# Mock AIME II 2012 Problems/Problem 9

## Problem

In , , , and . and lie on , and , respectively. If , and , the area of can be expressed in the form where are all positive integers, and and do not have any perfect squares greater than as divisors. Find .

## Solution

Here is a diagram (note that D should be on AC and F should be on BC): . Start out by finding . Remark that by the law of sines, . Therefore . We know because in , therefore is in the first quadrant. Use the Pythagorean Identity to give us .

Now, note that is the same thing as , from , therefore we have $\equal{} \frac{\sqrt3*\sqrt{73}}{20}-\frac{3\sqrt3}{20}=\frac{\sqrt3*\sqrt{73}-3\sqrt3}{20}$ (Error compiling LaTeX. ).

Next, use the law of sines to give us . This gives us . This gives us .

Now, we use coordinates to find that the coordinates of , , , and . is going to be the point from adding point to create triangle as shown in this diagram (where is the right angle, is the angle, and is the angle):

Now, note that we have to find the coordinates of and . Assume WLOG that is and be . is obviously , is going to be and the coordinates of is going to be . Since and have the same value, we can find by using . The base is going to be equal to the distance from to , which is the same thing as , and the height is the change in the coordinate from to . This is the same thing as . Hence, plugging these into give us . The answer is thus .

## Solution 2

Let be the area of the large triangle and let be the area we are trying to find. We have that is similar to with a ratio of while is similar to with a ratio of . Then the area of and the area of . Lastly, the area of . The area of is equal to the sum of the areas of the four smaller triangles so . Thus .

It remains to find . First we can use the Law of Cosines on to find which solves to . Then . Then substituting yields so the answer is