HOMC 2015-2017

Chu Xuan Dung, Nguyen Minh Tuan Nguyen Van Mau (editor-in-chief)

PROCEEDINGS OF THE HOMC’S (FROM 2015 TO 2017)

Ha Noi - 2018

2

Contents 1 Hanoi Op en Mathematics Competition 1.1 Junior section . . . . . . . . . . . . . . . . . . . . . 1.1. 1.1.11 Hano Hanoii Open Open Math Mathem emat atic icss Co Comp mpet etit itio ion n 2015 2015 1.1. 1.1.22 Hano Hanoii Open Open Math Mathem emat atic icss Co Comp mpet etit itio ion n 2016 2016 1.1. 1.1.33 Hanoi Hanoi Open Open Mathem Mathemati atics cs Competi Competiti tion on 2017 2017 1.2 Senior section . . . . . . . . . . . . . . . . . . . . . 1.2. 1.2.11 Hanoi Hanoi Open Open Mathem Mathemati atics cs Competi Competiti tion on 2015 2015 1.2. 1.2.22 Hanoi Hanoi Open Open Mathem Mathemati atics cs Competi Competiti tion on 2016 2016 1.2. 1.2.33 Hanoi Hanoi Open Open Mathem Mathemati atics cs Competi Competiti tion on 2017 2017

7 7 7 9 12 14 14 16 19

2 Answers and Solutions 2.1 Junior section . . . . . . . . . . . . . . . . . . . . . 2.1. 2.1.11 Hanoi Hanoi Open Open Mathem Mathemati atics cs Competi Competiti tion on 2015 2015 2.1. 2.1.22 Hanoi Hanoi Open Open Mathem Mathemati atics cs Competi Competiti tion on 2016 2016 2.1. 2.1.33 Hanoi Hanoi Open Open Mathem Mathemati atics cs Competi Competiti tion on 2017 2017 2.2 Senior section . . . . . . . . . . . . . . . . . . . . . 2.2. 2.2.11 Hanoi Hanoi Open Open Mathem Mathemati atics cs Competi Competiti tion on 2015 2015 2.2. 2.2.22 Hanoi Hanoi Open Open Mathem Mathemati atics cs Competi Competiti tion on 2016 2016 2.2. 2.2.33 Hanoi Hanoi Open Open Mathem Mathemati atics cs Competi Competiti tion on 2017 2017

23 23 23 32 42 50 50 59 72

Preface Founded in 2004 by the Hanoi Society of Mathematics, Hanoi Open Mathematics Competition (HOMC) is a mathematics competition held every year for school students, and it has two sections: contestants with 14 and 16 of age (equivalently, those in 8 and 10 grade by the education classification in Viet Nam). As the original regular of HOMC, all questions, problems, and contestants presentation should be presented in English. Through those fourteen events, HOMC were attracted participation of thousands of school students as well as attention of a great number of high and secondary schools in Viet Nam. From 2013 to 2016, hosting by the Department of Education and Training of Ha Noi City, the HOMCs were posited in three cities: Ha Noi, Dac Lac, and Cao Lanh with the participation of thousands of students coming from 50 high and secondary schools in Viet Nam. Taking some interaction between students as a part of competition, the 2017-HOMC took place of Hanoi city and time of two days with 900 contestants. Supporting by the administrative Committee of Hanoi city, 2018-HOMC will be organized with two groups: Group A is the international HOMC, and Group B is the home HOMC. On the occasion of the 2018-HOMC, we are very pleased to introduce the proceedings of the three last events. In particular, this material includes the problems and questions of junior and senior Sections for each one of the HOMC’s’ from 2015 to 2017. We would like to announce that the publication and releasing the proceedings are hosted by the Department of Education and Training of Ha Noi City, and the book is considered as a registered part among the activities of the 2018-HOMC, by opening and sharing with the purposes of HOMC. Moreover, we think that old participants of the

CONTENTS

5

HOMC’s will find their contribitions, and hope that new comers in 2018-HOMC will be pleased for new cooperation. The proofreading of this book is completed by the suggestions and comments from Prof. Do Ngoc Diep, Prof. Nguyen Viet Hai, Dr. Dinh Sy Dai, Mr. Do Minh Khoa, and Mr. Pham Dang Long. On this occasion, we would like to express our special thanks for their fruitful contributions.

The authors

Chapter 1 Hanoi Open Mathematics Competition 1.1 1.1.1

Junior section Hanoi Open Mathematics Competition 2015

Important: Answer to all 15 questions. Write your answers on the answer sheets provided. For the multiple choice questions, stick only the letters (A, B, C, D or E) of your choice. No calculator is allowed.

Question 1. What is the 7th term of the sequence

{−1, 4, −2, 3, −3, 2, . . .}?

(A): -1; (B): -2; (C): -3; (D): -4; (E): None of the above.

Question 2. The last digit of number 20172017

2015

− 2013

is

(A): 2; (B): 4; (C): 6; (D): 8; (E): None of the above.

Question 3. The sum of all even positive intergers less than 100 those are not divisible by 3 is

8

Chapter 1. Hanoi Open Mathematics Competition


Question 4. A regular hexagon and an equilateral triangle have equal perimeter. If the area of the triangle is 4 3 square units, the area of the hexagon is

√

√

√

√

√

(A): 5 3; (B): 6 3; (C):7 3; (D):8 3; (E): None of the above.

Question 5. Let a, b, c and m (0

≤ m ≤ 26) be integers such that a + b + c = (a − b)(b − c)(c − a) = m (mod 27)

then m is (A): 0; (B): 1; (C): 25; (D): 26; (E): None of the above.

Question 6. Let a, b, c [ 1, 1] such that 1 + 2abc Prove that 1 + 2 a2 b2 c2 a4 + b4 + c4.

∈−

2

≥a

+ b2 + c2 .

≥

Question 7. Solve equation x4 = 2x2 + [x],

where [x] is an integral part of x.

Question 8. Solve the equation (2015x

3

− 2014)

= 8(x

3

− 1)

+ (2013x

3

− 2012) .

Question 9. Let a,b,c be positive numbers with abc = 1. Prove that 3

3

2



3

3

3

a + b + c + 2 (ab) + (bc) + (ca)

≥

3(a2 b + b2 c + c2 a).

Question 10. A right-angled triangle has property that, when a square is drawn externally on each side of the triangle, the six vertices of the squares that are not vertices of the triangle are concyclic. Assume that the area of the triangle is 9 cm2 . Determine the length of sides of the triangle.

9

1.1 Junior section

Question 11. Give a convex quadrilateral ABCD. Let O be the intersection point of diagonals AC and BD, and let I , K , H be feet of perpendiculars from B , O , C to AD, respectively. Prove that AD

× BI × CH ≤ AC × BD × OK.

Question 12. Give a triangle ABC with heights ha = 3 cm, h b = 7 cm and hc = d cm, where d is an integer. Determine d. Question 13. Give rational numbers x, y such that x2 + y 2





−2

(x + y)2 + (xy + 1) 2 = 0.

√ 1 + xy is a rational number.

Prove that

Question 14. Determine all pairs of integers (x; y ) such that 2xy 2 + x + y + 1 = x 2 + 2y 2 + xy.

Question 15. Let the numbers a, b, c satisfy the relation a 2 + b2 + c2 8. Determine the maximum value of

≤

M = 4(a3 + b3 + c3)

1.1.2

− (a

4

+ b4 + c4 ).

Hanoi Open Mathematics Competition 2016


Question 1. If 2016 = 25 + 26 +

··· + 2

m

,

10


then m is equal to (A): 8

(B): 9 (C): 10 (D): 11 (E): None of the above.

Question 2. The number of all positive integers n such that n + s(n) = 2016,

where s(n) is the sum of all digits of n, is (A): 1


Question 3. Given two positive numbers a, b such that a 3 + b3 = a5 + b5 , then the greatest value of M = a 2 + b2 ab is

−

(A):

1 1 (B): 4 2

(C): 2 (D): 1 (E): None of the above.

Question 4. A monkey in Zoo becomes lucky if he eats three different fruits. What is the largest number of monkeys one can make lucky, by having 20 oranges, 30 bananas, 40 peaches and 50 tangerines? Justify your answer. (A): 30 (B): 35 (C): 40 (D): 45 (E): None of the above.

Question 5. There are positive integers x, y such that 3x2 + x = 4y2 + y, and (x y ) is equal to

−

(A): 2013 (B): 2014 (C): 2015 (D): 2016 (E): None of the above.

Question 6. Determine the smallest positive number a such that the number of all integers belonging to (a, 2016a] is 2016.

Question 7. Nine points form a grid of size 3 triangles are there with 3 vetices at these points?

× 3.

How many

11

1.1 Junior section

Question 8. Find all positive integers x, y,z such that x3

− (x + y + z )

2

= (y + z )3 + 34.

Question 9. Let x, y,z satisfy the following inequalities

| | − | − |

x + 2y

− 3z | ≤ 6 x 2y + 3 z | ≤ 6 x 2y − 3z | ≤ 6 x + 2 y + 3 z | ≤ 6 Determine the greatest value of M = |x| + |y | + |z |. Question 10. Let ha , hb , hc and r be the lengths of altitudes and radius of the inscribed circle of ∆ABC, respectively. Prove that ha + 4 hb + 9 hc > 36 r.

Question 11. Let be given a triangle ABC, and let I be the middle point of BC. The straight line d passing I intersects AB,AC at M, N , respectively. The straight line d ( d) passing I intersects AB, AC at Q, P , respectively. Suppose M, P are on the same side of BC and M P , N Q intersect BC at E and F, respectively. Prove that IE = I F. 

≡

Question 12. In the trapezoid ABCD, AB CD and the diagonals intersect at O. The points P, Q are on AD, BC respectively such that ∠AP B = ∠CP D and ∠AQB = ∠CQD. Show that OP = OQ.



Question 13. Let H be orthocenter of the triangle ABC. Let d1 , d2 be lines perpendicular to each-another at H. The line d1 intersects AB, AC at D, E and the line d 2 intersects B C at F. Prove that H is the midpoint of segment DE if and only if F is the midpoint of segment BC. Question 14. Given natural numbers a, b such that 2015a2 + a = 2016b2 + b. Prove that a b is a natural number.

√ −

Question 15. Find all polynomials of degree 3 with integer coefficients such that f (2014) = 2015, f (2015) = 2016, and f (2013) f (2016) is a prime number.

−

12

1.1.3




Question 1. Suppose x1 , x2 , x3 are the roots of polynomial P (x) = x 3

2

− 6x

+ 5x + 12.

The sum x1 + x2 + x3 is (A): 4 (B): 6 (C): 8 (D): 14 (E): None of the above.

| | | | | |

Question 2. How many pairs of positive integers (x, y ) are there, those satisfy the identity 2x (A): 1

−y

2

= 1?


Question 3. Suppose n 2 + 4n + 25 is a perfect square. How many such non-negative integers n’s are there? (A): 1 (B): 2 (C): 4 (D): 6 (E): None of the above. Question 4. Put S = 21 + 35 + 4 9 + 513 +

2013

·· · + 505

+ 5062017 .

The last digit of S is (A): 1 (B): 3 (C): 5 (D): 7 (E): None of the above.

Question 5. Let a,b,c be two-digit, three-digit, and four-digit numbers, respectively. Assume that the sum of all digits of number a + b, and the sum of all digits of b + c are all equal to 2. The largest value of a + b + c is

13

1.1 Junior section


Question 6. that

Find all triples of positive integers (m,p,q ) such 2m p2 + 27 = q 3,

and p is a prime.

Question 7. Determine two last digits of number Q = 22017 + 20172 .

Question 8. Determine all real solutions x, y,z of the following system of equations

 

x3

− 3x 2y − 6y 3z − 9z 3

3

=4 =6 =8

−y − z − x.

