Solutions to the Contest Problems of IMO 1960

4.2 Solutions to the Contest Problems of IMO 1960

1. Given the number acb, since 11 | acb, it follows that c = a + b or c = a + b − 11. In the first case, a 2 +b 2 + (a + b) 2 = 10a + b, and in the second case, a 2 +

b 2 + (a + b − 11) 2 = 10(a − 1) + b. In the first case the LHS is even, and hence

b ∈ {0,2,4,6,8}, while in the second case it is odd, and hence b ∈ {1,3,5,7,9}. Analyzing the 10 quadratic equations for a we obtain that the only valid solutions are 550 and 803.

2. The LHS term is well-defined for x √ ≥ −1/2 and x 6= 0. Furthermore, 4x √ 2 /(1 −

1 + 2x) 2 = (1 + 1 + 2x) 2 . Since √

f (x) = 1 + 1 + 2x − 2x − 9 = 2 1 + 2x − 7 is increasing and since f (45/8) = 0, it follows that the inequality holds precisely

for −1/2 ≤ x < 45/8 and x 6= 0.

be the middle of the n = 2k + 1 segments and let D be the foot of the perpendicular from A to the hypotenuse. Let us assume B (C, D,C ′ ,B ′ , B). Then from CD

3. Let B ′ C ′

< BD, CD + BD = a, and CD · BD = h 2 we have CD 2 √ 2 − a ·CD + h =

0 =⇒ CD = (a − a 2 − 4h 2 )/2 . Let us define ∡DAC ′ = γ and ∡DAB ′ = β ; then tan β √ = DB ′ /h and tan γ = DC ′ /h. Since DB ′ = CB ′ −CD = (k + 1)a/(2k +

√ )/2 and DC ′ = ka/(2k + 1) − (c − c 2 − 4h 2 )/2, we have

1 ) − (c − c 2 − 4h 2

tan a β tan α = tan( β γ )=

− (2k+1)h =

− tan γ

1 + tan β · tan γ

1 2 + a −4h 2 −

a 2 4h 2 4h 2 (2k+1) 2

The case B (C,C ′ , D, B ′ , B) is similar.

4. Analysis. Let A ′ and B ′

be the feet of the perpendiculars from A and B, respec- tively, to the opposite sides, A 1 the midpoint of BC, and let D ′

be the foot of the perpendicular from A 1 to AC. We then have AA 1 =m a , AA ′ =h a , ∠AA ′ A 1 = 90 ◦ ,

A 1 D ′ =h b /2, and ∠AD ′ A 1 = 90 ◦ .

Construction. We construct the quadrilateral AD ′ A 1 A ′ (starting from the circle with diameter AA 1 ). Then C is the intersection of A ′ A 1 and AD ′ , and B is on the

line A 1 C such that CA 1 =A 1 B and B (B, A 1 ,C).

Discussion. We must have m a ≥h a and m a ≥h b /2. The number of solutions is

0 if m a =h a =h b /2, 1 if two of m a ,h a ,h b /2 are equal, and 2 otherwise.

5. (a) The locus of the points is the square EFGH where these four points are the centers of the faces ABB ′ A ′ , BCC ′ B ′ , CDD ′ C ′ and DAA ′ D ′ . (b) The locus of the points is the rectangle IJKL where these points are on AB ′ , CB ′ , CD ′ , and AD ′ at a distance of AA ′ /3 with respect to the plane ABCD.

350 4 Solutions

6. Let E , F respectively be the midpoints of the bases AB,CD of the isosceles trape- zoid ABCD.

(a) The point P is on the intersection of EF and the circle with diameter BC. (b) Let x √ = EP. Since △BEP ∼ △PFC, we have x(h − x) = ab/4 ⇒ x 1 ,2 = (h ± h 2 − ab)/2 . (c) If h 2 > ab there are two solutions, if h 2 = ab there is only one solution, and if h 2 < ab there are no solutions.

7. Let A be the vertex of the cone, O the center of the sphere, S the center of the base of the cone, B a point on the base circle, and r the radius of the sphere. Let ∠SAB = α . We easily obtain AS = r(1 + sin α )/ sin α and SB = r(1 +

sin α ) tan α / sin α and hence V 1 = π SB 2 · SA/3 = π r 3 (1 + sin α ) 2 /[3 sin α (1 −

sin α )] . We also have V 2 =2 π r 3 and hence

(1 + sin α ) 2 k =

6 sin α α ⇒ (1 + 6k)sin + 2(1 − 3k)sin α (1 − sin +1=0. ) The discriminant of this quadratic must be nonnegative:

(1 − 3k) 2 − (1 + 6k) ≥

0 ⇒ k ≥ 4/3. Hence we cannot have k = 1. For k = 4/3 we have sin α = 1/3, whose construction is elementary.

4.3 Contest Problems 1961 351