IMC 2026 · Day 1 · P1

Assume, for contradiction, that there exists $x \in \mathbb{R}$ such that $\cos (\cos x) = \sin (\sin x)$. Using $\cos t = \sin \left( \frac{\pi}{2} - t \right)$, we may rewrite this as $$\sin \left( \frac{\pi}{2} - \cos x \right) = \sin (\sin x).$$ Set $$A = \frac{\pi}{2} - \cos x \quad \text{and} \quad B = \sin x.$$ Since $\sin x, \cos x \in [-1,1]$, we have $$A \in \left[ \frac{\pi}{2} - 1, \frac{\pi}{2} + 1 \right] \quad \text{and} \quad B \in [-1,1].$$ The general solutions of $\sin A = \sin B$ are $$A - B = 2 k \pi \quad \text{or} \quad A + B = (2 k + 1) \pi, \quad k \in \mathbb{Z}.$$ The above bounds on $A$ and $B$ force $k = 0$. Hence either $$\frac{\pi}{2} - \cos x = \sin x$$ or $$\frac{\pi}{2} - \cos x = \pi - \sin x.$$ After rearranging, these equations become, respectively, $$\sin x + \cos x = \frac{\pi}{2}$$ and $$\sin x - \cos x = \frac{\pi}{2}.$$ However, $$\sin x \pm \cos x \leq \sqrt{2},$$ as follows, for example, from $$\sin x \pm \cos x = \sqrt{2} \sin \left( x \pm \frac{\pi}{4} \right).$$ Since $$\sqrt{2} < \frac{\pi}{2},$$ neither equality is possible. Hence the equation $$\cos (\cos x) = \sin (\sin x)$$ has no real solutions.