Junior problems
J361. Solve Solve in positive positive integers integers the equation x2 8x
− y = y . − y2 x
Proposed by Titu Andreescu, University of Texas at Dallas, USA Solution by Alan Yan, Princeton Junction, NJ, USA Let gcd( Let gcd(x, x, y ) = g and g and x x = = ga ga,, y = g = gbb where gc where gcd d (a, b) = 1. Substituting, the equation simplifies to ga to ga 3 + gb 3 = 9ab. Taking ab. Taking modulo a and since a, b are relatively prime, we must have a g. By symmetry, symmetry, we have have b g. So we can write g = k = kab ab for for some positive integer k. So we have
|
kab( kab (a3 + b3 ) = 9ab 9 ab =
⇒
|
k (a3 + b3 ) = 9.
Note that a3 + b3 2, Obviously a3 + b3 = 3 has no solutions, so a3 + b3 = 2 , so a3 + b3 can only be 3 or 9. Obviously 9 = a = 2, b = 1 or a = 1, b = 2. Then Then k = 1, g = 2. So we have have the solution solutionss (4, (4, 2), 2), (2, (2, 4) but 4) but since 2 8x y = 0, the second ordered pair is extraneous. Thus, the only solution is (4 ( 4, 2). 2).
⇒ −
≥
Also solved by Arkady Alt, San Jose, CA, USA; Adnan Ali, Student in A.E.C.S-4, Mumbai, India; Georgios Tsapakidis, High School "Panagia Prousiotissa", Agrinio, Greece; Polyahedra, Polk State College, FL, USA; Michael Tang, USA; David E. Manes, Oneonta, NY, USA; Vincelot Ravoson, France, Paris, Lycée Henri IV; Arpon Basu, Mumbai, India; Nikos Kalapodis, Patras, Greece; Problem Solving Group, Department of Financial and Management Engineering, University of the Aegean; Brian Bradie, Department of Mathematics, Christopher Newport University, Newport News, VA; Yong Xi Wang, Affiliated High School of Shanxi University; Joel Schlosberg, Bayside, NY; Jorge Ledesma, Faculty of Sciences UNAM, Mexico City, Mexico; Alessandro Ventullo, Milan, Italy; Tolibjon Ismoilov, academic lyceum named after S.H.Sirojiddinov, Tashkent, Uzbekistan; Daniel Lasaosa, Pamplona, Spain; Paul Revenant, Lycee, Champollion, Grenoble, France; Nermin Hodzic, University of Tuzla at Tuzla, Bosnia and Herzegovina; Daniel López-Aguayo, Centro de Ciencias Matemáticas UNAM, Morelia, Mexico; Dimitris Avramidis, Evaggeliki Gymnasium, Athens, Greece; ce; Nicuşor Nicuşor Zlota Zlota ‚”T ‚”Tra raian ian Vuia” Technic echnical al College, College, Focşani, cşani, Romania; omania; Nishant Nishant Dhankhar, Dhankhar, Delhi, Delhi, India; India; Adithya Bhaskar, Mumbai, India; Albert Stadler, Herrliberg, Switzerland; Cherlyse Alexander-Reid, College at Brockpo Brockport, rt, SUNY; Joachim Studnia, Lycee Lycee Condorcet, Condorcet, Paris, France; France; Petros Petros Panigyrakis, Panigyrakis, Evaggeliki Gymnasium, Athens, Greece; WSA, L.T ”Orizont”, Moldova; Hyun Min Victoria Woo, Northfield Mount Hermon School, Mount Hermon, MA; Duy Quan Tran, University of Health Science, Ho Chi Minh City, Vietnam; Neculai Stanciu and Titu Zvonaru, Romania.
Mathematical Mathematical Reflections
6 (2015)
1
J362. Let a,b Let a,b,, c, d be real numbers such that abcd = abcd = 1. Prove that the following inequality holds: ab + bc + cd + da
≤ a12 + b12 + c12 + d12
Proposed by Mircea Becheanu, University of Bucharest, România Solution by Henry Ricardo, New York Math Circle, Tappan, NY, USA We use the Cauchy-Schwarz inequality to see that 1/2
ab + bc + cd + da = da =
ab = ab =
cyclic
=
cyclic
cyclic
abcd = cd
cyclic
· ≤
1 1 c d
cyclic
1 c2
cyclic
1/2
1 d2
1 1 1 1 1 = + + + . c2 a2 b2 c2 d2
Equality holds if and only if a a = b = b = = c c = = d d = = 1.
Also solved by Panigyraki Chrysoula, Evaggeliki Gymnasium, Athens, Greece; Hyun Min Victoria Woo, Northfield Mount Hermon School, Mount Hermon, MA; Adithya Bhaskar, Mumbai, India; Albert Stadler, Herrliberg, Switzerland; Rajarshi Kanta Ghosh, Calcutta Boy’s School, Kolkata, India; Nishant Dhankhar, Delhi, India; Joachim Studnia, Lycee Condorcet, Paris, France; WSA, L.T ”Orizont”, Moldova; Nicuşor Zlota ‚”Traian Vuia” Technical College, Focşani, Romania; Alok Kumar, Delhi, India; Mamedov Shatlyk, School of Young Physics and Maths No. 21. Dashoguz, Turkmenistan; Neculai Stanciu and Titu Zvonaru, Romania; Duy Quan Tran, University of Health Science, Ho Chi Minh City, Vietnam; Brian Bradie, Department of Mathematics, Christopher Newport University, Newport News, VA; Alan Yan, Princeton Junction, New Jersey, United States of America; Adnan Ali, Student in A.E.C.S-4, Mumbai, India; Paolo Perfetti, Università degli studi di Tor Vergata, Roma, Italy; Ángel Plaza, Department of Mathematics, University of Las Palmas de Gran Canaria, Spain; Joel Schlosberg, Bayside, NY; Tan Qi Huan, Universiti Sains Malaysia, Malaysia; Yong Xi Wang, Affiliated High School of Shanxi University; Vincelot Ravoson, France, Paris, Lycée Henri IV; Nermin Hodzic, University of Tuzla at Tuzla, Bosnia and Herzegovina; Alessandro Ventullo, Milan, Italy; Daniel Lasaosa, Pamplona, Spain; Jorge Ledesma, Faculty of Sciences UNAM, Mexico City, Mexico; Eeshan Banerjee, West Bengal, India; Tolibjon Ismoilov, academic lyceum S.H.Sirojiddinov, Tashkent, Uzbekistan; Francisco Javier Martínez Aguianga, student at UCM, Madrid, Spain; Michael Tang, USA; Nikos Kalapodis, Patras, Greece; Polyahedra, Polk State College, FL, USA; Georgios Tsapakidis, High School "Panagia Prousiotissa", Agrinio, Greece; David E. Manes, Oneonta, NY, USA.
Mathematical Mathematical Reflections
6 (2015)
2
J363. Solve Solve in integers integers the system of equations x2 + y2
− z(x + y) = 10 y2 + z 2 − x(y + z ) = 6 z 2 + x2 − y(z + x) = −2
Proposed by Titu Andreescu, University of Texas at Dallas, USA Solution by Michael Tang, USA Adding the equations, we get 2x2 + 2y 2y 2 + 2z 2z 2
zx = 14 =⇒ (x − y )2 + (y (y − z )2 + (z (z − x)2 = 14 14.. − 2xy − 2yz − 2zx = Thus, |x − y |, |y − z |, |z − x| ≤ 3. 3 . Now subtracting the second equation from the first equation, we get x2 − z 2 − yz + yz + xy = xy = 4 =⇒ (x − z )(x )(x + y + z ) = 4. Subtracting the third equation from the second equation, we get y2
− x2 − xz + xz + yz = yz = 8 =⇒ (y − x)(x )(x + y + z ) = 8. Thus, y − x = 2(x 2(x − z) z ) and y − x | 8, x − z | 4. Alon Alongg with with the fac factt | y − x |, |z − x | ≤ 3, this forces (y − x, x − z ) = (2, (2, 1) or (−2, −1). 1). If (y − x, x − z ) = (2, (2, 1), 1), then x + y + y + + z z = 4, and solving this system gives ( gives (x,y,z x,y,z)) = (1, (1, 3, 0), 0), which satisfies the equations. If (y ( y − x, x − z ) = (−2, −1), 1), then x then x + y + z = −4, and solving this system gives (x,y,z) x,y,z ) = (−1, −3, 0), 0), which which satisfies satisfies the equations. equations. Therefore, Therefore, the two solutions are ( are (x,y,z x,y,z)) = (1, (1, 3, 0) and 0) and (x,y,z ( x,y,z)) = (−1, −3, 0). 0). Also solved by Arkady Alt, San Jose, CA, USA; Dimitris Avramidis, Evaggeliki Gymnasium, Athens, Greec Greece; e; Joachim Joachim Studnia, Lycee Lycee Condorcet, Condorcet, Paris, France; rance; Panigyraki Panigyraki Chrysoula, Evaggeliki Gymnasium, Athens, Greece; Cherlyse Alexander-Reid, College at Brockport, SUNY; Hyun Min Victoria Woo, North field Mount Hermon School, School, Mount Hermon, MA; Albert Albert Stadler, Herrliberg, Herrliberg, Switzerland; Adithya Adithya Bhaskar, Mumbai, India; Daniel López-Aguayo, Centro de Ciencias Matemáticas UNAM, Morelia, Mexico; Polyahedra, Polk State College, FL, USA; Nermin Hodzic, University of Tuzla at Tuzla, Bosnia and Herzegovina; David E. Manes, Oneonta, NY, USA; Vincelot Ravoson, France, Paris, Lycée Henri IV; Adnan Ali, Student in A.E.C.S-4, Mumbai, India; Alan Yan, Princeton Junction, New Jersey, United States of America; Daniel Lasaosa Lasaosa,, Pamplona, Pamplona, Spain; Spain; Yong Xi Wang, Affiliate Affiliatedd High School School of Shanxi Shanxi Universit University; y; Neculai Neculai Stanciu and Titu Zvonaru, Romania; Alessandro Ventullo, Milan, Italy; Jorge Ledesma, Faculty of Sciences UNAM, Mexico City, Mexico.
