Geometry Problem

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

Click the figure below to see the problem 725 details.

## Monday, February 6, 2012

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

Labels:
dodecagon,
equilateral,
Kurschak,
regular polygon,
square,
triangle

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1) HF equal and perpendicular to EG => EFGH square

ReplyDelete2) 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)