Difference between revisions of "1999 AIME Problems/Problem 2"
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=== Solution 1 === | === 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>. | ||
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+ | === 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 <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>. | ||
== See also == | == See also == |
Revision as of 23:10, 21 August 2014
Contents
[hide]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 |
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