a positive integer, the discriminant of this quadratic equation in x must be non-negative, i.e. (y2 + z2 − 4yz)2 − (y2 + z2)2 + 1 ≥ 0, which is equivalent to −8yz(y − z)2 + 1 ≥ 0, which gives yz(y − z)2 ≤ 1/8. Since y and z are positive integers, it follows that

yz(y − z)2 = 0,

so y − z = 0, i.e. y = z. The given equation becomes (x − 2y2)2 = 1, which yields x = 2y2 ± 1. Therefore, all the positive integer solutions to the given equation are

2 2 + (2n − 1, n, n), (2n + 1, n, n), n ∈ Z .

Also solved by Albert Stadler, Herrliberg, Switzerland; Daniel Lasaosa, Pamplona, Navarra, Spain; Sima Shariﬁ, College at Brockport, SUNY, USA; Sayan Das, Indian Statistical Institute, Kolkata, India; Georgios Batzolis, Mandoulides High School, Thessaloniki, Greece; Viet Quoc Hoang, University of Auckland, New Zealand; Ioan Viorel Codreanu, Satulung, Maramures, Romania; Polyahedra, Polk State College, FL, USA.

Mathematical Reflections 1 (2014) 5 J294. Let a, b, c be nonnegative real numbers such that a + b + c = 3. Prove that

1 ≤ (a2 − a + 1)(b2 − b + 1)(c2 − c + 1) ≤ 7.

Proposed by An Zhen-ping, Xianyang Normal University, China

Solution by Daniel Lasaosa, Pamplona, Navarra, Spain Denote p = abc and s = ab + bc + ca, where clearly 0 ≤ p ≤ 1, with p = 0 iff at least one of a, b, c is zero, and p = 1 iff a = b = c = 1 by the AM-GM inequality, while 0 ≤ s ≤ 3, with s = 0 iff two out of a, b, c are zero, and s = 3 iff a = b = c = 1 by the scalar product inequality. Note that

(a2 − a + 1)(b2 − b + 1)(c2 − c + 1) = p2 + s2 − ps − 4s − p + 7.

The lower bound then rewrites as

(2 + p − s)2 + 2(1 − p) ≥ p(3 − s).

Now, 9p = 3abc(a + b + c) ≤ (ab + bc + ca)2 = s2, or 9(1 − p) ≥ (3 + s)(3 − s), or it suﬃces to show that 6 + 2s − 9p (2 + p − s)2 + (3 − s) ≥ 0. 9 √ 2 p3 2 3 Assume that 9p > 6, or p > 3 , hence by the AM-GM inequality, s ≥ 3 p ≥ 12 > 2, or 6 + 2s > 10 > 9p. It follows that both terms in the LHS are non-negative, being zero iﬀ s = 3 and simultaneously s = p + 2, for p = 1. On the other hand, the upper bound rewrites as

p(1 − p) + s(3 − s) + s + ps ≥ 0,

clearly true because s, ps, p(1 − p), s(3 − s) are all non-negative. Note that equality requires s = 0, and since a, b, c are non-negative, this requires at least two out of a, b, c to be zero, resulting in p = 0. The conclusion follows, equality holds in the lower bound iﬀ a = b = c = 1, and in the upper bound iﬀ (a, b, c) is a permutation of (3, 0, 0).

Also solved by Polyahedra, Polk State College, FL, USA; AN-anduud Problem Solving Group, Ulaanbaa- tar, Mongolia; Viet Quoc Hoang, University of Auckland, New Zealand; Arkady Alt, San Jose, California, USA; Nicuşor Zlota, “Traian Vuia” Technical College, Focşani, Romania; Paolo Perfetti, Università degli studi di Tor Vergata Roma, Italy; Shivang jindal, Jaipur, Rajasthan, India; Albert Stadler, Herrliberg, Switzerland.

Mathematical Reflections 1 (2014) 6 Senior problems

S289. Let x, y, z be positive real numbers such that x ≤ 4, y ≤ 9 and x + y + z = 49. Prove that 1 1 1 √ + √ + √ ≥ 1. x y z

Proposed by Marius Stanean, Zalau, Romania

Solution by Li Zhou, Polk State College, FL, USA √ Applying Jensen’s inequality to the convex function f(t) = 1/ t for t > 0, we get

1 1 1 1 x 1 y 1 z 1 x 1 y 1 z √ + √ + √ = f + f + f ≥ f · + · + · x y z 2 4 3 9 6 36 2 4 3 9 6 36 √ √ √ 216 216 216 = √ = √ ≥ = 1. 27x + 8y + z 26x + 7y + 49 p26(4) + 7(9) + 49

Also solved by Daniel Lasaosa, Pamplona, Navarra, Spain; Sima Shariﬁ, College at Brockport, SUNY, USA; Arkady Alt, San Jose, California, USA; AN-anduud Problem Solving Group, Ulaanbaatar, Mongolia; Ioan Viorel Codreanu, Satulung, Maramures, Romania; Shohruh Ibragimov, Lyceum Nr.2 under the SamIES, Samarkand, Uzbekistan; Paolo Perfetti, Università degli studi di Tor Vergata Roma, Italy; Sun Mengyue , Lansheng Fudan Middle School, Shanghai, China.

