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

$\triangle ADC$ 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 $30-60-90$ 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 $DF$ equal together, $AF = \frac{5 - AF}{\sqrt{3}}$ .

Solving gives $AF = DF = \frac{5\sqrt{3} - 5}{2}$ .

The area of $\triangle ADC = \frac12 \cdot DF \cdot AC = \frac{25\sqrt{3} - 25}{2}$ .

$\triangle ADC$ 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 $[ABC] = [ADC] + [BEC] + [CDE]$ .

$\frac{25}{2} = \frac{25\sqrt{3} - 25}{2} + \frac{25\sqrt{3} - 25}{2} + [CDE]$ .

Solving gives $[CDE] = \frac{50 - 25\sqrt{3}}{2}$ , 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