Solutions to the Contest Problems of IMO 1959

4.1 Solutions to the Contest Problems of IMO 1959

1. The desired result (14n + 3, 21n + 4) = 1 follows from

3 (14n + 3) − 2(21n + 4) = 1.

√ 2. For the square roots to be real we must have 2x − 1 ≥ 0 ⇒ x ≥ 1/2 and x ≥ p 2x −1⇒x 2 ≥ 2x − 1 ⇒ (x − 1) 2 ≥ 0, which always holds. Then we have √

(a) c 2 = 2. The equation holds for 1/2 ≤ x ≤ 1. (b) c 2 = 1. The equation has no solution. (c) c 2 = 4. The equation holds for 4x − 2 = 4 ⇒ x = 3/2.

3. Multiplying the equality by 4 (a cos 2 x − bcosx + c), we obtain 4a 2 cos 4 x +

2 (4ac − 2b 2 ) cos 2 x + 4c 2 = 0. Plugging in 2 cos 2 x = 1 + cos 2x we obtain (af- ter quite a bit of manipulation):

a 2 cos 2 2x + (2a 2 2 ) cos 2x + (a 2 + 4ac − 2b 2 + 4c 2 + 4ac − 2b ) = 0. For a

2 = 4, b = 2, and c = −1 we get 4cos 2 x + 2 cosx − 1 = 0 and 16cos 2x +

8 cos 2x − 4 = 0 ⇒ 4cos 2 2x + 2 cos2x − 1 = 0.

4. Analysis. Let a and b be the other two sides of the triangle. From the conditions

2 2 2 of the problem we have c √ =a +b and c /2 = ab 2 p

⇔ 3/2c =a 2 +b 2 + 2ab = (a + b) 2 ⇔ 3 /2c = a + b. Given a desired △ABC let D be a point on (AC such

that CD = CB. In that case, AD = a + b = 3 /2c, and also, since BC = CD, it follows that ∠ADB = 45 ◦ .

Construction. From a segment of length c we elementarily construct a segment p AD of length

3 /2 c. We then construct a ray (DX such that ∠ADX = 45 ◦

348 4 Solutions and a circle k (A, c) that intersects the ray at point B. Finally, we construct the

perpendicular from B to AD; point C is the foot of that perpendicular. Proof. It holds that AB = c, and, since CB = CD, it also holds that AC + CB = p √

AC + CD = AD = 3 /2 c. From this it follows that AC ·CB = c/2. Since BC is perpendicular to AD, it follows that ∡BCA = 90 ◦ . Thus ABC is the desired triangle.

Discussion. Since AB 2 = 2c > 3 /2 c = AD > AB, the circle k intersects the ray DX in exactly two points, which correspond to two symmetric solutions.

5. (a) It suffices to prove that AF ⊥ BC, since then for the intersection point X we have ∠AXC = ∠BXF = 90 ◦ , implying that X belongs to the circumcircles of both squares and thus that X = N. The relation AF ⊥ BC holds because from MA = MC, MF = MB, and ∠AMC = ∠FMB it follows that △AMF is obtained by rotating △BMC by 90 ◦ around M.

(b) Since N is on the circumcircle of BMFE, it follows that ∠ANM = ∠MNB =

45 ◦ . Hence MN is the bisector of ∠ANB. It follows that MN passes through the midpoint of the arc c AB of the circle with diameter AB (i.e., the circum- circle of △ABN) not containing N. (c) Let us introduce a coordinate system such that A = (0, 0), B = (b, 0), and M = (m, 0). Setting in general W = (x W ,y W ) for an arbitrary point W and denoting by R the midpoint of PQ, we have y R = (y P +y Q )/2 = (m + b − m )/4 = b/4 and x R = (x P +x Q )/2 = (m + m + b)/4 = (2m + b)/4, the parameter m varying from 0 to b. Thus the locus of all points R is the

closed segment R 1 R 2 where R 1 = (b/4, b/4) and R 2 = (b/4, 3b/4).

6. Analysis. For AB k CD to hold evidently neither must intersect p and hence constructing lines r in α through A and s in β through C, both being paral- lel to p, we get that B ∈ r and D ∈ s. Hence the problem reduces to a planar problem in γ , determined by r and s. Denote by A ′ the foot of the perpen- dicular from A to s. Since ABCD is isosceles and has an incircle, it follows

AD = BC = (AB + CD)/2 = A ′ C . The remaining parts of the problem are now obvious.

4.2 Contest Problems 1960 349