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D.M. BĂTINEŢU-GIURGIU MARIN CHIRCIU DANIEL SITARU NECULAI STANCIU, OCTAVIAN STROE ,
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OLYMPIAD PROBLEMS FROM ALL OVER THE WORLD VOLUME 5 9th GRADE CONTENT
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Dedicated to the International Mathematical Olympiad
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Table of Contents INDEX OF PROPOSERS AND SOLVERS .................................................. 7
Chapter I. Problems ....................................................................................... 11
Chapter II. Solutions Solutions...................................................................................... ...................................................................................... 65
BIBLIOGRAPHY................................................................ BIBLIOGRAP HY......................................................................................... ......................... 305
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INDEX OF PROPOSERS ND SOLVERS Abdallah Almalih, Syria Abdallah El Farissi, Algeria Abdelhak Maoukouf, Morocco
Bojan Bašić, Serbia Boris Colaković, Serbia Carlos Suarez, Ecuador
Abdelmalek Metidji,Turkey Algeria Abdilkadir Altintas, Abdul Aziz, Indonesia Abhay Chandra, India Abinash Mohapatra, India Adil Abdullayev, Azerbaijan Aditya Narayan Sharma, India Adrian Stan, Romania Agathoklis Theos, Greece Alexandar Ivanov, Bulgaria Ali Can Gullu, Turkey Anas Adlany, Morocco Andras Hrasko, United Kingdom Andrei Bara, Romania Andrei Bogdan Ungureanu, Romania Andriy Anikushin, Ukraine Andryi Gogolev, Ukraine Angel Plaza, Spain Anish Ray, Romania Anişoara Dudu, Romania Anouy Chakraborty, India Antonis Anastasiadis, Greece Athanasios Mplegiannis, Greece
Carmen Chirfot,Greece Romania Chris Kyriazis, Claudia Nanuti, Romania Clemens Heuberger, Austria Cornel Ioan Vălean, Romania Cristian Heuberger, Romania D. Basylev, Belarus D.M. Bătineţu-Giurgiu, Romania D. Odgerel, Mongolia Dan Griller, United Kingdom Dan Nanuti, Romania Dan Radu Seclaman, Romania Dana Heuberger, Romania Dang Thanh Tung, Vietnam Daniel Sitaru, Romania Danila Mysak, Ukraine Danilo Khilko, Ukraine Dat Vo, Vietnam David Monk, United Kingdom Dhanh Tang Tung, Vietnam Dimitris Kastriotis, Greece Dinh Tien Dung, Vietnam Do Huu Duc Thinh, Vietnam
Avishek Mishra, India B.Amarsanaa, Mongolia B. Bat-Od, Mongolia B. Battsengel, Mongolia B. Bayarjargal, Mongolia B. Bayasgalan, Mongolia B. Khoroldagva, Mongolia B. Sandagdorj, Mongolia B. Ulziinasan, Mongolia Babis Stergioiu, Greece Bedri Hajrizi, Kosovo Bernd Kreussler, Ireland
Do Quoc Chinh, Vietnam Dorin Marghidanu, Romania Duong Viet Hong, Vietnam Dušan Đjukić, Serbia E. Enkhzaya, Mongolia Ear Bunhpeng, Cambodia Erdene Natsagdorj, Mongolia Finbar Holland, Ireland Fotini Kaldi, Greece Francis Fregeau, Canada Francisco Javier Garcia Capitan, Spain G. Batzaya, Mongolia
Bogdan Rublyov, Ukraine
G. Bayrmagnai, Mongolia
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Geanina Tudose, Romania George Apostolopoulos, Greece Gerhard Woeginger, Austria Gheorghe Alexe, Romania Gheorghe Boroica, Romania Gheorghe Szollosy, Romania Gordon Lessels, Ireland Hamza Mahmood, Pakistan Hasan Bostanlik, Turkey Henry Ricardo, USA Hoang Le Nhat Tung, Vietnam Horia Nicolcea, Romania Hung Nguyen Viet, Vietnam I. Voronovich, Belarus Ibrahim Abdulazeez, Nigeria Igor Soposki, Macedonia Ilkin Guliyev, Azerbaijan Ilko Brnetić, Croatia Imad Zak, Lebanon Ior Nagel, Ukraine Iuliana Traşcă, Romania Jack Smith, United Kingdom Jan Kucharik, Slovakia Jan Mazak, Slovakia Jaromir Simsa, Slovakia Jaroslav Svrcek, Slovakia Jim Leahy, Ireland José Luis Díaz-Barrero, Spain Karl Czakler, Austria Kays Tomy, Tunisia Kevin Soto Palacios, Peru
Le Viet Hung, Vietnam Lelia Nicula, Romania Leonid Bedratyuk, Ukraine Long Nhat, Vietnam Lucian Stamate, Romania Madan Beniwal, India Manish Moryani, India Manish Tayal, India Marian Dincă, Romania Marian Ursărescu, Romania Marin Chirciu, Romania Marius Drăgan, Romania Marjan Milanović, Serbia Mark Flanagan, Ireland Marko Radovanovik, Serbia Martin Lukarevski, Macedonia Martin Panak, Slovakia Matija Bucić, Croatia Mehmet Sahin, Turkey Menghor Thuo, India Michel Rebeiz, Lebanon Mihalcea Andrei Ştefan, Romania Mihaly Bencze, Romania Miroslav Marinov, Bulgaria Mithun Chakraborty, India Mohammed Hijazi, Iordania Mohammed Lamine, Morocco Munkhuu Tomorbaatar, Mongolia Myagmarsuren Yadamsuren, Mongolia N. Argilsan, Mongolia Mon golia N. Naboj, Croatia
Khalef Ruhemi, Khanh Hung Vu,Iordania Vietnam Khung Long Xang, Vietnam Kiran Kedlaya, USA Koczinger Éva, Romania Kovács Béla, Romania Kristina Ana Skreb, Croatia Kunihiko Chikaya, Japan Laurențiu Panaitopol, Romania Lazaros Zachariadis, Greece Le Khanh Sy, Vietnam Le Minh Cuong, Vietnam Le Van, Vietnam
N. Plotnikov, Ukrai Ukraine Nawar Alasadi, Iraqne Neculai Stanciu, Romania Nela Ciceu, Romania Ngo Dinh Tuan, Vietnam Vie tnam Ngo Minh Ngoc Bao, Vietnam Vi etnam Nguyen Minh Triet, Vietnam V ietnam Nguyen Minh Tri, Vietnam Viet nam Nguyen Ngoc Tu, Vietnam Nguyen Phuc Tang, Vietnam Nguyen Thanh Nho, Vietnam Nguyen Van Nho, Vietnam Vietna m Nho Nguyen Van, Vietnam Vietna m
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Nicolae Nica, Romania Nicolae Papacu, Romania Nikola Adzaga, Croatia Nikola Petrović, Serbia Nirapada Pal, India Nitin Gubani, India Indi a Octavian Stroe, Romania Oles Dobosevycs, Ukraine Omar Raza, Pakistan Pakawut Jiradilok, Thailand Patrik Bak, Slovakia Paul Jeffreys, United Kingdom Pavel Calabek, Slovakia Peter Novotny, Slovakia Petre Daniel Alexandru, Romania Pham Quoc Sang, Vietnam Pham Quy, Vietnam Phan Loc So'n, Vietnam Pryanka Banerjee, India Quang Minh Tran, Vietnam Radek Horenski, Slovakia Rahim Shabazov, Azerbaijan Rajeev Rastogi, India Rajsekhar Azaad, India Ravi Prakash, India Redwane El Mellas, Morocco Regragui El Khammal, Morocco Richdad Phuc, Vietnam Robert Toth, Slovakia Rosie Cates, United Kingdom Rovsen Pirguliyev, Azerbaijan
