Toán Olympic quốc tế 1999 Tiếng Anh

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The Organizing Committee and the Problem Selection Committee thank the following 31 countries for sending in proposals of problems

Czech Republic Luxemburg United Kingdom

Poland

The Problem Selection Committee:

Marian Andronache Mihai Cipu Vasile Pop Mihai Baluna Nicusor Dan Dragos Popescu Barbu Berceanu Gheorghe Eckstem Dinu Serbanescu Dan Branzei Mihai Piticari

The 27 problems submitted to the Jury are classified under Number The- ory (NI-N6), Geometry (G1-G8), Algebra (A1-A6) and Combinatorics (C1-

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in conclusion the required solutions are (2,2), (3,3) and (1,p), where p is an arbitrary prime

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Solution We firstly claim that every rational number from the interval (1,2) can

be represented in the form

g3 + g1

Indeed, let m/n € (1,2), where m and n are natural numbers We will choose a,b, đ

such that 6 4 d and a? — ab+ 6? = a? —ad+ die b+d=a In that case

e+ ad t-—b Taking a+ = 3m, 2a—b = 3n, thal is a@ = m+n, 6 = 2m—n, the claim is proven

We can prove now the required conclusion If r > O is a rational number, take

3 positive integers p,g such that 1 < a < 2, There exists positive integers a, b,d such

that

q° as +

Hence

_ (a9)? + (bay

(ap)? + (dp)

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(g-+rilip+ ict yt) =b+s

that is we have to find integers x, y such that

(pr + g)z + (nạ — r) = I

It is casy to see that 1Ï ø, g,? are coprune then pr+q and pq —r are also copr.me:

if dis a prime divisor of pr + g and øq — r, then

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&) Prove that S is infinite

b) Find the highest value af Itk.p) fork >1andpe S

Solution a) The fundamental pericd of 1/p is the smallest integer d > i such that 108 — 1 is divisible by p

Let s be a prime and A, = 1075 4 108 + 1 Clearly NV, = 3{inod 9) Let p, be a prime divisor of N,/3; p, cannot be 3 Since N, is a divisor of 103% — I, the decimal representation of 1/p, has a period of length 3s, so its fundamental period is a divisor

of 3s The fundamental period cannot be s, because this would imply 10° = 1{mod ,), leading to N, = 3 # O(modp,) Also, the fundamental period can be 3 only in the case when p, is a divisor of 103 — 1 — 33.37, that is Ds = 37 We claim that p, can be

chosen 4 37: otherwise N, = 3-37" = 3(mod 4) and N, = 10° + 10° +1 = 1(mod4)

Hence, for every prime ¢ we can find a prime p, such that the decimal representation

of 1/p, has the fundamental period of length 3s

b) Let 3r(p) be the fundamental period for a prime pes Then pis a divisor of

10°) — 1 but not a divisor of 107) — 1, so p is a divisor of Np = 10?) 4 19°@) 4.1,

Let 1/p = 0, ayaa .,2; = 10°-1/p and Uj = {2} =O, ajajsiajgy

Clearly a; < 10y;, therefore

F(R, D) = Ge + Gis n(p) + Oerortp) < 10k + Uexxe) T ecsedp)):

Hence, the highest value for {{k,p) can be at most 19 From f(2,7) = 44847 = 19

we conclude that this is the required inaximum

Te + #y+r() TT k+ar(p) —

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Since (p,10) = 1, there exists k > 2 such that 10" = 1(modp) It follows that 10** = 1(mod p} and 10/*+! = 10(mod p) for every non-negative integers 7,7 We will look for integers u,v > 0 so that M = » 108 + > 107"! (if w or w is 0 then the

vg and uy = k — up we get the wanted M

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Bucharest 1999 9

Problem N6

Prove that for every real number M there exists an infinite arithmetic progression

such that:

~ each term is & posttive integer and the common difference is not divisible by 10;

~ the sum of the digits of each term (in decimal representation) cacceds M

Solution We will prove that it is possible to take a common difference of the form

10” + 1, where m is a positive integer,

Let ap be a positive integer and a, = ag + n(10™ + 1) = b,6,_) by, where s

and the digits 6y,6), ,6 depend on n [t is easy to see that if / = A(mad 2m) then

