Mathematical Excalibur, Volume 15, Number 4

Volume 15, Number 4 January 2011

Klamkin’s Inequality

Kin Y. Li

Olympiad Corner

Below are the problems of the 2011 Chinese Math Olympiad, which was held on January 2011.

Problem 1. Let a1,a2,…,an (n≥3) be

real numbers. Prove that

, ) ( 2 2 1 1 1 2 _aa n _{M m} a n i i i n i i ⎥⎦ − ⎤ ⎢⎣ ⎡ ≤ −

∑

∑

= + =

where , max , min ,

1 1 1 1 _{i n} i _{i n} i n a M a m a a ≤ ≤ ≤ ≤ + = = =

[x] denotes the greatest integer not exceeding x.

Problem 2. In the figure, D is the midpoint of the arc BC on the circumcircle Γ of triangle ABC. Point X is on arc BD. E is the midpoint of arc AX. S is a point on arc AC. Lines SD and BC intersect at point R. Lines SE and AX intersect at point T. Prove that if RT || DE, then the incenter of triangle ABC is on line RT. B A C D E X S R T (continued on page 4)

Editors: 張 百 康 (CHEUNG Pak-Hong), Munsang College, HK

高 子 眉 (KO Tsz-Mei)

梁 達 榮 (LEUNG Tat-Wing)

李 健 賢 (LI Kin-Yin), Dept. of Math., HKUST

吳 鏡 波 (NG Keng-Po Roger), ITC, HKPU

Artist: 楊 秀 英 (YEUNG Sau-Ying Camille), MFA, CU

Acknowledgment: Thanks to Elina Chiu, Math. Dept.,

HKUST for general assistance.

On-line:

http://www.math.ust.hk/mathematical_excalibur/ The editors welcome contributions from all teachers and students. With your submission, please include your name, address, school, email, telephone and fax numbers (if available). Electronic submissions, especially in MS Word, are encouraged. The deadline for receiving material for the next issue is February 28, 2011.

For individual subscription for the next five issues for the 09-10 academic year, send us five stamped self-addressed envelopes. Send all correspondence to:

Dr. Kin-Yin LI, Math Dept., Hong Kong Univ. of Science and Technology, Clear Water Bay, Kowloon, Hong Kong

Fax: (852) 2358 1643 Email: [email protected]

© Department of Mathematics, The Hong Kong University of Science and Technology

In 1971 Professor Murray Klamkin established the following

Theorem. For any real numbers x,y,z,

integer n and angles α,β,γ of any triangle, we have ). cos cos cos ( 2 ) 1 ( 1 2 2 2 γ β α zx n xy n n yz z y x n _{+ +} − ≥ + + +

Equality holds if and only if . sin sin sin α β nγ z n y n x _{= =}

The proof follows immediately from expanding

(

x₊(₋1)n(ycosnγ₊zcosnβ)

)

2 ₊(_ysin_{nγ −z}sin_nβ)2_≥0. There are many nice inequalities that we can obtain from this inequality. The following are some examples (see references [1] and [2] for more).

Example 1. For angles α,β,γ of any

triangle, if n is an odd integer, then 2 / 3 cos cos cosnα+ nβ+ nγ ≤ . If n is an even integer, then

. 2 / 3 cos cos cosnα+ nβ+ nγ ≥− (This is just the case x=y=z=1.)

Example 2. For angles α,β,γ of any

triangle, . 4 cos 3 2 cos 2 cos 3 α+ β+ γ ≤

(This is just the case n = 1, x = sin 90°, y = sin 60°, z = sin 30°.)

There are many symmetric inequalities in α,β,γ, which can be proved by standard identities or methods. However, if we encounter asymmetric inequality like the one in example 2, it may be puzzling in coming up with a proof.

