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.

3 comments:

  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.

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  2. http://s25.postimg.org/gymbfhbjz/pro_1054.png

    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

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  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..

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