Solutions to the Shortlisted Problems of IMO 1994

4.35 Solutions to the Shortlisted Problems of IMO 1994

1. Obviously a 0 >a 1 >a 2 > ··· . Since a k −a k +1 =1− 1 a k +1 , we have a n =a 0 + (a 1 −a 0 ) + ···+ (a n −a n −1 ) = 1994 − n + 1 a 1 0 +1 + ···+ a n −1 +1 > 1994 − n. Also, for 1 ≤ n ≤ 998,

a 997 +1 because as above, a 997 > 997. Hence ⌊a n ⌋ = 1994 − n.

2. We may assume that a 1 >a 2 > ··· > a m . We claim that for i = 1, . . . , m,

a i +a m +1−i ≥ n + 1. Indeed, otherwise a i +a m +1−i ,...,a i +a m −1 ,a i +a m are

i different elements of A greater than a i , which is impossible. Now by adding for

i = 1, . . . , m we obtain 2(a 1 + ··· + a m ) ≥ m(n + 1), and the result follows.

3. The last condition implies that f (x) = x has at most one solution in (−1,0) and at most one solution in (0, ∞). Suppose that for u ∈ (−1,0), f (u) = u. Then

putting x = y = u in the given functional equation yields f (u 2 + 2u) = u 2 + 2u. Since u

2 ∈ (−1,0) ⇒ u 2 + 2u ∈ (−1,0), we deduce that u + 2u = u, i.e., u = −1 or u = 0, which is impossible. Similarly, if f (v) = v for v ∈ (0,∞), we are led to

the same contradiction. However, for all x ∈ S we have

f (x + (1 + x) f (x)) = x + (1 + x) f (x),

1 +x for all x ∈ S. It is directly verified that this function satisfies all the conditions.

so we must have x

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

4. Suppose that α = β . The given functional equation for x = y yields f (x/2) = x −α f (x) 2 /2; hence the functional equation can be written as

f (x) f (y) = x α −α

y f (y) 2 + y α x −α f (x) 2 ,

i.e.,

f (x) = 0. Hence f (x)/x α = f (y)/y α for all x

(x/y) α/2 f (y) − (y/x) α/2

,y∈R + , so f (x) = λ x α for some λ . Substi- tuting into the functional equation we obtain that λ =2 1 −α or λ = 0. Thus either

f (x) ≡ 2 1 −α x α or f (x) ≡ 0. Now let α 6= β . Interchanging x with y in the given equation and subtracting these equalities from each other, we get (x α −x β ) f (y/2) = (y α −y β ) f (x/2), so for some constant λ ≥ 0 and all x 6= 1, f (x/2) = λ (x α −x β ). Substituting this into the given equation, we obtain that only λ = 0 is possible, i.e., f (x) ≡ 0.

5. If f (n)

(x) = (x) p n q n (x) for some positive integer n and polynomials p n ,q n , then

p n (n+1) 2 (x) +q n (x) 2

2p n (x)q n (x)

586 4 Solutions Note that f (0) (x) = x/1. Thus f (n)

q n (x) , where the sequence of polynomi- als p n ,q n is defined recursively by

p n (x) = (x)

p 0 (x) = x, q 0 (x) = 1, and p n +1 (x) = p n (x) 2 +q n (x) 2 ,q n +1 (x) = 2p n (x)q n (x).

Furthermore, p 0 (x) ± q 0 (x) = x ± 1 and p n +1 (x) ± q n +1 (x) = p n n (x) 2 +q n (x) 2 ± 2p n (x)q n (x) = (p n

n (x)) (x) ± q 2 , so p n (x) ± q n (x) = (x ± 1) 2 for all n. Hence

n (x)q n +1 (x)

2p n (x) 2 ((x + 1) 2 + (x − 1) 2 ) 2

(n+1)

f n (x) n (x)p n +1 (x) p n +1 (x) (x + 1) 2 +1 +1 + (x − 1) 2

6. Call the first and second player M and N respectively. N can keep A ≤ 6. Indeed, let 10 dominoes be placed as

shown in the picture, and whenever M

marks a 1 in a cell of some domino,

let N mark 0 in the other cell of that

domino if it is still empty. Since any

3 × 3 square contains at least three 12345 complete dominoes, there are at least

three 0’s inside. Hence A ≤ 6. We now show that M can make A = 6. Let him start by marking 1 in c3. By sym- metry, we may assume that N’s response is made in row 4 or 5. Then M marks

