# 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>. | ||

− | Because the side lengths of a <math>45-45-90</math> right triangle are in ratio <math>a:a:2a</math>, <math>DF = </ | + | Because the side lengths of a <math>45-45-90</math> right triangle are in ratio <math>a:a:2a</math>, <math>DF = AF</math>. |

− | Because the side lengths of a < | + | Because the side lengths of a <math>30-60-90</math> right triangle are in ratio <math>a:a\sqrt{3}:2a</math> and <math>AF</math> + <math>FC = 5</math>, <math>DF = \frac{5 - AF}{\sqrt{3}}</math>. |

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

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

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

− | < | + | <math>\triangle ADC</math> is congruent to <math>\triangle BEC</math>, so their areas are equal. |

− | A triangle's area can be written as the sum of the figures that make it up, so < | + | A triangle's area can be written as the sum of the figures that make it up, so <math>[ABC] = [ADC] + [BEC] + [CDE]</math>. |

− | < | + | <math>\frac{25}{2} = \frac{25\sqrt{3} - 25}{2} + \frac{25\sqrt{3} - 25}{2} + [CDE]</math>. |

− | Solving gives < | + | Solving gives <math>[CDE] = \frac{50 - 25\sqrt{3}}{2}</math>, so the answer is \boxed{\textbf{(D) }\frac{50 - 25\sqrt{3}}{2}}. |

## Revision as of 17:46, 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 $AF = DF = \frac{5\sqrt{3} - 5}$ (Error compiling LaTeX. ! Missing } inserted.).

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 \boxed{\textbf{(D) }\frac{50 - 25\sqrt{3}}{2}}.