Mathematical Reflections 1 (2014) 7 S290. Prove that there is no integer n for which

1 1 1 42 + + ··· + = 22 32 n2 5

Proposed by Ivan Borsenco, Massachusetts Institute of Technology, USA

Solution by Alessandro Ventullo, Milan, Italy Let p be the greatest prime number such that p ≤ n < 2p. Then the given equality can be written as

52k = (4 · n!)2,

n X (n!)2 (n!)2 where k = . Observe that k ≡ 6≡ 0 (mod p). Since p|52k and p does not divide k, it follows i2 p2 i=2 that p|52, i.e. p = 5. So, n ∈ {5, 6, 7, 8, 9}. An easy check shows that none of these values satisﬁes the equality.

Also solved by Daniel Lasaosa, Pamplona, Navarra, Spain; Yassine Hamdi, Lycée du Parc, Lyon, France; Sima Shariﬁ, College at Brockport, SUNY, USA; Li Zhou, Polk State College, FL, USA; Arghya Datta, Hooghly Collegiate School, Kolkata, India; Albert Stadler, Herrliberg, Switzerland.

Mathematical Reflections 1 (2014) 8 S291. Let a, b, c be nonnegative real numbers such that ab + bc + ca = 3. Prove that 5 2a2 − 3ab + 2b2 2b2 − 3bc + 2c2 2c2 − 3ca + 2a2 ≥ a2 + b2 + c2 − 4. 3

Proposed by Titu Andreescu, USA and Marius Stanean, Romania

Solution by Daniel Lasaosa, Pamplona, Navarra, Spain Note that

27 2a2 − 3ab + 2b2 2b2 − 3bc + 2c2 2c2 − 3ca + 2a2 + 27 · 4 − 45 a2 + b2 + c2 =

= 27 2a2 − 3ab + 2b2 2b2 − 3bc + 2c2 2c2 − 3ca + 2a2 + 4 (ab + bc + ca)3 − −5 a2 + b2 + c2 (ab + bc + ca)2 = = 211(a4b2 + b4c2 + c4a2 + a2b4 + b2c4 + c2a4) − 320(a3b3 + b3c3 + c3a3)− −334abc(a3 + b3 + c3) + 164abc(a2b + b2c + c2a + ab2 + bc2 + ca2) − 288a2b2c2, or it suﬃces to show that this last expression is non-negative. Note that it can be rearranged as X 3c4 + 160a2b2 + 48c2(a + b)2 − 164abc(a + b) (a − b)2 = cyc X = 3c4 + 5c2(a + b)2 + 39(bc + ca − 2ab)2 + 4(bc + ca − ab)2 (a − b)2, cyc clearly non-negative, and being zero iﬀ a = b = c. The conclusion follows.

Also solved by Paolo Perfetti, Università degli studi di Tor Vergata Roma, Italy; Nicuşor Zlota, “Traian Vuia” Technical College, Focşani, Romania; Arkady Alt, San Jose, California, USA.

Mathematical Reflections 1 (2014) 9 S292. Given triangle ABC, prove that there exists X on the side BC such that the inradii of triangles AXB and AXC are equal and ﬁnd a ruler and compass construction.

Proposed by Cosmin Pohoata, Princeton University, USA

Solution by Li Zhou, Polk State College, USA

A

Y

I J K

B E X F D C

As usual, let a, b, c, s, and r be the sides BC, CA, AB, semiperimeter, and inradius of 4ABC. Let I, J, K be the incenters of triangles ABC, ABX, AXC, and D, E, F be the feet of perpendiculars from A, J, K onto BC. Let h = AD, r1 = JE, and r2 = KF . As X moves from B to C, r1 increases from 0 to r while r2 t BE t(s−b) decreases from r to 0. Hence, there exists X such that r1 = r2 = t. Now r = s−b , so BE = r . Likewise, t(s−c) FC = r . Let [·] denote area. Then 1 ah = [ABC] = [ABX] + [AXC] = t(c + EX) + t(b + XF ) 2 t t = t(c + b + a − BE − FC) = (2rs − ta) = (ah − at). r r

2 1 1 p Hence, t − ht + 2 hr = 0, which yields t = 2 h − h(h − 2r) . This suggests an easy construction: Construct the length 2r on AD. Then construct the length AY = ph(h − 2r). Taking away the length AY from AD gives us h − ph(h − 2r). Halving this yields t, as in the ﬁgure.

Also solved by Titu Zvonaru, Comănes, ti, Romania, and Neculai Stanciu, ”George Emil Palade” School, Buzău, Romania; Arkady Alt, San Jose, California, USA.

