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|Regular N-gon (Posted on 2017-07-09)
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)
| Comment 1 of 2
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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