Difference between revisions of "1999 AIME Problems/Problem 2"
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Let the first point on the line <math>x=10</math> be <math>(10,45+a)</math> where a is the height above <math>(10,45)</math>. Let the second point on the line <math>x=28</math> be <math>(28, 153-a)</math>. For two given points, the line will pass the origin if the coordinates are [[proportion]]al (such that <math>\frac{y_1}{x_1} = \frac{y_2}{x_2}</math>). Then, we can write that <math>\frac{45 + a}{10} = \frac{153 - a}{28}</math>. Solving for <math>a</math> yields that <math>1530 - 10a = 1260 + 28a</math>, so <math>a=\frac{270}{38}=\frac{135}{19}</math>. The slope of the line (since it passes through the origin) is <math>\frac{45 + \frac{135}{19}}{10} = \frac{99}{19}</math>, and the solution is <math>m + n = \boxed{118}</math>. | Let the first point on the line <math>x=10</math> be <math>(10,45+a)</math> where a is the height above <math>(10,45)</math>. Let the second point on the line <math>x=28</math> be <math>(28, 153-a)</math>. For two given points, the line will pass the origin if the coordinates are [[proportion]]al (such that <math>\frac{y_1}{x_1} = \frac{y_2}{x_2}</math>). Then, we can write that <math>\frac{45 + a}{10} = \frac{153 - a}{28}</math>. Solving for <math>a</math> yields that <math>1530 - 10a = 1260 + 28a</math>, so <math>a=\frac{270}{38}=\frac{135}{19}</math>. The slope of the line (since it passes through the origin) is <math>\frac{45 + \frac{135}{19}}{10} = \frac{99}{19}</math>, and the solution is <math>m + n = \boxed{118}</math>. | ||
| − | === Solution 2 === | + | === 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 <math>(10,45)</math>, and <math>(28,153)</math> gives <math>(19,99)</math>, which is the center of the parallelogram. Thus the slope of the line must be <math>\frac{99}{19}</math>, and the solution is <math>\boxed{118}</math>. | 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 <math>(10,45)</math>, and <math>(28,153)</math> gives <math>(19,99)</math>, which is the center of the parallelogram. Thus the slope of the line must be <math>\frac{99}{19}</math>, and the solution is <math>\boxed{118}</math>. | ||
Revision as of 21:58, 6 April 2020
Contents
Problem
Consider the parallelogram with vertices 
, 
, 
, and 
. A line through the origin cuts this figure into two congruent polygons. The slope of the line is 
where 
and 
are relatively prime positive integers. Find 
.
Solution
Solution 1
Let the first point on the line 
be 
where a is the height above . Let the second point on the line 
be 
. For two given points, the line will pass the origin if the coordinates are proportional (such that 
). Then, we can write that 
. Solving for 
yields that 
, so 
. The slope of the line (since it passes through the origin) is 
, and the solution is 
.
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 , and gives 
, which is the center of the parallelogram. Thus the slope of the line must be 
, and the solution is 
.
Solution 3 (Area)
Note that the area of the parallelogram is equivalent to 
so the area of each of the two trapezoids with congruent area is 
Therefore, since the height is 
the sum of the bases of each trapezoid must be 
The points where the line in question intersects the long side of the parallelogram can be denoted as 
and 
respectively. We see that 
so 
Solution by Ilikeapos
See also
| 1999 AIME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 1 |
Followed by Problem 3 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
