Solutions to the Shortlisted Problems of IMO 1989

4.30 Solutions to the Shortlisted Problems of IMO 1989

1. Let I denote the intersection of the three internal bisectors. Then IA 1 =A 1 A 0 . 0 One way to prove this is to real-

BA ize that the circumcircle of △ABC is the nine-point circle of △A 0 B 0 0 C 0 , C B 1

so it bisects IA 0 , since I is the or-

thocenter of A 0 B 0 C 0 . Another way I

is through noting that IA 1 =A 1 B . This is a consequence of ∠A 1 IB =

∠IBA 1 = (∠A + ∠B)/2, and A 1 B =

A 1 A 0 which follows from ∠A 1 A 0 B = ∠A 1 BA 0 = 90 ◦ − ∠IBA 1 . Hence, we

A 0 obtain S IA 1 B =S A 0 A 1 B . Repeating this argument for the six triangles that have a vertex at I and adding them up gives us S A 0 B 0 C 0 = 2S AC 1 BA 1 CB 1 . To prove S AC 1 BA 1 CB 1 ≥ 2S ABC , draw the three altitudes in triangle ABC intersecting in H. Let X, Y , and Z be the symmetric points of H with respect the sides BC, CA, and AB, respectively. Then X ,Y, Z are points on

the circumcircle of △ABC (because ∠BXC = ∠BHC = 180 ◦ − ∠A). Since A 1 is the midpoint of the arc BC, we have S BA 1 C ≥S BXC . Hence

S AC 1 BA 1 CB 1 ≥S AZBXCY = 2(S BHC +S CHA +S AHB ) = 2S ABC .

2. Let the carpet have width x, length y. Suppose that the carpet EFGH lies in a room ABCD, E being on AB, F on BC, G on CD, and H on DA. Then

△AEH ≡ △CGF ∼ △BFE ≡ △DHG. Let x = k, AE = a and AH = b. In that case BE = kb and DH = ka.

Thus a + kb = 50, ka + b = 55, whence a = 55k −50 k 2 −1 and b = 50k −55 k 2 −1 . Hence x 2 =

a 2 +b 2 = 5525k 2 −11000k+5525 (k 2 −1) 2 , i.e.,

x 2 (k 2 − 1) 2 = 5525k 2 − 11000k + 5525. Similarly, from the equations for the second storeroom, we get x 2 (k 2 − 1) 2 = 4469k 2 − 8360k + 4469. Combining the two equations, we get 5525k 2 − 11000k + 5525 = 4469k 2 −

8360k + 4469, which implies k = 2 or 1/2. Without loss of generality we have √ y = 2x and a+ 2b = 50, 2a +b = 55; hence a = 20, b = 15, x = 15 2 + 20 2 = 25, and y = 50. We have thus shown that the carpet is 25 feet by 50 feet.

3. Let the carpet have width x, length y. Let the length of the storerooms be q. Let y /x = k. Then, as in the previous problem, (kq − 50) 2 + (50k − q) 2 = (kq −

38 ) 2 + (38k − q) 2 , i.e.,

kq = 22(k 2 + 1).

Also, as before, x 2 = kq −50 2 50k

−q k 2 −1

k 2 −1

, i.e.,

524 4 Solutions x 2 (q 2 − 1) 2 = (k 2 + 1)(q 2 − 1900),

(2) which, together with (1), yields x 2 k 2 (k 2 − 1) 2 = (k 2 + 1)(484k 4 − 932k 2 + 484). Since k is rational, let k = c/d, where c and d are integers with gcd(c, d) = 1.

Then we obtain x 2 c 2 (c 2 −d 2 ) 2 =c 2 (484c 4 − 448c 2 d 2 − 448d 4 ) + 484d 6 . We thus have c 2 | 484d 6 , but since (c, d) = 1, we have c 2 | 484 ⇒ c | 22.

Analogously, d | 22; thus k = 1,2,11,22, 1

2 , 1 11 , 1 22 , 2 11 , 11 2 . Since reciprocals lead to the same solution, we need only consider k ∈ 1 , 2, 11, 22, 11 2 , yielding

q = 44, 55, 244, 485, 125, respectively. We can test these values by substituting them into (2). Only k = 2 gives us an integer solution, namely x = 25, y = 50.

4. First we note that for every integer k > 0 and prime number p, p k doesn’t divide k !. This follows from the fact that the highest exponent r of p for which p r |k! is

< k. p p 2 + ··· = p −1 Now suppose that α is a rational root of the given equation. Then

from which we can conclude that α must be an integer, not equal to ±1. Let p be

a prime divisor of n and let r be the highest exponent of p for which p r |n!. Then p α . Since p k α k and p k ∤ k!, we obtain that p r +1

| k | | n! α /k! for k = 1, 2, . . . , n. But then it follows from (1) that p r +1 | n!, a contradiction.

