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Feb00 TN357 

{\bf Polycircles}

Three people cracked this one about the same time, Andaleeb Ahmed, Woodhouse 
Sixth Form College, London, Sue Liu, Madras College, St Andrews and Alexander 
Maryanovsky, Shevah-Mofet School, Tel Aviv.

Here is Alexander's solution.

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: $r_1>0$, $r_2>0$, $r_3>0$.
$$\eqalign{
r_1 + r_2 &= a \cr
r_2 + r_3 &= b \cr
r_3 + r_1 &= c}.$$

The solution for this system is:
$$\eqalign{
  r_1 = {(a + c - b)\over 2} \cr
  r_2 = {(a + b - c)\over 2} \cr
  r_3 = {(b + c - a)\over 2}}$$

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: $a_1$, $a_2$, $a_3$,...$a_n$. 
Suppose circles can be drawn with centres at the vertices of the polygon such 
that the circles just touch each other. Let $r_1$, $r_2$, $r_3$,... be the 
radii.  Then:
$$\eqalign{
  r_1 + r_2 &= a_1 \cr
  r_2 + r_3 &= a_2 \cr
     ... \cr
  r_n + r_1 &= a_n.}$$

Let's try to figure out what $r_1$ equals, and all the other solutions
will of course be symmetrical.
$$\eqalign{
r_1 &= a_n - r_n = a_n - a_{n-1} + r_{n-1}\cr 
    &= a_n - a_{n-1} + a_{n-2} - r_{n-2} = ...\cr 
    &= a_n - a_{n-1} + a_{n-2} - ...+(-1)^{n-1}a_1 + (-1)^nr_1.}$$

If $n$ is even then $r_1 = a_n - a_{n-1} + a_{n-2} - ... - a_1 + r_1$ 
which means that the condition for the existence of solutions is:
$$a_n - a_{n-1} + a_{n-2} - ... - a_1 = 0.$$
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 $n$ is odd then what we get is 
$$r_1 = {1\over 2}(a_n - a_{n-1} + a_{n-2} - ... + a_1)$$

and $r_1$ is positive when $(a_n - a_{n-1} + a_{n-2} - ... + a_1) > 0.$

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.

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possible if the following condition is true for all the
vertices of the polygon: the sum of the lengths of every other side starting 
and ending with the sides meeting at the vertex is bigger than the sum of the 
remaining lengths of the sides.