Junior problems - Phần 1

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Junior problems J163. Let a, b, c be nonzero real numbers such that ab + bc + ca ≥ 0. Prove that ab bc ca 1 ≥− . + + a2 + b2 b2 + c2 c2 + a2 2 Proposed by Titu Andreescu, University of Texas at Dallas, USA Solution by Ercole Suppa, Teramo, Italy We have (a + b)2 ab ab 1 3 3 − − = + = a2 + b2 a2 + b2 2 2 (a2 + b2 ) 2 2 cyc cyc cyc 2 a2 + b2 + c2 + 2(ab + bc + ca) 3 b)2 (a + 3 ≥ −= − 2 (a2 + b2 + c2 ) 2 2 (a2 + b2 + c2 ) 2 cyc ab + bc + ca 3 ab + bc + ca 1 1 −=2 − ≥− =1+ 2 + b2 + c2 2 + c2 a 2 a +b 2 2 where in the last step we have used the fact that ab + bc + ca ≥ 0. 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; Prithwijit De, HBCSE, India; Andrea Ligori, Universit` di Roma “Tor a Vergata”, Italy; Piriyathumwong P., Bangkok, Thailand. 1 Mathematical Reflections 4 (2010)
√ x2 + 1 y 2 + 1 = 2011, ﬁnd the J164. If x and y are positive real numbers such that x + y+ minimum possible value of x + y. Proposed by Neculai Stanciu, “George Emil Palade”, Buzau, Romania 2010 First solution by Michel Bataille, France The required minimum value is √2011 . √ Write x = sinh(a) and y = sinh(b) where a = ln(x + x2 + 1) > 0 and b = ln(y + y 2 + 1) > 0. From the hypothesis, we have a + b = ln(2011) and using a known formula, √ a−b a+b a+b ≥ 2 sinh x + y = sinh(a) + sinh(b) = 2 sinh cosh = 2 sinh(ln( 2011) 2 2 2 where the inequality follows from cosh(t) ≥ 1 for all t and sinh(u) > 0 for u > 0. √ √ 1 2010 Since 2 sinh(ln( 2011) = 2011 − √2011 = √2011 , we obtain 2010 x+y ≥ √ . 2011 Clearly equality holds when a = b (since cosh(0) = 1), that is, when x = y . The result follows. Second solution by the authors √ z 2 −1 2011 Let z = x + x2 + 1. We have z > 0 and (1) x = y2 + 1 = 2z . From hypothesis y + z, 2 2 we get (2) y = 2011 −·z . From (1) and (2), 2·2011 z z 2 − 1 20112 − z 2 2010 2011 2010 2011 ≥ z· x+y = + = z+ . 2 · 2011 · z 2 · 2011 2z z 2011 z The equality occurs for z 2011 or equivalently z 2 = 2011. Then from (1) and z (2) we obtain 2010 1005 x=y= √ =√ . 2 2011 2011 2010 So min(x + y ) = . √ 2011 Also solved by Arkady Alt, San Jose, California, USA; Francisco Javier Garcia Capitan, Spain; Ercole Suppa, Teramo, Italy; Daniel Lasaosa, Universidad P´blica de Navarra, Spain; Perfetti u Paolo, Dipartimento di Matematica, Universit` degli studi di Tor Vergata Roma, Italy. a 2 Mathematical Reflections 4 (2010)
J165. Find all triples (x, y, z ) of integers satisfying the system of equations z2 x2 + 1 y2 + 1 + = 2010 10 (x + y )(xy − 1) + 14z = 1985. Proposed by Titu Andreescu, University of Texas at Dallas, USA Solution by Arkady Alt, San Jose, California, USA z2 = 2010 − x2 + 1 y 2 + 1 is an integer. Let Note that z = 10k for some integer k because 10 p = x + y and q = xy − 1. Then y 2 + 1 = x2 y 2 + x2 + y 2 + 1 = (xy − 1)2 + (x + y )2 = p2 + q 2 x2 + 1 and the system becomes p2 + q 2 + 10k 2 = 2010 p2 + q 2 = 2010 − 10k 2 ⇐⇒ (1) pq = 1985 − 140k pq + 140k = 1985 Since (p − q )2 = 2010 − 10k 2 − 2 (1985 − 140k ) = −10 (k − 14)2 then only k = 14 can provide p2 + q 2 = 50 ⇐⇒ p = q = 5. solvability to (1). And for k = 14, (1) becomes pq = 25 x+y =5 x=4 x=1 ⇐⇒ Hence, or and triples (5, 1, 140) , (1, 5, 140) are all xy = 4 y=1 y=4 integer solutions of the original system in integers. Also solved by Daniel Lasaosa, Universidad P´blica de Navarra, Spain; Piriyathumwong P., u Bangkok, Thailand. 3 Mathematical Reflections 4 (2010)
