Solutions to the Longlisted Problems of IMO 1967

4.9 Solutions to the Longlisted Problems of IMO 1967

1. Let us denote the nth term of the given sequence by a n . Then

2. (n!) 2 /n

1 /n ) 2 1 2 n +1 2 1 2 1 = ((1 · 2···n) 1 ≤ +2+···+n

3. Consider the function f : [0, π /2] → R defined by f (x) = 1 − x 2 /2 + x 4 /16 − cos x . It is easy to calculate that f ′ (0) = f ′′ (0) = f ′′′ (0) = 0 and f ′′′′ (x) = 3/2 − cosx. Since f ′′′′ (x) > 0, f ′′′ (x) is increasing. Together with f ′′′ (0) = 0, this gives

f ′′′ (x) > 0 for x > 0; hence f ′′ (x) is increasing, etc. Continuing in the same way we easily conclude that f (x) > 0.

4. (a) Let ABCD be a parallelogram, and K , L the midpoints of segments BC and CD respectively. The sides of △AKL are equal and parallel to the medians of △ABC. (b) Using the formulas 4m 2 a = 2b 2 + 2c 2 −a 2 etc., it is easy to obtain that m 2 a +

m 2 b =m 2 c is equivalent to a 2 +b 2 = 5c 2 . Then

5 (a 2 +b 2 −c 2 ) = 4(a 2 +b 2 ) ≥ 8ab.

5. If one of x , y, z is equal to 1 or −1, then we obtain solutions (−1,−1,−1) and (1, 1, 1). We claim that these are the only solutions to the system.

Let f (t) = t 2 + t − 1. If among x,y,z one is greater than 1, say x > 1, we have x < f (x) = y < f (y) = z < f (z) = x, which is impossible. It follows that x, y, z ≤

1. Suppose now that one of x , y, z, say x, is less than −1. Since min t f (t) = −5/4, we have x = f (z) ∈ [−5/4,−1). Also, since f ([−5/4,−1)) = (−1,−11/16) ⊆ (−1,0) and f ((−1,0)) = [−5/4,−1), it follows that y = f (x) ∈ (−1,0), z =

f (y) ∈ [−5/4,−1), and x = f (z) ∈ (−1,0), which is a contradiction. Therefore −1 ≤ x,y,z ≤ 1. If −1 < x,y,z < 1, then x > f (x) = y > f (y) = z > f (z) = x, a contradiction. This proves our claim.

6. The given system has two solutions: (−2,−1) and (−14/3,13/3).

7. Let S k =x k 1 +x k 2 + ··· + x k n and let σ k ,k = 1, 2, . . . , n denote the kth elementary symmetric polynomial in x 1 ,...,x n . The given system can be written as S k =a k , k = 1, . . . , n. Using Newton’s formulas

k =S 1 σ k −1 −S 2 σ

k −2 + ··· + (−1) S k −1 σ 1 + (−1) −1 S k , k = 1, 2, . . . , n,

362 4 Solutions

the system easily leads to σ 1 = a and σ k = 0 for k = 2, . . . , n. By Vieta’s formulas, x 1 ,x 2 ,...,x n are the roots of the polynomial x n − ax n −1 , i.e., a , 0, 0, . . ., 0 in some order.

Remark. This solution does not use the assumption that the x j ’s are real.

8. The circles K A ,K B ,K C ,K D cover the parallelogram if and only if for every point

X inside the parallelogram, the length of one of the segments X A , XB, XC, XD does not exceed 1. Let O and r be the center and radius of the circumcircle of △ABD. For every point X inside △ABD, it holds that XA ≤ r or XB ≤ r or XD ≤ r. Similarly, for

X inside △BCD, XB ≤ r or XC ≤ r or XD ≤ r. Hence K A ,K B ,K C ,K D cover the parallelogram if and only if r ≤ 1, which is equivalent to ∠ABD ≥ 30 ◦ . However, this last is exactly equivalent to a = AB = 2r sin ∠ADB ≤ 2sin( α +

3 sin α + cos α .

9. The incenter of any such triangle lies inside the circle k. We shall show that every point S interior to the circle S is the incenter of one such triangle. If S lies on the segment AB, then it is obviously the incenter of an isosceles triangle inscribed in k that has AB as an axis of symmetry. Let us now suppose S does not lie on

AB . Let X and Y be the intersection points of lines AS and BS with k, and let Z

be the foot of the perpendicular from S to AB. Since the quadrilateral BZSX is cyclic, we have ∠ZX S = ∠ABS = ∠SXY and analogously ∠ZY S = ∠SY X , which implies that S is the incenter of △XY Z.

