# Lattice Equable Quadrilaterals III:
tangential and extangential cases

Christian Aebi and Grant Cairns  Collège Calvin, Geneva, Switzerland 1211
<EMAIL_ADDRESS>Department of Mathematics, La Trobe University,
Melbourne, Australia 3086<EMAIL_ADDRESS>
###### Abstract.

A lattice equable quadrilateral is a quadrilateral in the plane whose vertices
lie on the integer lattice and which is equable in the sense that its area
equals its perimeter. This paper treats the tangential and extangential cases.
We show that up to Euclidean motions, there are only 6 convex tangential
lattice equable quadrilaterals, while the concave ones are arranged in 7
infinite families, each being given by a well known diophantine equation of
order 2 in 3 variables. On the other hand, apart from the kites, up to
Euclidean motions there is only one concave extangential lattice equable
quadrilateral, while there are infinitely many convex ones.

### Introduction

A _lattice equable quadrilateral_ (LEQ for short) is a quadrilateral whose
vertices lie on the integer lattice $\mathbb{Z}^{2}$ and which is equable in
the sense that its area equals its perimeter. This paper is a continuation of
the work [2], which treated lattice equable parallelograms, and [3], which
treated lattice equable kites, trapezoids and cyclic quadrilaterals, but this
paper can be read independently of the previous two. Here we examine LEQs that
are _tangential_ , i.e., they have an incircle, or _extangential_ , i.e., they
have an excircle.

Before stating our main results, let us make some general remarks about the
importance and occurrence of tangential and extangential LEQs, up to Euclidean
motions. Remarkably, tangential and extangential LEQs apparently constitute a
large component of the overall set of LEQs. For example, apart from
parallelograms and trapezoids, we know of only one convex LEQ that is neither
tangential nor extangential. This is the LEQ with vertices
$(0,0),(2,0),(8,8),(8,15)$ and side lengths $2,10,7,17$. There seems to be
significantly more tangential LEQs than extangential LEQs, within a ball of
any given radius of sufficient size. The tangential LEQs are mainly concave;
indeed, as we show in Corollary 1, there are only 6 convex tangential LEQs.
The extangential LEQs are mainly convex; we show in Corollary 3 that there is
only one concave non-kite extangential LEQ. Kites which aren’t parallelograms
are both tangential and extangential, and they are the only LEQs with this
property.

Consider a tangential LEQ $OABC$ whose sides $OA,AB,BC,CO$ have length
$a,b,c,d$ respectively, which therefore are integers [2, Remark 2]. The key to
our results on tangential LEQs is the observation that a certain pair of
functions of the side lengths take a very restricted range of possible values.
The functions are:

######  Definition 1.

For a tangential LEQ $OABC$, let

$\sigma=\frac{ad+bc+2\delta\sqrt{abcd-4(a+c)^{2}}}{16+(a-b)^{2}},\quad\tau=\frac{ab+cd-2\delta\sqrt{abcd-4(a+c)^{2}}}{16+(a-d)^{2}},$

where $\delta=1$ if $B$ lies within the circumcircle of the triangle $OAC$,
and $\delta=-1$ otherwise.

In fact, as we show in Section 3, these functions can only take the seven
possible values $2,3,5,9,9/8,5/4,3/2$, and moreover
$\frac{1}{\sigma}+\frac{1}{\tau}=1$. In particular, in each case at least one
of $\sigma,\tau$ is an integer and belongs to $\\{2,3,5,9\\}$.

For each of the seven possibilities for the pair $(\sigma,\tau)$, we show that
the side lengths satisfy a certain corresponding diophantine equation, and
conversely, solutions to the equation, along with some auxiliary conditions,
lead to the existence of a corresponding tangential LEQ. There is a certain
redundancy both in the statement of the seven results and their proofs, so we
have been at pains to present the results in as compact a form as possible.
The statements of the resulting theorem and its converse are rather
cumbersome, but considerable saving is attained in the long run. Before
stating the results, note that for a tangential LEQ $OABC$ with consecutive
sides $a,b,c,d$, we see in Remark 9 that by making a reflection if necessary,
we may assume that $a$ and $c$ are even in the case $\sigma=\tau=2$. Our
classification result for tangential LEQs is then as follows.

######  Theorem 1.

Suppose that $OABC$ is a tangential LEQ with consecutive sides $a,b,c,d$ and
vertices $O,A,B,C$ in positive cyclic order. Suppose also that if $OABC$ is
concave, then its reflex angle is at $B$. Without loss of generality we also
assume that $a$ and $c$ are even in the case $\sigma=\tau=2$. Then the
following conditions hold:

  1. (i)

$|c-b|\tau<a+c$, (ii) $(a+d)\tau>a+c$, (iii) $(b+c)\tau\not=a+c$.

Moreover, $OABC$ is convex if and only if $(b+c)\tau>a+c$. Furthermore, there
are two cases:

  1. (I)

If $\tau\in\\{2,3,5,9\\}$, then $a,\tau b$ have the same parity and setting
$u=\frac{\tau b-a}{2},v=\frac{\tau b+a}{2}$, we have

(1) $(2\tau)^{2}+u^{2}=v^{2}-\left(v-\frac{\tau-1}{2}c\right)^{2}.$

  2. (II)

If $\sigma\in\\{3,5,9\\}$, then $a,\sigma d$ have the same parity and setting
$u=\frac{\sigma d-a}{2},v=\frac{\sigma d+a}{2}$, we have

(2) $(2\sigma)^{2}+u^{2}=v^{2}-\left(v-\frac{\sigma-1}{2}c\right)^{2}.$

We now state the converse result.

######  Theorem 2.

Let $x\in\\{2,3,5,9\\}$ and suppose we have an integer solution $(u,v,c)$ of
the diophantine equation

(3) $(2x)^{2}+u^{2}=v^{2}-\left(v-\frac{x-1}{2}c\right)^{2}$

for which $u+v\equiv 0\pmod{x}$ and $c>0$, and further that $c$ is even when
$x=2$ and that $c$ is not divisible by $3$ if $x=3$. Then we have:

  1. (I)

Let $t=x$, $a=v-u,b=(v+u)/t,d=a+c-b$, and suppose the following conditions
hold:

    1. (i)

$|c-b|t<a+c$, (ii) $(a+d)t>a+c$, (iii) $(b+c)t\not=a+c$.

Then there is a tangential LEQ $OABC$ with successive side lengths $a,b,c,d$
for which $(\sigma,\tau)=(\frac{t}{t-1},t)$.

  2. (II)

Let $s=x$, $a=v-u,d=(v+u)/s,b=a+c-d$ and suppose that the above conditions (i)
– (iii) hold for $t=\frac{s}{s-1}$ and that $b>0$. Then there is a tangential
LEQ $OABC$ with successive side lengths $a,b,c,d$ for which
$(\sigma,\tau)=(s,t)$.

Furthermore, in both of the above cases, if $OABC$ is concave, then the reflex
angle is at $B$.

######  Corollary 1.

Up to Euclidean motions, there are only six convex tangential LEQs:

  * •

the $4\times 4$ square,

  * •

the isosceles trapezoid of side lengths 5,2,5,8,

  * •

the right trapezoid of side lengths 5,3,4,6,

  * •

the equable rhombus of side length 5,

  * •

the equable kite of side lengths 3 and 15,

  * •

the LEQ with vertices $(0,0),(40,9),(36,12),(35,12)$, and sides 37,1,5,41.

######  Corollary 2.

The incenter of a tangential LEQ is an integer lattice point in the cases
where $\sigma,\tau\in\\{2,3,5,5/4,3/2\\}$.

Examples where $\sigma,\tau\in\\{9,9/8\\}$ and the incenter is not an integer
lattice point are given in Example 1.

We now turn to our results on extangential LEQs. Consider an extangential LEQ
$OABC$ whose sides $OA,AB,BC,CO$ have length $a,b,c,d$ respectively. We
introduce functions analogous to those of Definition 1. More precisely, it is
convenient to define functions $\Sigma,T$ analogous to $8\sigma,8\tau$, as
follows.

######  Definition 2.

For an extangential LEQ $OABC$, let

$\displaystyle\Sigma$
$\displaystyle=8\cdot\frac{ad+bc+2\delta\sqrt{abcd-4(a+b)^{2}}}{16+(a-c)^{2}},$
$\displaystyle T$
$\displaystyle=8(a+b)^{2}\cdot\frac{ab+cd+2\delta\sqrt{abcd-4(a+b)^{2}}}{16(a+b)^{2}+(a-c)^{2}(a-d)^{2}},$

where $\delta=1$ if $B$ lies within the circumcircle of the triangle $OAC$,
and $\delta=-1$ otherwise.

The functions $\Sigma,T$ are not constrained to take only a finite number of
possible values, as was the case with $\sigma,\tau$. So the study of
extangential LEQs is somewhat more complicated than that of tangential LEQs.
Our main result is as follows.

######  Theorem 3.

If a non-kite extangential LEQ $OABC$ has consecutive sides $a,b,c,d$, then
$\Sigma,T$ are integers and one the following holds:

  1. (a)

$(\Sigma,T)=(9,18)$ or $(18,50)$,

  2. (b)

$(\Sigma,T)=(5m^{2},5m^{2}+5)$ for some integer $m$ for which there exists
integers $n,Y,Z$ such that $m^{2}-10n^{2}=-1$ and $(5m^{2}-8)Y^{2}=5+8Z^{2}$.

  3. (c)

$(\Sigma,T)=(m^{2},m^{2}+1)$ for some integer $m$ for which there exists
integers $n,Y,Z$ such that $m^{2}-2n^{2}=-1$ and $(m^{2}-8)Y^{2}=1+8Z^{2}$.

The situation concerning case (a) of the above theorem is very satisfactory.
We examine the two possibilities for $(\Sigma,T)$ in Section 9, and study the
corresponding extangential LEQs up to Euclidean motions. We explicitly
classify all LEQs with $(\Sigma,T)=(9,18)$; there is a single infinite family
corresponding to solutions of the negative Pell equation $x^{2}-2y^{2}=-1$.
For $(\Sigma,T)=(18,50)$, we prove that there is precisely one extangential
LEQ; this isolated example has sides $(a,b,c,d)=(13,2,5,10)$ and is shown on
the right of Figure 10.

We don’t give a complete classification for case (b) of the above theorem.
However, in Section 9 we consider $m=3$, which is the smallest value of $m$
for which $m^{2}-10n^{2}=-1$ has a solution. Here $(\Sigma,T)=(45,50)$, and we
give explicit formulas for infinitely many such LEQs. The side lengths of the
first three members of this family are given in Table 5. One sees that the
lengths grow very rapidly. The next possible value of $m$ is $m=117$; see
Remark 29. Here $(\Sigma,T)=(5\cdot 117^{2},5\cdot 117^{2}+5)$. In Example 5,
we exhibit the smallest possible extangential LEQ with this $(\Sigma,T)$ pair;
it has perimeter $\cong 3\cdot 10^{27}$.

We do not know if there are any LEQs satisfying condition (c) of the above
theorem. Indeed, we have:

######  Open Problem.

Does there exist an integer solution $(m,n)$ of the negative Pell equation
$m^{2}-2n^{2}=-1$, for which the diophantine equation
$(m^{2}-8)Y^{2}=1+8Z^{2}$ has an integer solution for $(Y,Z)$.

Even if there were such a solution, it would still be necessary to prove that
there are lattice vertices that realise the corresponding side lengths. We
show at the very end of the paper that if there is an extangential LEQ
corresponding to case (c) of Theorem 3, then its perimeter is at least
$10^{718}$.

As a consequence of our study, we have the following.

######  Corollary 3.

Up to Euclidean motions, there is only one concave non-kite extangential LEQ;
it is the LEQ with vertices $(0,0),(12,5),(10,5),(6,8)$ and side lengths
$(13,2,5,10)$.

Theorem 3 is proved by reducing it to the following number theoretic result.

######  Theorem 4.

Let $y,z,k\in\mathbb{N}$ with $k>16$ and $k>yz$. Suppose that

  1. (a)

$\Sigma:=\frac{8(z^{2}+k)}{k-16}$ is an integer,

  2. (b)

$\Sigma^{\prime}:=\frac{y^{2}\Sigma}{k}$ is an integer,

  3. (c)

$x:=\sqrt{\frac{k(\Sigma+\Sigma^{\prime})}{8}}$ is an integer.

Then either

  1. (a)

$(\Sigma,\Sigma^{\prime})=(9,9),(12,24),(16,16),(24,12),(10,40),(40,10)$ or
$(18,32)$,

  2. (b)

$(\Sigma,\Sigma^{\prime})=(5m^{2},5)$ for some integer $m$ for which there
exists integers $n,Y,Z$ such that $m^{2}-10n^{2}=-1$ and
$(5m^{2}-8)Y^{2}=5+8Z^{2}$,

  3. (c)

$(\Sigma,\Sigma^{\prime})=(m^{2},1)$ for some integer $m$ for which there
exists integers $n,Y,Z$ such that $m^{2}-2n^{2}=-1$ and
$(m^{2}-8)Y^{2}=1+8Z^{2}$.

The proof of this theorem is established by writing the ratio
$\frac{\Sigma^{\prime}}{\Sigma}$ as $\frac{u}{v}$, with $\gcd(u,v)=1$, and
considering the 6 cases according to whether the pair $(u,v)$ is respectively
(odd,even), (odd,odd), or (even,odd), and whether the 2-adic order of the even
number (respectively $v,u+v$ or $u$) is even or odd. Each of the six cases is
conducted by a series of contradiction arguments.

The paper is organised in two parts. Part 1 covers tangential LEQs. Section 1
develops some general results true for all tangential quadrilaterals. Section
2 gives explicit examples: we present calculations of the incenters of LEQs
that are kites, and we give an infinite nested family of non-dart concave
tangential LEQs. Section 3 gives a series of lemmas on tangential LEQs leading
to the definition of the key functions $\sigma$ and $\tau$, and their
properties. In Section 4 we give the proof of Theorem 1 and Corollary 1.
Section 5 is the most substantial part of Part 1. Here we prove Theorem 2 and
Corollary 2. The final section of Part 1, Section 6, gives more examples. In
particular, we show that there are infinitely many LEQs for each of the seven
possible choices of $(\sigma,\tau)$.

Part 2 treats extangential LEQs. Sections 7 and 8 follow the general plan
adopted in Sections 1 and 3 of Part 1; Section 7 presents some general results
for all extangential quadrilaterals, and Section 8 gives a series of lemmas
leading to the definition of the functions $\Sigma$ and $T$, and their
properties. Section 9 treats extangential LEQs in the cases where
$(\Sigma,T)=(9,18),(18,50)$ and $(45,50)$. Section 10 shows how Theorem 3 can
be deduced from Theorem 4. Section 11 is the longest section in the paper;
here we prove Theorem 4. This section also contains the proof of Corollary 3,
see Remark 31. Finally, in Section 12 we discuss the Open Problem presented
above.

###### Contents

  1. I Tangential quadrilaterals
    1. 1 Basic notions for tangential LEQs
    2. 2 Examples of tangential LEQs
    3. 3 Lemmata for tangential LEQs
    4. 4 Proof of Theorem 1 and Corollary 1
    5. 5 Proof of Theorem 2
    6. 6 Infinite families of tangential LEQs
  2. II Extangential quadrilaterals
    1. 7 Basic notions for extangential LEQs
    2. 8 Lemmata for extangential LEQs
    3. 9 Explicit examples of extangential LEQs
    4. 10 Theorem 3 from Theorem 4
    5. 11 Proof of Theorem 4
    6. 12 Comments on the Open Problem

######  Notation.

In this paper, a quadrilateral $OABC$ is defined by four vertices $O,A,B,C$,
no three of which are colinear, such that the line segments $OA,AB,BC,CO$ have
no interior points of intersection; that is, our quadrilaterals have no self-
intersections. We always write the vertices $O,A,B,C$ in positive
(counterclockwise) cyclic order, and if $O,A,B,C$ is concave, then the
labelling is chosen so that the reflex angle is at $B$. We use the notation
$K(OABC)$ for area and $P(OABC)$ for perimeter. Throughout this paper, for
ease of expression, we often simply write $K$ for $K(OABC)$, and $P$ for
$P(OABC)$, and we abbreviate the triangle areas $K(COA),K(OAB),K(ABC),K(BCO)$
as $K_{O},K_{A},K_{B},K_{C}$ respectively. By abuse of notation, we write
$OA,AB,BC,CO$ for both the sides, and their lengths; the meaning should be
clear from the context. We also usually denote $OA,AB,BC,CO$ by the letters
$a,b,c,d$. The lengths of the diagonals $OB,AC$ are denoted $p,q$,
respectively. We use vector notation, such as $\overrightarrow{AB}$. But we
use the same symbol, $A$ say, for the vertex $A$ and its position vector
$\overrightarrow{OA}$. Finally, by _Euclidean motions_ , we mean both the
orientation preserving and orientation reversing kinds; that is, we consider
the group generated by translations, rotations and reflections. In this paper,
we employ the term _positive_ in the strict sense. So
$\mathbb{N}=\\{n\in\mathbb{Z}\ |\ n>0\\}$. Many of the arguments in this paper
are proofs by contradiction. To avoid having to write “which gives a
contradiction” countless times, we use the contradiction symbol
$\Rightarrow\\!\Leftarrow$ at the end of certain displayed formulas.

We used Mathematica and Maple for many of the calculations and algebraic
manipulations in this paper. The factorizations of large numbers conducted at
the end of the paper were performed using Dario Alpern’s integer factorization
calculator [4]. We remark that Alpern has a very nice continued fraction
calculator, and a quadratic diophantine equation solver that we also found
useful [5].

$O$$A$$B$$C$$d$$a$$b$$c$$p$$q$$K_{A}$$\theta$ Figure 1. Illustration of some
of the notation used

##  Part I Tangential quadrilaterals

###  1\. Basic notions for tangential LEQs

It is well known and easy to see that a triangle is equable if and only if its
incircle has radius 2. A quadrilateral that has an incircle is said to be
_tangential_ , or _circumscriptible_ [28, 17, 22, 25]. Obviously, a tangential
quadrilateral is equable if and only if its incircle has radius 2. Pitot’s
theorem says that a quadrilateral with consecutive side lengths $a,b,c,d$ is
tangential if and only if the following equation holds; see [34], [9, p.
62–64] and [24].

(4) $a+c=b+d.$

While Pitot’s Theorem is usually stated only for convex quadrilaterals, it
also holds in the concave case. Indeed, consider a concave quadrilateral
$OABC$ with reflex angle at $B$. Let $A^{\prime}$ denote the point of
intersection of the side $OA$ and the extension of side $BC$. Similarly, let
$C^{\prime}$ denote the point of intersection of the side $OC$ and the
extension of side $AB$. Let $a,b,c,d$ denote the lengths of $OA,AB,BC,CO$
respectively, and similarly, let $a^{\prime},b^{\prime},c^{\prime},d^{\prime}$
denote the lengths of $OA^{\prime},A^{\prime}B,BC^{\prime},C^{\prime}O$. Then
it is easy to see that (4) holds if and only if
$a^{\prime}+c^{\prime}=b^{\prime}+d^{\prime}$; see [9, Problem 262]. That is,
$OABC$ is tangential if and only if $OA^{\prime}BC^{\prime}$ is tangential.

Figure 2 gives an example of a concave tangential LEQ. Note that, as this
example shows, for a concave tangential LEQ $OABC$, while the associated
convex tangential quadrilateral $OA^{\prime}BC^{\prime}$ is equable, it may
fail to have integer sides or have its vertices on lattice points.

$2$$4$$6$$8$$10$$12$$14$$16$$1$$2$$3$$4$$5$$6$$O$$A$$A^{\prime}$$B$$C$$C^{\prime}$
Figure 2. A concave tangential LEQ with side lengths 16,5,2,13

For the rest of this section, $OABC$ denotes a tangential (convex or concave)
quadrilateral, with vertices in counterclockwise cyclic order, and $a,b,c,d$
denote the lengths of the sides $OA,AB,BC,CO$ respectively.

######  Proposition 1.

If $OABC$ is tangential, then $OABC$ is a kite if and only if one of the
diagonals divides $OABC$ into two triangles of equal area.

###### Proof.

Obviously, if $OABC$ is a kite, then its axis of symmetry diagonals divides
$OABC$ into two triangles of equal area. Conversely, applying Heron’s formula
to triangle $OAB$ gives

$\displaystyle 16K_{A}^{2}$ $\displaystyle=(a+b+p)(a+b-p)(a-b+p)(-a+b+p)$
$\displaystyle=-(a^{2}-b^{2})^{2}+2(a^{2}+b^{2})p^{2}-p^{4}.$

Similarly, from triangle $OBC$, we have
$16K_{C}^{2}=-(c^{2}-d^{2})^{2}+2(c^{2}+d^{2})p^{2}-p^{4}$. Hence,
subtracting,

(5)
$2(a^{2}-d^{2}+b^{2}-c^{2})p^{2}=16(K_{A}^{2}-K_{C}^{2})+(a^{2}-b^{2})^{2}-(c^{2}-d^{2})^{2}.$

Notice that

$\displaystyle a^{2}-d^{2}+b^{2}-c^{2}$ $\displaystyle=(a-d)(a+d)+(b-c)(b+c)$
$\displaystyle=(a-d)(a+d+b+c)=2(a+c)(a-d),$

and

$\displaystyle(a^{2}-b^{2})^{2}$
$\displaystyle-(c^{2}-d^{2})^{2}=(a-b)^{2}(a+b)^{2}-(d-c)^{2}(c+d)^{2}$
$\displaystyle=(a-b)^{2}(a+b+c+d)(a+b-c-d)=4(a-d)(a+c)(a-b)^{2}.$

So (5) gives

(6) $(a+c)(a-d)p^{2}=4(K_{A}^{2}-K_{C}^{2})+(a-d)(a+c)(a-b)^{2}.$

Now assume that $K_{A}=K_{C}$. Then (6) gives $(a-d)p^{2}=(a-d)(a-b)^{2}$.
Notice that $p=\pm(a-b)$ is impossible, as otherwise the triangle $OAB$ would
be degenerate. Hence $a=d$. Moreover, as $K_{A}=K_{C}$, the points $A,C$ are
equidistant from the line through $O,B$. So the triangles $OAB$ and $OBC$ are
congruent, and hence $OABC$ is a kite. Clearly, by considering triangles $OAC$
and $BCA$, the same argument would hold if $K_{O}=K_{B}$. ∎

It is well known that the incenter $I$ of a convex tangential quadrilateral
lies on the _Newton line_ $\mathcal{N_{L}}$, which is the line passing through
the midpoints of the two diagonals; see [7, Chap. 7.5], [8, Chap. 2.7] and
[12]. This is also true for concave tangential quadrilaterals, because the
midpoints of the three diagonals of a complete quadrilateral are colinear (see
[37] for 23 proofs of this fact). Let $M_{A},M_{O}$ denote the midpoint of the
diagonals $AC$, $OB$ respectively; see Figures 3 and 4. Notice that
$M_{A},M_{O}$ are distinct, and the Newton line unambiguously defined, if and
only if $OABC$ is not a parallelogram.

