 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Pentagon's constant (Posted on 2014-12-27) The sum of the lengths of the perpendiculars from a point inside a regular pentagon to the sides (or their extensions) is constant.

a. Prove it.
b. Derive a formula expressing it in terms of the circumradius.

 No Solution Yet Submitted by Ady TZIDON Rating: 4.0000 (1 votes) Comments: ( Back to comment list | You must be logged in to post comments.) Solution Comment 1 of 1
Let the point P be at distance d from the centre, O, and let the
angle between PO and one particular side, S1, have size A1
degrees, etc. Consider a triangle OPQ, with OP as hypotenuse
and PQ parallel to S1. It can be used as follows:

Perpendicular distance from P to S1 = p - d*sin(A1),
where p is the perpendicular distance from O to S1.

Adding the perpendicular distances to all 5 sides now gives:

Sum = 5p – d(sinA1 + sinA2 + sinA3 + sinA4 + sinA5)   (1)

If all 5 sides (of length s) are projected on to the straight line
through O and P, their sum (with consideration to direction)
will be zero, since the pentagon is closed.
Thus     s(sinA1 + sinA2 + sinA3 +sinA4 + sinA5) = 0
and (1) now gives:
Sum of perpendiculars    = 5p
= 5*r*cos 36 deg
= 5r(1 + sqrt(5))/4
where r is the circumradius.

 Posted by Harry on 2014-12-27 18:16:34 Please log in:

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