# 2015 AMC 10A Problems/Problem 19

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## Problem

The isosceles right triangle $ABC$ has right angle at $C$ and area $12.5$. The rays trisecting $\angle ACB$ intersect $AB$ at $D$ and $E$. What is the area of $\bigtriangleup CDE$?

$\textbf{(A) }\dfrac{5\sqrt{2}}{3}\qquad\textbf{(B) }\dfrac{50\sqrt{3}-75}{4}\qquad\textbf{(C) }\dfrac{15\sqrt{3}}{8}\qquad\textbf{(D) }\dfrac{50-25\sqrt{3}}{2}\qquad\textbf{(E) }\dfrac{25}{6}$

## Solution

Let $D$ and $E$ be the points at which the angle trisectors intersect $AB$.

$\triangle ADC$ can be split into a $45-45-90$ right triangle and a $30-60-90$ right triangle by dropping a perpendicular from $D$ to side $AC$. Let $F$ be where that perpendicular intersects $AC$.

Because the side lengths of a $45-45-90$ right triangle are in ratio $a:a:2a$, $DF =$AF$. Because the side lengths of a$ (Error compiling LaTeX. ! Missing $inserted.)30-60-90$right triangle are in ratio$a:a\sqrt{3}:2a$and$AF$+$FC = 5$,$DF = \frac{5 - AF}{\sqrt{3}}$.

Setting the two equations for$(Error compiling LaTeX. ! Missing$ inserted.)DF$equal together,$AF = \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$is congruent to$\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}}.