5. According to the Cauchy–Schwarz inequality,

Since r 1 + ··· + r n = −n, applying this inequality we obtain r 2 1 +...+r 2 n ≥ n, and applying it three more times, we obtain

r 16 1 16 + ··· + r n ≥ n,

with equality if and only if r 1 =r 2 =...=r n = −1 and p(x) = (x + 1) n .

6. Let us denote the measures of the inner angles of the triangle ABC by α , β , γ . Then P =r 2 (sin 2 α + sin 2 β + sin 2 γ )/2. Since the inner angles of the triangle

A ′ B ′ C ′ are ( β + γ )/2, ( γ + α )/2, ( α + β )/2, we also have Q = r 2 [sin ( β + γ )+

4.30 Shortlisted Problems 1989 525

sin ( γ + α ) + sin ( α + β )]/2. Applying the AM–GM mean inequality, we now obtain

3 16Q 16 = r 6 (sin ( β + γ ) + sin ( γ + α ) + sin ( α + β )) 3

8 ≥ 54r 6 sin ( β + γ ) sin ( γ + α )sin ( α + β )

= 27r 6 [cos( α − β ) − cos( α + β +2 γ )] sin( α + β ) = 27r 6 [cos( α − β ) + cos γ ] sin( α + β )

r 6 [sin( α + β + γ ) + sin( α + β

2 − γ ) + sin2 α + sin 2 β ]

r 6 [sin(2 γ ) + sin 2 α + sin 2 β ] = 27r 4 P .

2 This completes the proof.

7. Assume that P 1 and P 2 are points inside E, and that the line P 1 P 2 intersects the perimeter of E at Q 1 and Q 2 . If we prove the statement for Q 1 and Q 2 , we are done, since these arcs can be mapped homothetically to join P 1 and P 2 . Let V 1 ,V 2 be two vertices of E. Then applying two homotheties to the inscribed circle of E one can find two arcs (one of them may be a side of E) joining these two points, both tangent to the sides of E that meet at V 1 and V 2 . If A is any point of the side V 2 V 3 , two homotheties with center V 1 take the arcs joining V 1 to V 2 and

V 3 into arcs joining V 1 to A; their angle of incidence at A remains (1 − 2/n) π . Next, for two arbitrary points Q 1 and Q 2 on two different sides V 1 V 2 and V 3 V 4 , we join V 1 and V 2 to Q 2 with pairs of arcs that meet at Q 2 and have an angle of incidence (1 − 2/n) π . The two arcs that meet the line Q 1 Q 2 again outside E meet at Q 2 at an angle greater than or equal to (1 − 2/n) π . Two homotheties with center Q 2 carry these arcs to ones meeting also at Q 1 with the same angle of incidence.

8. Let A , B,C, D denote the vertices of R. We consider the set S of all points E of the plane that are vertices of at least one rectangle, and its subset S ′ consist- ing of those points in S that have both coordinates integral in the orthonormal coordinate system with point A as the origin and lines AB , AD as axes.

First, to each E ∈ S we can assign an integer n E as the number of rectangles R i with one vertex at E. It is easy to check that n E = 1 if E is one of the vertices

A , B,C, D; in all other cases n E is either 2 or 4.

Furthermore, for each rectangle R i we define f (R i ) as the number of vertices of R i that belong to S ′ . Since every R i has at least one side of integer length, f (R i ) can take only values 0, 2, or 4. Therefore we have

∑ f (R i ) ≡ 0 (mod 2).

i =1

On the other hand, ∑ n i =1 f (R i ) is equal to ∑ E ∈S ′ n E , implying that

∑ n E ≡ 0 (mod 2).

E ∈S ′

526 4 Solutions However, since n A = 1, at least one other n E , where E ∈S ′ , must be odd, and

that can happen only for E being B, C, or D. We conclude that at least one of the sides of R has integral length.

Second solution. Consider the coordinate system introduced above. If D is a rectangle whose sides are parallel to the axes of the system, it is easy to prove that

sin (2 π (x + α ))sin(2 π (y + β ))dxdy = 0, for all α , β ∈R

D if and only if at least one side of D has integral length. This holds for all R i ’s therefore, adding up these equalities for each α and β we get

sin (2 π (x + α ))sin (2 π (y + β )) dx dy = 0.

Thus, R also has a side of integral length.

