Solutions to the Contest Problems of IMO 1966

4.8 Solutions to the Contest Problems of IMO 1966

1. Let N a ,N b ,N c ,N ab ,N ac ,N bc ,N abc denote the number of students who solved ex- actly the problems whose letters are stated in the index of the variable. From the conditions of the problem we have

N a +N b +N c +N ab +N bc +N ac +N abc = 25,

N b +N bc = 2(N c +N bc ),

N a −1=N ac +N abc +N ab , N a =N b +N c . From the first and third equations we get 2N a +N b +N c +N bc = 26, and from the

second and fourth we get 4N b +N c = 26 and thus N b ≤ 6. On the other hand, we have from the second equation N b = 2N c +N bc ⇒N c ≤N b /2 ⇒ 26 ≤ 9N b /2 ⇒ N b ≥ 6; hence N b = 6.

2. Angles α and β are less than 90 ◦ , otherwise if w.l.o.g. α ≥ 90 ◦ we have tan ( γ /2) · (atan α + b tan β ) < b tan( γ /2) tan β ≤ btan( γ /2) cot( γ /2) = b < a +

b . Since a ≥ b ⇔ tana ≥ tan b, Chebyshev’s inequality gives a tan α + b tan β ≥ (a + b)(tan α + tan β )/2. Due to the convexity of the tan function we also have (tan α + tan β )/2 ≥ tan[( α + β )/2] = cot( γ /2). Hence we have

tan (a tan α + b tan β )≥ tan (a + b)(tan α + tan β )

γ ≥ tan (a + b) cot = a + b. 2 2

The equalities can hold only if a = b. Thus the triangle is isosceles.

3. Consider a coordinate system in which the points of the regular tetrahedron are placed at A (−a,−a,−a), B(−a,a,a), C(a,−a,a) and D(a,a, −a). Then the center of the tetrahedron is at O (0, 0, 0). For a point X (x, y, z) we see that the sum X A √ + XB + XC + XD by the QM–AM inequality does not exceed

2 XA 2 + XB 2 + XC 2 + XD 2 . Now, since X A 2 = (x + a) 2 + (y + a) 2 + (z + a) 2 etc., we easily obtain

XA 2 + XB 2 + XC 2 + XD 2 = 4(x 2 +y 2 +z 2 ) + 12a 2

2 = OA 2 + OB ≥ 12a 2 + OC 2 + OD 2 .

Hence X A + XB + XC + XD ≥ 2 OA 2 + OB 2 + OC 2 + OD 2 = OA + OB + OC + OD .

4. It suffices to prove 1 /sin 2 k x = cot 2 k −1 x − cot2 k x for any integer k and real x, i.e., 1 /sin 2x = cot x − cot2x for all real x. We indeed have

cot 2 x 2 cos x −1

cot x − cot2x = cotx −

5. We define L 1 = |a 1 −a 2 |x 2 + |a 1 −a 3 |x 3 + |a 1 −a 4 |x 4 and analogously L 2 ,L 3 , and L 4 . Let us assume w.l.o.g. that a 1 <a 2 <a 3 <a 4 . In that case,

Hence it follows that x 2 =x 3 = 0 and consequently x 1 =x 4 = 1/|a 1 −a 4 |. This solution set indeed satisfies the starting equations. It is easy to generalize this

result to any ordering of a 1 ,a 2 ,a 3 ,a 4 .

6. Let S denote the area of △ABC. Let A 1 ,B 1 ,C 1 be the midpoints of BC , AC, AB respectively. We note that S A 1 B 1 C =S A 1 BC 1 =S AB 1 C 1 =S A 1 B 1 C 1 = S/4. Let us assume w.l.o.g. that M ∈ [AC 1 ]. We then must have K ∈ [BA 1 ] and L ∈ [CB 1 ]. However, we then have S (KLM) > S(KLC 1 ) > S(KB 1 C 1 ) = S(A 1 B 1 C 1 ) = S/4. Hence, by the pigeonhole principle one of the remaining three triangles △MAL, △KBM, and △LCK must have an area less than or equal to S/4. This completes the proof.

4.9 Longlisted Problems 1967 361