Difference between revisions of "1999 AIME Problems/Problem 2"
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== Problem == | == Problem == | ||
| − | Consider the parallelogram with vertices <math>\displaystyle (10,45),</math> <math>\displaystyle (10,114),</math> <math>\displaystyle (28,153),</math> and <math>\displaystyle (28,84).</math> A line through the origin cuts this figure into two congruent | + | Consider the [[parallelogram]] with [[vertex|vertices]] <math>\displaystyle (10,45),</math> <math>\displaystyle (10,114),</math> <math>\displaystyle (28,153),</math> and <math>\displaystyle (28,84).</math> A [[line]] through the [[origin]] cuts this figure into two [[congruent]] [[polygon]]s. The [[slope]] of the line is <math>\displaystyle m/n,</math> where <math>\displaystyle m_{}</math> and <math>\displaystyle n_{}</math> are [[relatively prime]] [[positive]] [[integer]]s. Find <math>\displaystyle m+n.</math> |
== Solution == | == Solution == | ||
| − | Let the first point on the line x=10 be (10,45+ | + | 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>\displaystyle 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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== See also == | == See also == | ||
| − | + | {{AIME box|year=1999|num-b=1|num-a=3}} | |
| − | + | ||
| − | + | [[Category:Intermediate Geometry Problems]] | |
Revision as of 18:22, 8 March 2007
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
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 
.
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 | ||
