Solutions to the Shortlisted Problems of IMO 1975

4.17 Solutions to the Shortlisted Problems of IMO 1975

1. First, we observe that there cannot exist three routes of the form (A, B,C), (A, B, D), (A,C, D), for if E, F are the remaining two ports, there can be only one route covering A , E, namely, (A, E, F). Thus if (A, B,C), (A, B, D) are two routes, the one covering A ,C must be w.l.o.g. (A,C, E). The other roots are uniquely de- termined: These are (A, D, F), (A, E, F), (B, D, E), (B, E, F), (B,C, F), (C, D, E), (C, D, F ).

2. Since there are finitely many arrangements of the z i ’s, assume that z 1 ,...,z n is the one for which ∑ n i =1 (x i −z i ) 2 is minimal. We claim that in this case if i <j and x i 6= x j then z i ≥z j , from which the claim of the problem directly follows. Indeed, otherwise we would have

(x i −z j ) 2 + (x j −z i ) 2 = (x i −z i ) 2 + (x j −z j ) 2

+2(x i z i +x j z j −x i z j −x j z i ) = (x i −z i ) 2 + (x j −z j ) 2 + 2(x i −x j )(z i −z j )

2 ≤ (x 2

i −z i ) + (x j −z j ) ,

contradicting the assumption.

1 −k /3 = 1 and 3k 2 /3 < (k + 1) 2 /3 + (k + 1) 1 /3 k 1 /3 +k 2 /3 < 3(k + 1) 2 /3 we obtain

3. From (k + 1) 2 /3 + (k + 1) 1 /3 k 1 /3 +k 2 /3 (k + 1) 1 /3

1 −k /3 <k −2/3 <3 k − (k − 1) . Summing from 1 to n we get

1 −2 /3 < ∑ k −2/3 < 1 + 3(n 1 /3 − 1).

k =1

In particular, for n = 10 9 this inequality gives

1 −2 /3 < ∑ k −2/3 < 2998.

k =1

h Therefore ∑ 10 k 9 =1 k −2/3

i = 2997.

4. Put ∆ a n =a n −a n +1 . By the imposed condition, ∆ a n > ∆ a n +1 . Suppose that for some n, ∆ a n < 0: Then for each k ≥ n, ∆ a k < ∆ a n ; hence a n −a n +m = ∆ a n + ··· + ∆ a n +m−1 <m ∆ a n . Thus for sufficiently large m it holds that a n −a n +m < −1, which is impossible. This proves the first part of the inequality. Next one observes that

n ≥ ∑ a k = na n +1 + ∑ k ∆ a k ≥ (1 + 2 + ···+ n) ∆ a n =

n (n + 1)

k =1

2 Hence (n + 1) ∆ a n ≤ 2.

k =1

414 4 Solutions

5. There are exactly 8 ·9 k −1 k -digit numbers in M (the first digit can be chosen in 8 ways, while any other position admits 9 possibilities). The least of them is 10 k , and hence

= ∑ i −1 = 80 1 − k < 80.

i =1 10 10

6. Let us denote by C the sum of digits of B. We know that 16 16 ≡A≡B≡C

(mod 9). Since 16 16 =2 64 =2 6 ·10+4

≡2 4 ≡ 7, we get C ≡ 7 (mod 9). Moreover,

16 16 < 100 16 = 10 32 , hence A cannot exceed 9 ·32 = 288; consequently, B cannot exceed 19 and C is at most 10. Therefore C = 7.

7. We use induction on m. Denote by S m the left-hand side of the equality to be proved. First S

0 = (1 − y)(1 + y + ···+ y )+y n +1 = 1, since x = 1 − y. Further- more,

S m +1 −S m m +n+1

+1+j xy j

m +j −

j m +n+1

m ∑ +1+j

m +1+j

m +n+1 n +1

− +1 y j n

m +j

m +j+1

i.e., S m +1 =S m = 1 for every m. Second solution. Let us be given an unfair coin that, when tossed, shows heads

with probability x and tails with probability y. Note that x m +1 m+ j j y j is the prob- ability that until the moment when the (m + 1)th head appears, exactly j tails (j < n + 1) have appeared. Similarly, y n +1 n+i i x i is the probability that exactly

i heads will appear before the (n + 1)th tail occurs. Therefore, the above sum is the probability that either m + 1 heads will appear before n + 1 tails, or vice versa, and this probability is clearly 1.

