Junior problems. 2(a + b) 6c 3(d + e) = 6 2(a + b) 6c 3(d + e) = 6

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1 Junior problems J385. If the equalities a + b) 6c 3d + e) = 6 3a + b) c + 6d + e) = 6a + b) + 3c d + e) = 3 hold simultaneously, evaluate a b + c d + e. Proposed by Titu Andreescu, University of Texas at Dallas, USA Solution by AN-anduud Problem Solving Group, Ulaanbaatar, Mongolia Given equations are equivalent to the following equations: a + b) 6c 3d + e) = 6 a + b) 6c 3d + e) = 6 3a + b) c + 6d + e) = 7c + 0.5d + e) = 7 6a + b) + 3c d + e) = 3.5d + e) = 0 Hence we get, d + e = 0, c =, a + b = 0 a = b, d = e, c = a b + c d + e =. Also solved by Daniel Lasaosa, Pamplona, Spain; Alessandro Ventullo, Milan, Italy; Andreas Charalampopoulos, 4th Lyceum of Glyfada, Glyfada, Greece; Prithwijit De, HBCSE, Mumbai, India; Adnan Ali, A.E.C.S-4, Mumbai, India; Albert Stadler, Herrliberg, Switzerland; Alok Kumar, New Delhi, India; Andrianna Boutsikou, High School Of Nea Makri, Athens, Greece; Ángel Plaza, University of Las Palmas de Gran Canaria, Spain; David E. Manes, Oneonta, NY, USA; Duy Quan Tran, University of Health Science, Ho Chi Minh City, Vietnam; Orgilerdene Erdenebaatar, National University of Mongolia, Mongolia; Jennifer Johannes, College at Brockport, SUNY, NY, USA; Evgenidis Nikolaos, M.N.Raptou High School, Larissa, Greece; Robert Bosch, USA and Jorge Erick, Brazil; Tamoghno Kandar; Polyahedra, Polk State College, FL, USA; Dolsan Zheksheev, Karakol High School, Kyrgyzstan; Bekjol Joldubai, Kadamcay High School, Kyrgyzstan; Konstantinos Metaxas, st High School Ag. Dimitrios, Athens, Greece; Joehyun Kim, Bergen County Academies, Hackensack, NJ, USA; A.S.Arun Srinivaas, Chennai, India; Sushanth Sathish Kumar, Jeffery Trail Middle School, CA, USA; Hyun Min Victoria Woo, Northfield Mount Hermon School, Mount Hermon, MA, USA. Mathematical Reflections 5 06)

2 J386. Find all real solutions to the system of equations x + yzt = y + ztx = z + txy = t + xyz =. Proposed by Mohamad Kouroshi, Tehran, Iran Solution by Alessandro Ventullo, Milan, Italy Subtracting the second equation to the first equation, the third equation to the second equation, the fourth equation to the third equation and the first equation to the fourth equation, we obtain We have four cases. x y) zt) = 0 y z) tx) = 0 z t) xy) = 0 t x) yz) = 0. i) x y = 0 and z t = 0, i.e. x = y and z = t. Substituting these values into the second and the fourth equation, we get x z) zx) = 0. If x = z, then x + x 3 =, which gives x = y = z = t =. If zx =, then y + t =, i.e. x + z =, which gives x =, z =, so x = y = z = t =. ii) x y = 0 and xy = 0, i.e. x = y and xy =. We obtain x = y = ±. If x = y =, then zt = and z + t =, which gives z = t =. If x = y =, then zt = 3 and z + t =, which gives z = 3, t = or z =, t = 3. iii) zt = 0 and z t = 0, i.e. z = t and zt =. We obtain z = t = ±. If z = t =, then xy = and x + y =, which gives x = y =. If z = t =, then xy = 3 and x + y =, which gives x = 3, y = or x =, y = 3. iv) zt = 0 and xy = 0, i.e. zt = and xy =. We obtain x + y = and z + t =, which gives x = y = z = t =. In conclusion, the real solutions to the given system of equations are x, y, z, t) {,,, ),,, 3, ),,,, 3), 3,,, ),, 3,, )}. Also solved by Daniel Lasaosa, Pamplona, Spain; Polyahedra, Polk State College, FL, USA; Dolsan Zheksheev, Karakol High School, Kyrgyzstan; Bekjol Joldubai, Kadamcay High School, Kyrgyzstan; Konstantinos Metaxas, st High School Ag. Dimitrios, Athens, Greece; Joehyun Kim, Bergen County Academies, Hackensack, NJ, USA; Andreas Charalampopoulos, 4th Lyceum of Glyfada, Glyfada, Greece; David E. Manes, Oneonta, NY, USA; A.S.Arun Srinivaas, Chennai, India; Adnan Ali, A.E.C.S-4, Mumbai, India; Andrianna Boutsikou, High School Of Nea Makri, Athens, Greece; Erdenebayar Bayarmagnai, National University of Mongolia, Mongolia; Robert Bosch, USA; Albert Stadler, Herrliberg, Switzerland; Hyun Min Victoria Woo, Northfield Mount Hermon School, Mount Hermon, MA, USA. Mathematical Reflections 5 06)

