Geometry Problem

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

Click the figure below to see the complete problem 673.

## Friday, September 23, 2011

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

Labels:
chord,
circle,
metric relations,
tangent

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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)

ReplyDeleteAjit : ajitathle@gmail.com

http://img33.imageshack.us/img33/2368/problem673.png

ReplyDeleteConnect 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

If we denote BD by y,

ReplyDeleteby 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.

This answers

"If the bisector of angle A of

triangle ABC meets the opposite

side BC at D, find the radius

of the circle passing through A

and touching BC at D."