8. Denote by K and L the feet of perpendiculars from the points P and Q to the lines BC and AC respectively.

4.17 Shortlisted Problems 1975 415 Let M and N be the points on AB (or-

C dered A − N − M − B) such that the tri-

P angle RMN is isosceles with ∠R = 90 ◦ .

By sine theorem we have BM BA = BM BR ·

BR sin 15 ◦ BK

LK BA = sin 45 ◦ . Since BC = cos 15 ◦

sin 45 ◦ sin 30 ◦

sin 15 ◦ NM sin 45 ◦ , we deduce that MK k AC

A B and MK = AL. Similarly, NL k BC

−→ , and −→ LQ are the images of −→ RM , −→ KP , and −−→ MK respectively under a rotation of 90 ◦ , and consequently the same holds for their sums −→ RQ and − RP → . Therefore, QR = RP and ∠QRP = 90 ◦ . Second solution. Let ABS be the equilateral triangle constructed in the exterior

R and NL

= BK. It follows that the vectors RN , NL

of △ABC. Obviously, the triangles BPC, BRS, ARS, AQC are similar. Let f be the rotational homothety centered at B that maps P onto C, and let g be the rotational homothety about A that maps C onto Q. The composition h = g ◦ f is also a rotational homothety; its angle is ∠PBC + ∠CAQ = 90 ◦ , and the coefficient is

BP · AC = 1. Moreover, R is a fixed point of h because f (R) = S and g(S) = R. Hence R is the center of h, and the statement follows from h (P) = Q.

BC AQ

Remark. There are two more possible approaches: One includes using complex numbers and the other one is mere calculating of RP , RQ, PQ by the cosine the- orem.

Second remark. The problem allows a generalization: Given that ∠CBP = ∠CAQ = α , ∠BCP = ∠ACQ = β , and ∠RAB = ∠RBA = 90 ◦ − α − β , show that RP = RQ and ∠PRQ = 2 α .

9. Suppose n is the natural number with na ≤ 1 < (n + 1)a. Let f a (x) = f (x − a). If a function f with the desired properties exists, then f a (a) = 0 and let w.l.o.g.

f (a) > 0, or equivalently, let the graph of f a lie below the graph of f . In this case also f (2a) > f (a), since otherwise, the graphs of f and f a would intersect between a and 2a. Continuing in this way we are led to 0 = f (0) < f (a) <

f (2a) < ··· < f (na). Thus if na = 1, i.e., a = 1/n, such an f does not exist. On the other hand, if a 6= 1/n, then we similarly obtain f (1) > f (1 − a) > f (1 − 2a ) > ··· > f (1 − na). Choosing values of f at ia,1 − ia, i = 1,... ,n, so that they satisfy f (1 − na) < ··· < f (1− a) < 0 < f (a) < ··· < f (na), we can extend

f to other values of [0, 1] by linear interpolation. A function obtained this way has the desired property.

10. We shall prove that for all x , y with x + y = 1 it holds that f (x, y) = x − 2y. In this case f (x, y) = f (x, 1 − x) can be regarded as a polynomial in z = x − 2y = 3x − 2, say f (x,1 − x) = F(z). Putting in the given relation a = b = x/2, c =

1 −x, we obtain f (x,1− x)+2 f (1− x/2,x/2) = 0; hence F(z)+2F(−z/2) = 0. Now F (1) = 1, and we get that for all k, F((−2) k ) = (−2) k . Thus F (z) = z for infinitely many values of z; hence F (z) ≡ z. Consequently f (x,y) = x − 2y if x + y = 1.

416 4 Solutions For general x , y with x + y 6= 0, since f is homogeneous ,we have f (x,y) = (x +

x +y x +y = (x + y) x +y −2 x +y = (x + y) (x − 2y). The same is true for x + y = 0, because f is a polynomial.

n −1

11. Let (a k i ) be the subsequence of (a k ) consisting of all a k ’s that give remainder r upon division by a 1 . For every i > 1, a k i ≡a k 1 (mod a 1 ); hence a k i =a k 1 +ya 1 for some integer y > 0. It follows that for every r = 0, 1, . . . , a 1 − 1 there is exactly one member of the corresponding (a k i ) i ≥1 that cannot be represented as xa l + ya m , and hence at most a 1 + 1 members of (a k ) in total are not representable in the given form.

