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TRAN NAM DUNG (Chief Author)

VO THANH DAT – HOANG DINH HIEU – LUONG VAN KHAI – NGUYEN DUY TUNG

DO THUY ANH – DO TRAN NGUYEN HUY – PHAM THI HONG NHUNG – NGO HOANG ANH

PROBLEMS AND SOLUTIONS

FROM

WINTER SCHOOLS OF MATHEMATICS

IN VIETNAM

FEBRUARY, 2017

PREFACE

“THE MORE PEOPLE YOU GO WITH, THE FURTHER YOU REACH.”

Before Vietnamese Mathematics Olympiad (VMO), students in Vietnam join in some

Winter Schools of Mathematics, which are held in many areas of our country. In these schools,

many well-known Mathematics teachers try to help them review all the necessary materials

used in VMO, emphasize the most important theories, and give out mock examinations to

familiarize the students with solving the problems faster. Besides, students also receive

advice from the coaches, who participated in VMO in the past and now come back to convey

their experiences to the following generations. All these activities are to give competitors the

best preparation for their most important Mathematics Contest in their student lives.

With hope that students who want to participate in VMO next year have a useful

document to review, and international students have a good view about how Vietnamese

competitors study Math and prepare for VMO, we publish the book “Problems and Solutions

from Winter Schools of Mathematics in Vietnam”. In this attachment, we provide full

solutions to most of the exercises, but for some of them we just give simple instructions so

that readers can yourselves find out the solutions.

We really appreciate Mr. Luu Ba Thang (Ha Noi University of Education), Mr. Tran Quoc

Luat (Ha Tinh High School for the Gifted, Ha Tinh Province), Mr. Vu Nguyen Duy (Huynh Man

Dat High School for the Gifted, Kien Giang Province) for all their support. We also would like

to thank our brother Mr. Lê Phúc Lữ (FPT Software, Hồ Chí Minh City) for providing us the

LaTeX template, our friends Đỗ Hà Ngọc Anh (Lê Quý Đôn High School for the Gifted, Bà Rịa –

Vũng Tàu Province) for correcting all our grammar mistakes, and Lâm Quang Quỳnh Anh

(VNU-HCM High School for the Gifted) for designing the cover page.

This is the first time we have ever published a book in English, so we may have some

mistakes in this document. We really want to receive your comments, so that in other

projects, we will not make any mistakes at all. Please send your responses through email

[email protected]. For more information and attachments,

please visit our fanpage https://www.facebook.com/thcmn/ or our blog

https://thcmn.wordpress.com/.

Thanks a lot!

mailto:[email protected]

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Table of contents

1 PROBLEM STATEMENTS 91.1 Winter School of Vietnam Institute for Advanced Studying

Math . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101.1.1 Day 1 . . . . . . . . . . . . . . . . . . . . . . . . . 101.1.2 Day 2 . . . . . . . . . . . . . . . . . . . . . . . . . 11

1.2 Northern Winter School . . . . . . . . . . . . . . . . . . . . 111.2.1 Day 1 . . . . . . . . . . . . . . . . . . . . . . . . . 111.2.2 Day 2 . . . . . . . . . . . . . . . . . . . . . . . . . 12

1.3 Northern Central Winter School . . . . . . . . . . . . . . . 131.3.1 Day 1 . . . . . . . . . . . . . . . . . . . . . . . . . 131.3.2 Day 2 . . . . . . . . . . . . . . . . . . . . . . . . . 14

1.4 Southern Winter School . . . . . . . . . . . . . . . . . . . . 141.4.1 Day 1 . . . . . . . . . . . . . . . . . . . . . . . . . 141.4.2 Day 2 . . . . . . . . . . . . . . . . . . . . . . . . . 15

2 SOLUTIONS 172.1 Winter School of Vietnam Institute for Advanced Studying

Math . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 182.2 Northern Winter School . . . . . . . . . . . . . . . . . . . . 272.3 Northern Central Winter School . . . . . . . . . . . . . . . 352.4 Southern Winter School . . . . . . . . . . . . . . . . . . . . 43

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Problems and Solutions from Winter School of Mathematics inVietnam

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Part 1

PROBLEM STATEMENTS

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1.1. Winter School of Vietnam Institute for Ad-vanced Studying Math

1.1.1. Day 1

Problem 1. Denote xn is the unique positive root of the following equation:

xn + xn−1 + ...+ x = n+2

Prove that the sequence (xn) converges to a positive real number. Find thatlimit.Problem 2. Find all functions f :Q→Q satisfying these following conditionsfor all x ∈Q:

1. f (x+1) = f (x)+1.

2. f(x3)= f 3 (x).

Problem 3. Given two fixed point B,C lying on a circle (O). A variable pointA moves on that circle and always forms with B,C an acute-angled triangle.The interior bisector of ABC intersects line BC and (O) at D and E, respec-tively. A point F lies on the segment BC such that FD = FE.

1. Let H be the orthogonal projection of A on EF . Prove that H alwayslies on a fixed circle.

2. A circle (I) with center I is tangent to array AB, array AC, and line EFat M,N, and P, respectively (Center I and point A belong to the samehalf-plane in accordance to line EF). Let Q be the point on MN suchthat PQ⊥EF. Prove that line AQ always passes through a fixed point.

Problem 4. At the 2016 Winter School, all teachers presented 100 Math prob-lems (a hard problem is called “yam” ) for their students. Whether a problemis “yam” or not, there are exactly 20 students who can solve it. To preparefor the Ceremony, the School’s Committee will select qualified students forthe certificates. In each particular way of selection, a problem is considered"quacky" if it is a "yam" but 20 students who have solved it aren’t awardedthe certificates, or it is not a "yam" but all 20 students who have solved it areawarded the certificates. Prove that there exists a selection so that among 100given problems, there are no more than 4 "quacky" problems.


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1.1.2. Day 2Problem 5. Determine whether there exists 2016 distinct prime numbers p1, p2, . . . , p2016and positive integer n such that:

2016

∑i=1

1p2

i + 1=

1n2 .

Problem 6. Given an acute triangle ABC with incircle (I) is tangent to BC,CA,ABat D,E,F, respectively. Let H be the orthocenter of 4DEF and K is the footof the perpendicular from H to BC.

1. Let J be the midpoint of EF . Prove that AD intersects KJ at a pointlying on (T ), which is the circumcircle of triangle DHK.

2. AD intersects (I) at the second point P. Let M,N are the points lying onDE,DF , respectively, such that MPE = NPF = 90o. Prove that JD isthe radical axis of circle (T ) and the circumcircle of4MNP.

Problem 7. Find the smallest real constant c so that for all real numbers x,y,zthat x+ y+ z = 1, we have the following inequality:∣∣x3 + y3 + z3−1

∣∣≤ c∣∣∣x5 + y5 + z5− 1

∣∣∣When will the equality holds for that value of c?

1.2. Northern Winter School

1.2.1. Day 1Problem 1. Given a positive sequence {bn}∞

n=0 such that:{b0 = 1bn = 2+

√bn−1−2

√1+√

bn−1(n = 1,2,3, ..)

Denote Sn = ∑ni=1 bi2i. prove that the sequence {Sn} has a finite limit. Find

that limit.Problem 2. Find all functions f : (0;+∞)→ (0;+∞) such that

1. f (xyz)+ f (x)+ f (y)+ f (z) = f (√

xy) f (√

yz) f (√

zx) for all x,y,z > 0.

2. f (x)< f (y)∀1≤ x < y.

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Problem 3. Given an acute-angle triangle ABC (AB 6= AC) with circumcircle(O) and orthocenter H. D and E lie on segments AB and AC respectively suchthat AD = AE and D,H,E are collinear. Circle (ADE) intersects circle (ABC)at the second point L.

1. Prove that when A moves on major arc BC of circle (O) (such that4ABC is still a acute-angle triangle and AB 6= AC), then the tangentline in (ADE) at D and E intersect at a fixed point.

