Proposed Problem

See complete Problem 271 at:

gogeometry.com/problem/p271_tangent_circles_cube_common_external_tangent.htm

Level: High School, SAT Prep, College geometry

## Thursday, March 19, 2009

### Problem 271. Tangent Circles, the Cube of the Common external tangent

Labels:
circle,
common tangent,
cube,
tangent

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Extend FD & GE to meet in H. HFG is a rt. angled triangle where HC is perpendicular to FG. Also CH = DE = h both being diagonals of the rectangle CEHD. Now we've proven in Problem 269 that x^3 = CH^3 = h^3 = c*FD*GE.

ReplyDeleteHence, x^3 = c*a*b

Ajit: ajitathle@gmail.com

If you do not wish to refer to any earlier problem then with the construction as b4, a*FH=FC^2 & b*GH=CG^2 from where a*b=(FC*CG)^2/(FH*GH). But from Tr.FHG, FH*GH=HC*FG=x*c and FC*CG=x^2. Hence, a*b=(x^4)/(x*c) or x^3 =abc

ReplyDeleteAjit

Let

ReplyDeleteAC=r

BC =R

then c=2(r+R)

In problem 270 I fond the following equations for :

a^2=4r^3/(r+R)

b^2=4R^3/(r+R)

And in problem 277 we find

x^2=4rR

If we multiply a^2 , b^2 and c^2 together we get

(abc)^2=(4rR)^3(r+R)^2/(r+R)^2

(abc)^2=(4rR)^3

Where x^2=4rR

(abc)^2=x^6

Taking the square root of the equation above gives the desired result.