IMO 1991 SL 19

Origin: IRE

Problem

Let a be a rational number with 0 < a < 1 and suppose that cos 3\pia + 2 cos2\pia = 0. (Angle measurements are in radians.) Prove that a = 2/3.

Solution

Set x = cos(\pia). The given equation is equivalent to 4x3+4x2−3x−2 = 0, which factorizes as (2x + 1)(2x2 + x −2) = 0. The case 2x + 1 = 0 yields cos(\pia) = −1/2 and a = 2/3. It remains to show that if x satisfies 2x2 + x −2 = 0 then a is not rational. The polynomial equation 2x2 + x −2 = 0 has two real roots, x1,2 = −1\pm \sqrt , and since |x| \leq1 we must have x = cos \pia = −1+ \sqrt . We now prove by induction that, for every integer n \geq0, cos(2n\pia) = an+bn \sqrt for some odd integers an, bn. The case n = 0 is trivial. Also, if cos(2n\pia) = an+bn \sqrt , then cos(2n+1\pia) = 2 cos2(2n\pia) −1 = 1 a2 n + 17b2 n −8

• anbn \sqrt  = an+1 + bn+1 \sqrt . By the inductive step that an, bn are odd, it is obvious that an+1, bn+1 are also odd. This proves the claim. Note also that, since an+1 = 1 2(a2 n + 17b2 n −8) > an, the sequence {an} is strictly increasing. Hence the set of values of cos(2n\pia), n = 0, 1, 2, . . ., is infinite (because \sqrt 17 is irrational). However, if a were rational, then the set of values of cos m\pia, m = 1, 2, . . . , would be finite, a contradiction. Therefore the only possible value for a is 2/3.