Solutions to the Shortlisted Problems of IMO 1985

4.26 Solutions to the Shortlisted Problems of IMO 1985

9 = 23) less than 26, each number x

1. Since there are 9 primes (p 1 =2<p 2 = 3 < ··· < p

j ∈ M is of the form ∏ i =1 p i , where 0 ≤a ij . Now, x j x k is a square if a ij +a ik ≡ 0 (mod 2) for i = 1,...,9. Since the number of distinct ninetuples modulo 2 is 2 9 , any subset of M with at least 513 elements contains two elements

ij

with square product. Starting from M and eliminating such pairs, one obtains (1985 − 513)/2 = 736 > 513 distinct two-element subsets of M each having a square as the product of elements. Reasoning as above, we find at least one (in fact many) pair of such squares whose product is a fourth power.

2. The polyhedron has 3 · 12/2 = 18 edges, and by Euler’s formula, 8 vertices. Let v 1 and v 2 be the numbers of vertices at which respectively 3 and 6 edges meet. Then v 1 +v 2 = 8 and 3v 1 + 6v 2 = 2 · 18, implying that v 1 = 4. Let A, B,C, D be the vertices at which three edges meet. Since the dihedral angles are equal, all the edges meeting at A, say AE , AF, AG, must have equal length, say x. (If x =

AE = AF 6= AG = y, and AEF, AFG, and AGE are isosceles, ∠EAF 6= ∠FAG, in contradiction to the equality of the dihedral angles.) It is easy to see that at

E , F, and G six edges meet. One proceeds to conclude that if H is the fourth vertex of this kind, EFGH must be a regular tetrahedron of edge length y, and the other vertices A, B, C, and D are tops of isosceles pyramids based on EFG, EFH , FGH, and GEH. Let the plane through A , B,C meet EF, HF, and GF, at

= FA = FE ′ , we have E ′ G ′ = x and AE ′ = x/

E ′ ,H ′ , and G ′ . Then AE ′ BH ′ CG ′ is a regular hexagon, and since x √

3. From the isosceles triangles AEF and FAE ′ we obtain finally, with ∡EFA = α ,

= cos α = 1 − 2sin 2 ( α /2), x/(2x 3 ) = sin( α /2), 2x

and y /x = 5/3.

3. We shall write P ≡ Q for two polynomials P and Q if P(x) − Q(x) has even coefficients.

2 We observe that m (1 + x) ≡1+x 2 m for every m ∈ N. Consequently, for every polynomial p with degree less than k =2 m ,w (p · q k ) = 2w(p).

Now we prove the inequality from the problem by induction on i n . If i n ≤ 1, the inequality is trivial. Assume it is true for any sequence with i 1 < ··· < i n <2 m

<2 m (m ≥ 1), and let there be given a sequence with k = 2 +1 ≤i n . Consider two cases.

(i) i 1 ≥ k. Then w(q i 1 + ··· + q i n ) = 2w(q i 1 −k + ··· + q i n −k ) ≥ 2w(q i 1 −k )= w (q i 1 ). (ii) i 1 < k. Then the polynomial p = q i 1 + ··· + q i n has the form

∑ ≡ =0 ∑ (a i +b i )x +b i x .

i =0

Whenever some a i is odd, either a i +b i or b i in the above sum will be odd.

It follows that w (p) ≥ w(q i 1 ), as claimed.

484 4 Solutions The proof is complete.

4. Let hxi denote the residue of an integer x modulo n. Also, we write a ∼ b if a and

b receive the same color. We claim that all the numbers hi ji, i = 1,2,...,n − 1, are of the same color. Since j and n are coprime, this will imply the desired result. We use induction on i. For i = 1 the statement is trivial. Assume now that the statement is true for i = 1, . . . , k − 1. For 1 < k < n we have hk ji 6= j. If hk ji > j, then by (ii), hk ji ∼ hk ji − j = h(k − 1) ji. If otherwise hk ji < j, then by (ii) and (i), hk ji ∼ j − hk ji ∼ n − j + hk ji = h(k − 1) ji. This completes the induction.