10. Let n be the number of triangles and let b and i be the numbers of vertices on the boundary and in the interior of the square, respectively. Since all the triangles are acute, each of the vertices of the square belongs to at least two triangles. Additionally, every vertex on the boundary belongs to at least three, and every vertex in the interior belongs to at least five triangles. Therefore

3n ≥ 8 + 3b + 5i.

(1) Moreover, the sum of angles at any D C

vertex that lies in the interior, on the boundary, or at a vertex of the square

KL

is equal to 2 π , π , π /2 respectively. The sum of all angles of the triangles equals n π , which gives us n π =4· π /2 + b π + 2i π , i.e., n = 2 + b + 2i. This relation together with (1) easily yields that i ≥

2. Since each of the vertices inside the square belongs to at least five triangles, and at most two contain both, it follows that n ≥ 8.

A B It is shown in the figure that the square can be decomposed into eight acute triangles. Obviously one of them can have an arbitrarily small perimeter.

4.9 Longlisted Problems 1967 363

11. We have to find the number p n of triples of positive integers (a, b, c) satisfying

a ≤ b ≤ c ≤ n and a + b > c. Let us denote by p n (k) the number of such triples with c = k, k = 1, 2, . . . , n. For k even, p n (k) = k + (k − 2) + (k − 4) + ···+ 2 = (k 2 + 2k)/4, and for k odd, p n (k) = (k 2 ( + 2k + 1)/4. Hence n (n + 2)(2n + 5)/24,

for 2 | n, p n =p n (1) + p n (2) + ···+ p n (n) = (n + 1)(n + 3)(2n + 1)/24, for 2 ∤ n.

12. Let us denote by M n the set of points of the segment AB obtained from A and B by not more than n iterations of ( ∗). It can be proved by induction that

for some k ∈N .

Thus (a) immediately follows from M = S M n . It also follows that if a , b ∈ N and

a /b ∈ M, then 3 | a(b − a). Therefore 1/2 6∈ M. √

13. The maximum area is 3 3r 2 /4 (where r is the radius of the semicircle) and is attained in the case of a trapezoid with two vertices at the endpoints of the diameter of the semicircle and the other two vertices dividing the semicircle into three equal arcs.

14. We have that

q (p + q 2 ) because |p 2 − 2q 2 | ≥ 1.

q (p + q 2 )

The greatest p , q ≤ 100 satisfying the equation |p 2 − 2q 2 | = 1 are (p,q) = (99, 70). It is easy to verify using (1) that 99

2 among the fractions p

70 best approximates

99 √ /q with p, q ≤ 100. The numbers 70 = 1.41428 . . . and

√ 2 coincide up to the fourth decimal digit: indeed, (1) gives 7 · 10 −5 < p

q − 2 < 8 · 10 . Second solution. By using some basic facts about Farey sequences, one can find that 41

29 < 2 < 99 70 and that 41 29 < p q < 99 70 implies p ≥ 41 + 99 > 100 because √

99 · 29 − 41 · 70 = 1. Of the two fractions 41 29 and 99 70 , the latter is closer to 2.

15. Given that tan α ∈ Q, we have that tan β is rational if and only if tan γ is rational, where γ = β − α and 2 γ = α . Putting t = tan γ we obtain p

q = tan 2 γ = 1 −t 2 , which leads to the quadratic equation pt 2 + 2qt − p = 0. This equation has ra- tional solutions if and only if its discriminant 4 (p 2 +q 2 ) is a perfect square, and the result follows.

2t

16. First let us notice that all the numbers z m 1 ,m 2 =m 1 r 1 +m 2 r 2 (m 1 ,m 2 ∈ Z) are distinct, since r 1 /r 2 is irrational. Thus for any n ∈ N the interval [−n(|r 1 |+

2 1 2 |r 2 |),n(|r | + |r |)] contains (2n + 1) numbers z m 1 ,m 2 , where |m 1 |,|m 2 | ≤ n. Therefore some two of these (2n + 1) 2 numbers, say z m 1 ,m 2 ,z n 1 ,n 2 , differ by

at most 2n (|r 1 |+|r 2 |) = (|r 1 (2n+1) |+|r 2 2 −1 |) 2 (n+1) . By taking n large enough we can achieve that z q 1 ,q 2 = |z m 1 ,m 2 −z n 1 ,n 2 | ≤ p. If now k is the integer such that kz q 1 ,q 2 ≤x<

(k + 1)z q 1 ,q 2 , then z kq 1 ,kq 2 = kz q 1 ,q 2 differs from x by at most p, as desired.

