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

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== Problem ==
 
== Problem ==
<math>\displaystyle DEB</math> is a chord of a circle such that <math>\displaystyle DE=3</math> and <math>\displaystyle EB=5 .</math> Let <math>\displaystyle O</math> be the center of the circle. Join <math>\displaystyle OE</math> and extend <math>\displaystyle OE</math> to cut the circle at <math>\displaystyle C.</math> Given <math>\displaystyle EC=1,</math> find the radius of the circle
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<math>DEB</math> is a chord of a circle such that <math>DE=3</math> and <math>EB=5 .</math> Let <math>O</math> be the center of the circle. Join <math>OE</math> and extend <math>OE</math> to cut the circle at <math>C.</math> Given <math>EC=1,</math> find the radius of the circle
  
 
[[Image:CanadianMO_1971-1.jpg]]
 
[[Image:CanadianMO_1971-1.jpg]]
  
 
== 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>CO</math> to meet the circle at <math>P.</math> Let the radius be <math>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>  
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<math>(EP)(CE)=(BE)(ED)</math> and <math>2r-1=15.</math> Hence, <math>r=8.</math>  
  
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{{Old CanadaMO box|before=First question|num-a=2|year=1971}}
* [[1971 Canadian MO Problems/Problem 2|Next Problem]]
 
* [[1971 Canadian MO Problems|Back to Exam]]
 
  
 
[[Category:Intermediate Geometry Problems]]
 
[[Category:Intermediate Geometry Problems]]

Revision as of 21:47, 17 November 2007

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, $r=8.$

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