Question 9. Prove that the equilateral triangle of area 1 can be covered by five arbitrary equilateral triangles having the total area 2. Question 10. Find all non-negative integers a, b, c such that the roots of equations: x2 x2 x2

− 2ax + b = 0; − 2bx + c = 0; − 2cx + a = 0

are non-negative integers.

Question 11. Let S denote a square of the side-length 7, and let eight squares of the side-length 3 be given. Show that S can be covered by those eight small squares. Question 12. Does there exist a sequence of 2017 consecutive integers which contains exactly 17 primes?

14


Question 13. Let a, b, c be the side-lengths of triangle ABC with a + b + c = 12. Determine the smallest value of M =

a b+c

−a

+

4b c+a

−b

+

9c a+b

− c.

Given trapezoid ABCD with bases AB CD Question 14. (A B < C D ). Let O be the intersection of AC and BD . Two straight lines from D and C are perpendicular to AC and BD intersect at E , i.e. CE BD and DE AC . By analogy, AF BD and BF AC . Are three points E , O, F located on the same line?



⊥

⊥

⊥

⊥

Question 15. Show that an arbitrary quadrilateral can be divided into nine isosceles triangles.

1.2 1.2.1

Senior section Hanoi Open Mathematics Competition 2015


Question 1. The sum of all even positive intergers less than 100 those are not divisible by 3 is (A): 938; (B): 940; (C): 1634; (D): 1638; (E): None of the above.


√

15

1.2 Senior section

√

√

√

√

(A): 5 3; (B): 6 3; (C):7 3; (D):8 3; (E): None of the above.

Question 3. Suppose that a > b > c > 1. One of solutions of the equation (x (c

− a)(x − b) + (x − b)(x − c) + ( x − c)(x − a) = x − a)(c − b) (a − b)(a − c) (b − c)(b − a)

is (A): -1; (B): -2; (C): 0; (D): 1; (E): None of the above.



then m is (A): 0; (B): 1; (C): 25; (D): 26; (E): None of the above.


2015

− 2013

is



∈−

2

≥a

+ b2 + c2 .

≥


where [x] is an integral part of x.

Question 8. Solve the equation (x + 1)3 (x

− 2)

3

+ (x

3

− 1) (x + 2)

3

= 8(x2

3

− 2) .

Question 9. Let a,b,c be positive numbers with abc = 1. Prove that a3 + b3 + c3 + 2[(ab)3 + (bc)3 + (ca)3 ]

2

2

2

≥ 3(a b + b c + c a).

16


Question 10. A right-angled triangle has property that, when a square is drawn externally on each side of the triangle, the six vertices of the squares that are not vertices of the triangle are concyclic. Assume that the area of the triangle is 9 cm2 . Determine the length of sides of the triangle. Question 11. Give a convex quadrilateral ABCD. Let O be the intersection point of diagonals AC and BD and let I , K , H be feet of perpendiculars from B , O , C to AD, respectively. Prove that AD


Question 12. Give an isosceles triangle AB C at A . Draw ray C x being perpendicular to CA, BE perpendicular to Cx (E Cx). Let M be the midpoint of BE , and D be the intersection point of BC. AM and Cx. Prove that BD

∈

⊥

Question 13. Let m be given odd number, and let a, b denote the roots of equation x2 + mx 1 = 0 and c = a2014 + b 2014 , d = a2015 + b2015 . Prove that c and d are relatively prime numbers.

−

Question 14. Determine all pairs of integers (x; y ) such that 2xy 2 + x + y + 1 = x 2 + 2y 2 + xy.

Question 15. Let the numbers a,b,c,d satisfy the relation a2 + b2 + c2 + d2 12. Determine the maximum value of

≤

M = 4(a3 + b3 + c3 + d3)

1.2.2

− (a

4

+ b4 + c4 + d4 ).



17

1.2 Senior section

Question 1. How many are there 10-digit numbers composed from the digits 1, 2, 3 only and in which, two neighbouring digits differ by 1. (A): 48 (B): 64 (C): 72 (D): 128 (E): None of the above.

Question 2. Given an array of numbers A = (672, 673, 674, . . . , 2016)

on table. Three arbitrary numbers a,b,c A are step by step re1 placed by number min(a,b,c). After 672 times, on the table there 3 is only one number m, such that

∈

(A): 0 < m < 1 (B): m = 1 (E): None of the above.

(C): 1 < m < 2 (D): m = 2

Question 3. Given two positive numbers a, b such that the condition a 3 + b3 = a 5 + b5 , then the greatest value of M = a 2 + b2 ab is

−

(A):

1 1 (B): 4 2


Question 4. In Zoo, a monkey becomes lucky if he eats three different fruits. What is the largest number of monkeys one can make lucky having 20 oranges, 30 bananas, 40 peaches and 50 tangerines? Justify your answer. (A): 30 (B): 35 (C): 40 (D): 45 (E): None of the above.

Question 5. There are positive integers x, y such that 3x2 + x = 4y2 + y and (x y) is equal to

−


Question 6. Let A consist of 16 elements of the set 1, 2, 3, . . . , 106 , so that the difference of two arbitrary elements in A are different

{

}

18


from 6, 9, 12, 15, 18, 21. Prove that there are two elements of A for which their difference equals to 3.

Question 7. Nine points form a grid of size 3 3. How many triangles are there with 3 vertices at these points?

×

Question 8. Determine all 3-digit numbers which are equal to cube of the sum of all its digits. Question 9. Let rational numbers a, b, c satisfy the conditions a + b + c = a 2 + b2 + c2

∈ Z.

Prove that there exist two relative prime numbers m, n such that m2 abc = 3 . n

Question 10. Given natural numbers a, b such that 2015a2 + a = 2016b2 + b. Prove that a b is a natural number.

√ −

Question 11. Let I be the incenter of triangle ABC and ω be its circumcircle. Let the line AI intersect ω at point D = A. Let F and E be points on side BC and arc BDC respectively such 1 that ∠BAF = ∠CAE < ∠BAC . Let X be the second point of 2 intersection of line EI with ω and T be the point of intersection of segment DX with line AF . Prove that TF.AD = ID.AT .



Question 12. Let A be point inside the acute angle xOy. An arbitrary circle ω passes through O, A; intersecting Ox and Oy at the second intersection B and C, respectively. Let M be the midpoint of BC. Prove that M is always on a fixed line (when ω changes, but always goes through O and A). Question 13. Find all triples (a,b,c) of real numbers such that 2a + b 4 and

|

|≥

2

|ax

+ bx + c

| ≤ 1 ∀x ∈ [−1, 1].

19

1.2 Senior section

Question 14. Let f (x) = x2 + px + q, where p, q are integers. Prove that there is an integer m such that f (m) = f (2015).f (2016).

Question 15. Let a, b, c be real numbers satisfying the condition 18ab + 9ca + 29bc = 1. Find the minimum value of the expression T = 42a2 + 34b2 + 43c2 .

1.2.3



Question 1. Suppose x1 , x2 , x3 are the roots of polynomial P (x) = x 3 4x2 3x + 2. The sum x1 + x2 + x3 is (A): 4 (B): 6 (C): 8 (D): 10 (E): None of the above.

−

−

| | | | | |

Question 2. How many pairs of positive integers (x, y ) are there, those satisfy the identity 2x (A): 1 (B): 2

Question 3. equation

−y

2

= 4?

(C): 3 (D): 4 (E): None of the above. The number of real triples (x,y,z ) that satisfy the x4 + 4y 4 + z 4 + 4 = 8xyz

20


is (A): 0;

(B): 1; (C): 2; (D): 8; (E): None of the above.

Question 4. Let a,b,c be three distinct positive numbers. Consider the quadratic polynomial P (x) =

c(x (c

− a)(x − b) + a(x − b)(x − c) + b(x − c)(x − a) + 1. − a)(c − b) (a − b)(a − c) (b − c)(b − a)

The value of P (2017) is (A): 2015 (B): 2016 (C): 2017 (D): 2018 (E): None of the above.

Question 5. Write 2017 following numbers on the blackboard: 1007 1 1 2 1007 1008 − 1008 ,− ,...,− , 0, , ,..., , . 1008 1008 1008 1008 1008 1008 1008 One processes some steps as: erase two arbitrary numbers x, y on the blackboard and then write on it the number x + 7xy + y. After 2016 steps, there is only one number. The last one on the blackboard is 1 1 144 (A): (B): 0 (C): (D): (E): None of the 1008 1008 1008 above.

−

−

Question 6. Find all pairs of integers a, b such that the following system of equations has a unique integral solution (x,y,z )



x + y = a

−1 x(y + 1) − z = b. 2

Question 7. Let two positive integers x, y satisfy the condition . x2 + y 2 .. 44. Determine the smallest value of T = x 3 + y3 . Question 8. Let a, b, c be the side-lengths of triangle ABC with a + b + c = 12. Determine the smallest value of M =

a b+c

−a

+

4b c+a

−b

+

9c a+b

− c.

21

1.2 Senior section

Question 9. Cut off a square carton by a straight line into two pieces, then cut one of two pieces into two small pieces by a straight line, ect. By cutting 2017 times we obtain 2018 pieces. We write number 2 in every triangle, number 1 in every quadrilateral, and 0 in the polygons. Is the sum of all inserted numbers always greater than 2017? Consider all words constituted by eight letters Question 10. from C , H , M , O . We arrange the words in an alphabet sequence. Precisely, the first word is CCCCCCCC, the second one is CCCCCCCH, the third is CCCCCCCM, the fourth one is CCCCCCCO,. . . , and the last word is OOOOOOOO. a) Determine the 2017th word of the sequence? b) What is the position of the word HOMCHOMC in the sequence?

{

}

Question 11. Let ABC be an equilateral triangle, and let P stand for an arbitrary point inside the triangle. Is it true that



 P AB

−

 ≥ 

P AC



 P BC  P CB ?

−

Question 12. Let (O) denote a circle with a chord AB , and let W be the midpoint of the minor arc AB . Let C stand for an arbitrary point on the major arc AB . The tangent to the circle (O) at C meets the tangents at A and B at points X and Y, respectively. The lines W X and W Y meet AB at points N and M , respectively. Does the length of segment NM depend on position of C ? Question 13. Let ABC be a triangle. For some d > 0 let P stand for a point inside the triangle such that

|AB| − |P B| ≥ d, and |AC | − |P C | ≥ d. Is the following inequality true

|AM | − |P M | ≥ d,

22


for any position of M

∈ BC ?

Question 14. Put P = m 2003 n2017

2017

−m

n2003 , where

m, n

∈ N.

a) Is P divisible by 24? b) Do there exist m, n N such that P is not divisible by 7?

∈

Question 15. Let S denote a square of side-length 7, and let eight squares with side-length 3 be given. Show that it is impossible to cover S by those eight small squares with the condition: an arbitrary side of those (eight) squares is either coincided, parallel, or perpendicular to others of S .

Chapter 2 Answers and Solutions 2.1 2.1.1

Junior section Hanoi Open Mathematics Competition 2015

Question 1. What is the 7th term of the sequence

{−1, 4, −2, 3, −3, 2,...}? (A): -1; (B): -2; (C): -3; (D): -4; (E) None of the above. Solution.

Let a 1 =

... Then

−1, a = 4, a = −2, a 2

3

4

= 3, a5 =

−3, a = 2 6

− a = a − a = a − a = 5,.. a − a = a − a = −6. Hence, a − a = −6 and a = −4. a2

1

4

3

3

2

5

4

7

6

6

5

7

Answer: D.


2015

− 2013

is

(A): 2; (B): 4; (C): 6; (D): 8; (E) None of the above. We have 2017 = 7 (mod 10) and 20172017 = 74 504+1 (mod 10) = 7 (mod 10). On the other hand, 2015 = 4 503+ 3 = 3 (mod 4) and 2013 = 3 (mod 10) and then ×

Solution.

×

20132015 = 34

×503+3

=7

(mod 10).

