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
Subscribe to:
Post Comments (Atom)
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.