1999 AIME Problems/Problem 2

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

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 $m + n = \boxed{118}$ .

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 $\frac{99}{19}$ , and the solution is $\boxed{118}$ .

1999 AIME (Problems • Answer Key • Resources)
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1999 AIME Problems/Problem 2

Contents

Problem

Solution

Solution 1

Solution 2

See also