Shahlar Maharramov, Azerbaijan Shivam Sharma, India SK Abdul Halim, India SK Rejuan, India Sladjan Stanković, Macedonia Soumava Chakraborty, India Soumava Pal, India Soumitra Mandal, India Stanislav Chobanov, Bulgaria Steiner Barabbas, Tunisia Stephan Wagner, Austria Stephen Buckley, Ireland Stipe Vidak, Croatia Su Tanaya, India Subhajit Chattopadhyay, India Sumet Ketsri, Thailand T. Bazar, Mongolia T. Khulan, Mongolia Thanasis Xenos, Greece Theodoros Sampas, Greece Theresia Eisenkolbl, Austria Thomas Read, United Kingdom Titu Zvonaru, Romania Togrul Ehmedov, Azerbaijan Tom Bowler, United Kingdom Tom Laffey, Ireland Tonći Kokan, Croatia Tran Van Hien, Vietnam Ts. Bathuu, Mongolia Tuk Zaya, Mongolia U. Batzorig, Mongolia
Roxana Mihaela Stanciu, Romania Rozeta Atanasova, Macedonia Rustem Zeynalov, Azerbaijan Safal Das Biswas, India Said Ibnja, Morocco Sameer Shihab, Saudi Arabia Sanong Huayrerai, Thailand Saptak Bhattacharya, India Sarah El, Morocco Sarka Gergelitsova, Slovakia Şerban George Florin, Romania Seyran Ibrahimov, Azerbaijan Sh. Dorjsembe, Mongolia
Uche Okeke, Nigeria Trust UnitedEliezer Kingdom Mathematics V. Adiyasuren, Mongolia V. Kaskevich, Belarus V.N. Murty, USA Vadim Mitrofanov, Ukraine Vaggelis Stamatiadis, Greece Vasile Jiglău, Romania Vijay Rana, India Vitaly Senin, Ukraine Vladislav Yurushev, Ukraine Walter Janous, Austria Wasma Nayer, Pakistan
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Wijit Yangjit, Thailand Yasiinski Vyacheslav, Ukraine Yen Tung Chung, Taiwan Yubal Barrios, Spain
Yubian Andres Bedoya Henao, Colombia Zhang Yun, China Trinh Quoc Huy, Vietnam David Bonclay, Vietnam
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Chapter I Problems
be a fixed real number. Determine all functions f : satisfying y) for all x, y . f ( f f ( xx + y)f( x x – y)) = x2 + yf ( y W ALTER J ANOUS, A AUSTRIAN NMO, 2017 1. Let
2. Let M be be
a set of 2017 positive integers. For every non-empty A M M , we define f ( A A) = { x M M : : x is divisible by odd number of elements of A}. Find the minimum number of colors such that it is possible to paint all non-empty subset of M in in such a way that, whenever A f f ( A A), the sets A and f ( A A) are in different colors. ALEKSANDAR I VANOV, BULGARIAN NMO, 2017 3. Let
n be a positive integer and a1, a2, ..., a2n be 2n distinct integers. Given that the equation x x – a1 x x – a2 ... x x – a2n = ( n!)2 has an integer solution x = m, find m in terms
of a1, ..., a2n. 4. We
SINGAPORE, SMO, 2017
consider the real numbers x, y and z , which satisfy the system of equations:
x y z 3 . xy yz xz 2
Compute max( x x) + min( x x). D.M. B ĂTINEŢU-GIURGIU, NECULAI STANCIU, ROMANIA 5. Given
7 distinct positive integers, prove that there is an infinite arithmetic progression of positive integers a, a + d , a + 2d , ..., with a d , that contains exactly 3 or 4 of the 7 given integers. SINGAPORE, SMO, 2017
6. Solve
the equation x2(2 – x)2 = 1 + 2(1 – x)2. FINBAR HOLLAND, IRELAND SHL, 2017
7. Show
that, for all x, y, z , w, ( x x – w)( y y – z ) + ( y – w)( zz – x) + ( z – w)( x x – y) = 0 and sin( xx – w)sin( y y – z ) + sin( y y – w)sin( z z – – x) + sin( z z – – w)sin( x x – y) = 0. FINBAR HOLLAND, IRELAND SHL, 2017 = 4 n2 + 7n2 + 3 n + 6. Prove that if n is an integer, then f (n) is not the cube of an integer. TOM L AFFEY , IRELAND SHL, 2017
8. Let f (n)
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each positive integer n, let c p = 2017n. Suppose that a function f : : satisfies the following two conditions: (i) f (m + n) 2017 f f (m) f f (n + 325) for each positive integer m, n; (ii) 0 0, then:
6 3
a b
2 6 3 b a2
b c
A
.
2 D ANIEL SITARU, RMM, ROMANIA
c c2 2 6 3 2 15. c a a D ANIEL SITARU, RMM, ROMANIA b2
ABC , the following relationship holds: 1 (a 2 ab )(b 2
14
1 . r ab ) 18 D ANIEL SITARU, RMM, ROMANIA
29. If
in ABC :
a , y cos 1 b , z cos 1 c , then: bc ca ab x y z 3 tan tan tan . 2 2 2 9 D ANIEL SITARU, RMM, ROMANIA
x cos 1
30. In
ABC , the following relationship holds: (a 2 b 2 c 2 ) a 2 b 2 c 2 6abc 6 cos A cos B cos C . D ANIEL SITARU, RMM, ROMANIA
31. Let
be n * \ {1} și ak , k 1, n. Prove that: n
2
ak
ak21
ak ak 1
k 1
n
ak ; an 1 a1 . k 1
D.M. B ĂTINEŢU-GIURGIU, NECULAI STANCIU, ROMANIA, RMM SUMMER EDITION, 2016 32. Prove
that if a, b, c, d > > 0, then: a2 + b2 + c2 + d 2 = 1; abc + bcd + + cda + dab =
1 . 2
a2
4 1 2bcd 5. D ANIEL SITARU, ROMANIA, RMM SUMMER EDITION, 2016 33. If x, y, z
(0, ), then: x + y + z + +
1 x
3
1 y
3
1 3
z
12
.
3 3 D ANIEL SITARU, ROMANIA, RMM SUMMER EDITION, 2016
consider f ( x x) = x3 + bx2 + cx + d with with f (2014) (2014) = 2013 and g ( x x) = x2 – 2 x + 2014 such that the equation f ( g g ( x x)) = 0 doesn’t have real roots. Solve the equation f ( x x) = 0 knowing that it has three distinct natural numbers roots. D.M. B ĂTINEŢU-GIURGIU, NECULAI STANCIU, ROMANIA 34. We
35. If
a, b, c > 0, (a + b)(b + c)(c + a) = a2b2c2, then: ab bc
15
ca
6. D ANIEL SITARU, RMM, ROMANIA
36. If
a, b, c > 0, a2 + b2 + c2 = 3, then:
6 37. In
b2 a2
3
c2
b2
3
a b c )2 . ( c2 3 D ANIEL SITARU, RMM, ROMANIA a2
ABC , the following relationship holds: a4
b2
c4
2
tan 22 tan 22 tan 23 tan 23 48S . D ANIEL SITARU, RMM, ROMANIA 38. There
are 12 chairs which are aligned and labeled by numbers 1; 2; ...; 12 from left to right. A grasshopper can jump from one chair to another following the rule: from a chair with number k it it can jump to the chair with number n if and only if k – – n = = 5 or k –– n = 8. It is known that grasshopper managed to do the jump so that it visited all chairs exactly once. What chair could be the initial position for the grasshopper? UKRAINIAN NMO, 2016 39. Let x, y, z be be
real numbers from segment [0; 1]. Prove that: x – y)6 + ( y y – z )6 + ( z z – x)6 6. x5 + y5 + z 5) + ( x x4 + y4 + z 4) + ( x ( x Y ASINSKII V YACHESLAV, UKRAINIAN NMO, 2016
40. Find
all real numbers x satisfying the following equation: x + { x})2 – ( x x + { x}) = 6[ x x]{ x} – 1, ( x where [ x x] and { x} denote the integer part and fractional part of x, respectively. NGUYEN VIET HUNG, V VIETNAM, RMM A AUTUMN EDITION, 2016
41. A
convex quadrilateral ABCD is inscribed in a circle. The lines AD and BC meet meet at point E . Points M and N are taken on the sides AD and BC , respectively, so that AM : MD = BN : NC . Let the circumcircles of triangle EMN and and quadrilateral ABCD intersect at point X and and Y . Prove that either the lines AB, CD and XY have a common point, or they are all parallel. DUŠAN ĐJUKIĆ, SERBIAN NMO, 2017 42. Let p be
a prime. Show that
3
3 p 5 is irrational. THAILAND NMO, 2017
all functions f :: which satisfy: f ( f f ( x x) – y y) xf xf ( x x) + f ( y y) (1) for all real numbers x and y.
