Source: Chinese Mathematical Olympiad 1998 Problem 1
Tags: geometry, incenter, trigonometry, geometry unsolved
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Let $ABC$ be a non-obtuse triangle satisfying $AB>AC$ and $\angle B=45^{\circ}$. The circumcentre $O$ and incentre $I$ of triangle $ABC$ are such that $\sqrt{2}\ OI=AB-AC$. Find the value of $\sin A$.
\[2OI^{2} =(AB-AC)^{2} \]
\[2(R^{2}-2rR) =4R^{2}(\sin C-\sin B)^{2} \]
\[(R^{2}-2R^{2}(4\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}))=2R^{2}(\sin (A+B)-\sin B)^{2} \]
\[(R^{2}-2R^{2}(\cos A+\cos B+\cos C-1)) =2R^{2}(2\sin \frac{A}{2}\cos (45+\frac{A}{2}))^{2} \]
\[(3R^{2}-2R^{2}(\cos A+\cos B-\cos (A+B))) =2R^{2}(2\sin \frac{A}{2}(\cos \frac{A}{2}-\sin \frac{A}{2})\sin 45)^{2} \]
\[3R^{2}-2R^{2}(\cos A+2\sin \frac{A}{2}\sin (45+\frac{A}{2})) =R^{2}(\sinh+\cos A-1)^{2} \]
\[3-2(\cos A+\frac{1}{2}\sqrt{2}(1-\cos A+\sin A)) =(\sin A+\cos A-1)^{2} \]
\[3-2\cos A-\sqrt{2}(1-\cos A+\sin A) =2+2\sin A\cos A-2\sin A-2\cos A \]
\[1-\sqrt{2}+\sqrt{2}\cos A-\sqrt{2}\sin A) =2\sin A\cos A-2\sin A \]
\[1-\sqrt{2}\sin A =2\sin A(\cos A-1)-\sqrt{2}(\cos A-1) \]
\[0 =\sqrt{2}(\cos A-1)(\sqrt{2}\sin A-1)+(\sqrt{2}\sin A-1) \]
\[0 =(\sqrt{2}\sin A-1)(\sqrt{2}(\cos A-1)+1)\]
So
\[
\sin A=\frac{1}{\sqrt{2}}=\frac{1}{2}\sqrt{2}
\]
or
\[
\cos A=1-\frac{1}{2}\sqrt{2} \rightarrow \cos ^{2}A=\frac{3}{2}-\sqrt{2} \rightarrow \sin A=\sqrt{\sqrt{2}-\frac{1}{2}}=\frac{1}{2}\sqrt{4\sqrt{2}-2}
\]
Can't it be solved with elementary geometry?