Solutions to the Shortlisted Problems of IMO 1987

4.28 Solutions to the Shortlisted Problems of IMO 1987

1. By (ii), f (x) = 0 has at least one solution, and there is the greatest among them, say x 0 . Then by (v), for any x,

0 = f (x) f (x 0 ) = f (x f (x 0 )+x 0 f (x) − x 0 x ) = f (x 0 ( f (x) − x)). (1)

It follows that x 0 ≥x 0 ( f (x) − x).

Suppose x 0 > 0. By (i) and (iii), since f (x 0 )−x 0 < 0 < f (0) − 0, there is a number z between 0 and x 0 such that f (z) = z. By (1), 0 = f (x 0 ( f (z) − z)) =

f (0) = 1, a contradiction. Hence, x 0 < 0. Now the inequality x 0 ≥x 0 ( f (x) − x)

gives f (x) − x ≥ 1 for all x; so, f (1987) ≥ 1988. Therefore f (1987) = 1988.

2. Let d i denote the number of cliques of which person i is a member. Clearly

d i ≥ 2. We now distinguish two cases: (i) For some i, d i = 2. Suppose that i is a member of two cliques, C p and

C q . Then |C p | = |C q | = n, since for each couple other than i and his/her spouse, one member is in C p and one in C q . There are thus (n − 1)(n − 2) pairs (r, s) of nonspouse persons distinct from i, where r ∈ C p ,s ∈C q . We observe that each such pair accounts for a different clique. Otherwise, we find two members of C p or C q who belong to one other clique. It follows that k ≥ 2 + (n − 1)(n − 2) ≥ 2n for n ≥ 4.

(ii) For every i, d i ≥ 3. Suppose that k < 2n. For i = 1,2,...,2n assign to person

i an indeterminant x i , and for j = 1, 2, . . . , k set y = ∑ i ∈C j x i . From linear algebra, we know that if k < 2n, then there exist x 1 ,x 2 ,...,x 2n , not all zero,

such that y 1 =y 2 = ··· = y k = 0.

On the other hand, suppose that y 1 =y 2 = ··· = y k = 0. Let M be the set of the couples and M ′ the set of all other pairs of persons. Then

2n

0 = ∑ y 2 d j 2 = ∑ i x i +2 ∑ x i x j

j =1

i =1

(i, j)∈M ′

2n

= ∑ (d

i − 2)x i + (x 1 +x 2 + ··· + x 2n ) + ∑ (x i −x j )

i =1 (i, j)∈M 2n

≥ ∑ x 2 i > 0,

i =1

if not all x 1 ,x 2 ,...,x 2n are zero, which is a contradiction. Hence k ≥ 2n. Remark. The condition n ≥ 4 is essential. For a party attended by 3 couples {(1,4),(2,5),(3,6)}, there is a collection of 4 cliques satisfying the conditions:

3. The answer: yes. Set (k + m) 2 + 3k + m

p (k, m) = k + [1 + 2 + ···+ (k + m)] = .

2 It is obviously of the desired type.

4.28 Shortlisted Problems 1987 499

4. Setting x −→

−→ , we have to prove that

1 = AB ,x

2 = AD ,x 3 = AE

|x 1 +x 2 | + |x 2 +x 3 | + |x 3 +x 1 | ≤ |x 1 | + |x 2 | + |x 3 | + |x 1 +x 2 +x 3 |. We have

2 +x +x (|x 2

1 | + |x 2 | + |x 3 |) − |x 1 2 3 |

=2 ∑ (|x i ||x j | − hx i ,x j i) = ∑ (|x i | + |x j |) − i +x j

1 ≤i< j≤3

1 ≤i< j≤3

= ∑ (|x i | + |x j | + |x i +x j |)(|x i | + |x j | − |x i +x j |).

1 ≤i< j≤3

The following two inequalities are obvious:

|x i | + |x j | − |x i +x j | ≥ 0,

(1) |x i | + |x j | + |x i +x j | ≤ |x 1 | + |x 2 | + |x 3 | + |x 1 +x 2 +x 3 |.

