 # Polycircles

This comes from an article written by NRICH. Unfortunately their article contains several errors, which is why I have corrected and improved upon it here:

## Problem For any regular polygon it is always possible to draw circles with centres at the vertices of the polygon, and radii equal to half the length of the edges, to form a "polycircle" in which each circle just touches its neighbours.

Investigate and explain what happens in the case of non-regular polygons. Is it always possible to construct three circles with centres at the vertices of the triangle so that the circles just touch?

I recommend that you start by trying to find the radii for a specific triangle. For example, try finding the radii for the triangle having side lengths of 15, 13, and 12.

Now use the same method for a triangle having side lengths of $$a,$$ $$b,$$ and $$c.$$

Can you generalize your method to quadrilaterals? pentagons? $$n$$-gons?

## Solution

For the triangle of side $$c = 15,$$ $$b = 13,$$ and $$a = 12:$$ Split the longest side (c = 15) into lengths of 7 and 8, and draw 2 circles: the circle centered at $$A$$ with radius 8, and the circle centered at $$B$$ with radius 7. These 2 circles cut the 2 sides at $$P$$ and $$Q.$$ Thus, $$CP = 13 - 8 = 5,$$ and $$CQ = 12 - 7 = 5.$$ This allows us to draw a circle centered at $$C,$$ having radius 5, which just touches the other 2 circles. Notice that the difference between the 2 radii on 1 side of the triangle is equal to the difference of the 2 other sides. For example, $$AP - PC = 8 - 5 = 3,$$ and $$AB - CB = 15 - 12 = 3,$$ which means $$AP - PC = AB - CB.$$ Similarlly, $$BQ - CQ = BA - AC = 2,$$ and $$AM - MB = AC - BC = 1.$$

Also notice that when two sides are adjacent, the circle with the larger radius will sit on the vertex attached to the longer side. For example, $$AB > CB.$$ When we split $$AC$$ into $$AP$$ and $$CP,$$ $$AP$$ will be greater than $$CP.$$

These rules imply that we can always find the radii simply by splitting 1 side of the triangle such that the difference between the 2 numbers equals the difference between the 2 other sides and the larger number is corresponding to the larger side of the triangle.

Let's try applying what we've learned to a different triangle: one having sides of length 10, 8, and 4.

Because $$AB - CB = 10 - 8 = 2,$$ we split $$AC$$ such that $$AP - PC = 2.$$ But since we also know that $$AP + PC = 4,$$ we can solve the system to find $$AP = 3$$ and $$PC = 1.$$ So the circle centered at $$A$$ has radius 3, and the circle centered at $$C$$ has radius 1. And for the sake of completeness, $$B = 8 - 1 = 7.$$

We can also prove these rules. Assume that we can draw 3 circles which statisfy the conditions as shown by the graph.

Look at the 2 sides of the triangle $$AB$$ and $$BC:$$
\begin{align} & AB = AM + BM \\ & BC = BQ + CQ \\ & BM = BQ \end{align}

So $$AB - BC = AM - CQ\ (1)$$.

Look at the remaining side $$AC:$$

\begin{align} & AP = AM and CP = CQ (2) \end{align}

From (1) and (2) we have:

\begin{align} & AB - BC = AP - CP \end{align}

As long as $$AB \gt BC,$$ $$AP$$ will be greater than $$CP.$$

Now let's consider the more general solution:

Suppose it is possible for the triangle with sides 12, 13 and 15 and the radii of the circles are $$r_1,$$ $$r_2,$$ $$r_3.$$ Then

\begin{align} r_1 + r_2 = 12 \\ r_2 + r_3 = 13 \\ r_3 + r_1 = 15. \end{align}

This set of simultaneous equations is easily solved to give $$r_1 = 7,$$ $$r_2 = 5,$$ and $$r_3 = 8.$$ So we have proved that circles exist for this triangle.

Let's prove that it's possible for any triangle to construct circles with centres at the vertices so that the circles just touch. Let $$a,$$ $$b$$ and $$c$$ be the sides and $$r_1,$$ $$r_2,$$ $$r_3$$ the radii of the circles. The only constraints on the radii are: $$r_1 \gt 0,$$ $$r_2 \gt 0,$$ and $$r_3 \gt 0.$$

\begin{align} r_1 + r_2 = a \\ r_2 + r_3 = b \\ r_3 + r_1 = c. \end{align}

The solution for this system is:

\begin{align} r_1 = \dfrac{a + c - b}{2} \\ r_2 = \dfrac{a + b - c}{2} \\ r_3 = \dfrac{b + c - a}{2} \end{align}

and since the sum of the lengths of any 2 sides in a triangle is bigger than the 3rd side, there are solutions for any positive $$a,$$ $$b,$$ and $$c.$$

Now consider a convex polygon with $$n$$ sides: $$a_1,\,a_2,\,a_3, \ldots a_n.$$ Suppose circles can be drawn with centers at the vertices of the polygon such that the circles just touch each other. Let $$r_1,\,r_2,\,r_3,\,\ldots$$ be the radii. Then:

\begin{align} r_1 + r_2 = a_1 \\ r_2 + r_3 = a_2 \\ \ldots \\ r_n + r_1 = a_n. \end{align}

Let's try to figure out what $$r_1$$ equals, and all the other solutions will of course be symmetrical.

\begin{align} r_1 &= a_n - r_n = a_n - a_{n - 1} + r_{n - 1} \\ &= a_n - a_{n - 1} + a_{n - 2} - r_{n - 2} \\ &= \ldots \\ &= a_n - a_{n - 1} + a_{n - 2} - \ldots + (-1)^{n - 1}a_1 + (-1)^nr_1. \end{align}

If $$n$$ is even, then $$r_1 = a_n - a_{n - 1} + a_{n - 2} - \ldots - a_1 + r_1,$$ which means that the condition for existence of solutions is:

$$a_n - a_{n - 1} + a_{n - 2} - \ldots - a_1 = 0.$$

It also means that if this condition holds there are an infinite number of solutions and if it does not hold there are no solutions.

If, however, $$n$$ is odd, then we get

$$r_1 = \dfrac{1}{2}(a_n - a_{n - 1} + a_{n - 2} - \ldots + a_1)$$

and $$r_1$$ is positive when $$a_n - a_{n - 1} + a_{n - 2} - \ldots + a_1) \gt 0.$$

For polygons with an odd number of sides solutions always exist but 'negative' values of $$r_i$$ occur when, instead of touching externally, one circle surrounds its 'neighbour' which touches it internally so the length of the edge is given by the difference of the radii and not the sum of the radii.

## Going further

What about spheres centered at vertices of polyhedra? I imagine this would be tedious to do by hand. Maybe writing a program to accomplish this would be interesting. I imagine solving such problems would involve setting up a large number of simultaneous equations, then solving them by computer program. I wonder if any general formulas can be derived, such as the formulas for convex $$n$$-gons. Can those formulas, if they exist, be found by hand, or would a computer algebra system (CAS) be necessary to find them?