1987 AIME Problems/Problem 15
Contents
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 us We take the positive root, so , which means .
Messy Trig Solution
Let be the smaller angle in the triangle. Then the sum of shorter and longer leg is . We observe that the short leg has length . Grouping and squaring, we get . Squaring and using the double angle identity for sine, we get, . Solving, we get . Now to find , we find using the Pythagorean Identity, and then use the tangent double angle identity. Thus, . Substituting into the original sum, we get .
Solution 4 (Algebra)
Label points as above. Let , , be the side length of , and be the side length of .
Since , we have
.
Since , we have
Let . Repeatedly applying , we get
~rayfish
See also
1987 AIME (Problems • Answer Key • Resources) | ||
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