Difference between revisions of "2018 AIME II Problems/Problem 12"
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==Solution 3 (With yet another way to get the middle point)== | ==Solution 3 (With yet another way to get the middle point)== | ||
− | Using the formula for the area of a triangle, <cmath> | + | Denote <math>\angle APB</math> by <math>\alpha</math>. Then <math>\sin(\angle APB)=\sin \alpha = \sin(\angle APD)</math>. |
− | + | Using the formula for the area of a triangle, we get <cmath>\frac{1}{2} (AP\cdot BP+ CP\cdot DP)\sin\alpha=\frac{1}{2}(AP\cdot DP+ CP\cdot BP)\sin\alpha , </cmath> | |
+ | so <cmath>(AP-CP)(BP-DP)=0</cmath> | ||
Hence <math>AP=CP</math> (note that <math>BP=DP</math> makes no difference here). | Hence <math>AP=CP</math> (note that <math>BP=DP</math> makes no difference here). | ||
− | Now, assume that <math>AP=CP=x</math>,<math>BP=y</math>, and <math>DP=z</math>. Using the cosine rule for | + | Now, assume that <math>AP=CP=x</math>, <math>BP=y</math>, and <math>DP=z</math>. Using the cosine rule for <math>\triangle APB</math> and <math>\triangle BPC</math>, it is clear that <cmath>x^2+y^2-100=2 xy \cdot \cos{APB}=-(2 xy \cdot \cos{(\pi-CPB)})=-(x^2+y^2-196) </cmath> or <cmath>\begin{align}x^2+y^2=148\end{align}.</cmath> |
− | + | Likewise, using the cosine rule for triangles <math>APD</math> and <math>CPD</math>, <cmath>\begin{align}\tag{2}x^2+z^2=180\end{align}.</cmath> It follows that <cmath>\begin{align}\tag{3}z^2-y^2=32\end{align}.</cmath> Since <math>\sin\alpha=\sqrt{1-\cos^2\alpha}</math>, <cmath>\sqrt{1-\frac{(x^2+y^2-100)^2}{4x^2y^2}}=\sqrt{1-\frac{(x^2+z^2-260)^2}{4x^2z^2}}</cmath> which simplifies to <cmath>\frac{48^2}{y^2}=\frac{80^2}{z^2} \qquad \Longrightarrow \qquad 5y=3z.</cmath> Plugging this back to equations <math>(1)</math>, <math>(2)</math>, and <math>(3)</math>, it can be solved that <math>x=\sqrt{130},y=3\sqrt{2},z=5\sqrt{2}</math>. Then, the area of the quadrilateral is <cmath>x(y+z)\sin\alpha=\sqrt{130}\cdot8\sqrt{2}\cdot\frac{14}{\sqrt{260}}=\boxed{112}</cmath> | |
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--Solution by MicGu | --Solution by MicGu | ||
Revision as of 13:29, 29 November 2021
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)
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 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 , and , therefore, By Law of Cosines, Square and , and add them, to get Solve, , -Mathdummy
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.