364 4 Solutions

17. Using c r −c s = (r − s)(r + s + 1) we can easily get (c m +1 −c k ) ··· (c m +n −c k )

(m − k + n)!

(m + k + n + 1)!

c 1 c 2 ··· c n (m − k)!n! (m + k + 1)!(n + 1)! The first factor (m−k+n)! (m−k)!n! = m −k+n n is clearly an integer. The second factor is

also an integer because by the assumption, m + k + 1 and (m + k)!(n + 1)! are coprime, and (m + k + n + 1)! is divisible by both; hence it is also divisible by their product.

18. In the first part, it is sufficient to show that each rational number of the form m /n!, m, n ∈ N, can be written uniquely in the required form. We prove this by induction on n. The statement is trivial for n = 1. Let us assume it holds for n −1, and let there be given a rational number m /n!. Let us take a n ∈ {0,...,n − 1} such that m − a n =

nm 1 for some m 1 ∈ N. By the inductive hypothesis, there are unique a 1 ∈N 0 ,

i ∈ {0,...,i − 1} (i = 1,...,n − 1) such that m 1 /(n − 1)! = ∑ −1 i =1 a i /i!, and then

(n − 1)! n !

i =1 i !

i =1 a i /i!, multiplying by n! we see that m −a n must be a multiple of n, so the choice of a n was unique and therefore the

as desired. On the other hand, if m

/n! = ∑ n

representation itself. This completes the induction. In particular, since a i | i! and i!/a i > (i − 1)! ≥ (i − 1)!/a i −1 , we conclude that each rational q, 0 < q < 1, can be written as the sum of different reciprocals. Now we prove the second part. Let x > 0 be a rational number. For any integer

m > 10 6 , let n > m be the greatest integer such that y = x− 1 m − 1 m +1 −···− 1 n > 0. Then y can be written as the sum of reciprocals of different positive integers, which all must be greater than n. The result follows immediately.

19. Suppose n ≤ 6. Let us decompose the disk by its radii into n congruent regions, so that one of the points P j lies on the boundaries of two of these regions. Then one of these regions contains two of the n given points. Since the diameter of each of these regions is 2 sin π n , we have d n ≤ 2sin π n . This value is attained if P i are the vertices of a regular n-gon inscribed in the boundary circle. Hence

D n = 2 sin π n . For n = 7 we have D 7 ≤D 6 = 1. This value is attained if six of the seven points form a regular hexagon inscribed in the boundary circle and the seventh is at the center. Hence D 7 = 1.

20. The statement so formulated is false. It would be true under the additional as- sumption that the polygonal line is closed. However, from the offered solution, which is not clear, it does not seem that the proposer had this in mind.

21. Using the formula cos x cos 2x cos 4x ···cos2 n −1 x = sin 2 n x 2 n sin x , which is shown by simple induction, we obtain

1 cos

cos cos

cos

= − cos

cos

cos cos = ,

4.9 Longlisted Problems 1967 365

1 5 π 1 cos

cos

cos

15 15 4 15 2 Multiplying these equalities, we get that the required product P equals 1 /128.

22. Let O 1 and O 2 be the centers of circles k 1 and k 2 and let C be the midpoint of the segment AB. Using the well-known relation for elements of a triangle, we obtain

PA 2 + PB 2 = 2PC 2 + 2CA 2 ≥ 2O 1 C 2 + 2CA 2 = 2O 1 A 2 = 2r 2 . Equality holds if P coincides with O 1 or if A and B coincide with O 2 .

23. Suppose that a ≥ 0, c ≥ 0, 4ac ≥ b 2 . If a = 0, then b = 0, and the inequality reduces to the obvious cg 2 ≥ 0. Also, if a > 0, then

b 2 4ac

af 2 + b f g + cg 2 =a f + g + −b g 2 2a 4a ≥ 0.

Suppose now that a f 2 + b f g + cg 2 ≥ 0 holds for an arbitrary pair of vectors f ,g. Substituting f by tg (t ∈ R) we get that (at 2 + bt + c)g 2 ≥ 0 holds for any real

number t. Therefore a ≥ 0, c ≥ 0, 4ac ≥ b 2 .

