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:

blog_061315

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):

blog_061315_2

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:

blog_061315_3

\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.

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