J166. Let P be a point inside triangle ABC and let da , db , dc be the distances from point P to the sides of the triangle. Prove that K s ≥ da db dc Rr where K is the area of the pedal triangle of P and s, R, r are the semiperimeter, circumradius, and inradius of triangle ABC. Proposed by Andrei Razvan Baleanu, “George Cosbuc”, Motru, Romania Remark: The problem contains a typ and the inequality that needs to be proven is K s ≥ . da db dc 2Rr Many readers have solved the correct inequality. First solution by Daniel Lasaosa, Universidad P´blica de Navarra, Spain The proposed in- u equality is not true, since if P = I is the incenter of equilateral triangle ABC , then K = rs is 4 one quarter the area of ABC , while da = db = dc = r, and the proposed inequality would be equivalent to R ≥ 4r = 2R, absurd. We show that the correct inequality that always holds is 2K s = . da db dc Rr Now, denoting by PA , PB , PC the respective projections of P on sides BC, CA, AB , we have ∠CPA P = ∠CPB P = 90◦ , or ∠PA P PB = 180◦ − C , and the area of P PA PB is da db sin C da db dc c = . 2 4R dc Adding the analogous expressions for the areas of P PB PC and P PC PA , we ﬁnd da db dc a b c K= + + , 4R da db dc or the proposed inequality is equivalent to a b c 2s ≥. + + da db dc r d2 1 1 2 Now, is convex for positive x because = > 0, or by Jensen’s inequality, dx2 x3 x x 4s2 a b c a+b+c 2s ≥ (a + b + c) + + = =, da db dc ada + bdb + cdc S r where 2S = ada + bdb + cdc = 2rs is twice the area of ABC because ada is twice the area of BP C , and similarly for its cyclic permutations, and equality is reached iﬀ da = db = dc , ie iﬀ 2 2s P is the incenter of ABC , in which case we easily ﬁnd K = r (sin A+sin B +sin C ) = r R . 2 2 4 Mathematical Reflections 4 (2010)
Second solution by G.R.A.20 Math Problems Group, Roma, Italy Since 4KR = cda db + adb dc + bdc da and 2sr = ada + bdb + cdc , it follows that the inequality becomes a b c · (ada + bdb + cdc ) ≥ (a + b + c)2 + + da db dc which holds by Cauchy-Schwarz. Also solved by Arkady Alt, San Jose, California, USA; Michel Bataille, France; Ercole Suppa, Teramo, Italy. 5 Mathematical Reflections 4 (2010)
J167. Let a, b, c be real numbers greater than 1 such that b+c c+a a+b ≥ 1. +2 +2 2−1 b −1 c −1 a Prove that 2 2 2 bc + 1 ca + 1 ab + 1 10 ≥ + + . a2 − 1 b2 − 1 c2 − 1 3 Proposed by Titu Andreescu, University of Texas at Dallas, USA First solution by Prithwijit De, HBCSE, India Observe that 2 2 (b2 − 1)(c2 − 1) bc + 1 b+c − = ; a2 − 1 a2 − 1 (a2 − 1)2 2 2 (c2 − 1)(a2 − 1) ca + 1 c+a − = ; b2 − 1 b2 − 1 (b2 − 1)2 2 2 (a2 − 1)(b2 − 1) ab + 1 a+b − = . c2 − 1 c2 − 1 (c2 − 1)2 2 2 (b2 − 1)(c2 − 1) bc + 1 b+c Therefore = + .. . . (1) a2 − 1 a2 − 1 (a2 − 1)2 2 b+c b+c 2 a2 − 1 1 ≥ ≥ Now observe that . . . (2) a2 − 1 3 3 and by A.M-G.M inequality we get (b2 − 1)(c2 − 1) (a2 − 1)2 (b2 − 1)2 (c2 − 1)2 ≥33 2 = 3. . . . (3) (a2 − 1)2 (a − 1)2 (b2 − 1)2 (c2 − 1)2 2 bc + 1 1 10 ≥3+ By virtue of (1),(2) and (3) we obtain =. a2 − 1 3 3 Second solution by Perfetti Paolo, Dipartimento di Matematica, Universit` degli studi di Tor a Vergata Roma, Italy We observe that bc + 1 = b + c + (b − 1)(c − 1) and then (bc + 1)2 = (b + c)2 + 2(b + c)(b − 1)(c − 1) + (b − 1)2 (c − 1)2 . By power–means–inequality and the constraint on a, b, c we have 2 (b + c)2 1 b+c 1 ≥ ≥ 2 − 1)2 a2 − 1 (a 3 3 cyc cyc thus the inequality becomes (b + c)(b − 1)(c − 1) (b − 1)2 (c − 1)2 ≥3 2 + (a2 − 1)2 (a2 − 1)2 cyc 6 Mathematical Reflections 4 (2010)
