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BdMO National 2016 Secondary 3: Weird angle condition
Posted: Tue Dec 13, 2016 12:13 pm
by Thanic Nur Samin
In $\triangle ABC$, $AB=AC$. $P$ is a point inside the triangle such that $\angle BCP=30^{\circ}$ and $\angle APB=150^{\circ}$ and $\angle CAP=39^{\circ}$. Find $\angle BAP$
Re: BdMO National 2016 Secondary 3: Weird angle condition
Posted: Tue Dec 13, 2016 1:20 pm
by Thanic Nur Samin
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 $\angle BAP=13^{\circ}$
Re: BdMO National 2016 Secondary 3: Weird angle condition
Posted: Tue Dec 13, 2016 6:27 pm
by Thanic Nur Samin
On second thought, I am showing my calculation. Not that it is too long.
Re: BdMO National 2016 Secondary 3: Weird angle condition
Posted: Fri Dec 16, 2016 9:36 pm
by joydip