5. Let w.l.o.g. circle C have unit radius. For each m ∈ R, the locus of points M such that f (M) = m is the circle C m with radius r m = m/(m + 1), that is tangent to C at A. Let O m

be the center of C m . We have to show that if M ∈C m and N ∈C n , where m , n > 0, then the midpoint P of MN lies inside the circle C (m+n)/2 . This is trivial if m = n, so let m 6= n. For fixed M, P is in the image C ′ n of C n under the homothety with center M and coefficient 1 /2. The center of the circle C ′ n is at the midpoint of O n M . If we let both M and N vary, P will be on the union of circles with radius r n /2 and centers in the image of C m under the homothety with center O n and coefficient 1 /2. Hence P is not outside the circle centered at the midpoint O m O n and with radius (r m +r n )/2. It remains to show that r (m+n)/2 > (r m +r n )/2. But this inequality

is easily reduced to (m − n) 2 > 0, which is true.

+ 1 + ···+ n n ,

y n ,i =x i −1 n +1,i +x i −2

n +1,i x n ,i + ··· + x n ,i .

i −1

In particular, x n ,2 =x n and x n ,i = 0 for i > n. We observe that for n ≥ i ≥ 2,

x n +1,i+1 −x n ,i+1 x n +1,i −x n ,i =

n +1,i −x n ,i

y n ,i Since y ,i > ix i −1

y n ,i

n +1 √

and x n +1,n+1 −x n ,n+1 = n + 1, simple induction gives

n ,i ≥i

1 +(i−1)/i

The inequality for n = 2 is directly verified.

7. Let k i ≥ 0 be the largest integer such that p k i |x i ,i = 1, . . . , n, and y i =x i /p k i . We may assume that k =k 1 + ··· + k n . All the y i must be distinct. Indeed, if y i =y j and k i >k j , then x i ≥ px j ≥ 2x i ≥ 2x 1 , which is impossible. Thus y 1 y 2 ...y n = P /p k ≥ n!.

4.26 Shortlisted Problems 1985 485 If equality holds, we must have y i = 1, y j = 2 and y k = 3 for some i, j, k. Thus

p ≥ 5, which implies that either y i /y j ≤ 1/2 or y i /y j ≥ 5/2, which is impossible. Hence the inequality is strict.

8. Among ten consecutive integers that divide n, there must exist numbers divisible by 2 3 ,3 2 , 5, and 7. Thus the desired number has the form

n =2 α 1 3 α 2 5 α 3 7 α 4 11 α 5 ··· , where α 1 ≥ 3, α 2 ≥ 2, α 3 ≥ 1, α 4 ≥ 1. Since n has ( α 1 + 1)( α 2 + 1)( α 3 + 1) ··· distinct factors, and ( α 1 + 1)( α 2 +

1 )( α 3 + 1)( α 4 + 1) ≥ 48, we must have ( α 5 + 1) ··· ≤ 3. Hence at most one α j , j > 4, is positive, and in the minimal n this must be α 5 . Checking through the possible combinations satisfying ( α 1 + 1)( α 2 + 1) ···( α 5 + 1) = 144 one finds

that the minimal n is 2 5 ·3 2 · 5 · 7 · 11 = 110880.

9. Let − → − →

a , b ,− → c , d denote the vectors −→ OA , −→ OB , −→ OC , −→ OD respectively. Then |− → a |= − →

→ − → → − → − | b | = |− c |=| d | = 1. The centroids of the faces are ( b +− → c + d )/3, (− → a + − → − →

c + d )/3, etc., and each of these is at distance 1/3 from P = (− → a + b +− → c + − →

d )/3; hence the required radius is 1/3. To compute |P| as a function of the

2 − → → 2 → → edges of ABCD, observe that AB − =( b −− a ) = 2 − 2− a · b etc. Now

2 |− → a − + → b +− → c − + → d P 2 | =

16 − 2(AB 2 + BC 2 + AC 2 + AD 2 + BD 2 +CD 2 ) =

10. If M is at a vertex of the regular tetra- B hedron ABCD (AB = 1), then one can take M ′

M 5 at the center of the opposite

A C A C face of the tetrahedron.

M 6 M 4 Let M be on the face (ABC) of

the tetrahedron, excluding the ver- B D B tices. Consider a continuous map f

C onto the surface S of ABCD C M 1 M c ′ M 3 A

of

that maps m + ne i π/3 for m ,n∈Z A C onto A, B, C, D if (m, n) ≡ (1,1),

M 2 (1, 0), (0, 1), (0, 0) (mod 2) respect-

B ively, and maps each unit equilateral triangle with vertices of the form m +ne i π/3 isometrically onto the corresponding face of ABCD. The point M then has one preimage M j ,j = 1, 2, . . . , 6, in each of the six preimages of △ABC having two vertices on the unit circle. The M j ’s form

a convex centrally symmetric (possibly degenerate) hexagon. Of the triangles formed by two adjacent sides of this hexagon consider the one, say M 1 M 2 M 3 , with the smallest radius of circumcircle and denote by c M ′ its circumcenter. Then we can choose M ′ = f (c M ′ ). Indeed, the images of the segments M 1 c M ′ ,M 2 M c ′ ,

M 3 M c ′ are three different shortest paths on S from M to M ′ .