2. Let M be the midpoint of BC. Prove that MH⊥AL.

Problem 4. For each positive integer n, denote sn as the number of ways torepresent n into the sum(s) of ordered sequence(s) with each of its terms be-longs to the set {1,3,4}. Prove that s2016 is a perfect square.

1.2.2. Day 2

Problem 5. Given ABCD as the cyclic quadrilateral of circle (O) with twoperpendicular diagonals AC and BD. AD intersects BC at P. Segment PQ isthe diameter of circle (PCD). Points M,N are the midpoints of two types of arcCD in circle (PCD). QM intersects BD,CD at E,F respectively. QN intersectsAC,CD respectively at K,L. Prove that two circles (EDF) and (KCL) aretangent to each other.Problem 6. Given two positive integers m and n. The sequence a(m,n) of realnumbers has the following properties:

a(0,0) = 2,a(0,n) = 1,a(n,0) = 2,n≥ 1;

a(m,n) = a(m−1,n)+a(m,n−1),m,n ∈ N∗.

Prove that, for each positive integer k, all the roots of the following polynomial

Pk(x) =k

∑i=0

a(i,2k+1−2i)xi

are real numbers.Problem 7. A positive integer n is called beauty if for all positive integers aand b, the following statement is true: If n | a4b+1 then n | a12+b. Prove thatn is beauty if and only if n | 26 (216−1

)and 257 - n.


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1.3. Northern Central Winter School

1.3.1. Day 1

Problem 1. Find the largest positive real number k such that the followinginequality

a2 +b2−2√2(a+b)−2

≥ k

holds for all distinct real numbers a,b satisfying ab = 1.

Problem 2. Let f be a function satisfying the following conditions:{f : [1;+∞)→ [1;+∞)

x. f (x+1)≤ [ f (x)]2−1∀x≥ 1.

Prove that f (x)≥ x+1,∀x≥ 1.

Problem 3. Let ABC be a acute-angled triangle (AC > BC) with orthocenterH and circumcircle (O). CH intersects AB at D and intersects (O) the secondtime at AE. Let F be the intersection of AC and BE, I be the intersection of Aand BC, J be the intersection of AB and IF .

1. Let A′ be the reflection of A through CH. Prove that BABA′ =

BJBD .

2. Let K be the intersection of HA′ and BC. Prove that ODK = 90o.

Problem 4. Given a positive integer n. A square with side n is created from2n(n+ 1) unit segments. How many ways are there to divide those unit seg-ments into n(n+1) pairs, each of which have 2 unit segments satisfying all ofthe following conditions:

1. In a pair, there is a vertical and a horizontal segment with the samenode, creating a L shape.

2. There are no two pairs of the division having 4 segments with the sameend. In another word, there are no two L”s” with the same vertex.


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1.3.2. Day 2

Problem 5. Does there exist a function f : (0;+∞)→ (0;+∞) satisfying

f (2016+ x f (y)) = y f (x+ y+2016)x,y > 0?

Problem 6. Let ABC be a acute-angled triangle with AB < AC and a circum-circle (O). D is the projection of A on BC. The line passing through D; per-pendicular to AB intersects CA at E. The line which passes through D and isperpendicular to AC intersects AB at F . EF cuts the tangent line of (O) at Aat G. M,R are the midpoints of BC and AD, respectively.

1. Prove that GR⊥AM.

2. Let I,J be the intersections of DE,DF and GR, AG intersects BC at S;K,L be the midpoints of JE, IF , respectively. Prove that OS//KL.

Problem 7. Let p,q be positive integers. Write number 1 on the board andrepeat the following algorithm: Replace the current number on the board byits sum with p or with q. Find the conditions of p,q to assure that the algo-rithm can be repeated infinitely without creating a number on the board thatis neither a multiple of p nor a multiple of q.

1.4. Southern Winter School

1.4.1. Day 1

Problem 1. Let x,y be positive real numbers such that x+ y = 2.

1. Prove that1xn +

1yn ≥ xn+1 + yn+1 ∀n ∈ N∗.

2. For all n, can we replace n+ 1 with a positive integer k (k > n+ 1) sothat the inequality

1xn +

1yn ≥ xk + yk

holds true with all x,y > 0,x+ y = 2?


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Problem 2. Let (an) be a sequence defined by:{a1 = 1,a2 = 2,an+2 =

3an+1−an2 + 1

n2(n = 1,2,3, ..)

Prove that (an) has a defined limitation.

Problem 3. Let ABC be an acute-angled triangle with its circumcircle (I). (I)is tangent to BC,CA,AB at D,E,F respectively. J is the midpoint of EF . BJand CJ intersect CA and AB at H and K; EF intersects BC and HK at M andN, respectively. Prove that AM = AN.

Problem 4. Consider a 2n x 2n square. Each square is colored with only onecolor, white or black, such that

i. The number of black squares on each line are pairwise different.

ii. The number of black squares on each column are pairwise different.

1. How many black squares are there in the square?

2. How many pairs of consecutive color-different squares? (Two squaresare called consecutive if and only if they have a same side.)

1.4.2. Day 2Problem 5. Find all functions f: R→ R with the property that

f ( f (x)+2y) = 10x+ f ( f (y)−3x)

holds for all x,y ∈ R.

Problem 6. There are 23 positive integers (not necessarily distinct) writtenon a line. Prove that we can place the symbols “+”, “x” or the parenthesesreasonably so that the obtained expression is divisible by to 2000.

Problem 7. Let ABC be a triangle and P be a point inside. D is the pro-jection of P on BC. E,F are the points satisfying the following conditions:PE⊥AC,CE⊥BC,PF⊥AB, BF⊥BC, G is symmetric to D through BC.

1. Prove that AG⊥EF .

2. Let S,T be the intersection of EF and CA,AB, respectively. Q,R aresymmetric to P through the midpoints of GE,GF . (l1) be the line thatpasses through S and perpendicular to AQ, (l2) be the line that passesthrough T and perpendicular to AR. Prove that (l1) and (l2) intersect ata point on AG.


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Part 2

SOLUTIONS

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2.1. Winter School of Vietnam Institute for Ad-vanced Studying Math

Denote xn is the only root of the following equation:

xn + xn−1 + ...+ x = n+2

Prove that the sequence (xn) converges to a positive real number. Findthat limit.

Problem 1

Solution. Notice xn > 1 for all n∈N∗. We will prove that xn > xn+1. Considerthe polynomial Pn (x) = xn + xn−1 + · · ·+ x, we could see that Pn(x) is mono-tonically increasing on the interval (0,+∞) for all n≥ 1. From Pn (xn) = n+2and Pn+1 (xn+1) = (n+1)+2, we get

Pn+1 (xn+1) = Pn (xn)+ xn+1n > n+3.

So xn > xn+1. That means the sequence (xn) converges to a real limit. Thusthat limit is 1 cause for every a > 1, there exist a large enough integer n suchthat an > n+2.

Find all functions f : Q→ Q that satisfy these following conditions forall x ∈Q:

i. f (x+1) = f (x)+1.

ii. f(x3)= f 3 (x).

Problem 2

Solution. From the first condition, we can prove that:

f (x) = x+n∀x ∈Q,n ∈ Z. (2.1.2.1)

Also, as all x ∈ Q could be represented as x = pq (p,q ∈ Z,q ≥ 2). We fix a

positive integer k ≥ 2. On the one hand, by using (2.1.2.1) and the secondcondition, we have:

f

((pq+qk

)3)

= f(

p3

q3 +3p2.qk−2 +3p.q2k−1 +q6k)


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= f(

p3

q3

)+3p2.qk−2 +3p.q2k−1 +q6k

= f 3(

pq

)+3p2.qk−2 +3p.q2k−1 +q6k. (2.1.2.2)

On the other hand, using the second condition will result in:

f

((pq+qk

)3)

= f(

pq+qk

)3

=

(f(

pq

)+qk

)3

= f 3(

pq

)+3 f

(pq

)2

.qk +3 f(

pq

).q2k +q6k. (2.1.2.3)

Comparing (2.1.2.2) and (2.1.2.3) and dividing both sides byq2k, we get

f(

pq

)2

.1qk + f

(pq

)=

p2

qk+2 +pq

Now, by letting k→+∞ : f(

pq

)= p

q .Conversely, we can easily check that f (x) = x is a valid solution.