Example 3. Let a,b,c be sides of a

triangle with area Δ. If r,s,t are any real numbers, then prove that

. 4 2 ab rs ca tr bc st ct bs ar + + ≥ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ Δ + +

Solution. Let α,β,γ be the angles of the triangle. We first observe that

β α γ 2 2 2 2 2 2 2 2 2

2 _{sin sin sin}

4Δ =ab =bc =ca and cos 2θ = 1−2sin2 _{θ. So we can try to}

set n = 2, x=ar, y=bs, z=ct. Indeed, after applying Klamkin’s inequality, we get the result.

Example 4. Let a,b,c be sides of a

triangle with area Δ. Prove that . 4 2 2 2 2 2 2 2 2 2 2 a c c b b a c b a + + ≥ ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ Δ + +

Comment: It may seem that we can use example 3 by setting r=a, s=b, t=c, but unfortunately 2 2 2 2 2 2 3 a c c b b a ab rs ca tr bc st + + ≥ = + +

holds only when a=b=c by the AM-GM inequality.

Solution. To solve this one, we bring in the circumradius R of the triangle. We recall that 2Δ=bcsin α and by extended sine law, 2R=a/(sin α). So 4ΔR=abc. Now we set r=bcx, s=cay and t=abz. Then the inequality in example 3 becomes

. )

(_x+y+z 2_R2_≥yza2_+zxb2_+xyc2 _(*)

Next, we set yz=1/b2_{, zx=1/c}2_{, xy=1/a}2_,

from which we can solve for x,y,z to get . 4 , 4 , 4 2 R a z R c y R b ac b x Δ = Δ = Δ = = Then (*) becomes . 4 2 2 2 2 2 2 2 2 2 2 a c c b b a c b a + + ≥ ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ Δ + +

Mathematical Excalibur, Vol. 15, No. 4, Jan. 11 Page 2

Example 5. (1998 Korean Math

Olympiad) Postive real numbers a,b,c satisfy a+b+c=abc. Prove that

2 3 1 1 1 1 1 1 2 2 2 + ₊ + ₊ ≤ +a b c

and determine when equality holds. Solution. Let a = tan u, b = tan v and c = tan w, where u,v,w > 0. As a+b+c=abc,

tan u+tan v+tan w = tan u tan v tan w, which can be written as

). tan( tan tan 1 tan tan tan v w w v w v u = + − + = −

This implies u+v+w=nπ for some odd positive integer n. Let α = u/n, β = v/n and γ = w/n. Taking x = y = z = 1 in Klamkin’s inequality (as in example 1), we have 2 / 3 cos cos cosnα+ nβ+ nγ≤ , which is the desired inequality. Equality holds if and only if a = b = c = 3.

For the next two examples, we will introduce the following

Fact: Three positive real numbers x,y,z satisfy the equation

x2_+y2_+z2_{+xyz = 4 (**)}

if and only if there exists an acute triangle with angles α,β,γ such that

x = 2cos α, y = 2cos β, z = 2cos γ. Proof. If x,y,z > 0 and x2_+y2_+z2_{+xyz = 4,}

then x2_{, y}2_{, z}2 _{< 4. So 0 < x, y, z < 2.}

Hence, there are positive α,β,γ < π/2 such that

x = 2cos α, y = 2cos β and z = 2cos γ. Substituting these into (**) and simplifying, we get cos γ = −cos (α+β), which implies α+β+ γ = π. We can get the converse by using trigonometric identities.

Example 6. (1995 IMO Shortlisted

Problem) Let a,b,c be positive real numbers. Determine all positive real numbers x,y,z satisfying the system of equations

x+y+z = a+b+c, 4xyz−(a2_x+b2_y+c2_{z) = abc.} Solution. We can rewrite the second equation as . 4 2 2 2 = + ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ xyz abc xy c zx b yz a

By the fact, there exists an acute triangle with angles α,β,γ such that

. cos 2 , cos 2 , cos 2 α = β = γ = xy c zx b yz a

Then the first equation becomes

). cos cos cos ( 2 yz α zx β xy γ z y x+ + = + +

This is the equality case of Klamkin’s inequality. So . sin sin sinα β γ z y x = = As γ+β = π−α, so sin(γ+β)/sin α=1. Then _{cosβ cosγ} 2 2 x y x z x c x b _{+ = +} 1. sin cos sin cos sin + ₌ = α γ β β γ

So x = (b+c)/2. Similarly, y = (c+a)/2 and z = (a+b)/2.

