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Harvard-MIT Mathematics Tournament

Geometry Difficulty 4.9 AIME Prove it United States

Problem:
Let $a \star b = \sin a \cos b$ for all real numbers $a$ and $b$ . If $x$ and $y$ are real numbers such that $x \star y - y \star x = 1$ , what is the maximum value of $x \star y + y \star x$ ?

Solution

Solution:
We have $x \star y + y \star x = \sin x \cos y + \cos x \sin y = \sin(x + y) \leq 1$ .

Equality is achieved when $x = \frac{\pi}{2}$ and $y = 0$ . Indeed, for these values of $x$ and $y$ , we have $x \star y - y \star x = \sin x \cos y - \cos x \sin y = \sin(x - y) = \sin \frac{\pi}{2} = 1$ .

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