5.1 Inscribed Circles of Integer Angles

Proposition 5.1.1.

Consider an angle $\angle BAC$ and let $O$ be a centre of an inscribed circle of $\angle BAC$ . Then

\operatorname{lsin}\angle OAB=\operatorname{lsin}\angle OAC.

Proof.

From Proposition 2.3.6, we have the equalities:

\operatorname{lsin}\angle OAB=\begin{vmatrix}OA&AB\end{vmatrix};\quad% \operatorname{lsin}\angle OAC=\begin{vmatrix}OA&AC\end{vmatrix}.

If $O$ is the centre of an inscirbed circle of $BAC$ , then we have the equality:

\operatorname{ld}(AB,O)=\operatorname{ld}(AC,O).

Recall the following:

\operatorname{ld}(AB,O)=\begin{vmatrix}OA&AB\end{vmatrix}=\operatorname{lsin}(% \angle OAB);

\operatorname{ld}(AC,O)=\begin{vmatrix}OA&AC\end{vmatrix}=\operatorname{lsin}(% \angle OAC).

Hence $\operatorname{lsin}\angle OAB=\operatorname{lsin}\angle OAC.$ ∎

Each center of an inscribed circle is on the same ray as vertex $A$ . This observation leads to a definition of the bisector of integer angles.

5.1.1 Bisectors of Integer Angles

Definition 5.1.2.

The bisector of the angle $\alpha$ is the ray that contains all centres of inscribed circles of the angle. We call the angles obtained from this bisection $\alpha/2$ .

Proposition 5.1.3.

Let $\alpha=\angle BAC=\operatorname{larctan}(\frac{n}{m})$ . Then $\alpha/2=\operatorname{larctan}(\frac{n}{m+1})$ .

Proof.

Through integer congruency, there exists a pair of coprime integers $m,n$ such that $\angle BAC$ is integer congruent to $B^{\prime}OC^{\prime}$ with $B^{\prime}=(1,0)$ and $C^{\prime}=(m,n)$ .

Let $(x,y)$ be a point on the bisector line. Then we have the equality

{\begin{vmatrix}1&x\\ 0&y\end{vmatrix}}={\begin{vmatrix}x&m\\ y&n\end{vmatrix}}.

Then we have $y=nx-my.$ Hence we have bisector line

y=\frac{n}{m+1}x.

Thus, we find that the point

(x,y)=(m+1,n)

is on the bisector line, hence

\frac{\alpha}{2}=\operatorname{larctan}\frac{n}{m+1}.

∎

Corollary 5.1.4.

We have

\operatorname{lsin}(\alpha/2)=\operatorname{lsin}\alpha\quad\hbox{and}\quad% \operatorname{lcos}(\alpha/2)=\operatorname{lcos}\alpha+1\mod\operatorname{% lsin}\alpha

Proof.

If $\operatorname{ltan}(\alpha)=m/n$ and $\operatorname{ltan}(\alpha/2)=(m+1)/n$ then (from Definition 2.3.10) $\operatorname{lsin}(\alpha)=n=\operatorname{lsin}(\alpha/2)$ and $\operatorname{lcos}(\alpha)=m$ and $\operatorname{lcos}(\alpha/2)=m+1\mod n$ . ∎

This formula has similarities with the trigonometric formula

\cos(\alpha/2)=\sqrt{(1+\cos\alpha)/2}

with some restrictions on $\alpha$ , while the formula for $\operatorname{lsin}(\alpha/2)$ is similar to formula for the half of the adjacent angle.

Remark 5.1.5.

Notice that the bisectors are defined by the angles between the sides of the unit rhombi and its diagonal. These are classified by

(0,0),(1,0),(m,n),(m,n+1)

according to the computations of the above proposition.

Remark 5.1.6.

Iteratively bisecting angles, notice that for some integer $n$ we have the equality $\alpha/2^{n}=\operatorname{larctan}(1)$

Let look at some examples of $\alpha/2$ for different angles $\alpha$ .

Example 5.1.7.

