AIME 1999 · 第 2 题

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 by drawing.

Solution by maxamc