or (b − 1)(c − 1)(bc + b + c + 1) ≥3 (a2 − 1)2 cyc Since the inequality in the statement is symmetric, we can set a ≥ b ≥ c. Then we observe that (b − 1)(c − 1) (c − 1)(a − 1) (a − 1)(b − 1) , , (a − 1)2 (b − 1)2 (c − 1)2 and bc + b + c + 1 ca + c + a + 1 ab + a + b + 1 , , (a + 1)2 (b + 1)2 (c + 1)2 are equally sorted. This allows us to employ Chebyshev–inequality (b − 1)(c − 1)(bc + b + c + 1) (b − 1)(c − 1) 1 bc + b + c + 1 ≥ · 2 − 1)2 (a − 1)2 (a + 1)2 (a 3 cyc cyc cyc Moreover by AGM we have (b − 1)(c − 1) ≥3 (a − 1)2 cyc and bc + b + c + 1 (b + 1)(c + 1) ≥3 = (a + 1)2 (a + 1)2 cyc cyc Bringing together the last three inequalities we obtain (b − 1)(c − 1)(bc + b + c + 1) 1 ≥ ¨·3=3 3 2 − 1)2 (a 3 cyc and we are done. Also solved by Daniel Lasaosa, Universidad P´blica de Navarra, Spain; Prithwijit De, HBCSE, u India. 7 Mathematical Reflections 4 (2010)
J168. Let n be a positive integer. Find the least positive integer a such that the system x1 + x2 + · · · + xn = a x2 + x2 + · · · + x2 = a n 1 2 has no integer solutions. Proposed by Dorin Andrica, “Babe¸-Bolyai University”, Cluj-Napoca, Romania s Solution by Lorenzo Pascali, Universita di Roma La Sapienza, Italy First, we notice that if xi = 0, 1 for an integer component xi then x2 > xi and we have a i contradiction a = x2 + x2 + · · · + x2 > x1 + x2 + · · · + xn = a. 1 2 n Hence any component xi is 0 or 1 and the system has integer solutions for a = 1, . . . , n: take x1 = · · · = xa = 1 and xa+1 = · · · = xn = 0. Therefore the least positive integer a such that the system has no integer solutions is n + 1: x1 + x2 + · · · + xn ≤ x2 + x2 + · · · + x2 ≤ n < a = n + 1 . 1 2 n Also solved by Arkady Alt, San Jose, California, USA; Daniel Lasaosa, Universidad P´blica de u Navarra, Spain; Ercole Suppa, Teramo, Italy. 8 Mathematical Reflections 4 (2010)
Senior problems S163. (a) Prove that for each positive integer n there is a unique positive integer an such that √ √ √ (1 + 5)n = an + an + 4n . (b) When n is even, prove that an is divisible by 5 · 4n−1 and ﬁnd the quotient. Proposed by Dorin Andrica, “Babe¸-Bolyai University”, Cluj-Napoca, Romania s First solution by G. C. Greubel, Newport News, VA √ Let 2α = 1 + 5. With this we have √ √ 2n αn = an + an + 4n . (1) Squaring both sides leads to 4n (α2n − 1) = 2an + 2 an (an + 4n ). (2) Subtracting 2an from both sides and squaring the resulting value leads to [4n (α2n − 1) − 2an ]2 = 4an (an + 4n ). (3) This is reduced to 2 α2n − 1 an = 4n−1 αn = 4n−1 (αn − (−1)n β n )2 = 4n−1 α2n + β 2n − 2 = 4n−1 (L2n − 2) (4) where Lm is the mth Lucas number. Hence it has been shown that an is a positive integer and is given by an = 4n−1 (L2n − 2). B) If n is an even value, say n = 2m, then a2m = 42m−1 (L4m − 2) = 42m−1 · 5F2n 2 = 5 · 4m−1 · (2m F2m )2 . (5) From this relation it is shown that a2m is divisible by 5 · 4m−1 and has the quotient value (2m F2m )2 . 9 Mathematical Reflections 4 (2010)
