Difference between revisions of "2018 AIME II Problems/Problem 12"
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As with previous solutions, we find that we have a choice between <math>PA=PC</math> and <math>PD=PB</math>, opting for the latter. Applying Stewart's Theorem to <math>ABC</math>, <math>BCD</math>, <math>CDA</math>, and <math>DAB</math>, and (tediously) combining the equations that result, we find | As with previous solutions, we find that we have a choice between <math>PA=PC</math> and <math>PD=PB</math>, opting for the latter. Applying Stewart's Theorem to <math>ABC</math>, <math>BCD</math>, <math>CDA</math>, and <math>DAB</math>, and (tediously) combining the equations that result, we find | ||
− | < | + | <cmath>DP=PB=\sqrt{130}, CP=\sqrt{18}, PA=\sqrt{50}</cmath> |
− | Using the Law of Cosines, and | + | Using the Law of Cosines, and <math>\sin(\angle APB)^2+cos(\angle APB)^2=1</math>, we find |
− | \sin(\angle APB)=\frac{7}{\sqrt{65}} | + | <cmath>\sin(\angle APB)=\frac{7}{\sqrt{65}}</cmath> |
− | We can then use <math>ab\sin(\angle C)</math> to find <math>[APB]</math> , <math>[BPC]</math> , <math>[CPD]</math> , and <math>[DPA]</math>. Adding the areas up we find <math>[ABCD]=\boxed{112}</math> | + | Note that this is also the sine of all the "center" angles. |
+ | |||
+ | We can then use <math>[ABC]=\frac{ab\sin(\angle C)}{2}</math> to find <math>[APB]</math> , <math>[BPC]</math> , <math>[CPD]</math> , and <math>[DPA]</math>. Adding the areas up we find <math>[ABCD]=\boxed{112}</math> | ||
-aeai | -aeai |
Revision as of 23:12, 3 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 (Stewart)
As with previous solutions, we find that we have a choice between and , opting for the latter. Applying Stewart's Theorem to , , , and , and (tediously) combining the equations that result, we find
Using the Law of Cosines, and , we find
Note that this is also the sine of all the "center" angles.
We can then use to find , , , and . Adding the areas up we find
-aeai
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.