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

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Setting the two equations for <math>DF</math> equal together, <math>AF = \frac{5 - AF}{\sqrt{3}}</math>. | Setting the two equations for <math>DF</math> equal together, <math>AF = \frac{5 - AF}{\sqrt{3}}</math>. | ||

− | Solving gives <math>AF = DF = \frac{5\sqrt{3} - 5}</math>. | + | Solving gives <math>AF = DF = \frac{5\sqrt{3} - 5}{2}</math>. |

The area of <math>\triangle ADC = \frac12 \cdot DF \cdot AC = \frac{25\sqrt{3} - 25}{2}</math>. | The area of <math>\triangle ADC = \frac12 \cdot DF \cdot AC = \frac{25\sqrt{3} - 25}{2}</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

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 , .

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 .