2018 AIME II Problems/Problem 12
Problem
Let be a convex quadrilateral with , , and . Assume that the diagonals of intersect at point , and that the sum of the areas of triangles and equals the sum of the areas of triangles and . Find the area of quadrilateral .
Solution
For reference, , so is the longest of the four sides of . Let be the length of the altitude from to , and let be the length of the altitude from to . Then, the triangle area equation becomes
.
What an important finding! Note that the opposite sides and have equal length, and note that diagonal bisects diagonal . This is very similar to what happens if were a parallelogram with , so let's extend to point , such that is a parallelogram. In other words, and . Now, let's examine . Since , the triangle is isosceles, and . Note that in parallelogram , and , so and thus . Define , so . We use the Law of Cosines on and :
Subtracting the second equation from the first yields
This means that dropping an altitude from to some foot on gives and therefore . Seeing that , we conclude that is a 3-4-5 right triangle, so . Then, the area of is . Since , points and are equidistant from , so and hence . -kgator
2018 AIME II (Problems • Answer Key • Resources) | ||
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Followed by Problem 13 | |
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