2000 AIME I Problems/Problem 2
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 .
Solutions
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 notice pentagon fits into a rectangle of side lengths and , giving us two triangles, each with hypotenuse and . First, we can solve for the first triangle. Using the coordinates of and , we discover the side lengths are both , so the area of the triangle of hypotenuse is . Next, we can solve for the second triangle. Using the coordinates of and , we discover the side lengths are and , so the area of the triangle of hypotenuse is . Now, let’s subtract the area of these 2 triangles from the rectangle giving us . Next, we take note of the fact that and are both factors of 451, and since both and are positive integers, must be greater than , thus giving us two cases, where either or . After trying both, the only working pair of where both and are integers are and , thus meaning
~Aeioujyot
Solution 3
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 | ||
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