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

Labels:
circle,
diameter,
inscribed,
perpendicular,
radius,
semicircle

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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.

ReplyDelete