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Regular N-gon (Posted on 2017-07-09) Difficulty: 4 of 5

  
n≥=3 lines lie in the same plane and are concurrent at point O.
If y = mx is the equation of a line passing through the origin O
(where m is the slope), then mk labels the line y = tan(k*180°/n)*x
(for k = 0 to n-1). Note: If n is even and k = n/2, then line mk is
perpendicular to line m0. Point P (distinct from O) is an arbitrary point
in the plane of the n lines. Fi is the foot of the perpendicular from
point P to line mi (for i = 0 to n-1).

Prove that F0F1...Fn-1 is a regular n-gon.
  

No Solution Yet Submitted by Bractals    
Rating: 4.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution Solution | Comment 1 of 2
For i = 0 to n-1:

Since /PFiO = 900, all points Fi lie on a circle with diameter OP.

Since /FiOFi+1 = 1800/n, all chords FiFi+1 subtend that same angle

at O, on the circumference, so the chords are all equal.

Thus F0F1……Fn-1 is a regular n-gon



  Posted by Harry on 2017-07-09 12:02:32
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