Difference between revisions of "1999 AIME Problems/Problem 2"
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== Solution 2 == | == 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>. | + | 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 22:24, 24 March 2011
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 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 | ||
