Thursday, September 22, 2016

Geometry Problem 1262: Isosceles Triangle, Altitude, Medians, 90 Degrees, Perpendicular

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

This entry contributed by Omid Motahed.

Click on the figure below to view more details of problem 1262.


Geometry Problem 1262: Isosceles Triangle, Altitude, Medians, 90 Degrees, Perpendicular

9 comments:

  1. Let the centroid be G.
    Since BG:GH=2:1, thus AH=GH=CH.
    Therefore H is the Orthocenter if AGC, with diameter AC.
    Consequently, ∠AGC=90°

    ReplyDelete
  2. Palai Renato

    BH/AC=1,5 we put AC=2 BH=3, BC= √10, CD= √10/2 , CD2= 2,5
    The cross EC with AD =K; CK=X ; X2+X2=4, X=√2
    ED=1 , KD=y ; y2+y2=1 y=1/√2
    CD2= 2+1/2 = 2,5 therefore EC and AD are perpendicular
    Reply

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  3. Let AD, CE meet at M and let AC = 2p

    Then BH = 3p and MH = p
    So AH = MH = CH = p

    Hence < AMH = < CMH = 45 and so AD is perpendicular to CE

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  4. Problem 1262
    BH is altitude and median then K is centroid of the triangle ABC with BH=3.KH. Then
    3.KH=3.AC/2 or KH=AC/2 (AH=HC).Therefore AD is perpendicular CE.
    APOSTOLIS MANOLOUDIS 4 HIGH SCHOOL KORYDALLOS PIRAEUS GREECE

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  5. Comment sent by Sumith Peiris:

    Dear Omid and Antonio

    Just realized...,

    The triangle need not be isoceles for the medians AD and CE to be perpendicular.

    It is sufficient that the median BH (as opposed to altitude) be 1.5 times AC

    Hope u will adjust the problem accordingly.

    Rgds

    Sumith

    ReplyDelete
    Replies
    1. Thanks Sumith
      See problem 1263 at:
      http://www.gogeometry.com/school-college/3/p1263-triangle-medians-90-degree-perpendicular.htm

      Delete
  6. @Sumith, thanks that makes the problem even more interesting

    ReplyDelete
  7. extend AC to AZ such that HB=HZ
    then LBZH=45.......so....LECH=45 etc.

    ReplyDelete