Mathematical Excalibur, Volume 12, Number 3

Volume 12, Number 3 September 2007 – October 2007

Convex Hull

Kin Yin Li

Olympiad Corner

Below were the problems of the 2007 International Math Olympiad, which was held in Hanoi, Vietnam.

Day 1 (July 25, 2007)

Problem 1. Real numbers a1, a2, …, an are given. For each i (1 ≤ i ≤ n) define } 1 : min{ } 1 : max{a j i a j n d_i= _j ≤ ≤ − _j ≤ ≤

and let d = max{di : 1≤ i ≤ n}.

(a) Prove that, for any real numbers x1≤

x2≤…≤ xn, . 2 } 1 |: max{|xi−ai ≤i≤n ≥d (*) (b) Show that there are real numbers x1≤

x2≤…≤ xn such that equality holds in (*). Problem 2.

Consider five points A, B,

C, D and E such that ABCD is a

parallelogram and BCED is a cyclic quadrilateral. Let ℓ be a line passing through A. Suppose that ℓ intersects the interior of the segment DC at F and intersects line BC at G. Suppose also that EF=EG=EC. Prove that ℓ is the bisector of angle DAB.

(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 November 25, 2007.

For individual subscription for the next five issues for the 05-06 academic year, send us five stamped self-addressed envelopes. Send all correspondence to:

Dr. Kin-Yin LI Department of Mathematics

The Hong Kong University of Science and Technology Clear Water Bay, Kowloon, Hong Kong

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

A set S in a plane or in space is

convex if and only if whenever points X

and Y are in S, the line segment XY must be contained in S. The intersection of any collection of convex sets is convex. For an arbitrary set W, the convex hull of

W is the intersection of all convex sets

containing W. This is the smallest convex set containing W. For a finite set

W, the boundary of the convex hull of W

is a polygon, whose vertices are all in W. In a previous article (see pp. 1-2, vol. 5,

no. 1 of Math. Excalibur), we solved

problem 1 of the 2000 IMO using convex hull. Below we will discuss more geometric combinatorial problems that can be solved by studying convex hulls of sets.

Example 1. There are n > 3 coplanar points, no three of which are collinear and every four of them are the vertices of a convex quadrilateral. Prove that the

n points are the vertices of a convex n-sided polygon.

Solution. Assume one of these points, say P, is inside the convex hull of the n points. Let Q be a vertex of the convex hull. The diagonals from Q divide the convex hull into triangles. Since no three points are collinear, P is inside some △QRS, where RS is a side of the boundary. Then P,Q,R,S cannot be the vertices of a convex quadrilateral, a contradiction. So all n points can only be the vertices of the boundary polygon. Example 2. (1979 Putnam Exam) Let A be a set of 2n points in the plane, no three of which are collinear, n of them are colored red and the other blue. Prove that there are n line segments, no two with a point in common, such that the endpoints of each segment are points of A having different colors.

Solution. The case n = 1 is true. Suppose all cases less than n are true. For a vertex O on the boundary polygon of the convex hull of these 2n points, it

is one of the 2n points, say its color is red. Let P1, P2n−1 be adjacent vertices to

O. If one of them, say P1, is blue, then draw line segment OP1 and apply induction to the other 2(n−1) points to finish. Otherwise, d=1 n=4 d=0 d=1 d=-1 d=-2 d=0 O d=-1 P₁ P₄ P₇ P_{2 P5} P₆ P₃

let d = 1 and rotate the line OP1 toward line OP2n−1 about O hitting the other 2n−3 points one at a time. When the line hits a red point, increase d by 1 and when it hits a blue point, decrease d by 1. When the line hits P2n−1, d = (n−1) −

n = −1. So at some time, d = 0, say

when the line hits Pj. Then P1,…,Pj are on one side of line OPj and O, Pj+1,…,

P2n−1 are on the other side. The inductive step can be applied to these two sets of points, which leads to the case n being true.

Example 3. (1985 IMO Longlisted

Problem) Let A and B be finite disjoint

sets of points in the plane such that any three distinct points in A∪B are not collinear. Assume that at least one of the sets A, B contains at least five points. Show that there exists a triangle all of whose vertices are contained in A or in B that does not contained in its interior any point from the other set.

Solution. Suppose A has at least five points. Take a side A1A2 of the boundary of the convex hull of A. For any other Ai in A, let αi=∠A1A2Ai, say α3 < α4< ⋯ < 180°. Then the convex hull H of A1, A2,

A3, A4, A5 contains no other points of A. A₅ A₃ A₁ A₂ A₄ A6 A₇ (continued on page 4)

Mathematical Excalibur, Vol. 12, No. 3, Sept. 07 – Oct. 07 Page 2

Perpendicular Lines

Kin Yin Li

In geometry, sometimes we are asked to prove two lines are perpendicular. If the given facts are about right angles and lengths of segments, the following theorem is often useful.

