## Conceptual Solution to 2008 AIME I, Problem 14

Here’s a solution to problem 14 on the 2008 AIME I contest that attempts to avoid computation as much as possible. Sometimes, it’s an amusing exercise to try to solve a contest problem entirely in your head. Doing so often forces one to see deeper into the math.

First, the problem:

Let $\overline{AB}$ be a diameter of circle $\omega$. Extend $\overline{AB}$ through $A$ to $C$. Point $T$ lies on $\omega$ so that line $CT$ is tangent to $\omega$. Point $P$ is the foot of the perpendicular from $A$ to line $CT$. Suppose $AB = 18$, and let $m$ denote the maximum possible length of the segment $BP$. Find $m^2$.

Here’s an illustration of the situation:

We must maximize the length of the red line segment $\overline{BP}$.

As the diameter $\overline{AB}$ rotates, both $B$ and $P$ move. It’s typically easier to maximize the length of a line segment if one of its endpoints is stationary, so let’s instead maximize the length of the mid-segment of triangle $ABP$, specifically, the mid-segment that joins the center of $\omega$ and the midpoint of $\overline{AP}$ (which are labeled $O$ and $M$, respectively, in the following figure):

As diameter $\overline{AB}$ rotates, what path does $M$ trace? Since $M$ is the midpoint of $\overline{AP}$, the point $M$ traces out the curve obtained by squashing the circle $\omega$ by a factor of 1/2 in the vertical direction:

$\overline{OM}$ reaches its maximum length precisely when $\overline{OM}$ is perpendicular to the tangent to the squashed circle at $M$. The slope of the tangent to the squashed circle at $M$ is half the slope of the tangent to $\omega$ at point $A$. Therefore, we want the slope of $\overline{OM}$ to be twice the slope of $\overline{OA}$ (since $\overline{OA}$ is perpendicular to the tangent to $\omega$ at $A$), which is the same thing as saying that we want $XM$ to be twice $XA$.

Since $M$ is the midpoint of $\overline{AP}$, we see that $OM$ is maximized exactly when $XA = AM = MP$, i.e. when $A$ and $M$ trisect the segment $\overline{XP}$.

Thus, $OM^2 = OX^2 + XM^2$ is maximized when $OX^2 = 9^2 - 3^2$ and $XM = 6$. Hence, the maximal value of $OM^2$ is $9^2 - 3^2 + 6^2 = 108$.

Remembering that $OM$ is half that of $BP$, we conclude that the answer to the AIME problem is $4 \times 108 = 432$.