3 J387. Find all digits a, b, c, x, y, z for which abc, xyz, and abcxyz are all perfect squares no leading zeros allowed). Proposed by Adrian Andreescu, Dallas, Texas Solution by Polyahedra, Polk State College, USA Let m = abc, n = xyz, and k = abcxyz. Then 000m +n = k, with 0 m, n 3 and 37 k 999. There are three cases. Case I: Suppose that n and k are not divisible by 5. Since 5 3 k n ), we must have either 5 3 k n) or 5 3 k + n). Thus k = 5q ± n, where 3 q 8. Then 8m = 5q ± qn, so q must be even, that is, q {4, 6, 8}. If q = 4, then m = 50 ± n. But 50 + n [6, 8] which contains no perfect square; and 50 n [9, 39] \ {5} which contains no perfect square either. If q = 6, then m = 5 9 ± 3n. So m = 3b, n = 3a, and b = 5 ± a. But 5+a [65, 67] which contains no perfect square; and 5 a [58, 60] which contains no perfect square either. If q = 8, then m = 000 n. So m = d, n = c, and d = 50 c [36, 44] which contains no perfect square. Case II: Suppose that 5 n but 5 does not divide n. Then 5 k but 5 does not divide k. Write n = 5a and k = 5b, with a {, 3, 4, 6}. Then 40m + a = b. If a =, then b = c and c 0m =. This Pell s equation has fundamental solution c, m ) = 9, 6) and all solutions c i, m i ) given by c i + m i 0 = ) i. Hence no such m i [0, 3]. If a = 3, then b 0m) = 9. This Pell s equation has all solutions generated by three distinct sets of fundamental solutions b, m) = 7, ), 3, 4), and 57, 8). Since 7 + 0) ) = , we have either m 8 or m 80, thus no solution m [0, 3]. If a = 4, then b = 4d and m = e. So d 0e =, thus d, e) = 9, 6), and m, n, k) =, 0, 380). If a = 6, then b = f and f 0m = 9. So f, m) = 57, 8) and m, n, k) = 8, 30, 570). Case III: Finally, consider 5 n. Then n = 5, k = 5a, and m = 5b. So a 0b) =. Hence a, b) = 9, 6) and m, n, k) = 5, 5, 475). In conclusion, there are three solutions: 44400, 34900, and 565. Also solved by Daniel Lasaosa, Pamplona, Spain; Alessandro Ventullo, Milan, Italy; David E. Manes, Oneonta, NY, USA; Albert Stadler, Herrliberg, Switzerland; Joel Schlosberg, Bayside, NY, USA; Robert Bosch, USA. Mathematical Reflections 5 06) 3

4 J388. Let ABCD be a cyclic quadrilateral with AB = AD. Points M and N are taken on sides CD and BC, respectively, such that DM + BN = MN. Prove that the circumcenter of triangle AMN lies on segment AC. Proposed by Hayk Sedrakyan, Paris, France Solution by Polyahedra, Polk State College, USA As in the figure, locate E on MN such that EM = MD. Then BN = NE. So ABE + ADE = AEB + AED. If ABE > AEB, then ADE < AED, which imply that AB < AE < AD, a contradiction. Likewise, we cannot have ABE < AEB. Hence AB = AE = AD, thus ABN = AEN and AEM = ADM. Therefore, ANM = ANB = π ABN BAN = π ABD CAD BAN from which the conclusion follows. = π + BAD CAM MAD BAN = π CAM, A B D N E M C Also solved by Daniel Lasaosa, Pamplona, Spain; Dolsan Zheksheev, Karakol High School, Kyrgyzstan; Bekjol Joldubai, Kadamcay High School, Kyrgyzstan; Andrianna Boutsikou, High School Of Nea Makri, Athens, Greece; Erdenebayar Bayarmagnai,National University of Mongolia, Mongolia; Evgenidis Nikolaos, M.N.Raptou High School, Larissa, Greece; Robert Bosch, USA. Mathematical Reflections 5 06) 4

5 J389. Solve in real numbers the system of equations x y + ) y x + ) [ x = y ) + y x ) ] = 4. Proposed by Alessandro Ventullo, Milan, Italy Solution by Robert Bosch, USA Let x y = a, y x = b. The system becomes a + b =, ab + a + b = 3. After multiply by the second equation and adding we obtain a + b) + a + b) 8 = 0. So a + b = 4 or a + b =. In the first case a + 49 a =. This equation is equivalent to a4 a + 49 = 0, that can be considered a quadratic on a, its discriminant is negative, thus there are not real solutions. Now, if a+b =, the equation to be solved is a + a =, or a ) = 0, so a = ±. Hence the pairs a, b) to consider are, ) and, ). Anyways, x y = y x, or x y)x + y + ) = 0, so x = y or x + y =. If x = y, the resulting equation is x x = 0, so x = y = ± 5. If x + y =, and a = b =, we obtain x + y =, and after plugging the value y = x, the quadratic x + x = 0, and x = 0, y = or x =, y = 0. Finally if a = b =, then x + y = 3 < 0, impossible for real numbers. The solutions x, y) to the original system are ± 5, ± ) 5, 0, ),, 0). Also solved by Daniel Lasaosa, Pamplona, Spain; Polyahedra, Polk State College, FL, USA; Bekjol Joldubai, Kadamcay High School, Kyrgyzstan; Konstantinos Metaxas, st High School Ag. Dimitrios, Athens, Greece; Joehyun Kim, Bergen County Academies, Hackensack, NJ, USA; A.S.Arun Srinivaas, Chennai, India; Sushanth Sathish Kumar, Jeffery Trail Middle School, CA, USA; Hyun Min Victoria Woo, Northfield Mount Hermon School, Mount Hermon, MA, USA; Andreas Charalampopoulos, 4th Lyceum of Glyfada, Glyfada, Greece; Prithwijit De, HBCSE, Mumbai, India; Adnan Ali, A.E.C.S-4, Mumbai, India; Albert Stadler, Herrliberg, Switzerland; David E. Manes, Oneonta, NY, USA; Duy Quan Tran, University of Health Science, Ho Chi Minh City, Vietnam; Erdenebayar Bayarmagnai, National University of Mongolia, Mongolia; Evgenidis Nikolaos, M.N.Raptou High School, Larissa, Greece; Vincelot Ravoson, Lycée Henri IV, France, Paris; Wada Ali, Ben Badis College, Algeria. Mathematical Reflections 5 06) 5