$O$$A$$B$$C$$I$$M_{A}$$M_{O}$ Figure 3. The Newton line of a convex tangential
quadrilateral $O$$A$$B$$C$$I$$M_{A}$$M_{O}$ Figure 4. The Newton line of a
concave tangential quadrilateral

######  Proposition 2.

If $OABC$ is tangential and is not a parallelogram, then $OABC$ is a kite if
and only if the Newton line $\mathcal{N_{L}}$ contains one of the diagonals.

###### Proof.

It is obvious that if $OABC$ is a kite, then $\mathcal{N_{L}}$ is the axis of
symmetry of $OABC$ and hence contains a diagonal. Conversely, suppose
$\mathcal{N_{L}}$ coincides with one of the diagonals, say $OB$. As
$M_{A}\in\mathcal{N_{L}}$, we have $K(OAM_{A})=K(COM_{A})$ and
$K(ABM_{A})=K(BCM_{A})$, and hence

$K_{A}=K(OAM_{A})+K(ABM_{A})=K(COM_{A})+K(BCM_{A})=K_{C}.$

Thus the diagonal $OB$ divides $OABC$ into two triangles of equal area. Then
$OABC$ is a kite by Proposition 1. ∎

######  Remark 1.

For further equivalent conditions for a tangential quadrilateral to be a kite,
see [18].

The radius $r$ of the incircle, called the _inradius_ , is given by the
following obvious formula:

$r=\frac{K}{a+c}.$

We will be mainly interested in the equable case, where $r=2$, but in this
section we consider the general case as it provides a useful comparison for
results on the exradius of extangential quadrilaterals, which we will consider
below in Part 2.

######  Proposition 3.

If $OABC$ is tangential, we have the following two expressions for the
incenter $I$:

  1. (a)

$I=\frac{r}{2}\,\frac{aC+dA}{K_{O}}$, (b)
$I=A+\frac{r}{2}\,\frac{a(B-A)-bA}{K_{A}}$.

###### Proof.

Suppose $A,C$ have coordinates $(a_{1},a_{2}),(c_{1},c_{2})$ respectively, let
$I=(i_{1},i_{2})$ be the incenter. Considering the area of triangle $AIO$, we
have $ra=a_{1}i_{2}-a_{2}i_{1}$. Similarly, from the area of triangle $COI$,
we have $rc=-c_{1}i_{2}+c_{2}i_{1}$. Hence

$r\begin{pmatrix}a\\\ c\end{pmatrix}=\begin{pmatrix}-a_{2}&a_{1}\\\
c_{2}&-c_{1}\end{pmatrix}\begin{pmatrix}i_{1}\\\
i_{2}\end{pmatrix},\quad\text{and so}\quad\begin{pmatrix}i_{1}\\\
i_{2}\end{pmatrix}=\frac{r}{a_{1}c_{2}-a_{2}c_{1}}\begin{pmatrix}c_{1}&a_{1}\\\
c_{2}&a_{2}\end{pmatrix}\begin{pmatrix}a\\\ c\end{pmatrix}.$

That is, $I=\frac{r}{2}\,\frac{aC+dA}{K_{O}}$, which is expression (a) in the
statement of the proposition. Similarly, by considering triangles $BIA$ and
$OAI$ we obtain (b). ∎

Note that the above proposition holds in both convex and concave cases, but in
the latter case, with a reflex angle at $B$ for example, the signed area
$K_{B}$ is negative. For more on the incenter of tangential quadrilaterals,
see [6].

######  Proposition 4.

For a tangential equable quadrilateral $OABC$, one has

$(K_{A}-(a+b))(K_{O}-(a+d))=bd-ac.$

###### Proof.

Equating the two expressions for $I$ from the above proposition, with $r=2$,
and taking the vector cross product by $C$ on the right, gives

$d=K_{O}+\frac{a(K_{C}-K_{O})-bK_{O}}{K_{A}},$

so $K_{O}K_{A}-dK_{A}+aK_{C}-(a+b)K_{O}=0$. Thus, as $K_{C}=2(a+c)-K_{A}$, we
have $K_{O}K_{A}-(a+d)K_{A}-(a+b)K_{O}+2a(a+c)=0$. The required identity is
then obtained by factorizing, using the fact that $2a(a+c)=(a+b)(a+d)-(bd-ac)$
since $a+c=b+d$. ∎

Since the incenter $I$ lies on Newton line, $I$ is of the form $\lambda
M_{A}+(1-\lambda)M_{O}$, for some $\lambda\in[0,1]$. The following result will
use the fact that for a (arbitrary) quadrilateral $OABC$, one has the
following elementary vector equation:

(7) $K_{O}\,B=K_{C}\,A+K_{A}\,C.$

This equation is proved in [1], as an application of the vector triple
product. Alternately, one can simply notice that the vector products
$A\times(K_{C}\,A+K_{A}\,C-K_{O}\,B)$ and
$B\times(K_{C}\,A+K_{A}\,C-K_{O}\,B)$ are both zero, so (7) follows as $A,B$
are linearly independent in our case.

######  Proposition 5.

If $OABC$ is tangential but is neither a parallelogram nor a kite, we have the
following two expressions for the coordinate $\lambda$:

  1. (a)

$\lambda=\frac{r(a-b)}{2K_{O}-r(a+c)}$, (b)
$\lambda=1-\frac{r(b-c)}{2K_{A}-r(a+c)}$.

Furthermore, if $OABC$ is a kite, then the first of the above expressions for
$\lambda$ holds if $OABC$ is not a rhombus and we relabel the vertices if
necessary so that $OB$ is the axis of symmetry.

###### Proof.

By definition, $I=\lambda
M_{A}+(1-\lambda)M_{O}=\lambda\frac{A+C}{2}+(1-\lambda)\frac{B}{2}$, so using
(7) to eliminate $B$, we have

(8) $I=\frac{\lambda K_{O}+(1-\lambda)K_{C}}{2K_{O}}A+\frac{\lambda
K_{O}+(1-\lambda)K_{A}}{2K_{O}}C.$

Comparing with Proposition 3(a) gives $rd=\lambda K_{O}+(1-\lambda)K_{C}$ and
$ra=\lambda K_{O}+(1-\lambda)K_{A}$, so

(9) $\displaystyle\lambda(K_{O}-K_{C})$ $\displaystyle=rd-K_{C}$ (10)
$\displaystyle\lambda(K_{O}-K_{A})$ $\displaystyle=ra-K_{A}.$

Adding (10) to (9) and using $K_{A}+K_{C}=r(a+c)$ gives
$\lambda(2K_{O}-r(a+c))=r(d-c)$. If $OABC$ is not a kite, then by Proposition
1, $K_{O}\not=K_{B}$, so $2K_{O}\not=r(a+c)$. If $OABC$ is a kite, but not a
rhombus and we relabel the vertices if necessary so that $OB$ is the axis of
symmetry, then once again $2K_{O}\not=r(a+c)$. In either case,

$\lambda=\frac{r(d-c)}{2K_{O}-r(a+c)}=\frac{r(a-b)}{2K_{O}-r(a+c)},$

as required.

Subtracting (10) from (9) gives $\lambda(K_{A}-K_{C})=r(d-a)+K_{A}-K_{C}$. If
$OABC$ is not a kite, then once again $2K_{O}\not=r(a+c)$ by Proposition 1, so

$\lambda=1+\frac{r(d-a)}{K_{A}-K_{C}}=1-\frac{2(b-c)}{2K_{A}-r(a+c)}.$

∎

######  Remark 2.

As we mentioned above, it is well known that for a tangential quadrilateral
$OABC$, its incenter $I$ lies on the Newton line. It is less commonly
mentioned that $I$ lies _between_ $M_{A}$ and $M_{O}$; that is, it lies on the
closed line segment between $M_{A}$ and $M_{O}$. This can be proved by an easy
geometric argument. We will not require this fact, though for equable
tangential quadrilaterals, it follows from the above proposition and Remark 7
below.

###  2\. Examples of tangential LEQs

Of course, the lattice equable kites are tangential. For each of the four
families $K1-K4$ of [3, Theorem 1] we use Propositions 3 and 5 to compute the
incenter $I_{n,i}$ and the parameter $\lambda_{n,i}$ for which
$I_{n,i}=\lambda_{n,i}M+(1-\lambda_{n,i})\frac{B}{2}$, where $M=M_{A}$. We
omit the details, which are completely routine. The results are given in Table
1. Notice that in family K1, $n+i$ is even. Hence $I_{n,i}$ is a lattice point
for all the families.

Family | Equation | $M$ | $B$ | $I_{n,i}$ | $\lambda_{n,i}$
---|---|---|---|---|---
_K1_ | $n^{2}-5i^{2}=4$ | $\frac{1}{2}(n+5i)(2,1)$ | $n(2,1)$ | $\frac{n+i}{2}(2,1)$ | 1/5
_K2_ | $n^{2}-5i^{2}=1$ | $(2n+5i)(2,1)$ | $4n(2,1)$ | $2(n+2i)(2,1)$ | 4/5
_K3_ | $n^{2}-2i^{2}=1$ | $(n+2i)(2,2)$ | $4n(1,1)$ | $2(n+i)(1,1)$ | 1/2
_K4_ | $2n^{2}-i^{2}=1$ | $(4n+3i)(\frac{3}{2},\frac{3}{2})$ | $12n(1,1)$ | $2(3n+2i)(1,1)$ | 8/9

Table 1. The four families of kites

We will now exhibit an infinite nested family of non-dart concave tangential
LEQs. Let $(u_{i},v_{i})$ be the $i$-th solution to the Pell equation
$u^{2}-3v^{2}=1$, with initial solution $(u_{1},v_{1})=(2,1)$. From the
standard theory of Pell equations, one has the recurrences:

(11) $u_{i+1}=2u_{i}+3v_{i},\qquad v_{i+1}=u_{i}+2v_{i}.$

Let $A_{i}$ denote the point with coordinates
$(x_{i},y_{i})=(2u_{i}+4,6v_{i})$, and let $B$ be the point $(8,0)$. We will
consider the lattice quadrilateral $OA_{i}BA_{i+1}$. To verify that
$OA_{i}BA_{i+1}$ has no self-intersection, it suffices to calculate the vector
cross products $\overrightarrow{OA_{i}}\times\overrightarrow{OA_{i+1}}$ and
$\overrightarrow{BA_{i+1}}\times\overrightarrow{BA_{i}}$, using (11), and see
that they are both positive. We leave the details to the reader.

The distance $OA_{i}$ is given by

$OA_{i}^{2}=x_{i}^{2}+y_{i}^{2}=4u_{i}^{2}+16u_{i}+16+36v_{i}^{2}=16u_{i}^{2}+16u_{i}+4=(4u_{i}+2)^{2}.$

So $OA_{i}=4u_{i}+2$. Similarly, the distance $A_{i}B$ is given by

$A_{i}B^{2}=(2u_{i}-4)^{2}+(6v_{i})^{2}=4u_{i}^{2}-16u_{i}+16+36v_{i}^{2}=16u_{i}^{2}-16u_{i}+4=(4u_{i}-2)^{2}.$

So $A_{i}B=4u_{i}-2$. Thus $OA_{i}BA_{i+1}$ is tangential because

$OA_{i}-A_{i}B+BA_{i+1}-A_{i+1}O=(4u_{i}+2)-(4u_{i}-2)+(4u_{i+1}-2)-(4u_{i+1}+2)=0.$

The perimeter $P(OA_{i}BA_{i+1})$ of $OA_{i}BA_{i+1}$ is
$OA_{i}+A_{i}B+OA_{i+1}+A_{i+1}B=8(u_{i}+u_{i+1})$, while the area
$K(OA_{i}BA_{i+1})$ of $OA_{i}BA_{i+1}$ is
$4(y_{i+1}-y_{i})=24(v_{i+1}-v_{i})$. Hence, using (11),

$\displaystyle K(OA_{i}BA_{i+1})-P(OA_{i}BA_{i+1})$
$\displaystyle=24(v_{i+1}-v_{i})-8(u_{i}+u_{i+1})$
$\displaystyle=24(u_{i}+v_{i})-8(3u_{i}+3v_{i})=0.$

So $OA_{i}BA_{i+1}$ is LEQ. The vertices and side lengths of the first four
members of this family are given in Table 2. The first two members of the
family are shown in Figure 5.

$u_{i}$ | $v_{i}$ | $A_{i}$ | $B$ | $A_{i+1}$ | $OA_{i}$ | $A_{i}B$ | $BA_{i+1}$ | $A_{i+1}O$
---|---|---|---|---|---|---|---|---
2 | 1 | (8,6) | (8,0) | (18,24) | 10 | 6 | 26 | 30
7 | 4 | (18,24) | (8,0) | (56,90) | 30 | 26 | 102 | 106
26 | 15 | (56,90) | (8,0) | (198,336) | 106 | 102 | 386 | 390
97 | 56 | (198,336) | (8,0) | (728,1254) | 390 | 386 | 1446 | 1450

Table 2. The first four members of the tangential family $O$$B$ Figure 5. The
first two members of the family

By Proposition 3(b), the incenter $I_{i}$ of $OA_{i}BA_{i+1}$ is calculated to
be:

$\displaystyle I_{i}$
$\displaystyle=A_{i}+\frac{(4u_{i}+2)(B-A_{i})-(4u_{i}-2)A}{K(A_{i}BO)}$
$\displaystyle=(4+2u_{i}+2v_{i},2u_{i}+6v_{i})=A_{i}+(2v_{i},2u_{i}),$

using $u_{i}^{2}=1+3v_{i}^{2}$. In particular, the incenters $I_{i}$ are all
lattice points. From Proposition 5, for
$I_{i}=\lambda_{i}M_{O}+(1-\lambda_{i})M_{A_{i}A_{i+1}}$, one has

$\displaystyle\lambda_{i}$
$\displaystyle=\frac{4}{K(OA_{i}A_{i+1})-((4u_{i}+2)+(4u_{i+1}-2))}$
$\displaystyle=\frac{4}{6(1+2u_{i}+2v_{i})-((4u_{i}+2)+4(2u_{i}+3v_{i}-2))}=\frac{1}{3}.$

In particular, the family members all have the same value of the parameter
$\lambda_{i}$. The Newton line for the first member of the family is shown
(dotted) in Figure 6.

$2$$4$$6$$8$$10$$12$$14$$16$$18$$2$$4$$6$$8$$10$$12$$14$$16$$18$$20$$22$$24$$A_{1}$$B$$A_{2}$$I$$M_{O}$$M_{A_{1}A_{2}}$$O$
Figure 6. First member of the family

###  3\. Lemmata for tangential LEQs

For this section, $OABC$ denotes an equable tangential quadrilateral. In
particular, it has inradius $r=2$. Let $\theta$ denote the interior angle of
$OABC$ at $A$; see Figure 1. By the cosine rule,
$p^{2}=a^{2}+b^{2}-2ab\cos\theta$. As
$|ab\cos\theta|=\sqrt{a^{2}b^{2}-a^{2}b^{2}\sin^{2}\theta}=\sqrt{a^{2}b^{2}-4K_{A}^{2}}$,
so

(12) $p^{2}=a^{2}+b^{2}\pm 2\sqrt{a^{2}b^{2}-(2K_{A})^{2}},$

where the sign of the square root depends on whether $\theta$ is acute or
obtuse. Similarly,

(13) $q^{2}=a^{2}+d^{2}\pm 2\sqrt{a^{2}d^{2}-(2K_{O})^{2}}.$

The distances $p,q$ may fail to be integers (see [3, Theorem 4]), but as
$O,A,B,C$ are lattice points, $p^{2},q^{2}$ are integers. So the following
lemma is immediate from (12) and (13), and doesn’t require the equability or
tangential hypothesis.

######  Lemma 1.

The integers $a^{2}b^{2}-(2K_{A})^{2}$ and $a^{2}d^{2}-(2K_{O})^{2}$ are
squares.

######  Lemma 2.

If $OABC$ is not a kite, one has

$p^{2}=\frac{8(K_{A}-K_{C})}{a-d}+(a-b)^{2}\quad\text{and}\quad
q^{2}=\frac{8(K_{O}-K_{B})}{a-b}+(a-d)^{2}.$

Furthermore, if $OABC$ is a kite with $OB$ as its axis of symmetry, and if
$OABC$ is not a rhombus, then the above formula for $q^{2}$ still applies and
one has the following formula for $p^{2}$:

$p^{2}=\frac{2(a+c)^{2}(a-b)}{K_{O}-K_{B}}.$

###### Proof.

Arguing exactly as in Proposition 1 we reobtain (6):

$(a+c)(a-d)p^{2}=4(K_{A}^{2}-K_{C}^{2})+(a-d)(a+c)(a-b)^{2}.$

If $a=d$, then by the tangential hypothesis, $b=c$, so $OABC$ is a kite. Thus,
if $OABC$ is not a kite, $a\not=d$ and we have

$p^{2}=4(K_{A}^{2}-K_{C}^{2})+(a-b)^{2}.$

Then as $K_{A}^{2}-K_{C}^{2}=(K_{A}+K_{C})(K_{A}-K_{C})=2(a+c)(K_{A}-K_{C})$,
from which the required formula for $p^{2}$ follows. Similarly, the formula
for $q^{2}$ is obtained by applying Heron’s formula to triangles $OAC$ and
$BCA$.

If $OABC$ is a kite with $OB$ as its axis of symmetry, and is not a rhombus,
then $a=d,b=c$ and $a\not=b$, and the argument giving the formula for $q^{2}$
remains valid. For $p^{2}$, we use the standard formula for the area of a
kite: $pq=2K$. So
$p^{2}=16(a+c)^{2}/q^{2}=\frac{2(a+c)^{2}(a-b)}{K_{O}-K_{B}}$, as required. ∎

######  Remark 3.

If $OABC$ is not a kite, then from the above lemma, using (12),

$\frac{8(K_{A}-(a+c))}{a-d}=\frac{p^{2}-(a-b)^{2}}{2}=ab\pm\sqrt{a^{2}b^{2}-(2K_{A})^{2}},$

which is an integer by Lemma 1. Similarly, $\frac{8(K_{O}-(a+c))}{a-b}$ is an
integer. If $OABC$ is a kite with $OB$ as its axis of symmetry, and if $OABC$
is not a rhombus, then by the same reasoning, $\frac{8(K_{O}-(a+c))}{a-b}$ is
again an integer.

At this point we pause to explain the investigation we are about to perform.
The integers $\frac{8(K_{A}-(a+c))}{a-d}$ and $\frac{8(K_{O}-(a+c))}{a-b}$,
defined above for non-kites, will play a key role in what follows. Using
Proposition 5 (or Proposition 4) one could easily directly show that these
integers obey an important relation: their product is $8$ times their sum (see
Lemma 5 below). This enables us to show that these integers are restricted to
a small set of possibilities (see Lemma 6 below). However, we will follow a
somewhat more circuitous route to this result. We proceed by developing
results that will lead to Definition 3 which holds for all tangential LEQs
(kites as well as non-kites). This enables us to then progress in a more
natural manner, without having to appeal to the classification of kites in
[2]. Although it involves some unpleasant computations, this pathway forward
also has the advantage that it reveals certain important relations that will
be useful in what follows.

######  Lemma 3.

The integer $abcd-4(a+c)^{2}$ is a square, and

$\displaystyle K_{A}$ $\displaystyle=(a+c)+(a-d)\frac{ab+cd\pm
2\sqrt{abcd-4(a+c)^{2}}}{16+(a-d)^{2}}$ $\displaystyle K_{O}$
$\displaystyle=(a+c)+(a-b)\frac{ad+bc\mp
2\sqrt{abcd-4(a+c)^{2}}}{16+(a-b)^{2}},$

where the signs of the square roots in the formulas for $K_{O}$ and $K_{A}$
are opposite.

######  Remark 4.

In the statement of the above lemma, the terms

$ab+cd\pm 2\sqrt{abcd-4(a+c)^{2}}\quad\text{and}\quad ad+bc\mp
2\sqrt{abcd-4(a+c)^{2}}$

are strictly positive. Indeed, using $d=a-b+c$, by the arithmetic mean-
geometric mean inequality, $ab+cd\geq 2\sqrt{abcd}>\sqrt{abcd-4(a+c)^{2}}$. In
particular, $K_{A}<a+c$ if and only if $a<d$.

###### Proof of Lemma 3.

The formulas for $K_{A}$ and $K_{O}$ obviously hold when $OABC$ is a rhombus.
So, without loss of generality, we may assume that either $OABC$ is not a
kite, or is a kite that is not a rhombus and has axis of symmetry $OB$. Then,
from Lemma 2 and Equation (13),

$\frac{4(K_{O}-K_{B})}{a-b}-ad=\frac{q^{2}-a^{2}-d^{2}}{2}=\pm\sqrt{a^{2}d^{2}-(2K_{O})^{2}}$

so squaring, using $K_{O}+K_{B}=2(a+c)$ and rearranging gives

$16(K_{O}-(a+c))^{2}-4ad(a-b)(K_{O}-(a+c))=-(a-b)^{2}K_{O}^{2}.$

Let $s:=\frac{K_{O}-(a+c)}{a-b}$. Thus

(14) $16s^{2}+((a-b)s+(a+c))^{2}=4ads.$

Hence $\alpha s^{2}+2\beta s+\gamma=0$, where, using $a+c=b+d$,

$\alpha=16+(a-b)^{2},\quad\beta=-(ad+cb),\quad\gamma=(a+c)^{2}.$

Thus, as $\beta^{2}-\alpha\gamma=4(abcd-4(a+c)^{2})$ (using $a+c=b+d$ again),
we have

$s=\frac{ad+cb\pm 2\sqrt{abcd-4(a+c)^{2}}}{16+(a-b)^{2}},$

which gives the required formula for $K_{O}$. In particular, as $s$ is
rational, $abcd-4(a+c)^{2}$ is a square, as claimed. The formula for $K_{A}$
is similarly obtained by equating $p^{2}$ from Lemma 2 and Equation (12).