3 √ 3 √ 3 √ 3 n +1 +b n +1 2 +c n +1 4 = (a n +b n 2 +c n 4 )(1 + 4 2 −4 4 ) we

9. From a

obtain a n +1 =a n − 8b n + 8c n . Since a 0 = 1, a n is odd for all n. For an integer k > 0, we can write k = 2 l k ′ ,k ′ being odd and l a nonnegative integer. Let us set v (k) = l, and define β n = v(b n ), γ n = v(c n ). We prove the following lemmas:

Lemma 1. For every integer p ≥ 0, b 2 p and c 2 p are nonzero, and β 2 p = γ 2 p = p + 2. Proof. By induction on p. For p = 0, b 1 = 4 and c 1 = −4, so the assertion is true. Suppose that it holds for p. Then √ 3 √ p +1

4 2 p +2 ′ 3 2 ′ (1 + 4 3 −4 ) = (a + 2 (b +c 4 )) 2 with a ,b ′ , and c ′ odd.

√ p +1

√ Then we easily obtain that (1 + 4 3 2 3 3

√ 3 −4 4 ) 2 = A+ 2 p +3 (B 2 +C 4 ),

where A , B = ab ′ +2 p +1 E ,C = ac ′ +2 p +1 F are odd. Therefore Lemma 1 holds for p + 1. Lemma 2. Suppose that for integers n , m ≥ 0, β n = γ n = λ > β m = γ m = µ . Then b n +m ,c n +m are nonzero and β n +m = γ n +m = µ . Proof. Calculating (a ′ +2 λ (b ′

√ (b ′′

2 +c ′ 3 4 ))(a ′′

3 2 +c ′′ 3 √ 4 )), with a ′ ,

2 +C 3 4 ), where A ,B=a ′ b ′′ +2 λ −µ E , and C =a ′ c ′′ +2 λ −µ F are odd, which proves Lemma 2. Since every integer n > 0 can be written as n = 2 p r + ··· + 2 p 1 , with 0 ≤p 1 < ··· < p r , from Lemmas 1 and 2 it follows that c n is nonzero, and that γ n =p 1 + 2.

b ′ ,c ′ ,a ′′ ,b ′′ ,c ′′ odd, we easily obtain the product A +2 µ (B 3

Remark. b 1989 and c 1989 are divisible by 4, but not by 8.

10. Plugging in wz + a instead of z into the functional equation, we obtain

f (wz + a) + f (w 2 z + wa + a) = g(wz + a). (1) By repeating this process, this time in (1), we get

f (w 2 z + wa + a) + f (z) = g(w 2 z + wa + a). (2)

4.30 Shortlisted Problems 1989 527 Solving the system of linear equations (1), (2) and the original functional equa-

tion, we easily get

f (z) = + wa + a) − g(wz + a) .

g (z) + g(w 2 z

This function thus uniquely satisfies the original functional equation.

11. Call a binary sequence S of length n repeating if for some d | n, d > 1, S can

be the number of nonrepeating binary sequences of length n. The total number of binary sequences of length n is ob- viously 2 n . Any sequence of length n can be produced by repeating its unique longest nonrepeating initial block according to need. Hence, we obtain the re-

be split into d identical blocks. Let x n

cursion relation ∑ d |n x d =2 n . This, along with x 1 = 2, gives us a n =x n for all n . We now have that the sequences counted by x n can be grouped into groups of n , the sequences in the same group being cyclic shifts of each other. Hence, n |x n =a n .

12. Assume that each car starts with a unique ranking number. Suppose that while turning back at a meeting point two cars always exchanged their ranking num- bers. We can observe that ranking numbers move at a constant speed and direc- tion. One hour later, after several exchanges, each starting point will be occupied by a car of the same ranking number and proceeding in the same direction as the one that started from there one hour ago. We now give the cars back their original ranking numbers. Since the sequence of the cars along the track cannot be changed, the only possibility is that the original situation has been rotated, maybe onto itself. Hence for some d | n, after

d hours each car will be at its starting position and orientation.

13. Let us construct the circles σ 1 with center A and radius R 1 = AD, σ 2 with center

B and radius R 2 = BC, and σ 3 with center P and radius x. The points C and

D lie on σ 2 and σ 1 respectively, and CD is tangent to σ 3 . From this it is plain that the greatest value of x occurs when CD is also tangent to σ 1 and σ 2 . We shall show that in this case the required inequality is really an equality, i.e., that √ 1 x = √ 1

AD + √ 1 BC . Then the inequality will immediately follow.

Denote the point of tangency of CD with σ p 3 by M. By the Pythagorean theorem √ we have CD = (R 1 +R 2 ) 2 − (R 1 −R 2 ) 2 =2 R 1 R 2 √ . On the other hand, CD √ = CM + MD = 2 R 2 x +2 R 1 x . Hence, we obtain √ 1 = √ 1 + √ x 1 R 1 R 2 .

14. Lemma 1. In a quadrilateral ABCD circumscribed about a circle, with points of tangency P , Q, R, S on DA, AB, BC,CD respectively, the lines AC, BD, PR, QS concur.