Theorem. On a plane, for distinct points R, S, X, Y, we have RX2_−SX2 ₌

RY2 _{− SY}2 _{if and only if RS ⊥XY.} Proof. Let P and Q be the feet of the perpendicular from X and Y to line RS respectively. If RS ⊥XY, then P = Q and RX2 _{− SX}2 _{= RP}2 _{− SP}2 _{= RY}2 _{− SY}2_. Conversely, if RX2 _{− SX}2 _{= RY}2 _{− SY}2 ₌

m, then m = RP2 _{− (SR±RP)}2_{. So RP =} ∓(SR2_{+m)/2SR. Replacing P by Q, we} get RQ =_∓(SR2_{+m)/2SR. Hence, RP =}

RQ. Interchanging R and S, we also get SP=SQ. So P=Q. Therefore, RS ⊥XY.

Here are a few illustrative examples. Example 1. (1997 USA Math

Olympiad) Let ABC be a triangle, and

draw isosceles triangles BCD, CAE,

ABF externally to ABC, with BC, CA, AB as their respective bases. Prove the

lines through A, B, C, perpendicular to the lines EF, FD, DE, respectively, are concurrent. A B C F E D P

Solution. Let P be the intersection of the perpendicular line from B to FD with the perpendicular line from C to

DE. Then PB⊥FD and PC⊥DE. By

the theorem above, we have PF2_−PD2₌

BF2_−BD2 _{and PD}2_−PE2_{= CD}2_−CE2_. Adding these and using AF = BF, BD

= CD and CE = AE, we get PF2_−PE2 ₌

AF2_−AE2_{. So PA ⊥ EF and P is the} desired concurrent point.

Example 2. (1995 Russian Math

Olympiad) ABCD is a quadrilateral

such that AB = AD and ∠ABC and

CDA

∠ are right angles. Points F and

E are chosen on BC and CD

respectively so that DF⊥AE. Prove that AF⊥ BE.

A B C D E F

Solution. We have AE⊥DF, AB⊥BF and AD⊥DE, which areequivalent to

AD2_−AF2 _{= ED}2 _−EF2 _{, (a)} AB2_−AF2 _{= −BF}2 _{, (b)}

AD2_−AE2 _{= −DE}2_{. (c)} Doing (a) − (b) + (c) and using AD = AB, we get AB2 _{− AE}2 _{= BF}2 _{− EF}2 _{, which} implies AF⊥BE.

Example 3. In acute △ABC, AB = AC and P is a point on ray BC. Points X and Y are on rays BA and AC such that PX||AC and PY||AB. Point T is the midpoint of minor arc BC on the circumcircle of △

ABC. Prove that PT⊥XY.

T B C P X Y A

Since AT is a diameter, ∠ABT = 90˚ = ∠ACT. Then TX2 _{= XB}2 _{+ BT}2 _{and TY}2 ₌

TC2 _{+ CY}2_{. So TX}2_−TY2 _{= BX}2_−CY2_. Since PX || AC, we have ∠ABC = ∠ACB

= ∠XPB, hence BX = PX. Similarly, CY = PY. Therefore, TX2 _{− TY}2 _{= PX}2 _{− PY}2_, which is equivalent to PT⊥XY.

Example 4. (1994 Jiangsu Province Math Competition) For △ABC, take a point M by extending side AB beyond B and a point N by extending side CB beyond B such that AM = CN = s, where s is the semiperimeter of △ ABC. Let the inscribed circle of △ABC have center I and the circumcircle of △ ABC have diameter BK. Prove that KI⊥MN.

K A C I P Q M N

Solution. Let the incircle of △ABC touch side AB at P and side BC at Q. We will show KM 2 _−KN 2_=IM 2_−IN 2_. Now since ∠MAK = ∠BAK = 90˚ and ∠NCK = ∠BCK = 90˚, we get KM 2 _−KN 2_=(KA2_+AM2_{) −(KC}2_+CN2₎ = KA2 −KC2

=(KA2_−KB2_)+(KB2_−KC2₎ =BC 2 _−AB2_.

Also, since ∠MPI = ∠BPI = 90˚ and ∠NQI = ∠BQI = 90˚, we get

IM 2 _−IN 2_=(IP2_+PM2_{) −(IQ}2_+QN2₎ = PM 2 _−QN2 =(AM−AP)2_+(CN−QC)2_. Now BC BC CA AB s AP AM− = − + − = 2 and . 2 AB AB BC CA s QC CN− = − + − = So IM 2 _−IN 2_=BC2_−AB2 _{= KM} 2 _−KN 2_. Example 5. (2001 Chinese National

Senior High Math Competition) As in

the figure, in △ ABC, O is the circumcenter. The three altitudes AD,

BE and CF intersect at H. Lines ED

and AB intersect at M. Lines FD and

AC intersect at N. Prove that (1) OB⊥ DF and OC⊥DE; (2) OH⊥MN.

