Solutions to the Shortlisted Problems of IMO 1991

4.32 Solutions to the Shortlisted Problems of IMO 1991

1. All the angles ∠PP 1 C , ∠PP 2 C , ∠PQ 1 C , ∠PQ 2 C are right, hence P 1 ,P 2 , Q 1 ,Q 2 lie on the circle with diame-

A ter PC. The result now follows im- mediately from Pascal’s theorem ap-

X P 1 Q 2 tells us that the points of intersection

plied to the hexagon P 1 PP 2 Q 1 CQ 2 . It

of the three pairs of lines P 1 C (in- , PQ P

1 tersection A), P 1 Q 2 ,P 2 Q 1 (intersection

2 ,P 2 C (intersection B) are Q 1 collinear.

B X P ) and PQ 2 C

2. Let HQ meet PB at Q ′ and HR meet PC at R ′ . From MP = MB = MC we have ∠BPC = 90 o . So PR ′ HQ ′ is

a rectangle. Since PH is perpendicular to BC, it follows that the circle with di-

X Y ameter PH, through P ,R ′ , H, Q ′ , is tan-

gent to BC. It is now sufficient to show P that QR is parallel to Q ′

R ′ . Let CP meet

AB at X , and BP meet AC at Y . Since P Q ′ R is on the median, it follows (for exam-

ple, by Ceva’s theorem) that AX /XB = B H M C AY /YC, i.e. that XY is parallel to BC.

Therefore, PY /BP = PX/CP. Since HQ is parallel to CX, we have QQ ′ /HQ ′ = PX /CP and similarly RR ′ /HR ′ = PY /BP. It follows that QQ ′ /HQ ′ = RR ′ /HR ′ , hence QR is parallel to Q ′ R ′ as required.

Second solution. It suffices to show that ∠RHC = ∠RQH, or equivalently RH : QH = PC : PB. We assume PC : PB = 1 : x. Let X ∈ AB and Y ∈ AC be points such that MX ⊥ PB and MY ⊥ PC. Since MX bisects ∠AMB and MY bisects AMC , we deduce

AX :XB = AM : MB = AY : YC ⇒ XY k BC ⇒ ⇒ △XY M ∼ △CBP ⇒ XM : MY = 1 : x.

Now from CH : HB 2 =1:x 2 we obtain RH : MY = CH : CM = 1 : 1 +x

2 and

QH : MX = BH : BM = x 2 : 1 +x 2 2 . Therefore

3. Consider the problem with the unit circle on the complex plane. For convenience, we use the same letter for a point in the plane and its corresponding complex number.

Lemma 1. Line l (S, PQR) contains the point Z = P +Q+R+S 2 .

4.32 Shortlisted Problems 1991 551 Proof. Suppose P ′ ,Q ′ ,R ′ are the feet of perpendiculars from S to QR, RP, PQ

respectively. It suffices to show that P ′ ,Q ′ ,R ′ , Z are on the same line. Let us first represent P ′ ′ by Q , R, S. Since P ′ P ′ −Q ∈ QR, we have

R −Q = P −Q R −Q , that is, (P ′ − Q)(R − Q) = (P ′ − Q)(R − Q).

(1) On the other hand, since SP ′ ⊥ QR, the ratio P ′ −S R −Q is purely imaginary. Thus

(P ′ − S)(R − Q) = −(P ′ − S)(R − Q). (2) Eliminating P ′ from (1) and (2) and using the fact that X =X −1 for X

on the unit circle, we obtain P ′ = (Q + R + S − QR/S)/2 and analogously Q ′ = (P + R + S − PR/S)/2 and R ′ = (P + Q + S − PQ/S)/2. Hence Z − P ′ = (P + QR/S)/2, Z − Q ′ = (Q + PR/S)/2 and Z − R ′ = (R + PQ/S)/2.

Setting P =p 2 ,Q =q 2 ,R =r 2 ,S =s 2 we obtain Z −P ′ = pqr ps 2s qr + qr ps , Z

pq + pq rs . Since x +x −1 = 2Re x is real for all x on the unit circle, it follows that the ratio of every pair of these differences is real, which means that Z ,P ′ ,Q ′ ,R ′ belong to the same line.

Lemma 2. If P , Q, R, S are four different points on a circle, then the lines l (P, QRS), l(Q, RSP), l(R, SPQ), l(S, PQR) intersect at one point.

Proof. By Lemma 1, they all pass through P +Q+R+S 2 .

Now we can find the needed conditions for A, B, . . . , F. In fact, the lines l (A, BDF), l(D, ABF) meet at Z 1 = A +B+D+F 2 , and l (B, ACE), l(E, ABC) meet at Z 2 = A +B+C+E 2 . Hence, Z 1 ≡Z 2 if and only if D − C = E − F ⇔ CDEF is a rectangle.

Remark. The line l (S, PQR) is widely known as Simson’s line; the proof that the feet of perpendiculars are collinear is straightforward. The key claim, Lemma 1, is a known property of Simson’s lines, and can be shown elementarily:

∗ l(S,PQR) passes through the midpoint X of HS, where H is the orthocenter of PQR.