Second solution by the authors √ √ (a) Let (1 + 5)n = xn + yn 5, where xn , yn are positive integers, n = 1, 2, . . . Then √ √ (1 − 5)n = xn − yn 5, n = 1, 2, . . . , hence x2 − 5yn = (−4)n , n = 1, 2, . . . 2 (1) n If n is even, consider an = x2 − 4n and we have n √ √ an + an + 4n = x2 − 4n + x2 = 2 x2 5yn + n n n √ √n = yn 5 + xn = (1 + 5) . If n is odd, consider an = 5yn − 4n and we have 2 √ √ an + an + 4n = 5yn − 4n + 5yn = 2 2 x2 + 2 5yn n √ √ = xn + yn 5 = (1 + 5)n . (b) If n is even, then we have an = x2 − 4n = 5yn , where 2 n √ √ 1 yn = √ [(1 + 5)n − (1 − 5)n ] 25 √ √ n n 2n 1− 5 1+ 5 = 2n−1 Fn , =√ − 2 2 25 where Fn is the nth Fibonacci number. In this case we get an = 5 · 4n−1 Fn , hence 5 · 4n−1 |an 2 2 and the quotient is Fn . Also solved by Arkady Alt, San Jose, California, USA; Daniel Lasaosa, Universidad P´blica u de Navarra, Spain. 10 Mathematical Reflections 4 (2010)
S164. Let ABCD be a cyclic quadrilateral whose diagonals are perpendicular to each other. For a point P on its circumscribed circle denote by P the line tangent to the circle at P. Let U = A ∩ B , V = B ∩ C , W = C ∩ D , K = D ∩ A . Prove that U V W K is a cyclic quadrilateral. Proposed by Ivan Borsenco, Massachusetts Institute of Technology, USA Solution by Michel Bataille, France Let U1 , V1 , W1 , K1 be the midpoints of AB, BC, CD, DA, respectively. The Varignon paral- lelogram U1 V1 W1 K1 of the quadrilateral ABCD is a rectangle (because AC ⊥ BD), hence U1 , V1 , W1 , K1 lie on a circle γ centered at the centre of the rectangle. Note that O, U1 , U are collinear (on the perpendicular bisector of AB ) and that AB is the polar of U with respect to Γ. Similar results hold for V1 , W1 , K1 and it follows that the inverses of U1 , V1 , W1 , K1 in the circle Γ are U, V, W, K , respectively, so that U, V, W, K all lie on the inverse of the circle γ . Since U, V, W, K clearly cannot be collinear, the inverse of γ is a circle and so U V W K is a cyclic quadrilateral. K A U U1 K1 O B D V1 W 1 V C W Also solved by Ercole Suppa, Teramo, Italy; Daniel Lasaosa, Universidad P´blica de Navarra, u Spain; Prithwijit De, HBCSE, India. 11 Mathematical Reflections 4 (2010)
S165. Let I be the incenter of triangle ABC. Prove that AI · BI · CI ≥ 8r3 , where r is the inradius of triangle ABC. Proposed by Dorin Andrica, “Babe¸-Bolyai University”, Cluj-Napoca, Romania s Solution by Piriyathumwong P.,Bangkok, Thailand Since r r r AI = , BI = , CI = , A B sin C sin 2 sin 2 2 the inequality above is equivalent to sin A sin B sin C ≤ 1 ,which is immediately true because 2 2 2 8 of the two well-known facts below: r A B C , R ≥ 2r = 4 sin sin sin R 2 2 2 ,where R is the circumradius of triangle ABC . Also solved by Arkady Alt, San Jose, California, USA; Michel Bataille, France; Scott H. Brown, Auburn University Montgomery, USA; Ercole Suppa, Teramo, Italy; Daniel Lasaosa, Univer- sidad P´blica de Navarra, Spain; G.R.A.20 Math Problems Group, Roma, Italy. u 12 Mathematical Reflections 4 (2010)