A B C D E H F M _N O

Solution. (1) Since ∠AFC = 90°=

∠ADC, so A,C,D,F are concyclic.

Then ∠BDF =∠BAC. Also,

∠OBC = ½(180˚−∠BOC)

= 90°−∠BAC = 90°−∠BDF. So OB⊥DF. Similarly, OC⊥DE. (2) Now CH⊥MA, BH⊥NA, DA⊥

BC, OB⊥ DF=DN and OC⊥DE = DM. So

MC2_−MH2 _{= AC}2_−AH2 _(a)

NB2_−NH2 _=AB2_−AH2 _(b)

DB2_−DC2 _{= AB}2_−AC2 _(c)

BN2_−BD2 _=ON2_−OD2 _(d)

CM2_−CD2_=OM2_−OD2_{. (e)} Doing (a) − (b) + (c) + (d) − (e), we get

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

submitting solutions is November 25, 2007.

Problem 281. Let N be the set of all positive integers. Prove that there exists a function f : N → N such that

f ( f (n)) = n2 _{for all n in N. (Source:}

1978 Romanian Math Olympiad)

Problem 282. Let a, b, c, A, B, C be real numbers, a ≠ 0 and A ≠ 0. For every real number x,

|ax2_{+bx+c| ≤ |Ax}2_+Bx+C|. Prove that |b2_{−4ac| ≤ |B}2_−4AC|.

Problem 283. P is a point inside ∆ABC. Lines AC and BP intersect at

Q. Lines AB and CP intersect at R. It is

known that AR=RB=CP and CQ=PQ. Find ∠BRC with proof. (Source: 2003

Japanese Math Olympiad)

Problem 284. Let p be a prime number. Integers x, y, z satisfy 0 < x < y < z < p. If x3_{, y}3_{, z}3 _{have the same remainder} upon dividing by p, then prove that x2₊

y2 _{+ z}2 _{is divisible by x + y + z. (Source:}

2003 Polish Math Olympiad)

Problem 285. Determine the largest positive integer N such that for every way of putting all numbers 1 to 400 into a 20×20 table (1 number per cell), one can always find a row or a column having two numbers with difference not less than N. (Source: 2003 Russian

Math Olympiad)

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

Solutions

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

Problem 276. Let n be a positive integer. Given a (2n−1)×(2n−1) square board with exactly one of the following arrows ↑, ↓, →, ← at each of its cells. A beetle sits in one of the cells. Per year the beetle creeps from one cell to another in accordance with the arrow’s

direction. When the beetle leaves the cell, the arrow at that cell makes a counterclockwise 90-degree turn. Prove that the beetle leaves the board in at most 23n−1_{(n−1)! −3 years.}

(Source: 2001 Belarussian Math

Olympiad)

Solution. Jeff CHEN (Virginia, USA), GRA20 Problem Solving Group (Roma, Italy), PUN Ying Anna (HKU, Math Year 1) and Fai YUNG.

Let a(n) be the maximum number of years that the beetle takes to leave the (2n − 1) × (2n − 1) board. Then a(1) = 1. For n > 1, apart from 1 year necessary for the final step, the beetle can stay

(1) in each of the 4 corners for at most 2 years (two directions that do not point outside)

(2) in each of the other 4(2n−3) cells of the outer frame for at most 3 years (three directions that do not point outside) (3) in the inner (2n−3)×(2n−3) board for at most a(n−1) years (when the starting point is inside the inner board) plus 4(2n−3)a(n−1) years (when the arrow in a cell of the outer frame points inward the beetle enters the inner board).

Therefore, a(n) ≤ 1 + 4·2 + 3·4(2n − 3) + (4(2n − 3)+1)a(n−1). Since a(n−1) ≥ 0,

a(n)+3 ≤ 8(n−1)(a(n−1)+3) −3a(n−1)

≤ 8(n−1)(a(n−1)+3).

Since a(1) + 3 = 4, we get a(n) + 3 ≤ 23n−1_{(n−1)! and so a(n) ≤ 2}3n−1_{(n−1)!−3.} Problem 277. (Due to Koopa Koo, Univ.

of Washington, Seattle, WA, USA) Prove

that the equation

x2 _{+ y}2 _{+ z}2

+ 2xyz = 1

has infinitely many integer solutions (then try to get all solutions – Editors).

Solution. Jeff CHEN (Virginia, USA), FAN Wai Tong and GRA20 Problem Solving Group (Roma, Italy).

It is readily checked that if n is an integer, then (x, y, z) = (n, −n, 1) is a solution.

