Wednesday, May 21, 2014

Math Geometry Problem 1017: Square, Center, Midpoints, Circle, Concyclic Points, Cyclic Quadrilateral

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

Click the figure below to enlarge.

Online Math Geometry Problem 1017: Square, Center, Midpoints, Circle, Concyclic Points, Cyclic Quadrilateral.

6 comments:

  1. ∠ECD = ∠EFD = ∠EGD = ∠EHD = 90° because of square symmetry. Hence C,E,F,G,H,D are concyclic. ED is a diameter. So is CH by symmetry.

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    Replies
    1. Problem 1017
      To Anonymous Why angle EGD = 90 degree?

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  2. http://s2.postimg.org/r4m3ldq6x/pro_1017.png
    Draw points L, I, L as per attached sketch
    Note that triangles ILC congruent to triangle LFI… (case SAS)
    So IC=IF=IG=IE=ID=IH => so all these points are concyclic

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  3. If you draw a 4-by-4 grid, GE is a 1x3 move and GD is 3x-1 move, so ∠ECD=90°

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  4. ECDH es un rectángulo, por lo tanto es inscriptible en una circunferencia C1.
    EFGH es un trapecio isósceles, por lo tanto es inscriptible en un circunferencia C2.
    Como el rectángulo y el trapecio tienen el segmento EH en común C1=C2.
    Entonces C, D, E, F, G y H pertenecen a la misma circunferencia, son concíclicos.

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  5. To Anonymous Problem 1017
    Please, send your solution in Spanish to
    (Enviar su solucion en Espanol a:)
    Problema de Geometría 1017
    Thanks

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