1 in c2. If N puts 0 in c1, then M can always mark two 1’s in b × {1,2,3} as well as three 1’s in {a,d} × {1,2,3}. Thus either {a,b,c} × {1,2,3} or {b,c,d} × {1,2,3} will contain six 1’s. However, if N does not play his sec- ond move in c1, then M plays there, and thus he can easily achieve to have six 1’s either in {a,b,c} × {1,2,3} or {c,d,e} × {1,2,3}.

be the ages of the male citizens (m ≥ 1). We claim that the age of each female citizen can be expressed in the form c 1 a 1 + ··· + c m a m for some constants c i ≥ 0, and we will prove this by induction on the number n of female citizens. The claim is clear if n = 1. Suppose it holds for n and consider the case of n + 1 female citizens. Choose any of them, say A of age x who knows k citizens (at least one male). By the induction hypothesis, the age of each of the other n

7. Let a 1 ,a 2 ,...,a m

females is expressible as c 1 a 1 + ··· + c m a m +c 0 x , where c i ≥ 0 and c 0 +c 1 + ··· + c m = 1. Consequently, the sum of ages of the k citizens who know A is

4.35 Shortlisted Problems 1994 587 kx =b 1 a 1 + ···+b m a m +b 0 x for some constants b i ≥ 0 with sum k. But A knows

at least one male citizen (who does not contribute to the coefficient of x), so

b b 1 0 a ≤ k − 1. Hence x = 1 +···+b m a k m −b 0 , and the claim follows.

8. (a) Let a , b, c, a ≤ b ≤ c be the amounts of money in dollars in Peter’s first, second, and third account, respectively. If a = 0, then we are done, so sup- pose that a > 0. Let Peter make transfers of money into the first account as

follows. Write b = aq + r with 0 ≤ r < a and let q = m 0 + 2m 1 + ···+ 2 k m k

be the binary representation of q (m i ∈ {0,1}, m k = 1). In the ith transfer,

i = 1, 2, . . . , k + 1, if m i = 1 he transfers money from the second account, while if m i = 0 he does so from the third. In this way he has transferred

exactly (m 0 + 2m 1 + ··· + 2 k m k )a dollars from the second account, thus leaving r dollars in it, r < a. Repeating this procedure, Peter can diminish the amount of money in the smallest account to zero, as required.

(b) If Peter has an odd number of dollars, he clearly cannot transfer his money into one account.

9. (a) For i = 1, . . . , n, let d i

be 0 if the card i is in the ith position, and 1 otherwise. Define b =d 1 + 2d 2 +2 2 d 3 + ···+ 2 n −1 d n , so that 0

≤b≤2 n −1, and b = 0 if and only if the game is over. After each move some digit d l changes from

1 to 0 while d l +1 ,d l +2 , . . . remain unchanged. Hence b decreases after each move, and consequently the game ends after at most 2 n − 1 moves. (b) Suppose the game lasts exactly 2 n − 1 moves. Then each move decreases

b for exactly one, so playing the game in reverse (starting from the final configuration), every move is uniquely determined. It follows that if the configuration that allows a game lasting 2 n − 1 moves exists, it must be unique. Consider the initial configuration 0 , n, n −1,...,2,1. We prove by induction that the game will last exactly 2 n − 1 moves, and that the card 0 will get to the 0th position only in the last move. This is trivial for n = 1, so suppose that the claim is true for some n = m − 1 ≥ 1 and consider the case n = m . Obviously the card 0 does not move until the card m gets to the 0-th position. But if we ignore the card 0 and consider the card m to be the card 0, the induction hypothesis gives that the card m will move to the 0th position only after 2 m −1 − 1 moves. After these 2 m −1 − 1 moves, we come to the configuration 0 , m − 1,...,2,1,m. The next move yields m,0,m −

1 , . . . , 2, 1, so by the induction hypothesis again we need 2 m −1 − 1 moves more to finish the game.

