1999 AIME Problems/Problem 2

Problem

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

Solution

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 $\displaystyle 1530 - 10a = 1260 + 28a$ , 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 $m + n = 118$ .

1999 AIME (Problems • Answer Key • Resources)
Preceded by Problem 1	Followed by Problem 3
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1999 AIME Problems/Problem 2

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