Junior problems - Phần 2

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Junior problems J175. Let a, b ∈ (0, π ) such that sin2 a + cos 2b ≥ 1 1 sec a and sin2 b + cos 2a ≥ sec b. Prove that 2 2 2 1 cos6 a + cos6 b ≥ . 2 Proposed by Titu Andreescu, University of Texas at Dallas, USA Solution by Prithwijit De, HBCSE, India We will use the following well-known trigonometric identities (a) sin2 x = 1 − cos2 x, (b) cos 2x = 2 cos2 x − 1, 1 (c) sec x = . cos x The inequalities can be written as 1 2 cos2 b cos a − cos3 a ≥ (1) 2 and 1 2 cos2 a cos b − cos3 b ≥ . (2) 2 π The signs of the inequalities are preserved because cos x is positive when x ∈ 0, . Now by 2 squaring both sides of (1) and (2) and adding them we get 1 cos6 a + cos6 b ≥ . 2 Also solved by Arkady Alt, San Jose, California, USA; Daniel Lasaosa, Universidad P´blica u de Navarra, Spain; Perfetti Paolo, Dipartimento di Matematica, Universit` degli studi di Tor a Vergata Roma, Italy; Tigran Hakobyan, Armenia. 1 Mathematical Reflections 6 (2010)
J176. Solve in positive real numbers the system of equations x1 + x2 + · · · + xn = 1 1 1 1 1 3 x1 + x2 + · · · + xn + x1 x2 ···xn = n + 1. Proposed by Neculai Stanciu, George Emil Palade Secondary School, Buzau, Romania Solution by Tigran Hakobyan, Armenia We have n2 1 1 1 = n2 + ··· + ≥ + x1 + x2 + · · · + xn x1 x2 xn and 1 1 = nn . ≥ x1 +x2 +···+xn n x1 x2 · · · xn n Thus, n3 + 1 ≥ nn + n2 which implies that n ≤ 2. If n = 1 we get a contradiction. For n = 2 we get x1 + x + 2 = 1 1 1 1 x1 + x2 + x1 x2 = 9 2, 1 , 3 , 2, 2 , 1 2 which is (n, x1 , x2 ) ∈ . 3 33 Also solved by Arkady Alt, San Jose, California, USA; Daniel Lasaosa, Universidad P´blica u de Navarra, Spain; Perfetti Paolo, Dipartimento di Matematica, Universit` degli studi di Tor a Vergata Roma, Italy; Lorenzo Pascali, Universit` di Roma “La Sapienza”, Roma, Italy. a 2 Mathematical Reflections 6 (2010)
J177. Let x, y, z be nonnegative real numbers such that ax + by + cz ≤ 3abc for some positive real numbers a, b, c. Prove that z+x √ x+y y+z 1 + 4 xyz ≤ (abc + 5a + 5b + 5c). + + 2 2 2 4 Proposed by Titu Andreescu, University of Texas at Dallas, USA Solution by the author From the given condition, ax y z 3a ≥ ++ bc c b x by z 3b ≥ + + c ca a x y cz 3c ≥ + + . b a ab Then x+y y+z z+x ax by cz 3(a + b + c) ≥ + + + + + c a b bc ca ab hence x+y y+z z+x ax by cz abc + 5(a + b + c) ≥ + 2c + + 2a + + 2b + abc + + + c a b bc ca ab √ ≥ 2 2(x + y ) + 2 (y + z ) + 2 (z + x) + 4 4 xyz and the conclusion follows. The equality holds if and only if x + y = 2c2 , y + z = 2a2 , z + x = 2b2 by and ax = ca = ab = abc. This implies b2 c2 = c2 a2 = 2c2 , c2 a2 + a2 b2 = 2a2 , a2 b2 + b2 c2 = 2b2 , cz bc that is b2 + a2 = c2 + b2 = a2 + c2 = 2, implying a = b = c = 1 and x = y = z = 1. If a = b = c = x = y = z = 1, the equality, as well as the condition of the problem, hold 3 Mathematical Reflections 6 (2010)
