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The circumcircle of triangle $ABC$ has radius $R$ satisfying

$$AB^2+AC^2=BC^2-R^2$$

Prove that the angles in the triangle are uniquely determined, and state the values for the angles.

So far I have used the Extended Law of Sines and the fact that $\alpha+\beta+\gamma=180$ to show that $\sin{\beta}\sin{\gamma}\cos{\alpha}=-\frac{1}{8}$ (where $\alpha$, $\beta$, $\gamma$ are the angles opposite $BC$, $CA$, $AB$), but I cannot seem to get any further.

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$b^2+c^2=a^2-R^2$ leads to $$ 4\sin^2 B+4\sin^2 C=4\sin^2(B+C)-1 \tag{1}$$ that can be written as $$ \left(2\cos(B+C)-\cos(B-C)\right)^2 + \sin^2(B-C) = 0.\tag{2} $$ It follows that $B=C$ and $2\cos(2B)-1=0$, from which $$ A=\frac{2\pi}{3},\quad B=C=\frac{\pi}{6}.\tag{3} $$

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  • $\begingroup$ Thank you! Very elegant. Is it true that $\sin{\beta}\sin{\gamma}\cos{\alpha}>=-\frac{1}{8}$ given that the angles add to 180? I am just trying to see if I can salvage my solution by proving that this is a special case of the above inequality...? $\endgroup$ – Plato Jul 6 '17 at 17:12
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    $\begingroup$ @Fermat: that is true and can be proved by factoring $\sin\beta\sin\gamma\cos(\beta+\gamma)-\frac{1}{8}$, reaching a proof very similar to mine. $\endgroup$ – Jack D'Aurizio Jul 6 '17 at 17:15

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