# 2015 AMC 10A Problems/Problem 19

(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

## 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}}.