BdMO National 2016 Secondary 3: Weird angle condition

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Thanic Nur Samin
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BdMO National 2016 Secondary 3: Weird angle condition

Unread post by Thanic Nur Samin » Tue Dec 13, 2016 12:13 pm

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 $\angle BAP$
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Thanic Nur Samin
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Re: BdMO National 2016 Secondary 3: Weird angle condition

Unread post 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 $\angle BAP=13^{\circ}$
Hammer with tact.

Because destroying everything mindlessly isn't cool enough.

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Thanic Nur Samin
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Re: BdMO National 2016 Secondary 3: Weird angle condition

Unread post by Thanic Nur Samin » Tue Dec 13, 2016 6:27 pm

On second thought, I am showing my calculation. Not that it is too long.
$$\sin 2x \sin (60^{\circ}-x-y) \sin (60^{\circ}+x-y)=\sin 2y \sin 30^{\circ} \sin(30^{\circ}-2x)$$

$$\Rightarrow \sin 2x(\cos 2x-\cos (120^{\circ}-2y))=\sin 2y \sin(30^{\circ}-2x)$$

$$\Rightarrow \sin 4x+2\sin 2x\cos (60^{\circ}+2y)=2\sin 2y\cos (60^{\circ}+2x)$$

$$\Rightarrow \sin 4x+\sin (2x+2y+60^{\circ})+\sin (2x-2y-60^{\circ})=\sin (2y+2x+60^{\circ})+$$
$$\sin (2y-2x-60^{\circ})$$

$$\Rightarrow \sin 4x=\sin (2y-2x-60^{\circ})+\sin (2y-2x+60)$$

$$\Rightarrow \sin 4x=2\sin (2y-2x)\cos 60^{\circ}$$

$$\Rightarrow \sin 4x=\sin (2y-2x)$$

$$\Rightarrow 4x=2y-2x$$

$$\Rightarrow 6x=2y$$

$$\Rightarrow 2x=\dfrac{2y}{3}$$
Hammer with tact.

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joydip
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Re: BdMO National 2016 Secondary 3: Weird angle condition

Unread post by joydip » Fri Dec 16, 2016 9:36 pm

A synthetic solution :
Let $ Q$ be a point such that $\triangle PQB $ is equilateral ,$C$ and $Q$ are on the same side of $PB$.Then $Q$ is the center of $\odot BPC$. So, $AQ$ is the perpendicular bisector of $BC$ . As,$\angle APB=150^{\circ}$, so $AP \perp QB \Rightarrow AP$ bisect $\angle QAB$ .So ,$4\angle BAP=\angle BAC \Rightarrow 3\angle BAP =\angle CAP \Rightarrow \angle BAP=13^{\circ}$.
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