The good news is that it is possible to express the ratio in terms of the number of sides for any regular polygon. The bad news is that the formula looks quite ugly and complicated. All the simple formulas like E=mc

^{2}are already taken!

The secret to working out the ratio starts with drawing a diagonal that connects two alternate vertices of the exterior polygon. By alternate vertices, I mean two vertices that are separated by one intermediate vertex. In the figure to the left, consider segments AB and BC to be part of the exterior polygon (I have not drawn the rest of the polygon because I want to derive a formula that applies to polygons with any number of sides. So, my figure only shows two sides of a polygon with an arbitrary number of sides). They are equal to each other because the polygon is regular. Let us assume they are each 1 unit in length.

Segment AC is a diagonal that connects two alternate vertices of the exterior polygon (B is the intermediate vertex that is skipped over by the diagonal). ABC is now an isosceles triangle. Angle ABC is an interior angle of the polygon, and it is a function of the number of sides the polygon has. In particular, we know that angle ABC is equal to (n-2)*180/n degrees, where n is the number of sides of the regular polygon. In the figure, I have represented this angle by x.

We also know that angles BCA and BAC are equal, and because they are part of triangle ABC, we know that each is equal to (180–x)/2. Since x = (n-2)*180/n degrees, we can substitute that in the expression above, and get the measure of angles BAC and BCA to be 360/2n degrees. In the figure, I have represented this angle by y.

Surprisingly, that is all the information we need to get the ratio we are after. The first step in calculating the ratio is the calculation of the length of segment AC. Why? Because, we know from my previous post that each side of the midpoint polygon is half the length of the diagonal that connects alternate vertices of the exterior polygon.

So, how do we go about calculating the length of segment AC? We apply the law of sines to triangle ABC to calculate the length of segment AC. We know that angle x is opposite to the segment AC whose length we want, and angle y is opposite the segment AB whose length is 1. Thus, 1/sin(x) = AC/sin(y). Therefore, AC = sin(y)/sin(x).

At this point, we can express x and y in terms of n, and we will have AC in terms of n. Divide the length of AC by 2, and that is the ratio we are after (the length of one of the sides of the midpoint polygon). Doing these substitutions, we have:

AC = sin(360/2n)/sin((n-2)*180/n)

Therefore, the ratio of the perimeter of the midpoint polygon to the perimeter of the regular exterior polygon (which is half of AC) would be:

R = sin(360/2n)/(2*sin((n-2)*180/n))

Well, it may not be pretty, or roll off the tongue quite like E=mc

^{2}, but it is a function of n, the number of sides of the polygon, and that is what we started out seeking. So, pretty or not, we have achieved what we wanted to achieve, and anyway, half the fun of getting there is the traveling itself, right? So what if the destination was not stunning, the route was plenty scenic, right?!

Now, how do I know this is correct? Below is a table with the value of R calculated for regular polygons with different number of sides. The first 3 lines correspond to polygons that we dealt with in the previous post (squares, pentagons and hexagons). I had to start with squares because triangles do not have diagonals (unless you consider each side to be a diagonal also). As you can see, the value of the ratio as calculated using the above formula is identical to the value of the ratio which we derived using other methods in the previous post.

Number of Sides (n) | Interior Angle ABC (x) (in degrees) | Angle with Diagonal (y) (in degrees) | Length of Diagonal AC | Ratio of Perimeters (R) |

4 | 90 | 45 | 1.414213562 | 0.707106781 |

5 | 108 | 36 | 1.618033989 | 0.809016994 |

6 | 120 | 30 | 1.732050808 | 0.866025404 |

7 | 128.5714286 | 25.71428571 | 1.801937736 | 0.900968868 |

8 | 135 | 22.5 | 1.847759065 | 0.923879533 |

9 | 140 | 20 | 1.879385242 | 0.939692621 |

10 | 144 | 18 | 1.902113033 | 0.951056516 |

100 | 176.4 | 1.8 | 1.999013121 | 0.99950656 |

1,000 | 179.64 | 0.18 | 1.99999013 | 0.999995065 |

10,000 | 179.964 | 0.018 | 1.999999901 | 0.999999951 |

1,000,000 | 179.99964 | 0.00018 | 2 | 1 |

It is interesting to note that, as predicted, the ratio does approach 1 as the number of sides goes up. I have added rows for 100, 1000, 10000 and 1000000 sides to the table above just to illustrate how the ratio gets closer to 1 as the number of sides goes up. There are so many 9's in the answers for the polygon with a million sides that I decided to do away with them, and put down the answer rounded to a billionth!

By the way, a polygon with 100 sides is called a hectogon. A polygon with 1,000 sides is called a chiliagon, one with 10,000 sides is called a myriagon, and one with a million (1,000,000) sides is called a megagon!

Since we know that the ratio of the area of an inscribed midpoint polygon to the area of the exterior regular polygon is the square of the ratio of perimeters, you can just take the numbers from the table in this post and square them to get the ratios for the areas without having to explicitly calculate the areas of either the exterior polygons or the inscribed midpoint polygons.

There exist formulas for the area of a regular polygon based on the number of sides, and the length of each side, so if you are curious, you can calculate the area of the exterior polygon using such a formula, calculate the area of the midpoint polygon using the ratio in the table to compute the length of each side, and then calculate the ratio of the areas. The amazing thing about mathematics is that you can derive general results (such as: the ratio of areas is the square of the ratio of the perimeters), and after that you don't have to do all the hard work involved in calculating specific results. But the tools exist to do so if you want to.

Well, I consider this another example of using very simple, well-known concepts, and stringing them together in just the right order to derive something a little more interesting and complicated. It is what makes and keeps mathematics so interesting and challenging. And hopefully, it will keep our brains young and fresh! Good luck in your mathematical explorations!!

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