S166. If a1 , a2 , . . . , ak ∈ (0, 1), and k, n are integers such that k > n ≥ 1, prove that the following inequality holds nn min{a1 (1 − a2 )n , a2 (1 − a3 )n , . . . , ak (1 − a1 )n } ≤ . (n + 1)n+1 Proposed by Marin Bancos, North University of Baia Mare, Romania First solution by Arkady Alt, San Jose, California, USA Let M = min {a1 (1 − a2 )n , a2 (1 − a3 )n , . . . , ak (1 − a1 )n } and for any function f (x, y ) let k f (a1 , a2 ) = f (a1 , a2 ) + f (a2 , a3 ) + · · · + f (ak , a1 ) . cyc Since for any x, y ∈ (0, 1) by the AM–GM inequality nx + n − ny n (x − y ) + n nx (1 − y )n ≤ n+1 = . n+1 n+1 Then √ a1 (1 − a2 )n , a2 (1 − a3 )n , . . . , ak (1 − a1 )n n+1 n+1 n+1 n+1 M = min k 1 a1 (1 − a2 )n n+1 ≤ k cyc k 1 na1 (1 − a2 )n n+1 √ = n+1 k n cyc k 1 √ ≤ (n (a1 − a2 ) + n) n+1 k (n + 1) n cyc nk n √= √ = n+1 n+1 k (n + 1) n (n + 1) n √ n+1 nn = . (n + 1) nn Then M ≤ . (n + 1)n+1 Second solution by the author “Reductio ad absurdam” Let’s suppose that the inequality doesn’t hold. Therefore nn a1 (1 − a2 )n > (n + 1)n+1 13 Mathematical Reflections 4 (2010)
nn a2 (1 − a3 )n > (n + 1)n+1 ..... nn ak (1 − a1 )n > (n + 1)n+1 Multiplying these relations up, we get k nn a1 · a2 · ... · ak · (1 − a1 )n · (1 − a2 )n · ... · (1 − ak )n > (∗) (n + 1)n+1 But, for a ∈ (0, 1), we have nn a(1 − a)n ≤ (n + 1)n+1 Let’s prove this inequality. n+1 n times  n+1 nn 1  na+ (1 − a) + · · · + (1 − a)  AM −GM 1 1 n a·(1−a)n = ·n·a·(1−a)n ≤ · · = = (n + 1)n+1 n n n+1 n n+1 1 The equality holds for: na = 1 − a ⇔ a = ∈ (0, 1) n+1 Using the proved inequality for a1 , a2 , . . . , ak , we get: nn a1 (1 − a1 )n ≤ (n + 1)n+1 nn a2 (1 − a2 )n ≤ (n + 1)n+1 ..... nn ak (1 − ak )n ≤ (n + 1)n+1 Multiplying these relations up, we get k nn a1 · a2 · · · · · ak · (1 − a1 )n · (1 − a2 )n · · · · · (1 − ak )n ≤ (n + 1)n+1 This inequality contradicts (∗), which follows from the initial assumption. Therefore, that assumption is false. Also solved by Michel Bataille, France; Daniel Lasaosa, Universidad P´blica de Navarra, Spain; u Perfetti Paolo, Dipartimento di Matematica, Universit` degli studi di Tor Vergata Roma, Italy. a 14 Mathematical Reflections 4 (2010)
S167. Let Ia be the excenter corresnponding to the side BC of triangle ABC. Denote by A , B , C the tangency points of the excircle of center Ia with the sides BC, CA, AB, respectively. Prove that the circumcircles of triangles AIa A , BIa B , CIa C have a common point, diﬀerent from Ia , situated on the line Ga Ia , where Ga is the centroid of triangle A B C . Proposed by Dorin Andrica, “Babe¸-Bolyai University”, Cluj-Napoca, Romania s First solution by Michel Bataille, France A C A' B B' B1 C 1 G1 A1 C' I1 For typographical reasons, Ia and Ga are denoted by I1 and G1 on the ﬁgure above. Let γ be the excircle. Since Ia A = Ia C an BA = BC , the line Ia B is the perpendicular bisector of A C and intersects A C in its midpoint B1 . Since A C is the polar of B with respect to γ , the inversion in the circle γ exchanges B1 and B . Since B is invariant under this inversion, the circumcircle of ∆Ia BB inverts into the median B B1 of triangle A B C . Similarly, the circumcircles of ∆Ia AA , ∆Ia CC invert into the medians A A1 , C C1 . As a result, the three circumcircles all pass through Ia and through the inverse of Ga (because Ga lies on the three medians A A1 , B B1 , C C1 ). The