24

Chapter 2. Answers and Solutions

Hence, the last digit of number 20172017

− 2013

2015

is 0.

Answer: E.

Question 3. The sum of all even positive intergers less than 100 those are not divisible by 3 is (A): 938; (B): 940; (C): 1634; (D): 1638; (E) None of the above. We find the sum of all positive intergers less than 50 and subtract the sum of multiples of 3 between 0 and 50: Solution.

50

× 49 − 3 16 × 17 = 1225 − 408 = 817. 2

2

Hence the sum of all even positive intergers less than 100 those are not divisible by 3 is 2 817 = 1634.

×

Answer: C.


√

√

√

√

√

(A): 5 3; (B): 6 3; (C):7 3; (D):8 3; (E) None of the above. Let a, S be the length and area of the hexagon. Then the equilateral triangle has side-length 2a. We subdivide the hexagon into 6 small equilateral of side-length a then each of them must have Solution.

S

area . Since the equilateral triangle of side-length 2a has twice 6 the base and twice the altitude of the smaller triangle, it must have 4 times the area. Hence 4

S

6

√

√

= 4 3, it follows S = 6 3.

Answer: B.



25

2.1 Junior section

then m is (A): 0; (B): 1; (C): 25; (D): 26; (E) None of the above. - If a, b, c have three different remainder terms modulo 3 then a + b + c is divisible by 3 but a b, b c, c a are not divisible by 3. - If a, b, c have two different remainder terms modulo 3 then a + b + c is not divisible by 3 but one of three differences a b, b c, c a are divisible by 3. It follows the product (a b)(b c)(c a) is divisble by 3. - Hence, three numbers a,b,c have the same remainder terms, i.e. a = 3 p + s, b = 3q + s, c = 3r + s with s 0, 1, 2 , p, q,r are natural numbers. Hence, a b = 3( p q ), b c = 3(q r), c a = 3(r p) and then (a b)(b c)(c a) = 0 (mod 27). Solution.

− − −

− − − − −

−

∈ {

− − −

− −

−

−

}

−

−

Answer: A.


∈−

2

≥a

+ b2 + c2 .

≥

Solution.

The constraint can be written as (a

2

2

2

− bc) ≤ (1 − b )(1 − c ).

(2.1)

Using the Cauchy inequality, we have (a + bc)2

2

2

2

2

≤ (|a| + |bc|) ≤ (1 + |b||c|) ≤ (1 + b )(1 + c ).

Multiplying by (2.1), we get 2

(a

2

2

2

2

2

− bc) (a + bc) ≤ (1 − b )(1 + b )(1 − c )(1 + c ) ⇔ (a − b c ) ≤ (1 − b )(1 + b )(1 − c )(1 + c ) ⇔ (a − b c ) ≤ (1 − b )((1 − c ) ⇔ 1 + 2a b c ≥ a + b + c . 2

2 2 2

2

2

2

2 2 2

2 2 2

2

4

4

4

4

4

2

26



(2.2)

where [x] is an integral part of x. Solution.

We have 2

(2.2)

2

 −

⇔ [x] = x x 2 ≤ 2, then −√ 2 ≤ x ≤ √ 2 and [x] ≤ 0. It

Consider the case x2 follows [x] 1; 0 . If [x] = 0, then from (2.2) we find x = 0 as a solution. If [x] = 1, then from (2.2) we find x = 1 as a solution. Now we suppose that x2 > 2 . It follows from (2.2), [x] > 0 and [x] 1 then x > 2. Hence x2 (x2 2) = 1 and x2 2 < 1,

∈ {− }

− √ −√ √ √ then x < 3 and 2 < x < 3.

−

x

≤

It means that [x] = 1 and then x = the equation.

− ≤ x



1+

√ 2 is a solution of

Question 8. Solve the equation (2015x

3

− 2014)

Solution. Rewrite

(2015x

= 8(x

− 1)

3

+ (2013x

3

(2.3)

3

(2.4)

− 2012) .

equation (2.3) in the form 3

− 2014)

= (2x

− 2)

3

+ (2013x

− 2012) .

Factoring the sum of cubes on the right side of the equation (2.4), we find that one factor is 2015x 2014, thus, one solution of the 2014 equation is x = . 2015 Now we rewrite the equation (2.4) as

−

(2015x

3

− 2014) − (2x − 2)

3

= (2013x

3

− 2012) .

(2.5)

Factoring the difference of cubes on the left side of the equation (2.5), we find that one factor is 2013x 2012, thus, one solution of 2012 . the equation is x = 2013

−

27

2.1 Junior section

Finally, we rewrite (2.4) as (2015x

3

3

− 2014) − (2013x − 2012)

= (2x

3

− 2) .

Factoring again, we see that x = 1 is a solution. Since (2.3) is cubic in x, it has at most 3 roots, and we have them.

Question 9. Let a,b,c be positive numbers with abc = 1. Prove that 3

3



3

3

3

≥

3

a + b + c + 2 (ab) + (bc) + (ca) Solution. Assume

3(a2 b + b2 c + c2 a).

that a = max a,b,c . We then have

{ } a ≥ b ≥ c > 0 ,

or a

Hence, a3 + b3 + c3

2

2

≥ c ≥ b > 0.

2

2

2

2

2

− (a b + b c + c a) = (a − b)(a − c ) +(b − c)(b − c ) ≥ 0.

This implies a3 + b3 + c3

Due to the facts that 1 c

and

1 b

we deduce

1 c3

+

1 b3

= max

= max +

2

2

2

≥ a b + b c + c a.

(2.6)

1 1 1

   

, , , a b c

1 1 1

, , , a b c

1

1 1 1 + + ≥ . a c b b a a c 3

2

2

2

By abc = 1, this can be written as (ab)3 + (bc)3 + (ca)3

2

2

2

≥ a b + b c + c a.

(2.7)

Inequalities (2.6) and (2.7) together imply the proposed inequality.

28


Question 10. A right-angled triangle has property that, when a square is drawn externally on each side of the triangle, the six vertices of the squares that are not vertices of the triangle are concyclic. Assume that the area of the triangle is 9 cm2 . Determine the length of sides of the triangle. Solution. We

have OJ = OD = OG = radius of the circle. Let a,b,c denote the sides of ∆ABC. We have 1 = b 2 + ab + (a2 + b2 ); 2 2 4 2 2 1 b a OD 2 = ON 2 + ND 2 = a + + = a 2 + ab + (a2 + b2 ); 2 2 4 2 5 5 c OG2 = OL2 + LG2 = c 2 + = c2 = (a2 + b2 ) 2 4 4 1 = a 2 + b2 + (a2 + b2). 4 2

2

2

a

2

b

2

      

OJ = OM + M J = b +

+

Comparing these right-hand sides, we get b2 + ab = a 2 + ab = a 2 + b2

⇔ a = b.

It means that the given triangle with the desired property is the isosceles right triangle. Therefore, 1 ab = 9, 2 and

√

a = b = 3 2, c = 6.

Question 11. Given a convex quadrilateral ABCD. Let O be the intersection point of diagonals AC and BD, and let I , K , H be feet of perpendiculars from B , O , C to AD, respectively. Prove that

× BI × CH ≤ AC × BD × OK. Solution. Draw AE ⊥BD (E ∈ BD ). We have BI × AD AE × BD S . = = AD

ABD

2

2

29

2.1 Junior section

Then BI

× AD = AE × BD. It follows BI × AD ≤ AO × BD (AE ≤ AO),

and BI.AD

≤ AC × BD × AO . AC

Moreover, by OK CH it follows



AO OK = , and AC CH

× AD ≤ AC × BD × OK . CH Hence, BI × AD × CH ≤ AC × BD × OK. BI

The equality occurs if quadrilateral ABCD has two perpendicular diagonals.

Figure 2.1: Figure of Question 11

Question 12. Give a triangle ABC with heights ha = 3 cm, h b = 7 cm and hc = d cm, where d is an integer. Determine d. Solution.

Since 2S ABC = a.h a = b.h b = c.h c it follows a

=

1 ha

b

1

=

c

1

hb

hc

1

1

.

On the other hand,

| a − b| < c < a + b

 ⇔

ha

−h

b



<

1 hc

<

1 ha

+

1 hb

.

30


Hence

1 3

− 71 < h1

1 1 + 3 7 c 4 1 10 < < 21 21 hc 20 20 20 < < 105 20hc 42 105 > 20 hc > 42

⇔

Since hc

⇔ ⇔

<

∈ N , we direive h ∈ {3, 4, 5}. ∗

c

Question 13. Give rational numbers x, y such that x2 + y 2





−2

(x + y)2 + (xy + 1) 2 = 0.

(2.8)

√ 1 + xy is a rational number.

Prove that

−y = t. The equality (2.8) is of the form (t − 1) = 0 ⇔ t = ±1. √ It follows 1 + xy = 0, and 1 + xy is a rational number.  y. We have Consider the case x = Solution.

Let x =

2



x2 + y 2

⇔x

2

2

(x + y)2 + (xy + 1)2 = 0

  

−2 2

2

xy + 1 x+y

+y +

=2

2

⇔ (x + y) − 2.(xy + 1) +

 ⇔

x+y

−

xy + 1 x+y



2

  xy + 1 x+y

=0

It follows

2

xy + 1 = (x + y ) ,

and



xy + 1 = x + y .

The proof is completed.

|

|

2

=0

31

2.1 Junior section

Question 14. Determine all pairs of integers (x; y ) such that 2xy 2 + x + y + 1 = x 2 + 2y 2 + xy. Solution.

We have 2xy 2 + x + y + 1 = x 2 + 2y2 + xy ; 2y 2(x 1) y(x 1) x(x 1) = (x 1)(2y2 y x) = 1.

− − − − − − − − −

Since x; y are integers, we deduce that x divisors of -1. Case 1. We have

−

1=1

2y 2

− y − x = −1.



x = 2 y = 1,

or

  − − x = 2 y =

Case 2. We have

1 (absurd). 2

1=

−1 − y − x = 1.

x

2y 2

Equivalently, either



x = 0 y = 1,

or

 

x = 0 y =

− 12 (absurd).

2

− 1 and 2y − y − x are

x

Then, either

−1;

32


We deduce that all integer pairs (x; y ) are (2; 1); (0 ; 1).

Question 15. Let the numbers a, b, c satisfy the relation a 2 + b2 + c2 8. Determine the maximum value of

≤

M = 4(a3 + b3 + c3)

− (a

4

+ b4 + c4 ).

Note that x2 (x 2)2 0 for each real x. This inequality can be rewritten as 4x3 x4 4x2 . It follows that Solution.

(4a3

4

− a ) + (4b

3

− ≥ − ≤ − b ) + (4c − c ) ≤ 4(a 4

3

4

2

+ b2 + c2 ) = 32,

The equality holds for (a,b,c) = (2, 2, 0), (2, 0, 2), (0, 2, 2). Hence, max M = 32.

2.1.2


Question 1. If 2016 = 25 + 26 +

··· + 2

m

,

then m is equal to (A): 8


Anwser. (C).

Question 2. The number of all positive integers n such that n + s(n) = 2016,

where s(n) is the sum of all digits of n, is (A): 1


Anwser. (B):

n = 1989, 2007.

Question 3. Given two positive numbers a, b such that a 3 + b3 = a5 + b5 , then the greatest value of M = a 2 + b2 ab is

−

33

2.1 Junior section

(A):

1 1 (B): 4 2


Anwser. (D). Solution. We

ab(a2

have

2 2

3 3

5

5

3

3 2

5

5

− b ) ≥ 0 ⇔ 2a b ≤ ab + a b ⇔ (a + b ) ≤ (a + b)(a + b ).

Combining this and a3 + b3 = a 5 + b5 , we find a3 + b3

≤a+b⇔a

2

+ b2

− ab ≤ 1.

The equality holds if a = 1, b = 1.

