Difference between revisions of "1999 AIME Problems/Problem 2"
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Consider the [[parallelogram]] with [[vertex|vertices]] <math>(10,45)</math>, <math>(10,114)</math>, <math>(28,153)</math>, and <math>(28,84)</math>. A [[line]] through the [[origin]] cuts this figure into two [[congruent]] [[polygon]]s. The [[slope]] of the line is <math>m/n,</math> where <math>m_{}</math> and <math>n_{}</math> are [[relatively prime]] [[positive]] [[integer]]s. Find <math>m+n</math>. | Consider the [[parallelogram]] with [[vertex|vertices]] <math>(10,45)</math>, <math>(10,114)</math>, <math>(28,153)</math>, and <math>(28,84)</math>. A [[line]] through the [[origin]] cuts this figure into two [[congruent]] [[polygon]]s. The [[slope]] of the line is <math>m/n,</math> where <math>m_{}</math> and <math>n_{}</math> are [[relatively prime]] [[positive]] [[integer]]s. Find <math>m+n</math>. | ||
| − | == Solution 1 == | + | == Solution == |
| + | === Solution 1 === | ||
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 iff 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 = 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 iff 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 = 118</math>. | ||
Revision as of 23:09, 21 August 2014
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 iff 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
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 
.
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. 
