tag:blogger.com,1999:blog-6933544261975483399.post734357030217960985..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Problem 635: Semicircle, Diameter, Perpendicular, Inscribed Circle, RadiusAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger3125tag:blogger.com,1999:blog-6933544261975483399.post-52125252437305245052011-07-20T16:14:26.430-07:002011-07-20T16:14:26.430-07:00To Henkie (problem 635): Your conclusion about the...To Henkie (problem 635): Your conclusion about the Golden Ratio PHI is great.Antonio Gutierrezhttps://www.blogger.com/profile/04521650748152459860noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-17606963219396321712011-07-20T09:12:43.596-07:002011-07-20T09:12:43.596-07:00I found a nice addition to this problem!
What is t...I found a nice addition to this problem!<br />What is the ratio R/r when OF is perpendicular to AB?<br /><br />Solution:<br />Then OC = x<br />OC = 2r-R and x = r(R-r)/R<br />This gives 2rR - R² = rR - r²<br />R² - Rr - r² = 0<br />abc-formula: R = r * (1 + √5)/2 = r * phi<br />So R/r = phi where phi is the Golden Ratio !!!Henkiehttps://www.blogger.com/profile/04279523252566471532noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-55412360783528926662011-07-17T10:46:46.992-07:002011-07-17T10:46:46.992-07:00Draw FG perpendicular to CB
Note FD=r+x,GD=r–x and...Draw FG perpendicular to CB<br />Note FD=r+x,GD=r–x and OF=R–x <br />FG^2=(r+x)^2-(r-x)^2 = 4rx<br />Next OG=OB-GB=R-(CB-CG) <br />=R-(2r-x)=(R + x)-2r<br />So 4rx=FG^2 <br />=(R-x)^2-[(R+x)-2r]^2 <br />= -4Rx-4r^2+4r(R+x)<br />Follows 4rR=4r^2+4Rx, <br />rR=r^2+Rx<br />x=r(R–r)/RPravinhttps://www.blogger.com/profile/05947303919973968861noreply@blogger.com