(2) It follows that

1 2 3 2 (|x 2 | + |x | + |x |) − |x 1 +x 2 +x 3 |

≤ ∑ |x i |+ ∑ x i 2 ∑ |x i |− ∑ |x i +x j | ,

and dividing by the positive number ∑ 3 i =1 |x i |+ 3 i =1 x i

∑ |x i |− ∑ x i ≤2 ∑ |x i |− ∑ |x i +x j |.

The inequality is proven. Let us analyze the cases of equality. If one of the vec- tors is null, then equality obviously holds. Suppose that x i 6= 0, i = 1,2,3. For every i , j, at least one of (1) and (2) is equality. Equality in (1) holds if and only if x i and x j are collinear with the same direction, while in (2) it holds if and only

if −x k and x 1 +x 2 +x 3 are collinear with the same direction. If not all the vectors are collinear, then there are at least two distinct pairs x i ,x j ,i < j, for which (2) is an equality, so at least two of x i are collinear with x 1 +x 2 +x 3 , but then so is the third; hence, the sum x 1 +x 2 +x 3 must be 0. Thus the cases of equality are (a) the vectors are collinear with the same direction; (b) the vectors are collinear, two of them have the same direction, say x i ,x j , and |x k | ≥ |x i | + |x j |; (c) one of the vectors is 0; (d) their sum is 0.

Second solution. The following technique, although not quite elementary, is often used to effectively reduce geometric inequalities of first degree, like this one, to the one-dimensional case. Let σ

be a fixed sphere with center O. For an arbitrary segment d in space, and any line l, we denote by π l (d) the length of the projection of d onto l. Consider the integral of lengths of these projections on all possible directions of OP, with

500 4 Solutions

P moving on the sphere: σ π OP (d) d σ . It is clear that this value depends only on the length of d (because of symmetry); hence

π OP d σ = c · |d|

for some constant c 6= 0. (1)

Notice that by the one-dimensional case, for any point P ∈ σ , π OP (x 1 )+ π OP (x 2 )+ π OP (x 3 )+ π OP (x 1 +x 2 +x 3 ) ≥ π OP (x 1 +x 2 )+ π OP (x 1 +x 3 )+ π OP (x 2 +x 3 ).

By integration on σ , using (1), we obtain

c (|x 1 | + |x 2 | + |x 3 | + |x 1 +x 2 +x 3 |) ≥ c(|x 1 +x 2 | + |x 1 +x 3 | + |x 2 +x 3 |).

5. Assuming the notation a = BC, b = AC, c = AB; x = BL, y = CM, z = AN, from the Pythagorean theorem we obtain

2 2 2 2 2 (a − x) 2 + (b − y) + (c − z) =x +y +z x 2 + (a − x) 2 +y 2 + (b − y) 2 +z 2 + (c − z) 2

Since x 2 + (a − x) 2 =a 2 /2 + (a − 2x) 2 /2 ≥ a 2 /2 and similarly y 2 + (b − y) 2 ≥

b 2 /2 and z 2 + (c − z) 2 ≥c 2 /2, we get

2 2 2 a +b 2 +c x 2 +y +z

Equality holds if and only if P is the circumcenter of the triangle ABC, i.e., when x = a/2, y = b/2, z = c/2.

6. Suppose w.l.o.g. that a ≥ b ≥ c. Then 1/(b + c) ≥ 1/(a + c) ≥ 1/(a + b). Cheby- shev’s inequality yields

≥ (a n +b n +c n )

3 b +c a +c a +b By the Cauchy-Schwarz inequality we have

b +c a +c a +b

b +c a +c a +b ≥ 9, and the mean inequality yields

2 (a + b + c)

(a n n +b n +c n )/3 ≥ [(a + b + c)/3] . We obtain from (1) that

a +b+c n

3 a +b+c n −1

2 n −2

S n −1 .

4.28 Shortlisted Problems 1987 501

7. For all real numbers v the following inequality holds:

∑ 2 (v

i −v j ) 2 ≤5 ∑ (v i − v) .

∑ (v i −v j ) 2 = ∑ [(v i

i − v) − ∑ i − v) ≤5 ∑ i − v) .