Mathematical Reflections 1 (2014) 10 S293. Let a, b, c be distinct real numbers and let n be a positive integer. Find all nonzero complex numbers z such that c a b azn + bz¯ + = bzn + cz¯ + = czn + az¯ + . z z z Proposed by Titu Andreescu, University of Texas at Dallas, USA

Solution by Daniel Lasaosa, Pamplona, Navarra, Spain Since z 6= 0, we may multiply by z all terms in the proposed equation, or

azn+1 + b|z|2 + c = bzn+1 + c|z|2 + a = czn+1 + a|z|2 + b,

yielding (a − b)zn+1 + (b − c)|z|2 + (c − a) = (b − c)zn+1 + (c − a)|z|2 + (a − b) = 0, and eliminating the terms with zn+1, we obtain

(a − b)2 − (b − c)(c − a) a2 + b2 + c2 − ab − bc − ca |z|2 = = = 1, (b − c)2 − (a − b)(c − a) a2 + b2 + c2 − ab − bc − ca

since a2 + b2 + c2 > ab + bc + ca because of the Cauchy-Schwarz Inequality, a, b, c being distinct. Inserting this result in the original equation and rearranging terms, we obtain

a zn+1 − 1 = b zn+1 − 1 = c zn+1 − 1 ,

1 n or z must be one of the n + 1-th roots of unity. For any one of those n + 1 roots, we have z¯ = z = z , and all n + 1-th roots of unity are clearly solutions of the proposed equation.

Also solved by Arkady Alt, San Jose, California, USA; Moubinool Omarjee Lycée Henri IV, Paris, France; Ángel Plaza, University of Las Palmas de Gran Canaria, Spain; Albert Stadler, Herrliberg, Switzerland; Li Zhou, Polk State College, FL, USA.

Mathematical Reflections 1 (2014) 11 2 S294. Let s(n) be the sum of digits of n + 1. Deﬁne the sequence (an)n≥0 by an+1 = s(an), with a0 an arbitrary positive integer. Prove that there is n0 such that an+3 = an for all n ≥ n0.

Proposed by Roberto Bosch Cabrera, Havana, Cuba

Solution by Alessandro Ventullo, Milan, Italy We have to prove that the given sequence is 3-periodic. Since f(5) = 8, f(8) = 11 and f(11) = 5, it suﬃces to prove that for every positive integer a0 there exists some n ∈ N such that an ∈ {5, 8, 11}. Let m be the number of digits of a0. We prove the statement by induction on m. For m ≤ 2 we proceed by a direct check. 2 If a0 ∈ {5, 8, 11} there is nothing to prove. If a0 is a two-digit number, then a0 ≤ 10000, so a1 ≤ 37 and we reduce to analyze the cases for a0 ≤ 37.

(i) If a0 ∈ {2, 7, 20}, then a1 = 5. If a0 ∈ {1, 10, 26, 28}, then a1 ∈ {2, 20}, so a2 = 5. Finally, if a0 ∈ {3, 6, 9, 12, 15, 18, 27, 30, 33}, then a3 = 5.

(ii) If a0 ∈ {4, 13, 23, 32}, then a1 = 8.

(iii) If a0 ∈ {17, 19, 21, 35, 37}, then a1 = 11. If a0 ∈ {14, 22, 24, 31, 36}, then a1 ∈ {17, 19}, so a2 = 11. Finally, if a0 ∈ {16, 25, 29, 34}, then a3 = 11.

Thus, we have proved that if a0 is a one or two-digit number, then an ∈ {5, 8, 11} for some n ∈ N, i.e. the sequence is 3-periodic. Let m ≥ 2 and suppose that the statement is true for all k ≤ m. Let a0 be an m m+1 2m 2 2(m+1) (m + 1)-digit number. Then, 10 ≤ a0 < 10 which implies 10 ≤ a0 < 10 . Hence,

m a1 = f(a0) ≤ 9 · 2(m + 1) + 1 < 10 ,

and by the induction hypothesis, the sequence (an)n≥1 is 3-periodic, which implies that the sequence (an)n≥0 is 3-periodic, as we wanted to prove.

Also solved by Daniel Lasaosa, Pamplona, Navarra, Spain; Arghya Datta, Hooghly Collegiate School, Kolkata, India; Li Zhou, Polk State College, FL, USA; AN-anduud Problem Solving Group, Ulaanbaatar, Mongolia.

Mathematical Reflections 1 (2014) 12 Undergraduate problems

n 1 U289. Let a ≥ 1 be such that (ba c) n ∈ Z for all suﬃciently large integers n. Prove that a ∈ Z.

Proposed by Mihai Piticari, Campulung Moldovenesc, Romania

Solution by Li Zhou, Polk State College, FL, USA For the purpose of contradiction, suppose that a = m+h for some positive integer m and some real h ∈ (0, 1). 1 Then for all n ≥ h , mn + 1 ≤ mn + nh < an < (m + 1)n,

1 and thus mn < banc < (m + 1)n. Hence, m < (banc) n < m + 1 for all suﬃciently large n, a desired contradiction.