486 4 Solutions

11. Let −x 1 , . . . , −x 6 be the roots of the polynomial. Let s k ,i (k ≤ i ≤ 6) denote the sum of all products of k of the numbers x 1 ,...,x i . By Vieta’s formula we have

a k =s k ,6 for k = 1, . . . , 6. Since s k ,i =s k −1,i−1 x i +s k ,i−1 , one can compute the a k by the following scheme (the horizontal and vertical arrows denote multiplica- tions and additions respectively):

12. We shall prove by induction on m that P m (x, y, z) is symmetric and that (x + y)P m (x, z, y + 1) − (x + z)P m (x, y, z + 1) = (y − z)P m (x, y, z)

(1) holds for all x , y, z. This is trivial for m = 0. Assume now that it holds for m =

n − 1. Since obviously P n (x, y, z) = P n (y, x, z), the symmetry of P n will follow if we prove that P n (x, y, z) = P n (x, z, y). Using (1) we have P n (x, z, y) − P n (x, y, z) =

(y + z)[(x + y)P n −1 (x, z, y + 1) − (x + z)P n −1 (x, y, z + 1)] − (y 2 −z 2 )P n −1 (x, y, z) = (y + z)(y − z)P n −1 (x, y, z) − (y 2 −z 2 )P n −1 (x, y, z) = 0. It remains to prove (1) for m = n. Using the already established symmetry we have

(x + y)P n (x, z, y + 1) − (x + z)P n (x, y, z + 1) = (x + y)P n (y + 1, z, x) − (x + z)P n (z + 1, y, x) = (x + y)[(y + x + 1)(z + x)P

n −1 (y + 1, z, x + 1) − x P n −1 (y + 1, z, x)] −(x + z)[(z + x + 1)(y + x)P 2

n −1 (z + 1, y, x + 1) − x P n −1 (z + 1, y, x)] = (x + y)(x + z)(y − z)P 2

n −1 (x + 1, y, z) − x (y − z)P n −1 (x, y, z) = (y − z)P n (z, y, x) = (y − z)P n (x, y, z),

as claimed.

13. If m and n are relatively prime, there exist positive integers p , q such that pm = qn + 1. Thus by putting m balls in some boxes p times we can achieve that one box receives q + 1 balls while all others receive q balls. Repeating this process sufficiently many times, we can obtain an equal distribution of the balls. Now assume gcd (m, n) > 1. If initially there is only one ball in the boxes, then after k operations the number of balls will be 1 + km, which is never divisible by n . Hence the task cannot be done.

4.26 Shortlisted Problems 1985 487

14. It suffices to prove the existence of a good point in the case of exactly 661 −1’s. We prove by induction on k that in any arrangement with 3k + 2 points k of which are −1’s a good point exists. For k = 1 this is clear by inspection. Assume that the assertion holds for all arrangements of 3n + 2 points and consider an arrangement of 3 (n + 1) + 2 points. Now there exists a sequence of consecutive −1’s surrounded by two +1’s. There is a point P which is good for the arrange- ment obtained by removing the two +1’s bordering the sequence of −1’s and one of these −1’s. Since P is out of this sequence, clearly the removal either leaves a partial sum as it was or diminishes it by 1, so P is good for the original arrangement.

Second solution. Denote the number on an arbitrary point by a 1 , and the num- bers on successive points going in the positive direction by a 2 ,a 3 , . . . (in partic-

ular, a k +1985 =a k ). We define the partial sums s 0 = 0, s n =a 1 +a 2 + ··· + a n for all positive integers n; then s k +1985 =s k +s 1985 and s 1985 ≥ 663. Since s 1985m ≥ 663m and 3 · 663m > 1985(m + 2) + 1 for large m, not all values 0,1,2,...663m can appear thrice among the 1985 (m + 2) + 1 sums s −1985 ,s −1984 ,...,s 1985 (m+1) (and none of them appears out of this set). Thus there is an integral value s >0 that appears at most twice as a partial sum, say s k =s l = s, k < l. Then either a k or a l is a good point. Actually, s i > s must hold for all i > l, and s i < s for all

i < k (otherwise, the sum s would appear more than twice). Also, for the same reason there cannot exist indices p , q between k and l such that s p > s and s q < s; i.e., for k < p < l, s p ’s are either all greater than or equal to s, or smaller than or equal to s. In the former case a k is good, while in the latter a l is good.

