Friday, April 10, 2009

Problem 27: Right triangle, Incenter, Parallel, Incircles, Inradius

Proposed Problem

Problem 27: Right triangle, Incenter, Parallel, Incircles, Inradius.

See also:Complete Problem 27, Collection of Geometry Problems

Level: High School, SAT Prep, College geometry

3 comments:

  1. Hints: Two tangent segments to a circle are congruent. Square...
    See also Problem 668

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  2. http://img98.imageshack.us/img98/3497/problem27.png
    Connect CO and AO
    Note that CO and CO2 are angle bisector of angle (BCA) so
    C, O, O2 are collinear
    Triangle OGC congruence with triangle OKC ( right triangles with common hypotenuse and CO is angle bisector)
    So OC is the angle bisector of angle (GOK) >>Incircle O2 tangent to OG also tangent to OK and O2N=r2
    Similarly AO is angle bisector of angle HOK) and incircle O1 tangent to both OH and OK and O1P=r1
    DE=DK+KE=O1P+O2N=r1+r2
    Peter Tran

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  3. Using Antonio's hint above....

    If X is the point of tangency of circle O with AC

    AK = c - r and KC = a - r

    Similarly AD = c-r-r1 and CE = a-r-r2

    So AC - AD - EC = r1+r2 + (b-c-a + 2r) from using the above expressions for AD and CE.

    The expression within the bracket = 0 and the result follows

    Sumith Peiris
    Moratuwa
    Sri Lanka

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