6 J390. Let ABC be a triangle. Points D, D lie on side BC, points E, E lie on side AC and points F, F lie on side AB such that AD = AD = BE = BE = CF = CF. Prove that if AD, BE, CF are concurrent, then so are AD, BE, CF. Proposed by Josef Tkadlec, Vienna, Austria Solution by Robert Bosch, USA and Jorge Erick, Brazil Let P be the feet of altitude from A, then P D = P D and BD BD = BP P D)BP + P D), = BP P D, = BP + AP AD, = AB AD, = AB BE. By the same argument AB BE = AE AE. In a similar way we obtain CE CE = BF BF, AF AF = CD CD. Therefore BD DC CE EA AF F B BD D C CE E A AF BD BD CE CE AF AF F = B AE AE BF BF CD CD =. The conclusion is BD DC CE EA AF F B = BD D C CE E A AF F B =. Thus by Ceva s theorem the result follows. Also solved by Daniel Lasaosa, Pamplona, Spain. Mathematical Reflections 5 06) 6

7 Senior problems S385. Let a, b, c be positive real numbers. Prove that a 3 + 8abc + b 3 + 8abc + c 3 + 8abc 3abc. Proposed by Nguyen Viet Hung, Hanoi University of Science, Vietnam Solution by Evgenidis Nikolaos, M.N.Raptou High School, Larissa, Greece Since abc > 0 the given inequality is equivalent to abc a 3 + 8abc + Then, we can rewrite the inequality as follows: abc b 3 + 8abc + abc c 3 + 8abc 3. a 3 +8abc abc + b 3 +8abc abc + c 3 +8abc abc 3 a bc b ca c ab Now, we set a b c bc = x, ca = y, ab = z with xyz =. Therefore, it suffices to prove that x y z Doing the maths we transform the above mentioned inequality to 3[xy + yz + zx) + 6x + y + z) + 9] xyz + 8xy + yz + zx) + 64x + y + z) + 5 or equivalently because xyz =, 5xy + yz + zx) + 6x + y + z) 63). By AM-GM inequality, we obtain that xy + yz + zx 3 3 xyz ) = 3 and x + y + z 3 3 xyz = 3. Hence, ) holds. Equality holds if and only if x = y = z = a = b = c. Also solved by Daniel Lasaosa, Pamplona, Spain; Joehyun Kim, Bergen County Academies, Hackensack, NJ, USA; A.S.Arun Srinivaas, Chennai, India; Sushanth Sathish Kumar, Jeffery Trail Middle School, CA, USA; Hyun Min Victoria Woo, Northfield Mount Hermon School, Mount Hermon, MA, USA; Alessandro Ventullo, Milan, Italy; Andreas Charalampopoulos, 4th Lyceum of Glyfada, Glyfada, Greece; Arkady Alt, San Jose, CA, USA; Henry Ricardo, New York Math Circle, NY, USA; Prithwijit De, HBCSE, Mumbai, India; Adnan Ali, A.E.C.S-4, Mumbai, India; Albert Stadler, Herrliberg, Switzerland; AN-anduud Problem Solving Group, Ulaanbaatar, Mongolia; Andrianna Boutsikou, High School Of Nea Makri, Athens, Greece; Ángel Plaza, University of Las Palmas de Gran Canaria, Spain; David E. Manes, Oneonta, NY, USA; Duy Quan Tran, University of Health Science, Ho Chi Minh City, Vietnam; Erdenebayar Bayarmagnai, National University of Mongolia, Mongolia; Jamal Gadirov, Istanbul University, Istanbul, Turkey; Robert Bosch, USA. Mathematical Reflections 5 06) 7

