Difference between revisions of "2018 AIME II Problems/Problem 12"
(→Solution 3 (With yet another way to get the middle point)) |
(→Solution 3 (With yet another way to get the middle point)) |
||
Line 37: | Line 37: | ||
Now, assume that <math>AP=CP=x</math>,<math>BP=y</math>, and <math>DP=z</math>. Using the cosine rule for triangles <math>APB</math> and <math>BPC</math>, it is clear that <math>x^2+y^2-100=-(x^2+y^2-196)</math>, or <cmath>x^2+y^2=148...(1)</cmath> Likewise, using the cosine rule for triangles <math>APD</math> and <math>CPD</math>, <cmath>x^2+z^2=180...(2)</cmath>. It follows that <cmath>z^2-y^2=32...(3)</cmath>. Now, denote angle <math>APB</math> by <math>\alpha</math>. 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}</cmath>, giving <cmath>5y=3z</cmath>. Plugging this back to equations (1), (2), and (3), 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> | Now, assume that <math>AP=CP=x</math>,<math>BP=y</math>, and <math>DP=z</math>. Using the cosine rule for triangles <math>APB</math> and <math>BPC</math>, it is clear that <math>x^2+y^2-100=-(x^2+y^2-196)</math>, or <cmath>x^2+y^2=148...(1)</cmath> Likewise, using the cosine rule for triangles <math>APD</math> and <math>CPD</math>, <cmath>x^2+z^2=180...(2)</cmath>. It follows that <cmath>z^2-y^2=32...(3)</cmath>. Now, denote angle <math>APB</math> by <math>\alpha</math>. 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}</cmath>, giving <cmath>5y=3z</cmath>. Plugging this back to equations (1), (2), and (3), 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> | ||
--Solution by MicGu | --Solution by MicGu | ||
+ | ==Solution 4 == | ||
+ | As in all other solutions, we can first find that either <math>AP=CP</math> or <math>BP=DP</math>, but it's an AIME problem, we can take <math>AP=CP</math>, and assume the other choice will lead to the same result (which is true). | ||
+ | |||
+ | From <math>AP=CP</math>, we have <math>[DAP]=[ACP]</math>, <math>[BAP]=[BCP]</math> => <math>[ABD] = [CBD]</math>, therefore, | ||
+ | <cmath>1/2AB*AD\sin A = 1/2BC*CD\sin C => 7\sin C = \sqrt{65}\sin A ... (1)</cmath> | ||
+ | By Law of Cosine, | ||
+ | <cmath>10^2+14^2-2*10*14\cos C=10^2+4*65-2*10*2\sqrt{65}\cos A</cmath> | ||
+ | <cmath>(-8/5)-7\cos C = \sqrt{65}\cos A ...(2) </cmath> | ||
+ | Square (1) and (2), add them, we get | ||
+ | <cmath>(8/5)^2 +2(8/5)7\cos C + 7^2 = 65 </cmath> | ||
+ | Solve, <math>\cos C = 3/5</math> => <math>\sin C = 4/5</math>, | ||
+ | <cmath>[ABCD] = 2[BCD] = BC*CD*\sin C = 14*10*(4/5) = \boxed{112}</cmath> | ||
+ | -Mathdummy | ||
{{AIME box|year=2018|n=II|num-b=11|num-a=13}} | {{AIME box|year=2018|n=II|num-b=11|num-a=13}} | ||
+ | |||
+ | |||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 09:34, 10 March 2019
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 .
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 middle point of
~Solution by (Frank FYC)
Solution 3 (With yet another way to get the middle point)
Using the formula for the area of a triangle, But , so Hence (note that makes no difference here). Now, assume that ,, and . Using the cosine rule for triangles and , it is clear that , or Likewise, using the cosine rule for triangles and , . It follows that . Now, denote angle by . Since , which simplifies to , giving . Plugging this back to equations (1), (2), and (3), 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 , => , therefore, By Law of Cosine, Square (1) and (2), add them, we get Solve, => , -Mathdummy
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.