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

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<math>(EP)(CE)=(BE)(ED)</math> and <math>2r-1=15.</math> Hence, <math>r=8.</math>  
 
<math>(EP)(CE)=(BE)(ED)</math> and <math>2r-1=15.</math> Hence, <math>r=8.</math>  
  
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== See Also ==
 
{{Old CanadaMO box|before=First question|num-a=2|year=1971}}
 
{{Old CanadaMO box|before=First question|num-a=2|year=1971}}
  
 
[[Category:Intermediate Geometry Problems]]
 
[[Category:Intermediate Geometry Problems]]

Revision as of 14:48, 12 September 2012

Problem

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

CanadianMO 1971-1.jpg

Solution

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

See Also

1971 Canadian MO (Problems)
Preceded by
First question
1 2 3 4 5 6 7 8 Followed by
Problem 2
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