Mathematical Mathematical Reflections
6 (2015)
3
J364. Consider Consider a triangle triangle ABC ABC with with circumcircle ω circumcircle ω.. Let O Let O be be the center of ω and ω and let D let D,E, ,E, F be be the midpoints of minor arcs BC,CA,AB respectiv respectively ely.. Let DO intersect ω again at a point A . Defin Definee B and C similarly. Prove that [ABC ] ABC ] 1. 1 . [A B C ]
≤
Note that [X ] denotes the area of figure X . X .
Proposed by Taimur Khalid, Coral Academy of Science, Las Vegas, USA Solution by Daniel Lasaosa, Pamplona, Spain Note that A B C and DE are the result of a reflection upon O, hence their areas are the same. We will DEF F are show that the proposed result is true if if D,E,F are taken as the midpoints of arcs BC,CA,AB which do not contain A contain A,B, ,B, C , the result result being otherwise otherwise false for obtuse triangles. triangles. If ABC is ABC is rectangular wlog at A, then there is no minor arc BC , BC , and again the result is true iff we chose the arc BC which BC which does not contain A. We do so by considering three cases:
Case 1: If ABC is ABC is acute-angled, the minor arc BC and BC and the arc B arc B C which which does not contain A coincide. In this case, and since OD,OE since OD,OE are are the perpendicular bisectors of BC of BC,, CA, CA, we have ∠DOE = DOE = ∠DOC + + ∠COE CO E = =
1 1 ∠BOC + BOC + ∠COA CO A = A = A + B, 2 2
and similarly for the other angles, or [A B C ] = [DE DEF F ]] = [DOE [ DOE ] + [EOF [ EOF ]] + [F [ F OD] OD ] = R2 = (sin(A (sin(A + B ) + sin(B sin(B + C ) + sin(C sin(C + + A)) , 2 where R where R is the radius of ω. Similarly R2 [ABC ] ABC ] = [AOB] AOB ] + [BO [ BOC C ] + [CO [ COA A] = (sin(2A (sin(2A) + sin(2B sin(2B ) + sin(2C sin(2C )) )) , 2 and the conclusion follows by trivial application of Jensen’s inequality because the sine function is strictly concave in (0◦ , 180◦ ), equality holding iff A A = B = B = = C , C , ie iff ABC AB C is equilateral. AB C is is either right- or obtuse-angled, wlog at A, and D and D , E , F are are taken as the midpoints of Case 2: If AB arcs BC,CA,AB arcs BC,CA,AB which which do not contain A, A, B,C , then DOE = ∠DOC + + ∠COE CO E = = ∠DOE = and similarly for
OD , ∠F OD,
whereas
EO F is is ∠EOF
180◦
1 BOC + ∠COA CO A = A = A + B, − 21 ∠BOC + 2
calculated as in Case 1, yielding again
R2 [A B C ] = (sin(A (sin(A + B ) + sin(B sin(B + C ) + sin(C sin(C + + A)) , 2
whereas
2
[ABC ] ABC ] = [AOB] AOB ]
R − [BOC ] BOC ] + [CO [ COA A] = (sin(2B (sin(2B ) + sin(2C sin(2C ) − sin(360◦ − 2A)) = 2 =
R2 (sin(2A (sin(2A) + sin(2B sin(2B ) + sin(2C sin(2C )) )) . 2
Now, sin(2A sin(2A) 0 because A is right or obtuse, while 12 sin(2B sin(2B ) = sin B cos B < sin B = sin(A sin(A + C + C )), and 1 1 1 similarly 2 sin(2C sin(2C ) < sin(A sin(A + B + B)). Applying Applying finally Jensen’s Jensen’s inequality inequality we obtain 2 sin(2B sin(2B ) + 2 sin(2C sin(2C ) sin(B sin(B + C ), or the proposed inequality holds strictly in this case.
≤
Mathematical Mathematical Reflections
≤
6 (2015)
4
Case 3: If ABC ABC is obtuse-angled, wlog at A, and D , E , F are taken as the midpoints of minor arcs if ABC is right-angled, and D is taken as the midpoint of arc B C which which contains A, BC,CA,AB, BC,CA,AB , or if ABC A , then C BO D = ∠COD CO D = B = B + C , ∠COE CO E = B, B , and ∠DOE = DOE = ∠COD CO D ∠COE CO E = C , C , or ∠DF E = = 2 , and similarly ∠BOD B DEF F = 2 . Therefore, ∠DE
−
[A B C ] = [DE DEF F ]] = 2R2 sin
B C B + B + C B C A sin sin = 2R2 sin sin cos 2 2 2 2 2 2
whereas [ABC ] ABC ] = 2R2 sin A sin B sin C , or [ABC ] ABC ] A B C 2 A = 8sin cos cos = 4sin + 2 cos cos B + 2 cos cos C > 1, 1 , [A B C ] 2 2 2 2 since as B as B + + C 90◦ , either cos B or cos or cos C is C is at least cos45 least cos45◦ =
≤ ≤
√ 2 2
> 12 .
We conclude that the proposed inequality is true, with equality iff AB iff ABC C is is equilateral, as long at D at D,, E,F are defined as the midpoints of arcs BC,CA,AB which do not contain A, A, B,C . It is fals falsee when when ABC is obtuse if D, D, E,F are are defined as the midpoints of minor arcs BC,CA,AB. BC,CA,AB .
Also solved by Polyahe P olyahedra, dra, Polk State College, FL, USA; Tolibjon Ismoilov, academic lyceum S.H.Sirojiddinov, S.H.Sirojiddinov, Tashkent, Uzbekistan; Nermin Hodzic, University of Tuzla at Tuzla, Bosnia and Herzegovina; Nikos Kalapodis, podis, Patras, Patras, Gree Greece ce;; Adnan Ali, Ali, Student Student in A.E.C.S-4 A.E.C.S-4,, Mumbai, Mumbai, India; India; Joel Joel Schlosb Schlosber erg, g, Bayside, Bayside, NY; Alan Yan, Princeton Junction, New Jersey, United States of America; Neculai Stanciu and Titu Zvonaru, Romania; Hyun Min Victoria Woo, Northfield Mount Hermon School, Mount Hermon, MA, USA.
Mathematical Mathematical Reflections
6 (2015)
5
J365. Let x1 , x2 , , xn be nonnegative real numbers such that x1 + x + x2 + + xn = 1. Find the minimum possible value of x1 + 1 + 2x2 + 1 + + nxn + 1. 1.
···
√
√
· · · √
···
Proposed by Nguyen Viet Hung, Hanoi University of Science, Vietnam Solution by Daniel Lasaosa, Pamplona, Spain Claim: Let u, v ≥ 0 and k ≥ 2 a positive integer. Then,
√ u + 1 + √ kv + √ kv + 1 ≥ u + v + 1 + 1, 1,
with equality iff v v = 0. Proof: Squaring both sides yields the equivalent inequality (k
− 1)v 1)v + 2
Now,
(u + 1)(kv 1)(kv + + 1)
≥ 2√ u + v + 1.1.
