Relationships Between Circles Inscribed in Triangles and Curvilinear Triangles
Sangaku Journal of Mathematics (SJM) ©SJM
ISSN 2534-9562
Volume 4 (2020), pp.9-27
Received 19 November 2019. Published on-line 16 January 2020
web: http://www.sangaku-journal.eu/
©The Author(s) This article is published with open access111This article is
distributed under the terms of the Creative Commons Attribution License which
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original author(s) and the source are credited..

Relationships Between Circles Inscribed in Triangles and Related Curvilinear
Triangles

Stanley Rabinowitz

545 Elm St Unit 1, Milford, New Hampshire 03055, USA

e-mail<EMAIL_ADDRESS>
web: http://www.StanleyRabinowitz.com/

Abstract. If $P$ is a point inside $\triangle ABC$, then the cevians through
$P$ extended to the circumcircle of $\triangle ABC$ create a figure containing
a number of curvilinear triangles. Each curvilinear triangle is bounded by an
arc of the circumcircle and two line segments lying along the sides or cevians
of the original triangle. We give theorems about the relationships between the
radii of circles inscribed in various sets of these curvilinear triangles.

Keywords. circles, cevians, curvilinear triangles, sangaku.

Mathematics Subject Classification (2010). 51M04.

##  1\. Introduction

A curvilinear triangle is a geometric figure bounded by three curves. The
curves are typically line segments and arcs of circles, in which case there is
a unique circle tangent to each of the three boundary curves. This circle is
called the incircle of the curvilinear triangle.

Wasan geometers loved to find relationships between the radii of circles
inscribed in curvilinear triangles. An example is shown in Figure 1 which
comes from an 1841 book of Mathematical Formulae written by Yamamoto [12]. It
is also given as problem 5.3.9 in [4]. In the figure, $AH\perp BC$ and
$BA\perp AC$. There are three curvilinear triangles of interest in the figure.
The first curvilinear triangle is bounded by $BH$, $HA$, and arc
$\stackrel{{\scriptstyle\mbox{\large$\frown$}}}{{AB}}$. The second curvilinear
triangle is bounded by $AH$, $HC$, and arc
$\stackrel{{\scriptstyle\mbox{\large$\frown$}}}{{CA}}$. The third curvilinear
triangle is bounded by $CA$, $AB$, and arc
$\stackrel{{\scriptstyle\mbox{\large$\frown$}}}{{BC}}$. The radii of the
circles inscribed in these curvilinear triangles are $r_{1}$, $r_{2}$, and
$r_{3}$, respectively. Then the nice relationship that was found is
$r_{1}+r_{2}=r_{3}$.

Figure 1. $r_{1}+r_{2}=r_{3}$

In this paper, we will find some other nice relationships between the inradii
of curvilinear triangles.

If a curvilinear triangle is convex and bounded by two straight line segments
and one circular arc, then we will call the resulting figure a skewed sector
(see Figure 2).

Figure 2. a skewed sector

Anatomy of a skewed sector.

  * •

The two straight line segments are called the sides of the skewed sector.

  * •

The point of intersection of the two sides is called the vertex of the skewed
sector.

  * •

The angle between the two sides is called the vertex angle.

  * •

The circular arc is referred to as the arc of the skewed sector.

  * •

The circular measure of the arc of a skewed sector is called the arc angle.

  * •

The circle to which the arc belongs will be called the circle associated with
the skewed sector.

  * •

The triangle formed by the vertex of a skewed sector and the endpoints of its
arc will be referred to as the triangle associated with the skewed sector.
This would be $\triangle APB$ in Figure 2.

  * •

When naming a skewed sector, the vertex will always be the middle letter.
Thus, the skewed sector in Figure 2 is named skewed sector $APB$.

  * •

When the vertex of a skewed sector lies inside the associated circle, if the
sides of the skewed sector are extended back through the vertex, they will
intercept an arc of the associated circle. This arc is called the opposite arc
of the skewed sector. It is shown in red in Figure 3.

  * •

The vertex of a skewed sector and the opposite arc form another skewed sector
called the opposite skewed sector. This is skewed sector
$A^{\prime}PB^{\prime}$ in Figure 3.

Figure 3. opposite arc

A segment of a circle is the figure bounded by an arc of a circle and the
chord joining the endpoints of that arc. The height (or sagitta) of the
segment is the distance from the midpoint of the chord to the midpoint of the
arc.

If $P$ is a point inside $\triangle ABC$, then the cevians through $P$
extended to the circumcircle of $\triangle ABC$ create a figure containing a
number of skewed sectors. We will find relationships between the radii of the
circles inscribed in some of these skewed sectors.

Notation.

  * •

If $X$ and $Y$ are points, then we use the notation $XY$ to denote either the
line segment joining $X$ and $Y$ or the length of that line segment, depending
on the context.

  * •

A cevian of a triangle is a line segment from a vertex to the opposite side.

  * •

We use the notation $\angle XYZ$ to denote either the angle between $XY$ and
$YZ$ or the measure of that angle, depending on the context.