24. Let m be the total number of coins and suppose that the kth child receive x k coins. By the condition of the problem, the number of coins that remain after him was

6 (x k − k). This gives us a recurrence relation

6 6 x k +1 =k+1+

6 (x k − k) − k − 1

7 7 7 which, together with the condition x 1 = 1 + (m − 1)/7, yields

6 k −1

7 k (m − 36) + 6 for 1 ≤ k ≤ n. Since we are given x n = n, we obtain 6 n −1

(m − 36) = 7 n (n − 6). It follows that

6 n −1 | n − 6, which is possible only for n = 6. Hence, n = 6 and m = 36. √

25. The answer is R = (4 + 3 )d/6.

26. Let L be the midpoint of the edge AB. Since P is the orthocenter of △ABM and ML is its altitude, P lies on ML and therefore belongs to the triangular area LCD. Moreover, from the similarity of triangles ALP and MLB we have

LP

· LM = LA · LB = a 2 /4,

where a is the side length of tetrahedron ABCD. It easily follows that the lo- cus of P is the image of the segment CD under the inversion of the plane LCD with center L and radius a /2. This locus is the arc of a circle with center L and endpoints at the orthocenters of triangles ABC and ABD.

366 4 Solutions

27. Regular polygons with 3, 4, and 6 sides can be obtained by cutting a cube with

a plane, as shown in the figure. A poly- gon with more than 6 sides cannot be obtained in such a way, for a cube has

6 faces. Also, if a pentagon is obtained by cutting a cube with a plane, then its sides lying on opposite faces are paral- lel; hence it cannot be regular.

28. The given expression can be transformed into

− cos2x

It does not depend on x if and only if cos 2u = −1/2, i.e., u = ± π /3 + k π for some k ∈ Z.

29. Let arc l a be the locus of points A lying on the opposite side from A 0 with respect to the line B 0 C 0 such that ∠B 0 AC 0 = ∠A ′ . Let k a be the circle containing l a , and let S a be the center of k a . We similarly define l b ,l c ,k b ,k c ,S b ,S c . It is easy to show that circles k a ,k b ,k c have a common point S inside △ABC. Let A 1 ,B 1 ,

C 1 be the points on the arcs l a ,l b ,l c diametrically opposite to S with respect to S a ,S b ,S c respectively. Then A 0 ∈B 1 C 1 because ∠B 1 A 0 S = ∠C 1 A 0 S = 90 ◦ ; similarly, B 0 ∈A 1 C 1 and C 0 ∈A 1 B 1 . Hence the triangle A 1 B 1 C 1 is circumscribed

about △A 0 B 0 C 0 and similar to △A ′ B ′ C ′ . Moreover, we claim that △A 1 B 1 C 1 is the triangle ABC with the desired proper- ties having the maximum side BC and hence the maximum area. Indeed, if ABC is any other such triangle and S ′ b ,S ′ c are the projections of S b and S c onto the line BC , it holds that BC = 2S ′ b S ′ c ≤ 2S b S c =B 1 C 1 ,

which proves the maximality of B 1 C 1 .

30. We assume without loss of generality that m ≤ n. Let r and s be the numbers of pairs for which i − j ≥ k and of those for which j − i ≥ k. The desired number is r + s. We easily find that

r = (m − k)(m − k + 1)/2, k < m,

k ≥ m,

  m (2n − 2k − m + 1)/2, k < n − m,

s = (n − k)(n − k + 1)/2, n − m ≤ k < n,

k ≥ n.

31. Suppose that n 1 ≤n 2 ≤ ··· ≤ n k . If n k < m, there is no solution. Otherwise, the solution is

1 + (m − 1)(k − s + 1) + ∑ n i ,

i <s

where s is the smallest i for which m ≤n i holds.

4.9 Longlisted Problems 1967 367

32. Let us denote by V the volume of the given body, and by V a ,V b ,V c the volumes of the parts of the given ball

that lie inside the dihedra of the given

trihedron. It holds that V a = 2R 3 α /3,

V b = 2R 3 β /3, V c = 2R 3 γ /3. It is easy

O to see that 2

(V ′

a +V b +V c ) = 4V +

4 π R 3 /3, from which it follows that

V = R 3 ( α + β + γ π ).

33. If m 6∈ {−2,1}, the system has the unique solution

,z= . (2 + m)(1 − m)

,y=

(2 + m)(1 − m) The numbers x , y, z form an arithmetic progression if and only if a, b, c do so.

(2 + m)(1 − m)

For m = 1 the system has a solution if and only if a = b = c, while for m = −2 it has a solution if and only if a + b + c = 0. In both these cases it has infinitely many solutions.