J178. Find the sequences of integers (an )n≥0 and (bn )n≥0 such that √ √n 1+ 5 (2 + 5) = an + bn 2 for each n ≥ 0. Proposed by Dorin Andrica, Babes-Bolyai University, Cluj-Napoca, Romania First solution by Arkady Alt, San Jose, California, USA Let √ √ n n + 2− 2+ 5 5 pn = 2 and √ √ n n − 2− 2+ 5 5 √ qn = 25 √ √ n for n = 1, 2, . . . . Then 2 + 5 = pn + qn 5, n = 0, 1, 2, . . . and both obtained sequences satisfy the same recurrence xn+1 = 4xn + xn−1 , n ∈ N (1) with initial conditions p0 = 1, p1 = 2, q0 = 0, q1 = 1. It is clear that (pn )n≥0 and (qn )n≥0 are sequences of nonnegative integers and since √ √ bn bn √ √ 1+ 5 = pn + qn 5 ⇐⇒ an + an + bn + 5 = p n + qn 5 2 2 2  bn  an + = pn  2 ⇐⇒ bn = qn   2 an = pn − qn ⇐⇒ , n ∈ N ∪ {0} bn = 2qn we have that (an )n≥0 and (bn )n≥0 are sequences of integers and can be deﬁned independently by recurrence (1) with initial conditions a0 = 1, a1 = 1, b0 = 0, b1 = 2. In explicit form √n √n √ √n √ √n 2+ 5 − 2− 5 5−1 2+ 5 + 5+1 2− 5 √ √ bn = , an = . 5 25 Second solution by Daniel Lasaosa, Universidad P´blica de Navarra, Spain u √ √ Using Newton’s binomial formula, exchanging 5 by − 5 in the ﬁrst term results in exchanging √ √ √n √n √n √n √5 by − 5 in the second term, i.e., (2+ 5) +(2 − 5) = 2an + bn and (2+ 5) − (2 − 5) = 5bn , yielding √ √ √ √ √ √ (2 + 5)n − (2 − 5)n ( 5 − 1)(2 + 5)n + ( 5 + 1)(2 − 5)n √ √ bn = , an = . 5 25 4 Mathematical Reflections 6 (2010)
The fact that an , bn are integers for all n may be easily proved considering that they are solutions of the recursive equations an = 4an−1 + an−2 and bn = 4bn−1 + bn−2 , with initial conditions a0 = a1 = 1 and b0 = 0, b1 = 2. Also solved by Ivan Dinkov Gerganov, P. R. Slaveikov Secondary School, Bulgaria; Lorenzo Pascali, Universit` di Roma “La Sapienza”, Roma, Italy; Tigran Hakobyan, Armenia. a 5 Mathematical Reflections 6 (2010)
J179. Solve in real numbers the system of equations  (x + y )(y 3 − z 3 ) = 3(z − x)(z 3 + x3 )  (y + z )(z 3 − x3 ) = 3(x − y )(x3 + y 3 )  (z + x)(x3 − y 3 ) = 3(y − z )(y 3 + z 3 )  Proposed by Titu Andreescu, University of Texas at Dallas, USA Solution by Lorenzo Pascali, Universit` di Roma “La Sapienza”, Roma, Italy a Without loss of generality assume that x = 0, then it is clear that y = z = 0. Because the given system is symmetric we can assume that x, y, z = 0. Assume that x = y, then x = z or y = −z. If we assume that y = −z then the ﬁrst equation becomes 4x4 = 6x4 , contradiction. Now x = y = z = 0 and after multiplying all three equations we get 3(x2 − xy + y 2 )3(y 2 − yz + z 2 )3(z 2 − zx + x2 ) =1 (x2 + xy + y 2 )(y 2 + yz + z 2 )(z 2 + zx + x2 ) which can be written as 2xy 2yz 2zx 1 1− 1− 1− = . (x − y )2 + 3xy (y − z )2 + 3yz (z − x)2 + 3zx 27 1 It is not diﬃcult to see that each factor of the LHS is greater than which leads to a contra- 3 diction. So the only possible solution is x = y = z. Also solved by Arkady Alt, San Jose, California, USA; Daniel Lasaosa, Universidad P´blica de u Navarra, Spain; Tigran Hakobyan, Armenia. 6 Mathematical Reflections 6 (2010)