4. Assume the contrary, that ∠MAB, ∠MBC, ∠MCA are all greater than 30 ◦ . By the sine Ceva theorem, it holds that

sin ∠MAC sin ∠MBA sin ∠MCB

1 = sin ∠MAB sin ∠MBC sin ∠MCA > sin 30

8 On the other hand, since ∠MAC + ∠MBA + ∠MCB < 180 ◦ − 3 · 30 ◦ = 90 ◦ ,

Jensen’s inequality applied on the concave function ln sin x (x ∈ [0, π ]) gives us sin ∠MAC sin ∠MBA sin ∠MCB < sin 3 30 ◦ , contradicting (∗).

Second solution. Denote the intersections of PA , PB, PC with BC,CA, AB by

A 1 ,B 1 ,C 1 , respectively. Suppose that each of the angles ∠PAB, ∠PBC , ∠PCA is greater than 30 o and denote PA = 2x, PB = 2y, PC = 2z. Then PC 1 > x, PA 1 > y, PB 1 > z. On the other hand, we know that

552 4 Solutions PC 1 PA 1 PB 1 S ABP S PBC S APC

+ = 1. PC + PC 1 PA + PA 1 PB + PB 1 S ABC S ABC S ABC

Since the function t is increasing, we obtain x + y p +t 2z +x 2x +y + z 2y +z < 1. But on the contrary, Cauchy-Schwartz inequality (or alternatively Jensen’s inequality) yields

+x 2x +y 2y +z x (2z + x) + y(2x + y) + z(2y + z)

5. Let P 1 be the point on the side BC such that ∠BFP 1 = β /2. Then ∠BP 1 F = 180 o

BP 1 =

sin (β /2) = 3 − 4sin ( β /2) = 1 +

−3 sin β /2, and the sine law gives us BF (3β /2)

2 cos β . Now we calculate BF . We have ∠BIF = 120 BP o − β /2, ∠BFI = 60 o and ∠BIC = 120 o , ∠BCI = γ /2 = 60 o − β /2. By the sine law,

1 sin 120 o BF = BI

sin (120 o − β /2)

BP = BC = BI

sin 60

3 3 sin (60 − β /2)

BP =

This implies o BF 3 sin (60 o +β /2)

−β /2)sin(60

= 4 sin(60 o − β /2) sin(60 o + β /2) =

sin 2 60 o

2 (cos β − cos120 o ) = 2 cos β +1= BF BP 1 . Therefore P ≡P 1 .

6. Let a , b, c be sides of the triangle. Let A 1 be the intersection of line AI with BC. By the known fact, BA 1 :A 1 C = c : b and AI : IA 1 = AB : BA 1 , hence BA 1 = ac b +c

and AI AB b +c . Consequently AI b IA +c 1 = BA 1 = a l A = a +b+c .

Put a = n + p, b = p + m, c = m + n: it is obvious that m, n, p are positive. Our inequality becomes

(2m + n + p)(m + 2n + p)(m + n + 2p)

27 The right side inequality immediately follows from the inequality between arith-

(m + n + p)

metic and geometric means applied on 2m + n + p, m + 2n + p and m + n + 2p. For the left side inequality, denote by T = m + n + p. Then we can write (2m + n + p)(m + 2n + p)(m + n + 2p) = (T + m)(T + n)(T + p) and

(T + m)(T + n)(T + p) = T 3 + (m + n + p)T 2 + (mn + np + pn)T + mnp > 2T 3 . Remark. The inequalities cannot be improved. In fact, AI l ·BI·CI A l B l C is equal to 8 /27

for a = b = c, while it can be arbitrarily close to 1/4 if a = b and c is sufficiently small.

7. The given equations imply AB = CD, AC = BD, AD = BC. Let L 1 ,M 1 ,N 1 be the midpoints of AD , BD,CD respectively. Then the above equalities yield

L 1 M 1 = AB/2 = LM, L 1 M 1 k AB k LM; L 1 M = CD/2 = LM 1 ,L 1 M k CD k LM 1 .

4.32 Shortlisted Problems 1991 553 Thus L , M, L 1 ,M 1 are coplanar and

D LML 1 M 1 is a rhombus as well as

MNM 1 N 1 and LNL 1 N 1 . Then the seg- N 1 ments LL 1 , MM 1 , NN 1 have the com-

mon midpoint Q and QL ⊥ QM, Q QL ⊥ QN, QM ⊥ QN. We also in-

MC fer that the line NN 1 is perpendicular

B to the plane LML 1 M 1 and hence to the line AB. Thus QA = QB, and similarly, QB = QC = QD, hence Q is just the center O, and ∠LOM = ∠MON = ∠NOL =

8. Let P 1 (x 1 ,y 1 ), P 2 (x 2 ,y 2 ), . . . , P n (x n ,y n ) be the n points of S in the coordinate plane. We may assume x 1 <x 2 < ··· < x n (choosing adequate axes and renum- bering the points if necessary). Define d to be half the minimum distance of P i from the line P j P k , where i , j, k go through all possible combinations of mutually distinct indices. First we define a set T containing 2n − 4 points:

T = {(x i ,y i − d),(x i ,y i + d) | i = 2,3,...,n − 1}.