Question 4. A monkey in Zoo becomes lucky if he eats three different fruits. What is the largest number of monkeys one can make lucky, by having 20 oranges, 30 bananas, 40 peaches and 50 tangerines? Justify your answer. (A): 30 (B): 35 (C): 40 (D): 45 (E): None of the above. Anwser. (D).

First we leave tangerines on the side. We have 20 + 30 + 40 = 90 fruites. As we feed the happy monkey is not more than one tangerine, each monkey eats fruits of these 90 at least 2. Hence, the monkeys are not more than 90/2 = 45. We will show how you can bring happiness to 45 monkeys: 5 monkeys eat: orange, banana, tangerine; 15 monkeys eat: orange, peach, tangerine; 25 Monkeys eat peach, banana, tangerine. At all 45 lucky monkeys - and left five unused tangerines! Solution.

Question 5. There are positive integers x, y such that 3x2 + x = 4y2 + y, and (x y ) is equal to

−

(A): 2013 (B): 2014 (C): 2015 (D): 2016

34


(E): None of the above. Anwser. (E).

Since x

− y is a square. Solution. We have 3x + x = 4y + y ⇔ (x − y )(3x +3y +1) = y . We prove that (x − y ; 3x + 3y + 1) = 1. Indeed, if d = (x − y ; 3x + 3y + 1) then y is divisible by d and 2

2

2

2

2

y is divisible by d; x is divisible by d, i.e. 1 is divisible by d, i.e. d = 1.

Since x y and 3x + 3y + 1 are prime relative then x perfect square.

−

− y is a

Question 6. Determine the smallest positive number a such that the number of all integers belonging to (a, 2016a] is 2016. Solution. The

smallest integer greater than a is [a] + 1 and the largest integer less than or is equal to 2016a is [2016a]. Hence, the number of all integers belonging to (a, 2016a] is [2016a] [a]. Now we difine the smallest positive number a such that

−

[2016a]

− [a] = 2016. If 0 < a ≤ 1 then [2016a] − [a] < 2016. If a ≥ 2 then [2016a] − [a] > 2016.

Let a = 1 + b, where 0 < b < 1 . We have [a] = 1, and [2016a] = 2016 + [2016b]. This follows [2016a]

− [a] = 2015 + [2016b] = 2016,

which gives [2016b] = 1. Hence, the smallest positive number b 1 1 such that [2016b] = 1 is b = Thus, a = 1 + is a smallest 2016. 2016 positive number such that the number of all integers belonging to (a, 2016a] is 2016.

Question 7. Nine points form a grid of size 3 triangles are there with 3 vetices at these points?

× 3.

How many

35

2.1 Junior section

We divide the triangles into two types: Type 1: Two vertices lie in one horizontal line, the third vertice lies in another horizontal lines. For this type we have 3 possibilities to choose the first line, 2 posibilities to choose 2nd line. In first line we have 3 possibilities to choose 2 vertices, in the second line we have 3 possibilities to choose 1 vertex. In total we have 3 2 3 3 = 54 triangles of first type. Type 2: Three vertices lie in distinct horizontal lines. We have 3 3 3 triangles of these type. But we should remove degenerated triangles from them. There are 5 of those (3 vertical lines and two diagonals). So, we have 27 - 5 = 22 triangles of this type. Total we have 54 + 22 = 76 triangles. For those students who know about C nk this problem can be also solved as C 93 8 where 8 is the number of degenerated triangles. Solution.

×××

× ×

−

Question 8. Find all positive integers x, y,z such that x3 Solution.

− (x + y + z )

2

Putting y + z = a, a x3

−a

3

= (y + z )3 + 34.

∈ Z, a ≥ 2, we have

= (x + a)2 + 34.

(2.9)

Equivalently, (x

− a)



x2 + xa + a2 = x 2 + 2ax + a2 + 34;

(x

− a − 1)



x2 + xa + a2 = xa + 34.





Since x, a are integers, we have x2 + xa + a2 That follow x a 1 > 0, i.e. x a 2. This and (2.10) together imply

− −

x2 + 2 ax + a2 + 34

− ≥

≥2



x2 + xa + a2

≥ 0 and xa + 34 > 0.

⇔

x 2 + a2

Hence x2 < 34 and x < 6 . On the other hand, x a + 2 4 then x 4, 5 . If x = 5, then from x2 + a 2 34 it follows 2 a 2, 3 . a

≥

∈{ }

(2.10)

≥ ≤

≤ 34.

∈{ } ≤ ≤ 3. Thus

36


The case of x = 5, a = 2 does not satisfy (2.9) for x = 5, a = 3, from (1) we find y = 1, z = 2 or y = 2, z = 1, If x = 4, then from the inequality x a 2 we find a 2, which contradicts to (2.9).

− ≥

≤

Conclusion: (x,y,z ) = (5, 1, 2) and (x,y,z ) = (5, 2, 1).

Question 9. Let x, y,z satisfy the following inequalities

| | − | − |

x + 2y

− 3z | ≤ 6 x 2y + 3 z | ≤ 6 x 2y − 3z | ≤ 6 x + 2 y + 3 z | ≤ 6 Determine the greatest value of M = |x| + |y | + |z |. Solution. Note

that for all real numbers a,b, c, we have

|a| + |b| = max{|a + b|, |a − b|} and

|a| + |b| + |c| = max{|a + b + c|, |a + b − c|, |a − b − c|, |a − b + c|}. Hence M = x + y + z

| | | | | | ≤ |x| + 2|y| + 3|z | = |x| + |2y| + |3z | = max{|x + y + z |, |x + y − z |, |x − y − z |, |x − y + z |} ≤ 6. Thus max M = 6 when x = ±6, y = z = 0. Question 10. Let ha , hb , hc and r be the lengths of altitudes and radius of the inscribed circle of ∆ABC, respectively. Prove that ha + 4 hb + 9hc > 36 r.

37

2.1 Junior section

Let a, b, c be the side-lengths of ∆ABC corresponding to ha , hb , hc and S be the area of ∆ABC. Then Solution.

aha = bh b = ch c = (a + b + c)

× r = 2S.

Hence ha + 4 hb + 9 hc =

= 2S

2S a



8S

=

b

2

1

a

=

2

+

2

b

+

18S c

3

2

c

≥

(1 + 2 + 3)2 2S a+b+c

(1 + 2 + 3)2 = (a + b + c) r = 36r. a+b+c

The equality holds iff a : b : c = 1 : 2 : 3 (it is not posible for a + b > c).

Question 11. Let be given a triangle ABC and let I be the middle point of BC. The straight line d passing I intersects AB,AC at M, N , respectively. The straight line d ( d) passing I intersects AB, AC at Q and P , respectively. Suppose M, P are on the same side of BC and M P , N Q intersect BC at E and F, respectively. Prove that IE = I F. 

Solution.

≡

Since IB = I C then it is enough to show

EB F C = . EC FB

By Menelaus theorem: - For ∆ABC and three points E,M, P, we have EB EC

MA × PP C × = 1. A MB

It follws

EB PA MB = . EC P C M A - For ∆ABC and three points F,N, Q, we have

×

(2.11)

(2.12)

F C FB

NA × QB × = 1. QA NC

This implies

F C NC = FB NA

QA × QB .

38


Figure 2.2: Figure of Question 11 - For ∆ABC and three points M , I , N , we have MB MA

A IC = 1. × NN C × IB

Compare with IB = I C we find MB NC = . MA NA

(2.13)

- For ∆ABC and three points Q, I, P, we have PA P C

IC QB × IB × QA = 1.

Hence, PA QA . = P C QB

(2.14)

Equalities (2.11), (2.12), (2.13), and (2.14), toghether imply IE = IF.

Question 12. In the trapezoid ABCD, AB CD and the diagonals intersect at O. The points P, Q are on AD, BC respectively such that ∠AP B = ∠CP D and ∠AQB = ∠CQD. Show that OP = OQ.



39

2.1 Junior section

Figure 2.3: Fugure of Question 12 Extending DA to B such that BB = BA , we find ∠P B B = ∠B AB = ∠P DC and then triangles DP C and B P B are similar. 

Solution. 







DP CD CD DO = = = and so P O P B BB  BA BO Since triangles DP O and DB  B are similar, we have

It follows that

OP = BB 

 BB . 

DO DO = AB × × DB . DB

Similarly, we have OQ = AB

and it follows OP = OQ. × CO CA

Question 13. Let H be orthocenter of the triangle ABC. Let d1 , d2 be lines perpendicular to each-another at H. The line d1 intersects AB, AC at D, E and the line d 2 intersects B C at F. Prove that H is the midpoint of segment DE if and only if F is the midpoint of segment BC. Since HD HF,HA F C and HC DA, ∠DAH = ∠HCF and ∠DHA = ∠HFC, therefore the triangles DHA, HFC are similar. So Solution.

⊥

⊥

HA

HD

=

F C . F H

⊥

(2.15)

40


Similarly,

EH A  HFB,

or

HE HA

=

F H . FB

(2.16)

Figure 2.4: Fugure of Question 13 HE

F C

Thanks to (2.15) and (2.16) we obtain = . Therefore, HD FB H is midpoint of the segment DE iff F is midpoint of the segment BC.


√ −

Solution. From equality 2015a2 + a = 2016b2 + b, we find a b. If a = b , then a = b = 0 and If a > b, we have

≥

b2 = 2015(a2 b2 = (a

√ a − b = 0. 2

− b ) + (a − b);

− b)(2015a + 2015b + 1).

Let (a, b) = d then a = md, b = nd, where (m, n) = 1. Since a > b then m > n, and put m n = t > 0. Let (t, n) = u then n is

−

41

2.1 Junior section

divisible by u, t is divisible by u and m is divisible by u. That follows u = 1 and then (t, n) = 1. Putting b = nd, a b = td in the above identity, we find

−

n2 d = t (2015dt + 4030dn + 1).

Hence, n2 d is divisible by t and compaire with (t, n) = 1, it follows d is divisible by t. Moreover, we have n2d = 2015dt2 + 4030dnt + t and then t = n 2 d 2015dt2 4030dnt. Hence t = d(n2 2015t2 4030nt), i.e. t is divisible by d, i.e. t = d and then a b = td = d 2 and a b = d is a natural number.

−

−

−

− −

√ −

Question 15. Find all polynomials of degree 3 with integer coefficients such that f (2014) = 2015, f (2015) = 2016, and f (2013) f (2016) is a prime number.

−

Solution.

Hence,

Let g (x) = f (x)

− x − 1. We have g (2014) = f (2014) − 2014 − 1 = 0, g (2015) = 2016 − 2015 − 1 = 0.

g (x) = (ax + b)(x

and f (x) = (ax + b)(x

− 2014)(x − 2015),

− 2014)(x − 2015) + x + 1,

a,b

∈ Z, a = 0.

We derive f (2013) = 2(2013a + b) + 2014,

and f (2016) = 2(2016a + b) + 2017.

Therefore, f (2013)

− f (2016) = 2(2013a + b) + 2014 − [2(2016a + b) + 2017] = −6a − 3 = 3(−2a − 1). Thus, f (2013) − f (2016) is prime if and only if −2a − 1 = 1, i.e. a = −1.

42


Conlusion: All polynomials of degree 3 with integer coefficients such that f (2014) = 2015, f (2015) = 2016 and f (2013) f (2016) is a prime number are of the form

−

f (x) = (b

2.1.3

− x)(x − 2014)(x − 2015) + x + 1, b ∈ Z.


Question 1. Suppose x1 , x2 , x3 are the roots of polynomial P (x) = x 3

2

− 6x

+ 5x + 12.

The sum x1 + x2 + x3 is (A): 4 (B): 6 (C): 8 (D): 14 (E): None of the above.

| | | | | |

Solution. The choice is (C).

Question 2. How many pairs of positive integers (x, y ) are there, those satisfy the identity 2x (A): 1

−y

2

= 1?


Solution. The choice is (A).

