1998 AHSME Problems/Problem 28
Problem
In triangle , angle is a right angle and . Point is located on so that angle is twice angle . If , then , where and are relatively prime positive integers. Find .
Solution
Let , so and . Then, it is given that and
Now, through the use of trigonometric identities, . Solving yields that . Using the tangent addition identity, we find that , and
and . (This also may have been done on a calculator by finding directly)
Solution 3
By the application of ratio lemma for , we get , where we let . We already know hence the rest is easy (latex edit by aopspandy)
Solution 2
Let and . By the Pythagorean Theorem, . Let point be on segment such that bisects . Thus, angles , , and are congruent. Applying the angle bisector theorem on , we get that and . Pythagorean Theorem gives .
Let . By the Pythagorean Theorem, . Applying the angle bisector theorem again on triangle , we have The right side simplifies to. Cross multiplying, squaring, and simplifying, we get a quadratic: Solving this quadratic and taking the positive root gives Finally, taking the desired ratio and canceling the roots gives . The answer is .
See also
1998 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 27 |
Followed by Problem 29 | |
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