10! = 10* (mod (10™ + 1)) Therefore ay = ay = Bybe2) ‹la = 3} e¢10'(mod 10" +

i=0 1), where ec, = By + bomse + Bamag Ho fori =0,1, ,2m— 1

Let N > M be a positive integer The number of 2m-uples (€9,€1, ,C2m-1) of

non-negative integers with co+c,-+ 4+¢Cam—1 < N is equal to the number of strictly

Mor sufficiently large m we have Kam < 10" Taking ag € {1,2 10} such

that ap is not congruent (mod 10” +1) with any of the numbers belonging to the set

{Gam —1Cam 3G | Co + ey + + Coma < NY}

we get the required sequence

Remark For large M a “smali” common difference cannot do the job, because

wn such a case the sequence would hate at least one of tis terms in an interval of the

form {10°,10" + d] and all the integers from such an interual have the sum of their

digits at most 1+ 9-log,,d

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Proof of the lemma (Figure 1} The ray (AM intersects the quadrilateral in N;

i suppose, for instance, that N € [CD| Then MA+ MB < MA+MN+NB <

AN+NC+CB< AD+ DN+NO+CB=AD+DC+4+CB

Solution of the problem (Figure 2) The median triangle DEF divides triangle ABC into four regions Each region is covered by at least two of the convex quadrilaterals ABDE, BCEF, CAFD If, for instance, M belongs to [AB DE] and [BCEF| then MA+MB < BD+DE+FAand MB4+MC <CE+EF-+FB By adding these two inequalities we get MB + (MA+MB+MC)< AB+ BC+CA, which implies the required conclusion

Another solution Let O be the circumcenter of AABC

Case 1: O € |ABC| (Figure 3) Suppose M € |ADOE] Then MA = min{ MA, MB, MC and MA4+MB< NA+NB,MAiMC < NA4INC Since NA+NB<OA+0OB and NA+NC < AD+ DC (for OUANC); otherwise NA+ NC < BD + DC) it is enough to prove that m, + (c/2) + 2R <a+6+c¢ (with the usual notations) Since m, < {a+ 6)/2, it is enough to prove that 4R < a+6+c (in a nonobtuse-angled triangle)

This reduces to sin A+sn B+smC' > 2, or

1+ cos A) therefore it is greater than (1/2)(1 4+ 1) = 1

Case 2: ABC is obtuse-angled (Figure 4) Suppose for instance that mB) > 90° Let P and Q be the intersections of the perpendicular bisectors of the sides [AB] and [BC] with AC

For M € [ADP], min{MA,MB,MC} — MA and, as above, MA+ MB < PA+PB=2-AP, MA+MC <m,4+({ce/2) < (a+b+4c)/2 It is enough to prove thal 4-AP <a+6-+c, which is implied by 4AP < 2b <a+b+c A similar argument works in the case M € (CEQ)

For M € [BEQPD), let BM NAC = N

Then min{MA, MB, MC} = MB and MB + MC < NB+NC, MB+ MAX NB+NA, NB < max{PB,QB}, therefore it is enough to prove that 2-PB < AB+ BC Since PB < BD+ DP < BD + P/", the needed relation is obvious

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12 40 IMO

Problem G2

A eircle is calied a separator for a set of five points in a plane if it passes through three of these ports, it contains a fourth point inside and the fifth point is outside the cirele

Prove that every set of five points such that no three are collinear and no four are concyelic has ezactly four separators

Solution Let {A, B,C, D} be the inverses of four of the points of the set through ail inversion having as pole the fifth point Notice that a separator which passes through the pole is transformed into a straight line which passes through two of the poims A,B,C, D and separates the other two Notice also that a separator which does not pass through the pole is transformed into a circle which passes through three

of the points A, B,C, D and contains the fourth point inside

Considering A-the convex hull of the set {A,B,C,D}, two cases may appear

Case t K =quadrilateral (for mstance ABOD) In this case we have as scparators the two diagonals of the quadrilateral and one circle from each pair {(ABC), (ABD)}}, {(C.DA), (CDB)} (the one corresponding to the smailer angle from {ACB, ADB}

and {CAD, CBD} respectively)

Case 2 K=triangle (for instance AABC) In this case we have as separators, with obvious arguments, the straight lines DA, DB, DC and the circle (ABC)