It remains to see that the signs of the square roots in the formulas for
$K_{O}$ and $K_{A}$ are opposite. Let $R=2\sqrt{abcd-4(a+c)^{2}}$. Obviously,
we may assume that $R\not=0$. Let us write

$\displaystyle K_{A}$
$\displaystyle=(a+c)+(a-d)\frac{ab+cd+\delta_{A}R}{16+(a-d)^{2}}$
$\displaystyle K_{O}$
$\displaystyle=(a+c)+(a-b)\frac{ad+bc+\delta_{O}R}{16+(a-b)^{2}},$

where $\delta_{A},\delta_{O}$ are each $\pm 1$. Using $a+c=b+d$,

$\displaystyle K_{A}-(a+b)$
$\displaystyle=(d-a)+(a-d)\frac{ab+cd+\delta_{A}R}{16+(a-d)^{2}}=(d-a)\frac{16-(ac+bd)+\delta_{A}R}{16+(a-d)^{2}}$
$\displaystyle K_{O}-(a+d)$
$\displaystyle=(b-a)+(a-b)\frac{ad+bc+\delta_{O}R}{16+(a-b)^{2}}=(b-a)\frac{16-(ac+bd)+\delta_{O}R}{16+(a-b)^{2}}.$

Notice also that $(d-a)(b-a)=bd-ac$. Hence, by Proposition 4,

(15)
$\frac{16-(ac+bd)+\delta_{A}R}{16+(a-d)^{2}}\cdot\frac{16-(ac+bd)+\delta_{O}R}{16+(a-b)^{2}}=1.$

Now,

$\displaystyle\frac{16-(ac+bd)+\delta_{A}R}{16+(a-d)^{2}}\cdot\frac{16-(ac+bd)-\delta_{A}R}{16+(a-b)^{2}}$
$\displaystyle\quad=\frac{(16-(ac+bd))^{2}-R^{2}}{(16+(a-d)^{2})(16+(a-b)^{2}}=\frac{(16-(ac+bd))^{2}-4(abcd-4(a+c)^{2})}{(16+(a-d)^{2})(16+(a-b)^{2})},$

and substituting $d=a+c-b$ one finds that this expression reduces to 1. Hence,
if $\delta_{O}=\delta_{A}$, (15) gives

$\frac{16-(ac+bd)+\delta_{A}R}{16+(a-d)^{2}}\cdot\frac{2\delta_{A}R}{16+(a-b)^{2}}=0,$

which can only happen if $16-(ac+bd)+\delta_{A}R=0$, as $R\not=0$. But in that
case, one would have $(16-(ac+bd))^{2}=R^{2}$, which is impossible, since we
have already seen that substituting $d=a+c-b$ one has

$(16-(ac+bd))^{2}-R^{2}=(16+(a-d)^{2})(16+(a-b)^{2})>0.$

So $\delta_{O}=-\delta_{A}$, as claimed. ∎

A tangential quadrilateral is cyclic if and only if its area is given by
$K=\sqrt{abcd}$ [19, Theorem 4]. Hence the integer $abcd-4(a+c)^{2}$ in the
above proposition is zero if and only if $OABC$ is cyclic. This motivates the
following result.

######  Lemma 4.

The sign of the square root in the formulas for $K_{O}$ is positive if and
only if $B$ lies within the circumcircle of the triangle $OAC$; in particular,
the sign for $K_{O}$ is positive if $OABC$ is concave.

###### Proof.

In the notation of the above proof, let $x=\delta_{O}2\sqrt{abcd-4(a+c)^{2}}$,
so

$\displaystyle K_{O}$ $\displaystyle=(a+c)+(a-b)\frac{ad+bc+x}{16+(a-b)^{2}}$
$\displaystyle K_{A}$ $\displaystyle=(a+c)+(a-d)\frac{ab+cd-x}{16+(a-d)^{2}}.$

From a standard criteria for a point to be within the circumcircle of a
triangle (see [15]), $B$ is inside the circumcircle of the triangle $OAC$ if
and only if

(16) $p^{2}K_{O}<d^{2}K_{A}+a^{2}K_{C}.$

First suppose that $OABC$ is not a kite. Now
$d^{2}K_{A}+a^{2}K_{C}=K_{A}(d^{2}-a^{2})+2a^{2}(a+c)$, and by Lemma 2,

$K_{O}p^{2}=K_{O}\left(\frac{16(K_{A}-(a+c))}{a-d}+(a-b)^{2}\right).$

Also, by Proposition 4, $K_{O}K_{A}=(a+d)K_{A}+(a+b)K_{O}-2a(a+c)$. So
condition (16) can be written as $E>0$ where

$\displaystyle E$ $\displaystyle=K_{A}(d^{2}-a^{2})+2a^{2}(a+c)$
$\displaystyle\quad-\left(\frac{16((a+d)K_{A}+(a+b)K_{O}-2a(a+c)-(a+c)K_{O})}{a-d}+(a-b)^{2}K_{O}\right).$

Substituting the formulas for $K_{O}$ and $K_{A}$, one has

$\displaystyle E=$
$\displaystyle(d^{2}-a^{2})\left(a+c+\frac{(a-d)(ab+cd-x)}{16+(a-d)^{2}}\right)+2a^{2}(a+c)$
$\displaystyle-\frac{16}{a-d}\left((a+d)\left(a+c+\frac{(a-d)(ab+cd-x)}{16+(a-d)^{2}}\right)-2a(a+c)\right.$
$\displaystyle\left.+(b-c)\left(a+c+\frac{(a-b)(bc+ad+x)}{16+(a-b)^{2}}\right)\right)$
$\displaystyle-(a-b)^{2}\left(a+c+\frac{(a-b)(bc+ad+x)}{16+(a-b)^{2}}\right).$

Substituting $d=a+c-b$ one finds that the above expression reduces (rather
miraculously) to $E=(a+c)x$. Hence, as claimed, $x>0$ if and only if $B$ is
inside the circumcircle of the triangle $OAC$.

Now consider the case where $OABC$ is a kite with axis of symmetry $OB$. Then
$a=d,b=c,K_{A}=K_{C}=a+c$ and condition (16) is: $p^{2}K_{O}<2a^{2}(a+c)$. By
Lemma 2,
$p^{2}=\frac{2(a+c)^{2}(a-c)}{K_{O}-K_{B}}=\frac{(a+c)^{2}(a-c)}{K_{O}-(a+c)}$.
So the required condition is $E>0$, where

$E=\frac{2a^{2}(K_{O}-(a+c))}{(a+c)(a-c)}-K_{O}=\frac{(a^{2}+c^{2})K_{O}-2a^{2}(a+c)}{a^{2}-c^{2}}.$

Substituting for $K_{O}$, and using $d=a,b=c$, one finds that the above
expression reduces to

$E=\frac{-16(a+c)^{2}+4a^{2}c^{2}+x(a^{2}+c^{2})}{(a+c)(16+(a-c)^{2})}.$

Notice that the denominator of $E$ is positive, and in the numerator,
$-16(a+c)^{2}+4a^{2}c^{2}=x^{2}$, so the numerator is $x(a^{2}+c^{2}+x)$. Now
$a^{2}+c^{2}+x>0$ since
$(a^{2}+c^{2})^{2}-x^{2}=(a^{2}-c^{2})^{2}+16(a+c)^{2}>0$. Hence $x>0$ if and
only if $B$ is inside the circumcircle of the triangle $OAC$. ∎

We now restate the definition given in the introduction.

######  Definition 3.

Let

$\sigma=\frac{ad+bc+2\delta\sqrt{abcd-4(a+c)^{2}}}{16+(a-b)^{2}},\quad\tau=\frac{ab+cd-2\delta\sqrt{abcd-4(a+c)^{2}}}{16+(a-d)^{2}},$

where $\delta=1$ if $B$ lies within the circumcircle of the triangle $OAC$,
and $\delta=-1$ otherwise.

######  Remark 5.

From the above definition and Lemma 3 and 4,

(17) $\displaystyle K_{O}$ $\displaystyle=a+c+(a-b)\sigma,$ (18)
$\displaystyle K_{A}$ $\displaystyle=a+c+(b-c)\tau.$

######  Remark 6.

By Lemma 3 and 4 and Proposition 5, if $OABC$ is a tangential LEQ that is not
a rhombus, then $\lambda=\frac{1}{\sigma}$.

######  Remark 7.

By Remark 4, $\sigma$ and $\tau$ are both strictly positive.

######  Remark 8.

We saw in the proof of Lemma 3, in Equation (14), that for $a\not=b$, one has,
using $a+c=b+d$,

(19) $16\sigma^{2}+((d-c)\sigma+(a+c))^{2}=4ad\sigma.$

It is easy to verify directly that this equation also holds when $a=b$.
Similarly, the following equation holds in all cases:

(20) $16\tau^{2}+((b-c)\tau+(a+c))^{2}=4ab\tau.$

######  Remark 9.

Suppose $\sigma=\tau=2$. Then (19) gives

$16\cdot 2^{2}+(2(d-c)+(a+c))^{2}=8ad.$

In particular, $(2(d-c)+(a+c))^{2}$ is divisible by 8, and hence, being a
square, it is divisible by 16. In particular, $a+c$ is even. Furthermore $ad$
must be even. Hence, by a reflection in the line $y=x$ if necessary, we may
assume that $a$ is even. Then as $a+c$ is even, $c$ is also even.

Suppose $\tau=3$. Then (20) gives

$16\cdot 3^{2}+(a+3b-2c)^{2}=12ab.$

In particular, $a+3b-2c$ is even so $a$ and $3b$ have the same parity, and we
can pose $(3b-a)/2=u$ and $(3b+a)/2=v$. This gives $36+(u+a-c)^{2}=3ab$, so
$36+(u+c)^{2}=3ab-a^{2}-2ua+4uc+2ac=6bc$. Hence $u+c$ is divisible by 3, say
$u+c=3k$, so $bc$ is divisible by 3. But $4+k^{2}$ is not divisible by 3, so
$bc$ is not divisible by $9$. Thus precisely one of the numbers $b,c$ is
divisible by 3. Hence, by a reflection in the line $y=x$ is necessary, we may
assume that $c$ is not divisible by 3, and that $b$ is divisible by 3. By the
same reasoning, for $\sigma=3$, we may assume that $c$ is not divisible by 3,
and that $d$ is divisible by 3.

######  Lemma 5.

One has ${\sigma}+{\tau}=\sigma\tau$.

###### Proof.

As in the proof of Lemma 4, let $x=2\delta\sqrt{abcd-4(a+c)^{2}}$. Then,
cross-multiplying, the required identity is $E=0$, where

$E=(ad+bc+x)(16+(a-d)^{2})-(ab+cd-x)(16+(a-b)^{2})-(ad+bc+x)(ab+cd-x).$

Expanding and using $d=a+c-b$, one has

$E=x^{2}+4(4a^{2}+8ac-a^{2}bc+ab^{2}c+4c^{2}-abc^{2}).$

Then replacing $x^{2}$ by $4(abcd-4(a+c)^{2})$ and using $d=a+c-b$ again gives
$E=0$, as required. ∎

######  Remark 10.

Observe that $8\sigma$ and $8\tau$ are integers. Indeed, if $OABC$ is not a
kite, then from Lemma 3,

$8\sigma=\frac{8(K_{O}-(a+c))}{a-b},\qquad 8\tau=\frac{8(K_{A}-(a+c))}{a-d},$

which are integers by Remark 3. If $OABC$ is a kite but not a rhombus, with
for example, axis of symmetry $OB$ so $a=d,b=c$, then $\sigma$ is still given
by the above formula and is an integer by Remark 3, while
$8\tau=ac-\delta\sqrt{a^{2}c^{2}-4(a+c)^{2}}$, which is an integer by Lemma 3.
In fact, if $OABC$ is a kite that is not a rhombus, then by [3, Theorem 1],
$OABC$ appears in Table 2, at the beginning of Section 2. Its $\lambda$ value
is thus either $1/5,4/5,1/2$ or $8/9$, and so here $(\sigma,\tau)$ is either
$(5,5/4),(5/4,5),(2,2)$ or $(9/8,9)$ respectively, by Remark 6. If $OABC$ is a
rhombus, then by [3, Corollary 1], $OABC$ is either the $4\times 4$ square or
the equable rhombus of side length 5. Furthermore, $8\sigma$ and $8\tau$ are
$a^{2}\pm\sqrt{a^{4}-16a^{2}}$, which are also integers by Lemma 3. For the
$4\times 4$ square, this gives $(\sigma,\tau)=(2,2)$. For the rhombus of side
length 5, if one chooses $OB$ to be the longest diagonal, then
$(\sigma,\tau)=(\frac{5}{4},5)$, while if $OB$ is the shortest diagonal, then
$(\sigma,\tau)=(5,\frac{5}{4})$.

######  Lemma 6.

The only possibilities for the unordered pairs $\\{\sigma,\tau\\}$ are
$\\{9,\frac{9}{8}\\}$, $\\{5,\frac{5}{4}\\}$, $\\{3,\frac{3}{2}\\}$ and
$\\{2,2\\}$.

###### Proof.

By Remarks 7 and 10, $\sigma^{\prime}=8\sigma,\tau^{\prime}=8\tau$ are
positive integers and by Lemma 5,
$\sigma^{\prime}\tau^{\prime}=8(\sigma^{\prime}+\tau^{\prime})$ which can be
written as

(21) $(\sigma^{\prime}-8)(\tau^{\prime}-8)=2^{6}.$

The only positive integer solutions of the above equation are then

$\\{\sigma^{\prime},\tau^{\prime}\\}\in\\{\\{9,72\\},\\{10,40\\},\\{12,24\\},\\{16,16\\}\\},$

giving the result announced. ∎

As mentioned in Remark 6, if $OABC$ is a tangential LEQ that is not a rhombus,
then $\lambda=\frac{1}{\sigma}$. So Lemma 6 has the following corollary.

######  Corollary 4.

There are only seven possibilities for the barycentric coordinate parameter
$\lambda$, namely
$\frac{1}{2},\frac{1}{3},\frac{2}{3},\frac{1}{5},\frac{4}{5},\frac{1}{9},\frac{8}{9}$,
corresponding to $\sigma=2,3,\frac{3}{2},5,\frac{5}{4},9,\frac{9}{8}$
respectively.

######  Remark 11.

Consider a reflection in the line $y=x$, followed by a relabelling of the
vertices so they are positively oriented; that is, the vertices $O,A,B,C$ are
permuted to $O,C,B,A$ respectively. It is easy to see that under this
operation, $\sigma$ and $\tau$ are left unchanged, and the side lengths
$a,b,c,d$ are permuted to $d,c,b,a$ respectively.

Notice that for convex tangential LEQs (where we are not concerned about
having the reflex angle at $B$), under the rotation for which the vertices
$O,A,B,C$ are permuted to $A,B,C,O$ respectively, $\sigma$ and $\tau$ are
interchanged, and the side lengths $a,b,c,d$ are permuted to $d,a,b,c$
respectively. So, for the study of convex tangential LEQs, up to Euclidean
motions, we may assume that $\sigma\leq\tau$; that is, $\tau\in\\{2,3,5,9\\}$.

Notice also for convex tangential LEQs, under the rotation for which the
vertices $O,A,B,C$ are permuted to $B,C,O,A$ respectively, $\sigma$ and $\tau$
are also left unchanged, and the side lengths $a,b,c,d$ are permuted to
$c,d,a,b$ respectively. Note that the two permutations
$\sigma_{1}:(a,b,c,d)\mapsto(d,c,b,a)$ and
$\sigma_{2}:(a,b,c,d)\mapsto(c,d,a,b)$ are involutions and their compositions
give a group of order 4, under which each letter can be moved to any of the
four positions. So, for example, without changing $\sigma$ and $\tau$, we may
assume in the convex case that $b$ is the smallest of the side lengths. Note
however that when $\sigma=\tau=2$, this potentially conflicts with the
requirement in Theorem 1 (and in the proof of Corollary 1 which uses Theorem
1) that we also require $a$ and $c$ to be even. As we saw in Remark 9, $a+c$
and $ad$ are even when $\sigma=\tau=2$. So by reflection we may suppose that
$a,c$ to be even. Then by applying $\sigma_{2}$ if necessary, we may assume
that $b\leq d$.

In summary, for convex tangential LEQs we may assume:

  1. (a)

$\tau\in\\{2,3,5,9\\}$,

  2. (b)

$a,c$ are even and $b\leq d$ when $\tau=2$,

  3. (c)

$b$ is the smallest of the side lengths when $\tau\in\\{3,5,9\\}$.

###  4\. Proof of Theorem 1 and Corollary 1

###### Proof of Theorem 1.

Lemma 6 gives 7 possibilities for the ordered pair $(\sigma,\tau)$. Using
$\frac{1}{\tau}+\frac{1}{\sigma}=1$ and $a+c=b+d$, let us restate (18) and
(17):

(22) $\displaystyle K_{A}$ $\displaystyle=a+c+(b-c)\tau,$ (23) $\displaystyle
K_{O}$ $\displaystyle=a+c+(a-b)\frac{\tau}{\tau-1}.$

We now consider the area restrictions:

  1. (a)

As $K_{A}>0$, so (22) gives $a+c+(b-c)\tau>0$, which gives part of (i).

  2. (b)

As $K_{C}>0$ and $K_{C}=2(a+c)-K_{A}$, so (22) gives $a+c-(b-c)\tau>0$, which
gives the other part of (i).

  3. (c)

As $K_{O}>0$, so (23) gives $(a+c)(\tau-1)+(a-b)\tau>0$, which gives (ii).

  4. (d)

We also have $K_{B}\not=0$ as otherwise $ABC$ would be colinear. Thus
$K_{O}\not=2(a+c)$ and (23) gives $(a-b)\frac{\tau}{\tau-1}\not=a+c$, which
gives (iii).

Further, $OABC$ is convex if and only if $K_{O}<2(a+c)$. As we have just seen
in part (d), this occurs when $(a-b)\frac{\tau}{\tau-1}<a+c$; that is, when
$(b+c)\tau>a+c$.

Recall that from Remark 8,

(24) $\displaystyle 16\tau^{2}+((b-c)\tau+(a+c))^{2}$ $\displaystyle=4ab\tau,$
(25) $\displaystyle 16\sigma^{2}+((d-c)\sigma+(a+c))^{2}$
$\displaystyle=4ad\sigma.$

(I). If $\tau\in\\{2,3,5,9\\}$, then (24) gives $16\tau^{2}+(\tau
b+a-(\tau-1)c)^{2}=4\tau ab$. So $\tau b+a-(\tau-1)c$ is even. If $\tau=2$,
then $a,\tau b$ have the same parity since $a$ is even by assumption. If
$\tau\in\\{3,5,9\\}$, then $(\tau-1)c$ is even, so $a,\tau b$ again have the
same parity. Thus $v=\frac{\tau b+a}{2}$ is an integer, and we have
$(2\tau)^{2}+(v-\frac{\tau-1}{2}c)^{2}=\tau ab$. Then since $\tau
ab=v^{2}-u^{2}$, we have (1) as required.

(II). This case is completely analogous to case (I). Let
$\sigma\in\\{3,5,9\\}$, then (25) gives $16\sigma^{2}+(\sigma
d+a-(\sigma-1)c)^{2}=4\sigma ab$. So $\sigma d+a$ is even and $v=\frac{\sigma
d+a}{2}$ is an integer. We have
$(2\sigma)^{2}+(v-\frac{\sigma-1}{2}c)^{2}=\sigma ad$. Then since $\sigma
ad=v^{2}-u^{2}$, we have (2) as required. ∎

###### Proof of Corollary 1.

We use the notation of Theorem 1. By Remark 11, we may assume that
$\tau\in\\{2,3,5,9\\}$, that $a,c$ are even and $b\leq d$ when $\tau=2$, and
that $b$ is the smallest of the side lengths when $\tau\in\\{3,5,9\\}$.

As in the statement of Theorem 1, let $u=\frac{\tau b-a}{2},v=\frac{\tau
b+a}{2}$. Rewriting the conditions (a), (b), (c) of Theorem 1, we have

(26) $\displaystyle(\tau-1)c$ $\displaystyle<2v,$ (27) $\displaystyle 2u$
$\displaystyle<(\tau+1)c,$

and the convexity condition is

(28) $2u>-c(\tau-1).$

So by (27) and (28), we have $-\frac{\tau-1}{2}c<u<\frac{\tau+1}{2}c$. When
$\tau=2$, as $b\leq d$, we have $2b\leq b+d=a+c$, so $u\leq\frac{1}{2}c$. When
$\tau\in\\{3,5,9\\}$, as $b$ is the smallest of the side lengths, we have
$\tau b\leq(\tau-1)c+a$, so $u=(\tau b-a)/2\leq\frac{\tau-1}{2}c$. Thus, in
all cases, we have

(29) $-\frac{\tau-1}{2}c<u\leq\frac{\tau-1}{2}c.$

Assume $\tau=2$. By (29), we have $u^{2}\leq\frac{1}{4}c^{2}$. Thus by Theorem
1, $16+u^{2}=v^{2}-(v-\frac{1}{2}c)^{2}$ gives

$16+\frac{1}{4}c^{2}\geq 16+u^{2}=vc-\frac{1}{4}c^{2},$

from which it follows that

(30) $32\geq c(2v-c).$

From (26), we have $2v>c$. So (30) has only a finite number of solutions.
Indeed, one finds readily there are just 20 such pairs $c,v$ with $c$ even and
$2v>c$ for which (30) holds. For only three of these pairs does the equation
$16+u^{2}=v^{2}-(v-\frac{1}{2}c)^{2}$ have an integer solution for $u$ with
$u+v$ even; these are $(c,v,u)=(4,2,6),(2,1,9),(8,4,6)$, corresponding to the
sides $(a,b,c,d)=(4,4,4,4),(8,5,2,5),(2,5,8,5)$ respectively. The last two
cases correspond to the same LEQ, up to Euclidean motion.

