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 [[ | + | 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 15:17, 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,