Question 3. Suppose n 2 + 4n + 25 is a perfect square. How many such non-negative integers n’s are there? (A): 1 (B): 2 (C): 4 (D): 6 (E): None of the above. Solution. The choice is (B).

Question 4. Put S = 21 + 35 + 4 9 + 513 +

The last digit of S is

2013

·· · + 505

+ 5062017 .

43

2.1 Junior section

(A): 1


Solution. The choice is (E).

Question 5. Let a,b,c be two-digit, three-digit, and four-digit numbers, respectively. Assume that the sum of all digits of number a + b, and the sum of all digits of b + c are all equal to 2. The largest value of a + b + c is (A): 1099 (B): 2099 (C): 1199 (D): 2199 (E): None of the above. Solution. The choice is (E).

Question 6. that

Find all triples of positive integers (m,p,q ) such 2m p2 + 27 = q 3,

and p is a prime.

Solution. By the assumption it follows that q is odd. We have 2m p2 = (q

2

− 3)(q + 3q + 9).

Remark that q 2 + 3q + 9 is always odd. There are two cases: Case 1. q = 2m p + 3. We have q 3 = (2m p + 3)3 > 2 m p2 + 27,

which is impossible. Case 2. q = 2m + 3. We have q 3 = 23m + 9

2m

×2

+ 27

m

×2

+ 27 = 2m p2 + 27,

which implies p2 = 22m + 9

m

×2

+ 27.

If m 3, then 22m + 9 2m + 27 3 (mod 8), but p 2 1 (mod 8). We deduce m 3. By simple computation we find m = 1, p = 7, q = 5.

≥

≤

×

≡

Question 7. Determine two last digits of number Q = 22017 + 20172 .

≡

44


Solution. We have 201

22017 = 27 210 = 128 1024201 128 ( 1)201 = 128 22 (mod 25); 20172 14 (mod 25).

≡ ≡

×

 

×−

−

×

≡

It follows P 11 (mod 25), by which two last digits of P are in the set 11, 36, 61, 86 . In other side, P 1 (mod 4). This implies P 61 (mod 100). Thus, the number 61 subjects to the question.



≡

≡



≡

Question 8. Determine all real solutions x, y,z of the following system of equations

 

x3

− 3x 2y − 6y 3z − 9z

=4 =6 =8

3

3

−y − z − x.

Solution. From x3 + y = 3x + 4 it follows x3 Then (x 2)(x + 1)2 = 2 y

−

By 2y 3

− 2 − 3x = 2 − y . (1)

−

− 4 − 6y = 2 − z , we have 2(y − 2)(y + 1) = 2 − z. Similarly, by 3z − 3 − 3 − 9z = 2 − x we have 3(z − 2)(z + 1) = (2 − x). 2

(2)

3

2

(3)

Combining (1)-(2)-(3) we obtain (x

1 2) (x + 1) (y + 1) (z + 1) + = 0. 6



2

2

2



− 2)(y − 2)(z − Hence, (x − 2)(y − 2)(z − 2) = 0. Comparing this with (1), (2) and (3), we find the unique solution x = y = z = 2.

Question 9. Prove that the equilateral triangle of area 1 can be covered by five arbitrary equilateral triangles having the total area 2.

45

2.1 Junior section

Solution. Let S denote the triangle of area 1. It is clearly that b then triangle of area a can cover triangle of area b. It if a suffices to consider the case when the areas of five small triangles are all smaller than 1. Let 1 A B C D E stand for the areas. We will prove that the sum of side-lengths of B and C is not smaller than the side-length of triangle of area 1. Indeed, suppose B + C < 1 = 1. It follows

≥

≥ ≥ ≥ ≥ ≥

√ √

√

√

√ √

2 = A + B + C + D + E < 1 + B + C +2 BC = 1+( B + C )2 < 2 , which is impossible.

Figure 2.5: For Question 9 We cover S by A, B,C as Figure 2.5. We see that A, B,C will have common parts, mutually. Suppose X = B

∩ C ;

Y = A

∩ C ;

Z = A

X + Y

≤ C ;

Y + Z

≤ A;

Z + X

∩ B.

It follows

≤ B.

46


We deduce A, B,C cover a part of area:

− − − ≥ A+B +C − 12 (X +Y )+( Y +Z )+(Z +X ) ≥ 21 (A + B + C ) = 1 − D +2 E ≥ 1 − D.



A+B + C X Y Z



Thus, D can cover the remained part of S.

Question 10. Find all non-negative integers a, b, c such that the roots of equations: x2 x2 x2

− 2ax + b = 0; − 2bx + c = 0; − 2cx + a = 0

(2.17) (2.18) (2.19)

are non-negative integers. Solution. Namely,

We see that a2 a2

2

− b = p ;

2

2

− b, b − c, c − a are perfect squares. b2

2

− c = q ;

c2

2

− a = r .

There are two cases: Case 1. b = 0. We derive that b = c = 0. Thus (a,b,c) = (0, 0, 0) is unique solution. Case 2. a,b,c = 0. We have a2 b (a 1)2 = a2 2a + 1. This implies b 2a 1. Similarly, we can prove that c 2b 1, and a 2c 1. Combining three above inequalities we deduce a + b + c 3. By simple computation we obtain (a,b,c) = (1, 1, 1).

 ≥ − ≥ − ≤

− ≤ −

− ≥ −

Question 11. Let S denote a square of the side-length 7, and let eight squares of the side-length 3 be given. Show that S can be covered by those eight small squares. Solution. Figure 2.6 is a solution.

Question 12. Does there exist a sequence of 2017 consecutive integers which contains exactly 17 primes?

47

2.1 Junior section

Figure 2.6: For Question 11 Solution. It is easy to see that there are more than 17 primes in the sequence of numbers 1, 2, 3, 4, . . . , 2017. Precisely, there are 306 primes in that sequence. Remark that if the sequence k + 1 , k + 2 , . . . , k + 2017

was changed by the sequence k, k + 1 , . . . , k + 2016,

then the numbers of primes in the latter and former sequences are either equal, more or less by 1. In what follows, we say such change a back-shift with 1-step. First moment, we consider the sequence of 2017 consecutive integers: 2018! + 2, 2018! + 3, . . . 2018! + 2018 which contain no prime. After 2018!+1 times back-shifts with 1step, we obtain the sequence 1, 2, 3, 4, . . . , 2017.

48


The last sequence has 306 primes, while the first sequence has no prime. Reminding the above remark we conclude that there is a moment in which the sequence contains exactly 17 primes.

Question 13. Let a, b, c be the side-lengths of triangle ABC with a + b + c = 12. Determine the smallest value of M =

a b+c

−a

+

4b c+a

−b

+

9c a+b

− c.

Solution. Put x :=

b+c

2

− a , y := c + a − b , z := a + b − c . 2

2

Then x, y,z > 0, and x + y + z =

a+b+c

2

= 6, a = y + z, b = z + x, c = x + y.

We have y + z 4(z + x) 9(x + y) + + 2x 2y 2z 1 4x 4z 9 y y z 9x = + + + + + x y x z y z 2

M =

≥ 21

           2

y 4x +2 . x y

z 9x +2 . x z

4z 9y . x z

The equality occurs in the above if and only if

or

    

4x y = x y z 9x = x z 4z 9y = , y z y = 2x z = 3x

2z = 3y.

= 11.

49

2.1 Junior section

Since x + y + z = 6 we receive x = 1, y = 2, z = 3. Thus min S = 11 if and only if (a,b,c) = 5, 4, 3 .

 

Question 14. Given trapezoid ABCD with bases AB CD (A B < C D ). Let O be the intersection of AC and BD . Two straight lines from D and C are perpendicular to AC and BD intersect at E , i.e. CE BD and DE AC . By analogy, AF BD and BF AC . Are three points E , O, F located on the same line?



⊥

⊥

⊥

⊥

Solution. Since E is the orthocenter of triangle ODC, and F is the orthocenter of triangle OAB we see that OE is perpendicular to CD, and OF is perpendicular to AB . As AB is parallel to C D, we conclude that E , O, F are straightly lined.

Question 15. Show that an arbitrary quadrilateral can be divided into nine isosceles triangles. Solution. Remark that a quadrilateral can be divided into two triangles. Therefore, it suffices to cut an arbitrary triangle into for and five isosceles triangles. Figures 2.7, 2.8, and 2.9 shows some solution.

Figure 2.7: For Question 15

50




2.2 2.2.1

Senior section Hanoi Open Mathematics Competition 2015

Question 1. The sum of all even positive intergers less than 100 those are not divisible by 3 is (A): 938; (B): 940; (C): 1634; (D): 1638; (E): None of the above. We find the sum of all positive intergers less than 50 and subtract the sum of multiples of 3 between 0 and 50: Solution.

50

× 49 − 3 16 × 17 = 1225 − 408 = 817.

2 2 Hence the sum of all even positive intergers less than 100 those are

51

2.2 Senior section

not divisible by 3 is 2

× 817 = 1634.

Answer: C.


√

√

√

√

√

(A): 5 3; (B): 6 3; (C):7 3; (D):8 3; (E): None of the above. Let a, S be the length and area of the hexagon. Then the equilateral triangle has side-length 2a. We subdivide the hexagon into 6 small equilateral of side-length a then each of them must have Solution.

S

area . Since the equilateral triangle of side-length 2a has twice 6 the base and twice the altitude of the smaller triangle, it must have 4 times the area. Hence 4

S

6

√

√

= 4 3, it follows S = 6 3.

Answer: B.

Question 3. Suppose that a > b > c > 1. One of solutions of the equation (x (c

− a)(x − b) + (x − b)(x − c) + ( x − c)(x − a) = x − a)(c − b) (a − b)(a − c) (b − c)(b − a)

is (A): -1; (B): -2; (C): 0; (D): 1; (E): None of the above. Solution.

(x (c

From the identity

− a)(x − b) + (x − b)(x − c) + ( x − c)(x − a) ≡ 1, − a)(c − b) (a − b)(a − c) (b − c)(b − a)

it follows x = 1 is a solution of the equation. Answer: D.

52




then m is (A): 0; (B): 1; (C): 25; (D): 26; (E): None of the above. - If a, b, c have three different remainder terms modulo 3 then a + b + c is divisible by 3 but a b, b c, c a are not divisible by 3. - If a, b, c have two different remainder terms modulo 3 then a + b + c is not divisible by 3 but one of three differences a b, b c, c a are divisible by 3. It follows the product (a b)(b c)(c a) is divisble by 3. - Hence, three numbers a,b,c have the same remainder terms, i.e. a = 3 p + s, b = 3q + s, c = 3r + s with s 0, 1, 2 , p, q,r are natural numbers. Hence, a b = 3( p q ), b c = 3(q r), c a = 3(r p) and then (a b)(b c)(c a) = 0 (mod 27). Solution.

− − −

− − − − −

−

∈ {

− − −

− −

−

−

}

−

−

Answer: A.


2015

− 2013

is

(A): 2; (B): 4; (C): 6; (D): 8; (E): None of the above. We have 2017 = 7 (mod 10) and 20172017 = 74 504+1 (mod 10) = 7 (mod 10). On the other hand, 2015 = 4 503+ 3 = 3 (mod 4) and 2013 = 3 (mod 10) and then 20132015 = 34 503+3 = 7 (mod 10). Hence, the last digit of number 20172017 20132015 is 0. ×

Solution.

×

×

−

Answer: E.


∈−

≥

2

≥a

+ b2 + c2 .

53

2.2 Senior section Solution.

The constraint can be written as (a

2

2

2

− bc) ≤ (1 − b )(1 − c ).

(2.20)

Using the Cauchy inequality, we have (a + bc)2

2

2

2

2

≤ (|a| + |bc|) ≤ (1 + |b||c|) ≤ (1 + b )(1 + c ).