Given two fixed point B,C lying on a circle (O). A variable point A moveson that circle and always forms with B,C an acute-angled triangle. Theinterior bisector of ABC intersects line BC and (O) at D and E, respec-tively. A point F lies on the segment BC such that FD = FE.

1. Let H be the orthogonal projection of A on EF . Prove that H al-ways lies on a fixed circle.

2. A circle (I) with center I is tangent to array AB, array AC, and lineEF at M,N, and P, respectively (Center I and point A belong to thesame half-plane in accordance to line EF). Let Q be the point onMN such that PQ⊥EF. Prove that line AQ always passes through afixed point.

Problem 3


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Solution.

1. Since EF and BC are anti-parallel lines of BAC,A,O,H are collinear.Thus H lies on the fixed circle of diameter OE.

2. Assume that EF intersects AB,AC at K,L respectively. We will provethat AQ passes through the midpoint J of KL.

• Assume that PQ intersects (I) and line d passed through A inparallel to EF at R and S. Since d is the polar of Q in the cir-cle (I) (MN is the polar of point A and AS⊥IQ), we can inferthat (P,Q,R,S) =−1⇒ A(P,Q,R,S) =−1. Thus if EF intersectsAR,AQ at J,G respectively, then JG = JE.

• Also, KG = LE since G is the tangent point of the excircle rela-tive to angle A of 4AKL (from the homothecy (A, AG

AR )), J is the


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midpoint KL. Since BC and KL are anti-parallel lines, AQ is thesymmedian through A of 4ABC. So AQ passes through the in-tersection of tangents at B and C to circle (O), which is a fixedpoint.

Comment. Other solutions to prove AQ passes through midpoint J of KL.

• Solution 2: AD intersects MN at W . A line passing through Q and KL in-tersects AB,AC at U,V respectively. We can easily conclude that IUMQand INV Q are cyclic. Then IUV = IMN, IVU = INM. Thus 4IUV isan isosceles triangle with IV = IU . Therefore, Q is the midpoint ofUV ⇒ J is the midpoint of KL.

• Solution 3: M,Q,N are the foot of the perpendicular lines from I tothree sides of triangle AUV . According to Simson theorem, I lies on thecircumcircle of 4AUV . Furthermore, I lies on the interior bisector ofangle A. From Q is the midpoint of UV , J is the midpoint of KL.

At the 2016 Winter School, all teachers presented 100 Math problems (ahard problem is called "yam" ) for their students. Whether a problem is“yam” or not, there are exactly 20 students who can solve it. To preparefor the Ceremony, the School’s Committee will select qualified studentsfor the certificates. In each particular way of selection, a problem is con-sidered "quacky" if it is a "yam" but 20 students who have solved it aren’tawarded the certificates, or it is not a "yam" but all 20 students who havesolved it are awarded the certificates. Prove that there exists a selectionso that among 100 given problems, there are no more than 4 "quacky"problems.

Problem 4

Solution. Calling the number of students n, we can see that n ≥ 20 since aproblem was solved by exactly 20 students. We need to show that the Com-mittee could select some students for certificates so that no problems are"quacky". Assume that the opposite statement is true, which means every se-lection has at least one "quacky" problem. We will count the number of pairs(K,b), in which K is a way to give certificates and b is a "quacky" problemfrom the K way, by counting in two ways.


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1. Firstly, by using induction, we easily get there are total 2n ways to givecertificates with at least a "quacky" problem in each way. So the numberof pairs (K,b) is no less than 2n.

2. Secondly, an arbitrary problem b could only be "quacky" if it is a "yam"with 20 students are all awarded, or it is not a "yam" and none of thosestudents is awarded. Thus a problem b could only have 2n−20K ways tolet b become a "quacky" problem in the K way. Therefore, the numberof pairs (K,b) is equal to 100. 2n−20.

Comparing these two ways, we obtain 100.2n−20 ≥ 2n, or 100 ≥ 220, contra-diction.

Determine whether there exist 2016 distinct prime numbers p1, p2, . . . , p2016and positive integer n such that:

2016

∑i=1

1p2

i + 1=

1n2

Problem 5

Solution. Assume that there exist 2016 distinct prime numbers p1, p2, . . . ., p2016and a positive integer n such that

2016

∑i=1

1p2

i + 1=

1n2

Denote

P =2016

∏i=1

(p2

i +1)⇒ n2.

(2016

∑i=1

Pp2

i + 1

)= P. (2.1.5.1)

1. Solution 1: We solve the problem by using the congruence relation bymodulo 3 for equation (2.1.5.1).

• If pi 6= 3 for all i, then p2i +1≡−1 (mod 3). Then P≡ 1 (mod 3)

and each fraction Pp2

i +1≡ −1 (mod 3). So the left side of the

equality is divisible by 3 and the right side is not, contradiction.

• If there is a number j such that p j = 3, p2j +1 ≡ 1 (mod 3), and

for 2015 numbers i 6= j, p2i + 1 ≡ −1 (mod 3). Thus P ≡ −1


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(mod 3). In the sum ∑2016i=1

1p2

i + 1, there’s one term that is congru-

ent to −1 (mod 3), and 2015 terms are congruent to 1 (mod 3),thus the sum≡ 1 (mod 3). Combining with n2 ≡ 1,0 (mod 3), wehave a contradiction, since the left side of the equation is ≡ −1(mod 3).

Therefore, there do not exist 2016 prime numbers and positive integern that satisfy the problem’s condition.

2. Solution 2:

• If there is a number pi = 2, then v2

(p2

j +1)= 1for all odd num-

bers p j, so v2 (P) = 2015. Also, v2

(P

p2j+1

)= 2014 for all j 6= i

and v2

(P

p2j+1

)= 2015 if j = i. From there, if we put 22014 as the

common factor for the sum ∑2016i=1

Pp2

i +1, then we remain a sum of

2015 odd terms and 1 even term. Thus:

v2

(2016

∑i=1

Pp2

i + 1

)= 2014,

Therefore, v2(n2)= 1. Contradiction!

• Assume that all pi numbers are odd, then v2 (P) = 2016. Also,for all i, p2

i + 1 has the form of 8k + 2 = 2(4k+1). Thus, everyterm P

p2i +1

has the form of 22015(4k+1). After putting 22015 as the

common factor of the sum ∑2016i=1

Pp2

i +1, we will have a sum of 2016

terms with each of them≡ 1 (mod 4), and the sum≡ 0 (mod 4).Therefore, v2

(∑

2016i=1

Pp2

i + 1

)≥ 2015+2 > v2 (P) . Contradiction!

So our assumption is false, and the problem is solved.


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Given an acute-angled triangle ABC with incircle (I) is tangent to BC,CA,ABat D,E,F, respectively. Let H be the orthocenter of 4DEF and K is thefoot of the perpendicular from H to BC.

1. Let J be the midpoint of EF . Prove that AD intersects KJ at a pointlying on (T ), the circumcircle of triangle DHK,.

2. AD intersects (I) at the second point P. Let M,N are the points ly-ing on DE,DF , respectively, such that MPE = NDF = 90o. Provethat JD is the radical axis of circle (T ) and the circumcircle of4MNP.

Problem 6

Solution.

1. • Solution 1: Let E ′,F ′ be the feet of the attitudes from E,F of∆DEF respectively. As KDE ′F ′ is an isosceles trapezoid, K,D,E ′,F ′

and H lie on the circle with diameter HD. Also,4KF ′E ′=4DE ′F ′∼4DEF and KJ,DA are the symmedians inside4KF ′E ′ and4DEFrespectively ⇒ F ′KJ = EDA⇒ F ′KDL is a cyclic quadrilateral.So L ∈ (T ) .

• Solution 2: If DI meets (I) at S then H,J,S are collinear and HJ =


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JS. Also, HDA = JDS (DH,DI) is the isogonal pair) and JDS =

HKJ (HJ = JS)⇒ HDL = HKL⇒ L ∈ (T ) .

