## BdMO National 2016 Secondary 3: Weird angle condition

Thanic Nur Samin
Posts: 176
Joined: Sun Dec 01, 2013 11:02 am

### BdMO National 2016 Secondary 3: Weird angle condition

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 $\angle BAP$
Hammer with tact.

Because destroying everything mindlessly isn't cool enough.

Thanic Nur Samin
Posts: 176
Joined: Sun Dec 01, 2013 11:02 am

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

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.

Thanic Nur Samin
Posts: 176
Joined: Sun Dec 01, 2013 11:02 am

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

On second thought, I am showing my calculation. Not that it is too long.
Hammer with tact.

Because destroying everything mindlessly isn't cool enough.

joydip
Posts: 48
Joined: Tue May 17, 2016 11:52 am

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

A synthetic solution :
The first principle is that you must not fool yourself and you are the easiest person to fool.