Let us first take v i ’s, satisfying condition (1), so that w.l.o.g. v 0 ≤v 1 ≤v 2 ≤v 3 ≤ v 4 ≤1+v 0 . Defining v 5 =1+v 0 , we see that one of the differences v j +1 −v j , j = 0, . . . , 4, is at most 1/5. Take v = (v j +1 +v j )/2, and then place the other three v j ’s in the segment [v − 1/2,v + 1/2]. Now we have |v− v j | ≤ 1/10, |v− v j +1 |≤

1 /10, and |v − v k | ≤ 1/2, for any k different from j, j + 1. The v i ’s thus obtained have the required property. In fact, using the inequality (1), we obtain

2 1 ∑ 1 (v

Remark. The best possible estimate for the right-hand side is 2.

8. (a) Consider

a i = ik + 1,

i = 1, 2, . . . , m;

b j = jm + 1, j = 1, 2, . . . , k. Assume that mk |a i b j −a s b t = (ik + 1)( jm + 1) − (sk + 1)(tm + 1) =

km (i j − st) + m( j − t) + k(i − s). Since m divides this sum, we get that m | k(i − s), or, together with gcd(k,m) = 1, that i = s. Similarly j = t, which proves part (a).

(b) Suppose the opposite, i.e., that all the residues are distinct. Then the residue

0 must also occur, say at a 1 b 1 : mk |a 1 b 1 ; so, for some a ′ and b ′ ,a ′ |a 1 ,

b ′ |b 1 , and a ′ b ′ = mk. Assuming that for some i, s 6= i, a ′ |a i −a s , we obtain mk =a ′ b ′ |a i b 1 −a s b 1 , a contradiction. This shows that a ′ ≥ m and similarly b ′ ≥ k, and thus from a ′ b ′ = mk we have a ′ = m, b ′ = k. We also get (∗): all a i ’s give distinct residues modulo m =a ′ , and all b j ’s give distinct residues modulo k =b ′ . Now let p be a common prime divisor of m and k. By

(∗), exactly p −1 p m of a i ’s and exactly p −1 p k of b j ’s are not divisible by p. Therefore there are

precisely (p−1) 2 p 2 mk products a i b j that are not divisible by p, although from the assumption that they all give distinct residues it follows that the number

of such products is 2 p −1 p mk (p−1) 6= p 2 mk . We have arrived at a contradiction, thus proving (b).

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9. The answer is yes. Consider the curve

C = {(x,y,z) | x = t,y = t 3 ,z=t 5 , t ∈ R}. Any plane defined by an equation of the form ax + by + cz + d = 0 intersects

the curve C at points (t,t 3 ,t 5 ) with t satisfying ct 5 + bt 3 + at + d = 0. This last equation has at least one but only finitely many solutions.

10. Denote by r , R (take w.l.o.g. r < R) the radii and by A , B the centers of

v the spheres S 1 ,S 2 respectively. Let s

E 1 in the ring, C the center of one of

be the common radius of the spheres r

A them, say S, and D the foot of the

D CE perpendicular from C to AB. The cen-

s ters of the spheres in the ring form a

regular n-gon with center D, and thus sin ( π /n) = s/CD. Using the Heron’s

formula on the triangle ABC, we obtain (r + R) 2 CD 2 = 4rRs(r + R + s), hence B

CD 4 (r + R + s)rR Choosing the unit of length so that r + R = 2, for simplicity of writing, we write

(1) as 1 /sin 2 ( π /n) = rR (1 + 2/s). Let now v be half the angle at the top of the cone. Then clearly R − r = (R + r)sin v = 2sin v, giving us R = 1 + sin v,r =

1 − sinv. It follows that

We need to express s as a function of R and r. Let E 1 ,E 2 , E be collinear points of tangency of S √ 1 ,S √ 2 , and S with the cone. Obviously, E 1 E 2 =E 1 E +E 2 E , i.e.,

2 √ rs +2 Rs =2 Rr = (R + r) cosv = 2 cos v. Hence, cos 2

v = s( R + r ) 2 = s(R + r + 2 Rr ) = s(2 + 2 cosv). Substituting this into (2), we obtain 2 + cos v = 1/sin( π /n). Therefore 1/3 <

sin ( π /n) < 1/2, and we conclude that the possible values for n are 7, 8, and 9.