2. • Solution 1: First, we prove that (MNP) is the Appolonius circle(C) of vertex D ∈4DEF . As P is the intersection of the symme-dian and (DEF) ,P ∈ (C). Thus (I) and (C) have D,P as commonpoints. Both line DEM and circle (C) have the common point D,with EPM = 90o⇒M ∈ (C). Similarly, point N also lies on circle(C). Thus (MNP)≡ (C).Second, we prove that the intersection Q of DJ and (T ) lies on(C). By algebraic calculation

JE2 = FF ′2 = JQ.JD⇒4JFQ∼4JDE⇒ QE = ED.EJDJ

.

By the same argument, we get QF = FD.FJDJ ⇒

QEQF = ED

FD ⇒ Q ∈(C). Therefore, DJ is the radical axis of circle (C) and (T ).

• Solution 2:

– PND = 90o− NFP = 90o− PED = PMD⇒ D ∈ (MNP).Note that PQ//HD⇒ PQJ = HDQ = HDP+ PDJ. On theother hand, (DP,DJ) and (DH,DS) are isogonal pairs⇒ HDP=

JDS. Thus PQJ = PDS⇒ PQJ + PED = 90o. Also, DNP+NFP = 90o⇒ PQJ = DNP⇒ DNPQ is a concyclic quadri-lateral⇒ Q ∈ (MNP).

– Now let DJ,DH intersect (I) at U,V respectively. Note thatDP is the symmedian of4DFE and4FPE, so{4DFJ ∼4DPE4FPJ ∼4DPE ⇒

{FJD = PEDFJP = PED

⇒ FJD = FJP.

(2.1.6.1)Besides, V PJ = V PF + FPJ. Since V PF = EPS (due to theisogonal pairs) and FPJ = DPE (due to the similar triangles),V PJ = DPE + EPS = 90o. Thus V PJ = HQJ = 90o ⇒ Hand V are symmetric about FE. Combining with (2.1.6.1), Pand Q are symmetric about FE, which means PQ//HD.


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Find the smallest real constant c so that for all real numbers x,y,z thatx+ y+ z = 1, we have the following inequality:∣∣x3 + y3 + z3−1

∣∣≤ c∣∣∣x5 + y5 + z5− 1

∣∣∣When will the equality holds for that value of c?

Problem 7

Solution. Assume that constant c satisfies the given assertion. With x = y =z = 1

3 , we get89≤ c.

8081

or c≥ 910 . We prove that c = 9

10 is the minimum value satisfying the assertionof the problem.From x+ y+ z = 1, we have

x3 + y3 + z3−1 = x3 + y3 + z3− (x+ y+ z)3 =−3(x+ y)(y+ z)(z+ x),

x5 + y5 + z5−1 = x5 + y5 + z5− (x+ y+ z)5

=−52(x+ y)(y+ z)(z+ x) [(x+ y)2 + (y+ z)2 + (z+ x)2].

Therefore, if the denominator differs to 0,∣∣x5 + y5 + z5−1∣∣

|x3 + y3 + z3−1|=

56

[(x+ y)2 + (y+ z)2 + (z+ x)2

]≥ 5

6.[2(x+ y+ z)]2

3=

109

From this, we can conclude that for all real number x,y,z that satisfy x+ y+z = 1, ∣∣x3 + y3 + z3−1

∣∣≤ 910

∣∣∣x5 + y5 + z5− 1∣∣∣

By simple calculation, we obtain that the equality holds for x = y = z = 13 or

(x,y,z) as the permutation of (a, −a, 1) in which a is some real number.

Comment. The inequality is also true with x3 + y3 + z3 = 0.


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2.2. Northern Winter School

Given a positive sequence {bn}∞n=0 such that:{

b0 = 1bn = 2+

√bn−1−2

√1+√

bn−1(n = 1,2,3, ..)

Denote Sn = ∑ni=1 bi2i. prove that the sequence {Sn} has a finite limit.

Find that limit.

Problem 1

Solution. Denote an = 1 +√

bn,n ≥ 0. Then an > 1, a0 = 2. Also, an =√

an−1⇒ an = 22−n. Thus Sn = 2−2.22−n−1

2−n . Let n→ ∞, we obtain

limn→∞

Sn = limx→0

(2−2.

2x−1x

)= 2−2. ln2.

Find all functions f : (0;+∞)→ (0;+∞) such that

1. f (xyz)+ f (x)+ f (y)+ f (z) = f (√

xy) f (√

yz) f (√

zx)∀x,y,z > 0.

2. f (x)< f (y)∀1≤ x < y.

Problem 2

Solution.

• Replace x = y = z = 1 in (1), 4 f (1) = f (1)3 so f (1) = 2.

• Replace x = ts,y = ts ,z =

st in (1), we get f (t) . f (s) = f (t.s)+ f

( ts

).

• Replace s = 1 : f (t) = f(1

t

).

• Combine with condition (2), we get f (t)≥ f (1) = 2∀ t > 0. Therefore,there exist g(t)≥ 1 such that f (t ) = g(t)+ 1

g(t) .

• By induction and condition (1), we get g(tn) = gn (t) ∀n ∈ Z. Theng(tq) = gq (t) for all q ∈ Q. Thus for every fixed number t > 1, wehave f (tq ) = aq +a−q, a = g(t) , and q ∈Q.


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• Since f is a monotonic function on the range (0;1] and [1,+∞), thenf (tr ) = ar + a−r for all real number r. Therefore, if k satifies tk = a,then

f (x) = xk + x−k∀x ∈ R.

Given an acute-angle 4ABC(AB 6= AC) with circumcircle (O) and or-thocenter H. D and E lie on segments AB and AC respectively such thatAD = AE and D,H,E are collinear. Circle (ADE) intersects circle (ABC)at the second point L.

1. Prove that when A circulates on major arc BC of circle (O) (suchthat 4ABC is still a acute-angle triangle and AB 6= AC), then thetangents in (ADE) at D and E intersect at a fixed point.

2. Let M be the midpoint of BC. Prove that MH⊥AL.

Problem 3

Solution. Without loss of generality, we can assume that AB < AC. Let F andG be the feet of the altitudes at points A and B of4ABC, N is the midpoint of


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minor arc BC of circle (O). Then O,M,N are collinear, line AN is the interiorbisector of BAC and also the perpendicular bisector of segment DE. Let I bethe center of circle (ADE) and P be the intersection of DE and AN.

1. We will prove that all two tangents at D and E of (ADE) pass throughN. Equivalently, we need to prove that DEN = EDN = A (angle A oftriangle ABC). By algebraic calculation, we can obtain these equalities:

• HAP = B−C2 (Denote B and C as the angles of4ABC).

• AH = 2R.cosA (R is the radius of circle (O)).

• AP = AH.cos HAP = 2R.cosA.cos B−C2 .

• AN = 2R.sin(

C+ A2

)= 2R.cos

(B−C

2

).

• PE = AP. tan A2 = 2R.cosA.cos B−C

2 . tan A2 .

From these, we have:

tan DEM =PNPE

=AN−AP

PE=

2R.cos B−C2 −2RcosA.cos B−C

2

2RcosA.cos B−C2

=1− cosA

cosA. tan A2

=2sin2 A

2 .cos A2

cos A2 .sin A

2

= tanA2

Thus DEN = A. Similarly, EDN = A. Therefore, two tangents at D andE to the circle (ADE) intersect at the fixed point N when point A circu-lates on major arc BC of circle (O).

2. Because OI⊥AL, we need to prove that MH ‖ OI. It follows that weprove HMOJ is a parallelogram, with J is the intersection of OI andAH, and, specifically, OM = HJ. We already obtain that OM = 2AH =R.cosA. Thus proving AJ = R.cosA is sufficient to obtain the given as-sertion. Indeed,4AJI∼4NOI, so, with r is the radius of circle (ADE):

AJAI

=ONIN⇒ AJ = AI.

