tag:blogger.com,1999:blog-6933544261975483399.post4930464109867663774..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Problem 272. Tangent Circles, the Cube of the Common external tangentAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger3125tag:blogger.com,1999:blog-6933544261975483399.post-32822354067574056722018-02-12T08:51:17.152-08:002018-02-12T08:51:17.152-08:00Let
AC =r
BC =R
Then c=2(r+R)
From problem 27...Let <br /><br />AC =r <br />BC =R<br /><br />Then c=2(r+R)<br /><br />From problem 277 we have <br /><br />x^2=4rR<br /><br />And from problem 278 we have <br /><br />a^2=4r^2R/(r+R)<br />b^2=4R^2r(r+R)<br /><br />Multiplying a^2 , b^2 and c^2 we have <br /><br />(abc)^2=(4rR)^3(r+R)^2/(r+R)^2<br />(abc)^2=(4rR)^3<br /><br />With x^2=4rR<br /><br />(abc)^2=x^6<br /><br />Taking the square root in the equation above gives the desired result.Arifhttps://www.blogger.com/profile/16774426807135540272noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-77276358121689379182016-06-08T11:34:47.501-07:002016-06-08T11:34:47.501-07:00Join FD and GE.
Let FC + CG = c = x(a/b+b/a)
=...Join FD and GE.<br />Let FC + CG = c = x(a/b+b/a) <br />=> abc = x(a2+b2)<br />=> abc = x.(x2)<br />Hence x3 = abc<br /><br />Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-67753652200229805392009-03-31T01:40:00.000-07:002009-03-31T01:40:00.000-07:00In problem 271, we proved that x^3 = c*FD*GE (base...In problem 271, we proved that x^3 = c*FD*GE (based upon problem 269).<BR/>Now it's easy to see that triangles FDC & CEG are similar; hence, FD/b =a/GE or FD*GE=a*b.<BR/>Thus, x^3 = c*a*b<BR/>Ajit: ajitathle@gmail.comAjithttps://www.blogger.com/profile/00611759721780927573noreply@blogger.com