2017 AMC 12B Problems/Problem 24
Problem 24
Quadrilateral has right angles at and , , and . There is a point in the interior of such that and the area of is times the area of . What is ?
Solution 1
Let , , and . Note that . By the Pythagorean Theorem, . Since , the ratios of side lengths must be equal. Since , and . Let F be a point on such that is an altitude of triangle . Note that . Therefore, and . Since and form altitudes of triangles and , respectively, the areas of these triangles can be calculated. Additionally, the area of triangle can be calculated, as it is a right triangle. Solving for each of these yields: Therefore, the answer is
Solution 2
Draw line through , with on and on , . WLOG let , , . By weighted average .
Meanwhile, . This follows from comparing the ratios of triangle DEG to CFE and triangle AEG to FEB, both pairs in which the two triangles share a height perpendicular to FG, and have base ratio .
. We obtain , namely .
The rest is the same as Solution 1.
Solution 3
Let . Then from the similar triangles condition, we compute and . Hence, the -coordinate of is just . Since lies on the unit circle, we can compute the coordinate as . By Shoelace, we want Factoring out denominators and expanding by minors, this is equivalent to This factors as , so and so the answer is .
Notes
1) is the most relevant answer choice because it shares numbers with the givens of the problem.
2) It's a very good guess to replace finding the area of triangle AED with the area of the triangle DAF, where F is the projection of D onto AB(then find the closest answer choice).
See Also
2017 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 23 |
Followed by Problem 25 |
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