I’m heading out this morning to go camping for the weekend (Woohoo! Bug bites!) and I won’t have any way to access the Internet reliably, so if you enter this contest please be patient. I probably won’t be able to declare a winner or post a solution until Monday evening.

I’m told the place we’ll be camping is in a wireless dead zone, which inspired this question. Note that I’m pretty sure this isn’t really how cell carriers set up their networks, but it’s a fun problem. Because it’s a bit more involved than normal, I’m going to say this challenge is worth TWO Math Guides—one for you and one for a friend. They’ll both get shipped to you, and then if you don’t have any friends you can sell it on eBay or something. I dunno. I’m sure you have friends.

Annnnd here we go!

A wireless phone company’s cell towers are arranged in regular hexagonal structure, as shown in the figure above. If each tower’s coverage radius is exactly half the length of a side of the regular hexagon, what percentage of the hexagon’s area will not be covered by the towers?

Put your answers in the comments! The usual rules apply: you must not be anonymous to win, and although you may win if you’re not in the US, you will have to help me out with shipping costs via PayPal.

Good luck, and have a great weekend.

**UPDATE:** Nice work, Rex T95! Two books are on their way to you! Solution below the cut.

So you’ve got a couple of tricky things going on here, but this question is not NEARLY as gnarly as it looks at first.

We’re going to treat it like a shaded region question (because that’s what it *is*). First, we’ll find the area of the hexagon. Then, we’ll find the areas that *are* covered in the hexagon by the towers. Do a little subtraction, and voila! We’ll have our answer.

**Step 1: Find the area of the hexagon**

There are a lot of ways to do this, but the way I find most intuitive is to remember that all regular polygons are made up of congruent isosceles triangles that meet in the center, like so:

Since the center is 360º, you can figure out the measures of all the center-touching angles by dividing 360º by the number of triangles (which, of course, is the same as the number of sides):

360º ÷ 6 = 60º

So in a regular hexagon, the central angles of the isosceles triangles are all 60º, which means these triangles are not only isosceles, but *equilateral*. See how I did nice to you? You’re welcome.

So we need to find the area of an equilateral triangle, and multiply it by 6 to get the area of our hexagon.

**Step 1a: Find the area of an equilateral triangle**

At this point, I’m going to make an executive decision and plug in 2 for the length of the sides of my hexagon. We could do this without plugging in, of course, but it’d be more of a pain and I don’t subscribe to the school of thought that math should be difficult if it doesn’t have to be. So we’re dealing with an equilateral triangle with sides of length 2:

Surely, you know by heart that the area of a triangle is (½)*bh*. When we drop an altitude to find the height of an equilateral triangle, we create 2 30º-60º-90º triangles, so we don’t even need to Pythagorize since we know our special rights so well.

The base of the above triangle is 2, and its height is √3.

*A*_{△} = (½)2√3 = √3

*A*_{hexagon} = 6×*A*_{△}

*A*_{hexagon} = 6√3

**Step 2: Find the area that is covered by the towers**

This is actually a bit easier. We’ve already decided to make the sides of the hexagon equal 2, so the radii of all the circles in the diagram are 1, and therefore **each has an area of π**.

Furthermore, the 6 circles that are only partially included in the hexagon are all sliced the exact same way: in 120º chunks.

So let’s do a little part-whole ratio calculating to figure out the sector areas.

Of course, there are 6 of those sectors. So the total area covered by the cell towers is 2π (for the 6 sectors) + π (for the center circle) = **3π**.

**Step 3: Find the uncovered area**

This is the easy part! The area that’s not covered is simply *A*_{hexagon} – *A _{covered}*.

*A _{uncovered}* = 6√3 – 3π

**Step 4: Find the percentage of the hexagon that’s uncovered**

This last step, where we convert to a percentage, is why it doesn’t matter that I chose to plug in a value instead of working through this whole problem a bit more algebraically.

Throw that in your calculator, and you’ll get 9.31%.

Phew!

## Comments (7)

9.31% of the hexagon’s area will be left uncovered by the towers.

YES! Nice work. I’ll post a solution soon.

32% not covered?

14.3% I guessed 😉

9.31% is correct

9.31% too! Darn.. I have to start learning to check the blog and not my email for these :/

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