 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  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) Comments: ( Back to comment list | You must be logged in to post comments.) 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 Please log in:

 Search: Search body:
Forums (0)