DyingLoveGrape.

Circles Drawn Around Polygons: The basics and a bit more.

What is this?

Recently, I've been looking through some Math Olympiad problems just for kicks, and I've found that certain tricks come up over and over again. My favorite kinds of tricks, though, are the geometric ones: they're just visually satisfying! There's a few questions which required the reader to look at polygons circumscribed inside a circle — because these are particularly nice, I'll focus on them in this post. We'll start with the absolute basics for those who haven't seen this stuff in a while: the circumference and area of a circle. Along the way, we'll talk a bit about triangles. After that, we'll discuss some neat things one can do inside circles.

Prereqs.

This post requires only basic knowledge of high school algebra and high school geometry, but nothing too crazy. We'll derive most of what we need.

What is a Circle?

This might seem a bit obvious: a circle is a round thing. Well, maybe that's not specific enough: a circle is all of the points which are of some distance (called a radius) away from some center point. If you just had a pen, a piece of paper, and a ruler you could draw a circle by drawing the center point and then, after picking some radius, drawing all possible points which are that distance from the center.

What is a Circle's Circumference?

Two questions come up frequently when it comes to circles. First: suppose you want to make a circular fence made of chickenwire around your precious tomatoes so that wolves don't mess them up: how much chickenwire should you buy? In other, more formal words, what's the distance around the outside of the circle? This distance, by the by, is called the circumference of the circle.

If you've taken a geometry class before, the answer to the previous question should be obvious once you have the radius of the circle. But let's pretend for a moment that we don't know that there is a formula for circumference. We'll just have to be clever.

Let's do our best. Let's say we just have a ruler (so we can only measure straight lines). We could inscribe some figures inside our circle and measure their circumference and approximate the circle by measuring the outside of that figure. For example, let's suppose the radius of our circle is equal to 1. Let's inscribe a square:

We find that the edges of our square will be $\sqrt{2}$ by using some basic trigonometry. Hence, the perimeter of the square is $4\sqrt{2}\approx 5.65685$. But note that this doesn't closely model the circle: there's a lot of space where the square doesn't "touch" at the circle. Let's try another figure which looks more "circular"; one with more sides, perhaps. Let's try a pentagon:

Note that this figure is quite nice: some trigonometry gives us that each of the side lengths is approximately equal to $1.176$; hence, the perimeter of this figure is equal to 5.88. This is relatively close to what we got before, but a little larger. Let's try again and make our figure a bit closer. Maybe we ought to try a ten sided figure now, just to hurry things up.

Note that each side length is equal to approximately $0.618$, which gives the total perimeter as $6.18$; this is closer to the circle, so our circle circumference must be pretty close to $6.18$.

In general, note what we've done: we've made a triangle with two sides equal to $1$ (the radius of the circle) and found the other side in some mysterious way, then simply multiplied that side length by the number of sides there are. Because I don't have a ruler, the way that I've been finding the last side is by using a formula called the law of cosines which is useful enough for me to put it in a box below:

Using the law of cosines, we can talk about this a bit more formally. We know that two sides of our triangle will be of length $1$; we'll let these be our $a$ and $b$. The formula then gives us: \[c^2 = 2 - 2\cos(\theta)\] so we need only find $\theta$. Luckily, this isn't so hard: there are $360^{\circ}$ in a circle, and we're cutting it up into some number of equal parts with our triangle. Let's say we inscribe an $n$-sided (regular) figure, as above; then this gives us that each triangle will have $\theta = \dfrac{360}{n}^{\circ}$. Hence, \[c^{2} = 2 - 2\cos(\tfrac{360}{n}^{\circ})\] or, in other words, \[c = \sqrt{2 - 2\cos(\tfrac{360}{n}^{\circ})}.\] Since we have $n$ of these sides, we have that the perimeter of our regular $n$-sided inscribed polygon will be: \[n\cdot \sqrt{2 - 2\cos(\tfrac{360}{n}^{\circ})}.\] This, theoretically, should approximate the circumference of our circle and get better and better as $n$ gets quite large.

[If you think that using $\cos$ is a bit cheap, you can find the formula for the side lengths of these particular figures using other ways; I do it this way since I wanted to introduce the law of cosines for later.]

