Difference between revisions of "1999 AIME Problems/Problem 2"
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== Solution == | == Solution == | ||
| − | {{ | + | Let the first point on the line x=10 be (10,45+x) where x is the height above (10,45). Let the second point (on the line x=28) be (28, 153-x). Since for a line to pass through the origin, and for two points (x,y) and (w,z) then <math>\displaystyle\frac{y}{x}=\frac{z}{w}</math>. Then 10(153-x)=28(45+x) and |
| + | (needs to be finished) | ||
== See also == | == See also == | ||
Revision as of 17:41, 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 x=10 be (10,45+x) where x is the height above (10,45). Let the second point (on the line x=28) be (28, 153-x). Since for a line to pass through the origin, and for two points (x,y) and (w,z) then 
. Then 10(153-x)=28(45+x) and
(needs to be finished)