Proof. Follows immediately, for example, from Brianchon’s theorem. Lemma 2. Let a variable chord XY of a circle C (I, r) subtend a right angle at

a fixed point Z within the circle. Then the locus of the midpoint P of XY is a circle whose center is at the midpoint M of IZ and whose radius is p r 2 /2 − IZ 2 /4.

528 4 Solutions Proof. From ∠X ZY

IX − → IZ → − − IY → IZ → = 90 − · =( − )·( − ) = 0. There- fore,

◦ follows −→ ZX −→ ZY

Lemma 3. Using notation as in Lemma 1, if ABCD is cyclic, PR is perpendic- ular to QS. Proof. Consider the inversion in C (I, r), mapping A to A ′ etc. (P , Q, R, S are fixed). As is easily seen, A ′ ,B ′ ,C ′ ,D ′ will lie at the midpoints of PQ, QR, RS , SP, respectively. A ′ B ′ C ′ D ′ is a parallelogram, but also cyclic, since in- version preserves circles; thus it must be a rectangle, and so PR ⊥ QS. Now we return to the main result. Let I and O be the incenter and circumcenter, Z the intersection of the diagonals, and P , Q, R, S, A ′ ,B ′ ,C ′ ,D ′ points as defined in Lemmas 1 and 3. From Lemma 3, the chords PQ , QR, RS, SP subtend 90 ◦ at Z . Therefore by Lemma 2 the points A ′ ,B ′ ,C ′ ,D ′ lie on a circle whose center is the midpoint Y of IZ. Since this circle is the image of the circle ABCD under the considered inversion (centered at I), it follows that I , O,Y are collinear, and hence so are I , O, Z.

Remark. This is the famous Newton’s theorem for bicentric quadrilaterals.

15. By Cauchy’s inequality, 44 < 1989 <a+b+c+d≤ 2 · 1989 < 90. Since m 2 = a + b + c + d is of the same parity as a 2 +b 2 +c 2 +d 2 = 1989, m 2 is either

49 or 81. Let d = max{a,b,c,d}. Suppose that m 2 = 49. Then (49−d) 2 = (a +b + c) 2 >a 2 +b 2 +c 2 = 1989−d 2 , and so d 2 p − 49d + 206 > 0. This inequality does not hold for 5 ≤ d ≤ 44. Since

d ≥ 1989 /4 > 22, d must be at least 45, which is impossible because 45 2 > 1989. Thus we must have m 2 √ = 81 and m = 9. Now, 4d > 81 implies d ≥ 21. On the other hand, d < 1989, and hence d = 25 or d = 36. Suppose that d = 25 and put a = 25 − p, b = 25− q, c = 25− r with p,q,r ≥ 0. From a+ b+ c = 56 it

follows that p + q + r = 19, which, together with (25 − p) 2 + (25 − q) 2 + (25 − r ) 2 = 1364, gives us p 2 +q 2 +r 2 = 439 > 361 = (p + q + r) 2 , a contradiction. Therefore d = 36 and n = 6.

Remark.

A little more calculation yields the unique solution a = 12, b = 15,

c = 18, d = 36.

16. Define S

k =∑ i =0 a i (k = 0, 1, . . . , n) and S −1 = 0. We note that S n −1 =S n . Hence

4.30 Shortlisted Problems 1989 529 n −1

n −1 n −1

S n = ∑ a k = nc + ∑ ∑ a i −k (a i +a i +1 )

= nc + ∑ ∑ a i −k (a i +a i +1 ) = nc + ∑ (a i +a i +1 ) ∑ a i −k

∑ 2 (S

i +1 −S i −1 )S i = nc + S n ,

i =0

i.e., S 2 n −S n + nc = 0. Since S n is real, the discriminant of the quadratic equation must be positive, and hence c

4n .

17. A figure consisting of 9 lines is shown below. ❜

Now we show that 8 lines are not sufficient. Assume the opposite. By the pigeon- hole principle, there is a vertex, say A, that is joined to at most 2 other vertices. Let B ,C, D, E denote the vertices to which A is not joined, and F, G the other two vertices. Then any two vertices of B ,C, D, E must be mutually joined for an edge to exist within the triangle these two points form with A. This accounts for

6 segments. Since only two segments remain, among A, F, and G at least two are not joined. Taking these two and one of B ,C, D, E that is not joined to any of them (it obviously exists), we get a triple of points, no two of which are joined;

a contradiction. Second solution. Since (i) is equivalent to the fact that no three points make a

“blank triangle,” by Turan’s theorem the number of “blank edges” cannot exceed [7 2 /4] = 12, leaving at least 7 · 6/2− 12 = 9 segments. For general n, the answer is

[(n − 1)/2] 2 .