second result follows from the fact that the inverse of Ga is on the line through Ia and Ga . Second solution by Daniel Lasaosa, Universidad P´blica de Navarra, Spain Let D be the u midpoint of B C . Now AB ⊥ Ia B , while AIa ⊥ B C where B D = C D by symmetry around the internal bisector of angle A. Thus, triangles AB D and B Ia D are similar, hence B D · C D = B D 2 = Ia D · AD , and the power of D with respect to the circumcircles of A B C and AIa A is the same, or D lies on the radical axis of both circles, which is median AD. 15 Mathematical Reflections 4 (2010)
Let E be the midpoint of C A . BA ⊥ Ia A , while BIa ⊥ A C where A E = C E by symmetry around the external bisector of angle B . Thus, triangles BE A and A E Ia are similar, hence A E · C E = A E 2 = BE · Ia E , and median B E is the radical axis of the circumcircles of A B C and BIa B . Similarly, median C F (F is the midpoint of A B ) is the radical axis of the circumcircles of A B C and CIa C . Clearly, the point Ga where the medians A D , B E and C F meet, has the same power with respect to the four circumcircles; consider now the second point P where Ia Ga meets the circumcircle of AIa A . Since Ia Ga is the radical axis of the circumcircles of AIa A and BIa B because Ia , Ga have the same power with respect to both, then P also has the same power with respect to both, but since it is on the circumcircle of AIa A , it is also on the circumcircle of BIa B . Similarly, it is also on the circumcircle of CIA C . The conclusion follows. 16 Mathematical Reflections 4 (2010)
S168. Let a0 ≥ 2 and an+1 = a2 − an + 1, n ≥ 0. Prove that n loga0 (an − 1) loga1 (an − 1) · · · logan−1 (an − 1) ≥ nn , for all n ≥ 1. Proposed by Titu Andreescu, University of Texas at Dallas, USA Solution by Perfetti Paolo, Dipartimento di Matematica, Universit` degli studi di Tor Vergata a Roma, Italy Proof Induction loga0 (an+1 − 1) loga1 (an+1 − 1) · · · logan−1 (an+1 − 1) logan (an+1 − 1) ≥ (n + 1)n+1 an+1 − 1 = an (an − 1) ≥ (an − 1)2 =⇒ logx (an+1 − 1) ≥ 2 logx (an − 1) This implies loga0 (an+1 − 1) loga1 (an+1 − 1) · · · logan−1 (an+1 − 1) logan (an+1 − 1) ≥ 2n loga0 (an − 1) loga1 (an − 1) · · · logan−1 (an − 1) logan (an+1 − 1) ≥ (1) 2n nn logan (an+1 − 1) = 2n nn logan (an (an − 1)) = (2n)n (1 + logan (an − 1)) ≥ (n + 1)n+1 For n ≥ 4 we have (2n)n ≥ (n + 1)n+1 . Indeed n 1 2n ≥ (n + 1)e ≥ (n + 1) 1 + ∀n≥4 , n We need to show yet the validity of our inequality for n = 1, 2, 3. For n = 1 the inequality is loga0 (a0 (a0 − 1)) = 1 + loga0 (a0 − 1) ≥ 1 being a0 − 1 ≥ 1. For n = 2 we have loga0 (a2 − 1) loga1 (a2 − 1) ≥ 4 or loga0 a1 (a1 − 1) loga1 a1 (a1 − 1) ≥ 4 namely loga0 a1 a0 (a0 − 1) loga1 a1 a0 (a0 − 1) ≥ 4 (2) We rewrite (2) as 1 + loga0 a1 + loga0 (a0 − 1) 1 + loga1 a0 + loga1 (a0 − 1) ≥ 4 17 Mathematical Reflections 4 (2010)