J180. Let a, b, c, d be distinct real numbers such that 1 1 1 1 √ +√ +√ +√ = 0. 3 3 3 3 a−b b−c c−d d−a √ √ √ √ Prove that 3 a − b + 3 b − c + 3 c − d + 3 d − a = 0. Proposed by Dorin Andrica, Babes-Bolyai University, Cluj-Napoca, Romania Solution by Anthony Erb Lugo √ √ √ √ Let 3 a − b, 3 b − c, 3 c − d, 3 d − a by w, x, y, z , respectively and let S = w + x + y + z . So that, 1 111 w3 + x3 + y 3 + z 3 = 0(∗) + + + = 0(∗∗) and wxyz Since a, b, c and d are distinct, we can conclude that w, x, y, z = 0, and thus, 1 111 = 0(∗ ∗ ∗) wxy + wxz + wyz + xyz = wxyz +++ wxyz Furthermore, using (*), we have that, S 3 = 6(wxy + wxz + wyz + xyz ) + 3(w2 + x2 + y 2 + z 2 )S S (S 2 − 3(w2 + x2 + y 2 + z 2 )) = 6(wxy + wxz + wyz + xyz ) Which implies that S = 0, otherwise the right hand side would be 0 which is a contradiction of (∗ ∗ ∗), and we’re done. Also solved by Arkady Alt, San Jose, California, USA; Daniel Lasaosa, Universidad P´blica u de Navarra, Spain; Lorenzo Pascali, Universit` di Roma “La Sapienza”, Roma, Italy; Perfetti a Paolo, Dipartimento di Matematica, Universit` degli studi di Tor Vergata Roma, Italy; Tigran a Hakobyan, Armenia; Daniel Campos Salas, Costa Rica. 7 Mathematical Reflections 6 (2010)
Senior problems S175. Let p be a prime. Find all integers a1 , . . . , an such that a1 + · · · + an = p2 − p and all solutions to the equation pxn + a1 xn−1 + · · · + an = 0 are nonzero integers. Proposed by Titu Andreescu, University of Texas at Dallas, USA and Dorin Andrica, Babes-Bolyai University, Cluj-Napoca, Romania Solution by Daniel Lasaosa, Universidad P´blica de Navarra, Spain u Let −r1 , −r2 , . . . , −rn be the n nonzero integral roots of the equation given in the problem statement, then the equation rewrites as p(x + r1 )(x + r2 ) . . . (x + rn ) = 0, or all its coeﬃcients are nonzero integral multiples of p. In particular, ak is p times the sum of all possible products of k of the ri ’s, hence p + a1 + a2 + · · · + an = p(1 + r1 )(1 + r2 ) . . . (1 + rn ) = p2 , yielding (1 + r1 )(1 + r2 ) . . . (1 + rn ) = p, where each one of the 1 + ri are integers other than 1, ie, n − 1 of them must be equal to −1, and the remaining one must be p when n is odd and −p when n is even; in other words, n − 1 of the ri ’s are equal to −2, and the other one is p − 1 when n is −1 odd and −p − 1 when n is even. Now, there are n−1 products of k of the ri such that exactly k one of them is not −2, and n−1 products of k of the ri such that all of them are −2, ie, when k n is odd, n−1 n−1 (n − 1)! ak (−2)k−1 (p − 1) + (−2)k−1 (pk + k − 2n), (−2)k = = k−1 (n − k )!k ! p k and when n is even, n−1 n−1 (n − 1)!