Saturday, November 1, 2014

Geometry Problem 1054: Triangle, Altitude, Orthocenter, Circumcenter, Perpendicular, Midpoints

Geometry Problem. Post your solution in the comments box below.
Level: Mathematics Education, High School, Honors Geometry, College.

Click the diagram below to enlarge it.

Online Math: Geometry Problem 1054: Triangle, Altitude, Orthocenter, Circumcenter, Perpendicular, Midpoints.


  1. Obviously H is the orthocenter of ΔABC, and circle O is its nine-point circle.

    Let P be the circimcenter of ΔABC, then BH=2×PF.
    Now H,O,P are the Euler Line with HO=OP.
    Also clearly, H,G,F are the median of ΔAHC, with HG=GF.

    Thus, PF=2×OG, and hence BH=2×PF=4×OG.


    Note that circumcircle of triangle MNF is 9 points circle of triangle ABC
    Since N and F are midpoints of HC and AC => NF=1/2 AH=MH
    And NF//MH => HNFM is a parallelogram
    Since G is the midpoint of MN => H, G, F are collinear and GF=GH
    Since ∠(LKF)= 90 degree => FOL is a diameter of circle O => GO= ½ GL= ¼ HB

  3. Let BH intersect circle O (9-pt circle of ΔABC) at L.
    First observe Δs LMN, FMN have same circumcircle.
    L bisects BH. So LM // BA , so perp to HN. already BH perp MN.
    So H is the orthocenter of ΔLMN.
    Hence BH = 2 LH = 2(2 OG) = 4 OG..

  4. Clearly O is center of 9 pt circle of ABC. Let P be mid-point of BH and Q foot of BH on AC. P,Q lie on the same circle.
    Since m(PQF)=90=> P,O,F are collinear
    PM // AB and m(PMF)=90 => MF // HN and similarly NF//MH => HMFN is parallelogram and G its centroid => HGF is one of its diagonal
    Since G mid-point of HF and OG // PH => OG=PH/2=BH/4