Proposed Problem
See complete Problem 272 at:
gogeometry.com/problem/p272_tangent_circles_cube_common_external_tangent.htm
Level: High School, SAT Prep, College geometry
Sunday, March 22, 2009
Problem 272. Tangent Circles, the Cube of the Common external tangent
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In problem 271, we proved that x^3 = c*FD*GE (based upon problem 269).
ReplyDeleteNow it's easy to see that triangles FDC & CEG are similar; hence, FD/b =a/GE or FD*GE=a*b.
Thus, x^3 = c*a*b
Ajit: ajitathle@gmail.com
Join FD and GE.
ReplyDeleteLet FC + CG = c = x(a/b+b/a)
=> abc = x(a2+b2)
=> abc = x.(x2)
Hence x3 = abc
Let
ReplyDeleteAC =r
BC =R
Then c=2(r+R)
From problem 277 we have
x^2=4rR
And from problem 278 we have
a^2=4r^2R/(r+R)
b^2=4R^2r(r+R)
Multiplying a^2 , b^2 and c^2 we have
(abc)^2=(4rR)^3(r+R)^2/(r+R)^2
(abc)^2=(4rR)^3
With x^2=4rR
(abc)^2=x^6
Taking the square root in the equation above gives the desired result.