## Friday, September 23, 2011

### Problem 673: Internally tangent circles, Chord, Tangent, Metric Relations

Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the complete problem 673.

1. Draw the common tangent at A and join F to E. By the alternate segment theorem, /_FEA=/_CBA, both being equal to the angle made by BA with the common tangent. Thus, EF//BC which means f/CF=e/BE or f/(CF+f)=e/(BE+e).Further, by the power point theorem,(a-x)^2=BE(BE+e) and x^2=CF(CF+f) so [(a-x)/x]^2=[BE(BE+e)]/[(CF+f)CF] or (a-x)/x = e/f or x = (a*f)/(e+f)
Ajit : ajitathle@gmail.com

2. http://img33.imageshack.us/img33/2368/problem673.png
Connect EF, O’D and AD ( see picture)
Note that circle O is the image of circle O’ in the dilation transformation center A
With dilation factor k= AO/AO’=AB/AE=AC/AF
And BC//EF , O’C perpen. To BC
AB is the angle bisector of (BAC)
We have DC/DB=AC/AB=AF/AE=f/e
Replace DC/DB=x/(a-x) in above expression we will get the result.
Peter Tran

3. If we denote BD by y,
by symmetry we have
y = a.e/(e + f) and hence (following Tran)
x/y = f/e = b/c = k = R/ρ
where ρ is the radius of circle O'
Follows ρ = cR/b.