Let $\alpha$ have an LLS sequence $(a_{0},a_{1})$ . Then we have two cases for $\alpha/2$ :

•

If $a_{1}$ is even, then we have integer $k$ such that:

$\operatorname{ltan}\alpha=a+\frac{1}{2k}\quad\operatorname{ltan}\frac{\alpha}{% 2}=a+\frac{2}{2k}=a+\frac{1}{k}.$

Hence we have LLS sequences:

$\alpha:(a_{0},a_{1}),\quad\alpha/2:(a_{0},a_{1}/2);$
•

If $a_{1}$ is odd, then we have integer $k$ such that:

$\operatorname{ltan}\alpha=a+\frac{1}{2k+1}\quad\operatorname{ltan}\frac{\alpha% }{2}=a+\frac{2}{2k+1}=a+\frac{1}{k+1/2}.$

Hence we have LLS sequences:

$\alpha:\bigg{(}a_{0},a_{1}\bigg{)},\quad\alpha/2:\bigg{(}a_{0},\frac{a_{1}-1}{% 2},2\bigg{)}.$

Example 5.1.8.

Let $\alpha$ have LLS sequence $(a,1,k)$ . Then $\alpha/2$ has LLS sequence $(a+1)=(a,1)$ .

Proof.

We have expression for $\operatorname{ltan}\alpha$ :

a+\frac{1}{1+\frac{1}{k}}=\frac{a+ak+k}{k+1}.

Hence, we express $\alpha/2$ as:

\frac{a+ak+k+1}{k+1}=\frac{(k+1)(a+1)}{k+1}=a+1.

∎

Example 5.1.9.

Angle $\alpha/2$ has LLS sequence $[a+1]=[a,1]$ if and only if one of the following is true:

•

$\alpha$ has LLS sequence $[a,1,k]$ ;
•

$\alpha$ has LLS sequence $[a]$ .

One direction of the proof is the same as Example 5.1.8. Now we will examine the other direction of the proof.

For $\alpha/2$ to have this LLS sequence, we have that $\operatorname{ltan}(\alpha/2)=a$ . Hence for some integer $m$ we have $\operatorname{ltan}(\alpha)=(am+a+m)/(m+1)$ . Then we have the LLS sequence for $\alpha$ is the continued fraction of this.

First let $m\neq 0$ . Then we have

(am+a+m)/(m+1)=a+\frac{m}{m+1}=a+\frac{1}{m/m+1/m}=a+\frac{1}{1+\frac{1}{m}}.

Now let $m=0$ . Then we have $\operatorname{ltan}(\alpha)=(a)/(1)=a$ .

Example 5.1.10.

Let $\alpha$ have LLS sequence $(a,2,k)$ . Then $\alpha/2$ has LLS sequence $(a,1,1,k)$ .

Proof.

We have expression for $\operatorname{ltan}\alpha$ :

a+\frac{1}{2+\frac{1}{k}}=a+\frac{k}{2k+1}.

Hence, we express $\alpha/2$ as:

a+\frac{k}{2k+1}=a+\frac{1}{1+\frac{k}{k+1}}=a+\frac{1}{1+\frac{1}{1+\frac{1}{% k}}}.

∎

Example 5.1.11.

Let $\alpha$ have LLS sequence $(a_{0},a_{1},a_{2})$ with $a_{2}\neq 1$ . Then we have the first two terms of the LLS sequence for $\alpha/2$ as follows:

a_{0},\quad a_{1}+\left\lfloor\frac{a_{1}-1}{a_{2}+1}\right\rfloor.

If $\frac{a_{1}-1}{a_{2}+1}$ is an integer, then we have the LLS sequence to be

\left(a_{0},a_{1}+\left\lfloor\frac{a_{1}-1}{a_{2}+1}\right\rfloor\right).

Example 5.1.12.

Let us consider the angle $AOB$ with $A=(1,0)$ and $B=(m,n)$ . Then we have:

\tan(\angle AOB)=n/m;

\tan(\angle AOB/2)=m/(m+1).

Let us consider the example where $\angle AOB$ has an LLS sequence of length 2, $[a_{0},a_{1}]$ . Then we have LLS sequence for $\angle AOB/2$ has two cases

•

If $a_{1}$ is even, then $\angle AOB/2$ has LLS sequence $[a_{0},a_{1}/2]$ ;
•

If $a_{1}$ is odd, then $\angle AOB/2$ has LLS sequence $[a_{0},\lfloor a_{1}/2\rfloor,2]$ .