43. Find
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THAILAND NMO, 2017
all functions f :: + + such that: f ( xf xf ( x x) + f ( y y)) = ( f f ( x x))2 + y (1) for all positive rational x, y.
44. Find
THAILAND NMO, 2017
all functions f : such that for every positive integer m the following is true: If we denote by d 1, d 2, ..., d n all the divisors of number m, then: f (d 1) f f (d n) = m. f (d 2) ... f P AVEL C ALABEK , CZECH & SLOVAK NMO, 2017
45. Find
46. Find
all triplets of integers (a, b, c) such that each of the fractions: a
b
,
c
,
is an integer.
bc ca ab J AROSLAV S VRCEK , CZECH & SLOVAK NMO, 2017 47. Let ABC be be
an acute triangle with altitude AD. The bisectors of angles BAD, CAD intersect side BC at E , F , respectively. The circumcircle of triangle AEF intersects intersects sides AB, AC at at G, H , respectively. Prove that lines EH , FG, and AD pass through a common point.
P ATRIK B AK , CZECH & SLOVAK NMO, 2017
48. Let
n
a x
x) = be f : , f ( x
k
k
, where ak 0, k = 0, n. If f (4) = 8 and
k 0
f (9) = 18, then find max( ff (6)) and the value for which this maximum is achieved. D.M.B ĂTINEŢU-GIURGIU, NECULAI STANCIU, ROMANIA 49. If
a, b, c > 0, a + b + c = 3, x , then:
3
a sin 2 x
3 b cos 2 x
3
b sin 2 x
3 c cos2 c
3
c sin 2 x 3 a cos 2 x
4
D ANIEL SITARU, RMM, ROMANIA 50. If x, y, z > >
0, then: x 3
y
3
y
x
y3
x
y
x3
2 2
x2 y2
12
y2 x2
.
D ANIEL SITARU, RMM, ROMANIA 51. If x, y, z > >
0, x + y + z = = 3, then:
3
x
3
y
3
3
y
3
z
3
3 z 3
x
3
24.
D ANIEL SITARU, RMM, ROMANIA
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52. If
in ABC S = = 2, then: (2 s a )( 2s b )( 2s c ) 1 . ( 2 a sin A )(2 b sin B )(2 c sin C ) sin A sin B sin C D ANIEL SITARU, RMM, ROMANIA
53. In
acute ABC , the following relationship holds: a2b2c2 sin 2 A sin 2 B sin 2C cos cos A cos B cos C S 3.
D ANIEL SITARU, RMM, ROMANIA 54. In
ABC , the following relationship holds: (ama bmb cmc )(am a ma3 bmb3 cmc3 ) (a b c)(ama4 bmb4 cmc4 ). D ANIEL SITARU, RMM, ROMANIA
55. If
a, b, c > 0, then:
56. If
3
(a 3b)( 2a 2b)(3a b) 4. ab bc ca D ANIEL SITARU, RMM, ROMANIA
a, b, c > 0, a2 + b2 + c2 = 3, then: a3
1 2 a a 1
b3
c3
1 2 b b 1
1 6. 2 c c 1 D ANIEL SITARU, RMM, ROMANIA
57. In
ABC , the following relationship holds: (a b)4 64s 2 . ab 3 D ANIEL SITARU, RMM, ROMANIA
58. If
a, b, c, d > > 0,
1 1 1 1 1, then:
a
b
c 3
d
a
3 b 3 c 3 d 3 abcd . D ANIEL SITARU, RMM, ROMANIA
59. If x, y, z
1, then: 1 3 1 ( x 1)( y 1) 3 xyz . ANIEL ITARU
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D
S
OMANIA
, RMM, R
60. In
B C
ABC , B > C . Let D be the point on side BC such such that DAC = =
. The
2 circumcircle of ACD meets side AB again at E . The circumcircle of ABD meets side AC again at F . The internal angle bisector of BDE meets meets side AB at P . The internal angle bisector of CDE meets meets side AC at at Q. Prove that PQ and AB are perpendicular. HONG K ONG ONG, PREIMO 2017, MOCK EXAM 61. Let 2a,
b, c2, d be2positive 2 numbers satisfying Prove (a b + b c + c d + + d 2a)(real ab2 + bc2 + cd + da2) abcd (a + =c)(1.b + d ))((acthat: + bd + + 2). When does equality hold? HONG K ONG ONG, PREIMO 2017, MOCK EXAM
62. Prove
the following inequality: 163 4 y + z ))(( z z + x + y + z )3 x3 y3 z 3, where x, y, z are [( xx + y)( y + x)] ( x are positive real numbers. 27 ANDREI BOGDAN UNGUREANU, RMM WINTER EDITION, 2016 63. Prove
that if a, b, c, d (0, ); 3 (ad bc ) = ac + bd 0, then: d (a b
3 ) c (b a 3 ) 4 4 abcd . D ANIEL SITARU, RMM WINTER EDITION, 2016
64. Prove
that in an ABC acute-angled acute-angled triangle the following relationship holds: 2 S cos A cos B cos C 2 . 4 4 4 R D ANIEL SITARU, RMM WINTER EDITION, 2017
65. Prove
that in ABC : a 2 (b 2 c 2 a 2 )3
2 2
bc
66. Let x1, x2,
64 S 2 (1 cos 2 A cos 2 B cos2 C ).
D ANIEL SITARU, RMM WINTER EDITION, 2016
..., xn be positive real numbers such that
Find the minimum possible value of
1
1 x1
2 x2
...
n xn
n(n 1)
2
.
x22 ... xnn .
NGUYEN VIET HUNG, RMM SPRING EDITION, 2017 67. Prove
that for all x we have cos(sin x) > sin(cos x). ABDALLAH EL F ARISSI, RMM SPRING EDITION, 2017
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68. Let
a, b such that a + b > 0, then: n
a b 1 n a k b n k a n b n . 2 2 n 1 k 0 ABDALLAH EL F ARISSI, RMM SPRING EDITION, 2017 69. Call
a function f : lively if f (a + b – 1) = f ( f (...) f (b)...)) for all a
a, b .
times
A + 2018) = g ( A A) + 1 holds for some A 2. Suppose that g is is a lively function such that g ( A 2017 a) Prove that g (n + 2017 ) = g (n) for all n A A + 2. b) If g ( A A + 20172017) g g ( A A), determine g (n) for n A A – 1. M ARKO R ADOVANOVIĆ, SERBIAN TST, 2017 70. A
n n square
is divided into unit squares. One needs to place a number of isosceles right triangles with hypotenuse 2, with vertices at grid points, in such a way that every side of every unit square belongs to exactly on triangle (i.e. lies inside it or on its boundary). Determine all numbers n for which this is possible.