Let's look at some values for the approximation we've just derived, just for fun. We'll call the perimeter for an $n$-sided inscribed polygon $P_{n}$. Then, \[\begin{align*} P_{4} &= 5.65685\dots\\ P_{10} &=6.18034\dots\\ P_{100} &=6.28215\\ P_{1000} &=6.28317\dots\\ P_{100000} &=6.28319\dots\\ P_{10000000} &=6.28327\dots\\ \end{align*}\] If we note that the actual circumference of the circle of radius 1 is equal to $6.283185307\dots$, we're not all that close off.

It was discovered that, when the radius increased, the arguments above still worked and, moreover, that the circumference of a circle of integer radius would be some multiple of the circumference of radius 1. Weird. Since, for whatever reason, it was thought that the diameter was more important than the radius (the diameter is any line going through the circle and stopping at two antipodal sides of the circle; its length is two times the radius), some people created a constant which was equal to the ratio between the circumference of the circle and its diameter; in other words, the perimeter of a circle of radius $\frac{1}{2}$. That constant was called $\pi \approx 3.14159265358979\dots$.

It was eventually found that, given some radius $r$ of a circle, the circumference of that circle is equal to $2\pi r$. Hence, our circle of radius $1$ above has circumference equal to $2\pi = 6.28\dots$.

What is a circle's area?

The other question one might ask is: if I enclose some amount of land with a circular fence, how much land is there? In other words: what is the area inside of a circle? Before calculus was created (so that all it took was to integrate an appropriate function) one way to find the area inside the circle was to approximate the cirlce using regular $n$-sided polygons, as we did above. This time we need to do a bit of work because...well, how do we find the area of a 10-sided polygon? Hm. The reader may want to consider this before looking below.

We'll work with a circle of radius 1 again, for simplicity. Let's start out with inscribing a regular $6$-sided figure, the hexagon. Let's try to break it up into pieces we know about: triangles.

We'll focus on finding the area of the blue triangle here. Well, what's the area of a triangle? We could derive it, but we'll just give it here: $A = \frac{1}{2}bh$ where $b$ is the "base" of the triangle and $h$ is the "height" of the triangle. The base is any of the sides, and the height goes from the base up to the highest point of the triangle but, it must be perpendicular to the base. We didn't draw the height yet, but our base will be the top part (the part on the boundary of the hexagon). Let's draw in the height now:

We found the length of the base in the previous section: it was \[b = \sqrt{2 - 2\cos(\tfrac{360}{n}^{\circ})},\] the length of one side of the inscribed $n$-sided regular polygon. The hard part, now, will be to find the height. Let's restrict our attention to only half of the blue triangle, as below:

If we focus on just this triangle, which is half of the triangle we'd like to find the area of, we have the following problem:

At this point, we can apply the Pythagorean Theorem, which says that given a triangle with one right angle (a $90^{\circ}$ angle; in this case, in the bottom left) then let $A$ and $B$ be the lengths of the two sides touching the right angle and $C$ be the lenght of the side on the opposite of the right angle; then \[A^{2} + B^{2} = C^{2}.\] In this case, $C = 1$, and we can make $A = \frac{b}{2}$. The last side $B$ is the side we're looking for (marked with an $x$ in the picture). Then we have: \[\left(\frac{b}{2}\right)^2 + x^{2} = 1\] or, in other words, \[x = \sqrt{1 - \left(\frac{b}{2}\right)^{2}}.\] Whew. Okay. So. The area of the little blue triangle is equal to \[\frac{1}{2}\cdot\frac{b}{2}\cdot(\sqrt{1 - \left(\frac{b}{2}\right)^{2}}).\] This looks fairly complicated, but what's nice is that we have made no real reference to the number of sides here; this formula is surprisingly general. To find the area of the entire figure, we simply multiply this by 2 (to get the bigger blue triangle) and then multiply this by the number of sides the figure has (since there will be, in total, that many big triangles that we can cut the figure up into).

Hence, the general formula for the area of an $n$-sided regular figure is given by \[n\cdot\frac{b}{2}\cdot(\sqrt{1 - \left(\frac{b}{2}\right)^{2}})\] where $b$ is the length of a single side of the $n$-sided polygon. This is the usual form that the area is given in; the complicated expression on the right-hand side is generally called the length of the apothem: this is the distance form the center of the polygon to the middle of one of the sides.

