5.2 Inscribed Circles of Integer Polygons

Proposition 5.2.1.

The centre and radii of an inscribed circle is determined by the integer polygon, up to any affine transformation.

Proof.

This holds since integer distance is integer invariant. ∎

Theorem 5.2.2.

Let $a,b,m,n$ be integers with $\gcd(m,n)=1$ , and $m\pm 1\neq 0$ and $\alpha$ , $\beta$ , $\gamma$ be lines such that the intersection of $\alpha,\beta$ is $(0,0)$ , $\alpha,\gamma$ is $(a,0)$ and $\beta,\gamma$ is $(bm,bn)$ .

Then we have four inscribed circles of these lines as follows:

Centre $(x,y)$	Radius
$\left(\frac{ab(m+1)}{\gcd(a-bm,bn)+a+b(m+1)-bm},\par\frac{abn}{\gcd(a-bm,bn)+a% +b(m+1)-bm}\right)$	$\Bigg{\|}\frac{abn}{\gcd(a-bm,bn)+a+b(m+1)-bm}\Bigg{\|}$
$\left(\frac{ab(m-1)}{\gcd(a-bm,bn)+a+b(m-1)-bm},\par\frac{abn}{\gcd(a-bm,bn)+a% +b(m-1)-bm}\right)$	$\Bigg{\|}\frac{abn}{\gcd(a-bm,bn)+a+b(m-1)-bm}\Bigg{\|}$
$\left(\frac{-ab(m+1)}{\gcd(a-bm,bn)-a-b(m+1)+bm},\par\frac{-abn}{\gcd(a-bm,bn)% -a-b(m+1)+bm}\right)$	$\Bigg{\|}\frac{-abn}{\gcd(a-bm,bn)-a-b(m+1)+bm}\Bigg{\|}$
$\left(\frac{-ab(m-1)}{\gcd(a-bm,bn)-a-b(m-1)+bm},\par\frac{-abn}{\gcd(a-bm,bn)% -a-b(m-1)+bm}\right)$	$\Bigg{\|}\frac{-abn}{\gcd(a-bm,bn)-a-b(m-1)+bm}\Bigg{\|}$

Remark 5.2.3.

Consider triangle $ABC$ , then it is integer congruent to a triangle with vertices of the form $(0,0),(a,0)$ and $(bm,bn)$ when $a=\operatorname{l\ell}(AB)$ , $b=\operatorname{l\ell}(AC)$ , $m=\operatorname{lcos}(\angle ABC)$ , $n=\operatorname{lsin}(\angle ABC)$ .

Proof.

Let $p=(x,y)$ be the centre of the inscribed circle. Then we have integer distance from the point to each line as

r=\begin{Vmatrix}1&x\\ 0&y\end{Vmatrix}=\begin{Vmatrix}m&x\\ n&y\end{Vmatrix}=\begin{Vmatrix}\frac{a-bm}{\gcd(a-bm,bn)}&x-a\\ \frac{-bn}{\gcd(a-bm,bn)}&y\end{Vmatrix}.

Using $\operatorname{ld}(p,\beta)=\operatorname{ld}(p,\alpha)$ , we get $|y|=|my-nx|$ , hence we have $r=y$ and

y=\frac{nx}{m\pm 1}.

now consider the equality $\operatorname{ld}(p,\alpha)=\operatorname{ld}(p,\gamma)$ , we get

\gcd(a-mb,bn)|y|=|(a-bm)y-bna+bnx|.

From now, we let $y$ be expressed in terms of $x$ . As there are $2$ values of $y$ dependent on the sign of $m\pm 1$ , let $m\pm 1=c$ . We remove the absolute value signs by replacing them with $\pm$ (or $\mp$ where necessary). Then we get

\left(\frac{\gcd(a-bm,bn)\mp a\mp bc\pm bm}{c}\right)x=\mp ab.

Hence

x=\frac{\pm abc}{\gcd(a-bm,bn)\pm(a+bc-bm)}.

∎

Now let us consider some examples of triangles and their inscribed circles.

Example 5.2.4.

Let $a=6,b=1,m=3,n=5$ , so we consider the triangle with vertices $(0,0),(6,0)$ and $(3,5)$ . Then we have the centres of circumscribed circles to be:

x	y
3	15/4
3	15/2
4	5
2	5

Figure 6: The centres of circumscribed circles are blue.

Here, exactly one point, $(3,15/4)$ , is in the interior of the triangle.

Example 5.2.5.

In this example, let $a=b=m=n=1$ . Then our triangle has vertices $(0,0)$ , $(1,0)$ and $(1,1)$ . Here we can’t use the formula.

Consider the equalities

\begin{Vmatrix}1&x\\ 0&y\end{Vmatrix}=\begin{Vmatrix}1&x\\ 1&y\end{Vmatrix}=\begin{Vmatrix}0&x-1\\ -1&y\end{Vmatrix}

solving the determinants, we get

|y|=|y-x|=|x-1|.

Solving $|y|=|x-1|$ , we get $x=1\pm y$ .

Let $x=1+y$ . Then $|y|=|y-(y+1)|=1$ . Hence $y=\pm 1$ , and $x=1\pm 1$ . So this gives pairs

(2,1)\text{ and }(0,-1).

Let $x=1-y$ . Then $|y|=|y-(1-y)|=|2y-1|.$ Let $y=2y-1$ , then

(x,y)=(0,1).

Let $y=1-2y$ , then

(x,y)=\left(\frac{2}{3},\frac{1}{3}\right).

In this example, precisely one point, $(2/3,1/3)$ , is in the interior of the triangle.

In both of the examples, there was one inscribed circle in each quadrant formed by the triangle (those being enclosed by $\alpha,\beta$ and $\gamma$ (i.e. on the interior of the triangle), and being enclosed by two of the lines while not being on the interior).