Comments: Trying to get all solutions, we

can first rewrite the equation as (x2 _{− 1)(y}2 _{− 1) = (xy + z)}2_. For any solution (x, y, z), we must have

x2_{− 1 = du}2_{, y}2 _{− 1 = dv}2_{, xy + z = ±duv for} some integers d, u, v. The cases d is negative, 0 or 1 lead to trivial solutions. For d > 1, we may suppose it is square-free (that is, no square divisor greater than 1). Then we can find all

solutions of Pell’s equation s2 _−dt2 _{= 1} (see vol. 6, no. 3 of Math Excalibur,

page 1). Any two solutions (s0, t0) and (s1, t1) of Pell’s equation yield a solution (x, y, z)=(s0, s1, ±dt0t1−s0s1) of

x2 _{+ y}2 _{+ z}2

+ 2xyz = 1.

Commended solvers: PUN Ying Anna

(HKU, Math Year 1) and WONG Kam Wing (TWGH Chong Ming Thien College).

Problem 278. Line segment SA is perpendicular to the plane of the square

ABCD. Let E be the foot of the

perpendicular from A to line segment

SB. Let P, Q, R be the midpoints of SD, BD, CD respectively. Let M, N be on

line segments PQ, PR respectively. Prove that AE is perpendicular to MN. Solution 1. Stephen KIM (Toronto, Canada).

Below when we write XY⊥IJK…, we mean line XY is perpendicular to the plane containing I, J, K,…. Also, we write XY ⊥WZ for vectors XY and WZ to mean their dot product is 0.

Since SA⊥ABCD, so SA ⊥BC. Since

AB⊥BC, so BC⊥SAB. Since A,E are

in the plane of SAB, AE⊥BC. This along with the given fact AE ⊥ SB imply AE⊥SBC.

Since P, Q are midpoints of SD, BD respectively, we get PQ||SB. Similarly, we have QR||BC. Then the planes SBC and PQR are parallel. Since MN is on the plane PQR, so MN is parallel to the plane SBC. Since AE⊥SBC from the last paragraph, so AE⊥MN.

Solution 2. Kelvin LEE (Winchester College, England) and PUN Ying Anna (HKU, Math Year 1).

Let A be the origin, AD be the x-axis, AB be the y-axis and AS be the z-axis. Let

B = (0, a, 0) and S = (0, 0, s). Then C =

(a, a, 0) and E = (0, rs, ra) for some r. The homothety with center D and ratio 2 sends P to S, Q to B and R to C. Let it send M to M’ and N to N’. Then M’ is on

SB, N’ is on SC and M’N’||MN. So M’ =

(0, 0, s) + (0, a, −s)u = (0, au, s(1−u)) for some u and N’ = (0, 0, s) + (a, a, −s)v = (av, av, s(1−v)) for some v. Now the dot product of AE and M’N’ is

(0, rs, ra)·(av, a(v−u), s(u−v)) = 0. So AE⊥M’N’ . Therefore, AE⊥MN.

Commended solvers: WONG Kam

Wing (TWGH Chong Ming Thien College).

Mathematical Excalibur, Vol. 12, No. 3, Sept. 07 – Oct. 07 Page 4 Problem 279. Let R be the set of all

real numbers. Determine (with proof) all functions f: R→R such that for all real x and y,

(

f(x) y

)

2x f

(

f(f(y)) x

)

.

f + = + −

Solution. Jeff CHEN (Virginia, USA), Salem MALIKIĆ (Sarajevo College, 3rd _{Grade, Sarajevo, Bosnia and} Herzegovina) and PUN Ying Anna (HKU, Math Year 1).

Setting y = 0, we get

f ( f (x)) = 2x + f ( f (f (0)) −x). (1) Then putting x = 0 into (1), we get

f ( f (0)) = f ( f ( f (0))). (2) In (1), setting, x = f ( f (0)), we get f ( f ( f ( f (0)))) = 2f ( f (0)) + f (0). Using (2), we get f ( f (0)) = 2f ( f (0)) + f (0). So f ( f (0)) = −f (0). Using (2), we see f (k)_{(0) = −f (0) for k = 2,3,4,….} In the original equation, setting x = 0 and y = −f (0), we get f (0) = −2f (0) + f ( f ( f (−f (0)))) = −2f (0) + f (5)₍₀₎ = −2f (0) −f (0) = −3f (0). So f (0) = 0. Then (1) becomes f ( f (x)) = 2x + f (−x). (3)

In the original equation, setting x = 0, we get f (y) = f ( f ( f (y))). (4) Setting x = f (y) in (3), we get

f (y) = f ( f ( f (y))) = 2f (y) + f (−f(y)).

So f (− f (y)) = −f (y). Setting y = −f (x) in the original equation, we get 0 = 2x+f ( f ( f (−f (x))) −x). For every real number w, setting x = −w/2, we see w = f ( f ( f (−f (x))) − x). Hence, f is surjective. Then by (4), w =

f ( f (w)) for all w. By (3), setting x =

−w, we get f (w) = w for all w. Substituting this into the original equation clearly works. So the only solution is f (w) = w for all w.