Consider any triangle P k P l P m , where k < l < m. Its interior contains at least one of the two points (x l ,y l ± d), so T is a set of 2n − 4 points with the required property. However, at least one of the points of T is useless. The convex hull of S is a polygon with at least three points in S as vertices. Let P j

be a vertex of that hull distinct from P 1 and P n . Clearly one of the points (x j ,y j ± d) lies outside the convex hull, and thus can be left out. The remaining set of 2n − 5 points satisfies the conditions.

9. Let A 1 ,A 2 be two points of E which are joined. In E \ {A 1 ,A 2 }, there are at most 397 points to which A 1 is not joined, and at most as much to which A 2 is not joined. Consequently, there exists a point A 3 which is joined to both A 1 and A 2 . There are at most 3 · 397 = 1191 points of E \ {A 1 ,A 2 ,A 3 } to which at least one of A 1 ,A 2 ,A 3 is not joined, hence it is possible to choose a point A 4 joined to

A 1 ,A 2 ,A 3 . Similarly, there exists a point A 5 which is joined to all A 1 ,A 2 ,A 3 ,A 4 . Finally, among the remaining 1986 points, there are at most 5 ·397 = 1985 which are not joined to one of the points A 1 ,...,A 5 . Thus there is at least one point A 6 joined to all A 1 ,...,A 5 . It is clear that A 1 ,...,A 6 are pairwise joined.

Solution of the alternative version. Let be given 1991 points instead. Number the points from 1 to 1991, and join i and j if and only if i − j is not a multiple of

5. Then each i is joined to 1592 or 1593 other points, and obviously among any six points there are two which are not joined.

10. We start at some vertex v 0 and walk along distinct edges of the graph, numbering them 1 , 2, . . . in the order of appearance, until this is no longer possible without reusing an edge. If there are still edges which are not numbered, one of them has a vertex which has already been visited (else G would not be connected). Starting from this vertex, we continue to walk along unused edges resuming the

554 4 Solutions numbering, until we eventually get stuck. Repeating this procedure as long as

possible, we shall number all the edges. Let v be a vertex which is incident with e ≥ 2 edges. If v = v 0 , then it is on the edge 1, so the gcd at v is 1. If v 6= v 0 , suppose that it was reached for the first time by the edge r. At that time there was at least one unused edge incident with v (as e ≥ 2), hence one of them was labelled by r + 1. The gcd at v again equals gcd (r, r + 1) = 1.

1 11. To start with, observe that h n −m

+ m −1 . For n [n/2] = 1, 2, . . . set S

1 n −m

n −m−1

n −m

n =∑ =0 (−1) mn −m m m . Using the identity k = −1 k +

mm

m −1

k −1 we obtain the following relation for S n : m S n −m+1

n +1 = ∑ (−1)

m n = −m m n −m ∑

n −S n −1 . −1

=S

Since the initial members of the sequence S n are 1 , 1, 0, −1,−1,0,1,1,..., we thus find that S n is periodic with period 6. Now the sum from the problem reduces to

(S 1991 −S 1989 )=

be the set of those elements of S which are divisible by m. By the inclusion-exclusion principle, the number of elements divisible by 2, 3, 5 or

7 equals

|A 2 ∪A 3 ∪A 5 ∪A 7 |

= |A 2 | + |A 3 | + |A 5 | + |A 7 | − |A 6 | − |A 10 | − |A 14 | − |A 15 | −|A 21 | − |A 35 | + |A 30 | + |A 42 | + |A 70 | + |A 105 | − |A 210 | = 140 + 93 + 56 + 40 − 46 − 28 − 20 − 18 −13 − 8 + 9 + 6 + 4 + 2 − 1 = 216.

Among any five elements of the set A 2 ∪A 3 ∪A 5 ∪A 7 , one of the sets A 2 ,A 3 ,A 5 ,A 7 contains at least two, and those two are not relatively prime. Therefore n > 216. We claim that the answer is n = 217. First notice that the set A 2 ∪A 3 ∪A 5 ∪

A 7 consists of four prime (2 , 3, 5, 7) and 212 composite numbers. The set S \ A contains exactly 8 composite numbers: namely, 11 2 , 11 · 13,11 · 17,11 · 19,11 ·

23 , 13 2 , 13 · 17,13 · 19. Thus S consists of the unity, 220 composite numbers and

59 primes. Let A be a 217-element subset of S, and suppose that there are no five pairwise relatively prime numbers in A. Then A can contain at most 4 primes (or unity and three primes) and at least 213 composite numbers. Hence the set S \ A contains

4.32 Shortlisted Problems 1991 555 at most 7 composite numbers. Consequently, at least one of the following 8 five-

element sets is disjoint with S \ A, and is thus entirely contained in A: {2 · 23,3 · 19,5 · 17,7 · 13,11 · 11},

{2 · 2,3 · 3,5 · 5,7 · 7,13 · 13}. As each of these sets consists of five numbers relatively prime in pairs, the claim

is proved.