Assume $\tau=3$. By (29), we have $-c<u\leq c$. So $36+u^{2}=v^{2}-(v-c)^{2}$
gives $36+c^{2}\geq 36+u^{2}=2vc-c^{2}$, from which it follows that

(31) $18\geq c(v-c).$

By (26), we have $v>c$, so $c,v-c\in\\{1,\dots,18\\}$. One finds there are
just 58 pairs $c,v$ with $v>c$ for which (30) holds. Of these, there is only
one where the equation $16+u^{2}=v^{2}-(v-c)^{2}$ has an integer solution $u$
for which $v+u\equiv 0\pmod{3}$, and such that for the resulting side lengths
$(a,b,c,d)$, one has $b=\min\\{a,b,c,d\\}$; this is the case
$(c,v,u)=(4,7,2)$, corresponding to the sides $(a,b,c,d)=(5,3,4,6)$.

Assume $\tau=5$. By (29), we have $-2c<u\leq 2c$. So
$100+u^{2}=v^{2}-(v-2c)^{2}$ gives $100+4c^{2}\geq 100+u^{2}=4vc-4c^{2}$, from
which it follows that

(32) $25\geq c(v-2c).$

By (26), we have $v>2c$, so $c,v-2c\in\\{1,\dots,25\\}$. One finds there are
just 86 pairs $c,v$ with $v>2c$ for which (32) holds. Of these, one finds
there is only one where the equation $100+u^{2}=v^{2}-(v-2c)^{2}$ has an
integer solution $u$ for which $(v+u)/5$ is an integer, and such that for the
resulting side lengths $(a,b,c,d)$, one has $b=\min\\{a,b,c,d\\}$; this is the
case $(c,v,u)=(5,15,10)$, corresponding to the sides $(a,b,c,d)=(5,5,5,5)$.

Assume $\tau=9$. By (29), we have $-4c<u\leq 4c$. So
$324+u^{2}=v^{2}-(v-4c)^{2}$ gives $324+16c^{2}\geq 324+u^{2}=8vc-16c^{2}$,
from which it follows that

(33) $41\geq c(v-4c).$

By (26), we have $v>4c$, so $c,v-4c\in\\{1,\dots,41\\}$. One finds there are
979 pairs $c,v$ with $v>4c$ for which (33) holds. Of these, one finds there is
only two where the equation $324+u^{2}=v^{2}-(v-4c)^{2}$ has an integer
solution $u$ for which $(v+u)/9$ is an integer, and such that for the
resulting side lengths $(a,b,c,d)$, one has $b=\min\\{a,b,c,d\\}$; these are
the cases $(c,v,u)=(3,21,6)$ and $(5,23,1)$, corresponding respectively to the
sides $(a,b,c,d)=(15,3,3,15)$ and $(37,1,5,41)$.

This completes the proof of the corollary. ∎

###  5\. Proof of Theorem 2

We follow the general strategy used in [40], but in our case we employ a
slightly different solution form for the diophantine equations that appear in
the statement of Theorem 2.

######  Lemma 7.

Suppose $z^{2}+w^{2}+u^{2}=v^{2}$ for integers $u,v,w,z$ and that the prime
decomposition of $\gcd(u,v,w,z)$ contains no term $\rho^{k}$ where $\rho$ is
congruent to $3$ modulo 4 and $k$ is odd. Then there are integers $p,q,m,n$
such that

$v-u=p^{2}+q^{2},\quad v+u=m^{2}+n^{2},\quad w=pm+qn,\quad z=pn-qm.$

Numbers $u,v,w,z$ for which $z^{2}+w^{2}+u^{2}=v^{2}$ are said to form a
_Pythagorean quadruple_ , and of course their study has a long history; see
[33]. The above lemma is essentially equivalent to a classical result which
says that if $z^{2}+w^{2}+u^{2}=v^{2}$ for integers $u,v,w,z$ with
$\gcd(u,v,w,z)=1$, then supposing $z,w$ are even, there are integers $p,q,m,n$
such that

$v-u=2(p^{2}+q^{2}),\quad v+u=2(m^{2}+n^{2}),\quad w=2(pm+qn),\quad z=2(pn-
qm).$

This result, sometimes attributed to V. A. Lebesgue, is proved in many places;
see [11, pp. 28-37], [14], [30, p.14] and [36]. We require the slightly
stronger formulation of Lemma 7, which is readily deduced from the treatment
given in [13, Section II].

We will also make use of a certain elementary fact which we give in the
following lemma. For convenience, let us make a definition.

######  Definition 4.

We say that a positive integer $k$ has the _lattice preservation property_ ,
or is a _lattice preserver_ , if for every lattice point $X$ for which
$\frac{1}{k}X$ has integer length, the point $\frac{1}{k}X$ is also a lattice
point.

For example, it is easy to see that $2$ and $3$ are lattice preservers. Notice
that the set of lattice preservers is closed under multiplication. Hence, for
example, $4$ and $6$ are lattice preservers. Recall that a _hypotenuse number_
is a positive integer that occurs as the length of the hypotenuse of some
Pythagorean triangle. It is well known that hypotenuse numbers are those
numbers that have a prime factor congruent to 1 modulo 4 [31].

######  Lemma 8.

A positive integer $k$ is a lattice preserver if and only if $k$ is not a
hypotenuse number. So $k$ is a lattice preserver if and only if $k$ has no
prime factor congruent to 1 modulo 4.

###### Proof.

If $k$ is a hypotenuse number, say $k^{2}=x^{2}+y^{2}$, then
$\frac{1}{k}(x,y)$ has length 1 but it not a lattice point. So hypotenuse
numbers are not lattice preservers. Conversely, if $k$ is not a lattice
preserver (so $k>2$), then there exists a lattice point $(x,y)$ such that
$(x,y)/k$ has integer length, $a$ say, but is not a lattice point. We may
assume without loss of generality that $\gcd(x,y,k)=1$. We have
$x^{2}+y^{2}=k^{2}a^{2}$. Write $x^{\prime 2}+y^{\prime 2}=k^{2}a^{\prime 2}$
where $x^{\prime}=x/\gcd(x,y,a)$, etc. So
$\gcd(x^{\prime},y^{\prime},ka^{\prime})=1$ and hence
$x^{\prime},y^{\prime},ka^{\prime}$ is a primitive Pythagorean triple. So by
[31, Theorem 3.20] for example, all the odd prime factors of $k$ are congruent
to 1 mod 4 and $k$ is not divisible by 4. So as $k>2$, we conclude that $k$
has at least one prime factor congruent to 1 mod 4 and so $k$ is a hypotenuse
number. ∎

Recall that by Remark 9, when working with tangential LEQs we may suppose
without loss of generality that $c$ is even when $\sigma=\tau=2$ and that $c$
is not divisible by 3 when $\sigma$ or $\tau$ equals 3.

###### Proof of Theorem 2.

(I). Suppose $t\in\\{2,3,5,9\\}$. Notice that $a+c=b+d$ and so from hypothesis
(i), $(d-a)t>-(a+c)$. Adding hypothesis (ii) gives $2dt>0$, so $d>0$.
Furthermore, (3) gives $(v+u)(v-u)=v^{2}-u^{2}>0$, so as
$v=(a+tb)/2>c(t-1)/2>0$ by condition (i) of our hypotheses, $v+u$ and $v-u$
are both necessarily positive. That is, $a,b>0$. So, in all cases, $a,b,c,d$
are all positive.

The basic idea of the proof is to apply Lemma 7 to obtain integers $p,q,m,n$
such that

(34) $a=p^{2}+q^{2},\ tb=m^{2}+n^{2},\ pm+qn=v-\frac{t-1}{2}c,\ pn-qm=-2t.$

Then we consider the Gaussian integers $z:=p+qi,\ w:=m+ni$, and let

(35) $A=z^{2},\qquad B=z^{2}-\frac{1}{t}w^{2},\qquad
C=\frac{1}{t(t-1)}(tz-w)^{2}.$

We call this the _general case_. Unfortunately, as we will see below, this
procedure is not always possible, and we will require two variations on this
approach.

(Let us explain, in parenthesis, how the proposal of vertices of (35) can be
understood. Obviously, $A,B$ are suggested by (34). For a tangential LEQ, the
areas $K_{O},K_{A}$ are determined by $\sigma,\tau$ and the side lengths, by
Remark 10. Then (7) enables one to express $C$ in terms of $A$ and $B$. This
gives a formula for $C$ that must hold if this construction is to produce a
tangential LEQ. We suppress this derivation of the formula for $C$, and focus
on showing that it has the required properties).

First suppose that $t=2$. Then (3) is
$16+u^{2}=v^{2}-\left(v-\frac{1}{2}c\right)^{2}$. Clearly $\gcd(4,u,v,c)$ is
either 1, 2 or 4, so we may apply Lemma 7, and obtain the general case of (34)
and (35).

Now suppose that $t=3$. Then (3) is $6^{2}+u^{2}=v^{2}-\left(v-c\right)^{2}$.
As $c$ is not divisible by $3$ by assumption, $\gcd(6,u,v,c)$ is $1$ or $2$,
and we may again apply Lemma 7 and obtain the general case of (34) and (35).

Now suppose that $t=5$. Then (3) is
$10^{2}+u^{2}=v^{2}-\left(v-2c\right)^{2}$, and as $\gcd(10,u,v,c)$ is $1,2,5$
or $10$, we could apply Lemma 7 in all cases. In fact, for reasons that will
become apparent later in the proof, we will directly apply Lemma 7, and obtain
the general case, only in the cases where $u,v,c$ are not all divisible by 5,
so $\gcd(10,u,v,c)=1$ or $2$. Note that $u,v,c$ are all divisible by 5
precisely when $a$ and $c$ are divisible by 5. In this case, let
$\frac{u}{5}=u^{\prime},\frac{v}{5}=v^{\prime},\frac{c}{5}=c^{\prime}$. Thus
(3) can be written as $4+u^{\prime 2}=v^{\prime
2}-\left(v^{\prime}-2c^{\prime}\right)^{2}$, and applying Lemma 7, we have
integers $p,q,m,n$ such that

(36) $a=5(p^{2}+q^{2}),\ b=m^{2}+n^{2},\ pm+qn=\frac{1}{5}(v-2c),\ pn-qm=-2.$

Then let $z:=p+qi,\ w:=m+ni$, and set

(37) $A=5z^{2},\qquad B=5z^{2}-w^{2},\qquad C=\frac{1}{4}(5z-w)^{2}.$

We call this the _first exceptional case_.

Now suppose that $t=9$. Then (3) is
$18^{2}+u^{2}=v^{2}-\left(v-4c\right)^{2}$. If $\gcd(18,u,v,c)$ is not $3$ or
$6$, we may apply Lemma 7 and obtain the general case. If instead
$\gcd(18,u,v,c)$ is $3$ or $6$, which occurs when $\gcd(a,c)$ is divisible by
3 but not 9, let
$\frac{u}{3}=u^{\prime},\frac{v}{3}=v^{\prime},\frac{c}{3}=c^{\prime}$. Thus
(3) can be written as $36+u^{\prime 2}=v^{\prime
2}-\left(v^{\prime}-4c^{\prime}\right)^{2}$, and applying Lemma 7, we have
integers $p,q,m,n$ such that

(38) $a=3(p^{2}+q^{2}),\ 3b=m^{2}+n^{2},\ pm+qn=\frac{1}{3}(v-4c),\ pn-qm=-6.$

Then let $z:=p+qi,\ w:=m+ni$, and set

(39) $A=3z^{2},\qquad B=3z^{2}-\frac{1}{3}w^{2},\qquad
C=\frac{1}{24}(9z-w)^{2}.$

We call this the _second exceptional case_.

We now proceed to show that the points $O,A,B,C$ define a tangential LEQ
$OABC$ with successive side lengths $a,b,c,d$ for which
$(\sigma,\tau)=(\frac{t}{t-1},t)$. We first treat the general case of (34) and
(35), and deal with the two exceptional cases later. So we are assuming that
for $t=5$, the integers $u,v,c$ are not all divisible by 5, and for $t=9$, we
have that $\gcd(18,u,v,c)$ is not $3$ or $6$.

Note that from (34), $OA$ has length $p^{2}+q^{2}=a$ and
$A-B=\frac{1}{t}w^{2}$, which has length $\frac{1}{t}(m^{2}+n^{2})=b$. It
remains to see that:

  1. (a)

$C-B$ has length $c$, and $OC$ has length $d$,

  2. (b)

the quadrilateral $OABC$ has no self-intersections,

  3. (c)

$OABC$ is equable,

  4. (d)

the points $A,B,C$ are not colinear,

  5. (e)

$B$ is the only point at which the angle may be reflex,

  6. (f)

for $OABC$, one has $\tau=t$,

  7. (g)

$B$ and $C$ are lattice points.

Let us make some preliminary calculations. Substituting $z=p+qi,w=m+ni$ and
using (34), one has

(40) $\displaystyle z\bar{w}-\bar{z}w$ $\displaystyle=2(qm-pn)i=4ti,$ (41)
$\displaystyle z\bar{w}+\bar{z}w$
$\displaystyle=2(pm+qn)=2v-(t-1)c=a+c+t(b-c).$

Consequently,

(42)
$z^{2}\bar{w}^{2}-\bar{z}^{2}w^{2}=(z\bar{w}-\bar{z}w)(z\bar{w}+\bar{z}w)=4t(a+c+t(b-c))i.$

Now consider the signed areas $K_{O},K_{A},K_{C},K_{B}$. Recall that if $Z,W$
are points in the complex plane, the triangle $OZW$ has signed area
$i(Z\bar{W}-\bar{Z}W)/4$. Using (42), we have
$4tK_{A}=i(-z^{2}\bar{w}^{2}+\bar{z}^{2}w^{2})=4t(a+c+t(b-c))$, so

(43) $K_{A}=a+tb-(t-1)c.$

Using $z\bar{z}=a,w\bar{w}=tb$ and (40), (41), one has

$\displaystyle 4(t-1)t^{2}K_{C}$
$\displaystyle=i((tz^{2}-w^{2})(\overline{tz-w})^{2}-(tz-w)^{2}\overline{(tz^{2}-w^{2})})$
$\displaystyle=it(z\bar{w}-\bar{z}w)(-2t(a+b)+(t+1)(z\bar{w}+\bar{z}w))$
$\displaystyle=-4t^{2}(-2t(a+b)+(t+1)(a+c+t(b-c))$
$\displaystyle=4t^{2}(t-1)(a-tb+(t+1)c),$

so

(44) $K_{C}=a-tb+(t+1)c.$

Using $z\bar{z}=a$ and (40), (41), one has

$\displaystyle 4t(t-1)K_{O}$
$\displaystyle=i(z^{2}(\overline{tz-w})^{2}-(tz-w)^{2}\bar{z}^{2})=i(z\bar{w}-\bar{z}w)(z\bar{w}+\bar{z}w-2ta)$
$\displaystyle=-4t(a+c+t(b-c)-2ta))=4t((2t-1)a-tb+(t-1)c),$

so

(45) $K_{O}=\frac{1}{t-1}((2t-1)a-tb+(t-1)c).$

Before calculating $K_{B}$, note that
$t(t-1)(C-B)=(tz-w)^{2}-(t-1)(tz^{2}-w^{2})=t(z-w)^{2}$, so

(46) $C-B=\frac{1}{t-1}(z-w)^{2}.$

Thus, using $w\bar{w}=tb$ and (40), (41), one has

$\displaystyle 4t(t-1)K_{B}$
$\displaystyle=i((z-w)^{2}\bar{w}^{2}-\overline{(z-w)^{2}}w^{2})=i(-z\bar{w}+\bar{z}w)(2tb-z\bar{w}-\bar{z}w)$
$\displaystyle=4t(2tb-(a+c+t(b-c)))=4t(-a+tb+(t-1)c),$

so

(47) $K_{B}=\frac{1}{t-1}(-a+tb+(t-1)c).$

We now prove the requirements (a) – (g).

(a). From (46), we have, using (41),

$\displaystyle(t-1)\|C-B\|$
$\displaystyle=\|z-w\|^{2}=z\bar{z}+w\bar{w}-z\bar{w}-\bar{z}w$
$\displaystyle=a+tb-(a+c+t(b-c))=(t-1)c,$

as required. Using (41) again, we also have

$\displaystyle t(t-1)$
$\displaystyle\|C\|=\|tz-w\|^{2}=t^{2}z\bar{z}+w\bar{w}-tz\bar{w}-t\bar{z}w=t^{2}a+tb-t(z\bar{w}+\bar{z}w)$
$\displaystyle=t^{2}a+tb-t(a+tb-(t-1)c)$
$\displaystyle=(t^{2}-t)a-(t^{2}-t)b+t(t-1)c=t(t-1)d,$

as required.

(b). To verify that the quadrilateral $OABC$ has no self-intersections, it
suffices to show that the respective signed areas $K_{A},K_{C}$ of triangles
$OAB,OBC$ are both positive. The hypothesis (i) gives $tb+a-(t-1)c>0$, so
$K_{A}>0$ by (43). Hypothesis (i) also gives $a-tb+(t+1)c>0$, so $K_{C}>0$ by
(44).

(c). From (43),(44), we have $K_{A}+K_{C}=2(a+c)$, as required.

(d). To verify that the points $A,B,C$ are not colinear, it suffices to show
that $K_{B}\not=0$. But by (47), $(t-1)K_{B}=-a+tb+(t-1)c)\not=0$, by
hypothesis (iii).

(e). To see that $B$ is the only point at which the angle may be reflex, it
remains to show that $K_{O}>0$. The hypothesis (ii) gives $(2a+c-b)t>a+c$,
from which we have $(2t-1)a-tb+(t-1)c>0$, so $K_{O}>0$ by (45).

(f). If $a\not=d$, then (18),(43) give
$\tau=\frac{K_{A}-(a+c)}{a-d}=\frac{a+tb-(t-1)c-a-c}{a-d}=\frac{t(b-c)}{a-d}=t$.
Thus by Lemma 6, $\sigma=\frac{t}{t-1}$.

If $a\not=b$, then (17),(45) give
$\sigma=\frac{K_{O}-(a+c)}{a-b}=\frac{(2t-1)a-tb+(t-1)c-(t-1)(a+c)}{(t-1)(a-b)}=\frac{ta-
tb}{(t-1)(a-b)}=\frac{t}{t-1}$. Thus by Lemma 6, $\tau=t$.

Finally, if $a=d$ and $a=b$, then $OABC$ is a rhombus and $a=b=c=d$. But if
$a=b$, then $u=\frac{t-1}{2}a,v=\frac{t+1}{2}a$, and so by (3), $a=c$ would
give $(2t)^{2}+\frac{(t-1)^{2}}{4}a^{2}=\frac{(t+1)^{2}}{4}a^{2}-a^{2}$, so
$a^{2}=\frac{4t^{2}}{t-1}$. For
$t=\frac{9}{8},\frac{5}{4},\frac{3}{2},2,3,5,9$, this would give respectively
$a^{2}=\frac{81}{2},25,18,16,18,25$, which is impossible for
$t=\frac{9}{8},\frac{3}{2},3,9$. For $t=2$ we have $a=4$, so $OABC$ is the
$4\times 4$ square, which has $(\sigma,\tau)=(2,2)$, by Remark 10. For $t=5$
and $\frac{5}{4}$, we have $a=5$, so $OABC$ is the rhombus of side length 5,
which has $(\sigma,\tau)=(\frac{5}{4},5)$ and $(5,\frac{5}{4})$ respectively,
again by Remark 10.

(g). We now come to the most delicate part of the proof. Note that parts
(a)-(e) were simply equations or inequalities, and did not use the values of
$t$, or the fact that certain variables are integers. Part (f) did use these
facts, but only in a very simple manner.

First suppose $t=2$. So $B=z^{2}-\frac{1}{2}w^{2}$ and
$C=\frac{1}{2}(2z-w)^{2}$. Now $z^{2},w^{2}$ are lattice points. And from
above, $\frac{1}{2}w^{2}$ has integer length $b$. So by Lemma 8,
$\frac{1}{2}w^{2}$ is a lattice point. Thus $B$ is a lattice point. Similarly,
$(2z-w)^{2}$ is a lattice point and $C=\frac{1}{2}(2z-w)^{2}$ has integer
length $c$, so by Lemma 8, $C$ is a lattice point.

Now suppose $t=3$. So $B=z^{2}-\frac{1}{3}w^{2}$ and
$C=\frac{1}{6}(3z-w)^{2}$. Now $z^{2},w^{2}$ are lattice points. And from
above, $\frac{1}{3}w^{2}$ has integer length $b$. So by Lemma 8,
$\frac{1}{3}w^{2}$ is a lattice point. Thus $B$ is a lattice point. Similarly,
$(3z-w)^{2}$ is a lattice point and $C=\frac{1}{6}(3z-w)^{2}$ has integer
length $c$, so by Lemma 8, $C$ is a lattice point.

Now suppose $t=5$. So $B=z^{2}-\frac{1}{5}w^{2}$ and
$C=\frac{1}{20}(5z-w)^{2}$, where $w=m+ni$. We claim that, in the general
case, $m,n$ are multiples of 5. First note that (3) can be written as
$10^{2}+(v-2c)^{2}=(v+u)(v-u)$. So as $5$ divides $v+u=5b$, it follows that
$v-2c\equiv 0\pmod{5}$. Hence from (34),

$ma=m(p^{2}+q^{2})=p(pm+qn)-q(pn-qm)=p(v-2c)+10q\equiv 0\pmod{5}.$

Similarly, $na\equiv 0\pmod{5}$. So if $a\not\equiv 0\pmod{5}$, we have
$m,n\equiv 0$ as required. If $a\equiv 0\pmod{5}$, then as
$v-2c=\frac{5b+a-4c}{2}=\frac{5(b-c)+a+c}{2}$, and $v-2c\equiv 0\pmod{5}$, so
$5$ divides $a+c$, and thus $c\equiv 0\pmod{5}$ and hence $v\equiv 0\pmod{5}$.
So, as $v+u=5b$, we have that $5$ divides $u,v,c$. But this is the first
exceptional case, contrary to our current assumption.

