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. Unknown error_msg)30-60-90a:a\sqrt{3}:2aAFFC = 5DF = \frac{5 - AF}{\sqrt{3}}$.
Setting the two equations for$ (Error compiling LaTeX. Unknown error_msg)DFAF = \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\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}}.