Saturday, August 24, 2013

Problem 914: Triangle, Orthocenter, Angle Bisector, Parallel Lines, Midpoint

Geometry Problem
Post your solution in the comment box below

Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the complete problem 914.

Online Geometry Problem 914: Triangle, Orthocenter, Angle Bisector, Parallel Lines, Midpoint.

4 comments:

  1. See sketch for location of points O, L, N
    Quadrilateral ANHL is cyclic with circumcenter O
    Since AD is an angle bisector of ∠ (BAC) => ∠ (NAG)=∠(GAL)
    And ∠ (NOG)= ∠ (GOL)= 2. ∠ (NAG)=2. ∠ (GAL) => OG is a perpendicular bisector of LN => MN=ML
    Let M’ is midpoint of BC.
    In right triangle BLC we have M’L=M’B
    In right triangle BNC we have M’N=M’B
    So M’L=M’N => M’ is on perpendicular bisector of LN.
    Both M and M’ on both BC and OG extension => M coincide with M’
    Result of this problem show that
    1. Perpendicular bisector of NL where N and L are the feet of altitudes of triangle ABC will pass through the midpoint of a side of triangle ABC or
    2. Properties of 9 points circle of triangle ABC: perpendicular bisector of NL, where N and L are the feet of altitudes of triangle ABC will pass through another 9 points.

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  2. Can you show the sketch, Peter? Thanks.

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  3. Sean N y L los puntos medios de los arcos BC y BAC, respectivamente, del circuncirculo de ABC. Sean J y K las intersecciones de HG y HF con NL, respectivamente.
    Sabemos que las bisectrices AD y AE pasan por N y L, respectivamente. Sea 2a=AH, es conocido el lema AH=2OM, entonces
    OM=a,
    LJ=NK=2a, (1)
    OL=ON=R (circunradio)
    MN=R-a,
    OJ=R-2a,
    MJ=OJ+OM=R-a,
    vemos que MN=MJ (2).
    Por (1) y (2) M es punto medio de JN y LK que son respectivamente las hipotenusas de los triangulos rectangulos semejantes (por catetos paralelos) JNG y LKF, por lo tanto GM y FM estan sobre una misma recta.

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    Replies
    1. See below for English translation of QueloP's solution.
      In triangle ABC, if we have AH=2.OM then M will be midpoint of BC. No further proof is required. how do you get AH=2OM ? please explain .

      Let N and L be the midpoints of the arcs BC and BAC, respectively, of the circumcircle of ABC. Let J and K be the intersections of HG and HF with NL, respectively.
      We know that the bisectors AD and AE pass through N and L, respectively. Let 2a = AH, the lemma AH = 2OM is known, then
      OM = a,
      LJ = NK = 2a, (1)
      OL = ON = R (circumrad)
      MN = R-a,
      OJ = R-2a,
      MJ = OJ + OM = R-a,
      We see that MN = MJ (2).
      By (1) and (2) M is the midpoint of JN and LK which are respectively the hypotenuses of triangles similar rectangles (by parallel hinges) JNG and LKF, therefore GM and FM are on the same line.

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