Another solution Consider coordinates and the points A;{z,, y;), 7 = 1,2,3,4,5

Let djju be the value of the determinant |22+ 4? tp z1 1ip—;;x¡ The circle (A1AzAs)

is a separator iff djg3q and djo35 have different signs Let us look at the ten pairs (diosa, dioss), (diea3, di2as), etc corresponding to the ten circles which pass through three of the five given points Denoting by a, the number dij (¢ < 7 < & < TU, fi,j,k,t,n} = {1,2,3,4,5}), the ten pairs are (a5, a4), (—Gs, a3), (—a4, —@3), (a5, ae),

(đa, —aạ}, (dạ, 2), (—ds, đ1), (—a¿, —đ1), (—ữa, 1), (—a2, -4))

We notice that the number of separators is equal to the number of pairs of terms with the same sign from the sequence 5 = (a1,—0@2,a2,—@4,@5) We also remark (using the determinant |#2 + Tp Mẹ L 1lp=tsaa) that a) — a2 +43 — ag + a5 = 0

This shows that S$ cannot have all its terms of the same sign

We claim also that S cannot have four terms of Lhe same sign, Indeed if four

; terms have the same sign then all the six circles passing through one point of the

given five are separators Taking this point as origin, the Ox axis through an other

of the given points and passing to polar coordinates we would get, for instance,

Tạ cose sing rf; cosa sing

rT, cosh sind | > 0, rg cosh sind | <0,

rz cose sinc |>0, and rs cose sine | < 0

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Bueharest 1999 13

This would lead to

r¡ sin(€ — b} +7 sin(ø — c} + rạ sin(b — a} > Ú,

—T\ 8ỉn ỗ + r sIn ø + #4 sin(Š — ứ)} < 0,

—T gìn # + 's Sìn đ + ?4 8in(đ — œ) > 0,

—rysinc + rysinb+ rqsin(e-~ b) < 0

Multiplying the last three relations by rs, —r2 and r, respectively and adding the

results we would get rg[r) sin{e — b) + rgsin(a — c) + rg sin(b — a)| < 0, impossible

Hence three terms of S have the same sign and the other two have the other sign, therefore there are exactly four separators

ee etre AEE Eh cele emerge eon on aig men aga RESULTS SN ESTUARIES cL ASCE TT te i EE

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14 40 IMO

Problem G3

A set S of points from the space will be called completely symmetric if it has

at least three elements and fulfils the condition:

for every two distinct points A,B from S the perpendicular bisector plane of the segment AB is a plane of symmetry for 5

Prove that af a completely symmetric set is finite then it consists of the vertices of either of a regular polygon, or a regular tetrahedron or a reqular octahedron

Solution Denote by rpg the reflection in the perpendicular bisector plane of a segment PC) and let G be the barycenter of S From rag(S) = S we get rag(G) = G for every A,B € S, therefore all the points of S are at the same distance from G This shows that S$ is included in a sphere &,

Case f: S is included in a plane x In this case S is included in the circle X Na and its points form a convex polygon A) Ag A4, The reflection in the perpendicular bisector of A,A3 transforms cach half-plane bordered by A Ag into itself, therefore the point r4,4,(Az2) can be only Ay Hence AyAz = AgAg In the same way AoA; — A3A4 = = A,Ay Since S$ is included in a circle, this proves that A) Ay A, is a regular polygon

Case 2: the points of S are not coplanar In this case the points of S are the vertices

of a convex polyhedron P (since they are on the sphere £) Each face 4)A9 A; of

P is invariant under every reflection r 4, 4,1 <i1<9<k, therefore it is a regular polygon (from case 1)

Notice also that every reflection rap, A,B € S transforms a face of P into a face

of P

Take now a vertex V of P and denote P's edges issuing from V by V4, VVa, , VV, such that (VW), VW¿),(V¿, VVẠ), , (VY2,VVI) are on the same face (Figure 1) Notice that the intersection of the half-planes [V V3, Va and [Vi V3, V with P are triangles VjV2V3 and Vi V3V respectively The reflection ry,y, transforms each of these half-planes into itself, therefore it can transform V2 and V only into themselves This

j shows that ry,y, must transform the face containing (V Va, V V3) into the face contain-

ing (V¥2, VV) Hence these faces are congruent

' In the same way every two faces of P having a common side are congruent This

shows that all P’s faces are congruent, because every two faces can be ‘linked’ by a chain of faces so that every two consecutive faces of the chain have a common edge