As $m,n$ are multiples of 5, let
$m=5m^{\prime},n=5n^{\prime},w^{\prime}=m^{\prime}+n^{\prime}i$. So
$B=z^{2}-5w^{\prime 2}$, which is obviously a lattice point, and
$C=\frac{5}{4}(z-w^{\prime})^{2}$, which is a lattice point by Lemma 8.

Now suppose $t=9$. So $B=z^{2}-\frac{1}{9}w^{2}$ and
$C=\frac{1}{72}(9z-w)^{2}$. Now $z^{2},w^{2}$ are lattice points. And from
above, $\frac{1}{9}w^{2}$ has integer length $b$. So by Lemma 8,
$\frac{1}{9}w^{2}$ is a lattice point. Thus $B$ is a lattice point. Similarly,
$(9z-w)^{2}$ is a lattice point and $C=\frac{1}{72}(3z-w)^{2}$ has integer
length $d$. Hence, as $72=2^{3}3^{2}$ is a lattice preserver, $C$ is a lattice
point by Lemma 8. This completes part (g).

We now treat the first exceptional case. So $t=5$ and $u,v,c$ are all
divisible by 5. In the preliminary calculations part of the argument,
analogous to (40), (41) and (42), one has:

(48) $\displaystyle z\bar{w}-\bar{z}w$ $\displaystyle=4i,$ (49) $\displaystyle
z\bar{w}+\bar{z}w$ $\displaystyle=\frac{1}{5}(a+c)+(b-c),$ (50) $\displaystyle
z^{2}\bar{w}^{2}-\bar{z}^{2}w^{2}$ $\displaystyle=(\frac{4}{5}(a+c)+4(b-c))i.$

For the areas, using the above three expressions and
$z\bar{z}=a/5,w\bar{w}=b$, we find exactly the same formulas for
$K_{O},K_{A},K_{B},K_{C}$ as before; that is, we obtain (45),(43),(47),(44)
respectively with $t=5$.

Analogous to (46), one has

(51) $C-B=\frac{5}{4}(z-w)^{2}.$

For the proof of part (a), we have from (51), using (49),

$\displaystyle 4\|C-B\|$
$\displaystyle=5\|z-w\|^{2}=5(z\bar{z}+w\bar{w}-z\bar{w}-\bar{z}w)$
$\displaystyle=a+5b-(a+c+5(b-c))=4c,$

as required. Using (49) again, we also have

$\displaystyle 4$
$\displaystyle\|C\|=\|5z-w\|^{2}=5^{2}z\bar{z}+w\bar{w}-5z\bar{w}-5\bar{z}w=5a+b-5(z\bar{w}+\bar{z}w)$
$\displaystyle=5a+b-((a+c)+5(b-c))=4a-4b+4c=4d,$

as required.

As parts (b)-(f) only rely on the expressions for $K_{O},K_{A},K_{B},K_{C}$,
and as these are unchanged, the proofs of these parts need no amendment. It
remains to verify part (g). But $B=5z^{2}-w^{2}$, which is obviously a lattice
point, and $C=\frac{1}{4}(5z-w)^{2}$, which is a lattice point by Lemma 8.

Finally, we treat the second exceptional case. So $t=9$ and $\gcd(18,u,v,c)$
is $3$ or $6$. In the preliminary calculations part of the argument, analogous
to (40), (41) and (42), one has:

(52) $\displaystyle z\bar{w}-\bar{z}w$ $\displaystyle=12i,$ (53)
$\displaystyle z\bar{w}+\bar{z}w$ $\displaystyle=\frac{1}{3}(a+9b-8c),$ (54)
$\displaystyle z^{2}\bar{w}^{2}-\bar{z}^{2}w^{2}$ $\displaystyle=4(a+9b-8c)i.$

For the areas, using the above three expressions and
$z\bar{z}=a/3,w\bar{w}=3b$, we find the exactly same formulas for
$K_{O},K_{A},K_{B},K_{C}$ as before; that is, we obtain (45),(43),(47),(44)
respectively with $t=9$.

Analogous to (46), one has

(55) $C-B=\frac{3}{8}(z-w)^{2}.$

For the proof of part (a), we have from (55), using (53),

$\displaystyle 8\|C-B\|$
$\displaystyle=3\|z-w\|^{2}=3(z\bar{z}+w\bar{w}-z\bar{w}-\bar{z}w)$
$\displaystyle=3(a/3+3b-(a+9b-8c)/3)=a+9b-(a+9b-8c)=8c,$

as required. Using (53) again, we also have

$\displaystyle 24$
$\displaystyle\|C\|=\|9z-w\|^{2}=9^{2}z\bar{z}+w\bar{w}-9z\bar{w}-9\bar{z}w=27a+3b-9(z\bar{w}+\bar{z}w)$
$\displaystyle=27a+3b-3(a+9b-8c)=24a-24b+24c=24d,$

as required.

As parts (b)-(f) only rely on the expressions for $K_{O},K_{A},K_{B},K_{C}$,
and as these are unchanged, the proofs of these parts need no amendment. It
remains to verify part (g). Now $A,z,w$ are lattice points. Thus, as
$A-B=\frac{1}{3}w^{2}$ is a lattice point by Lemma 8, so $B$ is a lattice
point. Finally, $C=\frac{1}{24}(9z-w)^{2}$ is a lattice point by Lemma 8 since
$24=2^{3}3$ is a lattice preserver.

This completes the proof of part (I).

(II). Suppose $s\in\\{3,5,9\\}$. By hypothesis, $b$ and $c$ are positive.
Further, Equation (3) gives $(v+u)(v-u)=v^{2}-u^{2}>0$, so as $v$ is positive
by condition (i) of our hypotheses, $v+u$ and $v-u$ are both necessarily
positive. That is, $a,d>0$. So, in all cases, $a,b,c,d$ are all positive.

We now proceed exactly as we did in case (I) by considering a general case and
two exceptional cases. The first exceptional case is where $s=5$ and the
integers $u,v,c$ are all divisible by 5. The second exceptional case is where
$s=9$ and $\gcd(18,u,v,c)$ is either $3$ or $6$. In the general case, we apply
Lemma 7 to obtain integers $p,q,m,n$ such that

(56) $a=p^{2}+q^{2},\ sd=m^{2}+n^{2},\ pm+qn=v-\frac{s-1}{2}c,\ pn-qm=2s.$

Then we consider the Gaussian integers $z:=p+qi,\ y:=m+ni$, and let

(57) $A=z^{2},\qquad B=z^{2}-\frac{1}{s(s-1)}(sz-y)^{2},\qquad
C=\frac{1}{s}y^{2}.$

Note that $A$ is a lattice point, and $OA$ has length $a$, while $OC$ has
length $d$ (we will see later that $C$ is a lattice point). For
$t=\frac{s}{s-1}$, let

$w:=\frac{sz-y}{s-1}.$

Notice that $w$ has length (i.e., norm) given by

$\displaystyle(s-1)^{2}\|w||^{2}$
$\displaystyle=s^{2}a+sd-s(z\bar{y}+\bar{z}y)=s^{2}a+sd-s(2v-(s-1)c)$
$\displaystyle=s^{2}a+sd-s(sd+a-(s-1)c)=s(s-1)(a-d+c)=s(s-1)b.$

So $\frac{s-1}{s}w^{2}$ has length $b$. Observe that for $t=\frac{s}{s-1}$, we
have $B=z^{2}-\frac{1}{t}w^{2}$ and $A-B=\frac{1}{t}w^{2}$ has length $b$,
exactly as in case (I). Moreover, we can write

$C=\frac{1}{s}(sz-(s-1)w)^{2}=\frac{(s-1)^{2}}{s}\left(\frac{s}{s-1}z-w\right)^{2}=\frac{1}{t(t-1)}(tz-w)^{2},$

which is the same formula as appeared in (35), in case (I). Now compute:

$\displaystyle z\bar{w}-\bar{z}w$
$\displaystyle=\frac{-1}{s-1}(z\bar{y}-\bar{z}y)=\frac{-1}{s-1}2(qm-
pn)i=\frac{s}{s-1}4i=4ti,$ $\displaystyle z\bar{w}+\bar{z}w$
$\displaystyle=\frac{1}{s-1}(z\overline{(sz-y)}+\bar{z}(sz-y))=\frac{1}{s-1}(2sa-z\bar{y}-\bar{z}y)$
$\displaystyle=\frac{1}{s-1}(2sa-(2v-(s-1)c))=\frac{1}{s-1}(2sa-sd-a+(s-1)c)$
$\displaystyle=a+c+\frac{s}{s-1}(a-d)=a+c+t(a-d)=a+c+t(b-c),$ $\displaystyle
z^{2}\bar{w}^{2}-\bar{z}^{2}w^{2}$
$\displaystyle=(z\bar{w}-\bar{z}w)(z\bar{w}+\bar{z}w)=4t(a+c+t(b-c))i,$

which are exactly as in Equations (40), (41), (41) of case (I). It follows
that the areas $K_{O},K_{A},K_{B},K_{C}$ are given by the case (I) formulas
(45),(43),(47),(44). Thus the proof of parts (a)-(f) hold by the arguments
used in case (I) and it remains to verify part (g).

Let us first suppose $s=5$. So $C=\frac{1}{5}y^{2}$, where $w=m+ni$. We claim
that $m,n$ are multiples of 5. First note that (3) can be written as
$10^{2}+(v-2c)^{2}=(v+u)(v-u)$. So as $5$ divides $v+u=5d$, it follows that
$v-2c\equiv 0\pmod{5}$. Hence from (34),

$ma=m(p^{2}+q^{2})=p(pm+qn)-q(pn-qm)=p(v-2c)-10q\equiv 0\pmod{5}.$

Similarly, $na\equiv 0\pmod{5}$. So if $a\not\equiv 0\pmod{5}$, we have
$m,n\equiv 0$ as required. If $a\equiv 0\pmod{5}$, then as
$v-2c=\frac{5d+a-4c}{2}=\frac{5(d-c)+a+c}{2}$, and $v-2c\equiv 0\pmod{5}$, so
$5$ divides $a+c$, and thus $c\equiv 0\pmod{5}$ and hence $v\equiv 0\pmod{5}$.
So, as $v+u=5d$, we have that $5$ divides $u,v,c$. But this is the first
exceptional case, contrary to our current assumption.

As $m,n$ are multiples of 5, let
$m=5m^{\prime},n=5n^{\prime},y^{\prime}=m^{\prime}+n^{\prime}i$. So
$C=5y^{\prime 2}$, which is obviously a lattice point. Furthermore, for $s=5$,
one has from (57) that

$A-B=\frac{1}{20}5(z-y^{\prime})^{2}=\frac{1}{4}(z-y^{\prime})^{2}.$

So as $A-B$ has integer length $b$, and $(z-y^{\prime})^{2}$ is a lattice, so
$A-B$ is a lattice point by Lemma 8, and hence $B$ is a lattice point.

Now suppose $s=3$ or $9$. As $y$ is a lattice point and $C=\frac{1}{s}y^{2}$
has length $d$, so $C$ is a lattice point when $s=3$ or $9$, by Lemma 8. For
$s=3$, we have $A-B=\frac{1}{6}(3z-y)^{2}$ and for $s=9$, we have
$A-B=\frac{1}{72}(9z-y)^{2}$. In both cases, $A-B$ is a lattice point by Lemma
8, and hence $B$ is a lattice point.

We now treat the first exceptional case. So $t=5$ and $u,v,c$ are all
divisible by 5. Let
$\frac{u}{5}=u^{\prime},\frac{v}{5}=v^{\prime},\frac{c}{5}=c^{\prime}$. Thus
(3) can be written as $4+u^{\prime 2}=v^{\prime
2}-\left(v^{\prime}-2c^{\prime}\right)^{2}$, and applying Lemma 7, we have
integers $p,q,m,n$ such that

(58) $a=5(p^{2}+q^{2}),\ d=m^{2}+n^{2},\ pm+qn=\frac{1}{5}(v-2c),\ pn-qm=2.$

Then let $z:=p+qi,\ y:=m+ni$, and set

(59) $A=5z^{2},\qquad B=5z^{2}-\frac{1}{4}(5z-y)^{2},\qquad C=y^{2}.$

We remark that $B$ has other useful expressions:

$B=y^{2}-\frac{5}{4}(z-y)^{2}=\frac{1}{4}(-y^{2}+10yz-5z^{2}).$

Note that $A$ has length $a$ and $C$ has length $d$. Moreover,
$A-B=\frac{1}{4}(5z-y)^{2}$ and has length

$\displaystyle\frac{1}{4}(5z-y)\overline{(5z-y)}$
$\displaystyle=\frac{1}{4}(5a+d-5(z\bar{y}+\bar{z}y))=\frac{1}{4}(5a+d-(5d+a-4c))$
$\displaystyle=\frac{1}{4}(4a-4d+4c)=b,$

and $C-B=\frac{5}{4}(z-y)^{2}$ has length

$\displaystyle\frac{5}{4}(z-y)\overline{(z-y)}$
$\displaystyle=\frac{5}{4}(\frac{1}{5}a+d-(z\bar{y}+\bar{z}y))=\frac{1}{4}(a+5d-(5d+a-4c))=c.$

So part (a) of the requirements (a)-(g) is satisfied. Moreover, $A=5z^{2}$ and
$C=y^{2}$ are obviously lattice points, and $-y^{2}+10yz-5z^{2}$ is a lattice
point and $B=\frac{1}{4}(-y^{2}+10yz-5z^{2})$ has integer length $b$, so $B$
is a lattice point by Lemma 8. So part (g) of the requirements (a)-(g) is
satisfied.

For our preliminary calculations we will use $z$ and $y$, rather than $z$ and
$w$ as we did before. We have

$\displaystyle z\bar{y}-\bar{z}y$ $\displaystyle=-2(pn-qm)i=-4i,$
$\displaystyle z\bar{y}+\bar{z}y$
$\displaystyle=\frac{1}{5}(5d+a-4c)=\frac{1}{5}(6a-5b+c),$ $\displaystyle
z^{2}\bar{y}^{2}-\bar{z}^{2}y^{2}$ $\displaystyle=\frac{-4}{5}(6a-5b+c)i.$

These relations are also different to the Equations (48), (49), (50) we
obtained in the first exceptional case of case (I). Nevertheless, using the
above three relations together with (59), we find exactly the same formulas
for $K_{O},K_{A},K_{B},K_{C}$ as before; that is, we obtain
(45),(43),(47),(44) respectively with $t=\frac{5}{4}$. As parts (b)-(f) only
rely on the expressions for $K_{O},K_{A},K_{B},K_{C}$, and as these are
unchanged, the proofs of these parts need no amendment.

Finally, we treat the second exceptional case. So $s=9$ and $\gcd(18,u,v,c)$
is $3$ or $6$. Let
$\frac{u}{3}=u^{\prime},\frac{v}{3}=v^{\prime},\frac{c}{3}=c^{\prime}$. Thus
(3) can be written as

$36+u^{\prime 2}=v^{\prime 2}-\left(v^{\prime}-4c^{\prime}\right)^{2},$

and applying Lemma 7, we have integers $p,q,m,n$ such that

(60) $a=3(p^{2}+q^{2}),\ 3d=m^{2}+n^{2},\ pm+qn=\frac{1}{3}(v-4c),\ pn-qm=6.$

Then let $z:=p+qi,\ y:=m+ni$, and set

(61) $A=3z^{2},\qquad B=3z^{2}-\frac{1}{24}(9z-y)^{2},\qquad
C=\frac{1}{3}y^{2}.$

Observe that

$B=\frac{1}{3}y^{2}-\frac{3}{8}(z-y)^{2}=\frac{1}{24}(-y^{2}+18yz-9z^{2}).$

Note that $A$ has length $a$ and $C$ has length $d$. Moreover,
$A-B=\frac{1}{24}(9z-y)^{2}$ and has length

$\displaystyle\frac{1}{24}(9z-y)\overline{(9z-y)}$
$\displaystyle=\frac{1}{24}(27a+3d-9(z\bar{y}+\bar{z}y))=\frac{1}{8}(9a+d-(9d+a-8c))$
$\displaystyle=\frac{1}{8}(8a-8d+8c)=b,$

and $C-B=\frac{3}{8}(z-y)^{2}$ has length

$\displaystyle\frac{3}{8}(z-y)\overline{(z-y)}$
$\displaystyle=\frac{3}{8}(\frac{1}{3}a+3d-(z\bar{y}+\bar{z}y))=\frac{1}{8}(a+9d-(9d+a-8c))=c.$

So part (a) of the requirements (a)-(g) is satisfied. Obviously $A$ is a
lattice point, and $B,C$ are lattice points by Lemma 8. So part (g) of the
requirements (a)-(g) is satisfied.

We have

$\displaystyle z\bar{y}-\bar{z}y$ $\displaystyle=-2(pn-qm)i=-12i,$
$\displaystyle z\bar{y}+\bar{z}y$
$\displaystyle=\frac{1}{3}(9d+a-8c)=\frac{1}{3}(10a-9b+c),$ $\displaystyle
z^{2}\bar{y}^{2}-\bar{z}^{2}y^{2}$ $\displaystyle=-4(10a-9b+c)i.$

Using these relations together with (59), we find exactly the same formulas
for $K_{O},K_{A},K_{B},K_{C}$ as before; that is, we obtain
(45),(43),(47),(44) respectively with $t=\frac{9}{8}$. As parts (b)-(f) only
rely on the expressions for $K_{O},K_{A},K_{B},K_{C}$, and as these are
unchanged, the proofs of these parts need no amendment.

This completes the proof of part (II), and the theorem. ∎

###### Proof of Corollary 2.

We use the terminology and results from the proof of Theorem 2. By Remark 6,
$\lambda=\frac{1}{\sigma}$, so

$I=\frac{(A+C)+(\sigma-1)B}{2\sigma}=\frac{(\tau-1)(A+C)+B}{2\tau}.$

First suppose $\tau=2$. Then $I=\frac{A+C+B}{4}$, so by (35),

$I=\frac{z^{2}+z^{2}-\frac{1}{2}w^{2}+\frac{1}{2}(2z-w)^{2}}{4}=z^{2}-\frac{zw}{2}.$

Now $z^{2}=A$ is a lattice point and $\frac{zw}{2}=\frac{1}{2}((pm-
qn)+(pn+qm)i)$. We have $zw=(pm-qn)+(pn+qm)i$ and by (34), we have
$2b=m^{2}+n^{2}$. So if $b$ is even, then $m^{2}+n^{2}\equiv 0\pmod{4}$ and
hence $m,n$ are both even. In this case $\frac{zw}{2}$ is a lattice point, and
hence $I$ is a lattice point. So we may assume that $b$ is odd and that $m,n$
are both odd. Then by (34), modulo 2 one has $pn+qm\equiv pm-qn\equiv pn-
qm=-4\equiv 0$. So once again, $I$ is a lattice point.

Now suppose $\tau=3$. Then $I=\frac{2(A+C)+B}{6}$, so by (35),

$I=\frac{2z^{2}+\frac{1}{3}(3z-w)^{2}+z^{2}-\frac{1}{3}w^{2}}{6}=z^{2}-\frac{zw}{3}.$

We have $zw=(pm-qn)+(pn+qm)i$ and by (34), we have $3b=m^{2}+n^{2}$. So
$m^{2}+n^{2}\equiv 0\pmod{3}$ and hence $m,n$ are divisible by $3$. So $zw$ is
divisible by $3$ and thus $I$ is a lattice point.

Similarly, if $\sigma=3$. Then $I=\frac{(A+C)+2B}{6}$, so by (57),

$I=\frac{z^{2}+\frac{1}{3}y^{2}+2z^{2}-\frac{1}{3}(3z-y)^{2}}{6}=\frac{zy}{3}.$

We have $zy=(pm-qn)+(pn+qm)i$ and by (56), we have $3d=m^{2}+n^{2}$. So
$m^{2}+n^{2}\equiv 0\pmod{3}$ and hence $m,n$ are divisible by $3$. So $zy$ is
divisible by $3$ and thus $I$ is a lattice point.

Now suppose $\tau=5$. Then $I=\frac{4(A+C)+B}{10}$. First assume that
$\frac{5d+2}{2},\frac{5d-2}{2},c$ are not all divisible by 5, so we are in the
general case. Then by (57),

$I=\frac{4z^{2}+\frac{1}{5}(5z-w)^{2}+z^{2}-\frac{1}{5}w^{2}}{10}=z^{2}-\frac{zw}{5}.$

We have $zw=(pm-qn)+(pn+qm)i$. It was proved in the proof of Theorem 2 that,
in the general case, $m,n$ are divisible by $5$. So $zw$ is divisible by $5$
and thus $I$ is a lattice point. Now consider the exceptional case. By (37),

$I=\frac{4(A+C)+B}{10}=\frac{20z^{2}+(5z-w)^{2}+5z^{2}-w^{2}}{10}=5z^{2}-zw,$

which is clearly a lattice point.

Now suppose $\sigma=5$. Then $I=\frac{(A+C)+4B}{10}$. First assume that
$\frac{5b+2}{2},\frac{5b-2}{2},c$ are not all divisible by 5, so we are in the
general case. Then by (35),

$I=\frac{z^{2}+\frac{1}{5}y^{2}+4z^{2}-\frac{1}{5}(5z-y)^{2}}{10}=\frac{zw}{5}.$

We have $zw=(pm-qn)+(pn+qm)i$. It was proved in the proof of Theorem 2 that,
in the general case, $m,n$ are divisible by $5$. So $zw$ is divisible by $5$
and thus $I$ is a lattice point. Now consider the exceptional case. By (59),

$I=\frac{(A+C)+4B}{10}=\frac{5z^{2}+y^{2}+20z^{2}-(5z-y)^{2}}{10}=zy,$

which is clearly a lattice point. ∎

######  Example 1.

The convex tangential LEQ with vertices $(0,0),(40,9),(36,12),(35,12)$ and
side lengths $41,5,1,37$, has incenter $(\frac{106}{3},10)$, by Proposition 3.
Similarly, the concave tangential LEQ with vertices
$(0,0),(16,63),(12,60),(11,60)$ and side lengths $65,5,1,61$, has a non-
lattice point incenter $(\frac{38}{3},58)$. Notice that by Proposition 5, one
finds that $\lambda=\frac{8}{9}$ in both of these examples; that is, $\tau=9$.