Also solved by Arkady Alt, San Jose, California, USA; Daniel Lasaosa, Pamplona, Navarra, Spain; Alessandro Ventullo, Milan, Italy; AN-anduud Problem Solving Group, Ulaanbaatar, Mongolia.

Mathematical Reflections 1 (2014) 13 1 U290. Prove that there are inﬁnitely many triples of primes (pn−1, pn, pn+1) such that 2 (pn+1 + pn−1) ≤ pn.

Proposed by Ivan Borsenco, Massachusetts Institute of Technology, USA

Solution by G.R.A.20 Problem Solving Group, Roma, Italy Assume for the sake of contradiction that there is an integer n0 such that for all n > n0,

pn−1 + pn+1 > 2pn

that is, dn > dn−1 ≥ 1 where dn = pn+1 − pn. Hence, for k ≥ 1,

dn0+k−1 ≥ dn0+k−2 + 1 ≥ · · · ≥ dn0 + k − 1 ≥ k

and k(k + 1) k2 p = d + d + ··· + d + p > k + (k − 1) + ··· + 1 = > . n0+k n0+k−1 n0+k−2 n0 n0 2 2 On the other hand, by the Prime Number Theorem, π(n) ∼ n/ ln(n) and since x/ ln(x) is strictly increasing for x ≥ e, it follows that

2 π(pn0+k) ln(pn0+k) (n0 + k) ln(pn0+k) (n0 + k) ln(k /2) 1 ← = < 2 → 0 pn0+k pn0+k k /2 which yields a contradiction.

Also solved by Julien Portier, Francois 1er, France; Daniel Lasaosa, Pamplona, Navarra, Spain; Arpan Sadhukhan, Indian Statistical Institute, Kolkata; AN-anduud Problem Solving Group, Ulaanbaatar, Mongolia; Li Zhou, Polk State College, FL, USA.

Mathematical Reflections 1 (2014) 14 U291. Let f : R → R be a bounded function and let S be the set of all increasing maps ϕ: R → R. Prove that there is a unique function g in S satisfying the conditions a) f(x) ≤ g(x) for all x ∈ R. b) If h ∈ S and f(x) ≤ h(x) for all x ∈ R then g(x) ≤ h(x) for all x ∈ R.

Proposed by Marius Cavachi, Constanta, Romania

Solution by Arkady Alt, San Jose, California, USA a) Since f is bounded then for any x ∈ R set G (x) := {f (t) | t ∈ R and t ≤ x} is bounded. Therefore for any x ∈ R we can deﬁne g (x) := sup G (x) and, obviously, that function g (x) deﬁned by such way satisfy to condition (a).

b) Let now h ∈ S and f(x) ≤ h(x) for all x ∈ R. Since f(t) ≤ h(t) for any t ≤ x then g (x) = sup f(t) ≤ sup h(t) = h (x) (since h is increasing in t ∈ (−∞, x] ). t≤x t≤x

Also solved by Paolo Perfetti, Università degli studi di Tor Vergata Roma, Italy.

Mathematical Reflections 1 (2014) 15 U292. Let r be a positive real number. Evaluate

π Z 2 1 r dx. 0 1 + cot x

Proposed by Ángel Plaza, Universidad de Las Palmas de Gran Canaria, Spain

Solution by Robinson Higuita, Universidad de Antioquia , Colombia We denote by π π r Z 2 1 Z 2 sin x I(r) := r dx = r r dx. 0 1 + cot (x) 0 sin x + cos x π If we make the substitution x = 2 − y, we have

Z 0 r π Z π r sin ( 2 − y) 2 cos y I(r) = r π π (−dy) = r . π sin ( − y) + cosr( − y) cosr y + sin y 2 2 2 0 Therefore

π r π r Z 2 sin x Z 2 cos y 2I(r) = r r dx + r r 0 sin x + cos x 0 cos y + sin y π r r π Z 2 sin y + cos y Z 2 π = r r dx = dx = . 0 cos y + sin y 0 2 Thus π I(r) = . 4 We note that the hypothesis on r is not important.

Also solved by Daniel Lasaosa, Pamplona, Navarra, Spain; Albert Stadler, Herrliberg, Switzerland; Sayak Mukherjee, Kolkata, India; Samin Riasat, Dhaka, Bangladesh; Paolo Perfetti, Università degli studi di Tor Vergata Roma, Italy; Antoine Barré and Dmitry Chernyak, Lycée Stanislas, Paris, France; Arkady Alt, San Jose, California, USA; Moubinool Omarjee Lycée Henri IV, Paris, France; Li Zhou, Polk State College, FL, USA; AN-anduud Problem Solving Group, Ulaanbaatar, Mongolia.

Mathematical Reflections 1 (2014) 16 U293. Let f : (0, ∞) → R be a bounded continuous function and let α ∈ [0, 1). Suppose there exist real Pk numbers a0, ..., ak, whith k ≥ 2, so that p=0 ap = 0 and

k α X lim x apf(x + p) = a. x→∞ p=0 Prove that a = 0.

