Difference between revisions of "2018 AIME II Problems/Problem 12"
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− | Now we prove P is the midpoint of <math>BD</math>. Denote the height from <math>B</math> to <math>AC</math> as <math>h_1</math>, height from <math>D</math> to <math>AC</math> as <math>h_2</math>.According to the problem, <math>AP*h_1+CP*h_2=CP*h_1+AP*h_2</math> implies <math>h_1(AP-CP)=h_2(AP-CP), h_1=h_2</math>. Then according to basic congruent triangles we get <math>BP=DP</math> | + | Now we prove P is the midpoint of <math>BD</math>. Denote the height from <math>B</math> to <math>AC</math> as <math>h_1</math>, height from <math>D</math> to <math>AC</math> as <math>h_2</math>.According to the problem, <math>AP* h_1 +CP* h_2 =CP* h_1 +AP* h_2 </math> implies <math> h_1 (AP-CP)= h_2 (AP-CP), h_1 = h_2 </math>. Then according to basic congruent triangles we get <math>BP=DP</math> |
Firstly, denote that <math>CP=a,BP=b,CP=c,AP=d</math>. Applying Stewart theorem, getting that <math>100c+196b=(b+c)(bc+a^2); 100b+260c=(b+c)(bc+c^2); 100c+196b=100b+260c, 3b=5c</math>, denote <math>b=5x,c=3x</math> | Firstly, denote that <math>CP=a,BP=b,CP=c,AP=d</math>. Applying Stewart theorem, getting that <math>100c+196b=(b+c)(bc+a^2); 100b+260c=(b+c)(bc+c^2); 100c+196b=100b+260c, 3b=5c</math>, denote <math>b=5x,c=3x</math> | ||
Applying Stewart Theorem, getting <math>260a+100a=2a(a^2+25x^2); 196a+100a=2a(9x^2+a^2)</math> solve for a, getting <math>a=\sqrt{130},AP=5\sqrt{2}; CP=3\sqrt{2}</math> | Applying Stewart Theorem, getting <math>260a+100a=2a(a^2+25x^2); 196a+100a=2a(9x^2+a^2)</math> solve for a, getting <math>a=\sqrt{130},AP=5\sqrt{2}; CP=3\sqrt{2}</math> |
Revision as of 21:19, 2 February 2022
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
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
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.