ONIN

=Rr

AN−AI=

RrAN− r

Also,

r =DE

2.sinA=

PEsinA

=2R.cosA.cos B−C

2

2cos2(A

2

)http://thcmn.wordpress.com 29

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Therefore,

AJ =Rr

2R.cos(

B−C2

)− r

=

R2R.cosA.cos B−C2

2cos2(

A2

)2R.cos

(B−C

2

)− 2R.cosA.cos B−C

2

2cos2(

A2

)

=R.cosA

2cos2(

A2

)− cosA

= R.cosA,

and we have proved the required assertion.

For each positive integer n, denote sn as the number of ways to representn into the sum(s) of ordered sequence(s) with each of its terms belongsto the set {1,3,4}. Prove that s2016 is a perfect square.

Problem 4

Solution. Notice that s1 = 1, s2 = 1, s3 = 3, s4 = 4 since:

1 = 12 = 1+1

3 = 1+1+1 = 34 = 1+1+1+1 = 1+3 = 3+1 = 4

For n = 5, we consider these following cases:

i. If the last term of the sum is 1 then the numbers of ways to represent issn−1.

ii. If the last term of the sum is 3 then the numbers of ways to represent issn−3.

iii. If the last term of the sum is 4 then the numbers of ways to represent issn−4.

From these cases, we obtain the recurrence relation as below:{s1 = 1, s2 = 1, s3 = 3, s4 = 4sn = sn−1 + sn−3 + sn−4, n≥ 5


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To prove s2016 is a perfect square, we prove that for all positive integer n,s2nis also a perfect square. By direct calculation, s2 = 1, s4 = 22,s6 = 32, s8 =52,s10 = 82, We can easily predict this statement: s2n = F2

n+1, with (Fn) is theFibonacci Sequence, and prove this by induction.

• With n = 1, the statement is true.

• Assume that the statement is true for all positive integer k ≤ n = 2m.Using the property of the Fibonacci Sequence, we have

s2m+2 = s2m+1 + s2m−1 + s2m−2

= (s2m + s2m−2 + s2m−3)+ s2m−1 + s2m−2

= s2m +2s2m−2 +(s2m− s2m−4)

= 2F2m+1 +2F2

m−F2m−1

= (Fm+1 +Fm)2 +(Fm+1−Fn)

2−F2m−1

= F2m+2 + F2

m−1−F2m−1 = F2

m+2.

Therefore, the statement is also true for n = 2m+ 2. By induction, we haveproven the statement, which means s2016 is a perfect square.

Given ABCD as the cyclic quadrilateral inscribed in circle (O) with twoperpendicular diagonals AC and BD. AD intersects BC at P. Segment PQis the diameter of circle (PCD). Points M,N are the midpoints of twotypes of arc CD in circle (PCD). QM intersects BD,CD at E,F respec-tively. QN intersects AC,CD respectively at K,L. Prove that two circles(EDF) and (KCL) are tangent to each other.

Problem 5


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Solution. Let S be the intersection of AC and BD. The bisector of ASD inter-sects AD,BC at R,U respectively.

• DRU = DAS+ ASR = CBD+ BSU = CUR⇒4PRU is isosceles at P.

• PM is the bisector of PRU⇒PM⊥RU⇒PM is parallel to the bisectorSx of DSC⇒ Sx⊥MQ. Besides, DSC = 90o⇒ SEQ = 45o. But ESKQis an inscribed quadrilateral⇒ LKC = SEQ = 45o.

• Let T 6=C is another intersection of (CD) and (KCL)⇒ LKC = LTC =

45o⇒ DT L = 45o.

• M,N are the midpoints of arc CD respectively ⇒ QM,QN are the bi-sectors of DQC⇒ (DCLF) =−1, but DTC = 90o⇒ TC is the bisectorof LT F⇒ LT F = 90o⇒ DT F = 135o⇒ DT F + DEF = 180o⇒ T ∈(DEF).

• Construct the tangent Ty of (LTC). yT D= yT L−DT L= TCL−CT F =

T FD⇒ Ty is also the tangent of (DEF)⇒ our two circles are tangentto each other.


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Given two positive integers m and n. The sequence a(m,n) of real num-bers has the following properties:

a(0,0) = 2,a(0,n) = 1,a(n,0) = 2,n≥ 1;

a(m,n) = a(m−1,n)+a(m,n−1),m,n ∈ N∗.

Prove that, for each positive integer k, all the roots of the following poly-nomial

Pk(x) =k

∑i=0

a(i,2k+1−2i)xi

are real numbers.

Problem 6

Solution. Let Qk(x) = ∑ki=0 a(i,2k−2i)xi. From the sequence a(m,n), we

have: {Pk (x) = xPk−1 (x)+Qk (x)

Qk (x) = xQk−1(x)+Pk−1 (x)

It follows that Qk (x) = Pk (x)− xPk−1 (x) and therefore:

Pk (x) = (2k+1)Pk−1 (x)− x2Pk−1 (x) , k ≥ 2.

with P1 (x) = 3x+1 and P2 (x) = 5x2 +5x+1.We prove the following statement: For each integer k ≥ 2, all the roots of thepolynomials Pk (x) and Pk−1 (x) are all real numbers and are distinct from eachother. Specifically, if a1 < a2 < a3 . . . . < ak−1 and b1 < b2 < b3 · · · < bk areall the roots of Pk−1 (x) and Pk (x) respectively, then:

b1 < a1 < b2 < a2 . . . . < ak−1 < bk

That statement can be proved by induction and by considering the sign ofPk+1 (x) based on considering that of Pk (x) and Pk−1 (x) .Comment.

1. 0 is not the root of Pk (x) , k ≥ 1. Therefore, the sign of Pk (x) and thatof x2.Pk (x) are always the same.

2. The coefficient of xk of Pk (x) is a(2k,1) > 0, thus Pk (+∞) > 0 , andthe sign of Pk (−∞) depends on whether k is odd or even.


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A positive integer n is called beauty if for all positive integers a and b,the following statement is true: If n | a4b+1 then n | a12 +b. Prove thatn is beauty if and only if n | 26 (216−1

)and 257 - n.

Problem 7

Solution. Consider a positive integer a such that (a,n) = 1. Then, there existsb such that a4b+ 1 ≡ 0 (mod n). As a16 ≡ 1 (mod n), we have just proventhat for all (a,n) = 1, a16 ≡−1 (mod n).

• If n is odd, choose a = 2, we get n |a16−1 = 3.5.17.257 = M.

• If n is even, rewrite n = 2t .n1 (t ∈ Z+,n1 ∈ Z+,n1 is odd). Choose a =n1 + 2, then a is odd and (a,n1) = 1. As a16− 1 ≡ 0 (mod n), a16−1 ≡ 0 (mod n)1 ⇒ n1 |M. On the other hand, choose a ≡ 5 (mod 8)and a≡ 1 (mod n)1. We have v2(a16−1) = v2(a2−1)+ v2(16)−1 =3+4−1 = 6⇒ t ≤ 6. So n |26(216−1). Choose a = 7, we get

n |716−1 = (72−1)(72 +1)(74 +1)(78 +1),

so n is not divisible by 257, which means n |26.3.5.17.

Now, assume that n |26.3.5.17, and a,b ∈ N∗ such that n |a4b+1.

• If 2t |n(t = 1,2,3,4,5,6), then from 2t |a4b+1 and 26 |a16−1, we get2t |a4(a8 +b)⇒ 2t |b+a8.

• If 3 |n, then a4≡ 1 (mod 3)⇒ b≡−1 (mod 3). So a8+b≡ 0 (mod 3).

– If 5 |n |a4b + 1, then a4 ≡ 1 (mod 5) ⇒ b ≡ −1 (mod 5). So,5 |a8 +b.

– If 17 |n |a4b+1, then a16≡ 1 (mod 1)7⇒ 17 |a4(a8+b)⇒ 17 |a8+b.

In conclusion, all n such that n |26(216−1) are beautiful numbers.


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2.3. Northern Central Winter School

Find the largest positive real number k such that the following inequality

a2 +b2−2√2(a+b)−2

≥ k

holds for all distinct real numbers a,b satisfying ab = 1.