(u + 1)(kv 1)(kv + + 1)
1)v ≥ 0, 0 , − u − v − 1 = (ku + k − 1)v with equality iff v v = 0, and clearly ( clearly (k k − 1)v 1)v ≥ 0 with equality iff v v = 0. The Claim follows. By trivial induction over n, we can now prove that
√ x + 1 + √ 2x + 1 + · · · + √ nx + 1 ≥ √ 2 + n − 1, 1
2
n
with equality iff x1 = 1 and x 2 = x 3 = = x = x n = 0. Indeed, the result for n = 2 is equivalent to the Claim with k with k = 2. If the result is true for n 1, note that by the Claim with k with k = n = n we have
··· −
√ x + 1 + √ nx + 1 ≥ 1
n
(x1 + xn ) + 1 + 1, 1,
with equality iff x x n = 0, or renaming x renaming x 1 + xn as x as x 1 and applying the hypothesis of induction, the conclusion follows.
Also solved by Hyun Min Victoria Woo, Northfield Mount Hermon School, Mount Hermon, MA; Adithya Bhaskar, Mumbai, India; Neculai Stanciu and Titu Zvonaru, Romania; Ángel Plaza, Department of Mathematics, University of Las Palmas de Gran Canaria, Spain; Adnan Ali, Student in A.E.C.S-4, Mumbai, India; Paul Revenant, Lycee, Champollion, Grenoble, France; Nermin Hodzic, University of Tuzla at Tuzla, Bosnia and Herzegovina; Polyahedra, Polk State College, FL, USA; Joel Schlosberg, Bayside, NY, USA.
Mathematical Mathematical Reflections
6 (2015)
6
J366. Prove Prove that in any triangle AB triangle ABC C , A B C sin + sin + sin 2 2 2
≤
r 2R
6+
− 1.
Proposed by Florin Stănescu, Găesti, România Solution by Polyahedra, Polk State College, FL, USA By squaring, we see that the inequality is equivalent to 6+ Let x Let x = = s s
r 2R
≥ 1 +
sin2
cyclic
A +2 2
sin
cyclic
A +2 2
cyclic
= s − b, and z and z = s − c. Then it is well known that − a, y = s r A A (s − b)(s )(s − c) =1− sin2 , sin = =
2R
2
2
yz + (z + x)(x )(x + y)
cyclic
sin
bc
A B sin . 2 2
yz . (z + x)(x )(x + y)
Hence, it suffices to show that 3
≥
cyclic
cyclic
yz + (z + x)(x )(x + y)
z x+y
xy , (y + z )(z )(z + x)
cyclic
which, upon multiplying by (x + y)(y )(y + z )(z )(z + x), becomes 3(x 3(x + y)(y )(y + z )(z )(z + x)
≥
cyclic
Now 3(x 3(x + y)(y )(y + z )(z )(z + x) and
xy( xy(y + z )(z )(z + x).
cyclic
≤
−
xy( xy (y + z )(z )(z + x)
cyclic
yz( yz (y + z ) + (x ( x + y + z )
yz( yz (y + z ) = 2(x 2(x + y + z )(xy )(xy + + yz + yz + zx) zx),
cyclic
3
xy( xy (y + z)(z )(z + x)
2( xy + + yz + yz + zx) zx ), ≤ 2(xy
cyclic
where the last inequality follows from 4(xy 4(xy + + yz + yz + zx) zx )2
−3
cyclic
xy( xy(y + z )(z )(z + x) =
1 (xy 2
yz )2 + (yz (yz − zx) zx )2 + (zx (zx − xy) xy )2 ≥ 0. 0 . − yz)
This completes the proof.
Also solved by Arkady Alt, San Jose, CA, USA; Nicusor Zlota ‚”Traian Vuia” Technical College, Focsani, Romania; WSA, L.T ”Orizont”, Moldova; Bobojonova Latofat, academic lycuem S.H.Sirojiddinov, Tashkent, Uzbekistan; Yong Xi Wang, Affiliated High School of Shanxi University; Adnan Ali, Student in A.E.C.S-4, Mumbai, India; Nikos Kalapodis, Patras, Greece; Nermin Hodzic, University of Tuzla at Tuzla, Bosnia and Herzegovina; Scott H. Brown, Auburn University Montgomery, AL, USA.
Mathematical Mathematical Reflections
6 (2015)
7
Senior problems problems
S361. Find all integers integers n n for which there are integers a integers a and b such that ( that (a a + bi) bi)4 = n + 2016i 2016i.
Proposed by Titu Andreescu, University of Texas at Dallas, USA Solution by Adnan Ali, Student in A.E.C.S-4, Mumbai, India Notice that by equating the real and imaginary parts we get, a4 + b4
− 6a2b2 = n
and 4a3 b
ab(a + b)(a )(a − b) = 2016. 2016. − 4ab3 = 4ab( So, we need integer pairs ( pairs (a, a, b) such that ab that ab((a + b)(a )(a − b) = 504. 504. Notice that if (a, ( a, b) is a solution, then so are (−a, −b), (b, −a), (−b, a). So, we’ll assume W.L.O.G that a that a > b > 0 > 0.. Let d Let d = = a a − b. Then db Then db((d + b)(d )(d + 2b) = 504.. Notice that the highest power of three dividing 504 504 dividing 504 is is 2 2.. So clearly, exactly one of the four factors must be divisible by 9 by 9.. Observe that db that db((d + b)(d )(d + 2b) = 504 < 504 < 546 546 = 1 · 6 · (1+6) · (1+12). (1+12). Thus b Thus b ≤ 5. 5 . Similarly, db( db(d + b + b)( )(d d + 2b) = 504 = 7 · 1 · (7 + 1) · (7 + 2). 2). Thus Thus d ≤ 7. It is eviden evidentt that that neith neither er d nor b can be a multiple of 9. So, eith either er d + b + b = 9 or d + 2b = 9 (because d + 2b ≤ 7 + 2 · 5 = 17 < 18 18). ). So, the possib possible le pairs are (7, (7, 2), 2), (5, (5, 4), 4), (4, (4, 5) for (d, b) when d + b + b = = 9. But 5 50 5044 and so we reject reject the last two two pairs. pairs. It is seen that for (d, b) = (7, (7, 2), 2), d − b = 5, but clearly 5 504 504.. So, no solutio solutions ns for the case d + b + b = = 9. Mo Movin vingg
on, for d for d + 2b = 9, we have the following possible pairs (7 pairs (7,, 1), 1), (5, (5, 2), 2), (1, (1, 4). 4). Since 5 Since 5 504, 504, we reject the middle pair, while d + b + b = 5 for (d, b) = (1, (1, 4) and 4) and since 5 504 504,, we reject reject the last last pair. It is clear that that the first pair (7, (7, 1) satis 1) satisfies fies the equation. equation. This gives gives us (a, b) = (8, (8, 1) and 1) and so in summary the possible solutions are (a, b) = (8, (8, 1), 1), ( 8, 1), 1), ( 1, 8), 8), (1, (1, 8). 8). Hence the only possible value of n n is 84 + 14 6(8 1)2 = 3713. 3713.
− − −
−
−
·
Also solved by Adithya Bhaskar, Mumbai, India; Albert Stadler, Herrliberg, Switzerland; Joel Schlosberg, Bayside, NY; Jorge Ledesma, Faculty of Sciences UNAM, Mexico City, Mexico; Khurshid Juraev, academic lyceum lyceum S.H.Sirojiddinov, S.H.Sirojiddinov, Tashkent, Uzbekistan; Daniel Lasaosa, Pamplona, Spain; Daniel López-Agua López-Aguayo, yo, Centro de Ciencias Matemáticas UNAM, Morelia, Mexico; Nermin Hodzic, University of Tuzla at Tuzla, Bosnia and Herzegovina; Li Zhou, Polk State College, Winter Haven, FL, USA; Alessandro Ventullo, Milan, Italy.
Mathematical Mathematical Reflections
6 (2015)
8
S362. Let 0 Let 0 < a, a, b, c, d
≤ 1. 1. Prove that 1 a+b+c+d
≥ 14 + 64 (1 − a)(1 − b)(1 − c)(1 − d). 27
Proposed by Proposed by An Zhen-ping, Xianyang Normal University, China Solution by Brian Bradie, Christopher Newport University, Newport News, VA, USA By the arithmetic mean - geometric mean inequality, (1
− a)(1 − b)(1 − c)(1 − d)
≤ − 4
a
−b−c−d 4
4
,
so it suffices to show that
1 1 1 + (4 a+b+c+d 4 108 Let x Let x = = a a + b + c + d. Then we must show that
≥
1 x for 0 for 0 < x
− a − b − c − d)4.
1 ≥ 41 + 108 (4 − x)4 ,
4. This is equivalent to ≤ 4. (x
− 1)2(x − 4)
(x
0 , − 5)2 + 2 ≤ 0,
which is clearly true for 0 for 0 < < x 4. 4. Equality Equality holds holds for x for x = = 1 and x and x = = 4, which translates to a to a = = b b = = c c = = d d = = and a = b = b = = c c = = d d = = 1, respectively.
