Limit of Perimeter/Area Ratio for PolygonsDate: 03/23/2000 at 19:56:02 From: Christy Walsh Subject: Perimeter minimization I need to find a formula for the perimeter of a regular polygon as a function of its area. It needs to be shown that as the number of sides of a polygon of given area increases, the perimeter of the figure decreases and also the limit of the perimeter is the circumference of the circle of that given area. Part of the argument is that if we take the limit as the number of sides, n, of a regular polygon increasing without bound, the polygon will become a circle. Date: 03/24/2000 at 08:26:40 From: Doctor Luis Subject: Re: Perimeter minimization Hi Christy, Obtaining the area of a regular polygon is not terribly hard. Essentially you have to note that you can obtain a regular polygon by stacking a bunch of identical triangles side to side. (Draw lines from the vertices of your polygon toward the center so you can see this.) If you can figure out the area of each of those triangles, you'll be able to obtain the area of the polygon by simply multiplying by n, the number of sides. Here's one of those triangles, where O is the center of the polygon, h is the height of the triangle, and B = 2*b is the length of the base. You can see that the area of the triangle is T = (1/2)*(2*b)*h = b*h O /|\ / | \ l / | \ l / h| \ / | \ /_____|_____\ M b b N Thus, the area of the polygon is: A = n*T = nbh and its perimeter is P = n*(2b) = 2nb Now, because the angle MON is 2pi/n (in radians), we have tan((1/2)2pi/n) = b/h or tan(pi/n) = b/h If we keep A fixed, we can express b in terms of n only by eliminating h, A = nb(b/tan(pi/n)) A*tan(pi/n) b^2 = ------------ n This allows us to express P in terms of n only. _____________ / A*tan(pi/n) P = 2n / ------------- \/ n ____________________________________ / sin(pi/n) 1 = / 4A * pi * ---------- * ------------ \/ pi/n cos(pi/n) You can see that in the limit as n -> +infinity, we'll have sin(pi/n) / (pi/n) -> 1 cos(pi/n) -> 1 so naturally we have 1/cos(pi/n) -> 1 Finally, we get that tan(pi/n)/(pi/n) -> 1 All this goes to show that P -> sqrt(4pi*A) as n -> +infinity Keep in mind that A = nbh = (pi*h^2)*tan(pi/n)/(pi/n) Since h -> R, the radius of the circle, you'll get A -> pi*R^2*1 = (pi)R^2 and so P -> sqrt(4pi*pi*R^2) = 2pi*R This just goes to show that you get the correct formulas for A and P for an "infinite-sided" polygon. Showing that P is decreasing is a bit trickier, as you have to show that the function (n/pi)tan(pi/n) is decreasing for n > 3. (Why 3?) If looking at a graph of (x/pi)tan(pi/x) is not convincing enough, I think the shortest way you can show this is by looking at the Taylor series for the tangent. You have something like: tan(x) = x + (1/3)x^3 + (2/15)x^5 + (17/315)x^7 + ... This is the Taylor series expansion about the point x = 0. The nice thing about the tangent series is that all the coefficients are positive, a fact which comes in handy when trying to prove inequalities. (You can prove they are all positive by examining the sign of the derivatives of the tangent at x = 0. Verifying it for a few cases should be enough to convince you, and you can then prove by induction.) Supposing that m > n, we get that pi/m < pi/n, so we get the following string of inequalities (1/3)(pi/m)^2 < (1/3)(pi/n)^2 (2/15)(pi/m)^4 < (2/15)(pi/n)^4 (17/315)(pi/m)^6 < (17/315)(pi/n)^6 ... ... Adding all these up, you'd get (1/3)(pi/m)^2 + (2/15)(pi/m)^4 + (17/315)(pi/m)^6 + ... < (1/3)(pi/n)^2 + (2/15)(pi/n)^4 + (17/315)(pi/n)^6 + ... or 1 + (1/3)(pi/m)^2 + (2/15)(pi/m)^4 + (17/315)(pi/m)^6 + ... < 1 + (1/3)(pi/n)^2 + (2/15)(pi/n)^4 + (17/315)(pi/n)^6 + ... or (m/pi)((pi/m)+(1/3)(pi/m)^3+(2/15)(pi/m)^5+(17/315)(pi/m)^7+...) < (n/pi)((pi/n)+(1/3)(pi/n)^3+(2/15)(pi/n)^5+(17/315)(pi/n)^7+...) You should now be able to recognize the tangent expansion on both sides of the inequality. (I make use of the fact that a series can be thought of as the limit of a sequence of partial sums, and also that if the each term of one converging sequence is less than the corresponding term of another converging sequence, then the limit of the former will also be less than the limit of the latter.) In other words, (m/pi)tan(pi/m) < (n/pi)tan(pi/n) Since this expansion certainly holds for 0 < pi/m < pi/n < pi/3, and since originally we said that m > n, it follows that the function (n/pi)tan(pi/n) is certainly decreasing for n > 3, as we suspected. Hence, if (n/pi)tan(pi/n) is decreasing, then certainly P = sqrt(4pi*A*(n/pi)tan(pi/n)) is decreasing. Although this last argument is somewhat advanced, I find it easier to understand than some of the other methods I used to prove that (n/pi)tan(pi/n) is decreasing for n > 3. Alternatively, you could also show that the derivative of f(x) = (x/pi)*tan(pi/x) is always negative for x > 3, although I find the algebra to be very tedious and less intuitive than the method I presented to you. Since we are restricting ourselves to integral values of x, you could also try to prove that f(x) is decreasing by induction, but that proof is just too long and the algebra is very involved. If you don't really care about being rigorous, you can always point to the graph and say that it is obvious that f(x) is decreasing for x > 3 :) I hope this helped! - Doctor Luis, The Math Forum http://mathforum.org/dr.math/ |
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