11. Let A 1 be the set that contains 1, and let the minimal element of A 2 be less than that of A 3 . We shall construct the partitions with required properties by allocating successively numbers to the subsets that always obey the rules. The number 1 must go to A 1 ; we show that for every subsequent number we have exactly two possibilities. Actually, while A 2 and A 3 are both empty, every successive number can enter either A 1 or A 2 . Further, when A 2 is no longer empty, we use induction

4.28 Shortlisted Problems 1987 503 on the number to be placed, denote it by m: if m can enter A i or A j but not A k ,

and it enters A i , then m + 1 can be placed in A i or A k , but not in A j . The induction step is finished. This immediately gives us that the final answer is 2 n −1 .

B 12. Here all angles will be oriented and ′ measured counterclockwise. Note that

∡CA ′ B = ∡AB ′ C = ∡BC ′ A = π /3.

Let a ′ ,b ′ ,c ′ denote respectively the in- A ner bisectors of angles A ′ ,B ′ ,C ′ in

triangle A ′ B ′ C ′ . The lines a ′ ,b ′ ,c ′ M X

PK and ∡ (a ′

meet at the centroid X of A ′ B ′ C ′ ,

,b C ′ ) = ∡(b ′ ,c ′ ) = ∡(c ′ ,a ′ )= L

B C such that KB = KC, LC = LA, MA =

2 π /3. Now let K, L, M be the points

MB , and ∡BKC = ∡CLA = ∡AMB = A ′

2 π /3, and let C 1 ,C 2 ,C 3 be the

circles circumscribed about triangles BKC , CLA, and AMB respectively. These circles are characterized by C 1 = {Z | ∡BZC =2 π /3}, etc.; hence we deduce that they meet at a point P such that ∡BPC = ∡CPA = ∡APB = 2 π /3 (Torricelli’s point). Points A ′ ,B ′ ,C ′ run over

C 1 r{P}, C 2 r{P}, C 3 r{P} respectively. As for a ′ ,b ′ ,c ′ , we see that K ∈a ′ ,L ∈

b ′ ,M ∈c ′ , and also that they can take all possible directions except KP , LP, MP respectively (if K = P, KP is assumed to be the corresponding tangent at K). Then, since ∡KX L =2 π /3, X runs over the circle defined by {Z | ∡KZL =

2 π /3}, without P. But analogously, X runs over the circle {Z | ∡LZM = 2 π /3}, from which we can conclude that these two circles are the same, both equal to the circumcircle of KLM, and consequently also that triangle KLM is equilateral (which is, anyway, a well-known fact). Therefore, the locus of the points X is the circumcircle of KLM minus point P.

13. We claim that the points P p i (i, i 2 ), i = 1, 2, . . . , 1987, satisfy the conditions. In fact: p (i) P i P j = (i − j) 2 + (i 2 −j 2 ) 2 = |i − j| 1 + (i + j) 2 √ . It is known that for each positive integer n, n is either an integer or an p p irrational number. Since i +j< 1 + (i + j) 2 < i + j + 1, 1 + (i + j) 2 is not an integer, it is irrational, and so is P i P j . (ii) The area A of the triangle P i P j P k , for distinct i , j, k, is given by

i 2 +j 2 j 2 +k 2 k 2 +i 2

2 (i − j) + 2 ( j − k) + 2 (k − i) (i − j)( j − k)(k − i)

also showing that this triangle is nondegenerate.

14. Let x n

be the total number of counted words of length n, and y n ,z n ,u n ,z n ,y n the numbers of counted words of length n starting with 0 , 1, 2, 3, 4, respectively (in- deed, by symmetry, words starting with 0 are equally numbered as those starting

504 4 Solutions with 4, etc.). We have the clear relations (1) y n =z n −1 ;

(2) z n =y n −1 +u n −1 ; (3) u n = 2z n −1 ;

(4) x n = 2y n + 2z n +u n . From (1), (2), and (3) we get z n =z n −2 + 2z n −2 = 3z n −2 , with z 1 = 1, z 2 = 2,

which gives z 2n

n =2·3 −1 ,

z 2n +1 =3 n .

Then (1), (3), and (4) obviously imply y 2n =3 n −1 ,

y 2n +1 =2·3 n −1 ; u 2n =2·3 n −1 ,

u 2n +1 =4·3 n −1 ; x 2n =8·3 n −1 ,

x 2n +1 = 14 · 3 n −1 ;

with the initial number x 1 = 5.

