# 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, | + | 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

is a chord of a circle such that and Let be the center of the circle. Join and extend to cut the circle at Given find the radius of the circle

## Solution

First, extend to meet the circle at Let the radius be Applying Power of a Point, and Hence,