Saturday, October 19, 2013

Geometry Problem 933: Square, Center, Transversal, Perpendicular, Metric Relations, Congruence

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

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Online Geometry Problem 933: Square, Center, Transversal, Perpendicular, Metric Relations, Congruence

Posted by Antonio Gutierrez at 10:22 PM
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6 comments:

  1. AnonymousOctober 20, 2013 at 5:25 AM

    http://bleaug.free.fr/gogeometry/933.png

    bleaug

    ReplyDelete
  2. Μιχάλης ΝάννοςOctober 21, 2013 at 2:07 PM

    http://www.mathematica.gr/forum/viewtopic.php?f=22&t=37369&start=700#p188400

    ReplyDelete
  3. AjitOctober 21, 2013 at 10:27 PM

    What a wonderful proof!
    Thank you, Bleaug

    ReplyDelete
  4. AnonymousOctober 22, 2013 at 5:05 PM

    thanks for this kind remark...but...
    my post does not meet the standards of what can be called a 'proof'. I guess you filled in the gaps yourself. So 50% of the proof is yours ;-)

    bleaug

    ReplyDelete
  5. MANOLISOctober 28, 2013 at 12:29 PM

    Αν Α(0,0), Β(Ο,4),C(4,4),D(4,0) και (ε1)=EO: y-2=λ1(x-2) , (ε2)=FO:y-2=λ2(x-2) τότε

    ReplyDelete
  6. AnonymousNovember 28, 2013 at 8:35 AM

    <HCO=<BOG=a because <BOC=90. Both triangles HCO and GOB share hypotenuse and the angle a so they are congruent. Similarly OCM and DON are congruent. From those 2 pairs of congruent triangles we have that BG=HO and DN=OM, so the answer is 2(4/sqrt2)^2=16

    Ivan Bazarov

    ReplyDelete

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