Multiplying by (2.20), we get 2

(a

2

2

2

2

2

− bc) (a + bc) ≤ (1 − b )(1 + b )(1 − c )(1 + c ) ⇔ (a − b c ) ≤ (1 − b )(1 + b )(1 − c )(1 + c ) ⇔ (a − b c ) ≤ (1 − b )((1 − c ) ⇔ 1 + 2a b c ≥ a + b + c . 2

2 2 2

2

2

2

2 2 2

2

4

2 2 2

2

4

4

4

4


(2.21)

where [x] is an integral part of x. Solution.

We have 2

(2.21)

2

 

⇔ [x] = x x − 2 ≤ 2, then −√ 2 ≤ x ≤ √ 2 and [x] ≤ 0. It

Consider the case x2 follows [x] 1; 0 . If [x] = 0, then from (2.21) we find x = 0 as a solution. If [x] = 1, then from (2.21) we find x = 1 as a solution. Now we suppose that x 2 > 2 . It follows from (2.21), [x] > 0 and [x] 1 then x > 2. Hence x2 (x2 2) = 1 and x2 2 < 1 . It

∈ {− }

− √ − √ √ √ follows x < 3, i.e. 2 < x < 3.

−

x

≤

It means that [x] = 1 and then x = the equation.

− ≤x



1+

√ 2 is a solution of

Question 8. Solve the equation (x + 1)3 (x

− 2)

3

+ (x

3

− 1) (x + 2)

3

= 8(x2

3

− 2) .

(2.22)

54


Solution. Rewrite

(x2

equation (2.22) in the form

− x − 2)

3

+ (x2 + x

− 2)

3

= (2x2

3

− 4) .

(2.23)

Factoring the sum of cubes on the right side of the equation (2.20), we find that one factor is (2x2 4), thus, two solutions of the equation is x = 2. Now we rewrite the equation (2.23) as

−

±√

(x2

− x − 2)

3

= (2x2

3

− 4) − (x

2

+x

3

− 2) .

(2.24)

Factoring the difference of cubes on the left side of the equation (2.2.1), we find that one factor is (x2 x 2), thus, two solution of the equation is x = 1, x = 2. Finally, we rewrite (2.23) as

− −

−

(x2 + x

− 2)

3

= (2x2

3

2

3

− 4) − (x − x − 2) . Factoring again, we see that one factor is (x + x − 2). Thus, two solutions of the equation are x = 1, x = −2.. 2

Since the left hand side of the equation 8(x2

3

3

3

3

− 2) − (x + 1) (x − 2) − (x − 1) (x + 2)

3

=0

is a polynomial of degree 6, then it has at most 6 roots, and we have them. Hence, 8(x2 2)3 (x+1)3 (x 2)3 (x 1)3 (x+2)3 = 6(x2 2)(x2 1)(x2 4).

− −

− − −

−

−

−

Question 9. Let a,b,c be positive numbers with abc = 1. Prove that a3 + b3 + c3 + 2[(ab)3 + (bc)3 + (ca)3 ]

Asume that a = max a,b,c then a b > 0 and

{

Solution.

a

≥c≥

2

a3 + b3 + c3

2

2

2

}

2

2

2

≥ 3(a b + b c + c a).

2

≥ b ≥ c > 0 or 2

2

− (a b + b c + c a) = (a − b)(a − c ) +(b − c)(b − c ) ≥ 0.

55

2.2 Senior section

Hence a3 + b3 + c3

Since

1 c

and

1 b

we have

1 c3

+

1 b3

= max = max +

2

2

2

≥ a b + b c + c a.

1

(2.25)

1 1 1

   

, , , a b c

1 1 1

, , , a b c

1 1 1 ≥ . + + a c b b a a c 3

2

2

2

By abc = 1, this can be written as (ab)3 + (bc)3 + (ca)3

2

2

2

≥ a b + b c + c a.

(2.26)

(2.25) and (2.26) together imply the proposed inequality.

Question 10. A right-angled triangle has property that, when a square is drawn externally on each side of the triangle, the six vertices of the squares that are not vertices of the triangle are concyclic. Assume that the area of the triangle is 9 cm2 . Determine the length of sides of the triangle. Solution. We

have OJ = OD = OG = radius of the circle. Let the sides of ∆ABC be a, b, c. We have 1 = b 2 + ab + (a2 + b2 ); 2 2 4 2 2 1 b a OD 2 = ON 2 + ND 2 = a + + = a 2 + ab + (a2 + b2 ); 2 2 4 2 5 c OG2 = OL2 + LG2 = c 2 + = c2 2 4 5 1 = (a2 + b2 ) = a 2 + b2 + (a2 + b2 ). 4 4 Comparing these right-hand sides, we get 2

2

2

a

2

b

2

      

OJ = OM + M J = b +

+

b2 + ab = a 2 + ab = a 2 + b2

⇔ a = b.

It means that the given triangle with the desired property is the 1 a = b = 3 2, c = 6 isosceles right triangle and then ab = 9 2 units.

⇔

√

56


Question 11. Given a convex quadrilateral ABCD. Let O be the intersection point of diagonals AC and BD and let I , K , H be feet of perpendiculars from B , O , C to AD, respectively. Prove that AD Solution.


Draw AE BD (E S ABD

Then BI

∈ BD ). We have BI × AD AE × BD . = =

⊥

2

2

× AD = AE × BD. It follows BI × AD ≤ AO × BD (AE ≤ AO),

and BI.AD

≤ AC × BD × AO . AC

Moreover, we have OK CH, then



AO OK , = AC CH

and BI

× AD ≤ AC × BD × OK . CH

Therefore, BI

× AD × CH ≤ AC × BD × OK.

The equality holds if quadrilateral ABCD has two perpendicular diagonals.

Question 12. Give an isosceles triangle AB C at A . Draw ray C x being perpendicular to CA, BE perpendicular to Cx (E Cx). Let M be the midpoint of BE , and D be the intersection point of BC AM and Cx. Prove that BD

∈

⊥

Let K be intersection point of DB and AC. Since BE CD; CK CD then BE CK .

Solution.

⊥

⊥



57

2.2 Senior section

In ∆DAC we see M E AC so



M E DM = AC AD

(2.27)

In ∆DAK we see M B

 AK so MB DM = . AK AD M E

(2.28)

MB

From (2.27) and (2.28), we get = . This and equality AC AK M B = M E together imply AK = AC = AB and then BA =

CK .

2

Note that BA is a median line ∆BK C and BA = ∆BK C is a right triangle at B. Hence BD BC.

⊥

CK

2

then

Question 13. Let m be given odd number, and let a, b denote the roots of equation x2 + mx 1 = 0 and c = a2014 + b 2014 , d = a2015 + b2015 . Prove that c and d are relatively prime numbers.

−

Solution. Since

a2 + ma

− 1 = 0 then a = 0 and = −ma + a ∀n ∈ N. a Similarly, b = −mb + b ; ∀n ∈ N. Hence, the sequence x , n ∈ N are defined as x = 2, x = −m x = −mx + x ∀n ∈ N. n+2

n+2

n+1

n+1

n

n

n



0

n+2

1

n+1

n

It is easy to see all xn are integers. Hence, c, d are integers. Now we prove (xn ; xn+1 ) = 1 for every n N. For n = 0, x0 = 2 and m is odd then (x0 ; x1 ) = (2; m) = 1. Suppose that (xk ; xk+1 ) = 1 for k 0 and (xk+1; xk+2 ) > 1. Let p be a prime factor of x k+1 and x k+2 , then from x k = x k+2 + mxk+1 , it follows p is a prime factor of x k . It means that p (xk ; xk+1 ) = 1, absurd. Hence (xk+1 ; xk+2) = 1 and (c, d) = 1.

∈

−

≥

|

58


Question 14. Determine all pairs of integers (x; y ) such that 2xy 2 + x + y + 1 = x 2 + 2y 2 + xy. Solution.

We have 2xy 2 + x + y + 1 = x 2 + 2y2 + xy 2

⇔ 2y (x − 1) − y(x − 1) − x(x − 1) = −1 ⇔ (x − 1)(2y − y − x) = −1 2

Since x ; y are integers then x Case 1. We have

−

2

− 1 and 2y − y − x are divisors of -1.

x

1=1

2y 2

− y − x = −1.

Equivalently, either



x = 2 y = 1,

or

  − − x = 2 y=

- Case 2. We have

1 (absurd) 2

1=

−1 − y − x = 1.

x

2y 2

This implies that



x = 0 y = 1,

or

 

x = 0 y =

− 12 (absurd).

59

2.2 Senior section

Hence, all integral pairs (x; y) are (2; 1); (0 ; 1).

Question 15. Let the numbers a,b,c,d satisfy the relation a2 + b2 + c2 + d2

≤ 12.

Determine the maximum value of M = 4(a3 + b3 + c3 + d3)

− (a

4

+ b4 + c4 + d4 ).

Note that x2 (x 2)2 0 for each real x. This inequality can be rewritten as 4x3 x4 4x2 . It follows that

− ≥ − ≤ (4a −a )+(4b −b )+(4c −c )+(4d −d ) ≤ 4(a +b +c +d ) = 48, Solution.

3

4

3

4

3

4

3

4

2

2

2

2

The equality holds for (a,b,c,d) = (2, 2, 2, 0).(2, 2, 0, 2), (2, 0, 2, 2), (0, 2, 2, 2). Hence, max M = 48.

2.2.2


Question 1. How many are there 10-digit numbers composed from the digits 1, 2, 3 only and in which, two neighbouring digits differ by 1. (A): 48 (B): 64 (C): 72 (D): 128 (E): None of the above. Anwser. (B).

We solve the problem by counting the numbers satisfying the condition, starting from 1-digit, 2-digit, 3-digit numbers Solution.

60


Digit

1 1 2 3



2 12 21 23 32

3 121 123 212 232 321 323

4

6

3

4 1212 1232 2121 2123 2321 2323 3212 3232 8

.

We can see that a number ending by 2 in previous column generates 2 numbers for next column (we can add 1 or 3 at the end), but a number ending by 1 or 3 generate 1 number for next column (we can add only 2 at the end). From this, we can make a table. The first row is number of digits, the second row is the number of k-digit numbers satisfying the condition and ending with 1, 3, the third row is the number of k-digit numbers satisfying the condition and ending with 2. 1 2 1

2 2 2

3 4 2

4 4 4

5 8 4

6 8 8

7 16 8

8 16 16

9 32 16

10 32 . 32

Answer: 64.

Question 2. Given an array of numbers A = (672, 673, 674, . . . , 2016)

on table. Three arbitrary numbers a,b,c A are step by step re1 placed by number min(a,b,c). After 672 times, on the table there 3 is only one number m, such that

∈

(A): 0 < m < 1 (B): m = 1 (E): None of the above.

(C): 1 < m < 2 (D): m = 2

61

2.2 Senior section Anwser. (A). Solution. Note

1 a

that 1

+

b

+

1

3 = ≤ 1 min{a,b,c} c 3

1

.

min a,b,c

{

}

So after each operation, the sum of all inverse elements of A is non decreasing. On the other hand, for every positive integer n, we have S :=

1 n

+

1 + n+1

··· + 31n > 1.

Indeed, 1 S = + 2n

n

 k=1

1 2n

−

1 > + 2n

n

1 1 4n + = + k 2n + k 2n k=1 4n2 k2



n

1

  k=1

n



−



1 n + > 1 . 2n n

>

1 Hence, if on the table there is one number m then > 1, i.e. m 0 < m < 1 .

Question 3. Given two positive numbers a, b such that the condition a 3 + b3 = a 5 + b5 , then the greatest value of M = a 2 + b2 ab is

−

(A):

1 1 (B): 4 2


Anwser. (D). Solution. We

have ab(a2

2 2

−b ) ≥0 ⇔ 2a b ≤ ab + a b ⇔ (a + b ) ≤ (a + b)(a 3 3

3

5

5

3 2

5

+ b5 ).

Combining this and the fact that a3 + b3 = a 5 + b5, we find a3 + b3

≤a+b⇔a

2

+ b2

− ab ≤ 1.