Because we're awesome, we've already derived a formula for $b$ in terms of the number of sides of the polygon inside of a radius 1 circle in the last section. For an $n$-sided polygon in the radius 1 circle, recall that the length of each side was \[b = \sqrt{2 - 2\cos(\tfrac{360}{n}^{\circ})}\] which gives us the incredibly complicated-looking formula \[n\cdot\frac{\sqrt{2 - 2\cos(\tfrac{360}{n}^{\circ})}}{2}\cdot\sqrt{1 - \frac{1 - \cos(\tfrac{360}{n}^{\circ})}{2}}\] which, despite looking incredibly complicated, should be computable (even though this might take a while...); for convenience, I'll call the area of the regular $n$-sided polygon inscribed in the radius $1$ circle by the name $A_{n}$. Let's list some values here: \[\begin{align*} A_{5}&= 2.37764\dots\\ A_{10}&= 2.93893\dots\\ A_{100}&=3.13953\dots\\ A_{1000} &=3.14157\dots\\ A_{100000} &=3.14159\dots\\ \end{align*}\] Note that this converges toward the constant $\pi$. That's fairly neat. Eventually, it was found that increasing the radius of the circle to $r$ will increase the area by a factor of $r^{2}$; that is, for a circle with radius $r$, the area is given by \[A = \pi r^{2}.\] Note that by doing the same process as we did above for circles of varying radii (plural of radius) one could "guess" this formula; we won't show this to save some time, but the reader may want to note where we used the radius in this formula (for example, we used it in the law of cosines, ...) and try to extend to a general formula.

[For those of you who know calculus, recall that the equation of a semi-circle with center $(0,0)$ and radius $r$ is given by $y = \sqrt{r^{2}-x^2}$. One way to find the area of a circle of radius $r$ would be to double this integral: note that $\int_{-r}^{r}\sqrt{r^{2} - x^{2}}\,dx$ can be done by trigonometric substitution; it would be equivalent to $r^{2}\int_{0}^{\pi}\sin^{2}(\theta)\,d\theta$. In turn (using a power reducing formula, for example), this is equivalent to $\frac{1}{2}\left|\theta - \sin(\theta)\cos(\theta)\right|_{-\pi}^{\pi}$. Evaluating this gives us $\frac{\pi r^{2}}{2}$. Recall that this was the area for a semicircle, so to get the area of the circle we simply multiply this by two: $\pi r^{2}$ as we'd expect. This should illustrate some of the power of calculus: we've essentially boiled down a whole bunch of work into a few introductory calculus substitutions.]

Arcs and Radians.

Suppose that you're sitting on a beach around a small disk of water. Your friend calls you up and asks where you are. How do you explain to him where you are on the circle of beach?

If your friend knows you're on the circle, you can do a few things. Here's one way: choose a starting point (for example, the east-most point of the circle) and then tell him how many degrees around (in a counter-clockwise, say) the circle you are.

At this point, we could say something like, "I'm $145^{\circ}$ from the east-most point of the circle, counter-clockwise." The system that we're using in this case is the system of degrees. Degrees are a system of angle measurement where a full circle is considered to be $360^{\circ}$. But why? What's so nice about $360$? There's nothing intuitively obvious about the cirlce that would make $360$ stand out; indeed, there are other systems which replace $360$ with $400$; it seems like we could replace this with any number and it would make just as much sense.

On the other hand, we just talked about circumference and noted that the circumference of a circle is $2\pi r$. We could say, then, that we're some fraction of the circumference around the circle (again, let's assume we're looking in a counter-clockwise direction, just for simplicity). For example, if we go a quarter of the way around the circle, we'd be at $\frac{2\pi r}{4} = \frac{\pi r}{2}$. If we went halfway around the circle, we'd be at $\frac{2\pi r}{2} = \pi r$. This pesky $r$ term is here, but notice that we're in the same "place" on the circle regardless of the size of the circle:

Because of this, we can just get rid of the $r$ (or, more reasonably, set $r = 1$). Then we notice the following: going around the entire circle is $2\pi$. Going around a third of the circle is $\frac{1}{3}\times 2\pi = \frac{2\pi}{3}$. Going around two thids of the circle is $\frac{2}{3}\times 2\pi = \frac{4\pi}{3}$. Going around, in general, an $m$-th of the circle, is $\frac{2\pi}{m}$.

