# Mock AIME I 2012 Problems/Problem 7

## Problem

Let two circles $O_1, O_2$ with radii $3, 5$ in the plane be centered at points $M, N$, respectively. Consider a point $P$ in the plane such that $PM = \sqrt{33}, PN = 7$. Denote the intersections of the line $PM$ with $O_1$ as $A, B$, and the intersections of the line $PN$ with $O_2$ as $C, D$. Let $O_1$ and $O_2$ intersect at points $X, Y$ such that $XY=2$. If $Z$ equals the area of $\triangle{PMN}$, $24Z$ can be written in the form $a\sqrt{b}+c\sqrt{d}+e$ where $b, d$ are distinct squarefree integers. Find $a+b+c+d+e$.

## Solution

Let $MN \cap XY=Q$. Clearly we have that $MN \perp XY$ and $XQ=YQ=1$. Let $P_{O_1}(P), P_{O_2}(P)$ denote the power of $P$ with respect to the circles $O_1, O_2$, respectively. Then $P_{O_1}(P) = 24$. Additionally, $P_{O_2}(P) = 24$, as well, so it follows that $P$ lies on the radical axis of the two circles. However, $X, Q, Y$ also lie on this radical axis, so $P, X, Q, Y$ are collinear.

Moreover, since the radical axis is perpendicular to $MN$, we have that $PQ$ is an altitude of the triangle $\triangle{PMN}$. Notice that by considering the right triangles $\triangle{MXQ}, \triangle{NXQ}$ and using the Pythagorean Theorem we obtain that $MN = 2\sqrt{2}+2\sqrt{6}$. It now suffices to find $PQ$.

Since $P$ lies on the radical axis, we have that $PX\cdot (PX+2)= PX \cdot (PX+XY) = PX \cdot PY = P_{O_1}(P) = 24$, so $PX=4 \implies PQ = PX+XQ = 5$. So then the area can be written as $\frac{1}{2}\cdot 5 \cdot \left(2\sqrt{2}+2\sqrt{6} \right) = 5\sqrt{2}+5\sqrt{6}$.

This can be rewritten into the desired answer form, which is $24Z = 120\sqrt{2}+120\sqrt{6}+0 \implies a+b+c+d+e= 120+2+120+6+0 = \boxed{248}$.