Difference between revisions of "2018 AIME II Problems/Problem 12"
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Let <math>AP=x</math> and let <math>PC=\rho x</math>. Let <math>[ABP]=\Delta</math> and let <math>[ADP]=\Lambda</math>. We easily get <math>[PBC]=\rho \Delta</math> and <math>[PCD]=\rho\Lambda</math>. | Let <math>AP=x</math> and let <math>PC=\rho x</math>. Let <math>[ABP]=\Delta</math> and let <math>[ADP]=\Lambda</math>. We easily get <math>[PBC]=\rho \Delta</math> and <math>[PCD]=\rho\Lambda</math>. | ||
− | We are given that <math>[ABP] +[PCD] = [PBC]+[ADP]</math>, which we can now write as <cmath>\Delta + \rho\Lambda = \rho\Delta + \Lambda \qquad \Longrightarrow \qquad \Delta -\Lambda = \rho (\Delta -\Lambda).</cmath> Either <math>\Delta = \Lambda</math> or <math>\rho=1</math>. The former would imply that <math>ABCD</math> is a parallelogram, which it isn't; therefore we conclude <math>\rho=1</math> and <math>P</math> is the midpoint of <math>AC</math>. Let <math>\angle BAD = \theta</math> and <math>\angle BCD = \phi</math>. Then <math>[ABCD]=2\cdot [BCD]=140\sin\phi</math>. On one hand, since <math>[ABD]=[BCD]</math>, we have <cmath> | + | We are given that <math>[ABP] +[PCD] = [PBC]+[ADP]</math>, which we can now write as <cmath>\Delta + \rho\Lambda = \rho\Delta + \Lambda \qquad \Longrightarrow \qquad \Delta -\Lambda = \rho (\Delta -\Lambda).</cmath> Either <math>\Delta = \Lambda</math> or <math>\rho=1</math>. The former would imply that <math>ABCD</math> is a parallelogram, which it isn't; therefore we conclude <math>\rho=1</math> and <math>P</math> is the midpoint of <math>AC</math>. Let <math>\angle BAD = \theta</math> and <math>\angle BCD = \phi</math>. Then <math>[ABCD]=2\cdot [BCD]=140\sin\phi</math>. On one hand, since <math>[ABD]=[BCD]</math>, we have <cmath> |
==Solution 2== | ==Solution 2== |
Revision as of 14:31, 27 December 2021
Contents
[hide]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
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.