Problem 1

Solution. We need to find the largest positive real number k such that

k ≤a2 +

1a2 −2√

2(

a+1a

)−2

∀a > 0,a 6= 1.

Consider the function

f (a) =a2 + 1

a2 −2√2(a+ 1

a

)−2∀a > 0,a 6= 1

=

(a2 + 1

a2 −2)(√

2(a+ 1

a

)+2)

2(a+ 1

a

)−4

=

(a+1)2(√

2(a+ 1

a

)+2)

2a,

so lima→1

f (a) = 8. In addition,

f (a)>4(a+1)2

2a> 8∀a > 0,a 6= 1,

so k = 8 satisfies the condition of the problem.Assume that there exists another positive real number k > 8 satisfying thecondition of the problem, which means k≤ f (a);∀a> 0,a 6= 1. As lim

a→1f (a) =

8, limn→+∞

f(1+ 1

n

)= 8. So for ε = k−8, there exists M ∈ N∗ such that∣∣∣∣ f (1+

1n

)−8∣∣∣∣< ε = k−8∀n > M.

⇒ f(

1+1n

)< k;∀n > M,


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contradiction. So our assumption is false, which means k = 8 is the largestpositive real number that makes our inequality hold true.

Let f be a function satisfying the following conditions:{f : [1;+∞)→ [1;+∞)

x. f (x+1)≤ [ f (x)]2−1∀x≥ 1.

Prove that f (x)≥ x+1,∀x≥ 1.

Problem 2

Solution.

• Firstly, by induction, we prove that

f (x)> x1− 12n ∀n ∈ N∗,x≥ 1. (2.3.2.1)

– f (x+1)≥ 1⇒ x f (x+1)≥ x⇒ ( f (x))2 ≥ x+1 > x. So (2.3.2.1)is true for n = 1.Suppose that (2.3.2.1) is true until n = k ≥ 1. We have

f (x+1)> (x+1)1− 12k > x1− 1

2k

⇒ ( f (x))2 > ( f (x))2−1≥ x f (x+1)> x2− 12k ⇒ f (x)> x1− 1

2k+1 .

So (2.3.2.1) is true for n = k+1, and by induction, it is true ∀n ∈N∗.

– From (2.3.2.1), let n→+∞, we get f (x)≥ x, ∀x≥ 1.

• Now, one more time by induction, we prove that

f (x)> x+1− 12n ∀n ∈ N∗,x≥ 1. (2.3.2.2)

– f (x+ 1) ≥ x+ 1⇒ ( f (x))2 ≥ x(x+ 1)+ 1 >(x+ 1

2

)2⇒ f (x) >x+ 1

2 . So (2.3.2.2) is true for n = 1.Suppose that (2.3.2.2) is true till n = k ≥ 1. We have

f (x+1)> (x+1)+1− 12k = x+2− 1

2k

⇒ ( f (x))2 > x(

x+2− 12k

)+1=

(x+1− 1

2k+1

)2

+12k −

14k+1


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>

(x+1− 1

2k+1

)2

.

So (2.3.2.2) is true for n = k+1, and by induction, it is true ∀n ∈N∗.

– From (2.3.2.2), let n→+∞, we get f (x)≥ x+1, ∀x≥ 1.

Let ABC be an acute triangle (AC > BC) with orthocenter H and circum-circle (O). CH intersects AB at D and interscts (O) the second time at E.Let F be the intersection of AC and BE, I be the intersection of A andBC, J be the intersection of AB and IF .

1. Let A′ be the reflection of A through CH. Prove that BABA′ =

BJBD .

2. Let K be the intersection of HA′ and BC. Prove that ODK = 90o.

Problem 3

Solution.


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1. F(I,D,B,C)=−1⇒ (JDBA)=−1⇒ JAJB

=DADB⇒ JA− JB

JB=

DA−DBDB

⇒ABJB

=A′BDB⇒ BA′

BA=

BDBJ

.

2. We can easily prove that AEA′H is a rhombus ⇒ HA′//AE ⇒ BKBI =

BA′BA =

BDBJ⇒DK//IJ. So, according to Brocard’s theorem, we get HA′//AE⇒

BKBI = BA′

BA = BDBJ ⇒DK//IJ. So, according to Brocard’s theorem, we get

OD⊥IJ⇒ ODK = 90o.

Let be given a positive integer n. A square with side n is created from2n(n+ 1) unit segments. How many ways are there to divide those unitsegments into n(n+1) pairs, each of which have 2 unit segments satisfy-ing all of the following conditions:

1. In a pair, there is a vertical and a horizontal segment with the sameend, creating a L shape.

2. There are no two pairs of the division having 4 segments with thesame end. In another word, there are no two L”s” with the samevertex.

Problem 4

Solution. The net has (n+ 1)2 vertices. We call those that are vertices of aletter L is a conjunction vertex.

• According to the conditions above, there are no two letters L that havethe same vertex. As a result, the number of conjunction vertices is equalto the number of pairs of letter L, which are both equal to n(n+1). Now,we can conclude that there are (n+1)2−n(n+1) = n+1 vertices thatare not conjunction vertices. We call them isolated vertices.

• Consider a row. Because there are only n horizontal segments, there areat most n conjunction vertices. As a result, there is at least an isolatedvertex on a row.

• From the result above, we can easily see that there is only an isolatedvertex on every column and every row. We call a set S containing n+1vertices of the net a “reasonable” set if there are no two vertices in the


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set lying in the same row or the same column.We have proved that with every division satisfying the condition, the setof n+ 1 isolated vertices of the division is reasonable. We now provethe reverse, which means every “reasonable” set will be correspondentwith one and only one division satisfying the condition.

• The principle of building the vectors is that every vector on a row willhave its arrow turning to the isolated vertex. There will be no two vec-tors creating a V shape because if that happens, the two vectors will havedifferent directions, which means there are 2 vertices on the line or ar-row containing the segment. The result contradicts with the assumptionabove.

• Assume that there is another division different from the division above.In this case, there is a vector with its arrow turning in a different di-rection with the standard vector. We call it “bad” vectors. Without lossof generalization, we assume that the division has a horizontal “bad’vector. This vector will have its arrow turning to a conjunction vertexQ. As Q is a conjunction vertex, the horizontal vector from Q will alsobe a “bad” vector. . . As a result, we will go to point R at the rear endhaving only one “bad” vector turning to. Then, we can conclude that Ris an isolated vertex, contradiction. So our assumption is false.

• In conclusion, the number of divisions satisfying the conditions is equalto the “reasonable” sets. We can easily see that the number of “reason-able” sets is equal to (n+1)!, so the result of this problem is (n+1)!.


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Does there exist a function f : (0;+∞)→ (0;+∞) satisfying

f (2016+ x f (y)) = y f (x+ y+2016)x,y > 0? (2.3.5.1)

Problem 5

Solution. We will prove that there does not exist any function satisfying thecondition above.

a) Replace x by 1f (y) at (2.3.5.1), we get f

(1

f (y) + y+2016)= f (2017)

y , ∀y>

0.. As f (2017)y receive all values in (0;+∞), f is surjective.

b) Suppose that f is not bijective, which means there exists a > b > 0 suchthat f (a) = f (b). Let T = a−b > 0.

• Replace x by a,b at (2.3.5.1) respectively, we get

a f (x+a+2016)= f (2016+ x f (a))= f (2016+ x f (b))= b f (x+b+2016),∀x> 0

⇒ f (x) =ab

f (x+T ),∀x > b+2016

⇒ f (x) =(a

b

)nf (x+nT ),∀n ∈ N∗,∀x > b+2016.

As f is surjective, there exists a real number y such that f (y)> 1.Take a big-enough value of n such that y f (y)+nT

f (y)−1 > b.