Proposed by Marcel Chirita, Bucharest, Romania

Solution by the author Pk Let us denote M = sup|f(x)| and g(x) = p=0 apf(x + p). g : (0, ∞) → R is a continuous bounded function. Assume that a > 0.

Then there is an N > 0, such that xα|g(x)| > a, for ∀x > N (1)

From (1) we conclude that function g has the same sign ∀x > N, since it is continuous, and therefore has the intermediate value property. First, suppose that function g is positive on the interval (N, ∞).

a From (1) we have |g(x)| > (2) xα

Denoting n1 = [N] + 1 and x = n1, n1 + 1, n1 + 2, ... and summing up yields to

X X 1 |g(k)| > a . nα n≥n1 n≥n1

P 1 P 1 Since the series α is divergent for α ∈ (0, 1) it follows that α is divergent as well n≥1 n n≥n1 n X ⇒ |g(k)| = ∞.

n≥n1

Now, let T = max{|a0|, |a1|, |a2|, ..., |ak|}. Taking into consideration positiveness of g on (N, ∞) and the same values of n we get X X X Sn = |g(k)| = |apf(n + p)| = | apf(n + p)| =

n≥n1 n≥n1 n≥n1

|akf(n + k) + (ak + ak−1)f(n + k − 1) + ... + (ak + ak−1 + ... + a1)f(n + 1) + (ak−1 + ak−2 + ... + a0)f(n1 + k − 1) + (ak−2 + ak−3 + ... + a0)f(n1 + k − 2) + ... + (a1 + a0)f(n1 + 1) + a0f(n1)| ≤

|ak|M + |ak + ak−1|M + ... + |ak + ak−1 + ... + a1|M+

|ak−1 + ak−2 + ... + a0|M + |ak−2 + ak−3 + ... + a0|M + ... + |a1 + a0|M + |a0|M ≤ MT + 2MT + ... + kMT + kMT + ... + 2MT + MT = k(k + 1)MT and the series converges. Therefore, we have reached a contradiction ⇒ a = 0. Similarly to prove for g(x) ≤ 0.

Mathematical Reflections 1 (2014) 17 U294. Let p1, p2, . . . , pn be pairwise distinct prime numbers. Prove that √ √ √ √ √ √ Q ( p1, p2,..., pn) = Q ( p1 + p2 + ··· + pn) .

Proposed by Marius Cavachi, Constanta, Romania

Solution by Alessandro Ventullo, Milan, Italy √ √ √ √ √ √ √ √ √ Clearly, ( p + p +...+ p ) is a subfield of ( p , p ,..., p ). Observe that ( p , p ,..., p ) Q 1 2 n Q 1 2 √n √ Q 1 2 n is Galois over , since it is the splitting field of the polynomial (x2− p ) ··· (x2− p ). Every automorphism Q √ √ 1 n √ √ σ is completely determined by its action on p ,..., p , which must be mapped to ± p ,..., ± p , √ √ √ 1 n 1 n respectively. Therefore, Gal(Q( p1, p2,..., pn)/Q) is the group generated by {σ1, σ2, . . . , σn}, where σi is the automorphism defined by ( √ √ − pj if i = j σi( pj) = √ pj if i 6= j. √ √ √ Clearly, the only automorphism that fixes ( p , p ,..., p ) is the identity. Moreover, it’s easy to see Q 1 √2 n √ that the only automorphism that fixes the element p + ... + p is the identity, which means that the √ √ 1 n only automorphism that fixes ( p + ... + p ) is the identity. Hence, by the Fundamental Theorem of √ √ Q 1√ √n √ √ Galois Theory, Q( p1 + p2 + ... + pn) = Q( p1, p2,..., pn).

Also solved by Daniel Lasaosa, Pamplona, Navarra, Spain.

Mathematical Reflections 1 (2014) 18 Olympiad problems

O289. Let a, b, x, y be positive real numbers such that x2 − x + 1 = a2, y2 + y + 1 = b2, and (2x − 1)(2y + 1) = 2ab + 3. Prove that x + y = ab.

Proposed by Titu Andreescu, University of Texas at Dallas, USA

Solution by Khakimboy Egamberganov, S.H.Sirojiddinov lyceum, Tashkent, Uzbekistan From the conditions (2x − 1)2 = 4a2 − 3 and (2y + 1)2 = 4b2 − 3 and

(2ab + 3)2 = (2x − 1)2(2y + 1)2 = (4a2 − 3)(4b2 − 3),

a2b2 − ab − a2 − b2 = 0 (1). We have (2x − 1)2 + (2y + 1)2 = 4(a2 + b2) − 6, and since (2x − 1)(2y + 1) = 2ab + 3, we get that

(2(x + y))2 = 4(a2 + b2) − 6 + (4ab + 6),

(x + y)2 = a2 + b2 + ab (2). Since (1), (2), we get that (x + y)2 = a2b2 and

x + y = ab.

and we are done.

