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Geometry ProblemLevel: Mathematics Education, High School, Honors Geometry, College.Click the figure below to see the complete problem 635.
Draw FG perpendicular to CBNote FD=r+x,GD=r–x and OF=R–x FG^2=(r+x)^2-(r-x)^2 = 4rxNext OG=OB-GB=R-(CB-CG) =R-(2r-x)=(R + x)-2rSo 4rx=FG^2 =(R-x)^2-[(R+x)-2r]^2 = -4Rx-4r^2+4r(R+x)Follows 4rR=4r^2+4Rx, rR=r^2+Rxx=r(R–r)/R
I found a nice addition to this problem!What is the ratio R/r when OF is perpendicular to AB?Solution:Then OC = xOC = 2r-R and x = r(R-r)/RThis gives 2rR - R² = rR - r²R² - Rr - r² = 0abc-formula: R = r * (1 + √5)/2 = r * phiSo R/r = phi where phi is the Golden Ratio !!!
To Henkie (problem 635): Your conclusion about the Golden Ratio PHI is great.