Monday, February 6, 2012

Problem 725: Kurschak's Dodecagon, Square, Equilateral Triangle, Midpoints, 30,60 Degrees

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

Click the figure below to see the problem 725 details.

Online Geometry Problem 725: Kurschak's Dodecagon, Square, Equilateral Triangle, Midpoints, 30,60 Degrees
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1 comment:

  1. c.t.e.o.February 7, 2012 at 10:47 AM

    1) HF equal and perpendicular to EG => EFGH square
    2) N2N8 meet HN7 at P
    Tr HN1N2 equilateral
    In right tr HN2P HN2 = HN1 = N1P & N1N2 = N1P
    => N1M3N8 rhombus => M3N8 = N1M3 = N1N2
    (M3N1 & M3N8 middle lines of GHP)

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