13. Call a sequence e 1 ,...,e n good if e 1 a 1 + ··· + e n a n is divisible by n. Among the sums s 0 = 0, s 1 =a 1 ,s 2 =a 1 +a 2 , ...,s n =a 1 + ··· + a n , two give the same

remainder modulo n, and their difference corresponds to a good sequence. To show that, permuting the a i ’s, we can find n − 1 different sequences, we use the following

Lemma. Let A be a k × n (k ≤ n− 2) matrix of zeros and ones, whose every row contains at least one 0 and at least two 1’s. Then it is possible to permute columns of A is such a way that in any row 1’s do not form a block.

Proof. We will use the induction on k. The case k = 1 and arbitrary n ≥ 3 is trivial. Suppose that k ≥ 2 and that for k − 1 and any n ≥ k + 1 the lemma is true. Consider a k × n matrix A, n ≥ k + 2. We mark an element a ij if either it is the only zero in the i-th row, or one of the 1’s in the row if it contains exactly two 1’s. Since n ≥ 4, every row contains at most two marked elements, which adds up to at most 2k < 2n marked elements in total. It follows that there is a column with at most one marked element. Assume w.l.o.g. that it is the first column and that a 1j isn’t marked for

j > 1. The matrix B, obtained by omitting the first row and first column from A, satisfies the conditions of the lemma. Therefore, we can permute columns of B and get the required form. Considered as a permutation of column of A, this permutation may leave a block of 1’s only in the first row

of A. In the case that it is so, if a 11 = 1 we put the first column in the last place, otherwise we put it between any two columns having 1’s in the first row. The obtained matrix has the required property.

Suppose now that we have got k different nontrivial good sequences e i 1 ,...,e i n ,

i = 1, . . . , k, and that k ≤ n − 2. The matrix A = (e i j ) fulfills the conditions of Lemma, hence there is a permutation σ from Lemma. Now among the sums

s 0 = 0, s 1 =a σ (1) ,s 2 =a σ (1) +a σ (2) , ...,s n =a σ(1) + ··· + a σ (n) , two give the same remainder modulo n. Let s p ≡s q (mod n), p < q. Then n | s q −s p =

a σ (p+1) + ··· + a σ (q) , and this yields a good sequence e 1 ,...,e n with e σ (p+1) = ··· = e σ (q) = 1 and other e’s equal to zero. Since from the construction we see that none of the sequences e σ ( j) i has all 1’s in a block, in this way we have got

a new nontrivial good sequence, and we can continue this procedure until there are n − 1 sequences. Together with the trivial 0,...,0 sequence, we have found n good sequences.

556 4 Solutions

14. Suppose that f (x 0 ), f (x 0 + 1), . . . , f (x 0 + 2p − 2) are squares. If p | a and p ∤ b, then f (x) ≡ bx + c (mod p) for x = x 0 ,...,x 0 + p − 1 form a complete system of residues modulo p. However, a square is always congruent to exactly one of the

p +1 numbers 0 ,1 2 ,2 2 ,...,( p −1 ) 2

2 2 and thus cannot give every residue modulo p. Also, if p | a and p | b, then p | b 2 − 4ac.

We now assume p ∤ a. The following identities hold for any quadric polynomial: 4a 2 · f (x) = (2ax + b) 2 − (b − 4ac)

(1) and

f (x + p) − f (x) = p(2ax + b) + p 2 a . (2) Suppose that there is an y, x 0 ≤y≤x 0 + p − 2, for which f (y) is divisible by

p . Then both f (y) and f (y + p) are squares divisible by p, and therefore both are divisible by p 2 . But relation (2) implies that p | 2ay + b, and hence by (1)

b 2 − 4ac is divisible by p as well. Therefore it suffices to show that such an y exists, and for that aim we prove that there are two such y in [x 0 ,x 0 + p − 1]. Assume the opposite. Since for x =x

0 ,x 0 + 1, . . . , x 0 + p − 1 f (x) is congruent modulo p to one of the −1 2 num- bers 1 2 ,2 2 ,..., p −1 2 2 , it follows by the pigeon-hole principle that for some mu-

tually distinct u , v, w ∈ {x 0 ,...,x 0 + p − 1} we have f (u) ≡ f (v) ≡ f (w) (mod p ). Consequently the difference f (u) − f (v) = (u − v)(a(u + v) + b) is divisi- ble by p, but it is clear that p ∤ u − v, hence a(u + v) ≡ −b (mod p). Similarly

a (u + w) ≡ −b (mod p), which together with the previous congruence yields p | a(v − w) ⇒ p | v − w which is clearly impossible. It follows that p | f (y 1 ) for

at least one y 1 ,x 0 ≤y 1 <x 0 + p.

If y 2 ,x 0 ≤y 2 <x 0 + p is such that a(y 1 +y 2 ) + b ≡ 0 (mod p), we have p |

f (y 1 ) − f (y 2 ) ⇒ p | f (y 2 ). If y 1 =y 2 , then by (1) p |b 2 − 4ac. Otherwise, among

y 1 ,y 2 one belongs to [x 0 ,x 0 + p − 2] as required.