(2n + pk − k ) ak (−2)k−1 (−p − 1) + (−2)k−1 . (−2)k = − = k−1 (n − k )!k ! p k Note now that, when n is odd, n−1 n n (n − 1)! ak n! (−2)k−1 + (−2)k − 1 = = (p + 1) (n − k )!(k − 1)! (n − k )!k ! p k−1=0 k=1 k=0 = (p + 1)(1 − 2)n−1 + (1 − 2)n − 1 = p − 1, while when n is even, n−1 n n (n − 1)! ak n! (−2)k−1 + (−2)k − 1 = = −(p − 1) (n − k )!(k − 1)! (n − k )!k ! p k−1= k=1 k=0 = −(p − 1)(1 − 2)n−1 + (1 − 2)n − 1 = p − 1. There can be no other solutions. Also solved by Anthony Erb Lugo; Tigran Hakobyan, Armenia. 8 Mathematical Reflections 6 (2010)
S176. Let ABC be a triangle and let AA1 , BB1 , CC1 be cevians intersecting at P . Denote by Ka = KAB1 C1 , Kb = KBC1 A1 , Kc = KCA1 B1 . Prove that KA1 B1 C1 is a root of the equation x3 + (Ka + Kb + Kc )x2 − 4Ka Kb Kc = 0. Proposed by Ivan Borsenco, Massachusetts Institute of Technology, USA First solution by Arkady Alt, San Jose, California, USA Without loss of generality assume that area of triangle ABC is 1. Let pa , pb , pb be the baricentric coordinates of P, that is pa , pb , pc > 0, pa + pb + pc = 1 and −→ −→ − − −→ pa P A + pb P B + pc P C = 0. AC1 pb [ACC1 ] AC1 Since = and = then BC1 pa [BCC1 ] BC1 pb pb [ACC1 ] = [ABC ] = . pa + pb pa + pb [AB1 C1 ] AB1 pc Also, = = yields [CB1 C1 ] CB1 pa [AB1 C1 ] pc = . [ACC1 ] pc + pa pb pc Hence, Ka = [AB1 C1 ] = . Similarly, (pa + pb ) (pc + pa ) pc pa pa pb Kb = , Kc = . (pb + pc ) (pa + pb ) (pc + pa ) (pb + pc ) Let K = [A1 B1 C1 ] , then K = [ABC ] − ([AB1 C1 ] + [BC1 A1 ] + [CA1 B1 ]) pb pc = 1 − (Ka + Kb + Kc ) = 1 − (pa + pb ) (pc + pa ) cyc 2pa pb pc = . (pa + pb ) (pb + pc ) (pc + pa ) On the other hand, since Ka pb + pc Kb pc + pa Kc pa + pb = , = , = K 2pa K 2pb K 2pc then Ka Kb Kc (pa + pb ) (pb + pc ) (pc + pa ) 1 = = . 3 K 8pa pb pc 4K 9 Mathematical Reflections 6 (2010)
Thus, K 2 = 4Ka Kb Kc and, therefore, K = 1 − (Ka + Kb + Kc ) ⇐⇒ K 3 = K 2 − (Ka + Kb + Kc ) K 2 ⇐⇒ K 3 + (Ka + Kb + Kc ) K 2 − 4Ka Kb Kc = 0. Second solution by Daniel Campos Salas, Costa Rica Let x, y, z be the areas of the triangles BP C , CP A, AP B , respectively. It’s easy to prove that y AB1 AC1 z B1 C = x and C1 B = x . Therefore we have that 1 1 zAC yAB yz (x + y + z ) Ka = AB1 · AC1 sin A = · · sin A = . 2 2 x+z x+y (x + y )(x + z ) The expressions for Kb and Kc are obtained analogously. It follows that KA1 B1 C1 equals yz zx xy (x + y + z ) 1 − − − (x + y )(x + z ) (y + z )(y + x) (z + x)(z + y ) 2xyz (x + y + z ) = . (x + y )(y + z )(z + x) Finally, note that 3 KA1 B1 C1 + KA1 B1 C1 (Ka + Kb + Kc ) 2 = KA1 B1 C1 (KA1 B1 C1 + Ka + Kb + Kc ) 2 2xyz (x + y + z ) = ) (x + y + z ) (x + y )(y + z )(z + x) 4(xyz )2 (x + y + z )3 = ((x + y )(y + z )(z + x))2 = 4Ka Kb Kc , which proves that KA1 B1 C1 is a root of the given polynomial. Also solved by Daniel Lasaosa, Universidad P´blica de Navarra, Spain. u 10 Mathematical Reflections 6 (2010)
