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Harvard-MIT Mathematics Tournament

Geometry Difficulty 4.8 AIME Prove it United States

Problem:

Let $ABC$ be a triangle with $\angle A = 45^{\circ}$ . Let $P$ be a point on side $BC$ with $PB = 3$ and $PC = 5$ . Let $O$ be the circumcenter of $ABC$ . Determine the length $OP$ .

Solution

Solution:

Using the extended Sine Law, we find the circumradius of $ABC$ to be $R = \frac{BC}{2 \sin A} = 4 \sqrt{2}$ .

By considering the power of point $P$ , we find that $R^2 - OP^2 = PB \cdot PC = 15$ .

So $OP = \sqrt{R^2 - 15} = \sqrt{16 \cdot 2 - 15} = \sqrt{17}$ .

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