18. Consider the triangle MA i M i . Obviously, the point M

i is the image of A i under the composition C of rotation R α/2−90 M and homothety H 2 sin (α/2) M . Therefore, the polygon M 1 M 2 ...M n is obtained as the image of A 1 A 2 ...A n under the rotational homothety C with coefficient 2 sin ( α /2). Therefore S

2 M 1 M 2 ...M n = 4 sin ( α /2)·S.

19. Let us color the board in a chessboard fashion. Denote by S b and S w respectively the sum of numbers in the black and in the white squares. It is clear that every allowed move leaves the difference S b −S w unchanged. Therefore a necessary

condition for annulling all the numbers is S b =S w . We now show it is sufficient. Assuming S b =S w let us observe a triple of (differ- ent) cells a , b, c with respective values x a ,x b ,x c where a and c are both adjacent to b. We first prove that we can reduce x a to be 0 if x a > 0. If x a ≤x b , we subtract x a from both a and b. If x a >x b , we add x a −x b to b and c and proceed as in the previous case. Applying the reduction in sequence, along the entire board, we

530 4 Solutions reduce all cells except two neighboring cells to be 0. Since S b =S w is invariant,

the two cells must have equal values and we can thus reduce them both to 0.

20. Suppose k ≥ 1/2 + 2n. Consider a point P in S. There are at least k points in S having all the same distance to P, so there are at least k 2 pairs of points A ,B with AP

= BP. Since this is true for every point P ∈ S, there are at least n k

triples of points (A, B, P) for which AP = BP holds. However, k

2 4 2 Since n 2 is the number of all possible pairs (A, B) with A, B ∈ S, there must ex-

ist a pair of points A , B with more than two points P i such that AP i = BP i . These points P i are collinear (they lie on the perpendicular bisector of AB), contradict- ing condition (i).

21. In order to obtain a triangle as the intersection we must have three points P , Q, R on three sides of the tetrahedron passing through one vertex, say T . It is clear that we may suppose w.l.o.g. that P is a vertex, and Q and R lie on the edges

TP 1 and T P 2 (P 1 ,P 2 are vertices) or on their extensions respectively. Suppose −→

that TQ = λ −−→ TP 1 and − TR → = µ −−→ TP 2 , where λ , µ > 0. Then −→ PQ −→

cos ∠QPR − 1)( µ − 1) + 1

· PR

2 λ 2 − λ +1 µ 2 − µ +1 In order to obtain an obtuse angle (with cos < 0) we must choose µ p < 1 and

PQ

· PR

> −µ 1 −µ > 1. Since λ 2 − λ +1> λ − 1 and µ 2 − µ +1>1− µ , we get that for ( λ − 1)( µ − 1) + 1 < 0,

1 cos ∠QPR > − (1 − µ )( λ − 1)

hence ∠QPR < 120 ◦ .

Remark. After obtaining the formula for cos ∠QPR, the official solution was as follows: For fixed µ 0 < 1 and λ > 1, cos ∠QPR is a decreasing function of λ : indeed, ∂ cos ∠QPR

4 ( λ 2 − λ + 1) 3 /2 ( µ 2 − µ + 1) 1 /2 Similarly, for a fixed, sufficiently large λ 0 , cos ∠QPR is decreasing for µ de-

creasing to 0. Since lim λ →0,µ→0+ cos ∠QPR = −1/2, we conclude that ∠QPR < 120 ◦ .

22. The statement remains valid if 17 is replaced by any divisor k of 1989 =3 2 · 13 ·

17, 1 < k < 1989, so let k be one such divisor. The set {1,2,...,1989} can be partitioned as

{1,2,...,3k} ∪ j =1 {(2 j + 1)k + 1,(2 j + 1)k + 2,...,(2 j + 1)k + 2k } = X ∪Y 1 ∪ ··· ∪Y L , where L = (1989 − 3k)/2k. The required statement will

be an obvious consequence of the following two claims.

4.30 Shortlisted Problems 1989 531 Claim 1. The set X = {1,2,...,3k} can be partitioned into k disjoint subsets,

each having 3 elements and the same sum. Proof. Since k is odd, let t = k − 1/2 and X = {1,2,...,6t + 3}. For l =

1 , 2, . . . ,t, define

X 2l −1 = {l,3t + 1 + l,6t + 5 − 2l},

X 2l = {t + 1 + l,2t + 1 + l,6t + 4 − 2l}

X 2t +1 =X k = {t + 1,4t + 2,4t + 3}.

It is easily seen that these three subsets are disjoint and that the sum of elements in each set is 9t + 6.

