Area of the figure formed- the edges being chipped off forever

Question

Consider a random figure. Let's say a regular hexagon ($6$- sided polygon) of area $6$. Chip off the $6$ corners to get a dodecagon ($12$ sided polygon). And repreat this chipping process forever. Basically the number of sides will get twice the previous one. (Just for enumeration, let's clarify, the $n^\text{th}$ step yields a $3\cdot 2^n$ sided polygon). If this continues forever, we get a circle I guess. But how do we calculate its (circle's) area?

This gives rise to another question actually. What if it wasn't a hexagon but an eight sided polygon (aka octagon)? It would've resulted into a circle as well, right? So what would be the area?

Do we get an unique circle for polygons of different area? Or maybe the area is same at the beginning (for two different polygons) but changes when they result into a circle?

Answer 1

Short answer

For regular polygons, the area of the resulting circle is $$\boxed{A = \frac{P^2}{4} \frac{\pi}{n^2}\cot^2{\left(\frac{\pi}{n}\right)} }$$

where $P$ and $n$ are the perimeter and number of sides of the original regular polygon.

Furthermore, this chipping away process multiplies the original area and the perimeter of the figure by a factor of $(\pi/n) \cot{(\pi/n)}$. (This is a factor that starts at around 0.60 for triangles, and gets closer and closer to 1 for figures with more sides.) Because this factor depends on $n$, different-sided polygons with the same initial area will end up as circles of different area.

Chipping away reveals a circle

For simplicity, let's consider what happens with regular polygons. You begin with a regular polygon, and chip away at its sides, producing another regular polygon with twice as many sides. You repeat the process many times; in the limiting case you obtain a circle.

The dimensions of the circle are special: it is the largest circle that fits inside of the original polygon. It is the incircle of the original polygon.

Computing the area of the incircle

We still have to prove that the result of the chipping process is the incircle of the original polygon. But suppose we take this result as an assumption— what will be the area of the incircle?

We can find the area of the circle by finding its radius. We divide the regular $n$-sided polygon into $n$ triangles, one per side. The orange angle is, in general, $2\pi/n$. By trigonometry, we know that if the side length is $s$, the height of the triangle is $\frac{s}{2}\cot{(\pi/n)}$. This height is also equal the radius of the circle. If we like, we can express the radius in terms of the perimeter $P=n\cdot s$:

$$r = \frac{s}{2}\cot{(\pi/n)} = \frac{P}{2n}\cot{(\pi/n)}$$

The area of the circle (and the answer to this question) is therefore

$$\boxed{A_\infty = \pi r^2 = \frac{P^2}{4} \frac{\pi}{n^2}\cot^2{\left(\frac{\pi}{n}\right)} }$$

where $P$ and $n$ are the perimeter and number of sides of the original regular polygon. ■

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Interestingly, the original area of the regular polygon was:

$$A_0 \equiv \frac{P^2}{4} \cot(\pi/n) \times \frac{1}{n}$$

Hence the cutting process multiplied its area by a factor of $$\frac{\pi}{n}\cot{\left(\frac{\pi}{n}\right)}.$$

and in fact the perimeter of the figure was multiplied by exactly the same factor (!). See below for details.

Showing that the incircle is the limit

1. Let's formalize the chipping process by saying that you divide each edge into three partitions with two symmetrically-spaced marks. Then those marked divisions become the vertices of the new polygon. Because each vertex participates in two edges, this transforms a polygon with $n$ vertices into a polygon with $2n$ vertices.

2. If we choose the spacing of the marks wisely, we can ensure that this chipping process preserves regularity. This means that the process will transform an $n$-sided regular polygon with sides of length $s$ into a $2n$-sided regular polygon with sides of length $p\times s$—some factor $p$ shorter.

3. In the next step, we'll compute the factor $p$ which ensures regularity. Here, let's note that if the side length $s$ becomes $p\times s$, then the perimeter changes from $ns$ to $(2n)(ps) = (2p)(ns)$. In other words, the perimeter becomes multiplied by $2p$.

   

4. Using some triangle geometry, we find that this factor of $p$ actually depends on $n$, the number of sides: $$p(n) = \frac{\tan{(\pi/2n)}}{\tan{(\pi/n)}}$$

   (See below for details.)

