# Difference between revisions of "2015 AMC 10A Problems/Problem 19"

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

## Revision as of 17:48, 4 February 2015

## Problem

The isosceles right triangle has right angle at and area . The rays trisecting intersect at and . What is the area of ?

## Solution

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 , .

Because the side lengths of a right triangle are in ratio and + , .

Setting the two equations for equal together, .

Solving gives .

The area of .

is congruent to , so their areas are equal.

A triangle's area can be written as the sum of the figures that make it up, so .

.

Solving gives , so the answer is .