2023 AIME II Problem 3

Question

Let

\triangle ABC

be an isosceles triangle with

\angle A = 90^\circ.

There exists a point

P

inside

\triangle ABC

such that

\angle PAB = \angle PBC = \angle PCA

and

AP = 10.

Find the area of

\triangle ABC.

Answer

Answer: 250

Concepts:angle chasing law of sines trigonometry

Difficulty rating: 2460

Small hint:

Let

\omega

be the common angle. In triangle

APC

the angles at

A

and

C

are

90^\circ - \omega

and

\omega,

so

\angle APC = 90^\circ

Big hint:

Compare the law of sines in triangle

ABP

with

AC = \frac{AP}{\sin\omega};

everything reduces to

\tan\omega = \frac{1}{2}

Solution:

Let

\omega

denote the common angle and

L = AB = AC.

Since

\angle PAB = \omega,

we have

\angle PAC = 90^\circ - \omega,

and with

\angle PCA = \omega

the angles of triangle

APC

give

\angle APC = 90^\circ.

Hence in right triangle

APC,

L = AC = \frac{AP}{\sin\omega} = \frac{10}{\sin\omega}.

In triangle

ABP,

the angle at

A

is

\omega

and the angle at

B

is

45^\circ - \omega,

so

\angle APB = 135^\circ.

The law of sines gives

\frac{AP}{\sin(45^\circ - \omega)} = \frac{AB}{\sin 135^\circ},

that is,

10 \sin 135^\circ = L \sin(45^\circ - \omega).

Substituting

L = \frac{10}{\sin\omega}

and expanding yields

\begin{aligned} \sin\omega &= \sqrt{2}\,\sin(45^\circ - \omega) \\ &= \cos\omega - \sin\omega, \end{aligned}

so

\tan\omega = \frac{1}{2}

and

\sin^2\omega = \frac{1}{5}.

Therefore

L^2 = \frac{100}{\sin^2\omega} = 500,

and the area is

\frac{1}{2}L^2 = 250.