DUŠAN ĐJUKIĆ, SERBIAN TST, 2017 71. Let ABCD be a convex quadrilateral with AC BD BD. Prove that there exist points P , Q, R, S on AB, BC , CD, DA, respectively, such that PR QS and the area of quadrilateral PQRS is is exactly half that of ABCD. THAILAND TST, 2017 72. Let
a and b be real numbers such that a + b = 1. Prove the following f ollowing inequality:
1 5a 2 5 2 b 2 9. B. B ATTSENGEL, MONGOLIAN NMO, 2017 73. Let ABCD be
AD = BC and AB CD. Let O be the isosceles intersection of the an diagonals andtrapezoid let M be bewith the midpoint of AD. Circumcircle of BCM intersects AD again at K . Prove that OK is is parallel to AB. B. B AT-OD, MONGOLIAN NMO, 2017 74. The
altitudes AD and BE of acute triangle ABC intersect at H . Let F be the intersection of AB and a line that is parallel to the side BC and and goes through the circumcenter of ABC . Let M be be the midpoint of AH . Prove that CMF = 90. G. B ATZAYA, MONGOLIAN NMO, 2017 75. Let ABC ABCD D be
a cyclic quadrilateral with circumcenter w, and E be be the intersection of the diagonals AC and BD. A line passing through E intersects lines AB, BC at P , Q,
20
R D respectively. Let R ( R D) be the intersection point of w and circle that passes through D, E and and tangents the line PQ at E . Prove that B, P , Q, R are cyclic. B. K HOROLDAGVA HOROLDAGVA, MONGOLIAN NMO, 2017
all functions f : satisfying: (a – b) f f (a + b) + (b – c) f f (b + c) + (c – a)f(c + a) = 0 for all a, b, c .
76. Find
MONGOLIAN NMO, 2017 77. Let I be be
the center of triangle ABC . Let D be a point on side BC , and E be a point
on ray BC such that C lies between E and D and
BD
BE
. Let H be the feet of DC EC perpendicular from D to line IE . Prove that AHE = = IDE . B. B ATTSENGEL, G. B ATZAYA, MONGOLIAN NMO, 2017 78. A
rectangle R is dissected into 2016 small rectangles with sides parallel to the sides of R. We call the vertices of those small rectangles nodes. A segment parallel to the sides of R is called basic if its two end points are nodes and A there are no other nodes in the interior of it. Find the maximum and minimum of the number of basic segments among all possible dissections of R. For example, in the figure on the right, R is B dissected into 5 small rectangles with a total of 16 segments. The C segments AB and BC are are basic, while the segment AC is is not. CHINA NMO, 2017 positive real numbers such that a2 + ab + b2 = 9. Find the maximal value of expression: (a + b)6 + (ab)5 + 2(ab)3 + (ab)2 – 17. GEORGE APOSTOLOPOULOS, RMM SUMMER EDITION, 2017 79. Let
a, b be
80. Let ABC be be
an equilateral triangle inscribed in the circle (O) whose radius is R. Prove that for an arbitrary point P lies lies on (O): 3 3 3 PA PB PC 4 6 2 3 216 . 3
R NGUYEN VIET HUNG, RMM SUMMER EDITION, 2017
81. If
a, b, c, n > 0; n(ab + bc + ca) + 2abc = n3, then:
1 1 1 1 . a b 2n b c 2 n c a 2 n n M ARIN CHIRCIU, RMM SUMMER EDITION, 2017
21
82. Prove
that if n *; a > 1, then: (n + a – 1)(a – 1)n – 1 an. D ANIEL SITARU, RMM SUMMER EDITION, 2017
83. Let ABC be be
an acute triangle. Prove that: (a cot A)a(b cot B)b(c cot C )c (2r )a + b + c. where a = BC , b = CA, c = AB, and r is is the in radius. NGUYEN VIET HUNG, RMM A AUTUMN EDITION, 2017 84. Let
a, b, c be positive real numbers. Prove that: a3
b3 b3 c3 c 3 a 3 9abc . c 2 ab a 2 bc b 2 ca ab bc bc ca ca NGUYEN NGOC TU, RMM A AUTUMN EDITION, 2017 85. Prove
that if a, b, c are the length’s sides in ABC , then: sin2 a + sin2 b + sin2 c 4 sin s sin( ss – a) sin( ss – b) sin( ss – c c). D ANIEL SITARU, RMM A AUTUMN EDITION, 2017
0, with 2u – v > 0 and , , are the measures of the angles of triangle sin 3 ABC , then . uv cyc u sin v s in sin D.M. B ĂTINEŢU-GIURGIU, NECULAI STANCIU, RMM A AUTUMN EDITION, 2017 86. If
u, v >
87. Let
a, b, c be positive real numbers, take: a b b c c X = = , Y = = , Z Z = = b a c b a 4
2
2
a
. c
2
Y Z 3)( X Y Z 3) . Prove that: X + Y + Z 2 ( X NGUYEN NGOC TU, RMM A AUTUMN EDITION, 2017
unit squares of an N N N board board are coloured black and white so that squares that share a side have different colours, and so that at least one corner square is coloured black. bla ck. In eeach ach step step we choo choose se a 2 2 square and change the colour of all four unit squares inside that square, so that white unit squares become black, black become grey, and grey become white. Determine all positive integers N > > 1 for which it is possible, using a finite number of steps, to achieve that all unit squares that were originally black become white, and all unit squares that were originally or iginally white become black. 88. The
22
CROATIAN NMO, 2017
89. If x, y, z and and
w are real numbers such that: x2 + y2 + z 2 + w2 + x + 3 y + 5 z + + 7w = 4, determine the largest possible value of x + y + z + + w.
CROATIAN NMO, 2017
90. Determine
the maximum value of the expression: sin x sin y sin z + + cos x cos y cos z .
CROATIAN NMO, 2017
all functions f :: such that: f ( x x + f ( y y)) = f ( f f ( y y)) + 2 xf ( y y) + x2 holds for all real numbers x and y. TONCI K OKAN OKAN, CROATIAN NMO, 2017 91. Determine
92. Let
a)
a, b, c be positive real numbers such that a + b + c = 1. Prove that:
1 1 1 1 ; a bc bc b ca ca c ab a b 4abc a
b)
a bc 93. Let
b
c
b ca
1
.
cN ab 2 abc HUNG, RMM WINTER EDITION, 2017 GUYEN VIET
a, b, c positive numbers such that a4 + b4 + c4 = 3. Prove that:
a 3 b3 c 3 b 3 c 3 a 3 b5 c5 a5 a 5 b 5 c5 9. NGUYEN NGOC TU, H A GIANG, V VIETNAM 94. Let
a, b, c > 0 such that (a + b)(b + c)(c + a) = 8. Prove that: a a 1
95. Let
2b 2c 7 24 . b 1 c 1 2 NGUYEN NGOC TU, H A GIANG, V VIETNAM
m > 0 and F be be the area of the triangle ABC . Then: 2 2 2 a m bm cm b
96. Solve
m
c
m
2 3 F.
a a b D.M. B ĂTINEŢU-GIURGIU, ROMANIA, M ARTIN LUKAREVSKI, RMM WINTER EDITION, 2017
c
m
m
m
m
for real numbers: nn
( x12 x2 )
2
2
n
n
2
n n ( x2 x3 ) ... n n ( x 1 x ) n n( x x1 ) n
n
4
n
.
n D ANIEL SITARU, RMM WINTER EDITION, 2017
23
97. Inside
a circle of radius 1 (or on the circumference), one marks n points in such a way that the minimal distance between two marked points is as large as possible. Let d n be this distance between the two closest points. Is it true that d n + 1 < d n for every natural number n 2? ESTONIAN NMO, 2017 98. Find all 2 2
x + y) f f (( xy xy) = functions f : that for all numbers x and y satisfy f (( x
= f ( xx – y + 1).
ESTONIAN NMO, 2017
the system of equations 3 x + 7 y + 14 z = = 252, xyz – u2 = 2016 for non-negative real numbers. ESTONIAN NMO, 2017 99. Solve
100. Find
all positive integers n such that a square can be cut into n square pieces. ESTONIAN NMO, 2017
1 1 1 . Find the largest x y z 3 and the smallest possible value of the expression x2 + y3 + z 3 + xyz . ESTONIAN NMO, 2017 101. Real
numbers x, y and z satisfy satisfy x + y + z = 4 and
102. Solve
1
for real numbers: tan x tan y tan z 6 tan x tan y tan x tan z tan y tan z 11. x y z D ANIEL SITARU, RMM, ROMANIA
103. Find x, y, z
(0, ) such that: x y z xyz x y z 1 . y 3 z 2 z 3 x 2 x 3 y 2 3 D ANIEL SITARU, RMM, ROMANIA
104. Solve
for real numbers: 1 2 y 2 x 1 2 y x 1 2 y ( x 1) 3 . 1 2 1 4 3 3 3 4 y x x y NGO MINH NGOC B AO, RMM, V VIETNAM
24
105. Solve
the system of equations:
x y z 1 4 xyz (1). 4 4 xy yz zx 3 2 ( x 4 y z ) HOANG LE NHAT TUNG, RMM, V VIETNAM 106.