S177. Prove that in any acute triangle ABC, A B C 5R + 2r + sin + sin ≥ sin . 2 2 2 4R Proposed by Titu Andreescu, University of Texas at Dallas, USA First solution by Arkady Alt, San Jose, California, USA A B C Since r = 4R sin sin sin , the inequality can be rewritten as 2 2 2 A B C A B C 5 + sin + sin − 2 sin sin sin ≥ . sin ( A) 2 2 2 2 2 2 4 Inequality (A) is an immediate corollary from more general inequality represented by the following theorem Theorem. Let k be any real number such that k ≥ k∗ where 4 ≈ 1.2835. k∗ = √ √ 2− 2− 2 2+3 π π Then for any α, β, γ ∈ 0, , such that α + β + γ = the following inequality holds 4 2 12 − k sin α + sin β + sin γ − k sin α sin β sin γ ≥ . (M ) 8 Proof. Assuming, due symmetry, that α ≤ β ≤ γ and, denoting ϕ = α + β we obtain π − ϕ, β = ϕ − α, where the new variables α and ϕ satisfy the inequalities 0 < α ≤ γ= 2 π π ϕ − α ≤ − ϕ ≤ or equivalently 2 4 π π  ≤ϕ≤  4π 3 . (1)  2ϕ − ≤ α ≤ ϕ 2 2 Since α−β α+β sin α + sin β + sin γ − k sin α sin β sin γ = 2 sin + sin γ (1 − k sin α sin β ) cos 2 2 ϕ ϕ − α + cos ϕ (1 − k sin α sin (ϕ − α)) = 2 sin cos 2 2 then inequality (M) can be equivalently rewritten as 12 − k ϕ ϕ − α + cos ϕ (1 − k sin α sin (ϕ − α)) ≥ 2 sin cos (2) 2 2 8 where variables α and ϕ are subject to the system (1). 11 Mathematical Reflections 6 (2010)
Let ϕ ϕ − α + cos ϕ (1 − k sin α sin (ϕ − α)) h (α) = 2 sin cos 2 2 ππ 2 and k∗ > √ = 1. 154 7. We will prove that h (α) is decreasing on for any ﬁxed ϕ ∈ , 43 3 πϕ 2ϕ − , . Indeed, 22 ϕ ϕ ϕ −α − k cos ϕ cos −α ≤0 h (α) = 2 sin sin 2 2 2 πϕ ϕ 2 − α ≥ 0, k∗ > √ and on 2ϕ − , since sin 22 2 3 ϕ ϕ ϕ ϕ − k cos ϕ cos − α ≤ sin − k cos ϕ cos sin 2 2 2 2 √ π π π 1k 3 ≤ sin − k cos cos = − · 6√ 3 6√ 2 2 2 1 3 k∗ 1 31 ·√ ≤− ·
The original inequality immediately follows from (G). Indeed, A B C A B C A B C 4 A B C + sin + sin − 2 sin sin sin = sin + sin + sin − sin sin sin sin 2 2 2 2 2 2 2 2 2 3 2 2 2 2 A B C − sin sin sin 3 2 2 2 42 A B C 4 21 ≥ − sin sin sin ≥ − · 33 2 2 2 3 38 5 =. 4 [RAGI]. Mitrinovi´ D.S., Peˇari´ J. E. , Volenec V. Recent Advances, Geometric Inequality, c cc p.269, inequality 5.10. Second solution by Daniel Lasaosa, Universidad P´blica de Navarra, Spain u Assume wlog that C ≥ B ≥ A, and denote A+B = α, B −A = δ . It is well known that 4 4 r = 4R sin A sin B sin C , or the proposed problem is equivalent to showing that 2 2 2 A B C A B C 5 + sin + sin − 2 sin sin sin ≥ . sin 2 2 2 2 2 2 4 Assume that C is known such that the LHS is minimum, or for that value of C , deﬁne C f (x, y ) = sin x + sin y − 2 sin x sin y sin , 2 = 90◦ − A+B C where x + y = is ﬁxed. Now, 2 2 AB C − f (α, α) = f (α − δ, α + δ ) − f (α, α) = 2 sin α(cos δ − 1) + 2 sin2 δ sin f , = 22 2 δ δ C = 4 sin2 2 cos2 sin − sin α . 2 2 2 Now, if δ > 0, and since δ = B −A < B < 45◦ , we have sin2 2 > 0, 2 cos2 2 > 2 cos2 (45◦ ) > 1. δ δ 4 4 C A+B Moreover, 2 > 4 , since equality would only hold iﬀ A = B = C , which is not true because δ > 0. Thus, f A , B ≥ f A+B , A+B , with equality iﬀ A = B = 90◦ − C . It therefore suﬃces 22 4 4 2 to show that, for all 90◦ > C ≥ 60◦ , we have 5 2u + (1 − 2u2 ) − 2u2 (1 − 2u2 ) ≥ . 