######  Example 2.

The proof of Theorem 2 was complicated by the two exceptional cases. Let us
show that such cases really do occur. Further, one might wonder whether it
might be possible to remove this inconvenience by taking a reflection in the
line $y=x$ and thus interchanging $a$ with $d$ and $b$ with $c$. Our examples
show that this is not always possible.

Consider the tangential LEQ with vertices $(0,0),(35,120)(32,116)(32,126)$. It
has $\tau=5$ and the side lengths $a,b,c,d$ are $125,5,10,130$ respectively.
Here $u=(5b-a)/2=-50,v=(5b-a)/2=75,c=10$. So $u,v,c$ are all divisible by 5,
which is the first exceptional case. Interchanging $a$ with $d$ and $b$ with
$c$ would give new values $(u,v,c)=(-40,90,5)$ but again $u,v,c$ are all
divisible by 5. Similarly, consider the tangential LEQ with vertices
$(0,0),(231,108),(228,108),(240,117)$. It has $\tau=9$ and the side lengths
$a,b,c,d$ are $255,3,15,267$ respectively. Here
$u=(9b-a)/2=-114,v=(9b-a)/2=141$. So $\gcd(18,u,v,c)=3$, which is the second
exceptional case. Interchanging $a$ with $d$ and $b$ with $c$ would give new
values $(u,v,c)=(-66,201,15)$ but again $\gcd(18,u,v,c)=3$.

###  6\. Infinite families of tangential LEQs

As we saw in Section 2, the four infinite families $K1-K4$ of kites from [3,
Theorem 1] gave examples of tangential LEQs with $(\sigma,\tau)$ equal to
$(5,5/4),(5/4,5),(2,2)$ and $(9/8,9)$ respectively. Also in Section 2, we
exhibited an infinite nested family of tangential LEQs with
$(\sigma,\tau)=(3,3/2)$. In this final section we exhibit infinite families
with $(\sigma,\tau)=(3/2,3)$ and $(9,9/8)$, thus showing that there are
infinitely many tangential LEQs in each of the seven cases of Theorem 1. The
method employed can also be used to give further examples in the cases where
$(\sigma,\tau)$ is $(5,5/4),(5/4,5),(2,2),(9/8,9)$ and $(3,3/2)$. While by no
means comprehensive, we hope the examples in this section will convey the
impression that tangential LEQs are quite abundant.

First observe that for $\tau=3$, Equation 1 of Theorem 1 is:

(62) $6^{2}+u^{2}=v^{2}-\left(v-c\right)^{2}.$

So this equation can be solved by fixing $u$, and then expressing
$6^{2}+u^{2}$ as a difference of two squares. Recall that an integer can be
written as a difference of two squares if and only if it is odd or a multiple
of 4; see sequence A100073 in OEIS [35]. Clearly $6^{2}+u^{2}\not\equiv
2\pmod{4}$, for all $u$. Thus for every integer value of $u$, there are
integers $v,c$ for which $6^{2}+u^{2}=v^{2}-(v-c)^{2}$. (Note that we are
interested in solutions $v\in\mathbb{N}$ and $u\in\mathbb{Z}$). However, we
must also impose the restrictions of Theorem 2. We require $u+v\equiv
0\pmod{3}$ and $c>0$, and $c$ is not divisible by $3$, as well as the three
conditions (i) – (iii):

  1. (i)

$3|c-b|<a+c$, (ii) $3(a+d)>a+c$, (iii) $3(b+c)\not=a+c$.

For convenience we separate (i) into two conditions:

  1. (ia)

$3(c-b)<a+c$, (ib) $3(b-c)<a+c$.

Notice that the conditions can be rewritten in terms of
$u=\frac{3b-a}{2},v=\frac{3b+a}{2}$ as:

  1. (ia)

$c<v$, (ib) $u<2c$, (ii) $3u<(2v+c)$, (iii) $u\not=-c$.

It is not true that for every integer value of $u$, there are integers $v,c$
for which $6^{2}+u^{2}=v^{2}-(v-c)^{2}$ and the above restrictions hold. For
example, for $u=4$, the only solutions are $v=14,c=2$ and $v=14,c=26$, but
(ib) fails for the first solution and (ia) fails for the second.

One infinite family of solutions is as follows: for negative integer $x$, let
$u=6x-1$. Then it is easy to check that $v=6(3x^{2}-x+3)+1,c=1$ is a solution
to (62). Note that $u+v\equiv 0\pmod{3}$. Condition (ia) is true for all $x$.
Conditions (ib) and (iii) hold as $x<0$. Condition (ii) can be written as
$6x^{2}-5x+7>0$, which is true for all real $x$. Notice that this infinite
family has the values:

$a=18x^{2}-12x+19,\quad b=6x^{2}+6,\quad c=1,\quad d=12x^{2}-12x+14.$

Another infinite family of solutions is obtained by taking $c=2$ and for
integer $x\leq-3$, letting $u=6x-2$ and $v=9x^{2}-6x+11$. This family has the
values:

$a=9x^{2}-12x+13,\quad b=3x^{2}+3,\quad c=2,\quad d=6x^{2}-12x+12.$

We now turn to $(\sigma,\tau)=(9,9/8)$. First observe that for $\sigma=9$,
Equation (2) of Theorem 1 is:

(63) $18^{2}+u^{2}=v^{2}-\left(v-4c\right)^{2}.$

We require $u+v\equiv 0\pmod{9}$ and $c>0$, as well as the four conditions:

  1. (ia)

$9(d-a)<8(a+c)$, (ib) $9(a-d)<8(a+c)$,

  2. (ii)

$9(a+d)>8(a+c)$, (iii) $9(a+2c-d)\not=8(a+c)$,

which can be rewritten in terms of $u=\frac{9d-a}{2},v\frac{9d+a}{2}$ as:

  1. (ia)

$8v+4c>9u$, (ib) $u>-4c$, (ii) $v>4c$, (iii) $u\not=5c$.

One infinite family of solutions is as follows: for integer $x>1$, let
$u=6x-1$, then it is easy to check that $v=6(3x^{2}-x+27)+1,c=1$ is a solution
to (63). Note that $u+v\equiv 0\pmod{9}$. Condition (ia) is true for all $x$.
Conditions (ib) and (iii) hold as $x>1$. Condition (ii) can be written as
$6x^{2}-2x+53>0$, which is true for all real $x$. Notice that this infinite
family has the values:

$a=2(9x^{2}-6x+82),\quad b=16x^{2}-12x+147,\quad c=1,\quad d=2x^{2}+18.$

##  Part II Extangential quadrilaterals

###  7\. Basic notions for extangential LEQs

An _extangential_ quadrilateral is a quadrilateral with an _excircle_ , that
is, a circle exterior to the quadrilateral that is tangent to the extensions
of all four sides [20, 21, 23]. Analogous to Pitot’s Theorem, one has the
following result [34]: a quadrilateral with consecutive side lengths $a,b,c,d$
is extangential if and only if it has no pair of parallel sides and

(64) $|a-c|=|b-d|.$

As for Pitot’s Theorem, the above criteria is usually only stated for convex
quadrilaterals, but also holds in the concave case. Indeed, if $OABC$ is a
concave quadrilateral with reflex angle at $B$, let $A^{\prime}$ (resp.
$C^{\prime}$) denote the point of intersection of the side $OA$ (resp. $OC$)
and the extension of side $BC$ (resp. $AB$). Let $a,b,c,d$ denote the lengths
of $OA,AB,BC,CO$ respectively, and similarly, let
$a^{\prime},b^{\prime},c^{\prime},d^{\prime}$ denote the lengths of
$OA^{\prime},A^{\prime}B,BC^{\prime},C^{\prime}O$. Then $a+b=c+d$ if and only
if $a^{\prime}+b^{\prime}=c^{\prime}+d^{\prime}$. This follows from a result
sometimes referred to Urquhart’s quadrilateral theorem, which has a long and
interesting history; see [34, 32, 16].

A quadrilateral $OABC$ with consecutive side lengths $a,b,c,d$ can have at
most one excircle and when one exists, its radius $r_{e}$, called the
_exradius_ , is given by the formula $r_{e}=\frac{K}{|a-c|}$ [20, Theorem 8].
By relabelling the vertices if necessary, we will suppose throughout this
paper that the excircle lies outside the vertex $B$. In particular, the
extangential hypothesis is now $a+b=c+d$, and from the proof of [20, Theorem
8], one has $a>c$ and

(65) $r_{e}=\frac{K}{a-c}.$

For an extangential quadrilateral, one of the diagonals $L$ separates the
sides into two pairs of equal sum, and the excircle is located outside one of
the vertices joined by $L$ (for kites $L$ is the axis of symmetry). In fact,
of these two vertices, the excircle is located outside the vertex at which the
quadrilateral makes the largest angle [20]. For concave extangential
quadrilaterals, the excircle is located outside the vertex with the reflex
angle. Obviously, extangential quadrilaterals cannot have a pair of parallel
sides; in particular no trapezoid is extangential and no parallelogram is
extangential even though parallelograms satisfy the $a+b=c+d$ condition.

We remark that there is a strong relationship between tangential and
extangential quadrilaterals. Recall from (4) that a quadrilateral $OABC$ is
tangential if and only if $a+c=b+d$. If $OABC$ is a convex quadrilateral, and
if $B^{\prime}$ denotes the reflection of $B$ in the perpendicular bisector of
$AC$, then the quadrilaterial $OAB^{\prime}C$ is extangential if and only if
$OABC$ is tangential. Moreover, equability is preserved by this construction.
However, notice that if $OABC$ is a LEQ, $OAB^{\prime}C$ may fail to have its
vertices on lattice points, as in Figure 7. When $OABC$ is concave, the same
construction can be made, but it can happen that $OAB^{\prime}C$ has self-
intersections, as in Figure 8.

$1$$2$$3$$4$$5$$6$$7$$8$$9$$10$$11$$12$$13$$14$$15$$16$$17$$18$$19$$20$$1$$2$$3$$4$$5$$6$$7$$8$$9$$10$$11$$12$$13$$14$$15$$16$$O$$A$$B$$B^{\prime}$$C$
Figure 7. Tangential to extangential convex quadrilaterals

$2$$4$$6$$8$$10$$12$$14$$16$$18$$2$$4$$6$$8$$10$$12$$14$$16$$18$$20$$22$$24$$B$$C$$A$$O$

$2$$4$$6$$8$$10$$12$$14$$16$$18$$2$$4$$6$$8$$10$$12$$14$$16$$18$$20$$22$$24$$B^{\prime}$$C$$A$$O$

Figure 8. Self-intersections can occur in the concave case

Analogous to Proposition 1, we have the following result; its proof is very
similar to that of Proposition 1.

######  Proposition 6.

If $OABC$ is extangential with excircle outside $B$, then $OABC$ is a kite if
and only if the diagonal $OB$ divides $OABC$ into two triangles of equal area.

###### Proof.

Obviously, if $OABC$ is a kite, then its axis of symmetry diagonal divides
$OABC$ into two triangles of equal area. Conversely, applying Heron’s formula
to triangle $OAB$ gives

$\displaystyle 16K_{A}^{2}$ $\displaystyle=(a+b+p)(a+b-p)(a-b+p)(-a+b+p)$
$\displaystyle=-(a^{2}-b^{2})^{2}+2(a^{2}+b^{2})p^{2}-p^{4}.$

Similarly, from triangle $OBC$

$16K_{C}^{2}=-(c^{2}-d^{2})^{2}+2(c^{2}+d^{2})p^{2}-p^{4}.$

Hence, subtracting,

(66)
$2(a^{2}+b^{2}-c^{2}-d^{2})p^{2}=16(K_{A}^{2}-K_{C}^{2})+(a^{2}-b^{2})^{2}-(c^{2}-d^{2})^{2}.$

Notice that, using $a+b=c+d$,

$a^{2}+b^{2}-c^{2}-d^{2}=(a-d)(a+d)+(b-c)(b+c)=2(a-c)(a-d),$

and

$\displaystyle(a^{2}-b^{2})^{2}-(c^{2}-d^{2})^{2}$
$\displaystyle=(a-b)^{2}(a+b)^{2}-(d-c)^{2}(c+d)^{2}$
$\displaystyle=4(a-c)(a-d)(a+b)^{2}.$

So (66) gives

(67) $(a-c)(a-d)p^{2}=4(K_{A}^{2}-K_{C}^{2})+(a-c)(a-d)(a+b)^{2}.$

Now assume that $K_{A}=K_{C}$. Then (67) gives
$(a-c)(a-d)p^{2}=(a-c)(a-d)(a-b)^{2}$. Notice that $p=\pm(a-b)$ is impossible,
as otherwise the triangle $OAB$ would be degenerate. Hence either $a=d$ or
$a=c$. If $a=c$, then by the extangential hypothesis, $b=d$ and so $OABC$ is a
parallelogram, which is impossible. So $a=d$. Note that as $K_{A}=K_{C}$, the
points $A,C$ are equidistant from the line through $O,B$. So the triangles
$OAB$ and $OBC$ are congruent, and hence $OABC$ is a kite. ∎

Like the incenter of a tangential quadrilateral, the excenter of an
extangential quadrilateral lies on the Newton line $\mathcal{N_{L}}$ joining
the midpoints of the two diagonals. We have not seen this stated explicitly in
the literature, but the proof in the tangential case is readily adapted. For
example, the vector proof of Anne’s Theorem given in [12, Lemma 1] is valid as
is, for signed areas, as the authors indicate, and then [12, Theorem 3] can be
easily modified, with two positive areas and two negative areas. Since the
excenter $I_{e}$ lies on Newton line, $I_{e}$ is of the form
$\lambda_{e}M_{A}+(1-\lambda_{e})M_{O}$ for some $\lambda_{e}\in[0,1]$, where
we recall $M_{A},M_{O}$ refer to the midpoints of the diagonals $AC,OB$
respectively.

For the rest of this section, $OABC$ denotes an extangential (convex or
concave) quadrilateral, with vertices in counterclockwise cyclic order, and
$a,b,c,d$ denote the lengths of the sides $OA,AB,BC,CO$ respectively. We
suppose furthermore that the excircle lies outside the vertex $B$. In
particular, the extangential hypothesis is now $a+b=c+d$, and from the proof
of [20, Theorem 8], one has $a>c$.

######  Remark 12.

By reflecting in the line $y=x$ if necessary, we may always assume that $a\geq
d$. In this case, we have $a-b\geq d-b=a-c>0$; thus $a>b$. Hence
$a=\max\\{a,b,c,d\\}$. Similarly, $b=\min\\{a,b,c,d\\}$.

For tangential quadrilaterals, equability is equivalent to the condition that
the inradius is 2. For extangential quadrilaterals, the equability condition
(65) is equivalent to the condition

(68) $r_{e}=\frac{2(a+b)}{a-c}.$

Note that if $OABC$ is not a kite, then by Proposition 6,
$K_{A}\not=\frac{K}{2}$ and $K_{O}\not=\frac{K}{2}$. So, for equable
extangential quadrilaterals that are not kites, $K_{A}\not=a+b$ and
$K_{O}\not=a+b$. In fact, one has $K_{O}\not=a+b$ even when $OABC$ is a kite
(with its excircle outside the vertex $B$). Indeed, otherwise $OABC$ would be
a rhombus, and no rhombus is extangential. This will be important in
Proposition 10 below.

Analogous to many results for tangential quadrilaterals, there are very
similar results for extangential quadrilaterals. Indeed, analogous to
Propositions 2, 3, 4 and 5 of Part 1, we have the following four analogous
propositions. We omit the proofs which are essentially the same as those of
the propositions of Part 1.

######  Proposition 7.

If $OABC$ is extangential, then $OABC$ is a kite if and only if the Newton
line $\mathcal{N_{L}}$ contains one of the diagonals.

######  Proposition 8.

If $OABC$ is extangential, we have the following two expressions for the
excenter $I_{e}$:

  1. (a)

$I_{e}=\frac{r_{e}}{2}\,\frac{aC+dA}{K_{O}}$, (b)
$I_{e}=A+\frac{r_{e}}{2}\,\frac{a(B-A)+bA}{K_{A}}$.

######  Proposition 9.

If $OABC$ is extangential, we have:

$(K_{A}-\frac{r_{e}}{2}(a-b))(K_{O}-\frac{r_{e}}{2}(a+d))=\frac{r_{e}^{2}}{4}(ac-
bd).$

######  Proposition 10.

If $OABC$ is extangential but is not a kite, we have the following two
expressions for the coordinate $\lambda_{e}$:

  1. (a)

$\lambda_{e}=\frac{r_{e}}{2}\cdot\frac{a+b}{K_{O}-(a+b)}$, (b)
$\lambda_{e}=1-\frac{r_{e}}{2}\cdot\frac{c-b}{K_{A}-(a+b)}$.

Furthermore, the first of the above expressions for $\lambda_{e}$ is valid if
$OABC$ is a kite.

######  Example 3.

Apart from the rhombus of side length 5 (K1, $n=2$) and the $4\times 4$ square
(K3, $n=1$), the lattice equable kites of [3, Theorem 1] are extangential. In
each case $n,j$, to determine the exradius $r_{e,n,j}$, the excenter
$I_{e,n,j}$ and the parameter $\lambda_{e,n,j}$, one can employ (68) and
Propositions 8 and 10. Here $I_{e}=\lambda_{e}M+(1-\lambda_{e})\frac{B}{2}$,
where $M=M_{A}$. We omit the details, which are completely routine. The
results are given in Table 3. Notice that unlike the incenters, the excenters
are not necessarily lattice points.

Case | Equation | $M$ | $B$ | $r_{e}$ | $I_{e,n,j}$ | $\lambda_{e,n,j}$
---|---|---|---|---|---|---
_K1_ | $n^{2}-5j^{2}=4$ | $\frac{1}{2}(n+5j)(2,1)$ | $n(2,1)$ | $\frac{2n}{j}$ | $\frac{n(n+j)}{2j}(2,1)$ | $\frac{n^{2}}{5j^{2}}$
_K2_ | $n^{2}-5j^{2}=1$ | $(2n+5j)(2,1)$ | $4n(2,1)$ | $\frac{n}{j}$ | $\frac{n(n+2j)}{j}(2,1)$ | $\frac{n^{2}}{5j^{2}}$
_K3_ | $n^{2}-2j^{2}=1$ | $(n+2j)(2,2)$ | $4n(1,1)$ | $\frac{2n}{j}$ | $\frac{2n(n+j)}{j}(1,1)$ | $\frac{n^{2}}{2j^{2}}$
_K4_ | $2n^{2}-j^{2}=1$ | $(4n+3j)(\frac{3}{2},\frac{3}{2})$ | $12n(1,1)$ | $\frac{3n}{j}$ | $\frac{3n(3n+2j)}{j}(1,1)$ | $\frac{2n^{2}}{j^{2}}$

Table 3. The four kite families
$2$$4$$6$$8$$10$$12$$14$$16$$18$$2$$4$$6$$8$$10$$12$$14$$16$$18$ Figure 9.
Kite with side lengths 3 and 15 (K4, $n=j=1$)

######  Example 4.

Consider the convex extangential LEQ shown on the left of Figure 10; its
vertices are (0,0),(21,20),(20,20),(0,5) and the side lengths are 29, 1, 25,
5. The exradius is $r_{e}=\frac{K}{a-c}=15$. By Proposition 8, the excenter
$I_{e}$ is $=(15,35)$, and by Proposition 10, the coordinate $\lambda_{e}$ of
the excenter is $10$.

Similarly, a concave extangential LEQ is shown on the right of Figure 10; its
vertices are (0,0),(12,5),(10,5),(6,8), and the side lengths are: 13, 2, 5,
10. The exradius is $r_{e}=\frac{K}{a-c}=\frac{15}{4}$. By Proposition 8, the
excenter $I_{e}$ is $\frac{5}{4}(9,7)$, and by Proposition 10, the coordinate
$\lambda_{e}$ of the excenter is $\frac{25}{16}$.

$5$$10$$15$$20$$25$$30$$5$$10$$15$$20$$25$$30$$35$$A$$B$$C$$I_{e}$$M_{A}$$M_{O}$

$3$$6$$9$$12$$15$$2$$4$$6$$8$$10$$12$$A$$B$$C$$I_{e}$$M_{A}$$M_{O}$

Figure 10. Extangential LEQs; one convex and one concave

###  8\. Lemmata for extangential LEQs

For this section, $OABC$ denotes a non-kite extangential quadrilateral with
consecutive sides $a,b,c,d$ and with its excircle outside the vertex $B$, so
$a>c$. In particular, it has exradius $r_{e}=\frac{2(a+b)}{a-c}$. As explained
in Remark 12, we may assume $a=\max\\{a,b,c,d\\}$ and $b=\min\\{a,b,c,d\\}$.

The approach adopted in this section is the same as that of Section 3, and the
results obtained are analogous, but the calculations are often more
complicated. For the convenience of the reader, we repeat Equations (12) and
(13):

(69) $\displaystyle p^{2}=a^{2}+b^{2}\pm 2\sqrt{a^{2}b^{2}-(2K_{A})^{2}},$
(70) $\displaystyle q^{2}=a^{2}+d^{2}\pm 2\sqrt{a^{2}d^{2}-(2K_{O})^{2}},$

where $p,q$ are the lengths of the diagonals $OB,AC$ respectively. As
$O,A,B,C$ are lattice points, $p^{2},q^{2}$ are integers, so by Lemma 1, the
integers $a^{2}b^{2}-(2K_{A})^{2}$ and $a^{2}d^{2}-(2K_{O})^{2}$ are squares.

######  Lemma 9.

One has

$p^{2}=\frac{8(a+b)(K_{A}-K_{C})}{(a-c)(a-d)}+(a+b)^{2}\quad\text{and}\quad
q^{2}=\frac{8(K_{O}-K_{B})}{a-c}+(a-d)^{2}.$

###### Proof.

As $OABC$ is not a kite, by hypothesis, so $a\not=d$. Arguing exactly as in
Proposition 6 we reobtain (67):

$(a-c)(a-d)p^{2}=4(K_{A}^{2}-K_{C}^{2})+(a-d)(a-c)(a+b)^{2},$

and since $OABC$ is extangential and hence not a parallelogram, $a\not=c$.
Thus, as $K_{A}^{2}-K_{C}^{2}=(K_{A}+K_{C})(K_{A}-K_{C})=2(a+b)(K_{A}-K_{C})$,
we obtain the required formula for $p^{2}$.