Second solution. Using Legendre’s symbols a p for quadratic residues we can prove a stronger statement for p ≥ 5. It can be shown that

p −1

ax 2 + bx + c

p ∤b ∑ 2

hence for at most p +3 values of x between x 0 and x 0 + p − 1 inclusive, ax 2 2 + bx +

c is a quadratic residue or 0 modulo p. Therefore, if p ≥ 5 and f (x) is a square for p +5

2 consecutive values, then p |b − 4ac.

15. Assume that the sequence has the period T . We can find integers k > m > 0, as large as we like, such that 10 k ≡ 10 m (mod T ), using for example Euler’s

theorem. It is obvious that a 10 k −1 =a 10 k and hence, taking k sufficiently large and using the periodicity, we see that

a 2 ·10 k −10 m −1 =a 10 k −1 =a 10 k =a 2 ·10 k −10 m .

4.32 Shortlisted Problems 1991 557 Since (2 · 10 k − 10 m )! = (2 · 10 k − 10 m )(2 · 10 k − 10 m − 1)! and the last nonzero

digit of 2 ·10 k −10 m is nine, we must have a 2 ·10 k −10 m −1 = 5 (if s is a digit, the last digit of 9s is s only if s = 5). But this means that 5 divides n! with a greater power than 2 does, which is impossible. Indeed, if the exponents of these powers are

α 2 , α 5 respectively, then α 5 = [n/5] + [n/5 2 ] + ··· ≤ α 2 = [n/2] + [n/2 2 ] + ···.

16. Let p be the least prime number that does not divide n: thus a 1 = 1 and a 2 = p. Since a 2 −a 1 =a 3 −a 2 = ··· = r, the a i ’s are 1 , p, 2p − 1,3p − 2,.... We have the following cases: p = 2. Then r = 1 and the numbers 1, 2, 3, . . . , n − 1 are relatively prime to n, hence n is a prime. p = 3. Then r = 2, so every odd number less than n is relatively prime to n, from which we deduce that n has no odd divisors. Therefore n =2 k for some k ∈ N.

p > 3. Then r = p − 1 and a k +1 =a 1 + k(p −1) = 1+ k(p− 1). Since n− 1 also must belong to the progression, we have p − 1 | n − 2. Let q be any prime divisor of p − 1. Then also q | n − 2. On the other hand, since q < p, it must divide n too, therefore q | 2, i.e. q = 2. This means that p − 1 has no prime divisors other than 2 and thus p =2 l + 1 for some l ≥ 2. But in order for p to be prime, l must be even (because 3 l |2 + 1 for l odd). Now we recall that 2p

l − 1 is also relatively prime to n; but 2p − 1 = 2 +1 + 1 is divisible by 3, which is a contradiction because 3 | n.

17. Taking the equation 3 x +4 y =5 z (x , y, z > 0) modulo 3, we get that 5 z ≡ 1 (mod 3), hence z is even, say z = 2z 1 . The equation then becomes 3 x =5 2z 1 −4 y = (5 z 1 −2 y )(5 z 1 +2 y ). Each factor 5 z 1 −2 y and 5 z 1 +2 y is a power of 3, for which the only possibility is 5 z 1 +2 y =3 x and 5 z 1 −2 y = 1. Again modulo 3 these equations reduce to (−1) z 1 + (−1) y = 0 and (−1) z 1 − (−1) y = 1, implying that z 1 is odd and y is even. Particularly, y ≥ 2. Reducing the equation 5 z 1 +2 y =3 x modulo 4 we get that 3 x ≡ 1, hence x is even. Now if y > 2, modulo 8 this equation yields 5 ≡5 z 1 ≡3 x ≡ 1, a contradiction. Hence y = 2, z 1 = 1. The only solution of the original equation is x = y = z = 2.

18. For integers a > 0, n > 0 and α ≥ 0, we shall write a α k n when a α | n and

a α+1 ∤ n. Lemma. For every odd number a ≥ 3 and an integer n ≥ 0 it holds that

a k (a + 1) n k (a − 1) + 1. Proof. We shall prove the first relation by induction (the second is analogous).

For n = 0 the statement is obvious. Suppose that it holds for some n, i.e. n that (1 + a) a = 1 + Na n +1 ,a ∤ N. Then

= 1 + a · Na

(1 + a) a n +1 = (1 + Na n +1 ) a n +1

N 2 a 2n +2 + Ma +3 3n

558 4 Solutions for some integer M. Since a 2 is divisible by a for a odd, we deduce that

the part of the above sum behind 1 + a · Na n +1 is divisible by a n +3 . Hence

(1 + a) a n +1 =1+N ′ a n +2 , where a ∤N ′ .

It follows immediately from Lemma that

1991 k 1992 1990 − 1. Adding these two relations we obtain immediately that k = 1991 is the desired

19. Set x = cos( π a ). The given equation is equivalent to 4x 3 + 4x 2 − 3x − 2 = 0,

which factorizes as (2x + 1)(2x 2 + x − 2) = 0.

