# 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 , AF$.

Because the side lengths of a$ (Error compiling LaTeX. ! Missing $ inserted.)30-60-90a:a\sqrt{3}:2aAFFC = 5DF = \frac{5 - AF}{\sqrt{3}}$.

Setting the two equations for$ (Error compiling LaTeX. ! Missing $ inserted.)DFAF = \frac{5 - AF}{\sqrt{3}}$.

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

The area of$ (Error compiling LaTeX. ! Missing $ inserted.)\triangle ADC = \frac12 \cdot DF \cdot AC = \frac{25\sqrt{3} - 25}{2}\triangle ADC\triangle BEC$, so their areas are equal.

A triangle's area can be written as the sum of the figures that make it up, so$ (Error compiling LaTeX. ! Missing $ inserted.)[ABC] = [ADC] + [BEC] + [CDE]\frac{25}{2} = \frac{25\sqrt{3} - 25}{2} + \frac{25\sqrt{3} - 25}{2} + [CDE]$.

Solving gives$ (Error compiling LaTeX. ! Missing $ inserted.)[CDE] = \frac{50 - 25\sqrt{3}}{2}$, so the answer is \boxed{\textbf{(D) }\frac{50 - 25\sqrt{3}}{2}}.