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,