Example 7. (2007 IMO Chinese Team

Training Test) Positive real numbers u,v,w satisfy the equation u+v+w+ uvw=4. Prove that . w v u w uv v uw u vw + + ≥ + +

Solution. By the fact, there exists an acute triangle with angles α,β,γ such that

. cos 2 , cos 2 , cos 2 α = β = γ = v w u

The desired inequality becomes γ β α β α γ α γ β cos cos cos 2 cos cos cos 2 cos cos cos 2 _{+ +} ). cos cos (cos 4 2_α+ 2_β+ 2_γ ≥

Comparing with Klamkin’s inequality, all we have to do is to take n = 1 and

β α γ α γ β cos cos cos 2 , cos cos cos 2 ₌ = y x . cos cos cos 2 γ β α = z

Example 8. (1988 IMO Shortlisted

Problem) Let n be an integer greater than 1. For i=1,2,…,n, αi > 0, βi > 0 and

∑

∑

= = = = n i i n i 1 i 1 . π β α Prove that

_∑

_∑

= = ≤ n i i n i i i 1 1 . cot sin cos _α α β

Solution. For n = 2, we have equality

1 1 1 1 2 2 1 1 sin cos sin cos sin cos sin cos α β α β α β α β _{+ = −} . cot cot 0= α₁+ α₂ =

For n = 3, α1, α2, α3 are angles of a

triangle, say with opposite sides a,b,c. Let Δ be the area of the triangle. Now 2Δ = bcsin α1 = casin α2 = absin α3.

Combining with the cosine law, we get Δ − + = = 4 sin cos cot 2 2 2 1 1 1 a c b α α α

and similarly for cot α2 and cot α3. By

Klamkin’s inequality, ) cos cos cos ( 2 sin cos 4 3 2 1 1 β β β α β ab ca bc n i i i_{= + +} Δ

∑

=

∑

= Δ = + + ≤ 3 1 2 2 2 _{4 cot .} i i c b a α

Cancelling 4Δ, we will finish the case n = 3. For the case n > 3, suppose the case n−1 is true. We have

⎥ ⎦ ⎤ ⎢ ⎣ ⎡ + + − + =

∑

= sin( ) ) cos( sin cos sin cos sin cos 2 1 2 1 2 2 1 1 1 α α β β α β α β α β n i i i _⎥ ⎦ ⎤ ⎢ ⎣ ⎡ + + + +

∑

= sin( ) ) cos( sin cos 2 1 2 1 3 α α β β α β n i i i ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ + − + − + + = )) ( sin( )) ( cos( sin cos sin cos 2 1 2 1 2 2 1 1 α α π β β π α β α β _⎥ ⎦ ⎤ ⎢ ⎣ ⎡ + + + +

∑

= sin( ) ) cos( sin cos 2 1 2 1 3 α α β β α β n i i i

≤

[

cotα₁+cotα₂+cot(π−(α₁+α₂))

]

⎥ ⎦ ⎤ ⎢ ⎣ ⎡ _{+ +} +

∑

= n i3 i 1 2 ) cot( cotα α α

_∑

= = n i i 1 . cotα

This finishes the induction.

References

[1] M.S.Klamkin, “Asymetric Triangle Inequalities,” Publ.Elektrotehn. Fak. Ser. Mat. Fiz. Univ. Beograd, No. 357-380 (1971) pp. 33-44.

[2] Zhu Hua-Wei, From Mathematical Competitions to Competition Mathe- matics, Science Press, 2009 (in Chinese).