≤
1 4
Also solved by Hyun Min Victoria Woo, Northfield Mount Hermon School, Mount Hermon, MA; Yong Xi Wang, Affiliated High School of Shanxi University; Adnan Ali, Student in A.E.C.S-4, Mumbai, India; Li Zhou, Polk State College, Winter Haven, FL, USA; Jorge Ledesma, Faculty of Sciences UNAM, Mexico City, Mexico; Daniel Lasaosa, Pamplona, Spain; Nermin Hodzic, University of Tuzla at Tuzla, Bosnia and Herzegovina; Nicuşor Zlota‚ ”Traian Vuia” Technical College, Focşani, Romania; Paolo Perfetti, Università degli studi di Tor Vergata, Roma, Italy.
Mathematical Mathematical Reflections
6 (2015)
9
S363. Determine Determine if there are distinct positive positive integers integers n n 1 , n2 ,...,nk−1 such that (3n (3n21 + 4n 4 n22 + ... + (k (k + 1)n 1) nk2 −1 )3 = 2016(n 2016(n31 + n32 + ... + nk3 −1 )2
Proposed by Titu Andreescu, University of Texas at Dallas, USA Solution by Nermin Hodzic, University of Tuzla at Tuzla, Bosnia and Herzegovina Using Holder’s inequality we have
33 + 43 + ... + (k (k + 1) 3
n31 + n32 + ... + nk3 −1
·
33 + 43 + ... + (k (k + 1) 3
(k + 1)(k 1)(k + 2) 2
2
≥
2
≥
3n21 + 4n 4 n22 + ... + (k (k + 1)n 1) nk2 −1
3
= 2016(n 2016(n31 +...+ ...+nk3 −1 )2
⇒
(k + 1) 3 ≥ 2025 ⇔ ≥ 2016 ⇔ 13 + ... + (k 1)(k + 2) ⇔ (k + 1)(k ≥ 45 ⇒ k ≥ 8 2
2025
Equality holds if and only if k k = 8, n1 = 3, n2 = 4,...,n7 = 9
Also Also solved solved by Hyun Min Victoria Victoria Woo, Woo, Northfield Northfield Mount Hermon School, School, Mount Hermon, Hermon, MA; Li Zhou, Polk State College, Winter Haven, FL, USA; Jorge Ledesma, Faculty of Sciences UNAM, Mexico City, Mexico.
Mathematical Mathematical Reflections
6 (2015)
10
S364. Let a Let a,, b, c be nonnegative real numbers such that a a b c + + b+c c+a a+b
≥ 1 ≥ b ≥ c and a and a + b + c = 3. Prove that 2
2
2
2(a 2(a + b + c ) 5 ≥ 3(ab + . 3(ab + bc + ca) ca) 6
Proposed by Marius Stănean, Zalău, România Solution by Hyun Min Victoria Woo, Northfield Mount Hermon School, Mount Hermon, MA, USA Assign variables s = ab = ab + bc + ca and ca and p = abc. = abc. So, So, we have to prove the inequality 3 p
2(a + b + c)s + (a ( a + b + c) 3 2(a 2(a + b + c)2 − 4s 5 − 2(a + , ≥ (a + b + c)s − p 3s 6
which is equivalent to
3 p
− 6s + 27 ≥ 18 − 4s + 5 . 3s − p 3s 6 9s(s − 2) After simplifying, this inequality becomes p ≥ . Since a ≥ 1 ≥ b ≥ c, we have that (a − 1)(1 − 5s + 12 b)(1 − c) ≥ 0. 0. This implies that a + b + c − 1 − ab − bc − ca + abc ≥ 0 ⇔ p ≥ s − 2. If s < 2, then then the result result is triviall trivially y true. true. Otherw Otherwise ise,, 3s = 3(ab 3(ab + + ac + bc) bc ) ≤ (a + b + c) c )2 = 9, hence 9s(s − 2) s ≤ 3 ⇒ 5s 5 s + 12 ≥ 9s 9s, so p so p ≥ s − 2 ≥ . 5s + 12 Also solved by Adnan Ali, Student in A.E.C.S-4, Mumbai, India; Nermin Hodzic, University of Tuzla at Tuzla, Bosnia and Herzegovina; Paolo Perfetti, Università degli studi di Tor Vergata, Roma, Italy; Daniel Lasaosa, Pamplona, Spain.
Mathematical Mathematical Reflections
6 (2015)
11
S365. Let (k 2 + 1)2 ak = , k4 + 4
k = 1, 2, 3,
···
Prove that for every positive integer n integer n,, an1 an2 1 an3 2
−
−
···
2n+1 an = 2 . n + 2n 2n + 2
Proposed by Nguyen Viet Hung, Hanoi University of Science, Vietnam Solution by Brian Bradie, Christopher Newport University, Newport News, VA, USA Let c Let c k = k 2 + 1 and 1 and d k = k 4 + 1. 1. Then ck−1 ck+1 = =
− 1)2 + 1 (k + 1)2 + 1 (k 2 − 2k + 2)(k 2)(k 2 + 2k 2k + 2) = k = k 4 + 4 = d k ,
(k
so
ck2 ak = . ck−1 ck+1
Then, for any positive integer n integer n,,
n
an1 an2 1 an3 2
−
−
· · · an
= =
n
·
akn+1 k
− =
k=1 2n (n 1) c1 cn0
− −
+1 k cn+1 k ckn+1 k=1 k 1 n ck2n+2 2k 1 cn+2 k ckn k cn+1 k=2 k
=
cn1 +1 n−1 1 1 cn0 cn+1
=
cn1 +1 cn0 cn+1
·
ck2n+2−k
− −
−
−
−
−
·
·
With c0 = 1, c1 = 2, and c and c n+1 = (n + 1)2 + 1 = n2 + 2n 2n + 2, 2, it follows that an1 an2 −1 an3 −2
n+1
· · · an = n2 +2 2n . 2n + 2
Also solved by Arkady Alt, San Jose, CA, USA; Hyun Min Victoria Woo, Northfield Mount Hermon School, Mount Hermon, MA; Adithya Bhaskar, Mumbai, India; Li Zhou, Polk State College, Winter Haven, FL, USA; Adnan Ali, Student in A.E.C.S-4, Mumbai, India; Ángel Plaza, Department of Mathematics, University of Las Palmas de Gran Canaria, Spain; G. C. Greubel, Newport News, VA; Joel Schlosberg, Bayside, NY; Jorge Ledesma, Faculty of Sciences UNAM, Mexico City, Mexico; Daniel Lasaosa, Pamplona, Spain; Moubinool Omarjee Lycée Henri IV, Paris, France; Nermin Hodzic, University of Tuzla at Tuzla, Bosnia and Herzegovina; Mamedov Mamedov Shatlyk, School of Young Physics and Maths No. 21. Dashoguz, Dashoguz, Turkmenistan; Turkmenistan; Albert Stadler, Herrliberg, Switzerland; Neculai Stanciu and Titu Zvonaru, Romania; Alessandro Ventullo, Milan, Italy; WSA, L.T ”Orizont”, Moldova.
Mathematical Mathematical Reflections
6 (2015)
12
S366. Let a,b Let a,b,, c, d be positive real numbers such that a + b + c + d = 4. Prove that 1 9+ 6
1 1 1 1 + + + a b c d
2
≥
70 . ab + bc + cd + da + ac + bd
Proposed by Marius Stănean, Zalău, România Solution by Nermin Hodzic, University of Tuzla at Tuzla, Bosnia and Herzegovina Let a Let a + b + c + d = 4 p, ab + ac + ad + bc + bd + cd = cd = 6q 2 ,abc + acd + bcd + cda = cda = 4r3 , abcd abcd = = s s 4 1 1 1 1 1 9 + ( + + + )2 6 a b c d 1 2 4r3 2 9 + p ( 4 ) 6 s
· ·
≥
≥ ab + ac + ad70+ bc + bd + cd ⇔
70 70 p p2 6q 2
27 q 2 s8 + 8 p 8 p2 · q 2 · r6 ≥ 35 p 35 p2 s8 ⇔ 27q
Using Newton’s inequalities, we have q 2
2
1
3
3
q s ⇒ q 4 ≥ p2 r 2 ≥ p2 qs ⇒ q 3 ≥ p 2 s ⇒ q ≥ ≥ pr, ∧r2 ≥ qs ≥ p · s r2
2
4
1
2
3
3
3
3
q s ≥ p · s ⇒ r ≥ p · s ≥ qs
Using these two we get 27 27q q 2 s8 + 8 p 8 p2 q 2 r6
4
26
3
3
· · ≥ 27 · p · s
+ 8 p
·
16 3
·s
14 3
Hence it is suffices to prove 27 p
4
26
3
3
· ·s
+ 8 p
·
16 3
14
35 p2 s8 · s ≥ 35 p 3
Using AmGm inequality we have 27 p
4
26
3
3
· ·s
+ 8 p
·
Hence it is sufficies to prove 35
35
p
236 3
16 3
14
· s ≥ 35 3
35
p
236 3
·s
814 3
814
35 p2 s8 ⇔ p26 ≥ s26 · s ≥ 35 p 3
Which is an immediate consequence of AmGm inequality. Equality holds if and only if a if a = b = b = = c c = = d d = = 1.