10. (a) The case n > 1994 is trivial. Suppose that n = 1994. Label the girls G 1 to

G 1994 , and let G 1 initially hold all the cards. At any moment give to each card the value i, i = 1, . . . , 1994, if G i holds it. Define the characteristic C of

a position as the sum of all these values. Initially C = 1994. In each move, if G i passes cards to G i −1 and G i +1 (where G 0 =G 1994 and G 1995 =G 1 ),

C changes for ±1994 or does not change, so that it remains divisible by

588 4 Solutions 1994. But if the game ends, the characteristic of the final position will be

C = 1 + 2 + ···+ 1994 = 997 · 1995, which is not divisible by 1994. (b) Whenever a card is passed from one girl to another for the first time, let the girls sign their names on it. Thereafter, if one of them passes a card to her neighbor, we shall assume that the passed card is exactly the one signed by both of them. Thus each signed card is stuck between two neighboring girls, so if n < 1994, there are two neighbors who never exchange cards. Consequently, there is a girl G who played only a finite number of times. If her neighbor plays infinitely often, then after her last move, G will con- tinue to accumulate cards indefinitely, which is impossible. Hence every girl plays finitely many times.

11. Tile the table with dominoes and num- 4 3 2 1 bers as shown in the picture. The sec-

1 2 3 4 ond player will not lose if whenever

2 1 1 2 3 the first player plays in a cell of a 4 1 2 3 4 4 3 domino, he plays in the other cell of

4 3 2 1 the same domino. However, if the first

1 2 3 4 4 3 2 1 player plays in a cell with a number,

2 1 1 2 3 4 the second plays in the cell with same

1 2 3 4 4 3 4 3 2 1 number that is diagonally adjacent.

12. Define S n recursively as follows: Let S 2 = {(0,0),(1,1)} and S n +1 =S n ∪T n , where T n = {(x + 2 n −1 ,y+M n ) | (x,y) ∈ S n }, with M n chosen large enough so that the entire set T n lies above every line passing through two points of S n . By definition, S n has exactly 2 n −1 points and contains no three collinear points. We claim that no 2n points of this set are the vertices of a convex 2n-gon. Consider an arbitrary convex polygon P with vertices in S n . Join by a diagonal

d the two vertices of P having the smallest and greatest x-coordinates. This diagonal divides P into two convex polygons P 1 ,P 2 , the former lying above

d . We shall show by induction that both P 1 ,P 2 have at most n vertices. Assume to the contrary that P 1 has at least n + 1 vertices A 1 (x 1 ,y 1 ), . . . , A n +1 (x n +1 ,y n +1 ) in S n , with x 1 < ··· < x n +1 . It follows that

By the induction hypothesis, not more than n − 1 of these vertices belong to S n −1 or T n −1 , so let A k −1 ,A k ∈S n −1 ,A k +1

∈T n −1 . But by the construction of T , k +1 −y k

x k +1 −x k > x k −x k , which gives a contradiction. Similarly, P −1 2 has no more than n vertices, and therefore P itself has at most 2n − 2 vertices.

y k −y k n −1

13. Extend AD and BC to meet at P, and let Q be the foot of the perpendicular from P to AB. Denote by O the center of Γ . Since △PAQ ∼ △OAD and △PBQ ∼

OC = BC . Therefore QB · CP · DA = 1, so by the converse Ceva theorem, AC, BD, and PQ are concurrent. It follows that Q ≡ F.

AQ

△OBC, we obtain AQ = PQ = PQ BQ BC PD

AD OD

4.35 Shortlisted Problems 1994 589 Finally, since the points O ,C, P, D, F are concyclic, we have ∠DFP = ∠DOP =

∠POC = ∠PFC.

14. Although it does not seem to have been noticed at the jury, the statement of the problem is false. For A (0, 0), B(0, 4),C(1, 4), D(7, 0), we have M(4, 2), P(2, 1), Q (2, 3) and N(9/2, 1/2) 6∈ △ABM. The official solution, if it can be called so, actually shows that N lies inside ABCD and goes as follows: The case AD = BC is trivial, so let AD > BC. Let L

be the midpoint of AB. Complete the parallelograms ADMX and BCMY . Now N = DX ∩ CY , so let CY and DX intersect AB at K and H respectively. From LX = LY and

AD AD AD BC LY we get HL < KL, and the statement follows.