34. Each vertex of the polyhedron is a vertex of exactly two squares and triangles (more than two is not possible; otherwise, the sum of angles at a vertex exceeds 360 ◦ ). By using the condition that the trihedral angles are equal it is easy to see that such a polyhedron is uniquely determined by its side length. The poly- hedron obtained from a cube by “cut- ting” its vertices, as shown in the fig- ure, satisfies the conditions.

Now it is easy to calculate that the ratio of the squares of volumes of that poly- hedron and of the ball whose boundary is the circumscribed sphere is equal to

35. The given sum can be rewritten as

2x ! n k

2 . k =0 k

∑ 2 tan

tan

2 ∑ =0 k 1 − tan k 2x 2

2 tan 2 Since (x/2)

1 −tan 2 (x/2) = 1 −cosx cos x , the above sum is transformed using the binomial for- mula into

1 + tan 2 x n

1 − cosx n

= sec 2n + sec n x .

2 cos x

36. Suppose that the skew edges of the tetrahedron ABCD are equal. Let K, L, M, P, Q , R be the midpoints of edges AB, AC, AD, CD, DB, BC respectively. Segments KP , LQ, MR have the common midpoint T .

368 4 Solutions We claim that the lines KP, LQ and A MR are axes of symmetry of the tetra-

hedron ABCD. From LM k CD k RQ

M and similarly LR k MQ and LM = CD /2 = AB/2 = LR it follows that

L T LMQR is a rhombus and therefore

B Q D LQ ⊥ MR. We similarly show that KP

is perpendicular to LQ and MR, and P thus it is perpendicular to the plane

C LMQR . Since the lines AB and CD are parallel to the plane LMQR, they are perpendicular to KP. Hence the points A and C are symmetric to B and D with respect to the line KP, which means that KP is an axis of symmetry of the tetra- hedron ABCD. Similarly, so are the lines LQ and MR. The centers of circumscribed and inscribed spheres of tetrahedron ABCD must lie on every axis of symmetry of the tetrahedron, and hence both must coincide with T . Conversely, suppose that the centers of circumscribed and inscribed spheres of the tetrahedron ABCD coincide with some point T . Then the orthogonal pro-

jections of T onto the faces ABC and ABD are the circumcenters O 1 and O 2 of these two triangles, and moreover, T O 1 =TO 2 . Pythagoras’s theorem gives AO 1 = AO 2 , which by the law of sines implies ∠ACB = ∠ADB. Now it easily follows that the sum of the angles at one vertex of the tetrahedron is equal to 180 ◦ . Let D ′ ,D ′′ , and D ′′′

be the points in the plane ABC lying outside △ABC such that △D ′ BC ∼ = △DBC, △D ′′ CA ∼ = △DCA, and △D ′′′ AB ∼ = △DAB. The angle D ′′ AD ′′′ is then straight, and hence A , B,C are midpoints of the segments

D ′′ D ′′′ ,D ′′′ D ′ ,D ′ D ′′ respectively. Hence AD =D ′′ D ′′′ /2 = BC, and analogously AB = CD and AC = BD.

37. Using the A–G mean inequality we obtain

8a 2 b 3 c 3 8 + 3b 8 + 3c ≤ 2a 8 , 8a 3 b 2 c 3 8 + 2b 8 + 3c ≤ 3a 8 , 8a 3 b 3 c 2 8 + 3b 8 + 2c ≤ 3a 8 .

By adding these inequalities and dividing by 3a 3 b 3 c 3 we obtain the desired one.

38. Suppose that there exist integers n and m such that m 3 = 3n 2 + 3n + 7. Then from m 3 ≡ 1 (mod 3) it follows that m = 3k + 1 for some k ∈ Z. Substituting into the initial equation we obtain 3k (3k 2 + 3k + 1) = n 2 + n + 2. It is easy to check that n 2 + n + 2 cannot be divisible by 3, and so this equality cannot be true. Therefore our equation has no solutions in integers.

39. Since sin 2 A + sin 2 B + sin 2 C + cos 2 A + cos 2 B + cos 2 C = 3, the given equality is equivalent to cos 2 A + cos 2 B + cos 2 C = 1, which by multiplying by 2 is trans- formed into

0 = cos 2A + cos2B + 2 cos 2 C = 2 cos(A + B) cos(A − B) + 2cos 2 C = 2 cosC(cos(A − B) − cosC).

4.9 Longlisted Problems 1967 369 It follows that either cosC = 0 or cos(A − B) = cosC. In both cases the triangle

is right-angled.

