Post Number:#2 by Thanic Nur Samin » Tue Dec 13, 2016 1:20 pm
My solution is quite bash-y, so I am omitting the details. You can work them out by yourselves.
Let $\angle BAP=2x$ and $\angle CAP=2y$. Now, use the isosceles condition and other informations and use trig ceva to arrive at the conclusion,
\[\sin 2x \sin (60^{\circ}-x-y) \sin (60^{\circ}+x-y)=\sin 2y \sin 30^{\circ} \sin(30^{\circ}-2x)\]
From there, with enough manipulation with product to sum formulas, we can show that,
\[2x=\dfrac{2y}{3}\]
Since $2y=39^{\circ}$, we can conclude that $\angle BAP=13^{\circ}$
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