Similarly, by applying Heron’s formula to triangles $OAC$ and $BCA$, we obtain

(71)
$2(d^{2}-c^{2}+a^{2}-b^{2})q^{2}=16(K_{O}^{2}-K_{B}^{2})+(d^{2}-a^{2})^{2}-(b^{2}-c^{2})^{2}.$

Simplifying as in Proposition 6 gives

$(a-c)(a+b)q^{2}=8(a+b)(K_{O}-K_{B})+(a+b)(a-c)(a-d)^{2},$

from which the required formula for $q^{2}$ follows. ∎

######  Remark 13.

From the above lemma, using (69),

$\frac{8(a+b)(K_{A}-(a+b))}{(a-c)(a-d)}=\frac{p^{2}-(a+b)^{2}}{2}=-ab\pm\sqrt{a^{2}b^{2}-(2K_{A})^{2}},$

which is an integer by Lemma 1. Similarly, $\frac{8(K_{O}-(a+b))}{a-c}$ is an
integer.

######  Lemma 10.

The integer $abcd-4(a+b)^{2}$ is a square, and

$\displaystyle K_{A}$ $\displaystyle=(a+b)+(a-c)(a-d)(a+b)\frac{-(ab+cd)\pm
2\sqrt{abcd-4(a+b)^{2}}}{16(a+b)^{2}+(a-c)^{2}(a-d)^{2}},$ $\displaystyle
K_{O}$ $\displaystyle=(a+b)+(a-c)\frac{ad+bc\mp
2\sqrt{abcd-4(a+b)^{2}}}{16+(a-c)^{2}},$

where the signs of the square roots in the formulas for $K_{O}$ and $K_{A}$
are opposite.

######  Remark 14.

In the statement of the above lemma, the terms

$ab+cd\pm 2\sqrt{abcd-4(a+b)^{2}}\quad\text{and}\quad ad+bc\mp
2\sqrt{abcd-4(a+b)^{2}}$

are strictly positive. Indeed, using $d=a+b-c$, by the arithmetic mean-
geometric mean inequality, $ab+cd\geq 2\sqrt{abcd}>\sqrt{abcd-4(a+b)^{2}}$. In
particular, $K_{A}<\frac{K}{2}=a+b$ if and only if $a>d$, which is opposite to
the situation for tangential LEQs; see Remark 4.

###### Proof of Lemma 10.

From Lemma 9 and Equation (70),

$\frac{4(K_{O}-K_{B})}{a-c}-ad=\frac{q^{2}-a^{2}-d^{2}}{2}=\pm\sqrt{a^{2}d^{2}-(2K_{O})^{2}},$

so setting $s:=\frac{K_{O}-(a+b)}{a-c}$, squaring, and using
$K_{O}+K_{B}=2(a+b)$ gives

$\alpha s^{2}-2\beta s+\gamma=0,$

where

$\alpha=16+(a-c)^{2},\quad\beta=ad+bc,\quad\gamma=(a+b)^{2}.$

Thus, as $\beta^{2}-\alpha\gamma=4(abcd-4(a+b)^{2})$ (using $a+b=c+d$ again),
we have

$s=\frac{ad+bc\pm 2\sqrt{abcd-4(a+b)^{2}}}{16+(a-c)^{2}},$

which gives the required formula for $K_{O}$. In particular, as $s$ is
rational, $abcd-4(a+b)^{2}$ is a square, as claimed.

The formula for $K_{A}$ is similarly obtained by equating $p^{2}$ from Lemma 9
and Equation (69). We have

$\frac{4(a+b)(K_{A}-K_{C})}{(a-c)(a-d)}+ab=\frac{p^{2}-a^{2}-b^{2}}{2}=\pm\sqrt{a^{2}b^{2}-(2K_{A})^{2}}.$

Define $t:=\frac{(a+b)(K_{A}-(a+b))}{(a-c)(a-d)}$. One obtains
$\bar{\alpha}t^{2}+2\bar{\beta}t+\bar{\gamma}=0$, where

$\bar{\alpha}=16(a+b)^{2}+(a-c)^{2}(a-d)^{2},\quad\bar{\beta}=(a+b)^{2}(ab+cd),\quad\bar{\gamma}=(a+b)^{4}.$

One has

$\bar{\beta}^{2}-\bar{\alpha}\bar{\gamma}=4(a+b)^{4}(abcd-4(a+b)^{2}),$

so

$t=\frac{-(a+b)^{2}(ab+cd)\pm
2(a+b)^{2}\sqrt{abcd-4(a+b)^{2}}}{16(a+b)^{2}+(a-c)^{2}(a-d)^{2}},$

which gives the required formula for $K_{O}$.

It remains to see that the signs of the square roots in the formulas for
$K_{O}$ and $K_{A}$ are opposite. Let $R=2\sqrt{abcd-4(a+b)^{2}}$. Obviously,
we may assume that $R\not=0$ and $a\not=c$. Let us write

$\displaystyle K_{A}$
$\displaystyle=(a+b)+(a-c)(a-d)(a+b)\frac{-(ab+cd)+\delta_{A}R}{16(a+b)^{2}+(a-c)^{2}(a-d)^{2}},$
$\displaystyle K_{O}$
$\displaystyle=(a+b)+(a-c)\frac{ad+bc+\delta_{O}R}{16+(a-c)^{2}},$

where $\delta_{A},\delta_{O}$ are each $\pm 1$. Using $a+b=c+d$,

$\displaystyle K_{A}-$ $\displaystyle\frac{(a+b)(a-b)}{a-c}$
$\displaystyle=(a+b)\left(\frac{b-c}{a-c}+(a-c)(a-d)\frac{-(ab+cd)+\delta_{A}R}{16(a+b)^{2}+(a-c)^{2}(a-d)^{2}}\right)$
$\displaystyle=\frac{(a+b)(d-a)}{a-c}\cdot\frac{16(a+b)^{2}+(a-c)^{2}(ac+bd)-(a-c)^{2}\delta_{A}R}{16(a+b)^{2}+(a-c)^{2}(a-d)^{2}},$
$\displaystyle K_{O}-$
$\displaystyle\frac{(a+b)(a+d)}{a-c}=-\frac{(a+b)(c+d)}{a-c}+(a-c)\frac{ad+bc+\delta_{O}R}{16+(a-c)^{2}}$
$\displaystyle=\frac{-1}{a-c}\cdot\frac{16(a+b)^{2}+(a-c)^{2}(ac+bd)-(a-c)^{2}\delta_{O}R}{16+(a-c)^{2}}.$

Notice also that $(a+b)(d-a)=bd-ac$. Hence, by Proposition 9,

(72)
$\frac{X-(a-c)^{2}\delta_{A}R}{16(a+b)^{2}+(a-c)^{2}(a-d)^{2}}\cdot\frac{X-(a-c)^{2}\delta_{O}R}{16+(a-c)^{2}}=(a+b)^{2},$

where $X=16(a+b)^{2}+(a-c)^{2}(ac+bd)$. Now, substituting $d=a+c-b$ one finds
that

$\frac{X-(a-c)^{2}\delta_{A}R}{16(a+b)^{2}+(a-c)^{2}(a-d)^{2}}\cdot\frac{X+(a-c)^{2}\delta_{A}R}{16+(a-c)^{2}}=(a+b)^{2}.$

Subtracting from (72) gives

(73) $(X-(a-c)^{2}\delta_{A}R)\cdot(a-c)^{2}(\delta_{A}+\delta_{O})R=0.$

Note that $X-(a-c)^{2}\delta_{A}R\not=0$ as otherwise $X^{2}=(a-c)^{4}R^{2}$
which would give

$(16(a+b)^{2}+(a-c)^{2}(ac+bd))^{2}-(a-c)^{4}(abcd-4(a+b)^{2})=0,$

and hence

$(a+b)^{2}(16+(a-c)^{2})(16(a+b)^{2}+(a-c)^{2}(b-c)^{2})=0,$

which is impossible. So from (73), we have $\delta_{A}=-\delta_{O}$, as
claimed. ∎

######  Lemma 11.

The sign of the square root in the formulas for $K_{O}$ is positive if and
only if $B$ lies within the circumcircle of the triangle $OAC$; in particular,
the sign for $K_{O}$ is positive if $OABC$ is concave.

###### Proof.

In the notation of the above proof, let $x=\delta_{O}2\sqrt{abcd-4(a+b)^{2}}$,
so

$\displaystyle K_{A}$
$\displaystyle=(a+b)+(a-c)(a-d)(a+b)\frac{-(ab+cd)-x}{16(a+b)^{2}+(a-c)^{2}(a-d)^{2}},$
$\displaystyle K_{O}$ $\displaystyle=(a+b)+(a-c)\frac{ad+bc+x}{16+(a-c)^{2}}.$

From a standard criteria for a point to be within the circumcircle of a
triangle (see [15]), $B$ is inside the circumcircle of the triangle $OAC$ if
and only if

(74) $p^{2}K_{O}<d^{2}K_{A}+a^{2}K_{C}.$

Now $d^{2}K_{A}+a^{2}K_{C}=K_{A}(d^{2}-a^{2})+2a^{2}(a+b)$, and by Lemma 9,

$K_{O}p^{2}=K_{O}\left(\frac{16(a+b)(K_{A}-(a+b))}{(a-c)(a-d)}+(a+b)^{2}\right).$

So condition (74) can be written as $E>0$ where

$E=K_{A}(d^{2}-a^{2})+2a^{2}(a+b)-K_{O}\left(\frac{16(a+b)(K_{A}-(a+b))}{(a-c)(a-d)}+(a+b)^{2}\right).$

Substituting the formulas for $K_{O},K_{A}$ and $x$, one finds upon
simplification that

$E=2\delta_{O}(a+b)\sqrt{abcd-4(a+b)^{2}}.$

Hence, as claimed, $\delta_{O}>0$ if and only if $B$ is inside the
circumcircle of the triangle $OAC$. ∎

######  Definition 5.

Let

$\displaystyle\Sigma$
$\displaystyle=8\cdot\frac{ad+bc+2\delta\sqrt{abcd-4(a+b)^{2}}}{16+(a-c)^{2}},$
$\displaystyle T$
$\displaystyle=8(a+b)^{2}\cdot\frac{ab+cd+2\delta\sqrt{abcd-4(a+b)^{2}}}{16(a+b)^{2}+(a-c)^{2}(a-d)^{2}},$

where $\delta=1$ if $B$ lies within the circumcircle of the triangle $OAC$,
and $\delta=-1$ otherwise.

######  Remark 15.

Observe that $\Sigma$ and $T$ are positive integers. Indeed, from Lemma 10,

(75) $\Sigma=\frac{8(K_{O}-(a+b))}{a-c},\qquad
T=\frac{8(a+b)((a+b)-K_{A})}{(a-c)(a-d)},$

which are integers by Remark 13, and they are positive by Remark 14.
Furthermore, from Lemma 2,

(76) $\Sigma=\frac{1}{2}(q^{2}-(a-d)^{2}),\qquad
T=\frac{1}{2}((a+b)^{2}-p^{2}).$

In the notation of the proof of Lemma 10,

(77) $\alpha\Sigma^{2}-16\beta\Sigma+64\gamma=0,$

where $\alpha=16+(a-c)^{2}$, $\beta=ad+bc$, $\gamma=(a+b)^{2}=(c+d)^{2}$, and

(78) $\bar{\alpha}T^{2}-16\bar{\beta}T+64\bar{\gamma}=0,$

where $\bar{\alpha}=16(a+b)^{2}+(a-c)^{2}(a-d)^{2}$,
$\bar{\beta}=(a+b)^{2}(ab+cd)$, and $\bar{\gamma}=(a+b)^{4}$.

######  Lemma 12.

The following relations hold:

  1. (a)

$\Sigma T=8\frac{(a+b)^{2}}{(a-c)^{2}}(T-{\Sigma})$,

  2. (b)

$2\Sigma T=(a+b)^{2}(\Sigma-8)-(b-c)^{2}T$.

###### Proof.

(a). As in the proof of Lemma 11 , let $x=2\delta\sqrt{abcd-4(a+b)^{2}}$.
Then, cross-multiplying, the required identity is $E=0$, where

$\displaystyle E=$
$\displaystyle(ad+bc+x)\big{(}16(a+b)^{2}+(a-c)^{2}(a-d)^{2}\big{)}$
$\displaystyle-(a+b)^{2}(ab+cd+x)\big{(}16+(a-c)^{2}\big{)}+(a-c)^{2}(ad+bc+x)(ab+cd+x).$

Expanding and using $d=a+c-b$, one has

$E=(a-c)^{2}\big{(}4(4(a+b)^{2}-abcd)+x^{2}\big{)}.$

Then replacing $x^{2}$ by $4(abcd-4(a+b)^{2})$ gives $E=0$, as required.

(b). From the definitions, since $(a-c)^{2}=(a-c)(d-b)=ad+bc-ab-cd$,

$\displaystyle(a+b)^{2}(16+(a-c)^{2})\Sigma$
$\displaystyle-(16(a+b)^{2}+(b-c)^{2}(a-c)^{2})T$
$\displaystyle=8(a+b)^{2}(ad+bc-ab-cd)=8(a+b)^{2}(a-c)^{2}.$

Part (b) follows by applying part (a). ∎

######  Remark 16.

As $\Sigma$ and $T$ are positive, Lemma 12(a) gives $\Sigma<T$. Furthermore,
as $T-\Sigma<T$, Lemma 12(a) gives $\Sigma<8\frac{(a+b)^{2}}{(a-c)^{2}}$, and
Lemma 12(b) gives $2T<(a+b)^{2}\frac{\Sigma-8}{\Sigma}<(a+b)^{2}$. In
particular, $\Sigma>8$.

######  Remark 17.

From (75),

(79) $\displaystyle K_{O}$ $\displaystyle=(a+b)+\frac{1}{8}(a-c)\Sigma,$ (80)
$\displaystyle K_{A}$ $\displaystyle=(a+b)-\frac{(a-c)(a-d)}{8(a+b)}T.$

Then by Proposition 10, the parameter $\lambda_{e}$ and the exradius $r_{e}$
are related to $\Sigma$ by

(81) $\lambda_{e}\cdot\Sigma=8\frac{(a+b)^{2}}{(a-c)^{2}}=2r_{e}^{2}.$

Using Lemma 12(a), we can also write

(82) $\lambda_{e}=\frac{T}{T-\Sigma}.$

######  Remark 18.

We have the non-degeneracy condition $K_{B}\not=0$ as otherwise $ABC$ would be
colinear. Thus $K_{O}\not=2(a+b)$ and (79) gives
$\Sigma\not=8\frac{a+b}{a-c}$. Hence, by Lemma 12(a), we have

(83) $\Sigma(T-\Sigma)\not=8T.$

Notice also that $OABC$ is concave if and only if $K_{O}>2(a+b)$, that is,
from (75), when $\Sigma>8\frac{a+b}{a-c}$.

######  Lemma 13.

$(c-b)T<(a-b)\Sigma$.

###### Proof.

From the assumption that the vertices $O,A,B,C$ are positively oriented and if
$OABC$ is concave, the reflex angle is at $B$, we have $K_{A}>0$. So (80)
gives $8(a+b)^{2}>(a-c)(a-d)T$. Lemma 12(a) gives
$8(a+b)^{2}=(a-c)^{2}\frac{\Sigma T}{T-\Sigma}$. Hence, as $T-\Sigma>0$ by
Remark 16, and using $a+b=c+d$, we obtain $(a-c)\Sigma>(c-b)(T-\Sigma)$.
Rearranging this gives the required result. ∎

######  Lemma 14.

$\Sigma$ and $T$ both divide $8(a+b)^{2}$.

###### Proof.

From 77, $\frac{1}{16}\alpha\Sigma^{2}-\beta\Sigma+4\gamma=0$, where
$\alpha=16+(a-c)^{2}$, $\beta=ad+bc$, $\gamma=(a+b)^{2}$. So
$\frac{1}{16}\alpha\Sigma^{2}$ is an integer, and hence
$\frac{1}{16}(a-c)^{2}\Sigma^{2}$ is an integer. So $\frac{1}{4}(a-c)\Sigma$
is an integer, and hence $\frac{1}{4}\alpha\Sigma$ is an integer. Hence, as

$16(a+b)^{2}=16\gamma=(-\frac{1}{4}\alpha\Sigma+4\beta)\Sigma,$

$\Sigma$ divides $16(a+b)^{2}$. So if $\Sigma$ is odd, then $\Sigma$ divides
$(a+b)^{2}$. If $\Sigma$ is even, then as
$\frac{1}{32}\alpha\Sigma^{2}-\frac{1}{2}\beta\Sigma+2\gamma=0$, so
$\frac{1}{32}\alpha\Sigma^{2}$ is an integer, and hence
$\frac{1}{32}(a-c)^{2}\Sigma^{2}$ is an integer. It follows that
$\frac{1}{64}(a-c)^{2}\Sigma^{2}$ is an integer. So $\frac{1}{8}(a-c)\Sigma$
is an integer, and hence $\frac{1}{8}\alpha\Sigma$ is an integer. Hence, as

$8(a+b)^{2}=(-\frac{1}{8}\alpha\Sigma+2\beta)\Sigma,$

$\Sigma$ divides $8(a+b)^{2}$.

By Lemma 12(a), we have $8(a+b)^{2}(\frac{1}{\Sigma}-\frac{1}{T})=(a-c)^{2}$.
Since $8(a+b)^{2}\frac{1}{\Sigma}$ is an integer, it follows that
$8(a+b)^{2}\frac{1}{T}$ is also an integer. ∎

###  9\. Explicit examples of extangential LEQs

In this section we exhibit non-kite extangential LEQs in the three cases with
$(\Sigma,T)$ equal to $(9,18),(18,50)$ and $(45,50)$ respectively. As before,
let us define $h:=\frac{a+b}{a-c}$, so $h=\sqrt{\frac{\Sigma
T}{8(T-\Sigma)}}$, by Lemma 12(a). Note that $h>1$, but as we will see, $h$
may fail to be an integer.

We have

(84) $b=(h-1)a-hc.$

In particular, $b>0$ gives $a>\frac{h}{h-1}c$ and since $b\leq c$, we have
$a\leq\frac{h+1}{h-1}c$. From (77),
$\alpha\Sigma^{2}-16\beta\Sigma+64\gamma=0$, where $\alpha=16+(a-c)^{2}$,
$\beta=ad+bc$, $\gamma=(a+b)^{2}=(c+d)^{2}$. Using (84), substituting
$d=a+b-c$ and solving for $a$ gives

$a=\frac{64ch^{2}-16c\Sigma+c\Sigma^{2}\pm
4\sqrt{2c^{2}\Sigma(8h^{2}-\Sigma)(\Sigma-8)-(\Sigma-8h)^{2}\Sigma^{2}}}{(\Sigma-8h)^{2}}.$

We claim that
$\frac{64ch^{2}-16c\Sigma+c\Sigma^{2}}{(\Sigma-8h)^{2}}>\frac{h+1}{h-1}c$.
Indeed, cross multiplying and simplifying, the claim is
$(8h^{2}-\Sigma)(\Sigma-8)>0$, which is true since $\Sigma<8h^{2}$ by Remark
16 and $\Sigma>8$, also by Remark 16. Hence, since $a\leq\frac{h+1}{h-1}c$, it
follows that

(85)
$a=\frac{64ch^{2}-16c\Sigma+c\Sigma^{2}-4\sqrt{2c^{2}\Sigma(8h^{2}-\Sigma)(\Sigma-8)-(\Sigma-8h)^{2}\Sigma^{2}}}{(\Sigma-8h)^{2}}.$

We first classify the extangential LEQs with $(\Sigma,T)=(9,18)$. Note that
this is one of two cases in Theorem 3(a). Suppose we have an extangential LEQ
with $a>c\geq b$ and $\Sigma=9,T=18$. So $h=\sqrt{\frac{\Sigma
T}{8(T-\Sigma)}}=\frac{3}{2}$, and from (84),

(86) $b=\frac{1}{2}(a-3c).$

Hence $a>3c$ and since $b\leq c$, we have $a\leq 5c$. From (85),

(87) $a=9c-4\sqrt{2c^{2}-9}.$

Working modulo 3, the fact that $2c^{2}-9$ is a square gives us that $c$ is
divisible by 3, say $c=3v$. Then $v$ satisfies the negative Pell equation

(88) $u^{2}-2v^{2}=-1,$

for some positive integer $u$. Then (86) and (87) give

(89) $(a,b,c,d)=3(9v-4u,3v-2u,v,11v-6u).$

It is well known that the solutions $(u_{j},v_{j})$ to (88) are given
recursively by

(90) $u_{j+1}=3u_{j}+4v_{j},\qquad v_{j+1}=2u_{j}+3v_{j},$

with initial values $(u_{1},v_{1})=(1,1)$.

We now define the vertices of our quadrilaterals. Let

$\displaystyle A_{j}$
$\displaystyle=\frac{3}{2}\left(9u_{j}-8v_{j}-7,9u_{j}-8v_{j}+7\right),$
$\displaystyle B_{j}$
$\displaystyle=6\left(3u_{j}-3v_{j}-2,3u_{j}-3v_{j}+2\right)$ $\displaystyle
C_{j}$
$\displaystyle=\frac{3}{2}\left(11u_{j}-12v_{j}-7,11u_{j}-12v_{j}+7\right).$

Note that $A_{j},B_{j},C_{j}$ are lattice points as $u_{j},v_{j}$ are odd, as
one can see from the recursive formula (90). We will consider the
quadrilateral $OA_{j}B_{j}C_{j}$. Let $a_{j},b_{j},c_{j},d_{j}$ denote the
lengths of the sides $OA_{j},A_{j}B_{j},B_{j}C_{j},C_{j}O_{j}$ respectively.
The distance $a_{j}$ is given by

$\displaystyle a_{j}^{2}$
$\displaystyle=\frac{9}{4}((9u_{j}-8v_{j}-7)^{2}+(9u_{j}-8v_{j}+7)^{2})=\frac{9}{2}(81u_{j}^{2}-144u_{j}v_{j}+64v_{j}^{2}+49)$
$\displaystyle=\frac{9}{2}(81u_{j}^{2}-144u_{j}v_{j}+64v_{j}^{2}+49(2v_{j}^{2}-u_{j}^{2}))\quad\text{(by
\eqref{P1})}$
$\displaystyle=\frac{9}{2}(32u_{j}^{2}-144u_{j}v_{j}+162v_{j}^{2})$
$\displaystyle=9(9v_{j}-4u_{j})^{2}.$

So $a_{j}=3(9v_{j}-4u_{j})$. Similarly, the other side lengths are as follows:

$b_{j}=3(3v_{j}-2u_{j}),\qquad c_{j}=3v_{j},\qquad d_{j}=3(11v_{j}-6u_{j}),$

as anticipated by (89). So the perimeter of $OA_{j}B_{j}C_{j}$ is
$a_{j}+b_{j}+c_{j}+d_{j}=36(2v_{j}-u_{j})$.