62


The equality holds if a = 1, b = 1.

Question 4. In Zoo, a monkey becomes lucky if he eats three different fruits. What is the largest number of monkeys one can make lucky having 20 oranges, 30 bananas, 40 peaches and 50 tangerines? Justify your answer. (A): 30 (B): 35 (C): 40 (D): 45 (E): None of the above. Anwser. (D).

First we leave tangerines on a side. We have 20 + 30 + 40 = 90 fruites. As we feed the happy monkey is not more than one tangerine, each monkey eats fruits of these 90 at least 2. Hence, the monkeys are not more than 90/2 = 45. We will show how you can bring happiness to 45 monkeys: 5 monkeys eat: orange, banana, tangerine; 15 monkeys eat: orange, peach, tangerine; 25 monkeys eat peach, banana, tangerine. At all 45 lucky monkeys - and left five unused tangerines! Solution.

Question 5. There are positive integers x, y such that 3x2 + x = 4y2 + y and (x y) is equal to

−

(A): 2013 (B): 2014 (C): 2015 (D): 2016 (E): None of the above. Anwser. (E).

Since x

− y is a square. Solution. We have 3x + x = 4y + y ⇔ (x − y )(3x +3y +1) = y . We prove that (x − y ; 3x + 3y + 1) = 1. Indeed, if d = (x − y ; 3x + 3y + 1) then y is divisible by d and 2

2

2

2

2

y is divisible by d; x is divisible by d, i.e. 1 is divisible by d, i.e. d = 1.

Since x y and 3x + 3y + 1 are prime relative then x perfect square.

−

− y is a

63

2.2 Senior section

Question 6. Let A consist of 16 elements of the set

{1, 2, 3, . . . , 106}, so that the difference of two arbitrary elements in A are different from 6, 9, 12, 15, 18, 21. Prove that there are two elements of A for which their difference equals to 3.

Divide numbers 1, 2, . . . , 106 into three groups X = 1, 4, 7, . . . , 106 , Y = 2, 5, 8, . . . , 104 and Z = 3, 6, 9, . . . , 105 . A has 16 elements, so one of the sets X , Y , Z contains at least 6 numbers from A. Without loss of generality, let X contains 6 < a6 106. Since numbers from A. Let they be 1 a1 < a2 < Solution.

{

}

{

}

{

≤

105

≥ a −a 6

1

···

}

≤

= (a6 a5 )+(a5 a4 )+(a4 a3)+(a3 a2 )+(a2 a1 ),

−

−

−

−

−

there is an index i for which 0 < ai+1 ai 21. By the choice of X, ai+1 ai is multiple of 3, so ai+1 ai 3, 6, 9, 12, 15, 18, 21 . Finally, apply the given condition, it follows that a i+1 ai = 3, which was to be proved.

{

}

−

− ≤

− ∈

−

Question 7. Nine points form a grid of size 3 3. How many triangles are there with 3 vertices at these points?

×

We divide the triangles into two types: Type 1: Two vertices lie in one horizontal line, the third vertex lies in another horizontal lines. For this type we have 3 possibilities of choosing the first line, 2 possibilities of choosing the 2-nd line. In total we have 3 2 3 3 = 54 triangles of first type. Type 2: Three vertices lie in distinct horizontal lines. We have 3 3 3 triangles of these type. But we should remove degenerated triangles from them. There are 5 of those (3 vertical lines and two diagonals). So, we have 27 - 5 = 22 triangles of this type. Solution.

×××

× ×

64


Total, we have 54 + 22 = 76 triangles. For those students who know about C nk this problem can be also solved as C 93 8 where 8 is the number of degenerated triangles.

−

Question 8. Determine all 3-digit numbers which are equal to cube of the sum of all its digits. Let abc, where a,b,c 1, 2, 3, 4, 5, 6, 7, 8, 9 , a = 0 3 and abc = (a + b + c) . a + b + c Note that 100 (a + b + c)3 999 and 100 999. Hence 5 a + b + c 9. If a + b + c = 5 then abc = (a + b + c)3 =53 = 125 and a + b + c = 8 (not suitable). If a + b + c = 6 then abc = (a + b + c)3 =63 = 216 and a + b + c = 9 (not suitable). If a + b + c = 7 then abc = (a + b + c)3 =73 = 343 and a + b + c = 10 (not suitable). If a + b + c = 8 then abc = (a + b + c)3 =83 = 512 and a + b + c = 8 (suitable). If a + b + c = 9 then abc = (a + b + c)3 =93 = 729 and a + b + c = 18 (not suitable). Conclusion: abc = 512.

∈ {

Solution.

√ 3

≤

≤

≤

} 

√ ≤ 3

≤

≤

Question 9. Let rational numbers a, b, c satisfy the conditions a + b + c = a 2 + b2 + c2

∈ Z.

Prove that there exist two relative prime numbers m, n such that m2 abc = 3 . n

Put a + b + c = a 2 + b2 + c2 = t. We have 3 (a2 + b2 + c2 ) (a + b + c)2 , then t Since t Z then t 0;1;2;3 . 0 If t = 0 then a = b = c = 0 and abc = 0 = . 1 If t = 3 then

Solution.

∈

(a

∈{

≥

}

− 1) + (b − 1) + (c − 1) = (a − 1)

2

+ (b

2

− 1)

∈ [0; 3] .

+ (c

− 1)

2

= 0.

65

2.2 Senior section

12 That follows a = b = c = 1 and abc = 1 = 3 . 1 If t = 1. Without loss of generality, assume that c > 0; a =

m1 m2 m3 ; b = ; c = ; d = n1 n2 n3 . n1 n2 n3

|

Put

|

 

x = ad y = bd z = cd.

∈ Z and z > 0. Moreover,

We have x; y; z



x + y + z = d (a + b + c) = d x2 + y2 + z 2 = d 2 (a2 + b2 + c2 ) = d 2

It follows xy + yz + zx = 0

2

⇔ (z + x) (z + y) = z

∈ Z

Hence, there exist r ; p; q

∗

= c 2 d2 .

such that

x + z = rp2 ; y + z = rq 2 ; z = r qp ; ( p; q ) = 1; p; q

||

On the other hand d = x + y + z = r ( p2 + q 2 ) r > 0 . Hence, y = rq (q p) x = rp ( p q ) z = rpq.

 

Equivalently,

∈ Z . ∗

− |r| pq > 0 then

− −

[ pq ( p q )]2 . ( p2 + q 2 pq )3

− − − We shall prove that ( pq ( p − q ); p + q − pq ) = 1. Suppose s = ( pq ( p − q ) ; p + q − pq ) ; s > 1 then s| pq ( p − q ) . Case 1. Let s| p. Since s| ( p + q − pq ) then s|q and s = 1 (not abc =

2

2

2

2

2

2

suitable). Case 2. Let s q. Similarly, we find s = 1 (not suitable).

|

66

Chapter 2. Answers and Solutions 2

Case 3. If s ( p

2

| − q ) then s|( p − q ) − ( p

 ⇒ |

+ q 2

s p (not suitable). s q

− pq ) ⇒ s| pq

|

If t = 2 then a + b + c = a 2 + b2 + c2 = 2. We reduce it to the case where t = 1, which was to be proved.


√ −

Solution. From equality 2015a2 + a = 2016b2 + b. we find a b. If a = b , then a = b = 0 and If a > b, we have

≥

b2 = 2015(a2

2

√ a − b = 0.

− b ) + ( a − b) ⇔ b

2

= (a

− b)(2015a + 2015b + 1).

Let (a, b) = d then a = md ; b = nd, where (m, n) = 1. Since a > b then m > n; and put m n = t > 0 . Let (t, n) = u then n is divisible by u; t is divisible by u and m is divisible by u. That follows u = 1 and then (t, n) = 1. By b = nd ; a b = td , we find

−

−

n2 d = t (2015dt + 4030dn + 1).

Hence, n 2 d is divisible by t and compaire with (t, n) = 1, it follows d is divisible by t. Moreover, we have n 2 d = 2015dt2 +4030dnt + t and then t = n 2 d 2015dt2 4030dnt. Hence t = d (n2 2015t2 4030nt), i.e. t is divisible by d, i.e. t = d and then a b = td = d2 and a b = d is a natural number.

√ −

−

−

− −

−

Question 11. Let I be the incenter of triangle ABC and ω be its circumcircle. Let the line AI intersect ω at point D = A. Let F and E be points on side BC and arc BDC respectively such 1 that ∠BAF = ∠CAE < ∠BAC . Let X be the second point of 2



67

2.2 Senior section

intersection of line EI with ω and T be the point of intersection of segment DX with line AF . Prove that TF.AD = ID.AT . Let the line AF intersect ω at point K = A and L be the foot of the bisector of angle BAC . Since ∠BAK = ∠CAE we   have BK = CE , hence K E BC . Notice that ∠IAT = ∠DAK = ∠EAD = ∠IXT , so the points I , A , X, T are concyclic. Hence, ∠IT A = ∠IXA = ∠EX A = ∠EK A, so IT KE BC . There-



Solution.



fore,



T F IL = . AT AI



Figure 2.10: Fugure of Question 11 IL CL = . Furthermore, AI AC 1 ∠DCL = ∠DCB = ∠DAB = ∠CAD = ∠BAC. Hence, the

Since C I is bisector of ∠ACL, we get

2 CL DC triangles DCL and DCA are similar. Therefore, = .

AC AD Finally, we have ∠DI C = ∠IAC + ∠ICA = ∠ICL + ∠LCD = DC = ∠ICD. It follows DI C is a isosceles triangle at D. Hence AD ID . AD

Summarizing all these equalities, we get T F IL CL DC ID , = = = = AT AI AC AD AD

68


or

T F ID , = AT AD

and TF.AD = ID.AT,

as desired.

Question 12. Let A be point inside the acute angle xOy. An arbitrary circle ω passes through O, A; intersecting Ox and Oy at the second intersection B and C, respectively. Let M be the midpoint of BC. Prove that M is always on a fixed line (when ω changes, but always goes through O and A). Let (Ox ), (Oy ) be circles passing throught O, A and tangent to Ox, Oy, respectively. Circle (Ox ) intersects the ray Oy at D, distinct from O and circle (Oy ) intersects the ray Ox at E, distinct from O. Let N and P be midpoint of OE and OD, respectively. Then N, P are fixed. We’ll show that M , N , P are Solution.

collinear. For this, it is sufficient to prove that

NO NB

=

PO P C

Figure 2.11: Fugure of Question 12 Since (Ox ) is tangent to Ox, ∠ADC = ∠AOB. Since OBAC is cyclic, ∠ABO = ∠ACD. So triangles AOB,ADC are similar. Therefore

AB AC

=

OB DC

(1)

69

2.2 Senior section

Similarly,

AB = ABE  ACO, so BE CO AC

(2)

From (1) and (2), we deduce that OB BE = CD OC

Hence

CD = ⇒ OB BE OC

OE OD = , BE OC ON OP = , BE OC ON ON OP OP = = = . NB BE NO OC OP CP

−

−

It follows, if NP intersects BC at M, then M B P C NO =1 M C P O N B

·

·

MB

(by Menelaus’ Theorem in triangle OBC ) conclusion = 1, it M C follows NP passes through M is midpoint of BC.

Question 13. Find all triples (a,b,c) of real numbers such that 2a + b 4 and

|

|≥

2

| ≤ 1 ∀x ∈ [−1, 1]. Solution. From the assumptions, we have |f (±1)| ≤ 1, |f (0)| ≤ 1 |ax

and

+ bx + c

 

f (1) = a + b + c f ( 1) = a

−

f (0) = c

Equivalently,

  

−b+c

1 [f (1) + f ( 1)] 2 1 b = [f (1) f ( 1)] 2 c = f (0) a =

− − f (0) − −

70


That follows 4

≤ |2a + b| =

 

[f (1) + f ( 1)]

− −

1 2f (0) + [f (1) 2

3 1 = f (1) + f ( 1) 2f (0) 2 2 3 1 f (1) + f ( 1) + 2 f (0) 2 2 Hence 2a + b = 4. Therefore,

≤ |

|

|

| ||

|

− − | − | |

− f (−1)]





| ≤ 23 + 21 + 2 = 4.

f (1) = a + b + c = 1

| | | f (−1)| = |a − b + c| = 1 f (0)| = |c| = 1.