Problem Corner

We welcome readers to submit their solutions to the problems posed below for publication consideration. The solutions should be preceded by the solver’s name, home (or email) address and school affiliation. Please send submissions to Dr. Kin Y. Li, Department of Mathematics, The Hong Kong University of Science & Technology, Clear Water Bay, Kowloon, Hong Kong. The deadline for sending solutions is February 28, 2011.

Problem 361. Among all real numbers a and b satisfying the property that the equation x4_+ax3_+bx2_{+ax+1=0 has a real}

root, determine the minimum possible value of a2_+b2 _{with proof.}

Problem 362. Determine all positive

rational numbers x,y,z such that z y x xyz z y x+ + , , 1+ 1+1 are integers.

Problem 363. Extend side CB of

triangle ABC beyond B to a point D such that DB=AB. Let M be the midpoint of side AC. Let the bisector of ∠ABC intersect line DM at P. Prove that ∠BAP =∠ACB.

Problem 364. Eleven robbers own a

treasure box. What is the least number of locks they can put on the box so that there is a way to distribute the keys of the locks to the eleven robbers with no five of them can open all the locks, but every six of them can open all the locks? The robbers agree to make enough duplicate keys of the locks for this plan to work.

Problem 365. For nonnegative real

numbers a,b,c satisfying ab+bc+ca = 1, prove that . 2 1 1 1 1 _≥ + + − + + + + +b b c c a a b c a *****************

Solutions

****************

Problem 356. A and B alternately color

points on an initially colorless plane as follow. A plays first. When A takes his turn, he will choose a point not yet colored and paint it red. When B takes his turn, he will choose 2010 points not

yet colored and paint them blue. When the plane contains three red points that are the vertices of an equilateral triangle, then A wins. Following the rules of the game, can B stop A from winning?

Solution. LI Pak Hin (PLK Vicwood K.

T. Chong Sixth Form College), Anna

PUN Ying (HKU Math) and The 7B Mathematics Group (Carmel Alison Lam

Foundation Secondary School).

The answer is negative. In the first 2n moves, A can color n red points on a line, while B can color 2010n blue points. For each pair of the n red points A colored, there are two points (on the perpendicular bisector of the pair) that can be chosen as vertices for making equilateral triangles with the pair. When n > 2011, we have

. 2010 ) 1 ( 2 2 n_⎟⎟=nn− > n ⎠ ⎞ ⎜⎜ ⎝ ⎛

Then B cannot stop A from winning. Other commended solvers: King’s College Problem Solving Team (Angus CHUNG, Raymond LO, Benjamin LUI), Andy LOO (St. Paul’s Co-ed

College),Emanuele NATALE (Università di Roma “Tor Vergata”, Roma, Italy) and

Lorenzo PASCALI (Università di Roma

“La Sapienza”, Roma, Italy), WONG Sze

Nga (Diocesan Girls’ School).

Problem 357. Prove that for every

positive integer n, there do not exist four integers a, b, c, d such that ad=bc and n2 _< a < b < c < d < (n+1)2_.

Solution. U. BATZORIG (National

University of Mongolia) and LI Pak Hin (PLK Vicwood K. T. Chong Sixth Form College).

We first prove a useful

Fact (Four Number Theorem): Let a,b,c,d be positive integers with ad=bc, then there exists positive integers p,q,r,s such that a=pq, b=qr, c=ps, d=rs.

To see this, let p=gcd(a,c), then p|a and p|c. So q=a/p and s=c/p are positive integers. Now p=gcd(a,c) implies gcd(q,s)=1. From ad=bc, we get qd=sb. Then s|d. So r=d/s is a positive integer and a=pq, b=qr, c=ps, d=rs.

For the problem, assume a,b,c,d exist as required. Applying the fact, since d > b > a, we get s>q and r>p. Then s≥q+1, r≥p+1 and we get 2 ) 1 ( ) 1 )( 1 ( + + ≥ + ≥ =rs p q pq d ₌( _a+1)2_>(_n+1)2, a contradiction.