8 S386. Evaluate cos π 4 + cos π cos nπ 4 n Proposed by Mohamad Kouroshi, Tehran, Iran Solution by Daniel Lasaosa, Pamplona, Spain Let ρ = e iπ 4 and let R{z} denote the real part of complex number z. Clearly, cos kπ 4 = R { ρ k}, or the sum that we need to evaluate rewrites as n { } ρ k R = R k= k { ρ } ρn n ρ = R { } ρ ρ ) ρn ρ ) n ρ) ρ. ) Now, ρ) ρ ) = 4 ρ + ρ ) + ρ = 4 4 cos π 4 + = 5, or the proposed sum equals R {ρ ρ )} = R{ρ} ρ =, R {ρ n ρ )} = R {ρ n } ρ R { ρ n } = cos nπ 4 cos nπ n 4 + cos n )π n 4 5. n )π cos, 4 Also solved by Joehyun Kim, Bergen County Academies, Hackensack, NJ, USA. Mathematical Reflections 5 06) 8

9 S387. Find all nonnegative real numbers k such that aa b)a kb) 0 for all nonnegative numbers a, b, c. Proposed by Mehtaab Sawhney, Commack High School, New York, USA Solution by Adnan Ali, Student in A.E.C.S-4, Mumbai, India We show that k = k 0 is the maximum possible value of k, where k 0 = +3+ ) ). Assume that the inequality holds for all nonnegative reals a, b, c for k k 0. Then let c = 0 and W.L.O.G. let a > b. Then the inequality is a 3 +b 3 a b kaba b). Let x = a/b >, then the inequality is same as k x+ xx ). Let fx) = x + xx ), then by routine calculus, we get f min = +3+ ) ) for x = +)+. Hence k f min = +3+ ) ) = k 0. Now to complete the proof we note that the inequality rearranges to a 3 + b 3 + c 3 ) + kab + bc + ca ) k + )a b + b c + c a) b + c)b c) + c + a)c a) + a + b)a b) k + )a b)b c)a c) ) Denote the left and right hand sides of ) by La, b, c) and Ra, b, c) respectively. W.L.O.G assume that a b c and let a = a c 0, b = b c 0, c = 0, then La, b, c) La, b, c ) and Ra, b, c) = Ra, b, c ). Thus, to prove ), it suffices to prove that La, b, c ) Ra, b, c ), which is the same as a 3 + b3 a b ka b a b ). But we have already proved this at the start. Thus, the inequality holds true for all a, b, c 0 for all k k 0, where k 0 = +3+ ) ). Also solved by Daniel Lasaosa, Pamplona, Spain; Albert Stadler, Herrliberg, Switzerland; Robert Bosch, USA and Jorge Erick, Brazil. Mathematical Reflections 5 06) 9

10 S388. Let a, b, c be positive real numbers such that a + b + c = 3. Prove that a 6 c + b 6 a + ac 6 b 5 abc. Proposed by Marius Stănean, Zalău, România Solution by Ángel Plaza, University of Las Palmas de Gran Canaria, Spain The given inequality may be written equivalently as aba 6) 5. by the rearrangement inequality cyclic aba 6) a a 6) = a 3 6a ) = cyclic cyclic cyclic cyclic ) a ) 3/ 6a ). Since function x 3/ 6x is convex, then ) a a ) 3/ 6a + b + c ) 3/ a + b + c ) ) ) 3 6 = cyclic Also solved by Arkady Alt, San Jose, CA, USA. Mathematical Reflections 5 06) 0

11 S389. Let n be a positive integer. Prove that for any integers a, a,..., a n+ there is a rearrangement b, b,..., b n+ such that n n! divides b b ) b 3 b 4 ) b n b n ). Proposed by Cristinel Mortici, Valahia University, Târgovişte, România Solution by Evgenidis Nikolaos, M.N.Raptou High School, Larissa, Greece Let S = {a, a,..., a n+ } be the set of the statement. Then, it is obvious that S = {b, b,..., b n+ } because b, b,..., b n+ is a rearrangement of the elements a, a,..., a n+. Define a sequence of sets S k+ = {b, b,..., b k+ } for k n and n N. Observe that there are k different residues mod k). Since S k+ has k + elements, applying the Pigeonhole Principle, we deduce that there are at least elements of this set, say b k, b k, that have the same residue modk). Then, k b k b k ) ). Hence, by ) we obtain the following relations Finally, we can conclude that which is obviously equivalent to as we desired. n b n b n ) for k = n n ) b n 3 b n ) for k = n n ) b n 5 b n 4 ) for k = n b 3 b 4 ) for k = b b ) for k =. ) [n )] [n )] n) b b )b 3 b 4 ) b n b n ) n n! b b )b 3 b 4 ) b n b n ) Also solved by Daniel Lasaosa, Pamplona, Spain; Bekjol Joldubai, Kadamcay High School, Kyrgyzstan; Hyun Min Victoria Woo, Northfield Mount Hermon School, Mount Hermon, MA, USA; Andreas Charalampopoulos, 4th Lyceum of Glyfada, Glyfada, Greece; AN-anduud Problem Solving Group, Ulaanbaatar, Mongolia; Bekjol Joldubai, Kadamcay High School, Kyrgyzstan; Joel Schlosberg, Bayside, NY, USA; Alessandro Ventullo, Milan, Italy; Robert Bosch, USA and Jorge Erick, Brazil. Mathematical Reflections 5 06)

