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


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

    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
    Peter Tran

  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
    Sri Lanka

  4. Define O1 and O2 the center of incirles with radius r1 and r2
    Points A, O1 and O are aligned on the angle bisector AD=AD’ and CE=CE’
    D’H=r1, E’G=r2 and HB=HG=r
    AB=AD’+r1+r and CB=CE’+r2+r
    Since r is the incircle of ΔABC right in B, then 2r=AB+BC-AC
    Therefore 2r=(AD+r1+r)+(CE+r2+r) – AC
    But AD+DE+EC=AC
    Therefore 2r=r1+r2+2r-DE
    => DE=r1+r2