Other commended solvers: King’s College Problem Solving Team (Angus CHUNG, Raymond LO, Benjamin LUI), Anna PUN Ying

(HKU Math), The 7B Mathematics

Group (Carmel Alison Lam Foundation

Secondary School) and WONG Sze

Nga (Diocesan Girls’ School).

Problem 358. ABCD is a cyclic

quadrilateral with AC intersects BD at P. Let E, F, G, H be the feet of perpendiculars from P to sides AB, BC, CD, DA respectively. Prove that lines EH, BD, FG are concurrent or are parallel.

Solution. U. BATZORIG (National

University of Mongolia), King’s

College Problem Solving Team (Angus CHUNG, Raymond LO, Benjamin LUI), Abby LEE Shing Chi (SKH Lam Woo Memorial

Secondary School), LI Pak Hin (PLK Vicwood K. T. Chong Sixth Form College), Anna PUN Ying (HKU Math), Anderson TORRES (São Paulo, Brazil) and WONG Sze Nga (Diocesan Girls’ School). C D A B P E G F H

Since ABCD is cyclic, _{∠BAC =∠CDB} and _{∠ABD =∠DCA, which imply ΔAPB} and ΔDPC are similar. As E and G are feet of perpendiculars from P to these triangles (and similarity implies the corresponding segments of triangles are proportional), we get AE/EB=DG/GC. Similarly, we get AH/HD=BF/FC. If EH || BD, then AE/EB = AH/HD, which is equivalent to _DG/GC=BF/FC, and hence _{FG || BD.}

Otherwise, lines EH and BD intersect at some point I. By Menelaus theorem and its converse, we have

, 1 − = ⋅ ⋅ HA DH ID BI EB AE which is equivalent to

Mathematical Excalibur, Vol. 15, No. 4, Jan. 11 Page 4 , 1 − = ⋅ ⋅ FB CF GC DG ID BI

and lines BD and FG also intersect at I. Other commended solvers: Lorenzo PASCALI (Università di Roma “La

Sapienza”, Roma, Italy).

Problem 359. (Due to Michel BATAILLE) Determine (with proof) all real numbers x,y,z such that x+y+z ≥ 3 and ). ( 2 2 2 2 4 4 4 3 3 3 _{y z x y z x y z} x + + + + + ≤ + +

Solution. LI Pak Hin (PLK Vicwood

K. T. Chong Sixth Form College),

Paolo PERFETTI (Math Dept,

Università degli studi di Tor Vergata Roma, via della ricerca scientifica, Roma, Italy) and Terence ZHU (Affilated High School of South China Normal University).

Let x,y,z be real numbers satisfying the conditions. For all real w, w2_{+3w+3 ≥}

(w+3/2)2 _{implies (w}2_{+3w+3)(w−1)}2 _{≥ 0.}

Expanding, we get (*) w4_+w3_−2w2 _≥

3w−3. Applying (*) to w=x,y,z and adding, then using the conditions on x,y,z, we get ) ( 2 0_≥x3_+y3_+z3_+x4_+y4_+z4_{− x}2_+y2_+z2 . 0 9 ) ( 3 + + − ≥ ≥ x y z

Thus, for such x,y,z, we must have equalities in the (*) inequality for x,y,z. So x = y = z = 1 is the only solution. Comments: For the idea behind this solution, we refer the readers to the article on the tangent line method (see Math Excalibur, vol. 10, no. 5, page 1). For those who do not know this method, we provide the

