IMO 1977 · Problem 4

Four real constants \(a,b,A,B\) are given, and \[ f(\theta)=1-a\cos\theta-b\sin\theta-A\cos2\theta-B\sin2\theta. \] Prove that if \(f(\theta)\geq0\) for all real \(\theta\), then \[ a^2+b^2\leq2\quad\text{and}\quad A^2+B^2\leq1. \]

Official text of the problem. Official problem paper (PDF)

Solution

By the trigonometric addition formulas, for every real \(t\), \[ 0\leq f\left(t-\frac{\pi}{4}\right)+f\left(t+\frac{\pi}{4}\right) =2-\sqrt2\,(a\cos t+b\sin t). \] Here the terms involving \(A\) and \(B\) cancel.

Let \(r=\sqrt{a^2+b^2}\). If \(r=0\), the first required inequality is immediate. Otherwise, choose \(t\) such that \[ \cos t=\frac a r,\qquad \sin t=\frac b r. \] Then \(a\cos t+b\sin t=r\), so the preceding inequality yields \[ 0\leq2-\sqrt2\,r. \] Thus \(r\leq\sqrt2\), proving \(a^2+b^2\leq2\).

Similarly, for every real \(t\), \[ 0\leq f(t)+f(t+\pi) =2-2(A\cos2t+B\sin2t). \] Let \(s=\sqrt{A^2+B^2}\). If \(s=0\), the second required inequality is immediate. Otherwise, choose \(t\) such that \[ \cos2t=\frac A s,\qquad \sin2t=\frac B s. \] The last inequality becomes \(0\leq2-2s\). Hence \(s\leq1\), proving \(A^2+B^2\leq1\).

The solution was checked against John Scholes' IMO solution archive (Kalva).