Private Coast

The Problem

Setup: There is a cluster of n perfectly circular, flat islands in the Pacific, each with the same radius of R. A point on the coast of an island is private if it cannot be seen from the coast of any other island. (We're assuming they're forested islands so you can't see across the land of an island to the coast on the other side). It's easy to see that if there are 2 islands, then exactly half of the coast-line of each island is private, so the total private coast-line of both islands is 2πR.

Question: What's the maximum total private coast-line possible with 10 islands? How must they be arranged?

Experimentation:

Our first step, as usual, was to play about with it. Since the 2 island case is obvious, we tinkered with 3 islands, starting with the simplest two configurations we could think of: All in a line, and as the corners of an equilateral triangle.

Clearly when all 3 islands are arranged in a line, there is no private coast on the middle island, and half a circle's worth on the ends. So, again, the total private coast is a full circle's worth: 2πR.

What about the equilateral triangle? Here it looks like the inward facing parts of the islands are not private, but the outward facing parts are.

The way I like to imagine which parts are private and which aren't, is I imagine little people (the red dots) standing at different places with binoculars, trying to see everywhere they can. Something like this:

It looks like the "outside" red dots are able to see the furthest around to the back side. But even so, no one can see the bold shaded section...and that looks like about 1/3 of the circle. Since the whole arrangement is symmetric, the other two islands will also have the same amount of private coast. It's looking like ~1/3 + ~1/3 + ~1/3 = ~1 whole circle's worth of private coast-line again! Can't be too sure yet, but that's what it's starting to look like.

"Ok", we thought, but maybe those answers looked so nice because the shapes were so simple. What if we make a triangle that's not so nice.....

If I imagine taking the private parts in bold and putting them together, they kind of look like they'll make a full circle of private coastline! This led us to make a...

Conjecture: No matter how you arrange 3 islands, there will always be 1 full circle's worth of private coastline.

Kind of seemed true, but how could we prove it? Imagine drawing a triangle connecting the centers of the islands, and another that's just tangent to the edges. Actually, you don't need to: here it is!

Let's see. I want to know how much private coastline there is. In the picture, that's however much coast is in the arcs of b1 + b2 + b3.

What I know already is that a1 + a2 + a3 = 180, because they're the angles of a triangle.

So, I need a way to connect what I know with what I want to know. I need to find a relationship between the a's and the b's.

Looking at the upper left-hand circle, we know all the angles must sum to 360, because they're in a circle. Since two of them are 90 degree angles, that means that a1 + b1 = 180. Really, that's true of all 3 circles.

a1 + b1 = 180

a2 + b2 = 180

a3 + b3 = 180

Since we know (a1 + a2 + a3), and we want to know (b1 + b2 + b3), let's add the 3 equations above together to get...

a1 + b1 + a2 + b2 + a3 + b3 = 180 + 180 + 180

Re-grouping a bit, we get...

(a1 + a2 + a3) + (b1 + b2 + b3) = 540

And since (a1 + a2 + a3) = 180, we get...

180 + (b1 + b2 + b3) = 540 or b1 + b2 + b3 = 360. A full circle!

So that proves it. But that was only for 3 islands, and the original problem asks about 10.

Feeling optimistic we made another...

Conjecture: Any number of islands in any arrangement will always have exactly one full circle's worth of private coast-line.

The same argument as before extends easily to larger numbers of islands that are the vertices of some larger convex polygon. For example...

Since an n-gon is made up of n-2 triangles (here, the 5-gon is made of 3 triangles), all the internal angles must sum to 180(n-2). That gives us...

a1 + a2 + ... + an = 180(n-2)

And the same argument with the circle gives us for all a's and b's that a + b = 180.

This time let's re-write the equation as a = 180 - b, and substitute in for each of the a's in the first equation.

(180 - b1) + (180 - b2) + ... + (180 - bn) = 180n - 360

How many 180's on the left side? n. Re-ordering a bit, we get...

180n - (b1 + b2 + ... + bn) = 180n - 360

Cancel the 180n on both sides, multiply by -1 and you end with...

b1 + b2 + ... + bn = 360 The sum of all the private areas is a full circle's worth!

But wait. This only proves it for configurations where the islands form the verticies of convex polygons! What about something like this, where there are islands scattered around?

If an island lies inside the convex polygon created by the centers of the others (inside their "convex hull"), then it clearly has no private coast at all, so we can safely ignore it in our calculation. Now the problem is only about the islands that are the verticies of the polygon--but that's exactly the situation we've already proven for. So that, finally, does it! Any situation that isn't of the kind our proof was for easily reduces to another one that our proof covers.

Solution: Any number of radius R islands arranged in any arrangement at all will have exactly 2πR distance of private coast-line!