12 S390. Let ABC be a triangle and G be its centroid. Lines AG, BG, CG meet the circumcircle of triangle ABC at A, B, C, respectively. Prove that a + b + c GA + GB + GC R + ) a + b + c. 6 m a m b m c Proposed by Nguyen Viet Hung, Hanoi University of Science, Vietnam Solution by AN-anduud Problem Solving Group, Ulaanbaatar, Mongolia Let AA BC = A, BB CA = B, CC AC = C. By the Power of the Pointing theorem, we get: Similarly, we get AA A A = BA A C m a GA 3 m a) = a 4 GA = 3 m a + a = 4m a + a + 4m a m a 6 a. m a GB = 3 m b + b, GC = 4m b 3 m c + c. 4m c Hence we get GA + GB + GC = 4m a + a + 4m b + b + 4m c + c ) + ) a + b + c. m a m b m c 6 m a m b m c From the other hand, or Similarly, we have Thus we have, R AA = AA + A A = m a + 4m b + b m b 8R, 4m a + a 8R. m a a = 4m a + a 4m a 4m a 4m c + c 8R. m c GA + GB + GC 8R + 8R + 8R) + ) a + b + c 6 m a m b m c = R + ) a + b + c. 6 m a m b m c Hence right side inequality is proved. Use the median length formula, we get 4m a = b + c a, 4m b = c + a b, 4m c = a + b c : b + c m a + c + a GA + GB + GC = + a + b m b m c = 6 a + b + c ) + + ). m a m b m c ) + ) a + b + c 6 m a m b m c Mathematical Reflections 5 06)

13 Applying AM-GM inequality three times and using following formula, we get m a + m b + m c = 3 4 a + b + c ) : GA + GB + GC 6 a + b + c ) 3 3 m a m b m c a + b + c ) 3 ma m b m c a + b + c 3 ) m a + m b + m c a + b + c 3 ) 3m a + m b + m c) LHS is proved. Equality holds only when a = b = c. = 3 3 a + b + c = a + b + c. Also solved by Daniel Lasaosa, Pamplona, Spain; Robert Bosch, USA; Nikos Kalapodis, Patras, Greece; Evgenidis Nikolaos, M.N.Raptou High School, Larissa, Greece; Adnan Ali, A.E.C.S-4, Mumbai, India; Arkady Alt, San Jose, CA, USA; Sushanth Sathish Kumar, Jeffery Trail Middle School, CA, USA. Mathematical Reflections 5 06) 3

14 Undergraduate problems U385. Evaluate ) n + ) n n lim n n n n n n ) n. Proposed by Ángel Plaza, Universidad de Las Palmas de Gran Canaria, Spain Solution by Daniel Lasaosa, Pamplona, Spain Note first that, since ln + x) = x x + x3 3..., we then have ) n + ) n ln n n = lnn) + n ln + ) = lnn) + n n + 3n..., or n + ) n n n = en exp ) ) exp n 3n = e n + ) ) + O 4n n, where Landau notation has been used. Similarly, n n n ) n = e n 3 ) + + O 4n ) ) n ) = e n 3 + ) ) + O 4n n. Now, or Similarly, or n + ) )) n 4n + O n = 4 n + O n, n n + ) n n n = en n n n ) n = en 3 ) e 4 n + O n. n e 4 n + O ) n + ) n n n n n n n ) n = ) n, n e + O ), n whose limit when n is clearly e, and we are done. Also solved by Arkady Alt, San Jose, CA, USA; Adnan Ali, A.E.C.S-4, Mumbai, India; Albert Stadler, Herrliberg, Switzerland; AN-anduud Problem Solving Group, Ulaanbaatar, Mongolia; Erdenebayar Bayarmagnai, National University of Mongolia, Mongolia; Moubinool Omarjee, Lycée Henri IV, Paris, France; Robert Bosch, USA. Mathematical Reflections 5 06) 4

15 U386. Given a convex quadrilateral ABCD, denote by S A, S B, S C, S D the area of triangles BCD, CDA, DAB, ABC, respectively. Determine the point P in the plane of the quadrilateral such that S A P A + S B P B + SC P C + SD P D = 0. Proposed by Dorin Andrica, Babeş-Bolyai University, Cluj-Napoca, România Solution by Daniel Lasaosa, Pamplona, Spain Note first that the area of ABCD is S = S A + S C = S B + S D. Note also that, denoting by O the point where the diagonals AC and BD meet, and by h A, h C the respective distances from A, C to BD, we have that the sine of the angle formed by the diagonals is h A OA = h C OC, or h A OC + hc OA = 0 becase OA, OC are vectors in opposite directions on line AC. Note finally that S A = BD h C and S C = BD h A, or S C OC + SA OA = 0, hence for any point P on the plane, Similarly, or for any point P in the plane, we have S A P A + S C P C = S OP. S B P B + SD P D = S OP, S A P A + S B P B + SC P C + SD P D = S OP, where S > 0 because ABCD is convex, hence P is the intersection of diagonals AC and BD, and no other point in the plane may satisfy the condition given in the problem statement. The conclusion follows. Also solved by Adnan Ali, A.E.C.S-4, Mumbai, India. Mathematical Reflections 5 06) 5

