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

Labels:
circle,
common tangent,
cube,
diameter,
tangency point,
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.