# 2013 Mock AIME I Problems/Problem 12

## Problem

In acute triangle , the orthocenter lies on the line connecting the midpoint of segment to the midpoint of segment . If , and the altitude from has length , find .

## Solution (easy coordinate bash)

Toss on the coordinate plane with , , and , where is a real number and .

Then, the line connecting the midpoints of and runs from to , or more simply the line .

The orthocenter of will be at the intersection of the altitudes from and .

The slope of the altitude from is the negative reciprocal of the slope of . The slope of is , and its negative reciprocal is . Since the altitude from passes through the origin, its equation is .

The altitude from is the vertical line running through which has equation .

Thus the lines and meet on the line . Substituting the first equation into the second, .

Multiplying both sides by , we have .

This rearranges to the quadratic , and completing the square by adding to each side gives us . Thus .

The cases where and are similar; they merely correspond to two triangles that can each be obtained by reflecting the other across the perpendicular bisector of , so we consider the case where .

So .

Thus

The cases where and are shown below, labeled and , respectively, where the dotted line is a midline in both triangles. As you can see, the orthocenter falls perfectly on that line for both triangles, and the value of is the same for both triangles.

## Solution 2 (no coordinates)

Let be the midpoint of and be the midpoint of . Further let be the foot of the altitude from , from , and from , as in the diagram.

Because is a midpoint connector of amd is on and , we know that is the midpoint of altitude . Thus, because, from the problem, , . Now we see that is a midpoint connector of , so .

Now, let . We know that , because they are vertical angles. Because is right (by the definition of an altitude), we know that . is also right, so .

From , we know that . From , we know that . Equating these two expressions for , we see that . From the problem, we know that .

Now, we can proceed as in Solution 1 by using the quadratic formula to solve for and the Pythagorean Theorem to find and . We do this to obtain our answer .