Difference between revisions of "1999 AIME Problems/Problem 2"

Revision as of 10:05, 27 April 2023

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 if 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 (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 $(10,45)$ , 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 $\boxed{118}$ .

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 $69 \cdot 18 = 1242,$ so the area of each of the two trapezoids with congruent area is $621.$ Therefore, since the height is $18,$ the sum of the bases of each trapezoid must be $69.$

The points where the line in question intersects the long side of the parallelogram can be denoted as $(10, \frac{10m}{n})$ and $(28, \frac{28m}{n}),$ respectively. We see that $\frac{10m}{n} - 45 + \frac{28m}{n} - 84 = 69,$ so $\frac{38m}{n} = 198 \implies \frac{m}{n} = \frac{99}{19} \implies \boxed{118}.$

Solution by Ilikeapos

Solution 4 (centroid) (fastest way)

$(\Sigma x_i /4, \Sigma y_i /4)$ is the centroid, which generates $(19,99)$ , so the answer is $\boxed{118}$ . This is the fastest way because you do not need to find the opposite vertices.

Solution by maxamc

@@ Line 21: / Line 21: @@
 === Solution 4 (centroid) (fastest way) ===
-<math>(\Sigma x_i /4; \Sigma y_i /4)</math> is the centroid, which generates <math>(19,99)</math>, so the answer is <math>\boxed{118}</math> This is the fastest way because you do not need to find the opposite vertices.
+<math>(\Sigma x_i /4, \Sigma y_i /4)</math> is the centroid, which generates <math>(19,99)</math>, so the answer is <math>\boxed{118}</math>. This is the fastest way because you do not need to find the opposite vertices.
 Solution by maxamc

1999 AIME (Problems • Answer Key • Resources)
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