The case 2x + 1 = 0 yields cos( π a ) = −1/2 and a = 2/3. It remains to show that if x satisfies 2x 2 + x − 2 = 0 then a is not rational. The polynomial equation √

2x 2 + x − 2 = 0 has two real roots, x 1 ,2 = −1± 17 4 , and since

|x| ≤ 1 we must

have x = cos π a = −1+ 17 4 .

a +b We now prove by induction that, for every integer n √

≥ 0, cos(2 n π a )= n n 17 for some odd integers a √ n ,b n . The case n = 0 is trivial. Also, if cos(2 n 4 π a )=

a n +b n 17

4 , then

cos (2 n +1 π a ) = 2 cos 2 (2 n π a )−1

4 2 4 By the inductive step that a n ,b n are odd, it is obvious that a n +1 ,b n +1 are also

odd. This proves the claim. Note also that, since a

n +1 = 2 (a n + 17b n − 8) > a n , the sequence {a n } is strictly increasing. Hence the set of values of cos √ (2 n π a ), n = 0, 1, 2, . . ., is infinite (be-

cause

17 is irrational). However, if a were rational, then the set of values of cos m π a ,m = 1, 2, . . . , would be finite, a contradiction. Therefore the only pos- sible value for a is 2 /3.

20. We prove the result with 1991 replaced by any positive integer k. For natural numbers p , q, let ε =( α p −[ α p ])( α q −[ α q ]). Then 0 < ε < 1 and

ε = α 2 pq − α (p[ α q ] + q[ α p ]) + [ α p ][ α q ]. Multiplying this equality by α − k and using α 2 =k α + 1, i.e. α ( α − k) = 1, we

get ( α − k) ε = α (pq + [ α p ][ α q ]) − (p[ α q ] + q[ α p ] + k[ α p ][ α q ]).

Since 0 <( α − k) ε < 1, we have [ α (p ∗ q)] = p[ α q ] + q[ α p ] + k[ α p ][ α q ]. Now (p ∗ q) ∗ r = (p ∗ q)r + [ α (p ∗ q)][ α r ]=

= pqr + [ α p ][ α q ]r + [ α q ][ α r ]p + [ α r ][ α p ]q + k[ α p ][ α q ][ α r ]. Since the last expression is symmetric, the same formula is obtained for p ∗ (q ∗

r ).

4.32 Shortlisted Problems 1991 559

21. The polynomial g (x) factorizes as g(x) = f (x) 2 − 9 = ( f (x) − 3)( f (x) + 3). If one of the equations f (x) + 3 = 0 and f (x) − 3 = 0 has no integer solutions, then the number of integer solutions of g (x) = 0 clearly does not exceed 1991. Suppose now that both f (x) + 3 = 0 and f (x) − 3 = 0 have integer solutions. Let

be distinct integer solutions of the latter equation. There exist monic polynomials p (x), q(x) with integer coefficients such that f (x) + 3 = (x − x 1 )(x − x 2 ) . . . (x − x k )p(x) and

x 1 ,...,x k

be distinct integer solutions of the former, and x k +1 ,...,x k +l

f (x) − 3 = (x − x k +1 )(x − x k +2 ) . . . (x − x k +l )q(x). Thus we obtain (x − x 1 )(x − x 2 ) . . . (x − x k )p(x) − (x − x k +1 )(x − x k +2 ) . . . (x − x k +l )q(x) = 6. Putting x =x k +1 we get (x k +1 −x 1 )(x k +1 −x 2 ) ···(x k +1 −x k ) | 6, and since the

product of more than four distinct integers cannot divide 6, this implies k ≤ 4. Similarly l ≤ 4; hence g(x) = 0 has at most 8 distinct integer solutions.

Remark. The proposer provided a solution for the upper bound of 1995 roots which was essentially the same as that of (IMO74-6).

22. Suppose w.l.o.g. that the center of the square is at the origin O (0, 0). We denote the curve y = f (x) = x 3 + ax 2 + bx + c by γ and the vertices of the square by

A , B,C, D in this order. At first, the symmetry with respect to the point O maps γ into the curve γ (y =

f (−x) = x 3 − ax 2 + bx − c). Obviously γ also passes through A , B,C, D, and thus has four different intersection points with γ . Then 2ax 2 + 2c has at least four distinct solution, which implies a = c = 0. Particularly, γ passes through O and intersects all quadrants, and hence b < 0. Further, the curve γ ′ , obtained by rotation of γ around O for 90 ◦ , has an equation

−x = f (y) and also contains the points A,B,C,D and O. The intersection points (x, y) of γ ∩ γ ′ are determined by −x = f ( f (x)), and hence they are roots of a polynomial p (x) = f ( f (x)) + x of 9-th degree. But the number of times that one cubic actually crosses the other in each quadrant is in the general case even (draw the picture!), and since ABCD is the only square lying on γ ∩ γ ′ , the intersection points A , B,C, D must be double. It follows that

(1) where r , s are the x-coordinates of A and B. On the other hand, p(x) is defined by

(x) = x[(x − r)(x + r)(x − s)(x + s)] 2 ,

(x 3 + bx) 3 + b(x 3 + bx) + x, and therefore equating of coefficients with (1) yields

3b = −2(r 2 +s 2 ),

3b 2 = (r 2 +s 2 ) 2 + 2r 2 s 2 ,

b (b 2 + 1) = −2r 2 s 2 (r 2 +s 2 ),

b 2 +1=r 4 s 4 . √

Straightforward solving this system of equations gives b =−

8 and r 2 +s √ 2 =

18. The line segment from O to (r, s) is half a diagonal of the square, and thus a side p √

of the square has length a = 2 (r 2 +s 2 )= 4 72.

