Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.
Click the figure below to see the complete problem 635.
Saturday, July 16, 2011
Problem 635: Semicircle, Diameter, Perpendicular, Inscribed Circle, Radius
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Draw FG perpendicular to CB
ReplyDeleteNote FD=r+x,GD=r–x and OF=R–x
FG^2=(r+x)^2-(r-x)^2 = 4rx
Next OG=OB-GB=R-(CB-CG)
=R-(2r-x)=(R + x)-2r
So 4rx=FG^2
=(R-x)^2-[(R+x)-2r]^2
= -4Rx-4r^2+4r(R+x)
Follows 4rR=4r^2+4Rx,
rR=r^2+Rx
x=r(R–r)/R
I found a nice addition to this problem!
ReplyDeleteWhat is the ratio R/r when OF is perpendicular to AB?
Solution:
Then OC = x
OC = 2r-R and x = r(R-r)/R
This gives 2rR - R² = rR - r²
R² - Rr - r² = 0
abc-formula: R = r * (1 + √5)/2 = r * phi
So R/r = phi where phi is the Golden Ratio !!!
To Henkie (problem 635): Your conclusion about the Golden Ratio PHI is great.
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