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Let A, B, and C be three points on a circle of radius 1.

Suppose that the magnitude of $\angle ABC$ is fixed. Then show that the area of the triangle ABC is maximized when $\angle BCA$ = $\angle CAB$.

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Let $\angle BAC=\alpha, \angle ABC=\beta,\angle BCA=\gamma$. We know that $\beta$ is fixed. Let $O$ the be center of the circle, then $\angle BOC=2\alpha,\angle AOC=2\beta,\angle BOA=2\gamma$. So, \begin{align} [ABC]&=\frac12\sin2\beta+\frac12(\sin2\alpha+\sin2\gamma)\\ &=\frac12\sin2\beta+\sin(\alpha+\gamma)\cos(\alpha-\gamma)\\ &=\frac12\sin2\beta+\sin(\pi-\beta)\cos(\alpha-\gamma)\\ &=\frac12\sin2\beta+\sin\beta\cos(\alpha-\gamma).\\ \end{align} Since $\beta$ is fixed and $\sin\beta>0$ then $[ABC]$ is maximum whenever $\cos(\alpha-\gamma)$ takes its maximum, i.e., $\cos(\alpha-\gamma)=1=\cos0\iff \alpha=\gamma$ as you wanted.

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