2015 AMC 10A Problems/Problem 19

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. Unknown error_msg)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. Unknown error_msg)DF $equal together,$ AF = \frac{5 - AF}{\sqrt{3}}$.

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

The area of$ (Error compiling LaTeX. Unknown error_msg)\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. Unknown error_msg)[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. Unknown error_msg)[CDE] = \frac{50 - 25\sqrt{3}}{2}$, so the answer is \boxed{\textbf{(D) }\frac{50 - 25\sqrt{3}}{2}}.

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2015 AMC 10A Problems/Problem 19

Problem

Solution