Wednesday, January 2, 2013

Problem 842: Triangle, Medians, Centroid, Four Circumcenters, Perpendicular, Congruence, Similarity

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

Click the figure below to see the complete problem 842.

Online Geometry Problem 842: Triangle, Medians, Centroid, Four Circumcenters, Perpendicular, Congruence, Similarity

Posted by Antonio Gutierrez at 5:52 PM
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3 comments:

  1. Jacob HA (EMK2000)January 2, 2013 at 7:51 PM

    By Problem 841,
    HM=BE/2, HN=AD/2

    ∴ HO4×HM=HO2×HN
    ⇒ HO4×BE/2=HO2×AD/2
    ⇒ HO4×BE=HO2×AD

    ReplyDelete
  2. W FungJanuary 3, 2013 at 12:26 AM

    From problem 841, HM=BE/2 ; HN=AD/2 and HO4*HM = HO2*HN
    It simply leads to the result HO4*BE = HO2*AD

    ReplyDelete
  3. Peter TranOctober 16, 2016 at 9:07 PM

    Note that the result is true for any point G inside triangle ABC.
    Peter Tran

    ReplyDelete

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