5. Hence suppose we have a certain regular polygon. Initially, it has $n$ sides, and a perimeter of $P$. After cleaving it several times, our formula in step (3) tells us that it will have perimeter: $$P_k \equiv P\times 2p(n) \times 2p(2n) \times 2p(4n) \times 2p(8n) \times \ldots \times 2p(2^k n)$$

6. Our formula from step (4) allows us to write this in a form which simplifie: the perimeter after $k$ chipping iterations becomes

   $$P \times \frac{2 \tan{(\pi/2n)}}{\tan{(\pi/n)}}\times \frac{2 \tan{(\pi/4n)}}{\tan{(\pi/2n)}}\times \frac{2 \tan{(\pi/8n)}}{\tan{(\pi/4n)}}\times\ldots \times \frac{2 \tan{(\pi/2^kn)}}{\tan{(\pi/2^{k-1}n)}}$$ which cancels significantly: $$P_k \equiv P \times \frac{2^k \tan{(\pi/2^k n)}}{\tan{(\pi/n)}}$$

7. Miraculously, the limit of this process as $k\rightarrow \infty$ is simply: $$P_{\infty} = P\cdot \frac{\pi/n}{\tan{(\pi/n)}} = P\,\frac{\pi}{n}\cot{(\pi/n)}$$

   (Indeed, $\lim_{b\rightarrow \infty} b\tan{(a/b)} = a$ by l'Hospital's rule: $\lim_{b\rightarrow\infty}b\tan{(a/b)}=\lim_{t\rightarrow 0}\tan{(at)}/t = \lim_{t\rightarrow 0} a\sec^2(at)/1 = a\cdot 1/ 1 = a$).

   A circle with such a circumference would have a radius of $(P/2n)\cot{(\pi/n)}$ and a corresponding area of $$A = \pi\;\frac{P^2}{4n^2}\cot^2(\pi/n)$$

   This is exactly the dimensions of the incircle of the original polygon, and we are done.

Proof of the formula for p(n)

In the previous section, we made use of a formula for a value of $p$ which ensure that the polygon remained regular when cleaved. In this section, we derive this formula.

$$p(n) = \frac{\tan{(\pi/2n)}}{\tan{(\pi/n)}}$$

• Let's take the edge of a regular $n$-sided polygon and form a triangle with the center of the polygon as the third vertex. The angle at that vertex is $2\pi/n$ because the polygon is regular.

  

• When we subdivide the edge into three parts, the middle part will subtend angle $\alpha$ and the two outer parts will both subtend equal angles $\beta$. Based on this partitioning, we know that $$\alpha + 2\beta = 2\pi/n$$
• Call the length of the side $s$, and call the length of its middle partition $p\times s$. Our goal is to find $p$ such that the resulting $2n$ sided polygon has regular sides. Now, there are two types of new edge: the one between the two new vertices we've just made, and the one between new vertices in adjacent triangles. If we draw in the adjacent triangle, we find that the angle between a vertex in the first triangle and the vertex in the adjacent triangle is $2\beta$. By the side-angle-side theorem, it follows that the two types of new edge will have the same length if the angle they subtend is the same. In other words, if $\alpha$ (the angle subtended by the first kind of edge) is equal to $2\beta$ (the angle subtended by the second kind of edge). So $$\alpha = 2\beta$$
• From these two equations it follows that $$\alpha = \pi/n\qquad \beta = \pi/2n$$
• Next, let $s$ denote the length of an edge, and let $h$ denote the distance from the center of the polygon to the midpoint of one of its edges (this is called the inradius of the polygon.) With some right-triangle math, we find that: $$\begin{align*}\frac{p}{2}s &= h\tan{(\alpha/2)} = h\tan{(\pi/2n)}\\ \frac{s}{2} &= h\tan{(\alpha/2 + \beta)} = h\tan{(\pi/n)}\\ \end{align*}$$
• Dividing the first equation by the second, we get some cancellation of lengths, yielding: $$p = \frac{\tan{(\pi/2n)}}{\tan{(\pi/n)}}$$