15. We shall prove that AD is a common tangent of ω and ω 2 . Denote by K , L the points of tangency of ω with l 1 and l 2 respectively. Let r ,r 1 ,r 2 be the radii of ω , ω 1 , ω 2 respectively, and set KA = x, LB = y. It will be enough if we show that xy = 2r 2 , since this will imply that △KLB and △AKO are similar, where O is the center of ω , and consequently that OA ⊥ KD (because D ∈ KB). Now if O 1 is the center of ω 1 , we have x 2 = KA 2 = OO 2 1 − (KO − AO 1 ) 2 = (r + r 1 ) 2 − (r − r 1 ) 2 = 4rr 1 and analogously y 2 = 4rr 2 . But we also have (r 1 +r 2 ) 2 =O 1 O 2 2 = (x − y) 2 + (2r − r 1 −r 2 ) 2 , so x 2 − 2xy + y 2 = 4r(r 1 +r 2 − r), from which we obtain xy = 2r 2 as claimed. Hence AD is tangent to both ω , ω 2 , and similarly BC is tangent to ω , ω 1 . It follows that Q lies on the radical axes of pairs of circles ( ω , ω 1 ) and ( ω , ω 2 ). Therefore Q also lies on the radical axis of ( ω 1 , ω 2 ), i.e., on the common tangent at E of ω 1 and ω 2 . Hence QC = QD = QE.

Second solution. An inversion with center at D maps ω and ω 2 to parallel lines, ω 1 and l 2 to disjoint equal circles touching ω , ω 2 , and l 1 to a circle externally tangent to ω 1 ,l 2 , and to ω . It is easy to see that the obtained picture is symmetric (with respect to a diameter of l 1 ), and that line AD is parallel to the lines ω and ω 2 . Going back to the initial picture, this means that AD is a common tangent of ω and ω 2 . The end is like that in the first solution.

16. First, assume that ∠OQE = 90 ◦ . Extend PN to meet AC at R. Then OEPQ and ORFQ are cyclic quadrilaterals; hence we have ∠OEQ = ∠OPQ = ∠ORQ = ∠OFQ. It follows that △OEQ ∼ = △OFQ and QE = QF. Now suppose QE = QF. Let S be the

A point symmetric to A with respect to Q, so that the quadrilateral AESF is a par-

allelogram. Draw the line E ′ F ′ through

Q Q so that ∠OQE ′ = 90 ◦ and E ′

∈ AB,

F ′ ∈ AC. By the first part QE ′ = QF ′ ;

hence AE ′ SF ′ is also a parallelogram. O It follows that E ≡E ′ ,F ≡F ′ , and

C ∠OQE = 90 ◦ .

590 4 Solutions

17. We first prove that AB cuts OE in

a fixed point H. Note that ∠OAH = ∠OMA = ∠OEA (because O, A, E, M

F C lie on a circle); hence △OAH ∼

△OEA. This implies OH · OE = OA A i.e., H is fixed. Let the lines AB and CD

meet at K. Since EAOBM and ECDM are cyclic, we have ∠EAK = ∠EMB =

O ∠ECK, so ECAK is cyclic.

B Therefore ∠EKA = 90 ◦ , hence EKBD is also cyclic and EK k OM. Then

∠EKF = ∠EBD = ∠EOM = ∠OEK, from which we deduce that KF = FE. However, since ∠EKH = 90 ◦ , the point F is the midpoint of EH; hence it is fixed.

18. Since for each of the subsets {1,4,9}, {2,6,12}, {3,5,15} and {7,8,14} the product of its elements is a square and these subsets are disjoint, we have |M| ≤ 11. Suppose that |M| = 11. Then 10 ∈ M and none of the disjoint subsets {1,4,9},{2,5},{6,15},{7,8,14} is a subset of M. Consequently {3,12} ⊂ M, so none of {1},{4},{9},{2,6},{5,15}, and {7,8,14} is a subset of M: thus |M| ≤ 9, a contradiction. It follows that |M| ≤ 10, and this number is attained in the case M = {1,4,5,6,7,10,11,12,13,14}.