Claim 2. Each Y j = {(2 j + 1)k + 1,...,(2 j + 1)k + 2k} can be partitioned into k disjoint subsets, each having 2 elements and the same sum. Proof. The obvious partitioning works: Y j = {(2 j + 1)k + 1,(2 j + 1)k + 2k} ∪ ··· ∪ {(2 j + 1)k + k,(2 j + 1)k + (k + 1)}.

23. Two numbers x , y ∈ {1,...,2n} will be called twins if |x − y| = n. Then the set {1,...,2n} splits into n pairs of twins. A permutation (x 1 ,...,x 2n ) of this set

is said to be of type T k if |x i −x i +1 | = n holds for exactly k indices i (thus a permutation of type T 0 contains no pairs of neighboring twins). Denote by F k (n) the number of T k -type permutations of {1,...,2n}. Let (x 1 ,...,x 2n ) be a permutation of type T 0 . Removing x 2n and its twin, we obtain a permutation of 2n − 2 elements consisting of n − 1 pairs of twins. This new permutation is of one of the following types:

(i) type T 0 :x 2n can take 2n values, and its twin can take any of 2n −2 positions; (ii) type T 1 :x 2n can take any one of 2n values, but its twin must be placed to

separate the unique pair of neighboring twins in the new permutation. The recurrence formula follows:

F 0 (n) = 2n[(2n − 2)F 0 (n − 1) + F 1 (n − 1)]. (1) Now let (x 1 ,...,x 2n ) be a permutation of type T 1 , and let (x j ,x j +1 ) be the unique

neighboring twin pair. Similarly, on removing this pair we get a permutation of 2n − 2 elements, either of type T 0 or of type T 1 . The pair (x j ,x j +1 ) is chosen out of n twin pairs and can be arranged in two ways. Also, in the first case it can be placed anywhere (2n − 1 possible positions), but in the second case it must be placed to separate the unique pair of neighboring twins. Hence,

F 1 (n) = 2n[(2n − 1)F 0 (n − 1) + F 1 (n − 1)] = F 0 (n) + 2nF 0 (n − 1). (2) This implies that F 0 (n) < F 1 (n). Therefore the permutations with at least one

neighboring twin pair are more numerous than those with no such pairs. Remark 1. As in the official solution, formulas (1) and (2) together give for F 0

the recurrence

F 0 (n) = 2n[(2n − 1)F 0 (n − 1) + (2n − 2)F 0 (n − 2)].

532 4 Solutions For the ratio p n =F 0 (n)/(2n)!, simple algebraic manipulation yields p n =

n −1 +

(2n−3)(2n−1) . Since p 1 = 0, we get

1 1 1 1 p n <p n −1 +

2 − 2 < ··· < (2n − 3)(2n − 1) . (2n − 3) (2n − 1) 2 Remark 2. Using the inclusion–exclusion principle, the following formula can

=p n −1 +

be obtained: n

One consequence is that in fact, lim n →∞ p n = 1/e. Second solution. Let f : T 0 →T 1 be the mapping defined as follows: if (x 1 ,x 2 ,

...,x 2n )∈T 0 and x k ,k > 2, is the twin of x 1 , then

f (x 1 ,x 2 ,...,x 2n ) = (x 2 ,...,x k −1 ,x 1 ,x k ,...,x 2n ). The mapping f is injective, but not surjective. Thus F 0 (n) < F 1 (n).

24. Instead of Euclidean distance, we will use the angles ∠A i OA j , O denoting the center of the sphere. Let {A 1 ,...,A 5 } be any set for which min i 6= j ∠A i OA j ≥ π /2 (such a set exists: take for example five vertices of an octagon). We claim that two of the A i ’s must be antipodes, thus implying that min i 6= j ∠A i OA j is exactly √

equal to π /2, and consequently that min i 6= j A i A j = 2. Suppose no two of the five points are antipodes. Visualize A 5 as the south pole. Then A 1 ,...,A 4 lie in the northern hemisphere, including the equator (but ex- cluding the north pole). No two of A 1 ,...,A 4 can lie in the interior of a quarter of this hemisphere, which means that any two of them differ in longitude by at least π /2. Hence, they are situated on four meridians that partition the sphere into quarters. Finally, if one of them does not lie on the equator, its two neigh- bors must. Hence, in any case there will exist an antipodal pair, giving us a contradiction.

25. We may assume w.l.o.g. that a > 0 (because a, b < 0 is impossible, and a, b 6= 0 from the condition of the problem). Let (x 0 ,y 0 ,z 0 ,w 0 ) 6= (0,0,0,0) be a solution of x 2 2 2 + abw − ay 2 − bz . Then

x 2 − ay 2 = b(z 2 − aw 0 2 0 0 0 ). Multiplying both sides by (z 2 0 2 − aw 0 ), we get

(x 2 2 2 2 2 2 0 2 − ay 0 )(z 0 − aw 0 ) − b(z 0 − aw 0 ) =0 z

w ) 2 2 2 ⇔ (x 2 0 0 − ay 0 0 − a(y 0 0 −x 0 0 − b(z 0 − aw 0 ) = 0. Hence, for x 1 =x 0 z 0 − ay 0 w 0 ,y 1 =y 0 z 0 −x 0 w 0 ,z 1 =z 2 0 − aw 2 0 , we have

4.30 Shortlisted Problems 1989 533

x 2 1 − ay 2 1 − bz 2 1 = 0.