Also solved by Daniel Lasaosa, Pamplona, Spain; Nicuşor Zlota ‚”Traian Vuia” Technical College, Focşani, Romania.
Mathematical Mathematical Reflections
6 (2015)
13
Undergraduate problems
U361. Consider Consider all possible possible ways one can assign the numbers numbers 1 1 through 10 through 10 with a nonnegative probability so that the probabilities sum to 1. Let Let X X be the number selected selected.. Suppose Suppose that E [X ]k = E [X k ] for a given integer k integer k 2. 2 . Find the number of possible ways of assigning these probabilities.
≥
Proposed by Mehtaab Sawhney, Commack High School, New York, USA Solution by Li Zhou, Polk State College, FL, USA Let p Let p 1 , p2 , . . . , p10 be the probabilities assigned to 1, 1 , 2, . . . , 10 10,, respectively. Then E [X ]k = ( p1 + 2 p 2 p2 +
10 p10 )k , · · · + 10 p
E [X k ] = 1k p1 + 2 k p2 +
· · · + 10k p10 ,
which are clearly equal if one of the p the pi ’s is 1 is 1 and and the others are 0 are 0.. We show that these are the only possibilities. p +···+9 p If p10 = 1, then then we are done. done. So conside considerr that that p10 < 1. Let Let x = p p+2 Then x 9. By the the + p +···+ p . Then k convexity of t for t for t 0, 0 , we have 1
2
1
2
9
9
≥
E [X ]k = [(1
≤
− p10)x + p10 · 10]k ≤ (1 − p10)xk + p10 · 10k ≤ E [X k ],
where the first inequality is equality if and only if p10 = 0. Thu Thus p10 = 0 and E [X ]k = xk . Repeati Repeating ng this argument, we get p9 = 1 or p9 = 0, and so on. We conclud concludee that there there are 10 10 ways ways to assign these probabilities.
Also solved by Daniel Lasaosa, Pamplona, Spain; Albert Stadler, Herrliberg, Switzerland.
Mathematical Mathematical Reflections
6 (2015)
14
U362. Let S n =
j +k q i+ j+ ,
1 i
≤
where q where q ( 1, 0)
∈ ∈ −
≤
∪ (0, (0, 1). 1). Evaluate l Evaluate lim imn→∞ S n . Proposed by Nguyen Viet Hung, Hanoi University of Science, Vietnam
Solution by Albert Stadler, Herrliberg, Switzerland lim S n =
n
=
1 1
→∞
− q 1≤i
q
1 i
≤
q i+1
i+ j+ j +k
2i+2
=
i+ j
q
1 i
≤
1 · 1q − q 2 = (1 − q )(1 )(1 − q 2 )
j +1 q j+1 1 = 1 q 1 q
· −
1 i
≤
q 3i+3 =
−
(1
j +1 q i+2 j+1 =
1 i
−
≤
q 6 )(1 q 2 )(1 q )(1
−
− q 3) .
Also solved by Daniel Lasaosa, Pamplona, Spain; Shohruh Ibragimov, National University of Uzbekistan, Tashkent, Uzbekistan; Adnan Ali, A.E.C.S-4, Mumbai, India; Li Zhou, Polk State College, FL, USA; Yong Xi Wang, High School of Shanxi University, China.
Mathematical Mathematical Reflections
6 (2015)
15
U363. Let a Let a be a positive number. Prove that there is a number θ = θ = θ((a), 1 < θ < 2, 2, such that
∞ − a j
j=0 j =0
Furthermore, prove that
a
= 2 +θ
a 1 a +1
−
a 1 a +1
≤ |
sin πa . πa
|
.
Proposed by Albert Stadler, Herrliberg, Switzerland
Solution by Li Zhou, Polk State College, FL, USA By the binomial series,
∞ − a j
j=0 j =0
Let n Let n = a and x = x = a a
θ 2
a
2a = 2
∞
a a + 2i 2i
i=1
[0, [0 , 1). 1). Then
.
− ∈∞ − − − − ··· − ··· − − − − − − ·−· · − − · · · − − − · · · − · · · · · · ··· − − − − − − − − − − −··· 1
=
a 1 a +1
a 2i a + 2i
i=1
n + x (n ( n + x)(2 x)(3 x) (n ( n + x)(2 x) (5 x) + + + n + 2 (n + 2)(n 2)(n + 3)(n 3)(n + 4) (n + 2) (n + 6) 2 x (2 x)(3 x) (2 x)(3 x)(4 x) = 1 + n+2 (n + 2)(n 2)(n + 3) (n + 2)(n 2)(n + 3)(n 3)(n + 4) (2 x) (5 x) (2 x) (6 x) + + (n + 2) (n + 5) (n + 2) (n + 6) 2 x 3 x (2 x)(3 x)(4 x) 5 x = 1 1 1 n+2 n+3 (n + 2)(n 2)(n + 3)(n 3)(n + 4) n+5 =
< 1 < 1,,
since n since n and x cannot both be 0. Thus θ Thus θ < 2. 2 . n+x 1 Next, if n n 2, 2 , then n+2 1 . 2 , so θ > 1. Now consider n consider n = 0 and 0 and 0 < < x < 1. 1 . Then
≥
≥
x x( x (x 2)(x 2)(x θ = 2 + 2! 4! (1 + 1)x + (1 1)x = x 1
−
− 3) + x( x (x − 2) · · · (x − 5) + ···
−
− 2 = 2 − 2x > 1, 1 , 1−x
−
from 2 x from 2
6!
where the inequality follows < 1 + x. It remains to consider n = 1 and 0 x < 1. 1. Then
≤
x + 1 (x ( x + 1)(x 1)(x 2)(x 2)(x 3) (x ( x + 1)(x 1)(x 2) (x θ = 4 + + 3! 5! 7! 2 (1 + 1)x+1 (1 1)x+1 2(x 2(x + 1) 4 (1 + x 2x ) = = . x(x 1) x(1 x)
−
− − −
− · · · − 5) + · · ·
−
−
− −
Let f ( f (x) = 4 (1 + x 2x ) x(1 x (1 x) x ). Then Then f (0) f (0) = f (1) f (1) = 0 and f (x) = 3 + 2x 2x 2 x+2 ln 2. No Note te tha thatt f (0) = 3 4 l n 2 > 2 > 0 0,, f (1) = 5 8 l n 2 < 2 < 0 0,, and f and f (x) has only one 0 for 0 < x < 1. 1 . Hence, for 0 < x < 1, 1 , f ( f (x) > 0 > 0 and and thus θ thus θ > 1. 1. For x For x = = 0, θ = 4(1 ln2) > ln2) > 1 1 by L’Hˆ L’Hopital’s ˆopital’s rule. Finally, for the second part, we start with sin πa = sin πx and use the well-known fact that sinππx = Γ(x Γ(x)Γ(1 x). Therefore,
−
−
−
− −
−
πa sin πx
−
a 1 a +1
completing completing the proof.
Mathematical Mathematical Reflections
6 (2015)
−
− |
|
− −
1
aΓ(a Γ(a)Γ(2 x) = = a Γ(a Γ(a + 2 x)
0
a 1
t − (1 − t)1−x dt ≤ a
1
ta−1 dt = dt = 1,
0
16
U364. Evaluate Evaluate
5x2 x 4 dx. x5 + x4 + 1
− −
Proposed by Titu Andreescu, University of Texas at Dallas, USA Solution by Alok Kumar, Delhi, India Note that P ( P (x) = x 5 + x4 + 1 factors 1 factors as x5 + x4 + 1 = (x (x2 + x + 1)(x 1)(x3
1). − x + 1).
An elegant way to see this is to observe that P that P ((w) = 0, where w where w is a primitive cube root of unity. 2 To split the numerator 5x 5 x x 4, as to render the integrand integrable, we rewrite it as:
− −
5x2
1)(5x + 4) − x − 4 = (x − 1)(5x = x 3 (5x (5x + 4) − (x3 − x + 1)(5x 1)(5x + 4) = (5x (5x4 + 4x 4x3 ) − (x3 − x + 1)(5x 1)(5x + 4)
Note that 5 that 5x x4 + 4x3 is the derivative of x x 5 + x4 + 1 with respect to x to x.. Now the stage is set for integration. integration.