23. From (i), replacing m by f ( f (m)), we get

560 4 Solutions

f ( f ( f (m)) + f ( f (n)) ) = − f ( f ( f ( f (m)) + 1)) − n; analogously f ( f ( f (n)) + f ( f (m)) ) = − f ( f ( f ( f (n)) + 1)) − m.

From these relations we get f ( f ( f ( f (m)) + 1)) − f ( f ( f ( f (n)) + 1)) = m − n. Again from (i),

f ( f ( f ( f (m)) + 1)) = f (−m − f ( f (2)) ) and f ( f ( f ( f (n)) + 1)) = f (−n − f ( f (2)) ).

Setting f ( f (2)) = k we obtain f (−m − k) − f (−n − k) = m − n for all integers m , n. This implies f (m) = f (0) − m. Then also f ( f (m)) = m, and using this in (i) we finally get

for all integers n. Particularly f (1991) = −1992.

f (n) = −n − 1

From (ii) we obtain g (n) = g(−n − 1) for all integers n. Since g is a polynomial, it must also satisfy g (x) = g(−x − 1) for all real x. Let us now express g as a polynomial on x + 1/2: g(x) = h(x + 1/2). Then h satisfies h(x + 1/2) = h(−x −

1 /2), i.e. h(y) = h(−y), hence it is a polynomial in y 2 ; thus g is a polynomial in (x + 1/2) 2 =x 2 + x + 1/4. Hence g(n) = p(n 2 + n) (for some polynomial p) is

the most general form of g.

24. Let y k =a k −a k +1 +a k +2 − ··· + a k +n−1 for k = 1, 2, . . . , n, where we define x i +n =x i for 1 ≤ i ≤ n. We then have y 1 +y 2 = 2a 1 ,y 2 +y 3 = 2a 2 ,...,y n +y 1 = 2a n . (i) Let n = 4k − 1 for some integer k > 0. Then for each i = 1,2,...,n we have that y i = (a i +a i +1 + ··· + a i −1 ) − 2(a i +1 +a i +3 + ··· + a i −2 )=1 + 2 + ···+ (4k − 1) − 2(a i +1 +a i +3 + ··· + a i −2 ) is even. Suppose now that a 1 ,...,a n is a good permutation. Then each y i is positive and even, so y i ≥ 2. But for some t ∈ {1,...,n} we must have a t = 1, and thus y t +y t +1 = 2a t =2 which is impossible. Hence the numbers n = 4k − 1 are not good.

(ii) Let n = 4k + 1 for some integer k > 0. Then 2, 4, . . . , 4k, 4k + 1, 4k −

1 , . . . , 3, 1 is a permutation with the desired property. Indeed, in this case y 1 =y 4k +1 = 1, y 2 =y 4k = 3, . . . , y 2k =y 2k +2 = 4k − 1, y 2k +1 = 4k + 1. Therefore all nice numbers are given by 4k + 1, k ∈ N.

25. Since replacing x 1 by 1 can only reduce the set of indices i for which the desired inequality holds, we may assume x 1 = 1. Similarly we may assume x n = 0. Now

we can let i be the largest index such that x i > 1/2. Then x i +1 ≤ 1/2, hence

x i (1 − x i +1 )≥ = x

4 4 1 (1 − x n ).

26. Without loss of generality we can assume b 1 ≥b 2 ≥ ··· ≥ b n . We denote by A i the product a 1 a 2 ...a i −1 a i +1 ...a n . If for some i < j holds A i <A j , then b i A i +

b j A j ≤b i A j +b j A i (or equivalently (b i −b j )(A i −A j ) ≤ 0). Therefore the sum ∑ n i =1 b i A i does not decrease when we rearrange the numbers a 1 ,...,a n so that

A 1 ≥ ··· ≥ A n , and consequently a 1 ≤ ··· ≤ a n . Further, for fixed a i ’s and ∑b i =

4.32 Shortlisted Problems 1991 561

1, the sum ∑ n i =1 b i A i is maximal when b 1 takes the largest possible value, i.e.

b 1 = p, b 2 takes the remaining largest possible value b 2 = 1 − p, whereas b 3 = ··· = b n = 0. In this case

∑ b i A i = pA 1 + (1 − p)A 2 =a 3 ...a n (pa 2 + (1 − p)a 1 )

i =1

≤ p(a 1 +a 2 )a 3 ...a n ≤

, using the inequality between the geometric and arithmetic means for a 3 , ...,a n ,

(n − 1) n −1

a 1 +a 2 .