Also solved by Daniel Lasaosa, Pamplona, Navarra, Spain; Moubinool Omarjee Lycée Henri IV, Paris, France; Shohruh Ibragimov, Lyceum Nr.2 under the SamIES, Samarkand, Uzbekistan; Daniel Văcaru, Piteşti, Romania; Alessandro Ventullo, Milan, Italy; Li Zhou, Polk State College, FL, USA; AN-anduud Problem Solving Group, Ulaanbaatar, Mongolia; Arkady Alt, San Jose, California, USA; Ayoub Haﬁd, Elaraki School, Morroco.

Mathematical Reflections 1 (2014) 19 O290. Let Ω1 and Ω2 be the two circles in the plane of triangle ABC. Let α1, α2 be the circles through A that are tangent to both Ω1 and Ω2. Similarly, deﬁne β1, β2 for B and γ1, γ2 for C. Let A1 be the second intersection of circles α1 and α2. Similarly, deﬁne B1 and C1. Prove that the lines AA1,BB1,CC1 are concurrent.

Proposed by Cosmin Pohoata, Princeton University, USA

Solution by Daniel Lasaosa, Pamplona, Navarra, Spain Note that we must assume that Ω1, Ω2 have different radii. Otherwise, the figure clearly has symmetry around the perpendicular bisector of the segment joining their centers, and lines AA1,BB1,CC1 would be parallel to the line joining their centers (which could be considered equivalent to AA1,BB1,CC1 meeting at infinity). We will assume in the rest of the problem that Ω1, Ω2 have distinct radii, hence a homothety with center O and scaling factor ρ with |ρ|= 6 1 can be found, which transforms Ω1 into Ω2. Claim 1: Let Ω1, Ω2 be two circles with different radii, and let ω be a circle simultaneously tangent to both, respectively at points T1,T2. Then, T1,T2 and the center O of the homothety that transforms Ω1 into Ω2 are collinear. 0 Proof 1: Let O1,O2 be the respective centers, and R1,R2 the radii, of Ω1, Ω2. Let O , r be the center 0 0 0 and radius of ω. Consider triangle O1O2O , where T1,T2 are clearly inside segments O1O and O2O . Let P be the second point where T1T2 intersects O1O2, where by Menelaus’ theorem, we have

0 O1P T2O T1O1 r R1 R1 = · 0 = · = , PO2 O2T2 O T1 R2 r R2

or indeed P is a center of a homothety that transforms O1 into O2 with scaling factor with absolute value R1 , hence P = O. The Claim 1 follows. R2 Claim 2: The power of O with respect to ω is invariant for all the possible circles ω which are simultaneously tangent to Ω1, Ω2. Proof 2: Since T1,T2 are points on ω which are collinear with O, the power of O with respect to ω is 2 OT1 · OT2. Let P1,P2 be the respective powers of O with respect to Ω1, Ω2, which clearly satisfy P2 = ρ P1. 0 0 Consider now points T1, resulting from applying the homothety to T1. Clearly, T1 is collinear with O,T1, hence 0 0 on line OT2; at the same time, T1 ∈ Ω1, or T1 ∈ Ω2 by construction, hence T1,T2 are the two points where a 0 0 P2 line through O intersects T2, or OT1 · OT2 = P2. But since OT1 = |ρ|OT1, we have OT1 · OT2 = |ρ| = |ρ|P1, independently on the choice of ω. The Claim 2 follows.

By the Claim 2, the power of O with respect to α1, α2 is the same, or since AA1 is their radical axis, then A, A1,O are collinear, or line AA1 passes through O. Similarly, BB1,CC1 pass through O. The conclusion follows.

Also solved by Saturnino Campo Ruiz, Salamanca, Spain; Shohruh Ibragimov, Lyceum Nr.2 under the SamIES, Samarkand, Uzbekistan; Khakimboy Egamberganov, S.H.Sirojiddinov lyceum, Tashkent, Uzbekistan; Alessandro Pacanowski, PECI, Rio de Janeiro, Brazil.

Mathematical Reflections 1 (2014) 20 O291. Let a, b, c be positive real numbers. Prove that

a2 b2 a2 a + b + c √ + √ + √ ≥ . 4a2 + ab + 4b2 4b2 + bc + 4c2 4c2 + ca + 4a2 3

Proposed by Titu Andreescu, University of Texas at Dallas, USA

First solution by Marius Stanean From Hölder’s Inequality, we have

!2 ! X a2 X √ a2(4a2 + ab + 4b2) ≥ (a2 + b2 + c2)3. (1) 2 2 cyc 4a + ab + 4b cyc

But X a2(4a2 + ab + 4b2) = 4(a4 + b4 + c4) + a3b + b3c + c3a + 4(a2b2 + b2c2 + c2a2) cyc = 4(a2 + b2 + c2)2 + a3b + b3c + c3a − 4(a2b2 + b2c2 + c2a2) 13(a2 + b2 + c2)2 − 12(a2b2 + b2c2 + c2a2) ≤ , 4 where the last line follows from Cartoaje’s Inequality,

(a2 + b2 + c2)2 ≥ 3(a3b + b3c + c3a).