Commended solvers: Kelvin LEE

(Winchester College, England), Problem 280. Let n and k be fixed positive integers. A basket of peanuts is distributed into n piles. We gather the piles and rearrange them into n + k new piles. Prove that at least k + 1 peanuts are transferred to smaller piles than the respective original piles that contained them. Also, give an example to show the constant k + 1 cannot be improved.

Solution. Jeff CHEN (Virginia, USA),

Stephen KIM (Toronto, Canada) and PUN Ying Anna (HKU, Math Year 1).

Before the rearrangement, for each pile, if the pile has m peanuts, then attach a label of 1/m to each peanut in the pile. So the total sum of all labels is n.

Assume that only at most k peanuts were put into smaller piles after the rearrangement. Since the number of piles become n + k, so there are at least n of these n + k piles, all of its peanuts are now in piles that are larger or as large as piles they were in before the rearrangement. Then the sum of the labels in just these n piles is already at least n. Since there are k > 0 more piles, this is a contradiction. For an example to show k + 1 cannot be improved, take the case originally one of the n piles contained k + 1 peanuts. Let us rearrange this pile into k +1 piles with 1 peanut each and leave the other n − 1 piles alone. Then only these k + 1 peanuts go into smaller piles.

Commended solvers: WONG Kam Wing

(TWGH Chong Ming Thien College).

Olympiad Corner

(continued from page 1) Problem 3. In a mathematical competition some competitors are friends. Friendship is always mutual. Call a group of competitors a clique if each two of them are friends. (In particular, any group of fewer than two competitors is a clique.) The number of members of a clique is called its size.

Given that, in this competition, the largest size of a clique is even, prove that the competitors can be arranged in two rooms such that the largest size of a clique contained in one room is the same as the largest size of a clique in the other room. Day 2 (July 26, 2007)

Problem 4. In triangle ABC the bisector of angle BCA intersects the circumcircle again at R, the perpendicular bisector of BC at P, and the perpendicular bisector of AC at Q. The midpoint of BC is K and the midpoint of AC is L. Prove that the triangles RPK and RQL have the same area. Problem 5. Let a and b be positive integers. Show that if 4ab − 1 divides (4a2 − 1)2_{, then a = b.}

Problem 6. Let n be a positive integer. Consider } 0 }, , , 1 , 0 { , , : ) , , {( ∈ + + > = xyz xyz n x y z S _K

as a set of (n + 1)3 _{− 1 points in the} three-dimensional space. Determine the smallest possible number of planes, the union of which contains S but does not include (0, 0, 0).

Convex Hull

(continued from page 1) A₁ A₂ A₃ A₄ A₅ B₁ B3 B₂

Case 1: (The boundary of H is the

pentagon A1A2A3A4A5.) If △A1A2A3 or

△

A1A3A4 or △A1A4A5 does not contain any point of B in its interior, then we are done. Otherwise, there exist B1, B2, B3 in their interiors respectively. Then we see △B1B2B3 is a desired triangle. A₁ A₂ A₄ A₅ A₃ B₁ B3 B₄ B₂

Case 2: (The boundary of H is a

quadrilateral, say A1A2A4A5 with A3 inside.) If △A1A3A2 or △A2A3A4 or

△

A4A3A5 or △A5A3A1 does not contain any point of B in its interior, then we are done. Otherwise, there exist B1, B2, B3, B4 in their interiors respectively. Then either △B1B2B3 or

△

B3B4B1 does not contain A3 in its interior. That triangle is a desired triangle.

A₁

A₂ A5

A₃ A4

Case 3: (The boundary of H is a

triangle, say A1A2A5 with A3, A4 inside, say A3 is closer to line A1A2 than A4.) If

△

A1A2A3 or △A1A3A4 or △A1A4A5 or

△

A2A3A5 or △A3A4A5 does not contain any point of B in its interior, then we are done. Otherwise, there exists a point of B in each of their interiors respectively. Then three of these points of B lie on one side of line

A3A4. The triangle formed by these three points of B is a desired triangle.

Solution 1. (a) Let 1

<

p

<

q

<

r 5 n be indices for which

d = d,,, a, = -(aj : 1 I j I q}, ='min{aj : q

<

j

<

n ) and thus d = a,

-

g. (These indices are not necessarily unique.)

For arbitrary real numbers zl 5 q

5

. .

.

<

z,, consider just the two quantities Izp

-

and lzT -&I.

since

d ' d -

we hrrve either a,

-

zp

2

-

or zr

-

g

2

5.

Hence,

2

d max{lzi

-

oil : 1 I i I n) >_ -{1zp

-

a,(,

I t r

-

4)

>

matop

-

zp, z,

-

%)

2

3

(b) Define the sequence (zk) as

for 2 5 k

l

n. We show- that we have equality in (1) for this sequence. .