It follows that P is a regular polyhedron (a similar argument shows that from each vertex emerges the same number of edges)

It remains to rule out the cube, the regular dodecahedron and the regular icosahedron The cube (Figure 2) is ruled out because of the reflection rac: (the rectangle ACCA’ should be invariant, but it isn’t) he dodecahedron (Figure 3) is excluded because of the reflection r4,g, (same argument for the rectangle A,A;838)) Finally, the icosahedron (Figure 4) is eliminated because of the reflection r47 (use rectangle

AA, BB,)

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16 40 IMO

Problem G4 Hor a triangle T — ABC we take the point X on the side (AB) such thai AX/XB — 4/5, the point Y on the segment (CX) such that CY = 2YX and, if possible, the point Z on the ray (CA such that GXZ = 180° — ABC We denote by

X the set of all triangles T for which XYZ = 45°

Prove that all the triangles from & are sumilar and find the measure of their smail- est angle

Setution A convenient way to describe the position of the points using triyonom- etry is to employ the cotangent.We firstly prove the fallowing

Lemma (Figure 1) Ina triangle ABC, X € (AB), XA: XB=m:n CXB=e

and AC X = G Then (mt njecota =neotA—meotB and meot f= (m |} n)cotC | acot A

Proof of the lemma Let CF = h be the altitude fram C Then using oriented

segments, AX = |AF + FX| = h-cotA—A- cote and BX = |BF 4 FX| =

h cot B+h.cote The first part of the conclusion follows now from n-XA=m-XB

For the second part take XT||BC, T ¢ (AB) Then XTA = C and CT: TA = n:m The required result follows from the first part applied n AAXC’ O

We have (Figure 2) by the lemma and the hypothesis 4cat AOGX = 0catŒ +5 cot A and

cot ACX — 2cot CXZ = 3cat XYZ = 3

We also have CXZ = 180° — B, therefore (8cot Ở + 5cot A)/4 + 2+ cot B = 3,

that is

45 cot A+ 8cot B+ GcotC = 12

We will prove that this equation specifies the angles of the triangle ABC Denoting eot A = 2, cot B = y, cot C = z we have 52+8y+9z = 12 and the well-known relation

# + z + zz = 1

Eliminating z we get (ety) [12—5x—8y) 4 Sry = 9, that is (4/#+z—3)?+9(z— 1)? =

0 This shows that x = 1, y = — and z = -, therefore all the triangles from %: are similar and their smallest angle is A = 48°

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Problem G5 Let ABC be a triangle, 2 its incircle and Q4,0%, 0 three circles orthogonal to Q nassing through (B,C), (A.C) and (A,B) respectively The circles Q and %l, meet again in CƠ in the same way we obtain the points B’ and A’ Prove that the radtus

of the circumeicle of A’ B’C" ts half the radius of 9

Solution Denote by J the incenter, by r the inradius, by D,&,/ the contacts

of the incircle with BC,CA,AB respectively and by P,@,R the midpoints of the segments [4 F'), [FP], [DE]

We will prove that Q, is the circle (B,C,Q, R) We firstly notice that from the right-angled triangles ! BD and 1 DQ we get [Q-1B = ID® = r* and in the same way

TR 1Œ = r*, therefore the points 8,C,R,@ are on a circle [ The points @ and

R belong to the segments (1B) and (JC), so TF is exterior to the circle (B,Q, B,C) and I’s power with respect to the circle (B,Q,R,C) is 1B 1Q = r?, which is the condition of 2 beeing orthogonal to the circle (B,Q, R,C)

It follows that A’, B’,C’ coincide with P,Q, RA and the required conclusion is now obvious

Another solution Let O, be the center of 2, and M be the midpoint of (BC)

Denote, using oriented segments, MO, = x (the positive sense on the perpendicular bisector of (BC) beeing TD)