Hence, considering (1), it follows that it is suﬃcient to prove that

27(a2 + b2 + c2)3 ≥ [13(a2 + b2 + c2)2 − 12(a2b2 + b2c2 + c2a2)](a + b + c)2 9(a2 + b2 + c2)2[3(a2 + b2 + c2) − (a + b + c)2] − 4(a + b + c)2[(a2 + b2 + c2)2 − 3(a2b2 + b2c2 + c2a2)] ≥ 0 9(a2 + b2 + c2)2[(a − b)2 + (a − c)(b − c)] − 2(a + b + c)2[(a + b)2(a − b)2 + (a + c)(b + c)(a − c)(b − c)] ≥ 0 [9(a2 + b2 + c2)2 − 2(a + b + c)2(a + b)2](a − b)2 + [9(a2 + b2 + c2)2 − 2(a + b + c)2(a + c)(b + c)](a − c)(b − c) ≥ 0.

Without loss of generality, we may assume a ≤ b ≤ c. Then it suﬃces to show that

9(a2 + b2 + c2)2 ≥ 2(a + b + c)2(a + c)(b + c),

but this is true because 3(a2 + b2 + c2) ≥ (a + b + c)2 by Cauchy-Schwarz and

3(a2 + b2 + c2) − 2(a + c)(b + c) ≥ 0 ⇐⇒ c2 − 2(a + b)c + 3(a2 + b2) − 2ab ≥ 0 ⇐⇒ (c − a − b)2 + 2(a − b)2 ≥ 0,

so we are done.

Mathematical Reflections 1 (2014) 21 Second solution by the author We will begin by ﬁnding x, y such that

a2 √ ≥ xa + yb 4a2 + ab + 4b2

1 for all a, b > 0. Letting a = b = 1, the ”best” such x, y will satisfy x + y = 3 . Note that the inequality is a homogeneous, so letting t = b , we have t2 1 √ ≥ xt + y = x(t − 1) + . (1) 4t2 + t + 4 3 The inequality clearly holds if the RHS is negative. Otherwise, squaring and multiplying both sides by 9 yields

9t4 ≥ [3x(t − 1) + 1]2 4t2 + t + 4 9t4 − (4t2 + t + 4)[3x(t − 1) + 1]2 ≥ 0.

Considering the LHS as a function of t, say f(t), we want it to have a double root at t = 1. This means

f 0(1) = 27 − 54x = 0,

1 1 so x = 2 and y = − 6 . This implies 1 f(t) = (15t − 4)(t − 1)2. 4 4 4 Clearly, f is positive for t > 15 . For t ≤ 15 , (1) becomes 3t2 3t − 1 √ ≥ , 4t2 + 4t + 4 2 which must be true since the RHS is negative. Thus,

a2 3a − b √ ≥ 4a2 + ab + 4b2 6 for all a, b > 0. Adding up the similar inequalities with a and c and b and c yields

X a2 X 3a − b √ ≥ 2 2 6 cyc 4a + ab + 4b cyc a + b + c = , 3 as desired.

Also solved by Arkady Alt, San Jose, California, USA; Paolo Perfetti, Università degli studi di Tor Vergata Roma, Italy; Batzolis, Mandoulides High School, Thessaloniki, Greece; Nicuşor Zlota, “Traian Vuia” Technical College, Focşani, Romania.

Mathematical Reflections 1 (2014) 22 O292. For each positive integer k let k X 1 T = . k j2j j=1 Find all prime numbers p for which

p−2 X Tk ≡ 0 (mod p). k + 1 k=1

Proposed by Gabriel Dospinescu, Ecole Normale Superieure, Lyon

Solution by G.R.A.20 Problem Solving Group, Roma, Italy It is known that the following identity holds

n n X n(−1)k(1 − x)j X xk X 1 = − . k jk k2 k2 1≤j≤k≤n k=1 k=1

Let p > 2 be a prime (for p = 2 the congruence trivially holds). p−1 k Then, k ≡ (−1) (mod p) and the above identity imply

p−2 p−2 p−2 j j p−1 k X Tk X 1 X 1 X (1/2) X (1/2) X (1/2) = = = − k + 1 j2j k + 1 jk jk k2 k=1 j=1 k=j 1≤j 3. k=1 k=1 k=1 k=1

Mathematical Reflections 1 (2014) 23 2 xyz O293. Let x, y, z be positive real numbers and let t = max(x,y,z) . Prove that

2 3 4 x3 + y3 + z3 + xyz ≥ x2 + y2 + z2 + t2 .