If (x 1 ,y 1 ,z 1 ) is the trivial solution, then z 1 = 0 implies z 0 =w 0 = 0 and simi- larly x 0 =y 0 = 0 because a is not a perfect square. This contradicts the initial assumption.

26. By the Cauchy–Schwarz inequality,

∑ x i ≤n ∑ x 2 i .

i =1

i =1

Since ∑ n x

0 and ∑ i n =1 i =a−x i =1 x 2 i 2 =b−x 0 , we have (a − x 0 ) 2 ≤ n(b − x 2 0 ), i.e., (n + 1)x 2 0 2 − 2ax 0 + (a − nb) ≤ 0. The discriminant of this quadratic is D = 4n(n + 1) b −a 2 /(n + 1) , so we con-

clude that

(i) if a 2 > (n + 1)b, then such an x 0 does not exist; (ii) if a 2 = (n + 1)b, then x 0 √ = a/n + 1;

(iii) if a 2 < (n + 1)b, then a − D a n D +1 ≤x 0 ≤ n +1 .

It is easy to see that these conditions for x 0 are also sufficient.

27. Let n be the required exponent, and suppose n =2 k q , where q is an odd integer. Then we have

2 − 1 = (m k − 1)[(m (q−1)

2 + ··· + m k + 1] = (m 2 k − 1)A, where A is odd. Therefore m k n − 1 and m 2 − 1 are divisible by the same power

of 2, and so n =2 k . Next, we observe that

2 k − 1 = (m −1 − 1)(m + 1) = . . .

2 k −1

2 2 4 2 k = (m −1 − 1)(m + 1)(m + 1) ···(m + 1). Let s be the maximal positive integer for which m

≡ ±1 (mod 2 s ). Then m 2 −1 is divisible by 2 s +1 and not divisible by 2 s +2 . All the numbers m 2 4

+ 1, m +

1 ,...,m −1 + 1 are divisible by 2 and not by 4. Hence m 2 k − 1 is divisible by

2 s +k and not by 2 s +k+1 . It follows from the above consideration that the smallest exponent n equals

2 1989 −s if s ≤ 1989, and n = 1 if s > 1989.

28. Assume w.l.o.g. that the rays OA 1 , OA 2 , OA 3 , OA 4 are arranged clockwise. Set- ting OA 1 = a, OA 2 = b, OA 3 = c, OA 4 = d, and ∠A 1 OA 2 = x, ∠A 2 OA 3 = y, ∠A 3 OA 4 = z, we have

1 1 S 1 = σ (OA 1 A 2 )= ab

2 |sinx|, S 2 = σ (OA 1 A 3 )= ac 2 |sin(x + y)|,

1 1 S 3 = σ (OA 1 A 4 )= ad |sin(x + y + z)|, S 4 = σ (OA 2 A 3 )= bc |siny|,

1 1 S 5 = σ (OA 2 A 4 )= bd |sin(y + z)|, S 6 = σ (OA 3 A 4 )= cd

2 2 |sinz|.

534 4 Solutions Since sin (x + y + z) sin y + sin x sin z = sin(x + y) sin(y + z), it follows that there

exists a choice of k , l ∈ {0,1} such that S 1 S 6 + (−1) k S 2 S 5 + (−1) l S 3 S 4 = 0. For example (w.l.o.g.), if S 3 S 4 =S 1 S 6 +S 2 S 5 , we have

1 ≥S 3 S 4 =S 1 6 +S 2 S 5 ≤i≤6 ≥ 1 + 1 = 2, √ i.e., max 1 ≤i≤6 S i ≥

max S i

2 as claimed.