16 U387. A polynomial with complex coefficients is called special if all its roots lie on the unit circle. Is any complex polynomial the sum of two special polynomials? Proposed by Gabriel Dospinescu, Ecole Normale Supérieure, Lyon, France Solution by Robert Bosch, USA The answer is yes. We understand on the unit circle) as inside or on the unit disk. The idea is to use Rouché s theorem. Let s see, let this polynomial can be written as pz) = a n z n + a n z n + + a z + a 0, pz) = az n+ ) + az n+ + a n z n + a n z n + + a z + a 0 ), = fz)) + gz)), where a > a n + a n + + a 0. Clearly the polynomial fz) has all its roots inside the unit disk, we shall prove gz) is special too. By Rouché s theorem it s enough to verify the following inequality pz) < fz), on the curve γ : z =, since Finally gz) = pz) + fz)). a n z n + a n z n + + a z + a 0 < az n+, is true by Triangle Inequality and by the condition imposed on the constant a. Also solved by Adnan Ali, Student in A.E.C.S-4, Mumbai, India. Mathematical Reflections 5 06) 6

17 U388. Evaluate n= sin n θ + cos n θ n. Proposed by Li Zhou, Polk State College, USA Solution by AN-anduud Problem Solving Group, Ulaanbaatar, Mongolia ) Let θ = π + πk, k Z or θ = πm, m Z. Then we have n= ) Let θ π + πk, k Z and θ πm, m Z. Substitute that sin θ = t, then we have: n= sin n θ + cos n θ n = Where Li t) is Dilogarithm function and sin n θ + cos n θ n = n= t n + t) n n= t = 0 n ln z) dz + z n = π 6. = Li t) + Li t) = π 6 t 0 ) ln z) dz z ln t ln t). equality is Landen s formula. Hence we have: Li t) + Li t) = π 6 ln t ln t) n= sin n θ + cos n θ n = π 6 ln sin θ ) ln sin θ ). Also solved by Arkady Alt, San Jose, CA, USA; Adnan Ali, Student in A.E.C.S-4, Mumbai, India; Albert Stadler, Herrliberg, Switzerland; Robert Bosch, USA and Jorge Erick, Brazil. Mathematical Reflections 5 06) 7

18 U389. Let P be a nonconstant polynomial whose zeros x, x,..., x n are all real. Prove that b P ) x)p x) exp a P dx < P a) P b) 3 x) P a) 3 P b), whenever a < b < minx, x,..., x n ). Proposed by Titu Andreescu, University of Texas at Dallas, USA Solution by the author Use the identity Take the derivative: = P x) x x x x x x n P x). Take the derivative one more time: P x)p x) P x) P x) = x x k ) P x)p x) + P x)p x) P x)p x))p x) P x)p x) P x) )P x)p x) P x) 4 = x x k ) 3 < 0 for a x b < minx,..., x n ). Then P x)p x) 3 + P x)p x)p x) P x)p x)p x) P x)p x)p x) + P x) 3 P x) < 0, which is equivalent to Dividing by P x) P x) yields P x)p x) 3 3P x)p x)p x) + P x) 3 P x) 0.. By integration, P x)p x) P x) 3P x) P x) P x) P x) hence as desired. b a P x)p x) P x) 3 ln P b) ln P a) ln P b) ln P a) = ln P a) P b) 3 P a) 3 P b), b exp a P ) x)p x) P x) P a) P b) 3 P a) 3 P b), Mathematical Reflections 5 06) 8

19 U390. Prove that there is a unique representation of sinπz) as a series of the form sinπz) = a k z k z) k k= that converges for all complex numbers z, wherein the coefficients a k are real number satisfying a k c πk k)! for some absolute constant c. Proposed by Albert Stadler, Herrliberg, Switzerland Solution by Daniel Lasaosa, Pamplona, Spain Note first that if such an expression exists, it must be unique. Indeed, let j be the lowest value of k such that a k is different in both expression, or the difference between both expressions would be a polynomial with nonzero coefficient for z j, thus nonzero. Define now v = z z), or sinπz) = sin π z ) + π ) = cos π z )) = = π n v 4 n)! n=0 ) n, ) n π n z n)! where we have used that z ) = z z+ 4 = 4 v. Note therefore that the coefficient a k of v k = z k z) k is π n ) n a k = ) n k = πk ) d π d ) k)! k + d n)! k 4 k)! 4 d = k + d)! k n=k = πk k)! d=0 d=0 ) d k )!! k + d )!! n=0 π ) d, d! 4 where we have defined d = n k, and m )!! = m )m 3) 3. Note now first that, since the series is unique and sinπz) = 0, we must have a 0 = 0, or we will henceforth obviate k = 0. Note next that for all k, we have < k +, k + 3,..., k + d, or d k )!! < k + d )!!, hence a k πk k)! d k )!! π ) d π k < k + d )!! d! 4 k)! d=0 The conclusion follows, and it suffices to take c = exp ) π 6. d=0 d! π 4 ) d = exp π 6 ) n ). = Mathematical Reflections 5 06) 9