5x2 x 4 dx = dx = x5 + x4 + 1
− −
(5x (5x4 + 4x 4x3 )
1)(5x + 4) − (x3 − x + 1)(5x dx
x5 + x4 + 1 5x4 + 4x 4x3 5x + 4 = dx dx x5 + x4 + 1 x2 + x + 1 1 5(2x 5(2x + 1) + 3 = ln x5 + x4 + 1 dx 2 x2 + x + 1 5 2x + 1 3 = ln x5 + x4 + 1 dx 2 2 x +x+1 2
− |− |−
| |
5
4
= ln x + x + 1
|−
= ln x5 + x4 + 1
|−
| |
5 ln x2 + x + 1 2 5 ln x2 + x + 1 2
| |
dx (x + 12 )2 + 34 3 2 2x + 1 tan−1 2 3 3 2x + 1 3tan−1 + λ, 3
−
| − · √ √ |−
√ √
where λ where λ is an arbitrary constant.
Also Also solved solved by Daniel Daniel Lasaosa Lasaosa,, Pamplona, Pamplona, Spain; Spain; Alessan Alessandr droo Ventullo, entullo, Milan, Milan, Italy; Italy; Albert Albert Stadler, Herrliberg, Herrliberg, Switzerland; Shohruh Ibragimov, Ibragimov, National University of Uzbekistan, Uzbekistan, Tashkent, Uzbekistan; Nermin Hodzic, University of Tuzla at Tuzla, Bosnia and Herzegovina; Moubinool Omarjee Lycée Henri IV , Paris France; Daniel López-Aguayo, Centro de Ciencias Matemáticas UNAM, Morelia, Mexico; Jorge Ledesma, Faculty of Sciences UNAM, Mexico City, Mexico; Joel Schlosberg, Bayside, NY, USA; Henry Ricardo, New York Math Circle, USA; G. C. Greubel, Newport News, VA, USA; David E. Manes, Oneonta, NY, USA; Cherlyse Alexander - Reid, College at Brockport, SUNY, USA; Behzod Kurbonboev, National University of Uzbekistan, Tashkent, Uzbekistan; Ángel Plaza, University of Las Palmas de Gran Canaria, Spain; Adnan Ali, A.E.C.S-4, Mumbai, India; Li Zhou, Polk State College, FL, USA; Brian Bradie, Christopher Newport University, Newport News, VA, USA; Ravoson Vincelot, Lycee Henri IV, Paris, France; Yong Xi Wang, High School of Shanxi University, China; Hyun Min Victoria Woo, Northfield Mount Hermon School, Mount Hermon, MA, USA.
Mathematical Mathematical Reflections
6 (2015)
17
U365. Let n Let n be a positive integer. Evaluate (a) (b)
n x 0 e n x 0 e
dx, dx, dx, dx,
where a denotes the integer part of a of a..
Proposed by Nguyen Viet Hung, Hanoi University of Science, Vietnam Solution by Daniel Lasaosa, Pamplona, Spain (a) Clearly, e Clearly, e x = en−1 when n when n n
n
ex dx = dx =
0
n
0
e dx = dx =
= N n
ex dx = dx =
k 1
−
k 1
n
k
− e
dx = dx =
k 1
−
k=1
ek−1 =
k=1
en e
− 1. −1
ln( N + + 1). 1). Note that ex = k = k for all l all ln( n(k k) ≤ x < ln(k ln( k + 1), 1) , or ≤ n < ln(N
N 1 x
n
k
k=1
(b) Let N = en , or l or ln( n(N N ))
∈ [n [ n − 1, n), or
− k=1
ln(k ln(k+1)
ln(k ln(k)
x
e dx +
N 1
n
x
dx = e dx =
ln N
−
ln(k ln(k+1)
k
k=1
ln(k ln(k)
n
dx + N
dx = dx =
ln N
ln(N ) − ln(N ln(N − = N n − ln(N ln(N !) !) = n en − ln (en !) . − ln(N − 1) − · · · − ln(1) = N
Also solved by Alessandro Ventullo, Milan, Italy; Albert Stadler, Herrliberg, Switzerland; Shohruh Ibragimov, National University of Uzbekistan, Tashkent, Uzbekistan; Nermin Hodzic, University of Tuzla at Tuzla, Bosnia and Herzegovina; Moubinool Omarjee Lycée Henri IV , Paris France; Daniel López-Aguayo, Centro de Ciencias Matemáticas UNAM, Morelia, Mexico; Jorge Ledesma, Faculty of Sciences UNAM, Mexico City, Mexico; Joel Schlosberg, Bayside, NY, USA; Henry Ricardo, New York Math Circle, USA; G. C. Greubel, Newport News, VA, USA; Problem Solving Group, Department of Financial and Management Engineering, University of the Aegean, Greece; Ángel Plaza, University of Las Palmas de Gran Canaria, Spain; Adnan Ali, A.E.C.S-4, Mumbai, India; Li Zhou, Polk State College, FL, USA; Brian Bradie, Christopher Newport University, Newport News, VA, USA; Ravoson Vincelot, Lycee Henri IV, Paris, France; Yong Xi Wang, High School of Shanxi University, China; Hyun Min Victoria Woo, Northfield Mount Hermon School, Mount Hermon, MA, USA; Adithya Bhaskar, Mumbai, India.
Mathematical Mathematical Reflections
6 (2015)
18
U366. If f f : [0, [0, 1]
→ R is R is a convex and integrable function with f (0) f (0) = 0, 0, prove that 1
1
≥
f ( f (x)dx
0
2
4
f ( f (x)dx.
0
Proposed by Florin Stănescu, Găeşti, România Solution by Alessandro Ventullo, Milan, Italy Since f Since f is is a convex function, we have 1
0
1 f ( f (x) dx = dx = 2
1
1
f ( f (x) dx +
0
1
f (1 f (1
0
− x) dx
≥
f ( f (x) + f (1 f (1 x) dx 2 0 1 x + (1 x) f dx 2 0 1 = f . 2 =
−
−
On the other hand, since f is f is convex and f and f (0) (0) = 0, 0, we have
1 f 2
f (0) f (0) + f =2 2
·
1 2
1
= 2
− − (1
0
≥
2
1
f (1
0
1
1 dx 2 1 x) 0 + x dx 2
x)f (0) f (0) + xf
2
= 4
·
·
f ( f (x) dx,
0
and the conclusion follows.
Also solved by Daniel Lasaosa, Pamplona, Spain; Albert Stadler, Herrliberg, Switzerland; Shohruh Ibragimov, National University of Uzbekistan, Tashkent, Uzbekistan; Nermin Hodzic, University of Tuzla at Tuzla, Bosnia and Herzegovina; Daniel López-Aguayo, Centro de Ciencias Matemáticas UNAM, Morelia, Mexico; Jorge Ledesma, Faculty of Sciences UNAM, Mexico City, Mexico; Joel Schlosberg, Bayside, NY, USA; Henry Ricardo, New York Math Circle, USA; Bekhzod Kurbonboev, National University of Uzbekistan, Tashkent, Uzbekistan; Ángel Plaza, University of Las Palmas de Gran Canaria, Spain; Li Zhou, Polk State College, FL, USA; Brian Bradie, Christopher Newport University, Newport News, VA, USA; Ravoson Vincelot, Lycee Henri IV, Paris, France; Yong Xi Wang, High School of Shanxi University, China; Hyun Min Victoria Woo, Northfield Mount Hermon School, Mount Hermon, MA, USA; Adithya Bhaskar, Mumbai, India.
Mathematical Mathematical Reflections
6 (2015)
19
Olympiad problems
O361. Determine Determine the least integer integer n n > 2 such 2 such that there are n consecutive integers whose sum of squares is a perfect square.
Proposed by Alessandro Ventullo, Milan, Italy Solution by Li Zhou, Polk State College, FL, USA Notice that 1 that 1882 + 192 +
11 . First, · · · + 282 = 772. We shall show that n cannot be < 11. (k − 1)2 + k 2 + (k (k + 1) 2 = 3k 2 + 2 ≡ 2 (mod 3), (k − 1)2 + k2 + (k (k + 1) 2 + (k (k + 2) 2 = 4k 2 + 4k 4k + 6 ≡ 2 (mod 4), (k − 2)2 + (k (k − 1)2 + · · · + (k (k + 3) 2 = 6k(k + 1) + 19 ≡ 3 (mod 4), (k − 3)2 + (k (k − 2)2 + · · · + (k (k + 4) 2 = 4[2k 4[2k (k + 1) + 11]. 11].
