Difference between revisions of "2000 AIME I Problems/Problem 2"
(→Solution) |
(→Solution 1) |
||
(One intermediate revision by one other user not shown) | |||
Line 14: | Line 14: | ||
=== Solution 2 === | === Solution 2 === | ||
− | We find the coordinates like in the solution above: <math>A = (u,v)</math>, <math>B = (v,u)</math>, <math>C = (-v,u)</math>, <math>D = (-v,-u)</math>, <math>E = (v,-u)</math> | + | We find the coordinates like in the solution above: <math>A = (u,v)</math>, <math>B = (v,u)</math>, <math>C = (-v,u)</math>, <math>D = (-v,-u)</math>, <math>E = (v,-u)</math>. Then we apply the [[Shoelace Theorem]]. <cmath>A = \frac{1}{2}[(u^2 + vu + vu + vu + v^2) - (v^2 - uv - uv - uv -u^2)] = 451</cmath> <cmath>\frac{1}{2}(2u^2 + 6uv) = 451</cmath> <cmath>u(u + 3v) = 451</cmath> |
This means that <math>(u,v) = (11, 10)</math> or <math>(1,150)</math>, but since <math>v < u</math>, then the answer is <math>\boxed{021}</math> | This means that <math>(u,v) = (11, 10)</math> or <math>(1,150)</math>, but since <math>v < u</math>, then the answer is <math>\boxed{021}</math> |
Revision as of 20:14, 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.