Wholly Whexagons!

Dear Dr. Math,
I've noticed that hexagons show up in a lot of different places. Now that I've started looking for them, I see them everywhere! What's the deal with hexagons?
Jules, Canton OH

Dear Jules,

I know I'm not supposed to play favorites with mathematical objects, but I have to confess, the hexagon is probably my favorite shape. (Sorry, rhombicuboctahedron.)

While I suspect that you may be experiencing a fair amount of confirmation bias, it's true that hexagons do make an appearance in a large variety of different contexts. (It's also possible that you have the "hexagon madness" and are seeing them when they're not actually there. You might want to get that checked out.)

Part of the reason hexagons are so ubiquitous is that they have so many useful properties, probably even more than "familiar" shapes like squares and trapezoids. Primarily, I'm referring to regular hexagons--hexagons with 6 equal sides--like this guy:


Probably these are the ones you're seeing, Jules. Next time, I'll talk a little about the properties of irregular hexagons and why you might expect to see those, too.

First of all, a regular hexagon has the property that its opposite sides are parallel to each other, making it an ideal shape for a nut or bolt, because it fits nicely into a wrench:



Squares, octagons, and some other n-gons [those with even n] have the same property, but with a hexagonal nut or bolt, you can grab it at a variety of different angles, which is useful if you're putting together Ikea furniture in a tiny Manhattan apartment, for example. Also, since the exterior angles of an n-gon add up to 360° and there are n of them, each one measures 360/n. Therefore, more sides aren't really so good, because as the number of sides gets larger, the sharpness of the corners decreases, allowing for a greater possibility of slippage. A hexagon seems to be a nice compromise between a 2-gon and an -gon for these purposes.

Another important property of regular hexagons (that I'm sure you're aware of if you've ever looked at the floor of a public bathroom) is that they tile the plane. In other words, by putting a bunch of identical hexagons together as tiles, we can cover an entire plane surface:



There are other tilings, of course, with squares or triangles, but this one has some very appealing aspects. (For one, it's made of hexagons!) It turns out that among all possible tilings of the plane of shapes with a fixed area, the hexagonal one has the smallest possible perimeter.

One way to think about this is that the perimeter:area ratio goes down as a shape gets closer to being a circle. So, if we're using n-gons to tile the plane, we want n to be as big as possible. On the other hand, we have to be able to glue them together so that at each point of intersection, the angles add up to 360°. By the same reasoning as before, we can show that the interior
angles on an n-gon are each 180 - 360/n, and since there have to be at least 3 of these angles meeting at each corner, the greatest this angle can be is 120°. In this case, , so 360/n = 60; therefore, n=6. Hexagon!

Now, why does all of that matter? Well, say you weren't cutting these tiles out of a piece of ceramic but instead were building up walls to section an area into a number of chambers. If the material in those walls was really expensive for you to produce, it would bee in your best interests to make the chambers in the shape of a regular hexagon:



I don't know how bees managed to figure this out and yet here we are still living in rectangular grids like chumps.

A slightly different, but related, property of the regular hexagonal tiling is that it shows up if you're trying to pack together some circles:



See the hexagons?

Once again, this way of packing circles has the property of being optimal, in the sense that it leaves the least amount of empty space between circles. In fact, using a little trigonometry, we can even work out the efficiency of this packing:



The triangle in the picture is equilateral with all sides equal to 2*r, where r is the radius of the circles we're packing. If we split one in half (where the two black circles intersect), we'll get a right triangle with hypoteneuse 2*r and one leg r. Therefore, by the Pythagorean Theorem, if h is the height, then . So , and therefore, . That means the area of the triangle is .

Inside each triangle are three pieces of a circle, which together make up half of a circle of radius r. Thus, the area of the circular pieces is . This means the ratio of circular area to total area of the triangle is , approximately 0.91. Since the whole plane is made up of these triangles, the proportion of circle-area to total-area is the same, meaning the circles take up about 91% of the space. Pretty efficient, and fun at parties, too!




To Be Continued...

-DrM

Trig or Treat

Dear Dr. Math,
This is a question I thought of while pondering the air intake of a wood stove. The air intake is a series of holes, covered or uncovered by a sliding metal plate with equal sized holes.
Imagine 2 circles with equal radius, R. Slide one circle over the other. Express, in terms of R, how far one circle has to occlude the other such that half of the area is covered.
Bob H., Ashland, OR

Dear Bob,

Here's a picture of the problem, if I understand it correctly:

For legal reasons, before we get to the solution, I feel I should warn all you readers out there: what follows may involve some high-school level trigonometry, which I understand many of you have intentionally purged from your brains to make room for Grey's Anatomy plots. Part of the reason I like this question so much is that it shows that these concepts may very well have some relevance (outside of the very important pursuit of measuring the heights of buildings using a sextant) despite your high school math teacher's best attempts to convince you otherwise. Those readers who are subject to trigonometry-induced seizures should turn back now.

OK, with that out of the way, let's blow up part of the picture and label some of the relevant objects. The goal is to get a handle on this shady part of town:



First, there's the radius, R, which we've assumed is the same for the two circles. Let's call the angle formed by the center and two points of intersection . Note: this has nothing to do with thetans (or does it?). Splitting this angle down the middle forms two right triangles with an angle of . According to the rules of trigonometry, the height of each triangle is and the width is , as I've labeled here:

Now, the strategy I'd like to employ to compute the area of that funny little almond-shaped region, which I'll call C, is to think of it as consisting of two pieces, each of which is the difference between a pie-slice of the circle, A, and a triangle, B. In pictures:



The reason this helps is that circles and triangles are shapes whose areas we know how to compute. "Funny little almond shapes," not so much.

The area of the circular slice is in proportion to the whole area of the circle as the angle is to the whole angle of a circle, 360° (a quarter of a circle takes up 90°, for example). So in terms of R and , that's and so .
Now, the area of the whole triangle is , which in this case is . So, . By a sneaky trick I learned in trigonometry class, I can rewrite this as .

If we throw all these things into the hopper, we get that the area of the almond-shaped piece, C, is , and so . I can feel some of you starting to panic out there, but just take a deep breath and try to relax. Put on some Enya or something--maybe that song she wrote about trigonometry.

What were we doing? Oh yeah, right; now we have a formula for computing the area of the overlapping part of the two circles, which only depends on the angle . The question was, When is this area equal to half the area of the circle?, so we need to solve for . Half the area of the circle is ; therefore, the equation we need to solve is:



which, after we divide through by and clean up a bit leaves us with:

.

It's interesting to pause here and note that R completely vanished from the equation. This means whatever configuration we come up with as an answer must have the same angle, independent of the radius.

OK. So, how do we solve this equation?

Actually... we don't.

The problem is that we have a and a , and the thing about those two is that they're like Sydney and Cristina on Grey's Anatomy; they just don't mix well. Unfortunately, there's no way to get any further with this equation using the rules of algebra. So, here's where we cheat and approximate the solution with a calculator (in my case, a TI-89). Maybe someday I'll tell you all about what goes on inside a calculator when it does these approximations, or about how we could use calculus to solve the problem if the shape were something else. But anyway, for now, the answer is .

Lastly, we should translate this answer into a more meaningful form, for example by figuring out what the distance is between the two centers of the circles. Using trigonometry one last time, we can write this distance as , which for the magic above is . The final picture, then, is:






Put that in your wood stove and smoke it.

-DrM