12. Since sin 2x i = 2 sin x i cos x i and sin (x i +x i +1 ) + sin(x i −x i +1 ) = 2 sin x i cos x i +1 , the inequality from the problem is equivalent to

(cos x 1 − cosx 2 ) sin x 1 + (cosx 2 − cosx 3 ) sin x 2 + ··· π

··· + (cosx ν−1 − cosx ν ) sin x ν−1 < . (1)

4 Consider the unit circle with center at O (0, 0) and points M i (cos x i , sin x i ) on

it. Also, choose the points N i (cos x i , 0) and M ′ i (cos x i +1 , sin x i ). It is clear that (cos x i −cosx i +1 ) sin x i is equal to the area of the rectangle M i N i N i +1 M ′ i . Since all these rectangles are disjoint and lie inside the quarter circle in the first quadrant

whose area is π 4 , inequality (1) follows.

13. Suppose that A k A k +1 ∩A m A m +1 6= /0 for some k, m > k + 1. Without loss of gen- erality we may suppose that k = 0, m = n − 1 and that no two segments A k A k +1 and A m A m +1 intersect for 0 ≤ k < m − 1 < n − 1 except for k = 0, m = n − 1. Also, shortening A 0 A 1 , we may suppose that A 0 ∈A n −1 A n . Finally, we may re- duce the problem to the case that A 0 ...A n −1 is convex: Otherwise, the segment

A n −1 A n can be prolonged so that it intersects some A k A k +1 ,0 < k < n − 2. If n = 3, then A 1 A 2 ≥ 2A 0 A 1 implies A 0 A 2 >A 0 A 1 , hence ∠A 0 A 1 A 2 > ∠A 1 A 2 A 3 ,

a contradiction. Let n = 4. From A 3 A 2 >A 1 A 2 we conclude that ∠A 3 A 1 A 2 > ∠A 1 A 3 A 2 . Using the inequality ∠A 0 A 3 A 2 > ∠A 0 A 1 A 2 we obtain that ∠A 0 A 3 A 1 > ∠A 0 A 1 A 3 implying

A 0 A 1 >A 0 A 3 . Now we have A 2 A 3 <A 3 A 0 +A 0 A 1 +A 1 A 2 < 2A 0 A 1 +A 1 A 2 ≤ 2A 1 A 2 ≤A 2 A 3 , which is not possible. Now suppose n ≥ 5. If α

i is the exterior angle at A i , then α 1 > ··· > α n −1 ; hence α −1 < 360 n n −1 ≤ 90 ◦ . Consequently ∠A n −2 A n −1 A 0 ≥ 90 ◦ and A 0 A n −2 >

A n −1 A n −2 . On the other hand, A 0 A n −2 <A 0 A 1 +A 1 A 2 + ··· + A n −3 A n −2 <

2 n −2 + 1 2 n −3 + ··· + 2 A n −1 A n −2 <A n −1 A n −2 , which contradicts the previous relation.

14. We shall prove that for every n ∈ N, 2n + 25 ≤ x n ≤ 2n + 25+ 0.1. Note that for n = 1000 this gives us exactly the desired inequalities. First, notice that the recurrent relation is equivalent to

2x k (x k +1 −x k ) = 2.

4.17 Shortlisted Problems 1975 417 Since x 0 <x 1 < ··· < x k < ···, from (1) we get x 2 k 2 +1 −x k = (x k +1 +x k )(x k +1 −

x k ) > 2. Adding these up we obtain x 2 n 2 ≥x 0 + 2n, which proves the first inequal- ity. On the other hand, x k +1 =x k + 1 x k ≤x k + 0.2 (for x k ≥ 5), and one also de-

duces from (1) that x 2 k +1 −x 2 k −0.2(x k +1 −x k ) = (x k +1 +x k −0.2)(x k +1 −x k ) ≤ 2. Again, adding these inequalities up, (k = 0, . . . , n − 1) yields

x 2 n 2 ≤ 2n + x 0 + 0.2(x n −x 0 ) = 2n + 24 + 0.2x n . Solving the corresponding quadratic equation, we obtain

x n < 0.1 + 2n + 24.01 < 0.1 + 2n + 25.

15. Assume that the center of the circle is at the origin O (0, 0), and that the points

A 1 ,A 2 ,...,A 1975 are arranged on the upper half-circle so that ∠A i OA 1 = α i ( α 1 = 0). The distance A

equals 2 sin α j −α i α j cos α i

2 = 2 sin

2 k − cos 2 sin α α i α k 2 2 , and it

will be rational if all sin 2 , cos 2 are rational.

Finally, observe that there exist infinitely many angles α such that both sin α , cos α are rational, and that such α can be arbitrarily small. For example, take α so that sin α = 2t

t 2 −1

t 2 +1 and cos α = t 2 +1 for any t ∈ Q.

418 4 Solutions