20 Olympiad problems O385. Let fx, y) = x3 y xy Let m and n be odd integers such that fm, n) mn Evaluate fm, n). Proposed by Titu Andreescu, University of Texas at Dallas, USA Solution by Robert Bosch, USA and Jorge Erick, Brazil If α = m n, then α is integer and m = n + α. Consider the following identity: m 3 n 3 + kmn = α + k) [n + α) + 3 ] 6 α k) 4 3 k3, due to a 3 + b 3 + c 3 3abc = a + b + c)a + b + c ab bc ca), for a = m, b = n, c = k. Thus fm, n) mn + 37) = α + ) [n + α) + 3 ] α 4) α + < 0, fm, n) + mn + 37 = α + 4) [n + α) + 3 ] α 8) 3 0 α + 4 > 0. We conclude that α = 3 and fm, n) = α + 3) [n + α) + 3 α 6) ] + = f 5, ) =. Also solved by Alessandro Ventullo, Milan, Italy; Adnan Ali, Student in A.E.C.S-4, Mumbai, India; Albert Stadler, Herrliberg, Switzerland; Prithwijit De, HBCSE, Mumbai, India. Mathematical Reflections 5 06) 0

21 O386. Find all pairs m, n) of positive integers such that 3 m n is a perfect square. Proposed by Alessandro Ventullo, Milan, Italy Solution by David E. Manes, Oneonta, NY, USA By direct computation if m 4, then, ),, 3), 3, ) and 4, 5) are solutions. We will show that there are no others. Assume m 5. If n =, then a simple induction argument shows that 3 m n is always strictly between two consecutive squares. Therefore, 3 m is not a perfect square. Thus, n. Assume 3 m n = k for some integer k and the integer m is odd. Then 3 m 3 mod 4 and n 0 mod 4 imply 3 m n 3 mod 4, a contradiction since k 0 or mod 4. Therefore, m is even so that m = r for some integer r. Then 3 r k = n. Factoring, we get 3 r + k)3 r k) = n so that 3 r + k = s and 3 r k = t for some integers s, t, s > t and s + t + n. Adding the two equations, one obtains 3 r = t s t +. Therefore 3 r = t s t + ) implies t =. Therefore, 3 r s =, an equation that has solutions only if r = or in which case m 4, a contradiction. Hence, the only solutions are the ones claimed above. Also solved by Adnan Ali, A.E.C.S-4, Mumbai, India; Albert Stadler, Herrliberg, Switzerland; Prithwijit De, HBCSE, Mumbai, India; Robert Bosch, USA and Jorge Erick, Brazil; Minh Pham Hoang, High School for the Gifted, Vietnam National University, Ho Chi Minh City, Vietnam. Mathematical Reflections 5 06)

22 O387. Are there integers n for which 3 6n n + is a perfect cube? Proposed by Titu Andreescu, University of Texas at Dallas, USA Solution by Daniel Lasaosa, Pamplona, Spain Note that 3 n + ) 3 = 3 6n n + 3 n + > 3 6n n + > 3 6n 3 = 3 n ) 3, or since 3 6n n + is always strictly between two consecutive perfect cubes, it can never be a perfect cube itself, and we are done. Also solved by Sushanth Sathish Kumar, Jeffery Trail Middle School, CA, USA; Minh Pham Hoang, High School for the Gifted, Vietnam National University, Ho Chi Minh City, Vietnam; Alessandro Ventullo, Milan, Italy; Evgenidis Nikolaos, M.N.Raptou High School, Larissa, Greece; Adnan Ali, A.E.C.S-4, Mumbai, India; Albert Stadler, Herrliberg, Switzerland; AN-anduud Problem Solving Group, Ulaanbaatar, Mongolia; Arpon Basu, AECS-4, Mumbai, India; Bekjol Joldubai, Kadamcay High School, Kyrgyzstan; David E. Manes, Oneonta, NY, USA; Jamal Gadirov, Istanbul University, Istanbul, Turkey; Ricardo Largaespada, Universidad Nacional de Ingeniería, Managua, Nicaragua; Robert Bosch, USA. Mathematical Reflections 5 06)

23 O388. Prove that in any triangle ABC with area S, m a m b m c m a + m b + m c ) ma m b + m b m c + m c m a S. Proposed by Nguyen Viet Hung, Hanoi University of Science, Vietnam Solution by Daniel Lasaosa, Pamplona, Spain We prove a stronger result, namely that the inequality also holds with 3S in the RHS instead of S. Note first that, since m a = b +c a 4 and similarly for m b, m c, we have or using Heron s formula, m a m b + m b m c + m c m a = 9 a b + b c + c a ), 6 m 4 a + m 4 b + m 4 c = 9 a 4 + b 4 + c 4), 6 9S = 8 a b + b c + c a ) 9 a 4 + b 4 + c 4) 6 = m a m b + m b m c + m c m a ) m a 4 + m b 4 + m c 4 ). Squaring both sides and multiplying by m a m b + m b m c + m c m a, the proposed inequality is equivalent to m a m b m c m a + m b + m c ) + m a 4 + m b 4 + m c 4 ) m a m b + m b m c + m c m a ) or after some algebra, to m a m b + m b m c + m c m a ), m a m b m a + m b ) m c ) m a m b ) + m b m c m b + m c ) m a ) m b m c ) + = Now, it is well known that +m c m a m c + m a ) m b ) m c m a ) 0. m a + m b ) m c = m a + m b + m c ) m a + m b m c ) 0, since the medians of triangle ABC are the sides of a triangle whose area is 3 4 the area of ABC. Or, all terms in the LHS are non-negative, being simultaneoulsy zero iff m a = m b = m c. The conclusion follows, equality holds iff ABC is equilateral. Also solved by Evgenidis Nikolaos, M.N.Raptou High School, Larissa, Greece; Arkady Alt, San Jose, CA, USA; Adnan Ali, A.E.C.S-4, Mumbai, India; AN-anduud Problem Solving Group, Ulaanbaatar, Mongolia; Pham Ngoc Khanh, Hanoi National University of Education, Vietnam; Robert Bosch, USA and Jorge Erick, Brazil. Mathematical Reflections 5 06) 3