.

By the delinition, sequence ( z k j is non-decreasing and q

-

4

2

-!

2 for d 1 5 k 5 n. Next we prove that

z k - a k 5 L 2 f o r d l < k < n . (2)

Consid= an arbitrary ind& 1

5

k

<

n. Let f

<

k be the smallest index such.that zk = z ~ . We . have either

L

= 1, or

( 2

2 and zt

>

z ~ - ~ . In both cases,

d zk = zl = at

-

' 2' (3) Since equality (3) implies ; d d d , z k - o k = a t - a k - - < d - - = - . 2 2 2 .. : : . .. . . d

We obtained &it

-g'i

2 z k

-

ak 9

-

2 for d,J 5 k 5 n, so d

ma~{lzi-a;l: l < i < n ) ~ - 2 '

d .

We have equality because Izl

-

all = -, 2

i

Mi = max{aj : 1 5 j 5 i) and mi = min{aj : i 5 j

5

n). For all 1 i

<

n, we have

and

mi = ~ n { o i , a i + 1 1 . ..,a,,}

5

~ { & + l , . . . , a , , ) =

m+1.

Therefore sequences (Mi) and (mi) are non-decreasing. Moreover, since ai is listed in both d ~ t i o n s ,

m i < q < M i . To achieve quality in (I), set

Mi +mi

xi =

-

2

-

Since sequences (Mi) and (w) are non-decreasing, this sequence is non-decreasing as well.

From

4

=

Mi

-

mi we obtain that

-

Therefore

4 d

mm{lzi-el: I i i < n ) S m a x { - : 2 l < i < n ) ' = - 2 '

Since the opposite inequality has been prcived in part (a), we must have equality.

-

Solution. If CFc= CG, then LFGC = LGFC, hence LGRB = LGFC = LFGC = <FAD,

and t is a bisector.

Assllme that C F

<

GC. Let EK and EL be the altitudes _{the isosceles triangles}

_ECF

and EGC, respectively. Then in the right triangles EKF and ELC ,we have EF = EC and

Since quadrilateral BCED is cyclic, we have LEDC = LEBC, so the right triangles B E L and

DEK

are similar.

Then

KE

>

L E implies DK > BL, and hence

AD

GC this contradicts our But triangles ADF and GCF are similar, so we have 1

>

-

=

--

DF C F '

. assumption.

The case

CF

>

GC is completely similar. We consequent1 obtain the converse inequalities

d

_{GC a contradiction.}

K F > L C , h ' E < L E , D K < B L , D F < D , h e n c e l < - = - -

D F C F '

Solution. We

and Room B.

present an algorithm to arrange the competitors. Let the two rooms be Room A We start with an initial arrangement, and then we modify it several times by ' send& one person to the other room. At anystate of the algorithm, A and B denote the sets '

of.the competitors in the rooms, and c(A) and c(B) denote the largest sizes of cliques in the rooms, respectively.

Step 1. Let M be one of the cli_~ues of largest size, [MI = 2m. Send all members of M to . Room A and all other competitors to Room B.

Since M is a clique of the-largest size, we have c(A) = IM(

2

.c(B).

Step 2. While c(A)

>

q B ) ,

send.one person from Room A to Room B.

h mA Room B

I

I

Note that c(A) > c(B) implies that Room A is not empty.

I n each step, c(A) decmses by one aid c(B) i n a w e s by at most one. So at the end we have c(A) 5 c(B) 5 c(A)

+

1.

We also have c(A) = I A l r m at the end. Otherwisk we wohd have at least m

+

1 members of M in Room B and at most m- 1 in Room A, implying c(B)

-

c(A)

1

(m

+

1)

-

(m

-

1) = 2. Step 3. Let k = c(A). If c(B) = k then STOP.

If we reached c(A) = c(B) = k then we have found the M e d arrangement.

In all other ca& -%e have c(B) = k

+

1.

From the estimate above we also know that k s ~ A I

=

] A n MI

1

m and IB

n

MI 5 m. Step 4. If there ezists a competitor z €. B f l M and a cliqw C C' B such that IC1 = k i- 1 and z

%

C, then move z to

Room

A and STOP.

h m A Room B

m

m

After moving x back to Room A, we will have k

+

1 members of M in Room A, thus c(A) = k

+

1. Due to z $ C, c(B) = ICI is not decreased, and after this step we have c(A) = c(B) = k + 1.

If there is no such competitor x, then in Room B , all cliques of size k

+

1 contain B

n

M as a subset.

Step 5. While c(B) = k

+

1, choose a clique C C B such that

(c)

= k

+

1 and move one member of C

\

M to Room A.

Room A

I Room B

-

-

I

Note that ICJ = k.+ 1

>

m

2

IB

n

MI, SO C

\

lli1m o t be empty.

