Difference between revisions of "1999 AIME Problems/Problem 2"

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== Solution ==
 
== Solution ==
{{solution}}
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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
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(needs to be finished)
  
 
== See also ==
 
== See also ==

Revision as of 18:41, 8 March 2007

Problem

Consider the parallelogram with vertices $\displaystyle (10,45),$ $\displaystyle (10,114),$ $\displaystyle (28,153),$ and $\displaystyle (28,84).$ A line through the origin cuts this figure into two congruent polygons. The slope of the line is $\displaystyle m/n,$ where $\displaystyle m_{}$ and $\displaystyle n_{}$ are relatively prime positive integers. Find $\displaystyle m+n.$

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 $\displaystyle\frac{y}{x}=\frac{z}{w}$. Then 10(153-x)=28(45+x) and (needs to be finished)

See also