• Replace x by y+nTf (y)−1 at (2.3.2.5), we get

f(

2016+y+nTf (y)−1

f (y))= y f

(y+nTf (y)−1

+ y+2016)

⇒ f(

2016+y f (y)+nT

f (y)−1+nT

)= y f

(2016+

y f (y)+nTf (y)−1

)⇒(

ba

)n

f(

2016+y f (y)+nT

f (y)−1

)= y f

(2016+

y f (y)+nTf (y)−1

)⇒(

ba

)n

= y. (2.3.5.2)

As a is fixed and a,b > 0, (2.3.5.2) will be wrong when n is bigenough. So f is bijective.


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Replace y = 1 at (2.3.5.1) and use the bijection of f , we get 2016+x f (1) = x+ 2017⇒ x f (1) = x+ 1∀x > 0, which is not able to hap-pen. So there does not exist function f satisfying the condition of theproblem.

Let ABC be a acute-angled triangle with AB<AC and a circumcircle (O).D is the projection of A on BC. The line which passes through D and isperpendicular to AB intersects CA at E. The line which passes through Dand is perpendicular to AC intersects AB at F . EF cuts the tangent line of(O) at A at G. M,R is the midpoints of BC and AD, respectively.

1. Prove that GR⊥AM.

2. Let I,J be the intersections of DE,DF and GR; K,L be the mid-points of JE, IF , respectively. Prove that OS//KL.

Problem 6

Solution.

1. Let N be the midpoint of EF . DF,DE intersect CA,AB at P,Q respec-tively. We can easily get D is the orthocenter of 4DEF ⇒ RN is theperpendicular bisector of PQ⇒ RN ⊥ PQ. Besides, DBQ = QDA =

QPA⇒ BCPQ is an inscribed quadrilateral⇒ AQ⊥AO⊥PQ⊥RN⇒ Ris the orthocenter of4AGN⇒ GR⊥ AM.


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2. ∠MAC = ∠RJD,∠RDJ = ∠ACM ⇒ 4AMC ∼ 4JRD. By the sameargument, we get4AMB∼4IRD⇒ RJ.MC = RD.AM = RI.MB, butMB = MC ⇒ R is the midpoint of IJ ⇒ AJDI is a parallelogram ⇒4AIE ∼4AJF⇒ DI

DJ =AJAI =

FJEI =

2RL2RK = RL

RK . Note that JDI = LRK⇒4DIJ∼4RLK⇒ BAT = RKL= DJI = MAC⇒AT is the symmedianof4ABC⇒ ST touches (O)⇒ OS⊥AT⊥KL⇒ OS//KL.

Let p,q be positive integers. Write a number 1 on the board and repeatthe following algorithm: Replace the current number on the board withits sum with p or with q. Find the conditions of p,q to assure that thealgorithm can repeat infinitely without creating a number on the boardthat is neither a multiple of p nor a multiple of q.

Problem 7

Solution. We will point out that p,q must not be relatively prime so as tosatisfy the condition of the problem.

• Suppose that (p,q) = d ≤ 2. The number on the board belongs to type1+mp+nq (m,n ∈ N∗). That number is entirely not the multiple of por q.

• Suppose that (p,q)= 1, we will prove that when we infinitely repeat thatalgorithm, sooner or later a multiple of p or q will appear. As (p,q) = 1,there exists the inversion of p modulo q and the inversion of q modulo

p. So, there exists a,b ∈ N∗ such that ap+ 1...q, bq+ 1

...p⇒ 1+ ap+

mq...q∀m,1+bq+np

...p∀n. Therefore, our number will be the multipleof p or q when our choice for p reaches a or our choice for q reaches b.


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2.4. Southern Winter School

Let x,y be positive real numbers such that x+ y = 2.

1. Prove that

1xn +

1yn ≥ xn+1 + yn+1 ∀n ∈ N∗. (2.4.1.1)

2. For all n, can we replace n+1 with a positive integer k (k > n+1)so that the inequality

1xn +

1yn ≥ xk + yk (2.4.1.2)

holds true with all x,y > 0,x+ y = 2?

Problem 1

Solution.

1. Pay attention that1xn +

1yn ≥ xn+1 + yn+1

⇔ (x+ y)2n+1(xn + yn)≥ 22n+1xnyn (x2n+1 + y2n+1) . (2.4.1.3)

Let a =xy

, (2.4.1.3) is equivalent to

an((an+1 +1

)an +1

≤(

a+12

)2n+1

. (2.4.1.4)

Now we will prove (2.4.1.4) by induction.

• For n = 1, we need to prove

a(a2 +1

)a+1

≤(

a+12

)3

.

But this is true because according to AM-GM inequality:

x(a2 +1

)=

12(2a)

(a2 +1

)≤ 1

2

(2a+a2 +1

2

)=

(a+1)4

8


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⇒a(a2 +1

)a+1

≤(

a+12

)3

,

and the inequality holds for n = 1.

• Suppose that the inequality holds for n = k, or

ak((ak+1 +1

)ak +1

≤(

a+12

)2k+1

. (2.4.1.5)

From (2.4.1.5), we deduce that(a+1

2

)2(k+1)+1

=

(x+1

2

)2

.

(x+1

2

)2k+1

≥(

x+12

)2

.xk (xk+1 +1

)xk +1

. (2.4.1.6)

• For n = k+1, we need to prove

ak+1((ak+1+1 +1

)ak+1 +1

≤(

a+12

)2(k+1)+1

.

Using (2.4.1.6), we see that the inequality above can be proved ifthe problem below is true:(

a+12

)2

.ak (ak+1 +1

)ak +1

≥ak+1 (ak+2 +1

)ak+1 +1

⇔ (a+1)2

4a≥(ak+2 +1

)(ak +1

)(ak+1 +1

)2

⇔ (a+1)2

4a−1≥

(ak+2 +1

)(ak +1

)(ak+1 +1

)2 −1

⇔ (a−1)2

4a≥ ak(a−1)2(

ak+1 +1)2 ⇔ (a−1)2

[(ak+1 +1

)2−4ak+1

]≥ 0,

which is always true according to AM-GM inequality. So the in-equality holds for n = k+1.

Therefore, our inequality is true for all n.


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2. Suppose that there exists x,y > 0 such that (2.4.1.2) is true for k > n+1.Note that (2.4.1.2) can be written as below:

(x+ y)(n+k)(xn + yn)≥ 2(n+k)xnyn(xk + yk). (2.1.4.7)

Let a =xy

and rewrite (2.1.4.7), we need to prove

f (a) =(

a+12

)n+k

−an (ak +1

)an +1

≥ 0.

After calculating, we get

f ′′(1) =(n+1)(n+1− k)

4.

• If f ′′(1)≥ 0, then k ≤ n+1, contradiction.

• If f ′′(1)< 0, then 1 is the maximal point of f (a), so in the neigh-borhood of 1, we get f (a)< f (1) = 0, contradiction.

So it is impossible to replace n+ 1 by any real number k > n+ 1 suchthat (2.4.1.2) is true.

Let (an) be a sequence defined by:{a1 = 1,a2 = 2,an+2 =

3an+1−an2 + 1

n2(n = 1,2,3, ..)

Prove that (an) has a finite limit.

Problem 2

Solution. By induction, we can easily prove that (an) is an increasing se-quence.Let (bn) be a sequence defined by:

b1 =32

bn = an+1−an

2bn+1 = bn +

1n2

(n = 1,2,3, ..)


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From that we get

bn =32+

112 +

122 + . . .+

1(n−1)2 ∀n ∈ N

∗

⇒ bn <32+2 ∀n ∈ N∗⇒ bn <

72∀n ∈ N∗

Hence:

an+1 <72+

an

2< .. . <

72+

722 + . . .+

72n +

a1

2n ∀n ∈ N∗

⇒ an+1 <72.1− 1

2n

1− 12

+12n ∀n ∈ N

∗⇒ an < 8 ∀n ∈ N∗

Therefore (an) has a finite limit due to the Weierstrass’s theorem.

Let ABC be an acute-angled triangle with its incircle (I). (I) is tangent toBC,CA,AB at D,E,F respectively. J is the midpoint of EF . BJ and CJintersect CA and AB at H and K; EF intersects BC and HK at M and N,respectively. Prove that AM = AN.

