1999 AIME Problems/Problem 2
Contents
Problem
Consider the parallelogram with vertices , , , and . A line through the origin cuts this figure into two congruent polygons. The slope of the line is where and are relatively prime positive integers. Find .
Solution
Solution 1
Let the first point on the line be where a is the height above . Let the second point on the line be . For two given points, the line will pass the origin if the coordinates are proportional (such that ). Then, we can write that . Solving for yields that , so . The slope of the line (since it passes through the origin) is , and the solution is .
Solution 2 (the best solution)
You can clearly see that a line that cuts a parallelogram into two congruent pieces must go through the center of the parallelogram. Taking the midpoint of , and gives , which is the center of the parallelogram. Thus the slope of the line must be , and the solution is .
Edit by ngourise: if u dont think this is the best solution then smh
Solution 3 (Area)
Note that the area of the parallelogram is equivalent to so the area of each of the two trapezoids with congruent area is Therefore, since the height is the sum of the bases of each trapezoid must be
The points where the line in question intersects the long side of the parallelogram can be denoted as and respectively. We see that so
Solution by Ilikeapos
See also
1999 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
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