2010 AIME I Problems/Problem 15

Problem

In $\triangle{ABC}$ with $AB = 12$ , $BC = 13$ , and $AC = 15$ , let $M$ be a point on $\overline{AC}$ such that the incircles of $\triangle{ABM}$ and $\triangle{BCM}$ have equal radii. Let $p$ and $q$ be positive relatively prime integers such that $\frac {AM}{CM} = \frac {p}{q}$ . Find $p + q$ .

Solution

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Let $AM = 2x$ and $BM = 2y$ so $CM = 15 - 2x$ . Let the incenters of $\triangle ABM$ and $\triangle BCM$ be $I_1$ and $I_2$ respectively, and their equal inradii be $r$ . From $r = \sqrt {(s - a)(s - b)(s - c)/s}$ , we find that

$\begin{align*}r^2 & = \frac {(x + y - 6)( - x + y + 6)(x - y + 6)}{x + y + 6} & (1) \\ & = \frac {( - x + y + 1)(x + y - 1)( - x - y + 14)}{ - x + y + 14}. & (2) \end{align*}$

Let the incircle of $\triangle ABM$ meet $AM$ at $P$ and the incircle of $\triangle BCM$ meet $CM$ at $Q$ . Then note that $I_1 P Q I_2$ is a rectangle. Also, $\angle I_1 M I_2$ is right because $MI_1$ and $MI_2$ are the angle bisectors of $\angle AMB$ and $\angle CMB$ respectively and $\angle AMB + \angle CMB = 180^\circ$ . By properties of tangents to circles $MP = (MA + MB - AB)/2 = x + y - 6$ and $MQ = (MB + MC - BC) = - x + y + 1$ . Now notice that the altitude of $M$ to $I_1 I_2$ is of length $r$ , so by similar triangles we find that $r^2 = MP \cdot MQ = (x + y - 6)( - x + y + 1)$ (3). Equating (3) with (1) and (2) separately yields

$2y^2 - 30 = 2xy + 5x - 7y \\ 2y^2 - 70 = - 2xy - 5x + 7y,$

and adding these we have

$4y^2 - 100 = 0\implies y = 5\implies x = 11/3 \\ \implies AM/MC = (22/3)/(15 - 22/3) = 22/23 \implies \boxed{045}.$

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2010 AIME I Problems/Problem 15

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