15. There is no loss of generality if we as- C ′

sume K = ABCD, K ′ = AB ′ C ′ D ′ , and

2 that K ′ is obtained from K by a clock-

wise rotation around A by φ ,0 ≤ φ ≤

1 to AB through D ′ meet the line BC at

π /4. Let C ′ D ′ ,B ′ C ′ , and the parallel

5 4 C D E ′′ G ′ choose points E ′

, F, and G respectively. Let us now

′ ,G ′ E ′ ∈ AB

2 3 AD ′ , and E ′′ ∈ AD such that the trian- gles AE ′ ′ and AC ′′ ′′ are translates

∈ AB, C ′′ ∈

D E G ′′ E A of the triangles D ′ EG and FC ′ E respectively. Since AE ′ =D ′ E and AC ′′ = FC ′ ,

we have C ′′ E ′′ =C ′ E =B ′ E ′ and C ′′ D ′ =B ′ F , which imply that △E ′′ C ′′ D ′ is a translate of △E ′ B ′ F , and consequently E ′′ D ′ =E ′ F and E ′′ D ′ kE ′ F . It follows that there exist points H ∈ CD, H ′ ∈ BF, and D ′′ ∈E ′ G ′ such that E ′′ D ′ HD is a translate of E ′ FH ′ D ′′ . The remaining parts of K and K ′ are the rectangles

D ′ GCH and D ′′ H ′ BG ′ of equal area. We shall now show that two rectangles with parallel sides and equal areas can be decomposed into translation invariant parts. Let the sides of the rectangles XY ZT and X ′ Y ′ Z ′ T ′ (XY kX ′ Y ′ ) satisfy X ′ Y ′ < XY , Y ′ Z ′ > Y Z, and X ′ Y ′ ·Y ′ Z ′ = XY ·

YZ . Suppose that 2X ′ Y ′ > XY (otherwise, we may cut off congruent rectangles from both the original ones until we reduce them to the case of 2X ′ Y ′ > XY ).

488 4 Solutions Let U ∈ XY and V ∈ ZT be points such that YU = TV = X ′ Y ′ and W ∈ XV

be a point such that UW k XT . Then translating △XUW to a triangle V ZR and △XV T to a triangle W RS results in a rectangle UY RS congruent to X ′ Y ′ Z ′ T ′ . Thus we have partitioned K and K ′ into translation-invariant parts. Although not all the parts are triangles, we may simply triangulate them.

16. Let the three circles be α (A, a), β (B, b), and γ (C, c), and assume c ≤ a,b. We denote by R X ,ϕ the rotation around X through an angle ϕ . Let PQR be an equilateral triangle, say of positive orientation (the case of negatively oriented △PQR is analogous), with P ∈ α ,Q ∈ β , and R ∈ γ . Then Q =R P ,−60 ◦ (R) ∈ R P ,−60 ◦ ( γ )∩ β .

Since the center of R P ,−60 ◦ ( γ ) is R P ,−60 ◦ (C) = R C ,60 ◦ (P) and it belongs to R C ,60 ◦ ( α ), the union of circles R P ,−60 ◦ ( γ ) as P varies on α is the annulus U with center A ′ =R C ,60 ◦ (A) and radii a − c and a + c. Hence there is a solution if and only if U ∩ β is nonempty.

17. The statement of the problem is equivalent to the statement that there is one and only one a such that 1 − 1/n < f n (a) < 1 for all n. We note that each f n is

a polynomial with positive coefficients, and therefore increasing and convex in R + . Define x n and y n by f n (x n ) = 1 − 1/n and f n (y n ) = 1. Since

− n n =1− n and f n +1 (y n ) = 1 + 1/n, it follows that x n <x n +1 <y n +1 <y n . Moreover, the

f n +1 (x n )= 1 −

convexity of f n together with the fact that f n (x) > x for all x > 0 implies that y n −x n <f n (y n )−f n (x n ) = 1/n. Therefore the sequences have a common limit

a , which is the only number lying between x n and y n for all n. By the definition of x n and y n , the statement immediately follows.