40. Suppose CD is the longest edge of the tetrahedron ABCD, AB = a, CK and DL are the altitudes of the triangles ABC and ABD respectively, and DM is the alti-

tude of the tetrahedron ABCD. Then CK 2 ≤1−a 2 /4, since CK is a leg of the right triangle whose other leg has length not less than a /2 and whose hypotenuse has length not greater than 1 (AKC or BKC). In the similar way we can show that

DL 2 ≤1−a 2 /4. Since DM ≤ DL, then DM 2 ≤1−a 2 /4. It follows that

V = CK DM

3 2 ≤ 6 − 4 24 (2 − a)(2 + a)

[1 − (a − 1) 2 ](2 + a) ≤ ·1·3= 24 . 24 8

41. It is well known that the points K, L, M, symmetric to H with respect to BC ,CA, AB respectively, lie on the circumcircle k of the triangle ABC. For K, this follows from an elementary calculation of angles of triangles HBC and noting that ∡KBC = ∡HBC = ∡KAC. For other points the proof is analogous.

Since the lines l a ,l b pass through K and l a

L and l is obtained from l a by rota-

tion about C for an angle 2 γ = ∠LCK,

H l it follows that the intersection point P c l b

of l a and l b is at the circumcircle of KLC , that is, k. Similarly, l b and l c B C meet at a point on k; hence they must pass through the same point P.

42. E = (1 − sinx)(1 − cosx)[3 + 2(sinx + cosx) + 2sinxcosx + sinxcosx(sin x + cos x )].

43. We can write the given equation in the form x 5 −x 3 − 4x 2 − 3x − 2 + λ (5x 4 + α x 2 − 8x + α ) = 0.

A root of this equation is independent of λ if and only if it is a common root of the equations

x 5 3 −x 2 − 4x − 3x − 2 = 0

and

5x 4 + α x 2 − 8x + α = 0. The first of these two equations is equivalent to √ (x − 2)(x 2 + x + 1) 2 = 0 and has

three different roots: x 1 = 2, x 2 ,3 = (−1 ± i 3 )/2.

(a) For α = −64/5, x 1 = 2 is the unique root independent of λ . (b) For α = −3 there are two roots independent of λ :x 1 = ω and x 2 = ω 2 .

44. (a) S (x, n) = n(n − 1) x 2 + (n + 1)x + (n + 1)(3n + 2)/12 .

370 4 Solutions (b) It is easy to see that the equation S (x, n) = 0 has two roots

x 1 ,2 = −(n + 1) ± (n + 1)/3 /2. They are integers if and only if n = 3k 2 − 1 for some k ∈ N.

45. (a) Using the formula 4 sin 3 x = 3 sin x − sin3x one can easily reduce the given equation to sin 3x = cos 2x. Its solutions are given by x = (4k + 1) π /10, k ∈ Z.

(b) (1) The point B corresponding to the solution x = (4k + 1) π /10 is a vertex of the regular dodecagon if and only if (4k + 1) π /10 = 2m π /12, i.e.,

3 (4k +1) = 5m for some m ∈ Z. This is possible if and only if 5 | 4k+1, i.e., k ≡ 1 (mod 5). (2) Similarly, if the point B corresponding to x = (4k + 1) π /10 is a vertex of a polygon P, then (4k + 1)n = 20m for some m ∈ N, which implies that 4 | n.

46. Let us set arctan x = a, arctan y = b, arctan z = c. Then tan(a + b) = x +y 1 −xy and tan (a + b + c) = x +y+z−xyz 1 −yz−zx−xy = 1, which implies that

(x − 1)(y − 1)(z − 1) = xyz − xy − yz − zx + x + y + z − 1 = 0. One of x , y, z is equal to 1, say z = 1, and consequently x + y = 0. Therefore

x 2n +1 +y 2n +1 +z 2n +1 =x 2n +1

2n + (−x) +1 +1 = 1.

2n +1

47. Using the A–G mean inequality we get

(n + k − 1)x n 1 x

2 ···x k ≤ nx +k−1 +x +k−1

2 + ··· + x 1 +k−1 ,

2 + ··· + x k , ......

n +k−1

(n + k − 1)x 2 ···x k ≤x 1 + nx n +k−1

n +k−1

k ≤x 1 +x 2 + ··· + nx k . By adding these inequalities and dividing by n + k − 1 we obtain the desired one.

(n + k − 1)x 1 x 2 ···x n

n +k−1

n +k−1

n +k−1

Remark. This is also an immediate consequence of Muirhead’s inequality.