27. Write F (x 1 ,...,x n )=∑ <j x i x j (x i +x j ). Choose an n-tuple (x 1 ,...,x n ), ∑ i n i =1 x i =

1, x i ≥ 0 with at least three nonzero components, and assume w.l.o.g. that x 1 ≥ ··· ≥ x k −1 ≥x k ≥x k +1 = ··· = x n = 0. We claim that replacing x k −1 ,x k with

x k −1 +x k , 0 the value of F increases. Write for brevity x k −1 = a, x k = b. Then

F (. . . , a + b, 0, 0, . . .) − F(... ,a,b,0,...)

k −2

k −2

= ∑ x i (a + b)(x i + a + b) − ∑ [x i a (x i + a) + x i b (x i + b)] − ab(a + b)

= ab 2 ∑ x i −a−b = ab(2 − 3(a + b)) > 0,

i =1

because x +x

k ≤ 3 1 k −1 +x k −2 )≤ 3 . Repeating this procedure we can reduce the number of nonzero x i ’s to two, increasing the value of F in each

2 (x +x

k −1

step. It remains to maximize F over n-tuples (x 1 ,x 2 , 0, . . . , 0) with x 1 ,x 2 ≥ 0, x

1 +x 2 = 1: in this case F equals x 1 x 2 and attains its maximum value 4 when x 1 =x 2 = 1 2 ,x 3 =...,x n = 0. √

28. Let x √ n = c(n 2 − [n 2 ]) for some constant c > 0. For i > j, putting p = [i 2 ]− [j 2 ], we have

c c |x i −x j | = c|(i − j)

2 |2(i − j) −p |c

because p < (i − j) 2 + 1. Taking c = 4, we obtain that for any i > j, (i − j)|x i − x j | ≥ 1. Of course, this implies (i − j) a |x i −x j | ≥ 1 for any a > 1.

Remark. The constant 4 can be replaced with 3 /2 + 2. Second solution. Another example of a sequence {x n } is constructed in the

following way: x 1 = 0, x 2 = 1, x 3 = 2 and x

3 k i +m =x m + 3 k for i = 1, 2 and

1 ≤m≤3 k . It is easily shown that |i − j| · |x i −x j | ≥ 1/3 for any i 6= j. Third solution. If n =b 0 + 2b 1 + ··· + 2 k b k ,b i ∈ {0,1}, then one can set x a n

=b 0 +2 −a b

1 + ··· + 2 −ka b k . In this case it holds that |i − j| a |x i −x j |≥ −2 2 a −1 .

562 4 Solutions

29. One easily observes that the following sets are super-invariant: one-point set, its complement, closed and open half-lines or their complements, and the whole real line. To show that these are the only possibilities, we first observe that S is super- invariant if and only if for each a > 0 there is a b such that x ∈ S ⇔ ax + b ∈ S.

(i) Suppose that for some a there are two such b’s: b 1 and b 2 . Then x ∈S⇔ ax +b 1 ∈ S and x ∈ S ⇔ ax + b 2 ∈ S, which implies that S is periodic: y

a ∈ S. Since S is identical to a translate of any stretching of S, all positive numbers are periods of S. Therefore S ≡ R.

b 1 ∈S⇔y+ −b 2

(ii) Assume that, for each a, b = f (a) is unique. Then for any a 1 and a 2 , x ∈S⇔a 1 x + f (a 1 )∈S⇔a 1 a 2 x +a 2 f (a 1 ) + f (a 2 )∈S

⇔a 2 x + f (a 2 )∈S⇔a 1 a 2 x +a 1 f (a 2 ) + f (a 1 ) ∈ S. As above it follows that a 1 f (a 2 )+ f (a 1 )=a 2 f (a 1 )+ f (a 2 ), or equivalently

f (a 1 )(a 2 − 1) = f (a 2 )(a 1 − 1). Hence (for some c), f (a) = c(a − 1) for all

a . Now x ∈ S ⇔ ax+c(a−1) ∈ S actually means that y−c ∈ S ⇔ ay−c ∈ S for all a. Then it is easy to conclude that {y − c | y ∈ S} is either a half-line or the whole line, and so is S.

30. Let a and b be the integers written by A and B respectively, and let x < y be the two integers written by the referee. Suppose that none of A and B ever answers “yes”. Initially, regardless of a, A knows that 0 ≤ b ≤ y and answers “no”. In the second step, B knows that A obtained 0 ≤ b ≤ y, but if a were greater than x, A would know that a + b = y and would thus answer “yes”. So B concludes 0 ≤ a ≤ x but answers “no”. The process continues. Suppose that, in the n-th step, A knows that B obtained r n −1 ≤a≤s n −1 . If b > x −r n −1 , B would know that a + b > x and hence a + b = y, while if b < y −s n −1 ,

B would know that a + b < y, i.e. a + b = x: in both cases he would be able to guess a. However, B answered “no”, from which A concludes y −s n −1 ≤b≤ x −r n −1 . Put r n =y−s n −1 and s n =x−r n −1 . Similarly, in the next step B knows that A obtained r n ≤b≤s n and, since A answered “no”, concludes y −s n ≤ a ≤ x−r n . Put r n +1 = y−s n and s n +1 = x−r n . Notice that in both cases s i +1 −r i +1 =s i −r i − (y − x). Since y − x > 0, there exists an m for which s m −r m < 0, a contradiction.

4.33 Shortlisted Problems 1992 563