Every time we move a single person from Room B to Room A, so c(B) decreases by at '

most 1.' Hence, at the end of this loop we have c(B) = k.

In Room A we have the clique A n M with size IAn MI = k thus c(A)

2%.

We prove that there is no clique of larger size there. Let Q

c

A be an arbitrary clique. We show that

191

$ k.

Room B

I I

!

!

In Room A, and specially in set Q, there can be two types of competitors:

-

Some members of M. Since M is a clique, they are friends with all members of B

n

M.

-

Competitors which were moved to Room A in Step 5. Each of them has been in a clique _i with B

n

M so they are also Mends with all members of B r l M.

Hence, a! members of Q are.friends with all members of B n M. Sets Q and B

n M

are I cliques themselves, so Q U (B

n

M) is also anclique. Since M , is a clique of the largest size,

i

I

therefore . .

IQI S ( A n M I = k .

!

. .

Finally, after Step we have c(A) = c ( 3 ) =

k.

C3Wblon

4

i

Solution 1. I f AC = BC then triangle ABC is isosceles, triangles

RQT

and

RPS

are m e t r i c about the bisector CR and the statement is trivial. If AC

#

BC then it cab be kumed without loss of generality that AC < BC.'

Denote the circumcenter by 0. The right triangles CTQ and CSP have equal angles at vertex C, so they are similar, ~ C P S = LC-W = LOQP and

QT

CQ

'hiangies RQT and RQA have RQ as a common side, so the ratio between their areas is area(RQT)

-

_--=-=- d(T, CR) . C T 1

area(RQA) d(A,CR) CA 2' (d(X, YZ) denotes the distance between point X and line YZ).

It can be obtained similarly that

Now the proof can be completed as

-

=

-.

PS CP (1)

M

f

b. the m d i & b i o r of chord CR; of course, f pasm through the circum-

.

bbl~-ur

center 0. Due to the equal angles at P and Q, triangle OPQ is isosceles ivith O P = OQ. CLU a pair (zl V ) of positive integers' bid if 4xy

-

1 divides ( 4 9

-

i)%ut t

#

y. _{order to}

Then line ! is the &g of symmetry in this triangle as well. Therefore, points P and Q Fe _{Prove that bad pairs do not}

_m,

_{we present two properties of them which provide an infinite}

symmetrically on line segment CR, descent. .,

R P = CQ and R Q = C P . (2)

%angles RQT and ~ ~ S ' h a v e equal angles at vertices

Q

and PI respectively- Then area(RQT) =

-

RQ

.

QT

-

sin LRQT RQ

QT

= - a - area(RPS)

$

-

RP

-

PS

-

sin L R P S R P PS' Substituting (1) and (2), area(RQT) RQ QT C P CQ '. . .

_-

_-(RPS) = . = - . - -

_RP

_{PS CQ cp-l-} Hence, area(Rm)

=

area(RSP1.

Solution 2. Assume again AC < BC. Denote the circumcenter by 0 , . and let 7 be the angle at C. S i l y to the first solution; from right triangles mQ and C S P we obtain. that LOPQ = LOQP = 90'

-

z.

Then tria@e. OPQ is isosceles, O P = OQ and moreover

LPOQ = 7.

As is well-known, point R is the midpoint of ari AB and LROA = LBOR = y.

R

Property (i). If (z, y) is a bad pair and z

<

y then- there exists a positive integer z < z such

,

that (I, z) is also bad. '

~ = - r . ( - l ) ~ - r ( 4 z y - l ) = - ( 4 2 ~ - 1 ) ~ ~ - 1 (mod&) I

and r = 422

-

1 with some positive integer z. From s

<

y we obtain that

and therefore z < z. By the construction, the number 4rr

-

1 is a divisor of (4z2

-

_{so (I, x)} is a bad pair.

Property (ii). If (2, y) is a bad pair then (y, x) is also bad.

Since 1 = 1' ( 4 ~ ) ' (mod 4xy

-

I), we have

(4y'

-

1 1 2

&

(4y2

-

(42y)2)2 =.16y4(&Z

-

1)' t 0 (mod 4ry

-

1)'.

Hence, the number 423

-

1 divides ( 4 ~ ~

-

1)2 as well.

I

. Now sqppose that there e.xists at least one bad pair. Take a bad pair (x, y) such that 22 4- y

,

attains its smallest possible value. If x

<

y then property (i) provides a bad pair (x, z). with z

<

y and thus 2z+z < 2x+ y. Otherwise, if y <'I, property (ii) yields that pair (y;x) is

(yr6b-a-

Solutioil. It is easy t o find 3n such planes. For example, planes x = i, y = i or z = i

(i I.1,2, :,..,n) cover the set S but none of them contains thwrigin. Another such collection

&mists of all planes ~ + ~ + z = k for k = 1,2,

...,

371.

wk

show that 3n is the smallest possible number.