18. Set y

x i +1 x i +2 , where x n +i =x i . Then ∏ i =1 y i = 1 and the inequality to be proved becomes ∑ n

i =1 1 +y i ≤ n − 1, or equivalently

i =1 1 +y

We prove this inequality by induction on n. Since 1 1 +y + 1 1 +y −1 = 1, the inequality is true for n = 2. Assume that it is true

for n − 1, and let there be given y 1 ,...,y n > 0 with ∏ n i =1 y i = 1. Then 1 1 +y n −1 +

1 +y n > 1 +y n −1 y n , which is equivalent to 1 +y n y n −1 (1 + y n +y n −1 ) > 0. Hence by the inductive hypothesis

1 n −2 1 1

1 1 i ≥ 1. =1 +y i i =1 +y i +y n −1 y n

4.26 Shortlisted Problems 1985 489 Remark. The constant n − 1 is best possible (take for example x i =a i with a

arbitrarily large).

19. Suppose that for some n > 6 there is a regular n-gon with vertices having integer coordinates, and that A 1 A 2 ...A n is the smallest such n-gon, of side length a. If O is the origin and B i the point such that −−→ OB −−−−→ i = A i −1 A i ,i = 1, 2, . . . , n (where

A 0 =A n ), then B i has integer coordinates and B 1 B 2 ...B n is a regular polygon of side length 2a sin ( π /n) < a, which is impossible. It remains to analyze the cases n ≤ 6. If P is a regular n-gon with n = 3,5,6, then its center C has rational coordinates. We may suppose that C also has integer coordinates and then rotate P around C thrice through 90 ◦ , thus obtaining a regular 12-gon or 20-gon, which is impossible. Hence we must have n = 4 which is indeed a solution.

20. Let O be the center of the circle touching the three sides of BCDE and let F ∈ (ED) be the point such that EF = EB. Then ∠EFB = 90 ◦ − ∠E/2 = ∠C/2 = ∠OCB, which implies that B ,C, F, O lie on a circle. It follows that ∠DFC = ∠OBC = ∠B/2 = 90 ◦ − ∠D/2 and consequently ∠DCF = ∠DFC. Hence ED =

EF + FD = EB + CD. Second solution. Let r be the radius of the small circle and let M , N be the points

of tangency of the circle with BE and CD respectively. Then

∠D ∠E EM = rcot E, DN = rcot D, MB = r cot

The statement follows from the identity cotx + tan(x/2) = 1/sin x.

21. Let B 1 and C 1 be the points on the rays AC and AB respectively such that X B 1 = XC = XB = XC 1 . Then ∠X B 1 C = ∠XCB 1 = ∠ABC and ∠XC 1 B = ∠XBC 1 =

∠ACB, which imply that B 1 , X ,C 1 are collinear and △AB 1 C 1 ∼ △ABC. More- over, X is the midpoint of B 1 C 1 because X B 1 = XC 1 , from which we conclude that △AXB 1 ∼ △AMB. Therefore ∠CAX = ∠BAM and

= cos α .

AX XB 1 BC

22. Assume that △ABC is acute (the case of an obtuse △ABC is similar). Let S and R be the centers of the circumcircles of △ABC and △KBN, respectively. Since ∠BNK = ∠BAC, the triangles BNK and BAC are similar. Now we have ∠CBR = ∠ABS = 90 ◦ − ∠ACB, which gives us BR ⊥ AC and consequently BR k OS . Similarly BS ⊥ KN implies that BS k OR. Hence BROS is a parallelogram. Let L be the point symmetric to B with respect to R. Then RLOS is also a par- allelogram, and since SR ⊥ BM, we obtain OL ⊥ BM. However, we also have LM ⊥ BM, from which we conclude that O,L,M are collinear and OM ⊥ BM.

490 4 Solutions Second solution. The lines BM, NK,

B and CA are the radical axes of pairs

M of the three circles, and hence they

RK intersect at a single point P. Also,

N the quadrilateral MNCP is cyclic. Let S L

OA = OC = OK = ON = r. We then O

have BM · BP = BN · BC = OB 2 −r 2 , P

PM · PB = PN · PK = OP 2 −r 2 . It follows that OB 2 − OP 2 = BP(BM−

PM ) = BM 2 − PM 2 , which implies that OM ⊥ MB.

4.27 Shortlisted Problems 1986 491