24 O389. Let a, b, c be positive real numbers such that abc =. Prove that a b + c) b + c + b c + a) c + a + c a + b) a + b 3a + b + c). Proposed by Bazarbaev Sardar, National University of Uzbekistan, Uzbekistan Solution by Erdenebayar Bayarmagnai, National University of Mongolia, Mongolia a b + c) c + a) a + b) aba b) + aca c) bcb c) + bab a) cac a) + cbc b) b + c a+b c + a b+c a c = + b b + c + c + a + a + b = ) ) ) aba b) aba b) aca c) aca c) bcb c) bcb c) = b + c c + a + b + c a + b + c + a a + b = = aba b)a b ) b + c )c + a ) + aca c)a c ) b + c )a + b ) + bcb c)b c ) c + a )a + b ) = = aba b) a + b) b + c )c + a ) + aca c) a + c) b + c )a + b ) + bcb c) b + c) c + a )a + b ) 0 a b + c) c + a) a + b) b + c +b c + a +c a + b a+b+c) = a + b + c)a + b + c) This equality holds only when a = b = c = 3 3 abca + b + c) = 3a + b + c) Also solved by Albert Stadler, Herrliberg, Switzerland; Emil Gasimov, Baku Istek Lyceum, Azerbaijan; Nguyen Viet Hung, Hanoi University of Science, Vietnam; Pham Ngoc Khanh, Hanoi National University of Education, Vietnam; Robert Bosch, USA. Mathematical Reflections 5 06) 4

25 O390. Let p > be a prime. Find the number of 4p element subsets of the set {,,, 6p} for which the sum of the elements is divisble by p. Proposed by Vlad Matei, University of Wisconsin, Madison, USA Solution by Adnan Ali, Student in A.E.C.S-4, Mumbai, India We shall be using the Root of Unity Filter to solve the problem, so we introduce it first: Theorem Root of Unity Filter): Define ε = e πi/n for a positive integer n. For any polynomial F x) = f 0 + f x + f x + where f k = 0 if k > deg F ), the sum f 0 + f n + f n + is given by f 0 + f n + f n + = n F ) + ε) + F ε ) + + F ε n ) ). Proof: We use a property of the sum s k = + ε k + ε k + + ε n )k. If n k, then ε k = and so s k = n, else ε k and so s k = εnk = 0. Thus ε k F ) + ε) + F ε ) + + F ε n ) = f 0 s 0 + f s + f s + = nf 0 + f n + f n + ). Now, to start with, we define a generating function Gx, y) as Gx, y) = n,k 0 g n,kx n y k where g n,k is the number of k-element subsets of {,,, 6p} having a sum n. So, the answer required by the problem is nothing but A := g p,4p + g 4p,4p + g 6p,4p +. Next we observe that if a number m is not in a subset, it doesn t affect the size or sum of the subset, while if it is present in the subset, it increases the sum by m and size by. Thus G must contain the term + x m y) and so Gx, y) = + xy) + x y) + x 6p y). To get the value of A we must extract two types of terms from G: y 4p and powers of x p. We can do the latter using Theorem and so we perform it first. Define ε = e πi/p. Then the filter tells us that g n,k y k = p n,k 0 p n ) G, y) + Gε k, y) + Gε k, y) + + Gε p )k, y), ) so we need to calculate Gε k, y) for 0 k p. For k = 0, we have G, y) = + y) 6p. For k = l, where l p, ε l = γ l, where γ = e πi/p. Since gcdl, p) = the set {l, l,, pl} is a complete residue set modulo p. Thus we have Gε l, y) = Gγ l, y) = + γ l y) + γ l y) + γ 6pl y) = + γ l y) + γ l y) + γ pl y) = + γy) + γ y) + γ p y) ) 6 = + y p ) 6. ) 6 Mathematical Reflections 5 06) 5

26 For k = p, ε =, and so we have G, y) = y ) 3p. Now for k p k and k p) gcdp, k) = and so the set {k, k,, pk} is a complete residue set modulo p. Thus Putting all these values back in ) gives Gε k, y) = + ε k y) + ε k y) + ε 6pk y) = + ε k y) + ε k y) + ε pk y) = + εy) + ε y) + ε p y) ) 3 = y p ) 3. ) 3 g n,k y k = + y) 6p + y ) 3p + p ) + y p ) 6 + p ) y p ) 3) p n,k 0 p n from which we can easily extract the coefficient of y 4p. Thus our answer i.e. A) is p 6p 4p ) + 3p p) + p ) 6 4) + p ) 3 ) ) = p, 6p 4p ) + 3p ) ) + 8p 8. p Mathematical Reflections 5 06) 6