15. Since x 2 +x 2 1 2 2 + ··· + x n = 1, we get by the Cauchy-Schwarz inequality

2 2 2 √ |x 1 | + |x 2 | + ··· + |x n |≤ n (x 1 +x 2 + ··· + x n )= n .

Hence all k n sums of the form e 1 x 1 +e 2 x 2 + ··· + e n x n , with e i √ ∈ {0,1,2,...,k −

1 }, must lie in some closed interval ℑ of length (k − 1) n . This interval can be covered with k n − 1 closed subintervals of length k

n . By the pi- geonhole principle there must be two of these sums lying in the same subin- terval. Their difference, which is of the form e 1 x 1 +e 2 x 2 + ··· + e n x n where

e i ∈ {0,±1,...,±(k − 1)}, satisfies √

16. We assume that S = {1,2,...,n}, and use the obvious fact

∑ p n (k) = n!

k =0

(a) To each permutation π of S we assign an n-vector (e 1 ,e 2 ,...,e n ), where e i is 1 if i is a fixed point of π , and 0 otherwise. Since exactly p n (k) of the assigned vectors contain exactly k “1”s, the considered sum ∑ n k =0 kp n (k) counts all the “1”s occurring in all the n! assigned vectors. But for each

i ,1 ≤ i ≤ n, there are exactly (n − 1)! permutations that fix i; i.e., exactly (n − 1)! of the vectors have e i = 1. Therefore the total number of “1”s is n · (n − 1)! = n!, implying

∑ kp n (k) = n!.

k =0

4.28 Shortlisted Problems 1987 505

(b) In this case, to each permutation π of S we assign a vector (d 1 ,...,d n ) instead, with d i = k if i is a fixed point of π , and d i = 0 otherwise, where k is the number of fixed points of π . Let us count the sum Z of all components d i for all the n! permutations. There are p n (k) such vectors with exactly k components equal to k, and

sums of components equal to k 2 . Thus, Z =∑ n k =0 k 2 p n (k). On the other hand, we may first calculate the sum of all components d i for fixed i. In fact, the value d i = k > 0 will occur exactly p n −1 (k − 1) times, so that the sum of the d i ’s is ∑ n

k =1 kp n −1 (k − 1) = ∑ −1 k =0 (k + 1)p n −1 (k) =

2 (n − 1)!. Summation over i yields

2 k =0 k p n (k) = 2n!. (2) From (0), (1), and (2), we conclude that

Remark. Only the first part of this problem was given on the IMO.

17. The number of 4-colorings of the set M is equal to 4 1987 . Let A be the number of arithmetic progressions in M with 10 terms. The number of colorings containing

a monochromatic arithmetic progression with 10 terms is less than 4A 1977 ·4 . So, if A <4 9 , then there exist 4-colorings with the required property. Now we estimate the value of A. If the first term of a 10-term progression is k and the difference is d, then 1

9 ; hence 1978 1987

≤ k ≤ 1978 and d ≤ 1987 −k

18. Note first that the statement that some a + x, a + y, a + x + y belong to a class C is equivalent to the following statement:

(1) There are positive integers p, q ∈ C such that p < q ≤ 2p. Indeed, given p , q, take simply x = y = q − p, a = 2p − q; conversely, if a,x,y

(x ≤ y) exist such that a + x,a + y,a + x + y ∈ C, take p = a + y, q = a + x + y: clearly, p < q ≤ 2p.

be an arbi- trary partition into r classes. By the pigeonhole principle, two among the r +1 numbers r , r + 1, . . . , 2r belong to the same class, say i, j ∈ C k . If w.l.o.g. i < j, then obviously i < j ≤ 2i, and so by (1) this C k has the required property. On the other hand, we consider the partition

We will show that h (r) = 2r. Let {1,2,...,2r} = C 1 ∪C 2 ∪ ··· ∪C r

r [ −t

{1,2,...,2r − t} = {k,k + r} ∪ {r − t + 1} ∪ ···∪ {r}

k =1

and prove that (1), and thus also the required property, does not hold. In fact, none of the classes in the partition contains p and q with p < q ≤ 2p, because k + r > 2k.

