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 be a diameter of circle . Extend through to . Point lies on so that line is tangent to . Point is the foot of the perpendicular from to line . Suppose , and let denote the maximum possible length of the segment . Find .

Here’s an illustration of the situation:

We must maximize the length of the red line segment .

As the diameter rotates, both and 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 , specifically, the mid-segment that joins the center of and the midpoint of (which are labeled and , respectively, in the following figure):

As diameter rotates, what path does trace? Since is the midpoint of , the point traces out the curve obtained by squashing the circle by a factor of 1/2 in the vertical direction:

reaches its maximum length precisely when is perpendicular to the tangent to the squashed circle at . The slope of the tangent to the squashed circle at is half the slope of the tangent to at point . Therefore, we want the slope of to be twice the slope of (since *is* perpendicular to the tangent to at ), which is the same thing as saying that we want to be twice .

Since is the midpoint of , we see that is maximized exactly when , i.e. when and trisect the segment .

Thus, is maximized when and . Hence, the maximal value of is .

Remembering that is half that of , we conclude that the answer to the AIME problem is .

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