Proposer’s solution. Suppose (x,y,z) is

a solution. Let s=x+y+z and S=x2_y+y2_z

+z2_x+xy2_+yz2_+zx2_{. By expansion, we}

have s(x2_+y2_+z2_{)−S= x}3_+y3_+z3_{. Hence,} s(x2_+y2_+z2_)−S+x4_+y4_+z4 _{≤ 2(x}2_+y2_+z2_),

which is equivalent to

(s−2)(x2_+y2_+z2_)+x4_+y4_+z4 _{≤ S. (*)}

Since S is the dot product of the vectors v =(x2_,y2_,z2_{,x,y,z) and w =(y,z,x,y}2_,z2_,x2_),

by the Cauchy Schwarz inequality, S ≤ x2_+y2_+z2_+x4_+y4_+z4 _{. (**)}

Combining (*) and (**), we conclude (s−3)(x2_+y2_+z2_{) ≤ 0. Since s ≥ 3, we get} s=3 and (*) and (**) are equalities. Hence, vectors v and w are scalar multiple of each other. Since x,y,z are

not all zeros, simple algebra yields x=y=z=1. This is the only solution. Comments: Some solvers overlooked the possibility that x or y or z may be negative in applying the Cauchy Schwarz inequality!

Other commended solvers: U. BATZORIG (National University of

Mongolia) and Shaarvdorj (11th _High

School of UB, Mongolia), King’s College

Problem Solving Team (Angus CHUNG, Raymond LO, Benjamin LUI), Thien NGUYEN (Luong The Vinh High School,

Dong Nai, Vietnam), Anna PUN Ying (HKU Math), The 7B Mathematics Group (Carmel Alison Lam Foundation Secondary School) and WONG Sze Nga (Diocesan Girls’ School).

Problem 360. (Due to Terence ZHU, Affiliated High School of Southern China Normal University) Let n be a positive integer. We call a set S of at least n distinct positive integers a n-divisible set if among every n elements of S, there always exist two of them, one is divisible by the other. Determine the least integer m (in terms of n) such that every n-divisible set S with m elements contains n integers, one of them is divisible by all the remaining n−1 integers.

Solution. Anna PUN Ying (HKU Math)

and the proposer independently.

The smallest m is (n−1)2_{+1. First choose}

distinct prime numbers p1, p2, …, pn−1. For

i from 1 to n−1, let

{

_, 2_{, ,} −1

}

= n i i i i p p p A _K

and let A be any nonempty subset of their union. Then A is n-divisible because among every n of the elements, by the pigeonhole principle, two of them will be in the same Ai, then one is divisible by the

other. However, among n elements, two of them will also be in different Ai’s and

neither one is divisible by the other. So m ≤ (n−1)2 _{will not work.}

If m ≥ (n−1)2_{+1 and S is a n-divisible set}

with m elements, then let k1 be the largest

element in S and let B1 be the subset of S

consisted of all the divisors of k1 in S. Let k2 be the largest element in S and not in B1.

Let B2 be the subset of S consisted of all

the divisors of k2 in S and not in B1. Repeat

this to get a partition of S.

Assume there are at least n of these Bi set.

For i from 1 to n, let ji be the largest

element in Bi. However, by the

definition of the Bi sets, {j1,j2,…,jn}

contradicts the n-divisiblity of S. So there are at most n−1 Bi’s.

Since m ≥ (n−1)2_{+1, one of the B}

i must

have at least n elements. Then for S, we can choose n elements from this Bi with

ki included so that ki is divisible by all

the remaining n−1 integers. Therefore, the least m is (n−1)2_+1.

Other commended solvers: WONG Sze Nga (Diocesan Girls’ School).

Olympiad Corner

(continued from page 1)

Problem 3. Let A be a finite set of real

numbers. A1,A2,…,An are nonempty

subsets of A satisfying the following conditions:

(1) the sum of all elements in A is 0; (2) for every xi∈Ai (i=1,2,…,n), we

have x1+x2+⋯+xn > 0.

Prove that there exist 1≤i1<i2<⋯<ik ≤ n

such that . 2 1 _n A k A A A_i ∪ _i ∪_L∪ _ik <

Here |X| denotes the number of elements in the finite set X.

Problem 4. Let n be a positive integer,

set S = {1,2,…,n}. For nonempty finite sets A and B of real numbers, find the minimum of |A Δ S|+|B Δ S|+|C Δ S|, where C =A+B ={a+b | a_{∈A, b∈B}, X} ΔY = {x | x belongs to exactly one of X or Y }, |X| denotes the number of elements in the finite set X.