4 where we have deﬁned u = sin 45◦ − C , and therefore sin C = cos 2 45◦ − C = 1 − 2u2 . 4 2 4 After some algebra, this last inequality is equivalent to (2u − 1)2 (4u2 + 8u + 1) ≥ 0. Since ◦ 90◦ > C ≥ 60◦ , we have 45 < 45◦ − C ≤ 30◦ , or u > 0. The conclusion follows, equality 2 4 holds iﬀ u = 2 , ie iﬀ 45◦ − C = 30◦ , or C = 60◦ . We conclude that equality holds iﬀ ABC is 1 4 equilateral. Third solution by Neculai Stanciu, George Emil Palade, Romania 13 Mathematical Reflections 6 (2010)
We have A B C 5R + 2r A r 3 + sin + sin ≥ ⇐⇒ 2 sin ≥ 1 + + sin 2 2 2 4R 2 R2 A+B B+C C +A A+B+C ⇐⇒ 2 cos ≥ cos A + cos B + cos C + 3 cos + 2 cos + 2 cos 2 2 2 3 which is true from Popoviciu’s Inequality for the concave function cos x on the interval 0, π . 2 Also solved by Daniel Campos Salas, Costa Rica 14 Mathematical Reflections 6 (2010)
S178. Prove that there are sequences (xk )k≥1 and (yk )k≥1 of positive rational numbers such that for all positive integers n and k , √ √n 1+ 5 (xk + yk 5) = Fkn−1 + Fkn , 2 where (Fm )m≥1 is the Fibonacci sequence. Proposed by Dorin Andrica, Babes-Bolyai University, Cluj-Napoca, Romania Solution by G.R.A.20 Problem Solving Group, Roma, Italy Take xk = 1 Lk and yk = 1 Fk (see problem J178) where Ln is the n-th Lucas number. Since 2 2 √k √k √k √k   1− 5  1− 5 1  1+ 5 1+ 5 Fk = √ − and Lk = + 2 2 2 2 5 it follows that √ √ kn √ 1+ 5 1+ 5 n (xk + yk 5) = = Fkn−1 + Fkn . 2 2 Also solved by Arkady Alt, San Jose, California, USA; Daniel Lasaosa, Universidad P´blica u de Navarra, Spain; Daniel Campos Salas, Costa Rica. 15 Mathematical Reflections 6 (2010)
(a2 +1)2 S179. Find all positive integers a and b for which is a positive integer. ab−1 Proposed by Valcho Milchev, Petko Rachov Slaveikov Secondary School, Bulgaria Solution by Daniel Campos Salas, Costa Rica Note ab − 1 divides (ab − 1)(2a2 + ab + 1) = (a2 + ab)2 − (a2 + 1)2 . This implies that ab − 1 divides a2 (a + b)2 , but ab − 1 and a are coprime, from where we conclude that ab − 1 divides a+b)2 (a + b)2 . Let k = (ab−1 . Note that it is not possible for a, b to be equal, since it would imply 4a2 = 4 + a24 1 to be an integer, and 4 has no divisor of the form a2 − 1. The equation can a2 −1 − also be written in the form a2 − (k − 2)ab + b2 + k = 0. (1) (a+b)2 2 +k the pair a, a Note that for every pair of integers (a, b) satisfying the equation k = ab−1 , b a2 +k and its permutations are also positive integer solutions (it’s easy to prove that is an b integer). Iterating this process we ﬁnd that every solution is part of a family of solutions. Let (a0 , b0 ) be a solution for any of these families with the minimal sum. We may assume without a2 +k a2 +k a2 +k loss of generality that b0 > a0 . This implies that 0b0 ≥ b0 since a0 , 0b0 and 0b0 , a0 are also solutions to the equation. It follows that k ≥ b2 − a2 = (b0 − a0 )(a0 + b0 ) ≥ a0 + b0 . 