which is implied by, use again a0 − 1 ≥ 1, 1 + loga1 a0 ≥ 4 1 + loga0 a1 and this holds true since it may be written as (1 + x)(1 + 1/x) ≥ 4, and x + 1/x ≥ 2, x > 0. The last integer still remaining is n = 3 and by (1) we need to show 63 (1 + lna3 (a3 − 1)) ≥ 44 The ﬁrst step is: lna3 (a3 − 1) increases with a3 ≥ 7. To prove this let’s write loga3 (a3 − 1) = ln(a3 −1) so that ln a3 d ln(a3 − 1) ln(a3 − 1) 1 1 − for a3 ≥ 7 = >0 (3) a3 − 1 da3 ln a3 ln a3 a3 ln a3 The monotonicity of lna3 (a3 − 1) for a3 ≥ 7 implies that it is greater than or equal to ln7 6 and this in turn implies that it suﬃces to show 63 (1 + ln7 6) ≥ 44 which evidently holds true and we are done. Also solved by Michel Bataille, France; Daniel Lasaosa, Universidad P´blica de Navarra, Spain; u Prithwijit De, HBCSE, India; Lorenzo Pascali, Universit` di Roma “La Sapienza”, Roma, a Italy; Piriyathumwong P., Bangkok, Thailand. 18 Mathematical Reflections 4 (2010)
Undergraduate problems U163. Find the minimum of f (x, y, z ) = x2 + y 2 + z 2 − xy − yz − zx over all triples (x, y, z ) of positive integers for which 2010 divides f (x, y, z ). Proposed by Titu Andreescu, University of Texas at Dallas, USA Solution by Daniel Lasaosa, Universidad P´blica de Navarra, Spain u If wlog z is odd and x, y are even, then z 2 is the only odd term, f (x, y, z ) is odd, hence not a multiple of 2010, while if wlog x, y are odd and z is even, then x2 , y 2 , xy are the only odd terms, f (x, y, z ) is again odd. Therefore, x, y, z have the same parity, and we may deﬁne u = x−y , 2 v = y−z , or 2 3s2 + d2 = 4(u2 + v 2 + uv ) = (x − y )2 + (y − z )2 + (x − y )(y − z ) = f (x, y, z ), where s = u + v and d = u − v , and if 2010 divides f (x, y, z ), then 4020 = 22 · 3 · 5 · 67 divides f (x, y, z ). Now, any perfect square leaves a remainder equal to −1, 0, 1 modulus 5, hence if d, s are not both multiples of 5, then 3s2 + d2 cannot be a multiple of 5, hence 52 divides 3s2 + d2 = f (x, y, z ), and 20100 divides f (x, y, z ). Deﬁning s = 5 and d = d , we ﬁnd that s 5 f (x, y, z ) 20100k 3s 2 + d 2 = = = 804k. 25 25 But taking s = 16, d = 6, we ﬁnd 3s 2 + d 2 = 768 + 36 = 804, or f (x, y, z ) ≥ 25 · 804 = 20100, with equality for example for s = 80 and d = 30, ie u = 55 and v = 25, or f (z +160, z +50, z ) = 20100 for all positive integer z as it is easily checked by direct calculation. Note: We have restricted ourselves to positive values of f (x, y, z ), since clearly f (x, x, x) = 0 is a multiple of 2010 for all positive integer x, making the problem trivial. 19 Mathematical Reflections 4 (2010)
U164. Prove that ϕ 22010! − 1 ends in at least 499 zeros. Proposed by Dorin Andrica, “Babe¸-Bolyai University”, Cluj-Napoca, Romania s Solution by G.R.A.20 Math Problems Group, Roma, Italy We will prove that ϕ(22010! − 1) ends with 501 zeros by showing that it is divisible by 5501 and 2501 . Since ∞ ∞ 2010 2010 = 501 and = 2002, 5k 2k k=1 k=1 5501 22002 it follows that 2010! = 4 · ·a= · b for some positive integers a and b. By Euler’s Theorem 501 ·a 502 22010! − 1 = 24·5 − 1 = (2a )ϕ(5 (mod 5502 ) )−1≡0 which implies that 5501 divides ϕ(22010! − 1). k Now we show that 2k divides ϕ(22 ·b − 1) for all k ≥ 1. For k = 1 we have that 22·b − 1 is odd k−1 k−1 and 2 divides ϕ(22·b − 1). Moreover, since gcd(22 ·b − 1, 22 ·b + 1) = 1, k−1 ·b k−1 ·b k ·b ϕ(22 − 1) = ϕ(22 − 1) · ϕ(22 + 1) k−1 k−1 ·b and 2k−1 divides ϕ(22 ·b − 1) by inductive hypothesis and 2 divides ϕ(22 + 1) because k−1 22 ·b + 1 is odd. Hence 22002 divides ϕ(22010! − 1). Also solved by Daniel Lasaosa, Universidad P´blica de Navarra, Spain; Neacsu Adrian, Pitesti, u Romania. 20 Mathematical Reflections 4 (2010)