Copyright © University of Cambridge. All rights reserved.
Four people cracked this one: Bella and
Andaleeb from Woodhouse Sixth Form College, Sue from Madras College
and Alexander from Shevah-Mofet School.
Bella started her answer by trial and
error for different triangles
For the triangle of side c = 15, b = 13, a = 12:
I split the longest side (c = 15) into7 and 8, and draw 2 circles:
(A, 8), that is circle with center A and radius8, and (B, 7). These
2 circles cut the 2 sides at P and Q.So CP = 13 - 8 =5 and CQ = 12
- 7 = 5. So I drew the cirle(C, 5), which just touches the other 2
cirles
From the result above, I realize important rules to tackle the
problem:
First, the difference between the 2 radii on 1 side of the triangle
is equal to the difference of the 2 other sides. More specifically,
AP - PC = 8 - 5 = 3 and AB - CB = 15 - 12 = 3, which means AP - PC
= AB - CB. Similarlly, BQ - CQ = BA - BC = 2 and AM - MB = AC - BC
= 1.
Second, the larger radius is corresponding to the larger side. For
example, AB > CB. When we split AC into AP and CP, AP will be
greater than CP.
These rules imply that I 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.
I tried it out for another triangle of different size and it worked
perfectly. For the triangle of side c = 10, b = 8, a = 4:
Because AB - CB = 10 - 8 = 2. I split the remaining side AC such
that AP - PQ = 2. There is only the pair (1,3) which satisfy that
condition. So AP = 3 and CP = 1 since AB > CB. So 3 circles are
(A, 3); (B, 7); and (C, 1).
I 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:
AB = AM + BM
BC = BQ + CQ
BM = BQ
So AB - BC = AM - CQ (1)
Look at the remaining side AC:
AP = AM and CP = CQ (2)
From (1) and (2) we have:
AB - BC = AP - CP
As long as AB > BC, AP will be greater than CP
In words, this implies the rules (Q.E.D.)
Alexander sent us a 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
$\begin{eqnarray} \\ r_1 + r_2 &=& 12\\ r_2 + r_3
&=& 13\\ r_3 + r_1 &=& 15. \end{eqnarray}$
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 polycircles
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:
$r_1> 0$, $r_2> 0$, $r_3> 0$. \begin{eqnarray} \\ r_1 +
r_2 &=& a \\ r_2 + r_3 &=& b \\ r_3 + r_1
&=& c. \end{eqnarray} The solution for this system is:
\begin{eqnarray} \\ r_1 &=& \frac{(a + c - b)}{2} \\ r_2
&=& \frac{(a + b - c)}{2} \\ r_3 &=& \frac{(b + c -
a)}{2} \end{eqnarray} 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:
\begin{eqnarray} \\ r_1 + r_2 &=& a_1 \\ r_2 + r_3
&=& a_2 \\ ... \\ r_n + r_1 &=& a_n. \end{eqnarray}
Let's try to figure out what $r_1$ equals, and all the other
solutions will of course be symmetrical. \begin{eqnarray} \\ 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} = ...\\ &=& a_n - a_{n-1} +
a_{n-2} - ...+(-1)^{n-1}a_1 + (-1)^nr_1. \end{eqnarray} 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 $$r_1 = \frac{1}{2}(a_n - a_{n-1}
+ a_{n-2} - ... + a_1)$$ 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.