Because 2 Because 2 is is not a quadratic residue (mod 3 (mod 3), ), 2 2 and and 3 3 are are not quadratic residues (mod 4 (mod 4), ), and 2 and 2k k(k +1)+11 (mod 4), 4), we see that n cannot be 3, 4, 6, 8. Next,
≡ 3
− 2)2 + (k (k − 1)2 + · · · + (k (k + 2) 2 = 5(k 5(k2 + 2), 2), (k − 3)2 + (k (k − 2)2 + · · · + (k (k + 3) 2 = 7(k 7(k2 + 4), 4), (k − 4)2 + (k (k − 3)2 + · · · + (k (k + 4) 2 = 3(3k 3(3k2 + 20), 20), (k − 4)2 + (k (k − 3)2 + · · · + (k (k + 5) 2 = 5(2k 5(2k 2 + 2k 2k + 17). 17). Because k2 + 2 5), k 2 + 4 7), 20 3), and 2 and 2k k2 + 2k 2k + 17 5), we see ≡ 0 (mod 5), ≡ 0 (mod 7), ≡ 0 (mod 3), ≡ 0 (mod 5), (k
that n that n cannot be 5 be 5,, 7, 9, 10 10.. Hence, the least such integer is n = 11 11..
Also solved by Daniel Lasaosa, Pamplona, Spain; Albert Stadler, Herrliberg, Switzerland; Khurshid Juraev, Lyceum S.H.Sirojiddinov, Tashkent, Uzbekistan; José Hernández Santiago; Jorge Ledesma, Faculty of Sciences UNAM, Mexico City, Mexico; David E. Manes, Oneonta, NY, USA; Adnan Ali, A.E.C.S-4, Mumbai, India; Michael Tang, USA; Hyun Min Victoria Woo, Northfield Mount Hermon School, Mount Hermon, MA, USA; Adithya Bhaskar, Mumbai, India.
Mathematical Mathematical Reflections
6 (2015)
20
O362. Let (F n ), n 0, with F 0 = 0, F 1 = 1, F n+1 = F n + F + F n−1 for all n identities hold:
≥
(a) (b)
F 3n 2 2 F n 1 F n+1 . F n = 2 F n 1 + F n+1 2n+1 F 2n+1 + 2n1+1 F 2n 1 + 2n2+1 F 2n 3 + 0
−
−
−
−
−
··· +
2n+1 n
≥ 1.
Prove Prove that the follow following ing
F 1 = 5n .
Proposed by Nguyen Viet Hung, Hanoi University of Science, Vietnam
Solution by Brian Bradie, Christopher Newport University, Newport News, VA, USA Let (F n ) denote the Fibonacci sequence; that is, F 0 = 0, F 1 = 1, and F n+1 = F n + F + F n−1 for all n Binet formula for the nth Fibonacci number is
≥ 1. 1 . The
F n =
αn
− β n , where √ 5
Note αβ =
−1,
and α2 +
α +
√
1+ 5 α = 2 1 = α
√
5,
and β =
β + +
1
− √ 5 . 2
√ − 5,
1 = β
1 1 2 = β + = α 2 + β 2 = 3. 2 2 α β
(a) Using the Binet Binet formula, formula, F 3n α3n = n F n α
− β 3n = α2n + αnβ n + β 2n = α2n + β 2n + (−1)n. − β n
Next, F n2−1 = F n2+1 = so that 2(F 2(F n2 1 +
−
F n2+1 )
= =
1 2n−2 α + β 2n−2 + 2( 1)n , and 5 1 2n+2 α + β 2n+2 + 2( 1)n , 5
− −
2 1 2 1 α2 + 2 α2n + β 2 + 2 5 α 5 β 6 2n 6 2n 8 α + β + ( 1)n . 5 5 5
8 β 2n + ( 1)n 5
−
−
Moreover, F n−1 F n+1 = = = Thus, 2(F 2(F n2−1 + F n2+1 )
Mathematical Mathematical Reflections
6 (2015)
1 2n α αn 1β n+1 αn+1 β n−1 + β 2n 5 1 2n α αn−1 β n−1 (α2 + β 2 ) + β 2n 5 1 2n 1 2n 3 α + β + ( 1)n . 5 5 5
− − −
−
−
− F n−1F n+1 = α2n + β 2n + (−1)n = F F 3nn .
21
(b) First First note 2n+1
k=0
2n + 1 k
n
F 2n+1−2k =
k=0 n
=
k=0 n
2n + 1 k 2n + 1 k
= 2
k=0
2n+1
F 2n+1−2k + F 2n+1−2k +
2n + 1 k
k=n+1 n k=0
2n + 1 k
2n + 1 2n + 1 k
−
F 2n+1−2k F 2k−(2n (2n+1)
F 2n+1−2k ,
where we have used the identities F −n = ( 1)n+1 F n
n k
and
−
n
=
n
−k
.
Now, 2n+1
k=0
= = = = Thus,
− − − √ − − − − √ − √ − √ · √ √ − − 2n + 1 k
1 5 1 5 1 5 1 5
F 2n+1
2n+1 k=0
2k
2n + 1 k 2n+1
(2n (2n+1)
α
1 + α2 α 2n+1
5
n
k=0
+
k=0 2n+1
2n+1
α
k=0
2n + 1 k
2n + 1 k
β 2n+1
2k
2n+1
2n + 1 k
α4n+2
2k
β
(2n (2n+1)
k=0
1 + β 2 β
2n+1
5
2n+1 2k
2n + 1 k
−
β 4n+2
2k
2n+1
= 2 5n .
F 2n+1
2k
1 = 2
2n+1 k =0
2n + 1 k
F 2n+1
2k
= 5 n.
Also solved by Arkady Alt, San Jose, CA, USA; Daniel Lasaosa, Pamplona, Spain; Nermin Hodzic, University of Tuzla at Tuzla, Bosnia and Herzegovina; Jorge Ledesma, Faculty of Sciences UNAM, Mexico City, Mexico; Ángel Plaza, University of Las Palmas de Gran Canaria, Spain; Adnan Ali, A.E.C.S-4, Mumbai, India; Li Zhou, Polk State College, FL, USA; Ravoson Vincelot, Lycee Henri IV, Paris, France.
Mathematical Mathematical Reflections
6 (2015)
22
O363. Solve Solve in integers integers the system of equations xyz xy z xyz x +y +z + = 2 xy + xy + yz + yz + zx + 3 3 2
2
2
= 2016. 2016.
Proposed by Titu Andreescu, University of Texas at Dallas, USA Solution by Li Zhou, Polk State College, FL, USA It is easy to verify that all six permutations of x,y,z = 12 12,, 12 12,, 24 are solutions. We show that they are the only solutions. solutions. By symmetry symmetry, it suffices to consider consider x y z . Subtractin Subtracting g the second equation equation 2 2 2 from twice the first, we get (x ( x y ) + (y (y z ) + (z (z x) = 2016. 2016. Because 0, 1, 4, 9 is the complete set of quadratic residues (mod 16 16), ), x y y z z x 0 (mod 4). 4). No Now w 2016 2016//16 = 126, 126, and we have only 2 2 2 2 2 2 2 2 2 126 = 11 +2 + 1 = 10 +5 +1 = 9 +6 +3 . Also, x Also, x z = (x y) +(y +( y z ). Thus we must have x have x z = 36 2 2 and x y, y z = 24 24,, 12 . If x x y = 12 12,, then 4032 then 4032 = (x (x + y + z ) + xyz = xyz = (3x (3x 48) + x(x 12)(x 12)(x 36), 36), 3 2 which implies x = 3u and yields 0 = u 13u 13 u + 16u 16 u 64. 64 . This This has no integer integer solutio solution n by the rational rational 2 root theorem. theorem. If x y = 24 24,, then 4032 then 4032 = (3x (3x 60) + x( x(x 24)(x 24)(x 36), 36), which implies x = 3u and yields 2 0 = (u 4)(u 4)(u 13 13u u + 4). 4). Hence, u = 4 and ( and (x,y,z x,y,z)) = (12, (12, 12 12,, 24), 24), completing the proof.
{ {
{ − −
−} { − −
} { − − } ≥ ≥ − − − { − ≡ − ≡ − ≡ − − − } − − − − − − − −
}
−
−
−
−
Also solved by Daniel Lasaosa, Pamplona, Spain; Albert Stadler, Herrliberg, Switzerland; Nermin Hodzic, University of Tuzla at Tuzla, Bosnia and Herzegovina; Adnan Ali, A.E.C.S-4, Mumbai, India.
Mathematical Mathematical Reflections
6 (2015)
23
O364. (a) If n = n = p p 1e p2e 1
2
· · · pke , where pi are distinct primes, find the value of k
dn
|
nφ( nφ(d) d
as a function of pi and ei .