Lemma 1. Consider a nonzero polynomial P(z1,.

.

.

,

zk) in k variables. Suppose that P vanishes at aIipoints (zl

,...,

26) such that zl ,..., z k E (O,l,

...,

n)-and x l i - . - - + x k , > 0,

while P(O,O,.

.

.,

0)

#

0. Then degP

1

kn.

Pmof. We use induction on k. The base case k = 0 is clear since P

#

0. ~ e n o t e for clarity

y = z*.

Let R(sl,.

. .

,2k-1, y) be the residue of P modulo Q(y) = y(y

-

I ) .

.

.

(y -.n). Polyn& mid Q(y) vanishes at each y = 0,1,.

. .

, n, hence P(z1,.

.

.

, zk-1, y) = R(z1,.

.

.

, zk-~, y) for all z l ,

.

. .

,

~ k - ~ , y E (0, 1,

. . . ,

n). . Therefore, R also satisfies the condition of the Lemma; moreover, deg, R 5 n. Clearly, d e R 5 deg P , so it d c e s to prove th@ deg R 2 nk.

Now, expand polynomial R in the powers of y:

We show that polynomial R,,(zl,.

. .

, zk-l) satisfies the condition of the induction hypothesis. Consider the polynomial T(y) = R(0,.

. . ,

0, y) of degree 5 n. This polynomial has n roots y = 1,.

. .

, n; on the other hand, T(y) f 0 since T(0)

#

0. Hence degT = n, and its leading coacient is &(O, 0,.

.

.

,0)

#

0. In particular, in the case k = 1 we obtain that coefficient R, is nonzero.

Similarly, take any numbers al,

.

. .

,

ak-l E {O,1,

. .

.

,

n) with al+.

.

>

0. Substituting zi = Q into R(zl,

. .

.

,

~ k - ~ , y), we get a polynomial in y which h h e s at all points y = 0, ...

.

,

n

and has degree

<

n. Therefore, this p o l ~ ~ m i a l is null, hence R,(al,

. . .

,

ak-1) = 0 for all

i = O,l,;.

.

,

n. In particular, %(al,.

. . ,

ak-1) = 0.

Thus, the polynomial &(zl,.

.

.

,

zk-l) sat- the condition of the induction hypothesis. So, we have deg R,,

2

(k

-

l)n and deg P

3

degR

1

deg

R,,

+

n 2 kn. 0

Now we can finish thesolution. Suppose that there are N planes cov- all the points of S but not containing the origin. Let their equations be ~z

+

biy

+

q z

+

di = 0. Consider the polynomial

N

F(x, y, z) = n ( a + z

+

biy

+

q z

+

4).

it1

It has total degree N. This polynomial has the property that P(z0, yo,o,lo) = 0 for any (zo, yo,

a)

E S, while P(O,O, 0)

#

0. Hence by Lemma 1 we get N = degP

2

3n, as de- sired.

I _{Solution 2. We present a &ent proof of the main Lemma 1. Here we confine ourselues to}

'

_{the case k} = 3, which is applied in the solution, and denote the variables by z, y and z. (The same proof mrka for

the

general statement as well.) .

The following fact is known with various proofs; we provide one possible proof for the completeness.

Lemma 2. For arbitrary integers 0 5 m < n and for an arbitrary polynomial P(z) of degree m,

Proof. We use an induction on n. If n = 1, then P(x) is a constant polynomial, hence P ( l )

-

P(0) = 0, and the base is proved.

For the induction step, define Pl (x) = P ( x

+

1)

-

P(x). Then clearly deg PI = deg P

-

1 = m

-

1 < n

-

1, hence by the induction hypothesis we get

n-l n-1

I

_o

₌

_-

_{c(-1)' k=O .}

("

;

')

_{p1(k) = ~ ( - 1 ) ~ (n ;.').(~(k)}

_-

_{p ( k}

₊

₁₎₎

k=o

: ' Now return to the proof of Lemma. 1. Suppose, - t o the contrary, that deg P = N < 3n. Consider the sum

n n n

E

= ~ ( - l ) ~ + j + ~

(?)

c)

c)

~ ( i , j, k).

i=O j r O k O 2

The only nonzero term in this sum is P(0,0,0) and its coefficient is

cs

= 1; therefore

c

= P(0, 0,O)

#

0.

On the other hand, if P(z, y, a) = _{paA7za#z7, then .}

a+8+71N

n n n

=

C

C

C(-l)'+j+k

i = O j = O I.lo

Consider an arbitrary term in this sum. We claim that it is zero. Since N < 3n, one of three _. inequalities a

<

n, /3 < n or 7 < n is valid. For the convenience, suppose that a < n. Applying

n

Lemma 2 to polynomial za, we get = 0, hence the term is zero as required. .

i=O