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.

Online Geometry Problem 673: Internally tangent circles, Chord, Tangent, Metric Relations.
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3 comments:

  1. AjitSeptember 23, 2011 at 8:03 PM

    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

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  2. Peter TranSeptember 23, 2011 at 9:21 PM

    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

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  3. PravinSeptember 24, 2011 at 4:37 AM

    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.
    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."

    ReplyDelete

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