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 .