48. Put f (x) = x ln x. The given equation is equivalent to f (x) = f (1/2), which has the solutions x 1 = 1/2 and x 2 = 1/4. Since the function f is decreasing

on (0, 1/e), and increasing on (1/e, +∞), this equation has no other solutions.

49. Since sin 1 , sin 2, . . . , sin(N + 1) ∈ (−1,1), two of these N + 1 numbers have dis- tance less than 2 /N. Therefore |sin n − sink| < 2/N for some integers 1 ≤ k,n ≤ N + 1, n 6= k.

50. Since ϕ (x, y, z) = f (x + y, z) = ϕ (0, x + y, z) = g(0, x + y + z), it is enough to put

h (t) = g(0,t).

51. If there exist two numbers ab , bc ∈ S, then one can fill a crossword puzzle as

ab . The converse is obvious. Hence the set S has property A if and only if

bc

4.9 Longlisted Problems 1967 371 the set of first digits and the set of second digits of numbers in S are disjoint.

Thus the maximum size of S is 25.

52. This problem is not elementary. The solution offered by the proposer was not quite clear and complete (the existence was not proved).

53. (a) We can construct two lines parallel to the rays of the angle, at equal dis- tances from the rays. The intersection of these two lines lies on the bisector of the angle.

(b) If the length of a segment AB exceeds the breadth of the ruler, we can construct parallel lines through A and B in two different ways. The diagonal in the resulting rhombus is the perpendicular bisector of the segment AB. If the segment AB is too short, we can construct a line l parallel to AB and centrally project AB onto l from a point C chosen sufficiently close to

the segment, thus obtaining an arbitrarily long segment A ′ B ′ k AB. Then we construct the midpoint D ′ of A ′ B ′ as above. The line D ′ C intersects the segment AB at its midpoint D. By means of lines parallel to DC the segment AB can be prolonged symmetrically, and then the perpendicular bisector can be found as above.

(c) follows immediately from part (b). (d) Let there be given a point P and a line l. We draw an arbitrary line through

P that intersects l at A, and two lines l 1 and l 2 parallel to AP, at equal distances from AP and on either side of AP. Line l 1 intersects l at B. We can construct the midpoint C of AP. If BC intersects l 2 at D, then PD is parallel to l.

54. Let S be the given set of points on the cube. Let x , y, z denote the numbers of points from S lying at a vertex, at the midpoint of an edge, at the midpoint of a face of the cube, respectively, and let u be the number of all other points from S. Either there are no points from S at the vertices of the cube, or there is a point from S at each vertex. Hence x is either 0 or 8. Similarly, y is either 0 or 12, and z is either 0 or 6. Any other point of S has 24 possible images under rotations of the cube. Hence u is divisible by 24. Since n = x + y + z + u and 6 | y,z,u, it follows that either 6 | n or 6 | n − 8, i.e., n ≡ 0 or n ≡ 2 (mod 6). Thus n = 200 is possible, while n = 100 is not, because n ≡ 4 (mod 6).

55. It is enough to find all x from (0, 2 π ] such that the given inequality holds for all integers n. Suppose 0 <x<2 π /3. If n is the maximum integer for which nx ≤ 2 √ π /3, we

have π /3 < nx ≤ 2 √ π /3, and consequently sin nx ≥ 3 /2. Thus sin x + sin 2x + ··· + sinnx > 3 /2.

Suppose now that 2 π /3 ≤ x < 2 π . We have

3 sin x + ···+ sin nx =

2 2 sin 2 2 2 For x =2 π the given inequality clearly holds for all n. Hence, the inequality

2 sin x

holds for all n if and only if 2 π /3 + 2k π ≤x≤2 π + 2k π for some integer k.

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56. We shall prove by induction on n the following statement: If in some group of interpreters exactly n persons, n ≥ 2, speak each of the three languages, then it is possible to select a subgroup in which each language is spoken by exactly two persons. The statement of the problem easily follows from this: it suffices to select six such groups.

