Saturday, December 11, 2010

Problem 547: Triangle, Transversal, Four Circumcircles, Concurrency

Geometry Problem
Click the figure below to see the complete problem 547 about Triangle, Transversal, Four Circumcircles, Concurrency.

 Problem 547: Triangle, Medians, Perpendicular, Measurement.

Level: High School, SAT Prep, College geometry



    Connect GD,GE,GC, GF
    G is the intersecting point of circle C1 and C2. We will prove that quadrilaterals GECF and ADGF are cyclic
    1 m(BGC)=180- m(A) ABGC cyclic
    m(BGE)=180-m(BDE) DBGE cyclic
    Replace m(BDE)=m(A)+m(EFC) we will get m(EGC)=m(EFC) so GECF is cyclic

    2 We have m(BGD)=m(BED)=m(CEF)=m(GCF)
    So m(BGC)=m(DGF)=180-m(A) so ADGF cyclic

    Peter Tran

  2. If G is intersection point of C1 and C4 then
    A + BGC = A + DGF or BGD = CGF ???
    Let see ▲DBG = ▲CGF
    GCF = DBG, see arcs on C1
    GDB = GFC see arcs on C4
    => BGD = CGF

  3. Denote <AFD by x
    Let circles C1 and C2 intersect at G
    Denote angles of the triangle ABC at the vertices A, B, C respectively by the same letters A, B, C.
    From the figure, we note that
    B,G,E,D (lying on C2) being concyclic,
    <BGD = <BED = <CEF = C – x,
    A,B,G,C (lying on C1)being concyclic,
    <AGB = <ACB = C,
    So <AGD = < AGB - <BGD = C – (C-x) = x = <AFD
    Thus G lies on the circle C4 (through A, D, F)
    (i.e) C1, C2, C4 pass through G
    Next <CGF = <AGF - < AGC = <ADF - <ABC
    = (Pi – A –x) – B = C – x = <ACE - <CFE = <CEF
    So G lies on the circle C3 (through C, E, F)
    Hence C1, C2, C3, C4 pass through G.

  4. Proove that the centers of C_1, C_2, C_3 and C_4 lies in the same cirle