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:
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.
Thuss(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.
Solution