19. Since mn

3 − 1 and m 3 are relatively prime, mn − 1 divides n + 1 if and only if it

divides m 3 (n 3 + 1) = (m 3 n 3 − 1) + m 3 + 1. Thus

hence we may assume that m ≥ n. If m = n, then n 3 n +1 2 −1 =n+ 1 n −1 is an integer, so m = n = 2. If n = 1, then 2 m −1 ∈ Z, which happens only when m = 2 or m

= 3. Now suppose m > n ≥ 2. Since m 3 + 1 ≡ 1 and mn − 1 ≡ −1 (mod n), we deduce n 3 mn +1 −1 = kn − 1 for some integer k > 0. On the other hand, kn − 1 <

n −1 ≤ 2n−1 gives that k = 1, and therefore n 3 + 1 = (mn −1)(n−1). This yields m = n 2 n +1 −1 =n+1+ 2 n −1 ∈ N, so n ∈ {2,3} and m = 5. The solutions with m < n are obtained by symmetry. There are 9 solutions in total: (1, 2), (1, 3), (2, 1), (3, 1), (2, 2), (2, 5), (3, 5), (5, 2), (5, 3).

n 3 +1 =n+ 1 n 2 −1

20. Let A be the set of all numbers of the form p 1 p 2 ...p p 1 , where p 1 <p 2 < ··· < p p 1 are primes. In other words, A = {2 · 3,2 ·5,...} ∪ {3 · 5 ·7,3 ·5·11,...} ∪ {5 ·

7 · 11 · 13 · 17,...} ∪ ···. This set satisfies the requirements of the problem. Indeed, for any infinite set of primes P = {q 1 ,q 2 , . . . } (where q 1 <q 2 < ··· ) we have

m =q 1 q 2 ···q q 1 ∈ A and n = q 2 q 3 ···q q 1 +1 6∈ A.

4.35 Shortlisted Problems 1994 591

21. Note first that y n =2 k (k ≥ 2) and z k ≡ 1 (mod 4) for all n, so if x n is odd, x n +1 will be even. Further, it is shown by induction on n that y n >z n when x n −1 is even and 2y n >z n >y n when x n −1 is odd. In fact, n = 1 is the trivial case, while if it holds for n ≥ 1, then y n +1 = 2y n >z n =z n +1 if x n is even, and 2y n +1 = 2y n >y n +z n =z n +1 if x n is odd (since then x n −1 is even).

If x 1 = 0, then x 0 = 3 is good. Suppose x n = 0 for some n ≥ 2. Then x n −1 is odd and x n −2 is even, so that y n −1 >z n −1 . We claim that a pair (y n −1 ,z n −1 ), where 2 k =y n −1 >z n −1 > 0 and z n −1 ≡ 1 (mod 4), uniquely determines x 0 =

f (y n −1 ,z n −1 ). We see that x n −1 = 1 2 y n −1 +z n −1 , and define (x k ,y k ,z k ) backwards as follows, until we get (y k ,z k ) = (4, 1). If y k >z k , then x k −1 must have been even, so we define (x k −1 ,y k −1 ,z k −1 ) = (2x k ,y k /2, z k ); otherwise x k −1 must have been odd, so we put (x k −1 ,y k −1 ,z k −1 ) = (x k −y k /2 + z k ,y k ,z k −y k ). We eventu-

ally arrive at (y 0 ,z 0 ) = (4, 1) and a good integer x 0 = f (y n −1 ,z n −1 ), as claimed. Thus for example (y n −1 ,z n −1 ) = (64, 61) implies x n −1 = 93, (x n −2 ,y n −2 ,z n −2 )= (186, 32, 61) etc., and x 0 = 1953, while in the case of (y n −1 ,z n −1 ) = (128, 1) we get x 0 = 2080. Note that y ′ > y ⇒ f (y ′ ,z ′ ) > f (y, z) and z ′ > z ⇒ f (y,z ′ ) > f (y, z). Therefore there are no y , z for which 1953 < f (y, z) < 2080. Hence all good integers less than or equal to 1994 are given as f (y, z), y = 2 k ≤ 64 and 0 < z ≡ 1 (mod 4), and the number of such (y, z) equals 1 + 2 + 4 + 8 + 16 = 31. So the answer is

