2018 AIME II Problems/Problem 12
Contents
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 1
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 are congruent, 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
Just to be complete -- and can actually be equal. In this case, , but must be equal to . We get the same result. -Mathdummy.
Solution 2 (Another way to get the middle point)
So, let the area of triangles , , , . Suppose and , then it is easy to show that Also, because we will have So So So So As a result, Then, we have Combine the condition we can find out that so is the midpoint of
~Solution by (Frank FYC)
Solution 3 (With yet another way to get the middle point)
Using the formula for the area of a triangle, But , so Hence (note that makes no difference here). Now, assume that ,, and . Using the cosine rule for triangles and , it is clear that
, or Likewise, using the cosine rule for triangles and , . It follows that . Now, denote angle by . Since , which simplifies to , giving . Plugging this back to equations (1), (2), and (3), it can be solved that . Then, the area of the quadrilateral is --Solution by MicGu
Solution 4
As in all other solutions, we can first find that either or , but it's an AIME problem, we can take , and assume the other choice will lead to the same result (which is true).
From , we have , => , therefore, By Law of Cosine, Square (1) and (2), add them, we get Solve, => , -Mathdummy
See Also
2018 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
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