Problem 3

Solution. AI⊥MN at J.Using Menelaus theorem for those triangles AEF with:

N ∈ EF,K ∈ AF,H ∈ AE :NENF

.KFKA

.HAHE

= 1

M ∈ EF,B ∈ AF,C ∈ AE :MFME

.CECA

.BABF

= 1


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J ∈ EF,B ∈ AF,H ∈ AE :JEJF

.BFBA

.HAHE

= 1

J ∈ EF,K ∈ AF,C ∈ AE :JENF

.KFKA

.CACE

= 1

From 4 expressions above and JE = JF we imply thatNENF

=MFME

. ThereforeJ is the midpoint of MN ⇒ AMN is an isosceles triangle at A, which meansAM = AN.

Consider a 2n x 2n square. Each square is colored with only one color,white or black, such that

i. The number of black squares on each line are pairwise different.

ii. The number of black squares on each column are pairwise differ-ent.

1. How many black squares are there in the square?

2. How many pairs of consecutive color-different squares? (Two squaresare called consecutive if and only if they have a same side.)

Problem 4

Solution.

1. The maximum and minimum number of black unit squares in each col-umn and row are 2n and 0, respectively.There are 2n+1 values in the range from 0 to 2n. Follow by condition(i), since we have 2n columns, there exist only a value k≤ 2n that is notthe number of black unit squares of any columns on the board.Thus the total number of black unit squares on the board is n(2n+1)−k.Also, since there are 2n columns, by combining condition (ii), it follows

that n(2n+1)− k...2n⇒ n− k

...2n⇒ n = k.Therefore, there are 2n2 black unit squares on the board.

2. Firstly, we consider the columns. Assume that a is the number of blackunit squares on a particular column. Since a white or a black unit squareon the same column can only be adjacent to at most 2 different coloredunit squares, there is no less than 2 ∗min(a,2n− a) pairs on that col-umn. Thus the maximum number of pairs such that two unit squares are


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on the same column is 4.0+ . . . +4.(n−1) = 2n(n−1).Secondly, we consider the rows. We can easily see that every row canbe alternate with exactly another column, which means a line with ablack square units can only be alternate with a column with 2n−a blacksquares. So there are at least n−1 pairs of columns that are not alternatewith each other.Finally, We deduce that the total number of pairs such that 2 two unitsquares are on the same row is 2n(2n−1)−(n−1). Therefore, the valuethat satisfies the required assertion is 6n2−5n+1.

An example configuration: We number the rows downward from 1 to 2n.The rows numbered even is colored such that it will be alternate withthe row numbered odd right above. Meanwhile, the rows numbered oddwill be colored by this law: The row numbered 2i+ 1 (0 ≤ i ≤ n− 1)will have exactly i black square units and the jth black square unit willbe the 2 jth square unit of the row, from left to right.

Find all functions f : R→ R with the property that

f ( f (x)+2y) = 10x+ f ( f (y)−3x) . (4.2.5.0)

holds for all x,y ∈ R.

Problem 5

Solution. Suppose that the function f : R→ R satisfies (4.2.5.0).

• In (4.2.5.0), replace x = 0, we get

f ( f (0)+2y) = f ( f (y)) ∀y ∈ R. (4.2.5.1)

• In (4.2.5.0), replace y =− f (x)

2, we get

f (0) = 10x+ f ( f (− f (x)

2)−3x) ∀x ∈ R.

Therefore f (x) is surjective.


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• We will prove that f (x) is injective. Indeed, if f (x) is not injective, thereexists two distinct real numbers a and b such that f (a) = f (b).After respectively replacing y = a and y = b in (4.2.5.0), we easily get

f ( f (x)+2a) = f ( f (x)+2b) ∀x ∈ R.

Because f (x) is surjective, we imply that f (x+a) = f (x+b) ∀x ∈ R.Replace x by x− a and let c = b− a, we get f (x) = f (x+ c) ∀x ∈ R.Therefore, by induction we easily get f (x) = f (x+nc) ∀x ∈ R,n ∈ Z.Replace x by x+ c in (4.2.5.0), we get f ( f (x+ c)+ 2y = 10x+ 10c+f ( f (y)− 3x− 3c) ∀x,y ∈ R⇒ f ( f (x)+ 2y) = 10x+ 10c+ f ( f (y)−3x) ∀x,y∈R⇒ c= 0, contradiction (a 6= b). Therefore f (x) is injective.So from (4.2.5.1) we get

f (y) = 2y+ f (0) ∀y ∈ R⇒ f (x) = 2x+α ∀x ∈ R

We see that all the functions f (x) = 2x+α ∀x ∈ R satisfy (4.2.5.0),and they are all solutions to the problem.

There are 23 positive integers (not necessarily distinct) written on a line.Prove that we can place the symbols “+”, “x” or the parentheses reason-ably so that the obtained expression is divisible by to 2000.

Problem 6

Solution. This lemma below is used in the solution.Lemma. Given n integers arranged on a line. There always exists a numberthat is divisible by n or some consecutive numbers in line that have their sumdivisible by n.Come back to our problem. Note that 2000 = 42.53.

• Now, let’s divide 23 integers into 5 consecutive groups containing con-secutive numbers on the given line. The first three groups, each contains5 numbers. The last two groups, each contains 4 numbers.

• In each of the first three groups, there always exist a number that isdivisible by 5 or some consecutive numbers in line that have their sumdivisible by 5. We put these numbers in parentheses and symbols "+"between them. Between the other numbers or groups of numbers, weput symbols "x". Between the three groups we put symbols "x". So thatthe obtained expression from the first 15 numbers is divisible by 53.


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• In each of the last two groups, there always exist a number that is di-visible by 4 or some consecutive numbers in line that have their sumdivisible by 4.We put these numbers in parentheses and symbols "+"between them. Between the other numbers or groups of numbers, weput symbols "x". Between the two groups we put symbols "x". So thatthe obtained expression from the last 8 numbers is divisible by 42.Then put a "x" between two obtained expressions.

We have already had the expression that satisfied the given conditions.

Let ABC be a triangle and P be a point inside. D is the projection of Pon BC. E,F are the points satisfying the following conditions: PE⊥AC,CE⊥BC,PF⊥AB, BF⊥BC, G is symmetric to D through BC.

1. Prove that AG⊥EF .

2. Let S,T be the intersection of EF and CA,AB, respectively. Q,Rare symmetric to P through the midpoints of GE,GF . (l1) be theline that passes through S and perpendicular to AQ, (l2) be the linethat passes through T and perpendicular to AR. Prove that (l1) and(l2) intersect at a point on AG.

Problem 7


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Solution.

1. Let K be the projection of A on EF and G′ is the common point of AKand BC. We will prove that G′ is symmetric to D through BC, hence theproblem is solved. Indeed, using the sine formula in triangle, we imply:

G′BG′C

=sin ACBsinABC

.sin G′AB

sin G′AC.

We also have G′AB = PFE, G′AC = PEF , ACB = CEP, ABC = BFPdue to the perpendicular lines above. Therefore,

G′BG′C

=sinCEP

sin BFP.sin PFE

sin PEF=

sinCEP

sin BFP.PEPF

=DCDB

.

Thus G′ is symmetric to D through BC⇒ G′ ≡ G.

2. PEQG and PFRG are both parallelograms so QE = PG = RF andQE//PG//RF . Therefore, EQRF is a parallelogram ⇒ QR//EF ⇒QR⊥AG. Let L be the intersection of QR and AG.Let X ,Y consecutively be the intersects of (l1) and AQ, (l2) and AR. LetM,N consecutively be the intersects of GQ and AC, GR and AB. We eas-ily see that SXQM,TY RN,SKGM,T KGN are inscribed quadrilaterals,so

AX .AQ = AS.AM = AK.AG = AT .AN = AY .AR

Let I be the intersection of (l1) and AG, I′ be the intersection of (l2) andAG. Easily, we get ILQX and I′LRY are inscribed quadrilaterals, then

AI.AL = AQ.AX = AR.AY = AI′.AL

Therefore I ≡ I′.

The problem is completely solved.


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