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 .
Diagram
Let and let . Let and let .
Solution 1
Let and let . Let and let . We easily get and .
We are given that , which we can now write as Either or . The former would imply that is a parallelogram, which it isn't; therefore we conclude and is the midpoint of . Let and . Then . On one hand, since , we have whereas, on the other hand, using cosine formula to get the length of , we get Eliminating in the above two equations and solving for we getwhich finally yields .
Solution 2
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 3 (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 4 (With yet another way to get the middle point)
Denote by . Then . Using the formula for the area of a triangle, we get so Hence (note that makes no difference here). Now, assume that , , and . Using the cosine rule for and , it is clear that or Likewise, using the cosine rule for triangles and , It follows that Since , which simplifies to Plugging this back to equations , , and , it can be solved that . Then, the area of the quadrilateral is --Solution by MicGu
Solution 5
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 , and , therefore, By Law of Cosines, Square and , and add them, to get Solve, , -Mathdummy
Solution 6
Either or . Let . Applying Stewart's Theorem on and , dividing by and rearranging, Applying Stewart on and , Substituting equations 1 and 2 into 3 and rearranging, . By Law of Cosines on , so . Using to find unknown areas, .
-Solution by Gart
Solution 7
Now we prove P is the midpoint of . Denote the height from to as , height from to as .According to the problem, implies . Then according to basic congruent triangles we get Firstly, denote that . Applying Stewart theorem, getting that , denote Applying Stewart Theorem, getting solve for a, getting Now everything is clear, we can find using LOC, , the whole area is
~bluesoul
Solution 8 (Simple Geometry)
as in another solutions.
Let be the point symmetrical with respect to Let points and be the foot of perpendiculars from and to respectively. The area of quadrilateral is equal to the area of triangle with sides . The semiperimeter is the area
vladimir.shelomovskii@gmail.com, vvsss
See Also
2018 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.