1987 AIME Problems/Problem 15
Contents
[hide]Problem
Squares and are inscribed in right triangle , as shown in the figures below. Find if area and area .
Solution
Because all the triangles in the figure are similar to triangle , it's a good idea to use area ratios. In the diagram above, Hence, and . Additionally, the area of triangle is equal to both and
Setting the equations equal and solving for , . Therefore, . However, is equal to the area of triangle ! This means that the ratio between the areas and is , and the ratio between the sides is . As a result, . We now need to find the value of , because .
Let denote the height to the hypotenuse of triangle . Notice that . (The height of decreased by the corresponding height of ) Thus, . Because , .
Easy Trig Solution
Let . Now using the 1st square, and . Using the second square, . We have , or Rearranging and letting gives Taking the positive root gives , so .
See also
1987 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Last Question | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.