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Nonagon (Posted on 2004-11-25) Difficulty: 4 of 5
Draw a regular nonagon. If we label the vertices of the polygon sequentially, starting from A, we can draw three line segments: AB, AC, and AE.

Show that AB + AC = AE.

See The Solution Submitted by SilverKnight    
Rating: 3.0000 (7 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution Solution | Comment 4 of 13 |
WOLOG let the circumradius of the nonagon be 1.

AE = AB + AC

2*sin(80) = 2*sin(20) + 2*sin(40)

4*sin(40)*cos(40) = 2*sin(20) + 4*sin(20)*cos(20)

8*sin(20)*cos(20)*cos(40) = 2*sin(20) + 4*sin(20)*cos(20)

4*cos(20)*cos(40) = 1 + 2*cos(20)

2*cos(20)*[2*cos(40) -1] = 1

2*cos(20)*[2*{2*cos(20)^2 -1} - 1] = 1

2*[4*cos(20)^3 - 3*cos(20)] = 1

2*cos(60) = 1

cos(60) = 1/2

QED


Edited on November 26, 2004, 5:05 pm
  Posted by Bractals on 2004-11-26 16:40:35

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