The case n = 2 is trivial. Let us assume n ≥ 2, and let N j ,N m ,N f ,N jm ,N jf ,N mf , N jm f

be the sets of those interpreters who speak only Japanese, only Malay, only Farsi, only Japanese and Malay, only Japanese and Farsi, only Malay and Farsi, and all the three languages, respectively, and n j ,n m ,n f ,n jm ,n jf ,n mf ,n jm f the cardinalities of these sets, respectively. By the condition of the problem, n j +n jm +n jf +n jm f =n m +n jm +n mf +n jm f =n f +n jf +n mf +n jm f = 24, and consequently

n j −n mf =n m −n jf =n f −n jm = c. Now if c < 0, then n jm ,n jf ,n mf > 0, and it is enough to select one interpreter

from each of the sets N jm ,N jf ,N mf . If c > 0, then n j ,n m ,n f > 0, and it is enough to select one interpreter from each of the sets N j ,N m ,N f and then use the induc- tive assumption. Also, if c = 0, then w.l.o.g. n j =n mf > 0, and it is enough to select one interpreter from each of the sets N j ,N mf and then use the inductive hypothesis. This completes the induction.

57. Obviously c n > 0 for all even n. Thus c n = 0 is possible only for an odd n. Let us assume a 1 ≤a 2 ≤ ··· ≤ a 8 : in particular, a 1 ≤0≤a 8 . If |a 1 | < |a 8 |, then there exists n 0 such that for every odd n >n 0 ,7 |a 1 | n <a n ⇒a n + ··· + a n +a n

1 7 8 > 7a n 1 +a n 8 > 0, contradicting the condition that c n =0 for infinitely many n. Similarly |a 1 | > |a 8 | is impossible, and we conclude that

a 1 = −a 8 . Continuing in the same manner we can show that a 2 = −a 7 ,a 3 = −a 6 and a 4 = −a 5 . Hence c n = 0 for every odd n.

58. The following sequence of equalities and inequalities gives an even stronger estimate than needed.

|l(z)| = |Az + B| = |(z + 1)(A + B) + (z − 1)(A − B)|

2 |(z + 1) f (1) + (z − 1) f (−1)|

≤ 2 (|z + 1| · | f (1)| + |z − 1| · | f (−1)|)

1 1 ≤ (|z + 1| + |z − 1|)M = ρ M . 2 2

59. By the arc AB we shall always mean the positive arc AB. We denote by |AB| the length of arc AB. Let a basic arc be one of the n + 1 arcs into which the circle is

partitioned by the points A 0 ,A 1 ,...,A n , where n ∈ N.

4.9 Longlisted Problems 1967 373 Suppose that A p A 0 and A 0 A q are the basic arcs with an endpoint at A 0 , and that

x n ,y n are their lengths, respectively. We show by induction on n that for each n the length of a basic arc is equal to x n ,y n , or x n +y n .

The statement is trivial for n = 1. Assume that it holds for n, and let A i A n +1 ,

be basic arcs. We shall prove that these two arcs have lengths x n ,y n , or x n +y n . If i , j are both strictly positive, then |A i A n +1 | = |A i −1 A n | and |A n +1 A j |= |A n A j −1 | are equal to x n ,y n , or x n +y n by the inductive hypothesis. Let us assume now that i = 0, i.e., that A p A n +1 and A n +1 A 0 are basic arcs. Then |A p A n +1 | = |A 0 A n +1−p | ≥ |A 0 A q |=y n and similarly |A n +1 A q |≥x n , but |A p A q |= x n +y n , from which it follows that |A p A n +1 | = |A 0 A q |=y n and consequently n + 1 = p + q. Also, x n +1 = |A n +1 A 0 |=y n −x n and y n +1 =y n . Now, all basic arcs have lengths y n −x n ,x n ,y n ,x n +y n . A presence of a basic arc of length x n +y n would spoil our inductive step. However, if any basic arc A k A l has length x n +y n , then we must have l − q = k − p because 2 π is irrational, and therefore the arc A k A l contains either the point A k −p (if k ≥ p) or the point A k +q (if k < p), which is impossible; hence, the proof is complete for i = 0. The proof for j = 0 is analogous. This completes the induction. It can be also seen from the above considerations that the basic arcs take only two distinct lengths if and only if n = p + q − 1. If we denote by n k the sequence of n’s for which this holds, and by p k ,q k the sequences of the corresponding p , q,

A n +1 A j

we have p 1 =q 1 = 1 and (

(p k +q k ,q k ), if {p k /(2 π )} + {q k /(2 π )} > 1, (p k +1 ,q k +1 )= (p k ,p k +q k ), if {p k /(2 π )} + {q k /(2 π )} < 1.

It is now “easy” to calculate that p 19 =p 20 = 333, q 19 = 377, q 20 = 710, and thus n 19 = 709 < 1000 < 1042 = n 20 . It follows that the lengths of the basic arcs for n = 1000 take exactly three different values.

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