{ { }
{ }
(b) Find the number of integral solutions to xx
97), 1 ≤ x ≤ 9312 9312.. ≡ 1 (mod 97),
Proposed by Mehtaab Sawhney, Commack High School, New York, USA Solution by Li Zhou, Polk State College, FL, USA
is a multiplicativ multiplicativee function function of n, so is f ( f (n) = d|n φ(dd) . (See (See G. H. Hardy & E. M. Wrigh Wright, t, An Intro. to the Theory of Numbers , 5th ed., Oxford, p. 235.) Now for any prime p, p ,
(a) Since
φ ( n) n
e e
f ( f ( p ) =
i=0
φ( pi ) = 1+ pi
e
i=1
p
− 1 = 1 + e 1 − 1 p p
.
Hence,
dn
|
nφ( nφ(d) = nf ( nf (n) = n d
k
− 1 + ei 1
i=1
1 pi
k
= p1e1 1
−
pkek 1
···
−
[ p pi (1 + ei )
i=1
− ei ] .
(b) Notice that 9312 = 97 96 and x 0 (mod 97). 97). So we cons consid ider er x = 97 97q q + r + r,, with 0 q 95 and x 97q 97 q + r q + r 1 r 96 96.. Then Then x r r (mod 97). 97). No Now w for any prime prime p, it is well known that if d ( p 1), 1), d then r 1 (mod p (mod p)) has exactly d roots. roots. (See (See G. H. Hardy & E. M. Wrigh Wright, t, An Intro. Intro. to the Theory Theory of r is the least positive integer such that r that r d 1 Numbers , 5th ed., Oxford, p. 85.) Also, recall that the order of r is (mod p (mod p)). Hence, there are exactly φ( φ (d) roots r of order d order d.. For each such r of order d order d,, there are 96 96/d /d values values of q such q such that d that d (q + + r). Therefore, the answer to the question is the special case of (a) for n = 96 = 25 3:
≤ ≤
≡
≡
·
≡
≡
≤ ≤ ≤ | − ≡ ·
|
d 96
|
96 96φ φ(d) = 24 [2(1 + 5) d
− 5] [3(1 [3(1 + 1) − 1] = 560. 560.
Also Also solved solved by Albert Albert Stadler, Stadler, Herrlib Herrliber erg, g, Switzerlan Switzerland; d; Joel Joel Schlosb Schlosber erg, g, Bayside, Bayside, NY, USA; David E. Manes, Oneonta, NY, USA; Adnan Ali, A.E.C.S-4, Mumbai, India.
Mathematical Mathematical Reflections
6 (2015)
24
O365. Prove Prove or disprov disprovee the following following statement: statement: there is a non-vanish non-vanishing ing polynomial P polynomial P ((x,y,z) x,y,z) with integer π coefficients such that P (sin whenever u + v + w = 3 . P (sin u, sin v, sin w) = 0 whenever u
Proposed by Albert Stadler, Herrliberg, Switzerland Solution by Daniel Lasaosa, Pamplona, Spain Let A Let A,, B,C be be the angles of a triangle (or for that matter, any three angles, positive or negative, such that their sum is π ), and for brevity denote p denote p = sin A, q = = sin B and r = sin C . Expanding Expanding sin( sin(A A + B + B + + C C ) = 0 as a function of the trigonometric functions of A of A,, B,C yields yields pqr
− p cos B cos C = (q ( q cos cos C + + r cos B )cos A, which after squaring and using cos 2 A = 1 − p2 , cos 2 B = 1 − q 2 and cos and cos 2 C = 1 − r 2 , and rearranging rearranging terms,
results results in
p2 q 2 r 2 + p2 (1
− q 2)(1 − r2) − q (1 (1 − p2 )(1 − r 2 ) − r (1 − p2 )(1 − q 2 ) = 2qr cos B cos C.
Squaring again allows us to find a non-vanishing expression equal to zero in terms of p, p, q,r (it q,r (it suffices to notice 2 that the coefficient of p p in the LHS is nonzero, and p and p does does not appear in the RHS). Now, for any u any u+ +v +w = π3 , we can take A = 3u, B = 3v and C = 3w, or p = sin(3u sin(3u) = 3cos2 u sin u sin3 u = 3 sin sin u 4sin3 u, and similarly for q, for q, r. Substitution into the previously found expression on p on p,, q,r allows q,r allows us to find a non-vanishing polynomial (it suffices to take v take v = w = w = = 0 and let u let u vary to obtain nonzero values, since we know that terms 2 containing p , and hence containing sin u, do not vanish). But this polynomial is zero when u when u + v + w = π3 . It follows that the answer is yes, we have proved that such a polynomial exists.
−
−
Also solved by Khurshid Juraev, Lyceum S.H.Sirojiddinov, Tashkent, Uzbekistan; Li Zhou, Polk State College, FL, USA.
Mathematical Mathematical Reflections
6 (2015)
25
O366. In triangle triangle ABC , ABC , let A1 , A2 be two arbitrary isotomic points on BC . BC . We define points points B1 , B2 C A and C and C 1 , C 2 AB similarly. similarly. Let Let a be the line passing through the midpoints of segments (B1 C 2 ) and (B2 C 1 ). We define lines b and c similarly. Prove that all three of these lines are concurrent.
∈
∈
Proposed by Marius Stănean, Zalău, România Solution by Andrea Fanchini, Cantù, Italy
We use barycentric coordinates and the usual Conway’s notations with reference to the triangle ABC AB C . Points A Points A 1 , B1 , C 1 have the followings absolute coordinates
A1 0,
a
− d , d a
a
,
B1
e b e , 0, b b
−
− ,
C 1
c
f f , ,0 c c
where d where d,, e, f are are parameters. Then points A 2 , B2 , C 2 have absolute coordinates A2
−
d a d 0, , , a a
−
B2
b
e
e , 0, b b
,
C 2
f c f , ,0 c c
−
• Midpoints of segments (B1C 2) and (B2C 1) and the others similarly . M B C (bf + ce : ce : b b((c − f ) f ) : c( c (b − e)), )), M B C (2bc (2bc − bf − ce : bf bf : ce) ce ) − ce : 1
M A
2
1
C 2 (af :
M A
B2 (a(b
1
2
2ac
− af − − cd : cd : cd cd)),
M A
b (a − d) : ae + bd) bd), − e) : b(
2
1
C 1 (a(c
M A
2
− f ) f ) : af + + cd : cd : c c((a − d))
ae : bd bd : : B1 (ae :
2ab
bd) − ae − bd)
• Equations of lines la, lb, lc. la : (ec
bf )x + (2bc (2bc − 3ce − bf )y + (3bf (3 bf + + ce − 2bc) bc)z = 0 − bf )
lb : (af + + 3cd lc : (2ab (2ab Mathematical Mathematical Reflections
6 (2015)
− 2ac) − cd) − cd) ac)x + (af (af − cd)y + (2ac (2 ac − 3af − cd)z = 0
bd)x + (bd (bd + 3ae 3ae − 2ab) ab)y + (bd ( bd − ae) ae)z = 0 − ae − 3bd) 26
Now the three lines la , lb , lc are concurrent if and only if
ec
− bf + 3cd − 2ac af + 2ab − ae − 3bd
2bc
− 3ce − bf af − cd bd + 3ae 3ae − 2ab
making calculations we obtain
3bf + ce
− 2bc 2ac − 3af − − cd bd − ae
=0
8a2 b2 c2 12 12a a2 b2 cf 16 16a a2 bc2 e+22 +22a a2 bcef +6 bcef +6a a2 c2 e2 8a2 ce2 f 20 20ab ab2 c2 d+26 +26ab ab2 cdf cdf +30 +30abc abc2 de 28 28abcdef abcdef 4ac2 de2
−
−
−
−
−
−
+12bb2 c2 d2 12 +12 12bb2 cd2 f 12 12bc bc2 d2 e 8a2 b2 c2 +12 +12a a2 b2 cf +20 cf +20a a2 bc2 e 26 26a a2 bcef 12 12a a2 c2 e2 +12 +12a a2 ce2 f +16 f +16ab ab2 c2 d 22 22ab ab2 cdf cdf
−
−
− − − − 30abc abc2 de+28 de+28abcdef abcdef +12 +12ac ac2 de2 −6b2 c2 d2 +8 +8bb2 cd2 f +4 f +4bc bc2 d2 e−4a2 bc2 e+4 +4a a2 bcef +6 bcef +6a a2 c2 e2 −4a2 ce2 f +4 f +4ab ab2 c2 d −30 4b2 cd2 f + + 8bc2 d2 e = 0 −4ab2cdf − 8ac2de2 − 6b2c2d2 + 4b cdf − as we can easy check.
Also solved by Daniel Lasaosa, Pamplona, Spain; Khurshid Juraev, Lyceum S.H.Sirojiddinov, Tashkent, Uzbekistan; Li Zhou, Polk State College, FL, USA.
Mathematical Mathematical Reflections
6 (2015)
27
