1996 AIME Problems/Problem 15
Problem
In parallelogram , let be the intersection of diagonals and . Angles and are each twice as large as angle , and angle is times as large as angle . Find .
Contents
Solution
Solution 1 (trigonometry)
Let . Then , , and . Since is a parallelogram, it follows that . By the Law of Sines on ,
Dividing the two equalities yields
Pythagorean and product-to-sum identities yield
and the double and triple angle () formulas further simplify this to
The only value of that fits in this context comes from . The answer is .
Solution 2 (trigonometry)
Define as above. Since , it follows that , and so . The Law of Sines on yields that
Expanding using the sine double and triple angle formulas, we have
By the quadratic formula, we have , so (as the other roots are too large to make sense in context). The answer follows as above.
Solution 3
We will focus on . Let , so . Draw the perpendicular from intersecting at . Without loss of generality, let . Then , since is the circumcenter of . Then .
By the Exterior Angle Theorem, and . That implies that . That makes . Then since by AA ( and reflexive on ), .
Then by the Pythagorean Theorem, . That makes equilateral. Then . The answer follows as above.
See also
1996 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Final Problem | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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