Proposed by Nairi Sedrakyan, Yerevan, Armenia

Solution by Paolo Perfetti, Università degli studi di Tor Vergata Roma, Italy The inequality is implied by

(a+b+c)3 !3 3 3 3 2 2 2 2 27 4(x + y + z + xyz) ≥ x + y + z + a+b+c 3 Now deﬁne

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

a2 + b2 + c2 = 9u2 − 6v2, a3 + b3 + c3 = 27u3 − 27uv2 + 3w3 The inequality reads as

3 4(27u3 − 27uv2 + 4w3)2 ≥ (9u2 − 6v2) + u2 that is

64w6 + 864w3(−uv2 + u3) + 1836u2v4 + 1916u6 − 4032u4v2 + 216v6 ≥ 0 This a convex parabola in w3 whose minimum has abscissa negative. Since w3 ≥ 0, it follows that the parabola is nonnegative if and only it is nonnegative its value for w3 = 0. The theory states that, ﬁxed the values of (u, v), the minimum of w3 occurs for a = 0 or a = b.

If a = 0 we have

479(b6 + c6) + 1156b3c3 ≥ 780(b4c2 + b2c4) + 150(c5b + cb5) which implied by

479(b6 + c6) + 1156b3c3 ≥ 930(c5b + cb5) This follows by AM-GM since r 479 479 (479)2578 b6 + c6 + 578b3c3 ≥ 3 3 b5c > 963b5c 2 2 4 and the same with (c, b) in place of (b, c).

If a = b we have

(254b4 + 1972b3c + 525b2c2 + 658c3b + 479c4)(b − c)2 ≥ 0 and the proof is complete.

Also solved by Ioan Viorel Codreanu, Satulung, Maramures, Romania; Daniel Lasaosa, Pamplona, Na- varra, Spain; Shohruh Ibragimov, Lyceum Nr.2 under the SamIES, Samarkand, Uzbekistan; Ángel Plaza, University of Las Palmas de Gran Canaria, Spain; Khakimboy Egamberganov, S.H.Sirojiddinov lyceum, Tashkent, Uzbekistan; Arkady Alt, San Jose, California, USA.

Mathematical Reflections 1 (2014) 24 O294. Let ABC be a triangle with orthocenter H and let D,E,F be the feet of the altitudes from A, B and C. Let X,Y,Z be the reﬂections of D,E,F across EF , FD, and DE, respectively. Prove that the circumcircles of triangles HAX, HBY, HCZ share a common point, other than H.

Proposed by Cosmin Pohoata, Princeton University, USA

Solution by Daniel Lasaosa, Pamplona, Navarra, Spain 0 Claim: Let ABC be any triangle, I its incenter, Ia its excenter opposite vertex A, and A the symmetric of 0 0 0 A with respect to side BC. Deﬁne similarly Ib,Ic and B ,C . Then, the circles through I,Ia,A , through 0 0 I,Ib,B and through I,Ic,C , pass through a second common point other than I.

0 Proof 1: In exact trilinear coordinates, I ≡ (r, r, r), Ia ≡ (−ra, ra, ra) and A ≡ (−ha, 2ha cos C, 2ha cos B), or using non-exact trilinear coordinates, we have

0 I ≡ (1, 1, 1),Ia ≡ (−1, 1, 1),A ≡ (−1, 2 cos C, 2 cos B).

The equation of a circle in trilinear coordinates is given by

(`α + mβ + nγ)(aα + bβ + cγ) + k(aβγ + bγα + cαβ) = 0,

where substitution of the coordinates of the three given points yields k = −(` + m + n) when applied to the incenter, and consequently m + n = 0 and k = −` when applied to the excenter, yielding ﬁnally 1 − 2 cos A m = −n = ` 2(cos C − cos B)

when applied to A0. Indeed, the circle equation

(1 − 2 cos A)(β − γ) aβγ + bγα + cαβ α + = 2(cos C − cos B) aα + bβ + cγ

0 0 is easily checked to be satisfied by I,Ia,A . Analogous equations may be found for the circles through I,Ib,B 0 0 0 and I,Ic,C . The intersection of the circles through I,Ib,B and I,Ic,C must clearly satisfy (1 − 2 cos B)(γ − α) (1 − 2 cos C)(α − β) β + = γ + , 2(cos A − cos C) 2(cos B − cos A) or 1 + 2 cos A − 2 cos B − 2 cos C 1 + 2 cos B − 2 cos C − 2 cos A α + )β+ (cos A − cos B)(cos A − cos C) (cos B − cos A)(cos B − cos C) 1 + 2 cos C − 2 cos B − 2 cos A + γ = 0. (cos C − cos A)(cos C − cos B) This line equation represents the radical axis of these two circles, and is clearly satified by I, and since it is invariant under cyclic permutation of A, B, C and α, β, γ, it is also therefore the equation of the radical axes of the other two pairs of circles, which intersect the three circles at I, and at another point. The Claim follows. Consider triangle DEF . Clearly, X,Y,Z are the reflections of its vertices with respect to its sides, H is its incenter, and A, B, C its excenters. The conclusion follows by direct application of the Claim.

Also solved by Khakimboy Egamberganov, S.H.Sirojiddinov lyceum, Tashkent, Uzbekistan; Sebastiano Mosca, Pescara, Italy

Mathematical Reflections 1 (2014) 25

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