Now let the corners be rounded; each "corner" is an arc of a circle of radius r. If r grows to 1/[2*tan(pi/N)], the figure becomes a complete circle, with an area/perimeter ratio of 1/[4*tan(pi/N)].
Find r to maximize the area/perimeter ratio.
(It might be easier to start with small values of N, and see if a pattern emerges.)
Source: Extension of previous puzzle.
Phillipe Fondanaiche sent:
Area A=N/(4xtan(pi/N)) - Nxr^2xtan(pi/N) + pixr^2 Perimeter P=N - 2xNxrxtan(pi/N) + 2xpixr A/P maximized for r = (N-sqr(Nxpixtan(pi/N)) / 2x(Nxtan(pi/N) - pi)Volkov Igor sent the following:
Let's divide a regular polygon with N sides into N isosceles triangles (each triangle has one vertex into the center of polygon and two others into two consecutive vertices of contour). Consider one of that N triangles. If we draw an arc of a circle of radius r then in the point where side and arc intersect, the tangent to arc coincides with the tangent to side of triangle. So in this point the radius is perpendicular to the side of triangle. We get a triangle with angles a=Pi/2, b=Pi/N, and c=Pi(1/2-1/N). Triangle area is equal to 0.5*r^2*tan(Pi/N). Area of sector under the arc is Pi*r^2/(2N). Finally A/P(r)=[1/(4*tan(Pi/N))-r^2*(tan(Pi/N)-Pi/N)]/[1-2*r*(tan(Pi/N)-Pi/N)] Denote Pi/N as b. After differentiating and the setting the result equal to zero we get r1=-0.5*(tan(b)-sqrt(tan(b)*b))/(tan(b)*(-tan(b)+b)); r2=-0.5*(tan(b)+sqrt(tan(b)*b))/(tan(b)*(-tan(b)+b)); If we get the second derivative on r we receive d^2[A/P]/dr^2 = 2*tan(b)*(-tan(b)+b)/sqrt(tan(b)*b)<0 if r=r1 d^2[A/P]/dr^2 =-2*tan(b)*(-tan(b)+b)/sqrt(tan(b)*b)>0 if r=r2 So the correct answer is r=-0.5*(tan(Pi/N)-sqrt(tan(Pi/N)*Pi/N))/(tan(Pi/N)*(-tan(Pi/N)+Pi/N))Nick Baxter sent:
r = .5 / ( tan(pi/n) + sqrt(pi*tan(pi/n)/n) ) Area and perimeter are functions of r. Find r such that P*dA = A*dP. The formulas are not practical for text email.