506 4 Solutions

19. The facts given in the problem allow us to draw a triangular pyramid with angles

2 α ,2 β ,2 γ at the top and lateral edges of length 1 /2. At the base there is a triangle whose side lengths are exactly sin α , sin β , sin γ . The area of this triangle does not exceed the sum of areas of the lateral sides, which equals (sin 2 α + sin 2 β + sin 2 γ )/8.

20. Let y be the smallest nonnegative integer with y ≤ p − 2 for which f (y) is a composite number. Denote by q the smallest prime divisor of f (y). We claim that y < q. Suppose the contrary, that y ≥ q. Let r be a positive integer such that y ≡ r (mod q). Then f (y) ≡ f (r) ≡ 0 (mod q), and since q ≤ y ≤ p − 2 ≤ f (r), we conclude that q | f (r), which is a contradiction to the minimality of y. Now, we will prove that q > 2y. Suppose the contrary, that q ≤ 2y. Since

f (y) − f (x) = (y − x)(y + x + 1), we observe that f (y) − f (q − 1 − y) = (2y − q + 1)q, from which it follows that

f (q − 1 − y) is divisible by q. But by the assumptions, q − 1 − y < y, implying that f (q − 1 − y) is prime and therefore equal to q. This is impossible, because

f (q − 1 − y) = (q − 1 − y) 2 + (q − 1 − y) + p > q + p − y − 1 ≥ q. Therefore q ≥ 2y + 1. Now, since f (y), being composite, cannot be equal to q,

and q is its smallest prime divisor, we obtain that f (y) ≥ q 2 . Consequently, y 2 2 +y+p≥q 2 ≥ (2y + 1) = 4y 2 + 4y + 1 ⇒ 3(y 2 + y) ≤ p − 1,

and from this we easily conclude that y < p /3, which contradicts the condition of the problem. In this way, all the numbers

f (0), f (1), . . . , f (p − 2)

must be prime.

21. Let P be the second point of inter- A section of segment BC and the cir-

cle circumscribed about quadrilateral AKLM . Denote by E the intersection point of the lines KN and BC and by

F the intersection point of the lines M MN and BC. Then ∠BCN = ∠BAN

and ∠MAL = ∠MPL, as angles on E P the same arc. Since AL is a bisector,

B F C ∠BCN = ∠BAL = ∠MAL = ∠MPL, and consequently PM

k NC. Similarly N we prove KP k BN. Then the quadrilaterals BKPN and NPMC are trapezoids;

hence S BKE =S NPE

S NPF =S CMF . Therefore S ABC =S AKNM .

and

4.28 Shortlisted Problems 1987 507

22. Suppose that there exists such function f . Then we obtain

f (n + 1987) = f ( f ( f (n))) = f (n) + 1987 for all n ∈ N, and from here, by induction, f (n + 1987t) = f (n) + 1987t for all n,t ∈ N.

Further, for any r ∈ {0,1,...,1986}, let f (r) = 1987k + l, k,l ∈ N, l ≤ 1986. We have

r + 1987 = f ( f (r)) = f (l + 1987k) = f (l) + 1987k,

and consequently there are two possibilities: (i) k = 1 ⇒ f (r) = l + 1987 and f (l) = r; (ii) k = 0 ⇒ f (r) = l and f (l) = r + 1987; in both cases, r 6= l. In this way, the set {0,1,...,1986} decomposes into pairs {a,b} such that

f (b) = a and f (a) = b + 1987. But the set {0,1,...,1986} has an odd number of elements, and cannot be de-

f (a) = b and f (b) = a + 1987,

or

composed into pairs. Contradiction.

23. If we prove the existence of p , q ∈ N such that the roots r,s of

f (x) = x 2 − kp · x + kq = 0

are irrational real numbers with 0 < s < 1 (and consequently r > 1), then we are done, because from r + s, rs ≡ 0 (mod k) we get r m +s m ≡ 0 (mod k), and

0 <s m < 1 yields the assertion. To prove the existence of such natural numbers p and q, we can take them such that f (0) > 0 > f (1), i.e.,

p > q > 0. The irrationality of r can be obtained by taking q = p − 1, because the discrimi-

kq > 0 > k(q − p) + 1

nant D = (kp) 2 − 4kp+ 4k, for (kp− 2) 2 < D < (kp − 1) 2 , is not a perfect square for p ≥ 2.

508 4 Solutions