HMMT February, 2022

Geometry Difficulty 5.8 AIME, harder Prove it United States

Problem:
Let $ABCD$ be a rectangle inscribed in circle $\Gamma$ , and let $P$ be a point on minor arc $AB$ of $\Gamma$ . Suppose that $PA \cdot PB = 2$ , $PC \cdot PD = 18$ , and $PB \cdot PC = 9$ . The area of rectangle $ABCD$ can be expressed as $\frac{a \sqrt{b}}{c}$ , where $a$ and $c$ are relatively prime positive integers and $b$ is a squarefree positive integer. Compute $100a + 10b + c$ .

Solution

Solution:
We have
$PD \cdot PA = \frac{(PA \cdot PB)(PD \cdot PC)}{PB \cdot PC} = \frac{2 \cdot 18}{9} = 4$
Let $\alpha = \angle DPC = 180^\circ - \angle APB$ and $\beta = \angle APD = \angle BPC$ . Note that $\alpha + \beta = 90^\circ$ . We have, letting $x = AB = CD$ and $y = AD = BC$ ,
$2[PAD] + 2[PBC] = y(d(P, AD) + d(P, BC)) = y \cdot x = [ABCD]$
Here $d(X, \ell)$ is used to denote the distance from $X$ to line $\ell$ . By the trig area formula, the left-hand side is
$PA \cdot PD \cdot \sin \beta + PB \cdot PC \cdot \sin \beta = 13 \sin \beta$
Similarly, we have $[ABCD] = 16 \sin \alpha$ . Thus, letting $K = [ABCD]$ ,
$1 = \sin^2 \alpha + \sin^2 \beta = \frac{K^2}{13^2} + \frac{K^2}{16^2} = \frac{425}{13^2 \cdot 16^2} K^2$
giving $K = \frac{208}{\sqrt{425}} = \frac{208 \sqrt{17}}{85}$ .

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