Difference between revisions of "2000 AIME I Problems/Problem 2"
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Since <math>A = (u,v)</math>, we can find the coordinates of the other points: <math>B = (v,u)</math>, <math>C = (-v,u)</math>, <math>D = (-v,-u)</math>, <math>E = (v,-u)</math>. If we graph those points, we notice that since the latter four points are all reflected across the x/y-axis, they form a rectangle, and <math>ABE</math> is a triangle. The area of <math>BCDE</math> is <math>(2u)(2v) = 4uv</math> and the area of <math>ABE</math> is <math>\frac{1}{2}(2u)(u-v) = u^2 - uv</math>. Adding these together, we get <math>u^2 + 3uv = u(u+3v) = 451 = 11 \cdot 41</math>. Since <math>u,v</math> are positive, <math>u+3v>u</math>, and by matching factors we get either <math>(u,v) = (1,150)</math> or <math>(11,10)</math>. Since <math>v < u</math> the latter case is the answer, and <math>u+v = \boxed{021}</math>. | Since <math>A = (u,v)</math>, we can find the coordinates of the other points: <math>B = (v,u)</math>, <math>C = (-v,u)</math>, <math>D = (-v,-u)</math>, <math>E = (v,-u)</math>. If we graph those points, we notice that since the latter four points are all reflected across the x/y-axis, they form a rectangle, and <math>ABE</math> is a triangle. The area of <math>BCDE</math> is <math>(2u)(2v) = 4uv</math> and the area of <math>ABE</math> is <math>\frac{1}{2}(2u)(u-v) = u^2 - uv</math>. Adding these together, we get <math>u^2 + 3uv = u(u+3v) = 451 = 11 \cdot 41</math>. Since <math>u,v</math> are positive, <math>u+3v>u</math>, and by matching factors we get either <math>(u,v) = (1,150)</math> or <math>(11,10)</math>. Since <math>v < u</math> the latter case is the answer, and <math>u+v = \boxed{021}</math>. | ||
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=== Solution 2 === | === Solution 2 === |
Latest revision as of 19:14, 9 October 2020
Problem
Let and be integers satisfying . Let , let be the reflection of across the line , let be the reflection of across the y-axis, let be the reflection of across the x-axis, and let be the reflection of across the y-axis. The area of pentagon is . Find .
Solution
Solution 1
Since , we can find the coordinates of the other points: , , , . If we graph those points, we notice that since the latter four points are all reflected across the x/y-axis, they form a rectangle, and is a triangle. The area of is and the area of is . Adding these together, we get . Since are positive, , and by matching factors we get either or . Since the latter case is the answer, and .
Solution 2
We find the coordinates like in the solution above: , , , , . Then we apply the Shoelace Theorem.
This means that or , but since , then the answer is
See also
2000 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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