Difference between revisions of "2015 AMC 10A Problems/Problem 19"
(→Solution) |
(→Solution) |
||
Line 6: | Line 6: | ||
==Solution== | ==Solution== | ||
− | |||
− | |||
<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 .