Solve for integers:
107. Solve
x( y z ) y 2 z 2 6 2 2 y ( z x ) z x 6 . z ( x y ) x 2 y 2 6 M ARIN CHIRCIU, RMM, ROMANIA
for real numbers:
cos 2 x cot 3 y tan 5 z cot 3 y cot 5 z tan 2 x . cot 5 z cot 2 x tan 3 y D ANIEL SITARU, RMM, ROMANIA 108. Suppose
a sequence of integers x1, x2, ... satisfies the following conditions: x x1 = 1, x2 = x3 = ... = x13 = 0. x xn + 13 = xn + 5 + 2 xn (n = 1, 2, ...). Find the value of x144. J APAN NMO, 2017
•
•
109. Solve
the question in : x
110. Solve
3
2
2x 2x 3
3
x
2
x 4 3x 3 x 1 2 4 x 3x 7. (1) 2 , RMM, V HOANG LE NHAT TUNG VIETNAM 4
4
for real numbers:
x 2 x . 2 ROVSEN PIRGULIEV, RMM, A AZERBAIJAN
arcsin[ x ] arccos[x ]
111. Solve
for real numbers: [tan x] (cot x – [cot x]) = (tan x – [tan x]) [cot x] [*] – great integer function.
ROVSEN PIRGULYEV, RMM, A AZERBAIJAN 25
112. Prove
that: sin 2
113. Find A, B,
7 5 7 5 3 sin sin 2 sin sin 2 sin . 18 18 18 18 18 18 4 GEORGE APOSTOLOPOULOS, RMM, GREECE
C (0, ), A + B + C = = such that: cos A | cos B | cos B | cos A | 1 cos 2C
cos B | cos C | cos C | cos B | 1 cos 2 A. cos C | cos A | cos A | cos C | 1 cos 2 B D ANIEL SITARU, RMM, ROMANIA
114. Solve
for real numbers:
27 x 2 3 y 1 y z 1 27 2 3 . 27 z 2 3x 1 SEYRAN IBRAHIMOV, RMM, A AZERBAIJAN 115. For
each positive real number x, we define { x} to be the greater of x and 1 , with
{1} = 1. Find, with proof, all positive real numbers y such that 5 y{8 y}{25 y} = 1. D AN GRILLER, BRITISH NMO, 2017 116. Naomi
and Tom play a game, with Naomi going first. They take it in turns to pick an integer from 1 to 100, each time selecting an integer which no one has chosen before. A player loses the game if, after their turn, the sum of all the integers chosen since the start of the game (by both of them) cannot be written as the difference of two square numbers. Determine if one of the players has a winning strategy, and if so, which. TOM BOWLER, BRITISH NMO, 2017 117. If
m [0, ); a, b, x, y, z (0, ), then: x 2 m 2 y 2m 2
(ay bz )2 m 2 sec2 m
18
z 2 m 2
3 . 4 m (a b)2 m 2
(ax by )2 m 2 csc2 m 9 9 D.M. B ĂTINEŢU-GIURGIU, D ANIEL SITARU, L A G ACETA DE L A RSME, 2017
(az bx )2 m 2 csc2 m
118. Given
a positive integer n, define f (0, j) = f (i, 0) = 0, f (1, (1, 1) = n, and f ((i, j) = f (i 1, j ) f (i , j 1) for all positive integers i and j, (i, j) (1, 1). How many = 2 2 ordered pairs of positive integers (i, j) are there for which f ((i, j) is an odd number? DUŠAN ĐJUKIĆ, SERBIAN NMO, 2016
26
119. Solve
for real numbers: 3 sin[ x ] cos( x [ x ]) 2 , [*] – great integer function. sin( x [ x ]) cos[x ] 3 2 ROVSEN PIRGULIYEV, RMM, A AZERBAIJAN
120. Solve
121. If
for x > 0 the equation: 1 1 1 1 . e x x x x 1 x 1 x cot (e ) cot ( ) e ROVSEN PIRGULIYEV, RMM, A AZERBAIJAN
a, r (0, ), then: n
k 1
122. Find
k
1 i 1 a i r ( 1 )
(2a (n 1)r )n , n *.
k
D ANIEL SITARU, RMM, ROMANIA
all functions f :: such that: f ( xf ( y) – f ( x)) = 2 f ( x x) + xy (1). ( xf ( y ( x VIETNAMESSE NMO, 2017
123. Prove
that in any a ny right-angled triangle the following inequality holds: ma mb mc 5 . 8 abc D.M. B ĂTINEŢU-GIURGIU, NECULAI STANCIU, ROMANIA
all the functions f :: * [1, ), which satisfy the conditions: (i) f (2) (2) = 2; (ii) f ( (n) f f ( (n + 1); (iii) f ( (nm) = f (n) f f ((m). D.M. B ĂTINEŢU-GIURGIU, NECULAI STANCIU, ROMANIA
124. Find
125. If x1, x2,
126. Let
be the sequences (an )n 2 , (bn )n 2 , (cn )n 2 defined: a 2
..., xn are real positive numbers, prove that: n xk 1 1. n 2 x1x2 ... xn k 1 ( x1 1)( x 2 1) ... ( xk 1) D.M. B ĂTINEŢU-GIURGIU, NECULAI STANCIU, ROMANIA
3, b 2, a 2
n 1
2
2
n
n
a 2b , b
bn2
2a b , c
n 1
27
n n
n
an2
, n 2 .
2
bn2
Prove that the sequence (cn )n2 has all the terms natural numbers. D.M. B ĂTINEŢU-GIURGIU, NECULAI STANCIU, ROMANIA 127. Solve
in the equation: x2 – 2 x + 2 xy – 2 y + y2 = 3.
M ARIN CHIRCIU, ROMANIA 128. Solve
129. Let
for real numbers: x 4 x 2 21 x 5 x 2 4 . M ARIN CHIRCIU, ROMANIA
be a > 0. Solve for real numbers: x a x
130. In ABCD cyclic
a3 a2
a ax (a 1) x 2 a . M ARIN CHIRCIU, ROMANIA
ABCD]. quadrilater, AB = a, BC = = b, CD = c, DA = d , S – – area [ ABCD 4S sin A + sin B + sin C + + sin D . abcd
D ANIEL SITARU, RMM, ROMANIA 131. In ABCD cyclic
quadrilater, AB = a, BC = = b, CD = c, DA = d , s – semiperimeter: s s s sin A sin B 1 1 1 1 . a b c d D ANIEL SITARU, RMM, ROMANIA
132. In ABCD convexes
quadrilateral: AB = a, BC = = b, CD = c, DA = d . Prove that:
a2
b 2 c 2 2 3 AC BD . D ANIEL SITARU, RMM, ROMANIA
133. If x, y, z
[0, ), then: x 2
134. If x, y, z
xy 3 y 3
y2
yz 2 z 2 x 2 xz z 2 . D ANIEL SITARU, RMM, ROMANIA
(0, ), xyz = = 1, then: x( x x – 3( y y + z ))))2 + (3 x – ( y y + z ))))2( y y + z ) 27. D ANIEL SITARU, RMM, ROMANIA
28
135. If
a, b, c > 0, then: a b
136. If
b
c
c
a
137. If
2c(a 3 1)
ab
bc
.
c4
a2
b2 c2
.