The area of $OA_{j}B_{j}$ is
$\frac{1}{2}\|\overrightarrow{OA_{j}}\times\overrightarrow{OB_{j}}\|=9(5v_{j}-3u_{j})$.
The area of $OB_{j}C_{j}$ is
$\frac{1}{2}\|\overrightarrow{OB_{j}}\times\overrightarrow{OC_{j}}\|=9(3v_{j}-u_{j})$.
Notice in passing that as the signed areas of $OA_{j}B_{j}$ and $OB_{j}C_{j}$
are both positive, as one can see recursively using (90), so
$OA_{j}B_{j}C_{j}$ has no self-intersection. The area of $OA_{j}B_{j}C_{j}$ is
$9(5v_{j}-3u_{j})+9(3v_{j}-u_{j})=36(2v_{j}-u_{j})$, so $OA_{j}B_{j}C_{j}$ has
equal area and perimeter, i.e., it is equable. Thus $OA_{j}B_{j}C_{j}$ is a
LEQ. Furthermore, $OA_{j}B_{j}C_{j}$ is extangential because
$a_{j}+b_{j}=c_{j}+d_{j}$.

The first member of this family, corresponding to the initial condition
$(u_{1},v_{1})=(1,1)$, is the kite with side lengths $15,3,3,15$ shown in
Figure 9. The vertices and side lengths of the next four members of this
family are given in Table 4. The case $(u_{2},v_{2})=(7,5)$ of the family is
shown in Figure 11.

$u_{j}$ | $v_{j}$ | $A_{j}$ | $B_{j}$ | $C_{j}$ | $a_{j}$ | $b_{j}$ | $c_{j}$ | $d_{j}$
---|---|---|---|---|---|---|---|---
7 | 5 | (24,45) | (24,48) | (15,36) | 51 | 3 | 15 | 39
41 | 29 | (195,216) | (204,228) | (144,165) | 291 | 15 | 87 | 219
239 | 169 | (1188,1209) | (1248,1272) | (891,912) | 1695 | 87 | 507 | 1275
1393 | 985 | (6975,6996) | (7332,7356) | (5244,5265) | 9879 | 507 | 2955 | 7431

Table 4. Four members of the $(\Sigma,T)=(9,18)$ family
$4$$8$$12$$16$$20$$24$$28$$32$$4$$8$$12$$16$$20$$24$$28$$32$$36$$40$$44$$48$$52$$56$$60$$O$
Figure 11. The case $j=2$ of the $(\Sigma,T)=(9,18)$ family

The exradius of $OA_{j}B_{j}C_{j}$ is
$r_{e,j}=\frac{K(OA_{j}B_{j}C_{j})}{a_{j}-c_{j}}$. Substituting the values
gives $r_{e,j}=3$ for each $j$. Then by Proposition 8, the excenter is
$I_{e,j}=\frac{3}{2}\,\frac{a_{j}C_{j}+d_{j}A_{j}}{K(OA_{j}C_{j})}$, which
simplifies to $15(1,-1)+21(u_{j}+7v_{j})(1,1)$. In particular, the excenter of
each family member is a lattice point. By Proposition 10, the coordinate
$\lambda_{e,j}$ of the excenter is
$\lambda_{e,j}=\frac{3(a_{j}+b_{j})}{2K(OA_{j}C_{j})-3(a_{j}-c_{j})}$.
Substituting the values gives $\lambda_{e,j}=2$ for each $j$.

######  Remark 19.

By Remark 18, an extangential LEQ $OABC$ is concave if and only if
$\Sigma>8\frac{a+b}{a-c}$; that is, $\Sigma>8h$. For the above family, with
$(\Sigma,T)=(9,18)$, we have $h=\frac{3}{2}$. So all members of this family
are convex.

Now suppose we have an extangential LEQ with $a>c\geq b$ and $\Sigma=18,T=50$.
Note that this is the other case in Theorem 3(a). So $h=\sqrt{\frac{\Sigma
T}{8(T-\Sigma)}}=\frac{15}{8}$ and (84) gives

(91) $b=\frac{1}{8}(7a-15c).$

Hence $a>15/7c$ and since $b\leq c$, we have $a\leq 23c/7$. From (85),

(92) $a=29c-12\sqrt{5c^{2}-4}.$

So $c$ satisfies the Pell-like equation

(93) $u^{2}-5c^{2}=-4,$

for some positive integer $u$. Then (91) and (92) give

(94) $(a,b,c,d)=(29c-12u,\frac{1}{2}(47c-21u),c,\frac{1}{2}(103c-45u)).$

From [27], the solutions to (93) are $(u_{j},c_{j})=(L_{2j-1},F_{2j-1})$,
where $L_{j}$ is the $j$-th Lucas number and $F_{j}$ is the $j$-th Fibonacci
number. (Recall that the Lucas and Fibonacci numbers satisfy the same
recurrence relation but with different initial conditions: $F_{1}=F_{2}=1$
while $L_{1}=1,L_{2}=3$). Hence (94) gives the potential solutions

(95)
$(a_{j},b_{j},c_{j},d_{j})=F_{2j-1}\big{(}29,\frac{47}{2},1,\frac{103}{2}\big{)}-L_{2j-1}\big{(}12,\frac{21}{2},0,\frac{45}{2}\big{)}.$

From Lemma 13, $T(c-b)<\Sigma(a-b)$. So $50(c-b)<18(a-b)$ and thus from (91)
and (92), $51c>23u$, which is

(96) $51F_{2j-1}>23L_{2j-1}.$

Now $F_{6}=8,L_{6}=18$ and $F_{7}=13,L_{7}=29$, and thus $51F_{6}<23L_{6}$ and
$51F_{7}<23L_{7}$. It follows from the Lucas and Fibonacci recurrence relation
that (96) only holds for $j=1,2,3$. For $i=1,2$ one finds using (95) that
$a_{j}<d_{j}$, contrary to our hypothesis. So the only possibility is $j=3$,
which gives the solution $(a,b,c,d)=(13,2,5,10)$, which is the concave LEQ
shown on the right of Figure 10.

Now suppose we have an extangential LEQ with $a>c\geq b$ and $\Sigma=45,T=50$.
Note that this is the case $m=3$ in Theorem 3(b). So $h=\sqrt{\frac{\Sigma
T}{8(T-\Sigma)}}=\frac{15}{2}$ and (84) gives

(97) $b=\frac{1}{2}(13a-15c).$

Hence $a>15/13c$ and since $b\leq c$, we have $a\leq 17c/13$. From (85),

(98) $a=\frac{1}{5}(109c-12\sqrt{74c^{2}-25}).$

So $c$ satisfies the Pell-like equation

(99) $W^{2}-74c^{2}=-25,$

for some positive integer $W$. The solutions to this equation are not readily
enumerated, and moreover, some solutions do not result in LEQs. For example,
for the solution $W=5927,c=689$, one obtains the non-integer value
$a=\frac{3977}{5}$ from (98). We will restrict ourselves to constructing a
particular infinite family of LEQs for which $c$ is divisible by $5$. Set
$w=5u,c=5v$, so that (98) gives the negative Pell equation $u^{2}-74v^{2}=-1$.
Let us denote its solutions $(u_{j},v_{j})$, where $(u_{1},v_{1})=(43,5)$ and
$(u_{2},v_{2})=(318157,36985)$. So

(100) $u_{j}^{2}-74v_{j}^{2}=-1.$

The solutions $(u_{j},v_{j})$ are well known; see entries A228546 and A228547
in [35]. In particular, they satisfy the second order recurrence relation

(101) $X_{j+2}=7398X_{j+1}-X_{j}.$

It follows from this recurrence relation and the initial conditions that
$u_{j}$ is divisible by 43 and $v_{j}$ is divisible by 5 for all $j$. Let
$x_{j}=u_{j}/43,y_{j}=v_{j}/5$, so

(102) $43^{2}x_{j}^{2}-74\cdot 25\,y_{j}^{2}=-1,$

and $(x_{1},y_{1})=(1,1)$ and $(x_{2},y_{2})=(7399,7397)$. Note that
$(x_{j},y_{j})$ also satisfies the recurrence relation (101). We have
$c_{j}=25y_{j}$ and from (98), $a_{j}=\frac{1}{5}(109c_{j}-60u_{j})=-12\cdot
43x_{j}+545y_{j}$. Then from (97),
$b_{j}=\frac{1}{2}(13a_{j}-15c_{j})=-3354x_{j}+3355y_{j}$. Thus, as
$d_{j}=a_{j}+b_{j}-c_{j}$,

(103)
$(a_{j},b_{j},c_{j},d_{j})=-x_{j}(516,3354,0,3870)+y_{j}(545,3355,25,3875).$

In particular, $(a_{j},b_{j},c_{j},d_{j})$ are determined by the recurrence
relation (101) with $(a_{1},b_{1},c_{1},d_{1})=(29,1,25,5)$ and
$(a_{2},b_{2},c_{2},d_{2})=(213481,689,184925,29245)$. The first three members
of this family are shown in Table 5.

$a_{j}$ | $b_{j}$ | $c_{j}$ | $d_{j}$
---|---|---|---
29 | 1 | 25 | 5
213481 | 689 | 184925 | 29245
1579332409 | 5097221 | 1368075125 | 216354505

Table 5. Side lengths of three members of the $(\Sigma,T)=(45,50)$ family

We have shown above that if an extangential LEQ with side lengths $a,b,c,d$
has $(\Sigma,T)=(45,50)$ and $c$ is divisible by 5, then
$(a,b,c,d)=(a_{j},b_{j},c_{j},d_{j})$ for some $j$, where
$(a_{j},b_{j},c_{j},d_{j})$ is given by (103). We now show that conversely,
for each $j$, the 4-tuples $(a_{j},b_{j},c_{j},d_{j})$ given by (103) are
realised by the side lengths of an extangential LEQ. We mention in passing
that the difficulty in determining suitable vertices is two-fold: firstly, the
quadrilaterals in this family grow so fast that we only had three examples to
base our study on, and secondly, the pattern of the vertex coordinates is
considerably more complicated than in the $(\Sigma,T)=(8,18)$ family exhibited
above.

To define the vertices, we will employ the following two first-order
recurrence relations in two variables

(104) $\displaystyle x_{j+1}$ $\displaystyle=78x_{j}+25y_{j},\qquad
y_{j+1}=25x_{j}+8y_{j},$ (105) $\displaystyle x_{j+1}$
$\displaystyle=68x_{j}+35y_{j},\qquad y_{j+1}=35x_{j}+18y_{j},$

under various initial conditions. We identify the 2-tuple
$(x,y)\in\mathbb{Z}^{2}$ with the Gaussian integer $x+yi\in\mathbb{C}$, and we
use the notations interchangeably, according to convenience. Consider the
following families, for all $j\geq 1$:

  1. (a)

$z_{a,j}=x_{a,j}+y_{a,j}i$ satisfies (104) with $z_{a,1}=5+2i$,

  2. (b)

$z_{b,j}=x_{b,j}+y_{b,j}i$ satisfies (104) with $z_{b,1}=i$,

  3. (c)

$z_{c,j}=x_{c,j}+y_{c,j}i$ satisfies (105) with $z_{c,1}=2+i$,

  4. (d)

$z_{d,j}=x_{d,j}+y_{d,j}i$ satisfies (105) with $z_{d,1}=1$.

Let $\rho:\mathbb{Z}^{2}\to\mathbb{Z}^{2}$ denote the reflection in the line
$y=x$, so $\rho(x,y)=(y,x)$ or equivalently $\rho(z)=i\bar{z}$, where
$\bar{z}$ denotes the complex conjugate of $z$. Let $\rho^{j}$ denote the
$j$-th iterate of $\rho$ under composition, so $\rho^{j}=\rho$ if $j$ is odd
and $\rho^{j}=\operatorname{id}$ otherwise. We now define the vertices. Set

$\displaystyle A_{j}$ $\displaystyle=\rho^{j+1}z_{a,j}^{2},\quad
B_{j}=A_{j}+\rho^{j+1}z_{b,j}^{2},\quad C_{j}=5\rho^{j}z_{d,j}^{2},\quad
B^{\prime}_{j}=C_{j}+5\rho^{j}z_{c,j}^{2}.$

The first three members of this family are shown in Table 6; the LEQ given in
the first row is shown on the left of Figure 10.

$A_{j}$ | $B_{j}$ | $C_{j}$
---|---|---
(21, 20) | (20, 20) | (0, 5)
(124080,173719) | (124480,174280) | (16995,23800)
(1285155641, 917968320) | (1289303420, 920931020) | (176054900, 125753505)

Table 6. Vertices of three members of the $(\Sigma,T)=(45,50)$ family

We will establish the following properties:

  1. (i)

$B_{j}=B^{\prime}_{j}$ for all $j$,

  2. (ii)

The areas $K_{A_{j}},K_{C_{j}}$ of triangles $OA_{j}B_{j}$ and $B_{j}C_{j}O$
are positive (and hence $OA_{j}B_{j}C_{j}$ is a non-self-intersecting,
positively oriented quadrilateral),

  3. (iii)

$OA_{j}B_{j}C_{j}$ has the side lengths $a_{j},b_{j},c_{j},d_{j}$ given by
(103),

  4. (iv)

$a_{j}+b_{j}=c_{j}+d_{j}$; i.e., $OA_{j}B_{j}C_{j}$ is extangential,

  5. (v)

$a_{j}+b_{j}+c_{j}+d_{j}=K_{A_{j}}+K_{C_{j}}$; i.e., $OA_{j}B_{j}C_{j}$ is
equable.

The proofs of these properties will use the following technical results.

######  Lemma 15.

Suppose the sequence $z_{j}=(x_{j},y_{j})$ satisfies either (104) or (105).
Then for all $j\geq 1$,

  1. (a)

$z_{j+2}=86z_{j+1}+z_{j}$,

  2. (b)

$|z^{2}_{j+2}|=7398\,|z^{2}_{j+1}|-|z^{2}_{j}|$ (i.e., $|z^{2}_{j}|$ satisfies
(101)),

  3. (c)

$z^{2}_{a,j+2}=7398\,z^{2}_{a,j+1}-z^{2}_{a,j}+(-1)^{j}(17500-24500i)$,

  4. (d)

$z^{2}_{b,j+2}=7398\,z^{2}_{b,j+1}-z^{2}_{b,j}+(-1)^{j}(2500-3500i)$,

  5. (e)

$z^{2}_{c,j+2}=7398\,z^{2}_{c,j+1}-z^{2}_{c,j}-(-1)^{j}(700-500i)$,

  6. (f)

$z^{2}_{d,j+2}=7398\,z^{2}_{d,j+1}-z^{2}_{d,j}-(-1)^{j}(4900-3500i)$.

###### Proof.

(a). It is easy to see that if $z_{j}$ (regarded as a 2-tuple) satisfies the
relation $z_{j+1}=Mz_{j}$, where

$M=\begin{pmatrix}\alpha&\beta\\\ \gamma&\delta\end{pmatrix},$

then $z_{j+2}=\operatorname{tr}(M)z_{j+1}-\det(M)z_{j}$. The result follows as
$68+18=78+8=86$ and $68\cdot 18-35^{2}=78\cdot 8-25^{2}=-1$.

(b). From (a) we have
$|z^{2}_{j+2}|=(86x_{j+1}+x_{j})^{2}+(86y_{j+1}+y_{j})^{2}$, so

$|z^{2}_{j+2}|=86^{2}\,|z^{2}_{j+1}|+|z^{2}_{j}|+172(x_{j+1}x_{j}+y_{j}y_{j}).$

Assume first that (104) holds. Then using (104) twice,

$\displaystyle|z^{2}_{j+1}|-|z^{2}_{j}|$
$\displaystyle=(x_{j+1}^{2}+y_{j+1}^{2})-(x_{j}^{2}+y_{j}^{2})$
$\displaystyle=(68^{2}+35^{2}-1)x_{j}^{2}+(35^{2}+18^{2}-1)y_{j}^{2}+2\cdot(68+18)\cdot
35x_{j}y_{j}$ $\displaystyle=86(68x_{j}^{2}+70x_{j}y_{j}+18y_{j}^{2})$
$\displaystyle=86(x_{j+1}x_{j}+y_{j+1}y_{j}),$

so, from above,
$|z^{2}_{j+2}|=(86^{2}+2)|z^{2}_{j+1}|-|z^{2}_{j}|=7398\,|z^{2}_{j+1}|-|z^{2}_{j}|$,
as required. A similar argument applies when (105) holds.

(c). From (a) we have
$z^{2}_{a,j+2}=7396\,z^{2}_{a,j+1}+172z_{a,j+1}z_{a,j}+z^{2}_{a,j}$. So we are
required to show that
$172z_{a,j+1}z_{a,j}+2z^{2}_{a,j}=2z^{2}_{a,j+1}+(-1)^{j}(17500-24500i)$, or
equivalently, using (104),
$5x_{a,j}^{2}-14x_{a,j}y_{a,j}-5y_{a,j}^{2}=35(-1)^{j}$, for all $j$. In
matrix notation, the condition is $z_{j}Qz_{j}^{t}=35(-1)^{j}$, where
$z_{j}=(x_{a,j},y_{a,j})$, $z^{t}$ denotes the transpose of $z$, and

$Q=\begin{pmatrix}5&-7\\\ -7&-14\end{pmatrix}.$

One verifies readily that this condition holds for $j=1$, where $z_{1}=(5,2)$.
Now using (104) and setting $M=\begin{pmatrix}78&25\\\ 25&8\end{pmatrix}$, we
have $z_{j+1}Qz_{j+1}^{t}=(z_{j}M)Q(Mz_{j}^{t})=-z_{j}Qz_{j}^{t}$, since
$MQM=-Q$. So the required result follows by induction. This proves (c).

Part (d) is proven in the same manner; only the initial condition is
different. For parts (e) and (f), one repeats the argument using the matrices

$Q=\begin{pmatrix}7&-5\\\ -5&-7\end{pmatrix},\qquad M=\begin{pmatrix}68&35\\\
35&18\end{pmatrix}.$

Once again, the argument works because $MQM=-Q$. ∎

######  Remark 20.

It is well known that if a sequence $r_{j}$ satisfies a second-order
homogenous recurrence relation, then $r_{j}^{2}$ satisfies a third-order
recurrence relation, but in certain exceptional circumstances, $r_{j}^{2}$ may
satisfy a second-order recurrence relation, which is typically non-homogeneous
[10]. In this respect, Lemma 15 is perhaps somewhat surprising.

(i). We have $B_{1}=(5+2i)^{2}+i^{2}=20+20i$ and
$B^{\prime}_{1}=5\rho((2+i)^{2}+1^{2})=B_{1}$, while $B_{2}=\rho((78\cdot
5+25\cdot 2+(25\cdot 5+8\cdot 2)i)^{2}+(25+8i)^{2})=124480+174280i$, and
$B^{\prime}_{2}=5((68\cdot 2+35+(35\cdot 2+18)i)^{2}+(68+35i)^{2})=B_{2}$.
From Lemma 15(c) and (d), we have for all $j$,

$\displaystyle B_{j+2}$
$\displaystyle=7398\rho(B_{j+1})-B_{j}+(-1)^{j}\rho^{j+1}(17500-24500i+2500-3500i)$
$\displaystyle=7398\rho(B_{j+1})-B_{j}+(-1)^{j}\rho^{j+1}(20000-28000i),$

and from Lemma 15(e) and (f), we have

$\displaystyle B^{\prime}_{j+2}$
$\displaystyle=7398\rho(B^{\prime}_{j+1})-B_{j}-(-1)^{j}5\rho^{j}(700-500i+4900-3500i)$
$\displaystyle=7398\rho(B_{j+1})-B_{j}-(-1)^{j}\rho^{j}(28000-20000i)=B_{j+2}.$

Hence $B_{j}=B^{\prime}_{j}$ for all $j$.

(ii). Regarding $A_{j}$ and $B_{j}$ as complex numbers, one has
$K_{A_{j}}=\frac{i}{4}(A_{j}\overline{B_{j}}-\overline{A_{j}}B_{j})$. Hence
when $j$ is odd,

$K_{A_{j}}=\frac{i}{4}(z^{2}_{a,j}\bar{z}^{2}_{b,j}-\bar{z}^{2}_{a,j}z^{2}_{b,j})=(x_{a,j}y_{b,j}-x_{b,j}y_{a,j})(x_{a,j}x_{b,j}+y_{a,j}y_{b,j}),$

while when $j$ is even,

$\displaystyle K_{A_{j}}$
$\displaystyle=\frac{i}{4}(\rho(z^{2}_{a,j})\rho(\bar{z}^{2}_{b,j})-\rho(\bar{z}^{2}_{a,j})\rho(z^{2}_{b,j})=\frac{i}{4}(\bar{z}^{2}_{a,j}z^{2}_{b,j}-z^{2}_{a,j}\bar{z}^{2}_{b,j})$
$\displaystyle=-(x_{a,j}y_{b,j}-x_{b,j}y_{a,j})(x_{a,j}x_{b,j}+y_{a,j}y_{b,j}).$

It is easy to see that $x_{a,j}x_{b,j}+y_{a,j}y_{b,j}>0$. Moreover,
$x_{a,j}y_{b,j}-x_{b,j}y_{a,j}$ is the area $K_{j}$ of the parallelogram
$P_{j}$ spanned by $z_{a,j}$ and $z_{b,j}$. So we are required to show that