0 0 (a0 +b0 )2 = k ≥ a0 + b0 , or equivalently, 2 ≥ (a0 − 1)(b0 − 1). Note that a0 < 3, This implies that a0 b0 −1 (b0 +1)2 4 since a0 < b0 . If a0 = 1, it follows that is an integer. This implies that b0 −1 , so it b0 −1 (b0 +2)2 follows that b0 can be equal to 2, 3 or 5. If a0 = 2, it follows that is an integer. This 2b0 −1 implies that 2b25 1 , or equivalently, b0 can equal 3 or 13, since b0 > a0 = 2. 0− For (a0 , b0 ) = (1, 2) it follows that k = 9. This initial solution generates the family of solutions given by the pairs, (cn , cn+1 ), n ≥ 0, which satisfy, c0 = 1, c1 = 2, cn+2 = 7cn+1 − cn , where the recursion holds by (1). The explicit expression for cn is given by the formula √ √n√ √ n 5−1 5+1 7−3 5 7+3 5 √ +√ cn = 2 2 25 25 √ 4n−1 √ 4n−1 1− 5 1 1+ 5 1 √ −√ = 2 2 5 5 = F4n−1 , where Fn is the n-th term of the Fibonacci sequence deﬁned by F0 = 0, F1 = 1 and Fn+2 = Fn+1 + Fn (abusing of the deﬁnition we set F−1 = 1). 16 Mathematical Reflections 6 (2010)
For (a0 , b0 ) = (1, 3) it follows that k = 8. This initial solution generates the family of solutions given by the pairs, (dn , dn+1 ), n ≥ 0, which satisfy, d0 = 1, d1 = 2, dn+2 = 6dn+1 − dn , where the recursion holds by (1). The explicit expression for dn is given by the formula √n1 √n 1 3−2 2 dn = 3+2 2 + 2 2 1√ √ ( 2 + 1)2n + ( 2 − 1)2n . = 2 For (a0 , b0 ) = (1, 5) it follows that k = 9. This initial solution generates the family of solutions given by the pairs, (en , en+1 ), n ≥ 0, which satisfy, e0 = 1, e1 = 2, en+2 = 7en+1 − en , where the recursion holds by (1). The explicit expression for en is given by the formula √ √n√ √ n 5−1 7−3 5 5+1 7+3 5 √ +√ en = 2 2 25 25 √ 4n+1 √ 4n+1 1− 5 1 1+ 5 1 √ −√ = 2 2 5 5 = F4n+1 . We may interpret this solution as an extension of the ﬁrst one for negative indexes since F4n+1 = F−4n−1 . For (a0 , b0 ) = (2, 3) it follows that k = 5. This initial solution generates the family of solutions given by the pairs, (fn , fn+1 ), n ≥ 0, which satisfy, f0 = 1, f1 = 2, fn+2 = 3dn+1 − fn , where the recursion holds by (1). The explicit expression for fn is given by the formula √ √ n n 3− 5 3+ 5 fn = + 2 2 √ √ 2n 2n 1− 5 1+ 5 = + 2 2 = L2n , where Ln is the n-th term of the Lucas sequence deﬁned by L0 = 2, L1 = 1 and Ln+2 = Ln+1 + Ln . Therefore, we conclude that all the possible pairs of positive integers are of the form (F4n−1 , F4n+3 ), (L2n , L2n+2 ) (for all integers n) and (dn , dn+1 ) (deﬁned above), and its permutations, and we’re done. Also solved by Tigran Hakobyan, Armenia 17 Mathematical Reflections 6 (2010)