Problem 5. Let n ≥ 4 be a given integer.

For nonnegative real numbers a1,a2,…,an, b1,b2,…,bn satisfying a1+a2+⋯+an = b1+b2 +⋯+bn > 0, find the maximum of . ) ( ) ( 1 1

∑

∑

= = + + n i i i i n i i i i b a b b a a

Problem 6. Prove that for every given

positive integers m,n, there exist infinitely many pairs of coprime positive integers a,b such that

Problem

1.

If

n=2k (k

a positive integer), then

t

a; -

t

a

j

al+

1

=

l

t

(a

j -

a

j

+1

l

~

n (M -

m)2.

If

_ 1=1 1=1

2

1=1

2

n=2k+

1

(k

a positive integer), then since in every cyclic listing of

2k+

1 numbers, there always exist three

2k+l

2 k + l _

consecutive terms that are monotone (because

IT

_{(al -a/-1)(al+}

₁ ~a;)

=

II

_{(al -al_1)2}

~

0

implies for

1=1 1=1

some i,

ai- ai-1

and

ai+1-ai

are of the same sign), we may assume

a2k, a2k+1, a1

are monotone. Then

(a2k - a2k+1)2

+

_{(a2k+1 - a1)2}

~

_{(a2k - ( 1)2 .}

Hence, we can reduce the

2k+

1 case to to the

2k

case to get

n n

1

n

1

2k-l

[n ]

La; - Lalal+

1 =-

2:(al -aI+1)2

~-(L(al-al+l)2

+

(a 2k -a1)2)

~

-

(M

_m)2.

1=1 1=1 2 1=1 2 1=1 - 2

Problem

2. Connect AD. Let AD and RT intersect at 1. Since D is the midpoint of arc BC, so AI is the angle bisector of

LBAC.

Connect AS, S1. Then since RTIIDE, so LSTI=LSED=LSAL Hence, A,T,I,S are points on some circle COl. Connect CE. Let CE and RT intersect at J. Connect SC. Then

LSRJ=LSDE=LSCE.

Hence, S,J,R,C are points on some circle CO2. Let COl intersect 0)2 at Sand

K.

We claim AJ and CI intersect at K. To see this, let COl intersect AJ at

A

and K l. Since E is the midpoint of arc AX, LSK1A=LSTA=(arc SA+arc XE)/2=(arc SA+arc

AE)/2=LSDE=LSRT=LSRJ.

So S, K-1,J,R are concyclic, hence K1 is on CO2. Similarly, let CO2 intersect CI at C and K2. Then "K~ is on 0)1. Hence, Kl and K2 coincide with

K.

Since LCAD=LCAI, so

LTJE=LCJR=LCED=LCAD.

Hence, A,I,J,C are concyclic. Then LACI=LAJI. Also, since C,K,J,R are concyclic, we get LBCI=LICRLAJ1. Then LACI=LBC1. Therefore, I is the incenter of 6.ABC.

Problem

3. Let A={a1," .,am} witha1> .. ·>am. From (1), we get a1+"·+am=O. Consider the least number in every Ai. Among A1, ... ,Am, let ki of them have aj as least number for i=l, ... ,m. Then kl+ .. +km=n and froni (2), klal+ .. ·+kmam>O. For s=1,2, ... ,m-l, there are k1+ .. ·+ks sets with least number at least as. Hence, the_union of these sets is contained in {al, ... ,as} and so has at most s elements.

Next, we prove there exists s in {1,2, ... ,m-1} such that k=k1+ ... +ks>snlm. Assume not, then k l+ ... +ks~snlmfor s=1,2, ... ,m -1. Applying Abel's summation by part, we get the contradiction that

m - ~ ~ - m

0<

_{Lkja j}

=

L(a

_{s -}

_as+1)(k1

+ ... +

_ks)

+

_am(k1

+ ... +

_{km)::;; L (as - as+1) sn +(;[mn}

=

~

_{La j}

=

O.

j=1 -

j=1

s=1

m -

m j=1

For this s, take those AI, ... ,An with least number at least as. Let them be

A," , ... , AI .