2a (b3 1) a 4 2b (c3 1) 2 HOANG LE NHAT TUNG, RMM, V VIETNAM
a, b, c > a6 + b6 + c6 = 9, then:
2 3 (a 138.
c4
cb ac b a GEORGE APOSTOLOPOULOS, RMM, GREECE
a, b, c > 0, a + b + c = 3, then: a4 b4 b4
ca
ab b b 2 a )2
1. b )2 (c3 a a 3 c )2 D ANIEL SITARU, RMM, ROMANIA
bc
ca
(b3 c c 3
Let be 0 < a < b < c. Prove that: (e
a b
e
ba
) 2a 2c 3
139. If
b
ab
.
a D ANIEL SITARU, SSMA M AGAZINE, 2017
a, b, c > 0, a b c a, then:
1 1 1 1 1 1 81 . (a b)2 (b c )2 (c a )2 ab bc ca 4(a 2 b 2 c 2 ) D ANIEL SITARU, RMM, ROMANIA 140. If
a, b, c > 0, a + b + c = 1, then: a3 + b3 + c3 + 6abc a a
141. Prove
2
2 bc
2
2
b b 2 ac c c 2 ab . D ANIEL SITARU, RMM, ROMANIA
that if x, y, z [1, 2], then in ABC : 2 3 a b c 16 s 3 abcxyz x y z 3 . D ANIEL SITARU, RMM, ROMANIA
142. If x, y, z
4 0 such that x3 + y3 + z 3 = 3 and n , prove that: 3 x n yn zn 3. nx 2 1 ny 2 1 nz 2 1
M ARIN CHIRCIU, ROMANIA
29
143. Solve
144. If
for real numbers: 2 x 2 a 2 x arcsin 2 a r c s i n , where a > 1, a is given. 2 2 a a M ARIN CHIRCIU, ROMANIA
a, b, c > 0 such that a + b + c = 1 and n *, prove that: n 1
n 1
n 1
a b c 1 a n 1 b n 1 c n 3n 1 1 . M ARIN CHIRCIU, OCTAVIAN STROE, ROMANIA 145. Let
O be circumcenter of ABC triangle. Prove that: a 2 b 2 AO AB BO BC CO CA
c2
.
2 M ARIN CHIRCIU, ROMANIA
146. Prove
that in any triangle ABC and for all positive real numbers x, y, z the following inequality holds: y z x y z x y z xyz 3 . ra rb rc rb rc ra rc ra rb r HUNG NGUYEN VIET, RMM, V VIETNAM 147. Prove
that in any triangle ABC the the following relationship holds: B
C
C
A
A
B
cot cot cot cot cot cot 2 2 2 2 2 2 2 cos A cos B cos C . A B C 2 2 2 cot cot cot 2 2 2 HUNG NGUYEN VIET, RMM, V VIETNAM 148. Let ABC be be
a triangle with a = BC , b = CA, and c = AB. Let A'B'C' be be another triangle with B'C' = = a , C'A' = = b , A'B' = c . Prove that: 1 1 1 sin A sin B sin C cos A cos B cos C . 2 2 2 MEHMET Ș AHIN, RMM, TURKEY acute-angled ABC , the following relationship holds: 1 1 (sin 2 A sinB ) sin 2 A sin 2B (tan A tan B )(cot A cot B ).
149. In
30
D ANIEL SITARU, RMM, ROMANIA
150.
Let be a, b, c (1, ). Solve for real numbers: x x1 x 1x 1x x x 1x 2 2 a b c a b c (a b c ) . M ARIN CHIRCIU, ROMANIA
151. Let
be n *. Find x 0 with the property: n 1
n 1
n (n 1) (n 1) x n . M ARIN CHIRCIU, ROMANIA 152. Prove
that the sequence (an)n 1 defined by an 1
an 1 an
2an 1 an
has a term that is
the reverse of a perfect square, then the sequence (bn)n 1 defined by bn
1 an
contains
an infinity of perfect squares. D.M. B ĂTINEŢU-GIURGIU, NECULAI STANCIU, ROMANIA 153. If
an = 1 + 2011n (n ), then prove that:
a) (an)n 0 contains an infinity of perfect squares; b) (an)n 0 contains an infinity of perfect squares; c) (an)n 0 contains an infinity of natural numbers having the form a2 +a + 1 (a ). D.M. B ĂTINEŢU-GIURGIU, NECULAI STANCIU, ROMANIA 154. If f :
,
such that for any 1 < x yy z z we we have f (( x x) + f (( y y) = z , f (( y y) +
+ f (( z z ) = x, f (( z z ) + f ( ( x x) = y and for any z > > 2 we have f (( z z ) > 1 + triplets ( x x, y, z ))..
z
2
, then find all the
D.M. B ĂTINEŢU-GIURGIU, NECULAI STANCIU, ROMANIA
that if a + b 0 such that a cos ax + b cos bx = c cos cx + d cos dx, xx , then a = c or a = d . D.M. B ĂTINEŢU-GIURGIU, NECULAI STANCIU, ROMANIA 155. Prove
156. Let
be n * and k , k 2. Solve the system:
2 n x x 2 n y y k 3 1 . 7 n x x x 7 n y y y k 1 M ARIN CHIRCIU, ROMANIA
31
157. The
real positive numbers x, y, z , n have the property that xy, yz , zx n. Prove
that if: (n x 2 )(n y 2 ) (n y 2 )(n z 2 ) (n z 2 )(n x 2 ) 3n xy yz zx . 2 2 2 2 2 2 2n x y 2n y z 2n z x 2 M ARIN CHIRCIU, ROMANIA 158. If f :: ,
such that f (ctg x) = sin 2 x + cos 2 x, x x (0, ), then find f ( x x). D.M. B ĂTINEŢU-GIURGIU, NECULAI STANCIU, ROMANIA
159. Prove
that for any triangle ABC and and all positive real numbers x, y, z the the following inequality holds: sin A sin B sin C x y z 3. y z sin B sin C z x sin C sin A x y sin A sin B NGUYEN VIET HUNG, RMM, V VIETNAM
160. In ABC , the following relationship holds:
sin2 A A cos4 A A + sin2 B B cos4 B B + sin2 C cos4 C < <
4
.
9 D ANIEL SITARU, RMM, ROMANIA 161. In ABC , the following relationship holds:
1 1 1 b 2 bc c 2 . 3 bcma ha hb hc D ANIEL SITARU, RMM, ROMANIA 162. Prove
that in any ABC the the following relationship holds: ma mb mc ra rb rc
wa wb wc ha hb hc
. r
D ANIEL SITARU, RMM, ROMANIA
163. In ABC , the following relationship holds:
2( AN AN 2 + BN 2 + CN 2) + 42 Rr 4 s2 + 3r ON N – – Nagel’s point, O – circumcentre, s – semiperimeter, r – – inradius, R – circumradius. D ANIEL SITARU, RMM, ROMANIA 164. In
acute triangle ABC , the following relationship holds: tan 4 A tan 4 B tan 4 C tan A tan B tan C . tan 3 B tan 3 C tan 3 C
D ANIEL SITARU, RMM, ROMANIA
32
165.
In ABC , the following relationship holds: 1 3 2 c o s c o s 1 6 s i n c o s a B C A A . 3 r
D ANIEL SITARU, RMM, ROMANIA 166. In ABC , the following relationship holds:
(ma + r a)(mb + r b)(mc + r c) 8wawbwc. D ANIEL SITARU, RMM, ROMANIA 167. In ABC , the following relationship holds:
ra2 a
168. In ABC , I –
rb2 b
r c2 81r 2
c
. 2p D ANIEL SITARU, RMM, ROMANIA
incentre, the following relationship relationship holds: 2 AI 2 ma mb wa wb mb mc wbwc mcma wcwa .