Equivalently,



(a,b,c) = (2, 0, 1) (a,b,c) = ( 2, 0, 1)

−

−

It is easely seen that both two triples (2, 0, 1) and ( 2, 0, 1) satisfy the required conditions.

−

−

Question 14. Let f (x) = x2 + px + q, where p, q are integers. Prove that there is an integer m such that f (m) = f (2015).f (2016). Solution. We

shall prove that f [f (x) + x] = f (x)f (x + 1).

Indeed, we have f [f (x) + x] = [f (x) + x]2 + p[f (x) + x] + q

= f 2 (x) + 2 f (x).x + x2 + pf (x) + px + q = f (x)[f (x) + 2 x + p] + x2 + px + q = f (x)[f (x) + 2 x + p] + f (x) = f (x)[f (x) + 2 x + p + 1] = f (x)[x2 + px + q + 2 x + p + 1] = f (x)[(x + 1)2 + p(x + 1) + q ] = f (x)f (x + 1),

(1)

71

2.2 Senior section

which proves (1). Putting m := f (2015) + 2015 gives f (m) = f [f (2015)+2015] = f (2015)f (2015+1) = f (2015)f (2016),

as desired.

Question 15. Let a, b, c be real numbers satisfying the condition 18ab + 9ca + 29bc = 1. Find the minimum value of the expression T = 42a2 + 34b2 + 43c2 . Solution.

We have

T 2(18ab + 9 ca + 29bc) =

−

= (5a 3b)2 + (4a 0, a,b,c R.

− ≥ ∀

∈

2

− 3c)

+ (4b

− 5c)

2

+ (a

That follows T

− 3b + 3c)

≥ 2. The equality occures if and only if 5a − 3b = 0 4a − 3c = 0 4b − 5c = 0 a − 3b + 3 c = 0

      

Equivalently,

and



We thus have

18ab + 9ca + 29bc = 1. 5a 3b = 0 4a 3c = 0 4b 5c = 0 18ab + 9ca + 29bc = 1,

− − −

a = 3t, b = 5t, c = 4t

(18

2

× 15 + 9 × 12 + 29 × 20)t

a =

3 √ ±958 ,

b =

5 √ ±958 ,

c =

= 1.

4 √ ±958 .

2

72


2.2.3


Question 1. Suppose x1 , x2 , x3 are the roots of polynomial P (x) = x 3 4x2 3x + 2. The sum x1 + x2 + x3 is (A): 4 (B): 6 (C): 8 (D): 10 (E): None of the above.

−

−

| | | | | |

Solution. The solution is (B).

Question 2. How many pairs of positive integers (x, y ) are there, those satisfy the identity 2x (A): 1 (B): 2

−y

2

= 4?


Solution. The solution is (A).

Question 3. equation

The number of real triples (x,y,z ) that satisfy the x4 + 4y 4 + z 4 + 4 = 8xyz

is (A): 0;

(B): 1; (C): 2; (D): 8; (E): None of the above.

Solution. The solution is (E).

Question 4. Let a,b,c be three distinct positive numbers. Consider the quadratic polynomial P (x) =

c(x (c

− a)(x − b) + a(x − b)(x − c) + b(x − c)(x − a) + 1. − a)(c − b) (a − b)(a − c) (b − c)(b − a)

The value of P (2017) is (A): 2015 (B): 2016 (C): 2017 (D): 2018 (E): None of the above. Solution. The solution is (D).

Question 5. Write 2017 following numbers on the blackboard: 1007 1 1 2 1007 1008 − 1008 ,− ,...,− , 0, , ,..., , . 1008 1008 1008 1008 1008 1008 1008

73

2.2 Senior section

One processes some steps as: erase two arbitrary numbers x, y on the blackboard and then write on it the number x + 7xy + y. After 2016 steps, there is only one number. The last one on the blackboard is 1 1 144 (A): (B): 0 (C): (D): (E): None of the 1008 1008 1008 above.

−

−

Solution. The solution is (D).

Question 6. Find all pairs of integers a, b such that the following system of equations has a unique integral solution (x,y,z )



x + y = a

−1 x(y + 1) − z = b. 2

Solution. Write the given system in the form



x + y + 1 = a x(y + 1)

2

− z

( )

∗

= b.

System (*) is symmetric by x, y + 1 and is reflect in z at 0 then the necessary condition for (*) to have a unique solution is (x, y +1 , z ) = (t,t, 0). Putting this in (*), we find a2 = 4b. Conversely, if a2 = 4b then (x

2

− (y +1))

+ 4z 2 = (x + y + 1)2 + 4z 2

2

− 4x(y + 1) = a − 4b = 0.

This implies the system has a unique solution (x, y + 1 , z ) =

a a , ,0 .

  2 2

Question 7. Let two positive integers x, y satisfy the condition . x2 + y 2 .. 44. Determine the smallest value of T = x 3 + y3 . . Solution. By the assumption we have x 2 + y 2 .. 11. One can prove . . that x .. 11 and y .. 11. Due to limitted space we left the proof

74

Chapter 2. Answers Answers and Solutions

for for the the reader reader.. In other side, side, by the assu assump mptio tion n we have have x and y are are even. even. Hence, Hence, x 0 (mod 22) and y 0 (mod 22). Thus, min A = (22)3 + (22)3 = 21296.

≡

≡

Question 8. Let a, a, b, c be the side-lengths of triangle ABC with a + b + c = 12. Determine the smallest value of M =

Solution.

a b+c

4b

+

+

9c a+b

−a c+a−b − c. b+c−a c+a−b a+b−c Let x = then ,y = , z = 2

x, y,z > 0 and x + y + z =

2

a+b+c

x + y. We have

2

2

= 6, a = y + z, b = z + + x, c =

+ x) 9( 9 (x + y) y + z 4(z + + + 2x 2y 2z y z 9x 1 4x 4z 9 y = + + + + + 2 x y x z y z

M =

≥ 21

           2

y 4x . +2 x y

z 9x . +2 x z

4z 9y y

.

z

= 11.

The equality yields if and only if 4x y = x y z 9x = x z 4z 9y = . y z

Equivelently,

      

y = 2x

z = 3x

2z = 3y.

By simple computation we receive x = 1, y = 2, z = 3. Therefore, min S = = 11 when (a,b,c) = 5, 4, 3 .

75

2.2 Senior section

square carton carton by a straigh straightt line into two two Question 9. Cut off a square pieces, then cut one of two two pieces into two two small pieces by a straight line, line, ect. By cutting cutting 2017 times we obtain obtain 2018 pieces. pieces. We write write number 2 in every triangle, number 1 in every quadrilateral, and 0 in the polygons. Is the sum of all inserted numbers always greater than 2017? Solution. After After 201 20177 cuts, cuts, we obta obtain in 2018 2018 n-convex -convex polygons with n 3. After each cut the total of all sides of those n-convex polygons increases at most 4. We deduce that the total number of sides of 2018 pieces is not greater than 4 2018. If the side of a piece is k j , then the number inserted on it is greater or equal to 5 k j . Therefore, the total of all inserted numbers on the pieces is greater or equal to

≥

×

−



(5 k j ) = 5 2018

−

j

×

−

The answer is positive.

 j

≥ 5 × 2018 − 4 × 2018 = 2018 > 2017.

k j

Question Question 10. Consid Con sider er all words words constitute constituted d by eight eight letters letters from C , H , M , O . We arrange the words in an alphabet sequence. Precisely, the first word is CCCCCCCC, the second one is CCCCCCCH, the third third is CCCCCCCM, CCCCCCCM, the the fourth one is CCCCCCC CCCCCCCO,. O,. . . , and the last word is OOOOOOOO. a) Determi Determine ne the 2017 2017th word of the sequence? b) What What is the positio position n of the wo word rd HOMCHO HOMCHOMC MC in the sequence?

{

}

Solution. We can associate the letters letters C , H, H, M , O with four numbers 0, 1, 2, 3, respectively respectively. Thus, Thus, the arrangement arrangement of those words as a dictionary is equivalent to arrangement of those numbers increasing. a) Number Number 2017 in quatern quaternary ary is 133201 4 = 00133201 4

{

CCHOOMCH.

} {

} ∼

b) Th Thee word word HOMCHOMC is corresponding to the number 13201320 4 which means the number 13201320 in quaternary. Namely,

{

}

{13201320}

4

= 47 + 3 46 + 2 45 + 0 44 + 1 43 + 3 42 + 2 4 + 0.

×

×

×

×

×

×

76

Chapter 2. Answers Answers and Solutions

A simple computation gives 13201320 4 = 30840. Thus, the word HOMCHOMC is is 30840th in the sequence.

{

}

b e an equilateral equilateral triangle, triangle, and let P stand Question 11. Let ABC AB C be for an arbitrary point inside the triangle. Is it true that

 Solution.

 P AB

−

 ≥ 

P AC



  P BC  P CB ?

−

If P lies on the symmetric straightline Ax of ∆ABC ,

Figure 2.12: For Question 11 then



 P AB

 



 − P AC =  P BC −  P CB .

We should consider other cases. Let P denote the symmetric point of P with respect to Ax. The straightline P P intersects AB and AC at M and N, respectively respectively.. Choose B that is symmetric point of B with respect to M N. Then 







 P AB



 − P AC = P AP , 

77

2.2 Senior section

and





  P BC  P CB = P BP  = P B  P  .

We will prove that

−

 P AP 

≥ P B P . 

( )

∗



Indeed, consider the circumscribed circle (O) of the equilateral triangle AMN. Since   M B  N = M BN

 ≤ M BC = M AN = 60 , 0

B  is outside (O). Consider the circumscribed circle (O ) of the equilateral triangle AP P  . It is easy to see that (O ) inside (O), by  which B  is outside (O ). Hence, P AP  P B  P  . The inequality

≥

(*) is proved.

Question 12. Let (O) denote a circle with a chord AB , and let W be the midpoint of the minor arc AB . Let C stand for an arbitrary point on the major arc AB . The tangent to the circle (O) at C meets the tangents at A and B at points X and Y, respectively. The lines W X and W Y meet AB at points N and M , respectively. Does the length of segment NM depend on position of C ? Solution. Let T be the common point of AB and CW . Consider circle (Q) touching XY at C and touching AB at T . Since 1  1    = AW = W B ACW = W AT 2 2   and AW T = CW A, we obtain that ∆AWT, ∆CW A are similar triangles. Then W A2 = W T W C.





×

It is easy to see that W X is the radical axis of A and (Q), thus it passes through the midpoint N of segment AT . Similarly, W Y passes through the midpoint M of segment BT . We deduce M N = AB

2

.

Question 13. Let ABC be a triangle. For some d > 0 let P stand for a point inside the triangle such that

|AB| − |P B| ≥ d, and |AC | − |P C | ≥ d.

78


Figure 2.13: For Question 12 Is the following inequality true

|AM | − |P M | ≥ d, for any position of M

∈ BC ?

Solution. Note that AM always intersects P B or P C of ∆PBC. Without loss of generality, assume that AM has common point with P B. Then ABMP is a convex quadrilateral with diagonals AM and P B. It is known that for every convex quadrilateral, we have

|AM | + |P B| ≥ |AB| + |P M |, that follows

|AM | − |P M | ≥ |AB| − |P B| ≥ d. Question 14. Put P = m 2003 n2017

2017

−m

a) Is P divisible by 24?

n2003 , where

m, n

∈ N.

HOMC 2015-2017

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