29. Let P i , sitting at the place A, and P j sitting at B, be two birds that can see each other. Let k and l respectively be the number of birds visible from B but not from

A , and the number of those visible from A but not from B. Assume that k ≥ l. Then if all birds from B fly to A, each of them will see l new birds, but won’t see k birds anymore. Hence the total number of mutually visible pairs does not increase, while the number of distinct positions occupied by at least one bird decreases by one. Repeating this operation as many times as possible one can arrive at a situation in which two birds see each other if and only if they are in the same position. The number of such distinct positions is at most 35, while the total number of mutually visible pairs is not greater than at the beginning. Thus the problem is equivalent to the following one:

(1) If x i ≥ 0 are integers with ∑ 35 j =1 x j = 155, find the least possible value of ∑ 35 j =1 (x 2 j −x j )/2. If x j ≥x i + 2 for some i, j, then the sum of (x 2 j −x j )/2 decreases (for x j −x i −

2) if x i ,x j are replaced with x i + 1, x j − 1. Consequently, our sum attains its minimum when the x i ’s differ from each other by at most 1. In this case, all the x i ’s are equal to either [155/35] = 4 or [155/35] + 1 = 5, where 155 =

20 ·4 + 15 · 5. It follows that the (minimum possible) number of mutually visible

pairs is 20 4 · 5 ·3 2 + 15 · ·4 2 = 270.

Second solution for (1). Considering the graph consisting of birds as vertices and pairs of mutually nonvisible birds as edges, we see that there is no complete 36- subgraph. Turan’s theorem gives the answer immediately. (See problem (SL89- 17).)

30. For all n such N exists. For a given n choose N = (n + 1)! 2 + 1. Then 1 + j is a proper factor of N + j for 1 ≤ j ≤ n. So if N + j = p m is a power of a prime p, then 1 +j=p r for some integer r, 1 ≤ r < m. But then p r +1 divides

both (n + 1)! 2 = N − 1 and p m = N + j, implying that p r +1 | 1 + j, which is impossible. Thus none of N + 1, N + 2, . . ., N + n is a power of a prime.

be distinct primes. By the Chinese remain- der theorem, there exists a natural number N such that p 1 p 2 | N +1, p 3 p 4 | N +2, ...,p 2n −1 p 2n | N + n, and then obviously none of the numbers N + 1,...,N + n can be a power of a prime.

Second solution. Let p 1 ,p 2 ,...,p 2n

4.30 Shortlisted Problems 1989 535

31. Let us denote by N pqr the number of solutions for which a p /x p ≥a q /x q ≥a r /x r , where (p, q, r) is one of six permutations of (1, 2, 3). It is clearly enough to prove

that N pqr +N qpr ≤ 2a 1 a 2 (3 + ln(2a 1 )).

First, from 3a p

x p we get a p +1≤x p ≤ 3a p . Similarly, for fixed x p we have

which gives max a q ·x p /a p ,a q ·x p /(x p −a p ) ≤x q ≤ 2a q ·x p /(x p −a p ), i.e., if a p +1≤x p ≤ 2a p there are at most a q ·x p /(x p −a p ) + 1/2 possible values for x q (because there are [2x] − [x] = [x + 1/2] integers between x and 2x), and if

2a p +1≤x p ≤ 3a p , at most 2a q ·x p /(x p −a p )−a q ·x p /a p + 1 possible values. Given x p and x q ,x r is uniquely determined. Hence

N pqr ≤ ∑

+a q ∑ 1 − +a

2 ∑ k + =1 k k +a p

≤a p a q −

2a q 2a p p

+ ln(2a )+

2 − ln2

Here we have used ∑ n k =1 (1/k + 2/(k + n)) ≤ ln(2n) + 2 − ln2 (this can be proved by induction). Hence,

N pqr +N qpr ≤ 2a p a q (1 + 0.5 + ln(2a p ) + 2 − ln2) < 2a 1 a 2 (2.81 + ln(2a 1 )). Remark. The official solution was somewhat simpler, but used that the interval

(x, 2x], for real x, cannot contain more than x integers, which is false in general. Thus it could give only a weaker estimate N ≤ 6a 1 a 2 (9/2 − ln2 + ln(2a 1 )).

32. Let CC ′

be an altitude, and R the circumradius. Then, since AH = R, we have AC ′ = |Rsin B| and hence (1) CC ′ = |Rsin BtanA|. On the other hand, CC ′ =

|BC sinB| = 2|RsinAsinB|, which together with (1) yields 2|sin A| = |tanA| ⇒ |cosA| = 1/2. Hence, ∠A is 60 ◦ . (Without the condition that the triangle is acute, ∠A could also be 120 ◦ .)

536 4 Solutions Second Solution. For a point X , let X denote the vector OX . Then |A| = |B| =

|C| = R and H = A + B +C, and moreover,

− (B −C) 2 = 4R 2 − BC 2 . It follows that sin A = BC

R 2 2 2 = (H − A) 2 = (B +C) 2 = 2B + 2C

2R = 3 /2, i.e., that ∠A = 60 . Third Solution. Let A 1 be the midpoint of BC. It is well known that AH = 2OA 1 , and since AH = AO = BO, it means that in the right-angled triangle BOA 1 the

relation BO = 2OA 1 holds. Thus ∠BOA 1 = ∠A = 60 ◦ .

4.31 Shortlisted Problems 1990 537