1

The radius of Q, is 2? + and

MN = ztan 5 2A sin 5 sin 5 CO8 2 5 2 3 2

Hence BN = = and, because MN < BM, N € (BM) The same holds for the common point of AB and O,O,, therefore the reflection of B in O,0, is on DF This proves that B’ — the second common point of 1, and 2, — is the midpoint of (DOF)

The conclusion follows now easily

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20 40 IMO

Problem G6

Two circles (2, and Qe touch internally the circle Q in M and N and the cenier

of 22 18 on 2) Fhe common chord of the circles Q, and tạ intersects Q in A and

B MA and MB intersect 1, in C and D Prove that Q, is tangent to CD, Solution

Lemma The circle ky touches internally circle k at A and teuches one of k's chords MN at B Let C be the mid-point of k’s arc MN which does nul contain A

Then the points A, B, C are collinear and CA-CB = CM?

Proof of the lemma (fig.1) The homothety with center A which transforms ky into & transforms MN inte a tangent at k parallel to MN, i.e into the tangent at C

to k, so A, B,C are collinear

For the second part notice that NMC = CAM, therefore AAC M ~ AMCB,

whence CA-CB=CM?, Solution of the problem Let O, and Os be the centers of Q, and OQ» respectively and í¡, é their common tangents (fig.2) Let a, @ be the arcs cut from Q by f¡ and

ég, positioned like in Lhe lemma

Their midpoints have, according to the lemma, equal powers with respect to 2, and §%9, therefore they are on the radical axis of the two circles ‘hus A and B are the midpoinis of a and 6 From the lemma we also conclude that (’ and D are the points in which the tangents ¢; and f touch 9 If H is the homothety with center M transforming Q) into M2 then H :CD++ AB whence AB|\C'D Therefore

CD 1 O10 and Os; is midpoint of one of the arcs CD from Qs _

Let X be the point in which t; touches Q2 We get XCOz = (1/2)CO,0, =

DCOs, 50 Oz lies on the bisector of the angle XCD, therefore C'D is tangent to the

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22 40 IMO

Another solution (Figure 3) An inversion of pole N transforms the figure as follows: 9 and {» in parallel straight lines, 2, in a circle which passes through the reflection of N in Q»y and is tangent to 2, AB in the circle (congruent with 2, ) passing through N and through 2; 993, AM and BM into the circles (congrucnt) passing through NV, M and A, 8 respectively, therefore CD becomes circle (NCD) We have

\ ~ to prove that (VCD) is tangent to Mo

The circle (CDN) has its center on MN and ND — BD, therefore it is enough to

| prove that B, D, T are collinear, that is the common point # of AT and MQ, is the

projection of N on BT This results from BR 2BT = BM? = BN? — NM?* = BN? + BT? — NT?, the last equality being justified by NM? = d’?, BT? — NT? =

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Bucharest 1999 23

Probiem G7 The poinW MỸ i4 inside the conuer guadrilateral ABCD, such that

MA=MC,AMB— MAD+ MCD and CMD = MCB + MAR

Prove that AB-CM = BC.MD and BM-AD=MA-CD

Solution Coustruct the convex quadrilateral PQRS and the interior point T such that APTQ = AAMB, AQTR~ AAMD and APTS ~ ACMD

and STH = BMC, therefore ARTS ~ ABMC The assumption on angles leads to

OPS + RSP = OPT + TPS +7SP 4 TER = PTS 4 TBs 4 TSP = 180°

and

ROP + SPQ = ROT + TOP + FPO + TRS = QTP + TOP + TPO = 180°,

30 PORS is a parallelogram, Hence PQ = RS and QR = PS, that is

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Solution The inequality is symmetric and homogeneous, so we can suppose that my>fe> >2, >0 and “2; = 1 In this case we have to maximize the sum

k—1 P(x!) Fg) =_ rre.i|3C + Øgạn) » Ti —x — Thay =

=

=_ #i#k+|3(2k + #k+ì)(Ï — #g — Tega) — Le - 21] =

#w#k+1|(#k + #k+l)(3 — 42g + Zk+l)) + 2x#eil|- From

1

13> 2, +244 p41 > 5 (ee + nai) + te + ps1

it follows that 2/3 > 2, + 2,11 and therefore

F(z’) — F(z) > 0

Applying the above replacements several times we obtain

F(z) < F(a,b,0, ,0) = abla? +54) = 5(2ab)(1 — 948) <

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