D ANIEL SITARU, RMM, ROMANIA 169. In ABC , the following relationship holds:
1 a
1
1
m7 m7 m7 m3 m3 m3 s 4 , s – semiperimeter. a
b
c
b
c
D ANIEL SITARU, RMM, ROMANIA 170. In ABC , the following relationship holds:
r a6 a3 16 3 2 9(a 2 b 2 c 2 )2 . a r a D ANIEL SITARU, RMM, ROMANIA 171. Prove that
172. Prove
in any triangle the following relationship holds: 2 p 3 ( r r )(r r )(r r ) . a b b c c a 3 ADIL ABDULLAYEV, RMM, A AZERBAIJAN
that in any triangle the following relationship holds:
ra2
1 6(4 R r ) .
cyc
ADIL ABDULLAYEV, RMM, A AZERBAIJAN 33
173. In ABC , the following relationship holds:
a 3 wa 2 3 4 wa . wa a D ANIEL SITARU, RMM, ROMANIA 174. Let S = =
{1, 2, ..., n}, n 2, and let f :: S S be be a bijective function distinct from n
the identity. Let u = | (k ) k | and let it be the number of ordered pairs (a, b) of k 1 elements of S such such that a > f (a) < f (b). Show that v < u 2v, and that u = 2v if and only if there do not exist positive integers a > b > c such that f (a) ... > xn 0, n 2, that satisfy a condition x1 + x2 + + ... + xn = n, prove an inequation: x2 + x3 + ... + xn). 2( xx1 x2 + x1 x3 + ... + x1 xn + x2 x3 + x2 x4 + ... + x2 xn + ... + xn – 1 xn) n ( x VITALY SENIN, UKRAINIAN NMO, 2017
175. For
176.
x, y the equation is true: Find all the functions f : f + f ( f f such that x x ( y y))) = y + f ( f f ( x x)). ( x ANDRIY ANIUKUSHIN, UKRAINIAN NMO, 2017
a prime number. Prove that equation ax2 + bx + c = 0 does not have
177. Let abc be
rational roots. JOSÉ LUIS DÍAZ-B ARRERO, B ARCELONA TECH, M ATH CONTEST, 2015 178. Let a, b, c be
three positive numbers such that ab + bc + ca = 3abc. Prove that: ab bc ca 3. 2 2 2 2 2 2 c(a b ) a (b c ) b (c a )
JOSÉ LUIS DÍAZ-B ARRERO, B ARCELONA TECH, M ATH CONTEST, 2017 179. If
, , 0, such that + + = and a 0, prove that: 2 2 a tg tg a tg tg a tg tg 9a 3 . M ARIN CHIRCIU, ROMANIA 2
180. If a, b, c a
+ b2 + c2 = 2, then: 2 3 ab csc bc csc ca csc 4. 7 7 7
D ANIEL SITARU, RMM, ROMANIA
34
*
181. If n
, n 2, a, b, c > 1, a + b + c = 3n + 1, then:
n
a
n
a
n a n a 18. D ANIEL SITARU, RMM, ROMANIA
182. If m , m 2,
then: m2
m t an 2
tan 2 11 tan 2 13 tan 2 21 2 3 m . 36 36 36 36 16 D ANIEL SITARU, RMM, ROMANIA
0, , then: 2 cos a sin a ccoos b sin a ssiin b 9 3 . 4 4 4 4 4 1 cos a 1 sin a cos b 1 sin a sin b 10 D ANIEL SITARU, RMM, ROMANIA
183. If a, b
184.
If a, b, c, x , then: a2 + b2 + c2 + (sin x + cos x + sin x cos x)(ab + bc + ca) 0. D ANIEL SITARU, RMM, ROMANIA a tetrahedron with AOB = BOC = = COA = 90 and let P be be any point inside the triangle ABC . Denote, respectively, by d a, d b, d c the distances from P to faces (OBC ), ), (OCA), (OAB). Prove that: 2 2 a) d a db d c2 OP 2 ; OA OB OC b) d a db d c ; 27 3 c) OA d a OB db3 OC d c3 OP 4 . 185. Let OABC be be
NGUYEN VIET HUNG, RMM, V VIETNAM 186. If x, y >
0; z , then: x y ( x y )2 6. 2 2 2 2 ( x sin z y cos z )( x cos z y sin z ) y x D ANIEL SITARU, RMM, ROMANIA
187. In ABC :
sin 2 A sin 2 B sin 2 C 2s 2 2 4 5 1 6 sin 1 sin 1 sin 1 5 13 65
35
sin 2 A sin 2 B sin 2 C 4s 2 2 . 1 1 1 tan 1 tan 1 tan 1 2 5 8
D ANIEL SITARU, RMM, ROMANIA 188. If a, b, c, d > >
0, then: (a + b)2(a + c)2(a + d )2(b + c)2(b + d )2(c + d )2
a
3
bcd
3
b
3
cda
3
c
3
dab
3
d
3
abc
3
.
MIHÁLY BENCZE, RMM, ROMANIA 189. If a, b, c, x, y, z > >
0, a + b + c 3, then: 729 . 3 1 1 1 y z D ANIEL SITARU, RMM, ROMANIA
(ax + by + cz ))((ay + bz + + cx)(az + + bx + cy)
190. If a, b, c >
3
0, a + b + c = 3, x , then:
a sin 2 x
3 b co s 2 x
3
b sin 2 x
3 c cos 2 c
3
c sin 2 x
3 a cos 2 x 4.
D ANIEL SITARU, RMM, ROMANIA 191. If x, y 0, n 1, n , AM = =
y , GM = = 2
xy , then:
2
x n y n 2n 2n AM GM . 2 UCHE ELIEZER OKEKE, RMM, NIGERIA , then: 2
192. If x, y, z 0,
2 2 2 2 x y l n ( 1 t a n ) l n ( 1 c o t ) l n sin2 z . D ANIEL SITARU, RMM, ROMANIA 193. Prove
cos x
that 2
sin x
+ 2
2
2 1 2
, x x .
IBRAHIM ABDULAZEEZ, RMM, NIGERIA 36
194. Prove that
in any triangle the inequality is true: 3 2n 3 (na – p)(nb – p)(nc – p) , where n 2. 3 M ARIN CHIRCIU, ROMANIA
195. Prove that
in any triangle the following inequality holds: 27n 1 2 2 n sin A , where 1 n 3. cos A 56 1 cos AM ARIN CHIRCIU, ROMANIA
196. Prove that
in any triangle the following inequality holds: n R k r p2 (6n 3k 27)r 2 , where n 4 and k 1. p 3 M ARIN CHIRCIU, ROMANIA
197. Prove that
in any triangle the following inequality is true: n(a 2 b 2 c 2 ) k (ab bc bc ca ca ) 2 (m m m ) , b c n k 3 a where n and k are are positive numbers with 2n 7k . M ARIN CHIRCIU, ROMANIA
198. Prove that
in any triangle the following inequality is true: 3p a b c 2 , where 0 n 1. 3 2n p na p nb p nc M ARIN CHIRCIU, ROMANIA
199. Prove that
in any triangle the following inequality holds: p n r (4 R r )3 (n 1) p 2 ( R r )( 4 R r ) n , where 0. , ROMANIA M ARIN CHIRCIU 4
200. Prove that
201. Prove
in any triangle the following inequality is true: A ( p b nc)( p c nb ) n sin , where n 0. bc 2 M ARIN CHIRCIU, OCTAVIAN STROE, ROMANIA
that if a, b, c, d , e, f , (0, ) and a + b + c = 2, d + + e + f = = 3, then: a b c d c f 9 . a b c 4 D ANIEL SITARU, RMM, ROMANIA
37
202.
If x, y, z 0, , then: 2 tan x tan y tan z 3 . sin y sin z sin z sin x sin x sin y 2 D.M. B ĂTINEŢU-GIURGIU, NECULAI STANCIU, RMM, ROMANIA
203. If n , n 3, x, y 0, 3
3
y 3 4
then:
x4
y 4 ... n
xn
y n (n 2) x 2 y 2 . D ANIEL SITARU, RMM, ROMANIA
204. If a, b, c 0,
then: a ab 3 abc a b a b c 3 a 3 . 3 2 K IRAN IRAN K EDLAYA EDLAYA, RMM, USA
205. If a, b, c, d > >
0, x, y , then: sin 2 x cos2 x sin 2 y cos 2 y a
b
2 . d abcd D ANIEL SITARU, RMM, ROMANIA
c
206. If a, b, c, d
> 0, then: 4 2 3 3 4 4 5 3 2 3 3 4 4 5 4 2 3 3 4 4 5 5 2 3 3 4 4 5 1 2a b c d 2 b c d a 3 c d a b 4 d a b c a 4672 . 3 UCHE ELIEZER OKEKE, RMM, NIGERIA
207. If a, b, c 0,
then: a + b + c
4
a
4 b 4 abc .
