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1999 AIME Problems/Problem 2

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Problem

Consider the parallelogram with vertices (10,45), (10,114), (28,153), and (28,84). A line through the origin cuts this figure into two congruent polygons. The slope of the line is m/n, where m_{} and n_{} are relatively prime positive integers. Find m+n.

Solution 1

Let the first point on the line x=10 be (10,45+a) where a is the height above (10,45). Let the second point on the line x=28 be (28, 153-a). For two given points, the line will pass the origin iff the coordinates are proportional (such that \frac{y_1}{x_1} = \frac{y_2}{x_2}). Then, we can write that \frac{45 + a}{10} = \frac{153 - a}{28}. Solving for a yields that 1530 - 10a = 1260 + 28a, so a=\frac{270}{38}=\frac{135}{19}. The slope of the line (since it passes through the origin) is \frac{45 + \frac{135}{19}}{10} = \frac{99}{19}, and the solution is m + n = 118.

Solution 2

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 (10,45), and (28,153) gives (19,99), which is the center of the parallelogram. Thus the slope of the line must be \frac{99}{19}.

See also

1999 AIME (ProblemsResources)
Preceded by
Problem 1
Followed by
Problem 3
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
Want to learn how to tackle those tough AMC/AIME/Olympiad counting and probability problems? Check out Art of Problem Solving's Intermediate Counting & Probability by David Patrick.
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