From above, k=kl +

I k

... +

ks>sn/m

and I All u .. · U Alk I~

s

<

km / n

=

k

I A I

In.

Problem

4. The minimum value is n+ 1. First, taking A=B=S, we get IA 6. SI+IB 6. SI+IC 6. SI=n+ 1. Next we show generally k=IA6. SI+IB 6. SI+IC6. SI2':n+ 1. Let X\Y={xl xEX; xE;tY}. Then k=IA\SI+IB\SI+IC\SI

+ IS\AI+IS\BI+IS\q.

It

suffices to show (i) IA\SI+IB\SI+IS\q 2': 1 and (ii) IC\SI+IS\AI+IS\BI2': n.

Now (i) is clear if IA\SI>O or IB\SI>O. Otherwise, IA\SI=IB\SI~O, then A,BS;S.Hence 1 is not in C. So IS\q2':1. Next, (ii) holds if AnS=0. Otherwise, let n-j be the largest number in AnS, where O::sj ::s

n

-1.

Then IS\AI;:::k. On the other hand, for i=k+l,k+2, ... ,n, either iEtB (that is iES\B) or iEB (then n -k+iEC\S). So IC\SI+IS\BI;:::n-k. Then (li) follows.

Problem

5. The maximum value is n~l. _{Without loss of generality, we may assume al+ ... +an} == 1

=

bj+

n n

.. ·+bn. It suffices to show

_{Ial(al +b}

_l) S;

(n-1)Ibl(al +b,).

After rearrangement, we can write it as

1=1 '=1

n n . n

(n

-1)

I

b,2

+

(n -

2)

I

alb,

~

I

a,2.

By symmetry, we may assume bi is the least among b l,'" ,bn' Then

1=1 1=1 1=1 n n

=

(n -1)b

I

2

_{+ (1-}

_b

1

)2

+

(n - 2)b

1

=

nb

1

2 +(n - 4)b

1 + 1 ~

1=

I al

~Ia;. 1=1 1=1

Problem

6. Ifmn=1, then the result follows. In case mn ;:::2, since na(ama+bnb)=(a+b)na+b+ a[(mn)a _na+b], it suffices to show there are infinitely many pairs of coprime positive integers a,b such that (a+b,n)=1 and a+b I (mnt - na+b. Let p=a+b. Then, it suffices to show there are infinitely many primes p and integers 1:S a :S p -1 such that p I (mn)a '- nP•

By Fermat's little theorem, if we have al=a2 (mod p-1), a I;::: 1 an'd a2;:::1, then

. (mnr'

==

(mnr2

(mod

p).

Hence, it suffices to show there are infinitely many primes p and positive

integers a such that p

I

(mn)a -

n.

Assume there are only finitely many such prime p. Let them be PI, P2, "', pro (Since mn;:::2, such prime exists.) Suppose

(mn)2 - n

=

P:'

p~2

...

p~r (call this (*)), where ai are nonnegative for 1 :S i :S r.

Next define

a

= p~' p~2

...

p~r (PI -1) .. '(Pr -1) + 2. Suppose

(mnt - n

=

pf'pf2 ... pfr

(call this (**)), where ~i are nonnegative for 1 :S i :S r.

If Pi I n, then by (* *) and a> 2, we see

pri In.

Hence,

pri

I

(mn)2 - n .

From (*), we see ~i :S ai.

If

Pi

t

n, then Pi

t

m.

Hence, (p~i+l,

mn)

=

1. Observe that rp(p~i+l)

=

p~1 (PI -1) is a divisor of a-2. By Euler's theorem, we see

(mnt

-11 ==

(mn)2

-n(modp~/+l). Since p~/+l

t

(mn)2 - n, we get,

p~/+l

t

(mnt- n. Hence ~i:S ai.