Saturday, October 19, 2013

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

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

Click on the figure below to enlarge it.

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

6 comments:

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

    bleaug

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

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  3. What a wonderful proof!
    Thank you, Bleaug

    ReplyDelete
  4. 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

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  5. 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

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  6. <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