# AoPS Wiki talk:Problem of the Day/June 18, 2011

## Problem

AoPSWiki:Problem of the Day/June 18, 2011

## Solution

Let . Since we know that the perimeter of is , . Construct a perpendicular to from point . Call the intersection of and this perpendicular point . Since angle is right, is a rectangle, and thus angle is right, and . ; by the Pythagorean Theorem, , so and . Again by the Pythagorean Theorem, . Since and are bases, is parallel to, and so angle = angle ; by vertical angles, angle = angle . By AA, triangle ~ triangle in a ratio of . Thus, , so is of . Note that the distance from to is really the length of the perpendicular to drawn from by definition. Drawing this perpendicular and letting the point of intersection of and be point , we have that triangle is similar to triangle by AA in a ratio of . Thus , so . Solving, we get .