Monday, May 19, 2008

Geometry Problem 92

See complete Problem 92
Similar Triangles, Circumcircles, Circumradii, Parallel. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.


  1. B, O1, O are collinear ( are on bisector of ang B )
    O1 BD = BDO1 ( DO1 = BO1 = R )
    DAO = ABO ( BO = AO = R )
    BDO1 = DAO => ( as corresponding ang, DE//AC )

    join O1 to O2 ( O1, E, O2 are collinear, on bisect E )
    O1DE = DEO1 = EO2F = FO2E ( DO1=O1E =R1,EO2=O2F=R2)
    BDO1 = O2FG (as alternate ang BDE=EFG, AB//FG )

    in the same way (O2FG = O3MG)

    Tg at B to BO1 is tg for BO (B,O1, O are collinear)
    circle O and circle O1 tg at B

  2. In solution proposed by c. t. e. o. I think that the sentence “B, O1, O are collinear (are on bisector of ang B)” is not true (am I wrong?). I suggest the solution above.
    By their constructions, ABC, DBE, FGE and MGC are similar, so ang(BAO) = ang(CDO1) =
    = ang(GFO2) = ang(GMO3). Thus R, R1, R2 and R3 are parallel.
    Let t1 be the straight line tangent to C0 at B. OB and t1 are perpendicular. From the similarity of ABC and DBE, ang(DBO1)=ang(ABO), so O, O1 and B are collinear and O1B and t1 are perpendicular too. Hence, t1 is tangent to O1 at B and C0 and C1 are tangent at B.
    The other cases are solved similarly.