Difference between revisions of "1998 AHSME Problems/Problem 28"
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== Solution == | == Solution == | ||
Let <math>\theta = \angle DAB</math>, so <math>2\theta = \angle CAD</math> and <math>3 \theta = \angle CAB</math>. Then, it is given that <math>\cos 2\theta = \frac{AC}{AD} = \frac{2}{3}</math> and | Let <math>\theta = \angle DAB</math>, so <math>2\theta = \angle CAD</math> and <math>3 \theta = \angle CAB</math>. Then, it is given that <math>\cos 2\theta = \frac{AC}{AD} = \frac{2}{3}</math> and |
Revision as of 11:22, 30 July 2017
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 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 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 | ||
All AHSME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.