2023 AIME II Problems/Problem 3

Problem

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 $\triangle ABC.$

Solution 1

Since the triangle is a right isosceles triangle, angles B and C are $45$

Let the common angle be $\theta$

Note that angle PAC is $90-\theta$ , thus angle APC is $90$ . From there, we know that AC is $\frac{10}{\sin\theta}$

Note that ABP is $45-\theta$ , so from law of sines we have: $\frac{10}{\sin\theta \cdot \frac{\sqrt{2}}{2}}=\frac{10}{\sin(45-\theta)}$

Dividing by 10 and multiplying across yields: $\sqrt{2}\sin(45-\theta)=\sin\theta$

From here use the sin subtraction formula, and solve for $\sin\theta$

$\cos\theta-\sin\theta=\sin\theta$ $2\sin\theta=\cos\theta$ $4\sin^2\theta=cos^2\theta$ $4\sin^2\theta=1-\sin^2\theta$ $5\sin^2\theta=1$ $\sin\theta=\frac{1}{\sqrt{5}}$

Substitute this to find that AC= $10\sqrt{5}$ , thus the area is $\frac{(10\sqrt{5})^2}{2}=\boxed{250}$ ~SAHANWIJETUNGA

2023 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 2	Followed by Problem 4
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2023 AIME II Problems/Problem 3

Problem

Solution 1

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