This system of measurement of angles is called the radian system, perhaps because if you travel $1$ radian around a circle you will have gone a distance of $r$, the radius of the circle.

Each of the three red pieces (the two radius pieces, and the arc between them) have a length $r$. Usually, though, when we talk about radians we will have them in terms of $\pi$; it's much more likely to see $\frac{3\pi}{5}$ or $\frac{11\pi}{7}$ than it is to see $40$ or $90$ when it comes to radians. Because you need a bit of time to get used to radians, most of what we do will be in degrees below; in future posts, we will switch to radians.

If you're not ready to give up degrees, you're in luck: it's relatively easy to convert back and forth between radians and degrees. Since a full circle is $360^{\circ}$ or $2\pi$ radians, we have that \[360^{\circ} = 2\pi\mbox{ rad}\] or, \[1^{\circ} = \frac{\pi}{180^{\circ}}\mbox{ rad}\] so, for example, $50^{\circ}$ is equal to $50\times \frac{\pi}{180}$ radians, which is $\frac{5\pi}{18}$ radians. You can similarly convert radians to degrees, but we won't do much of this.

One neat part about using radians is that it's super-easy to calculate arc-length using radians. If you have a circle of radius $r$ and you go around $\frac{3\pi}{4}$ radians, then the length of that arc (the distance traveled on the circle) is equal to $\frac{3\pi}{4}r$. Easy-peasy.

The Central Angle Theorem.

One of the more useful theorems when it comes to circles and angles is the Central Angle Theorem; we'll prove a slightly weaker version, but the idea is the same. First, we need some definitions.

A line that connects two points on the circle is called a secant line. Secant lines don't have to go through the center like the diameter does; in the above picture, there are two different secant lines which happen to intersect at the bottom of the circle: when two secant lines connect at a single point on the circle, like in the above picture, we'll call the angle that they make an inscribed angle.

On the other hand, you draw two radii from the center of the circle to the outside of the circle, like this:

then we call that a central angle. Notice that there are two angles that we make when we do this: a larger angle and a smaller angle, as shown below:

We will always take the central angle to be the smaller angle. We also call the red shaded-in part (which looks a bit like an slice of pizza) a sector of the circle.

Notice that, given two points on a circle, you can make an associated central and inscribed angle, like so:

The interesting thing is that, regardless of where you put the blue dot at the bottom to make the inscribed angle, so long as it's not in the sector associated to the central angle (in this case, the two red dots at the top), the central angle will be two times the inscribed angle. This is the Central Angle Theorem. Formally,

Theorem (Central Angle Theorem). Let $A,B$ be two points on a circle and $C$ be the center of the circle. Let $D$ be a point on the circle which is not in the sector associated with the central angle. Then $\angle ADB = 2\times(\angle ACB)$; that is, the central angle is two times the inscribed angle.

As an example, we can look at the following picture:

We've denoted the angles using degrees here for clarity, but we could very well have written them in radians.

"But wait!" You might say, "I can draw that internal angle thing a few different ways, like this:"

"Which one is the one the theorem above is talking about?" As we noted above (but it bears repeating), it doesn't matter which you draw since they'll each have half the degree measurement of the central angle! This is pretty amazing. It helps deal with problems like this:

One neat thing that comes from this (albeit from an extremely special case!) is the fact that a quadralateral inscribed in a circle has the property that diagonal angles add up to $180^{\circ}$, like this:

The reason for this follows from this idea:

The idea here is to use the central angle theorem (and, in fact, here we have to use the more powerful version where the central angle can be the "bigger angle"); we obtain, by the (stronger) central angle theorem that \[2\alpha + 2\beta = 2(\alpha + \beta) = 360^{\circ}\] since the angles at the center sum up to $360^{\circ}$. That is, \[\alpha + \beta = \frac{360^{\circ}}{2} = 180^{\circ}\] as required. Neat. Formally,

Proposition. A quadralateral inscribed in a circle has the property that the sum of diagonal angles is $180^{\circ}$.

You now have an alternate method of doing Problem 1. We sometimes use this proposition to "break up" figures in a circle so that we may work with easier figures that we know things about. For example,

Next Time...

Next time, we'll discuss some other geometric ideas about circles and things we can do with them. Inscribed figures (ones inside a circle) are neat, but there's also circumscribed figures (ones outside a circle) and even (gasp!) interactions between two or more circles. Too exciting.

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