S180. Solve in nonzero real numbers the system of equations 121x−122y x4 − y 4 = 4xy 2 y 2 + y 4 = 122x+121y . x4 + 14x x2 +y 2 Proposed by Titu Andreescu, University of Texas at Dallas, USA Solution by Daniel Lasaosa, Universidad P´blica de Navarra, Spain u Note ﬁrst that x4 + 14x2 y 2 + y 4 = 4(x2 + y 2 )2 − 3(x2 − y 2 )2 = s4 − s2 d2 + d4 , where we have 2 2 deﬁned x + y = s and x − y = d, while x4 − y 4 = sd(x2 + y 2 ) = sd(s 2+d ) , 4xy = s2 − d2 , 2 2 x2 + y 2 = s +d . Therefore, the system may be rewritten as 2 sd(s2 + d2 )(s2 − d2 ) = 243d − s, (s4 − s2 d2 + d4 )(s2 + d2 ) = 243s + d. We can then obtain (243d − s)(243s + d) = sd(s2 + d2 )(s2 − d2 )(243s + d) = (s4 − s2 d2 + d4 )(s2 + d2 )(243d − s). Since s2 + d2 > 0 (otherwise x = y = 0, in contradiction with the problem statement), it follows that 243s2 d(s2 − d2 ) + sd2 (s2 − d2 ) = 243d(s4 − s2 d2 + d4 ) − s(s4 − s2 d2 + d4 ), which after simpliﬁcation yields s5 = 243d5 , or s = 3d. Substitution in both equations yields d6 = d, or since x = y (if x = y = 0 the LHS of the ﬁrst equation would be zero, but the RHS would not), we ﬁnd that d5 = 1, ie s = 3 and d = 1 for x = 2, y = 1. These values can be clearly shown to satisfy the system by plugging them into the given equations, and no other solutions exist. Second solution by Arkady Alt, San Jose, California, USA Since x4 + 14x2 y 2 + y 4 x2 + y 2 = 122x + 121y, 4xy x4 − y 4 = 121x − 122y and x, y = 0 then x4 + 14x2 y 2 + y 4 x2 + y 2 (x − y ) − 4xy x4 − y 4 (x + y ) = (122x + 121y ) (x − y ) − (121x − 122y ) (x + y ) x2 + y 2 x4 + 14x2 y 2 + y 4 (x − y ) − 4xy x2 − y 2 (x + y ) ⇐⇒ = x2 + y 2 ⇐⇒ x4 + 14x2 y 2 + y 4 (x − y ) − 4xy x2 − y 2 (x + y ) = 1 ⇐⇒ (x − y )5 = 1 ⇐⇒ x − y = 1. Let t = x + y then t2 + 1 t−1 t−1 x2 − y 2 = t, x2 + y 2 = , 4xy = t2 − 1, y = , 121x − 122y = 121 − 2 2 2 18 Mathematical Reflections 6 (2010)
121x − 122y and the equation x4 − y 4 = becomes 4xy t t4 − 1 t−1 ⇐⇒ t t4 − 1 + t − 1 = 242 ⇐⇒ t5 = 243 ⇐⇒ t = 3. = 121 − 2 2 Hence, x−y =1 ⇐⇒ x = 2, y = 1. x+y =3 19 Mathematical Reflections 6 (2010)
Undergraduate problems U175. What is the maximum number of points of intersection that can appear after drawing in a plane l lines, c circles, and e ellipses? Proposed by Dorin Andrica, Babes-Bolyai University, Cluj-Napoca, Romania Solution by Andrea Ligori and Emanuele Natale, Universit` di Roma “Tor Vergata”, Roma, a Italy We note that c • the intersection of two circles yields 2 points, so the contribution is 2 ; 2 • the intersection of a line with a circle or an ellipse yields 2 points, so the contribution is 2l(c + e); • the intersection of an ellipse with a circle or another ellipse yields 4 points, so the contri- e bution is 4 2 + 4ec; l • the intersection of two incindent lines yields 1 point so the contribution is . 2 Therefore the ﬁnal formula is c e l 2 + 2l(c + e) + 4 + 4ec + . 2 2 2 It is easy to ﬁnd a conﬁguration of l lines, c circles, and e ellipses with such a number of intersection points. Also solved by Daniel Lasaosa, Universidad P´blica de Navarra, Spain. u 20 Mathematical Reflections 6 (2010)