D ANIEL SITARU, RMM, ROMANIA 208. If
0 a < 3, 0 b < 5, 0 c < 7, then: 3 a 1 5 b 1 7 c 1 6. D ANIEL SITARU, RMM, ROMANIA
209. If
0 a < b < c < d , then: 3 a 3 4 b 2 6 c 8 d 6 ab 12 abc 20 abcd . D ANIEL SITARU, RMM, ROMANIA
38
210. If x, y, z > >
1 , then: 2 y 3 z (4 x 2 y 2 1)(36 y 2 z 2 1)(9z 2 x 2 1) 1 2 2 2 2 . 2304 x y z ( x 2 y 3z ) D ANIEL SITARU, RMM, ROMANIA
0, 6 xyz = =
211.
2
x + f ( x x) f f ( y y)) = xf ( x x + y) for all Determine all y.functions f : satisfying f ( x x and real numbers MEMO, 2017 212. In ABC , the following relationship holds: 3
9ra rb r c 2 2 m m m m a a b b ra rb r c . D ANIEL SITARU, RMM, ROMANIA 213. In ABC , the following relationship holds:
b4c7
c4 a7
a12
a 4b7
b12
3
R . D ANIEL SITARU, RMM, ROMANIA
c 12
214. In ABC , the following relationship holds:
hb5 hc4
10
ha
215. In
hc5 ha4 10
hb
ha5 hb4 10
hc
33
. 2S 2 D ANIEL SITARU, RMM, ROMANIA
acute-angled ABC , the following relationship holds: 2 1 4 R r a b c s a (m w h ) 3 r 2 . D ANIEL SITARU, RMM, ROMANIA
216. In ABC , K – –
Lemoine’s point. Prove that: AK b2
c2
BK c2
a2
mb mc . a 2 b2 a 2 b2 c2 ADIL ABDULLAYEV, RMM, A AZERBAIJAN CK
ma
217. In ABC , the following equality holds:
3
a
3
b c 3
3
3 3a 3 3b 3 3 3c 2. D ANIEL SITARU, RMM, ROMANIA
39
218. Prove
that in any triangle the double inequality is true: 6 r a b c 3 R , where n 0. n 2 R na b c nb c a nc a b 2( n 2) r M ARIN CHIRCIU, ROMANIA
219. Prove that
in any triangle the following inequality is true: p 4 n r (4 R r )3 (n 1) p 2 ( R r )( 4 R r ) , where n 0. M ARIN CHIRCIU, ROMANIA
220. Prove
that in any ABC triangle triangle the following inequality is true: bc( p na ) ca ( p nb ) ab ( p nc ) 3R (3 2n ) p , where n 1. M ARIN CHIRCIU, ROMANIA
221. Prove
that in any ABC triangle triangle the following double inequality inequa lity is true:
2
2
2
a (b c na ) b (c a nb ) c (a b nc )
S 2
24(2 n )
, where n 1.
M ARIN CHIRCIU, ROMANIA 222. Prove
that in any ABC triangle triangle the following inequality holds: ha hb hh h h 9r b c c a . ha hb hb hc hc ha 2 M ARIN CHIRCIU, ROMANIA
223. Prove that
224. In
in any triangle the following inequality holds: 2 4 r nr n 3 , where n 9. p 4 R r 9 M ARIN CHIRCIU, ROMANIA
acute-angled ABC , the following relationship holds: 1 3 sin 2 B sin 2C sin 2 A 3 sin2 A 2 3. D ANIEL SITARU, RMM, ROMANIA
225. Let ma, mb, mc be
the lengths of the medians of a triangle, and let wa, wb, wc be the lengths of the internal bisectors of the angle opposite of the sides of lengths a, b, c, 2 ma2 mb2 mc2 a b c 9. respectively. Prove that: 2 4 wa wb2 wc2 GEORGE APOSTOLOPOULOS, RMM, GREECE
40
Let ABC be be an arbitrary triangle, I a, I b, I c are excenters, r is inradius, and R is circumradius of ABC . Prove that: 12r 3 P( I a I b I c ) 6 R 3 , where P ( I I a I b I c) is perimeter of ABC . MEHMET Ș AHIN, RMM, TURKEY 226.
227. Let a, b, and c be
the side lengths of a triangle ABC , with inradius r . Prove that: 4 4
228. In
4
4
b2 B c2 C 6 r . t an tan tan 2 2 2 GEORGE APOSTOLOPOULOS, RMM, GREECE a
2 A
any triangle ABC , the following relationship holds: A B C R 2R tan tan tan 1. 2 2 2 2 r r GEORGE APOSTOLOPOULOS, RMM, GREECE
229. In ABC ,
the following relationship holds: 3 1 1 1 a b c 2 . 3 a b c b c ca c a ab a b ( ) ( ) ( ) b c a c a b a b c D ANIEL SITARU, RMM, ROMANIA
230. In ABC ,
the following relationship holds: a
231. Prove
6
b c 8r s 6
6
2
a5
. bc c2 D ANIEL SITARU, RMM, ROMANIA
b2
that in any triangle ABC : ma
ma
b c
mc
c
a
a
b
,
2 a 2 b 2 c where l a, l b, l c are internal angle bisectors from A, B, C , respectively. NGUYEN VIET HUNG, RMM, V VIETNAM la
lb
l c
232. Prove
that in any ABC triangle triangle the following inequality holds: sin A sin B sin C 3 , m n m sin B n sin A sin B m sin C n sin B sin C m sin A n sin C sin A where 2m n > 0. ARIN HIRCIU OMANIA
M
41
C
,R
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Inter-county Inter-coun ty Contests of Mathematic Mathematics, s, 9 -12 grades, 1998, Paralela 45 Publishing House, Pitești, 1998. 46. Dan Brânzei, Niculae Ghiciu, Iaroslav Chebici, Marin Chirciu, Mathematics in th th Scholar Contests, 5 -7 grades, 1997, Paralela 45 Publishing House, Pitești, 1997. 47. Dan Brânzei, Niculae Ghiciu, Iaroslav Chebici, Marin Chirciu, Mathematics in th th Scholar Contests, 9 -12 grades, 1997, Paralela 45 Publishing House, Pitești, 1997. th 48. Dan Brânzei, Sebastian Anița, Marin Chirciu, Geometry, 9 grades, MATE-2000 Lyceum, Paralela 45 Publishing House, Pitești, 1997. 49. Ștefan Alexe, Marin Chirciu, Arge ș Olympiads of Mathematics, Lyceum 1990-1995, GIL Publishing House, Zalău, 1995. 50. Marin Chirciu, Dragoș Constantinescu, Dumitru Săvulescu, Mathematics in th th Scholar Contests, 5 -7 grades, 1995, Paralela 45 Publishing House, Pitești, 1995. 51. Marin Chirciu, Inequalities with Sides and Radiuses in Triangle, from Initiation to Performance, Paralela 45 Publishing House, Pitești, 2017. 52. Marin Chirciu, Trigonometrical Inequalities, Paralela 45 Publishing House, Pitești, 2016. 53. Marin Chirciu, Geometrical Inequalities, Paralela 45 Publishing House, Pitești, 2015. 54. Marin Chirciu, Algebraic Inequalities, Paralela 45 Publishing House, Pitești, 2014. th 55. Marin Chirciu, Marian Haiducu, Octavian Stroe, Mathematics 8 grade, Algebra and Geometry. Notebook. Part I , Paralela 45 Publishing House, Piteşti, 2017. th 56. Ioan Chera, Marin Chirciu, Nicolae Vizuroiu, Mathematics, 5 grade, Semester I , Tiparg Publishing House, Pitești, 2008. 57. Marin Chirciu, Sorin Ulmeanu, Daniel Jinga, Octavian Stroe, Daniel Văcaru, Viorel Popa, Ilie Dinulescu, Gheorghe F. Molea, Mathematics, Scholar Contests 2009, th
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