Example 5.1.13.

Let $\alpha$ be a straight angle. Here the bisector is not a ray but instead a halfplane. There is no obvious definition of $\alpha/2$ .

5.1.2 $k$ -secting integer angles

Definition 5.1.14.

We say that the set of rays passing through points $A_{1},\ldots,A_{k-1}$ is a $k$ -sector of the angle $\angle A_{0}OA_{k}$ if for every $i=1,\ldots,k-1$ , the ray $A_{i}O$ is a bisector of $A_{i-1}OA_{i+1}$ .

Definition 5.1.15.

A $k_{1}:k_{2}$ ratio-bisector of an angle $AOB$ is the ray $P$ generated by the rule:

p\in P\text{ if }k_{1}\cdot\operatorname{ld}(p,AO)=k_{2}\cdot\operatorname{ld}% (p,OB).

Remark 5.1.16.

A $1:1$ ratio-bisector of an angle is the bisector of said angle.

Unlike in Euclidean geometry, these definitions do not lead to the same thing. In the following example, we will observe that the $2:1$ ratio-bisector of an angle is not necessarily one of the 3-sectors (trisectors).

Example 5.1.17.

Let us consider the angle $\alpha=\operatorname{larctan}[1;1,1,2,1]$ . This angle trisects such that $\alpha/3=\operatorname{larctan}(1)$ .

Taking the line $y=x$ , which gives us one of the angles $\alpha/3$ , we get that this is a $4:1$ ratio-bisector of $\alpha$ , whereas in the Euclidean case we would expect a $2:1$ ratio-bisector.

Remark 5.1.18.

Palindromic LLS sequences provides splitting alpha and alpha transpose. For non-palindromic we experiment.

Theorem 5.1.19.

Let us trisect between $y=n/mx$ and $y=0$ . The trisecting lines are as follows:

y=\frac{n}{m+1}.

Proof.

Suppose we have trisector lines passing through points $(a_{1},b_{1})$ and $(a_{2},b_{2})$ respectively. Firstly, we will find values for $(a_{1},b_{1})$ . Then the integer distance between point $(a_{1},b_{1})$ and $(1,0)$ and $(a_{2},b_{2})$ is the same. Then using the determinant in Remark 2.1.11, we have that

b_{1}=\frac{b_{2}}{a_{2}+1}a_{1}.

Using this, we find points $(a_{1},b_{1})=(a_{2}+1,b_{2}).$

Now we find the bisector line between $y=m/nx$ and $y=b_{2}/(a_{2}+1)x.$ We have the distance between points on line 2 and the line 1 to be

\frac{\begin{vmatrix}a_{2}+1&a_{2}\\ b_{2}&b_{2}\end{vmatrix}}{\gcd(a_{2}+1,b_{2})}=\frac{b_{2}}{\gcd(a_{2}+1,b_{2}% )}.

Now, we solve for $(a_{2}.b_{2})$ when this distance is equal to the distance between points on line 2 and the line $y=n/mx$ :

\frac{\begin{vmatrix}a_{2}&m\\ b_{2}&n\end{vmatrix}}{=}\frac{b_{2}}{\gcd(a_{2}+1,b_{2})}

a_{2}n-b_{2}m=\frac{b_{2}}{\gcd(a_{2}+1,b_{2})}.

Re-aranging, we get

b_{2}=\frac{n\gcd(a_{2}+1,b_{2})}{1+m\gcd(a_{2}+1,b_{2})}a_{2}.

∎

5.1 Inscribed Circles of Integer Angles

Proposition 5.1.1.

Proof.

5.1.1 Bisectors of Integer Angles

Definition 5.1.2.

Proposition 5.1.3.

Proof.

Corollary 5.1.4.

Proof.

Remark 5.1.5.

Remark 5.1.6.

Example 5.1.7.

Example 5.1.8.

Proof.

Example 5.1.9.

Example 5.1.10.

Proof.

Example 5.1.11.

Example 5.1.12.

Example 5.1.13.

5.1.2 kk-secting integer angles

Definition 5.1.14.

Definition 5.1.15.

Remark 5.1.16.

Example 5.1.17.

Remark 5.1.18.

Theorem 5.1.19.

Proof.

5.1.2 $k$ -secting integer angles