# 2015 AMC 10A Problems/Problem 19

## Problem

The isosceles right triangle has right angle at and area . The rays trisecting intersect at and . What is the area of ?

## Solution

Let and be the points at which the angle trisectors intersect .

can be split into a right triangle and a right triangle by dropping a perpendicular from to side . Let be where that perpendicular intersects .

Because the side lengths of a right triangle are in ratio , .

Because the side lengths of a right triangle are in ratio and + , .

Setting the two equations for equal together, .

Solving gives $AF = DF = \frac{5\sqrt{3} - 5}$ (Error compiling LaTeX. ! Missing } inserted.).

The area of .

is congruent to , so their areas are equal.

A triangle's area can be written as the sum of the figures that make it up, so .

.

Solving gives , so the answer is .