Difference between revisions of "1971 Canadian MO Problems/Problem 1"

 
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== Solution ==
 
== Solution ==
First, extend <math>\displaystyle CO</math> to meet the circle at <math>\displaystyle P.</math> Let the radius be <math>\displaystyle r.</math> Applying Power of a Point,
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First, extend <math>\displaystyle CO</math> to meet the circle at <math>\displaystyle P.</math> Let the radius be <math>\displaystyle r.</math> Applying [[Power of a Point]],
 
<math>\displaystyle (EP)(CE)=(BE)(ED)</math> and <math>\displaystyle 2r-1=15.</math> Hence, <math>\displaystyle r=8.</math>  
 
<math>\displaystyle (EP)(CE)=(BE)(ED)</math> and <math>\displaystyle 2r-1=15.</math> Hence, <math>\displaystyle r=8.</math>  
  

Revision as of 15:16, 26 July 2006

Problem

$\displaystyle DEB$ is a chord of a circle such that $\displaystyle DE=3$ and $\displaystyle EB=5 .$ Let $\displaystyle O$ be the center of the circle. Join $\displaystyle OE$ and extend $\displaystyle OE$ to cut the circle at $\displaystyle C.$ Given $\displaystyle EC=1,$ find the radius of the circle

CanadianMO 1971-1.jpg

Solution

First, extend $\displaystyle CO$ to meet the circle at $\displaystyle P.$ Let the radius be $\displaystyle r.$ Applying Power of a Point, $\displaystyle (EP)(CE)=(BE)(ED)$ and $\displaystyle 2r-1=15.$ Hence, $\displaystyle r=8.$


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