Solutions to the Contest Problems of IMO 1964

4.6 Solutions to the Contest Problems of IMO 1964

1. Let n = 3k + r, where 0 ≤ r < 2. Then 2 n =2 3k +r =8 k ·2 r ≡2 r (mod 7). Thus the remainder of 2 n modulo 7 is 1 , 2, 4 if n ≡ 0,1,2 (mod 3). Hence 2 n − 1 is divisible by 7 if and only if 3 | n, while 2 n + 1 is never divisible by 7.

2. By substituting a = x + y, b = y + z, and c = z + x (x, y, z > 0) the given inequality becomes

6xyz

2 y + xy 2 +y 2 z + yz 2 +z 2 x + zx ≤x 2 , which follows immediately by the AM–GM inequality applied to x 2 y , xy 2 ,x 2 z ,

xz 2 ,y 2 z , yz 2 .

3. Let r be the radius of the incircle of △ABC, r a ,r b ,r c the radii of the smaller circles corresponding to A , B,C, and h a ,h b ,h c the altitudes from A , B,C respec- tively. The coefficient of similarity between the smaller triangle at A and the triangle ABC is 1 − 2r/h a , from which we easily obtain r a = (h a − 2r)r/h a = (s − a)r/s. Similarly, r b = (s − b)r/s and r c = (s − c)r/s. Now a straightforward computation gives that the sum of areas of the four circles is given by

(b + c − a)(c + a − b)(a + b − c)(a 2 +b 2 +c 2 ) π Σ =

(a + b + c) 3

4. Let us call the topics T 1 ,T 2 ,T 3 . Consider an arbitrary student A. By the pigeon- hole principle there is a topic, say T 3 , he discussed with at least 6 other students. If two of these 6 students discussed T 3 , then we are done. Suppose now that the 6 students discussed only T 1 and T 2 and choose one of them, say B. By the pigeonhole principle he discussed one of the topics, say T 2 , with three of these students. If two of these three students also discussed T 2 , then we are done. Otherwise, all the three students discussed only T 1 , which completes the task.

5. Let us first compute the number of intersection points of the perpendiculars passing through two distinct points B and C. The perpendiculars from B to the lines through C other than BC meet all perpendiculars from C, which counts to

3 · 6 = 18 intersection points. Each perpendicular from B to the 3 lines not con- taining C can intersect at most 5 of the perpendiculars passing through C, which counts to another 3 · 5 = 15 intersection points. Thus there are 18 + 15 = 33 intersection points corresponding to B ,C. It follows that the required total number is at most 10 · 33 = 330. But some of these points, namely the orthocenters of the triangles with vertices at the given points, are counted thrice. There are 10 such points. Hence the maximal number of intersection points is 330 − 2 · 10 = 310.

Remark. The jury considered only the combinatorial part of the problem and didn’t require an example in which 310 points appear. However, it is “easily”

verified that, for instance, the set of points A (1, 1), B(e, π ), C(e 2 , π 2 ), D(e 3 , π 3 ),

E (e 4 , π 4 ) works.

356 4 Solutions

6. We shall prove that the statement is valid in the general case, for an arbitrary point D 1 inside △ABC. Since D 1 belongs to the plane ABC, there are real num- bers a , b, c such that (a + b + c) DD −−→

1 =a DA +b DB +c DC . Since AA 1 k DD 1 ,

DA 1 −−→ =k 1 ∈ R. Now it is easy to get 1 = −→

it holds that −−→ AA −−→ DD for some k

−→ −(b DB +c DC )/a, −−→ DB 1 = −(a DA +c DC )/b, and DC 1 = −(a DA + b−→ DB )/c. This

By using

6V h−−−→ D A D −−−→ B −−−→

h−→ −→ −→

1 1 , 1 1 , D 1 C 1 DABC = DA , DB , DC we get

V D 1 A 1 B 1 C 1 = 6abc (a + b + c) 3

= 3V DABC .

4.7 Contest Problems 1965 357