22. (a) Denote by b (n) the number of 1’s in the binary representation of n. Since

b (2k + 2) = b(k + 1) and b(2k + 1) = b(k) + 1, we deduce that

f (k) + 1, if b(k) = 2;

f (k + 1) =

otherwise. The set of k’s with b (k) = 2 is infinite, so it follows that f (k) is unbounded.

f (k),

Hence f takes all natural values. (b) Since f is increasing, k is a unique solution of f (k) = m if and only if

f (k − 1) < f (k) < f (k + 1). By (1), this inequality is equivalent to b(k −

1 ) = b(k) = 2. It is easy to see that then k − 1 must be of the form 2 t +1 for some t. In this case, {k + 1,...,2k} contains the number 2 t +1 +3=

10 . . . 011 2 and t (t−1) 2 binary (t + 1)-digit numbers with three 1’s, so m =

f (k) = t (t−1) 2 + 1.

23. (a) Let p be a prime divisor of x i ,i > 1, and let x j ≡u j (mod p) where 0 ≤ u j ≤ p − 1 (particularly u i ≡ 0). Then u j +1 ≡u j u j −1 + 1 (mod p). The number of possible pairs (u j ,u j +1 ) is finite, so u j is eventually periodic. We claim that for some d p > 0, u i +d p = 0. Indeed, suppose the contrary and let (u m ,u m +1 ,...,u m +d−1 ) be the first period for m ≥ i. Then m 6= i. By the assumption u m −1 6≡ u m +d−1 , but u m −1 u m ≡u m +1 −1≡u m +d+1 −1≡ u m +d−1 u m +d ≡u m +d−1 u m (mod p), which is impossible if p ∤u m . Hence there is a d p with u i =u i +d p = 0 and moreover u i +1 =u i +d p +1 = 1, so the sequence u j is periodic with period d p starting from u i . Let m be the least

592 4 Solutions common multiple of all d p ’s, where p goes through all prime divisors of x i .

Then the same primes divide every x i +km ,k = 1, 2, . . . , so for large enough k and j = i + km, x i j i |x j .

(b) If i = 1, we cannot deduce that x i +1 ≡ 1 (mod p). The following example shows that the statement from (a) need not be true in this case. Take x 1 = 22 and x 2 = 9. Then x n is even if and only if n ≡ 1 (mod 3), but modulo 11 the sequence {x n } is 0,9,1,10,0,1,1,2,3,7,0,..., so 11 | x n (n > 1) if and only if n ≡ 5 (mod 6). Thus for no n > 1 can we have 22 | x n .

24. A multiple of 10 does not divide any wobbly number. Also, if 25 | n, then every multiple of n ends with 25, 50, 75, or 00; hence it is not wobbly. We now show that every other number n divides some wobbly number.

(i) Let n be odd and not divisible by 5. For any k ≥ 1 there exists l such that (10 k − 1)n divides 10 l − 1, and thus also divides 10 kl − 1. Consequently, v k = 10 10 kl k −1 −1 is divisible by n, and it is wobbly when k = 2 (indeed, v 2 =

101 . . . 01). If n is divisible by 5, one can simply take 5v 2 instead. (ii) Let n be a power of 2. We prove by induction on m that 2 2m +1 has a wobbly multiple w m with exactly m nonzero digits. For m = 1, take w 1 = 8. Sup- pose that for some m ≥ 1 there is a wobbly w m =2 2m +1 d m . Then the num- bers a

· 10 2m +w m are wobbly and divisible by 2 2m +1 when a ∈ {2,4,6,8}. Moreover, one of these numbers is divisible by 2 2m +3 . Indeed, it suffices to

choose a such that a 2 +d m is divisible by 4. This proves the induction step. (iii) Let n =2 m r , where m ≥ 1 and r is odd, 5 ∤ r. Then v 2m w m is wobbly and divisible by both 2 m and r (using notation from (i), r |v 2m ).

4.36 Shortlisted Problems 1995 593