  * •

The notation [XYZ] denotes the area of $\triangle XYZ$.

  * •

The notation $O(r)$ refers to the circle centered at point $O$ with radius
$r$. The circle may sometimes also be referred to as circle $O$.

  * •

If $\stackrel{{\scriptstyle\mbox{\large$\frown$}}}{{XY}}$ is an arc of a
circle, then $m(\stackrel{{\scriptstyle\mbox{\large$\frown$}}}{{XY}})$ denotes
the circular measure of that arc. The arc extends counterclockwise along the
circle from $X$ to $Y$.

  * •

Typically, we use $r$ for the inradius of a triangle and $w$ for the inradius
of a skewed sector.

##  2\. Inradius Formula

Formulas for the radius of the circle inscribed in a skewed sector and in a
triangle are known. Since these are not well-known, we review them here.

######  Theorem 2.1 (Inradius of skewed sector).

Let $APB$ be a skewed sector and let $\mathcal{C}$ be the circle associated
with arc $\stackrel{{\scriptstyle\mbox{\large$\frown$}}}{{AB}}$. Suppose $P$
lies inside $\mathcal{C}$. Let $R$ be the radius of $\mathcal{C}$, let $w$ be
the radius of the circle inscribed in the skewed sector, and let $r$ be the
radius of the circle inscribed in $\triangle APB$. Extend $BP$ to meet the
circle $\mathcal{C}$ at $C$ and draw $AC$. Let $\alpha$, $\beta$, $\gamma$,
$\delta$, and $\epsilon$ be the measures of five angles associated with the
skewed sector as shown in Figure 4. Then

$\displaystyle w$ $\displaystyle=\frac{\displaystyle 4R\sin{\beta\over
2}\sin{\gamma\over 2}\cos{\delta\over 2}\sin{\epsilon\over
2}}{\displaystyle\left(\cos{\alpha\over 2}\right)^{2}},$ $\displaystyle r$
$\displaystyle=\frac{\displaystyle 4R\sin{\beta\over 2}\sin{\gamma\over
2}\sin{\epsilon\over 2}\cos{\epsilon\over 2}}{\displaystyle\cos{\alpha\over
2}}.$

Figure 4. angles associated with a skewed sector

###### Proof.

See [4, pp. 96–97] or [11, p. 26]. ∎

An immediate consequence of this theorem is the following result.

######  Theorem 2.2.

Let $APB$ be a skewed sector and suppose $P$ lies inside the associated
circle. Let $w$ be the radius of the circle inscribed in the skewed sector,
and let $r$ be the radius of the circle inscribed in $\triangle APB$. Let
$\alpha$ be the vertex angle of the skewed sector, let $\theta_{1}$ be the arc
angle of the skewed sector, and let $\theta_{2}$ be the arc angle of the
opposite skewed sector. See Figure 5. Then

$\frac{w}{r}=\frac{\cos(\theta_{2}/4)}{\cos(\alpha/2)\cos(\theta_{1}/4)}.$

Figure 5.

We can also express $w/r$ without using $\theta_{2}$ as follows.

######  Theorem 2.3.

Using the same notation as in Theorem 2.2,

$\frac{w}{r}=1+\tan(\alpha/2)\tan(\theta_{1}/4).$

###### Proof.

See [11, pp. 26–27]. ∎

##  3\. Relationship Between Incircles of Skewed Sectors and Incircles of
Triangles

To prove a relationship between skewed sector inradii, Theorems 2.1, 2.2, or
2.3 could be used to find the length of each radius. This is a brute force
technique and better methods are available. One strategy for finding
relationships between the radii of circles inscribed in skewed sectors is to
relate these circles to circles inscribed in triangles, for which results are
already known.

Here are some theorems that relate circles in skewed sectors to circles in
triangles.

The following theorem appeared on a tablet in 1781. See [4, problem 4.0.3],
[3, problem 2.2.8], [5], and [9].

######  Theorem 3.1 (Ajima’s Theorem).

Let $AB$ be a chord of a circle and let $C$ be a point inside the circle, not
on the chord. See Figure 6. Let $W(w)$ be the incircle of skewed sector $ACB$
(the red circle) and let $O(r)$ be the incircle of $\triangle ACB$ (the yellow
circle). Then

$w=r+\frac{2d(s-a)(s-b)}{cs},$

where $d$ is the height of the segment formed by $AB$, $a=BC$, $b=AC$, $c=AB$,
and $s$ is the semiperimeter of $\triangle ABC$.

Figure 6. $w=r+2d(s-a)(s-b)/(cs)$

###### Proof.

See [4, pp. 96-97]. A more detailed proof can be found in [10, pp. 40-49]. ∎

######  Theorem 3.2.

Let $D$ be any point on side $BC$ of $\triangle ABC$. Cevian $AD$ extended
meets the circumcircle of $\triangle ABC$ at $D^{\prime}$. Let $W_{1}(w_{1})$
be the incircle of skewed sector $BDD^{\prime}$ and let $W_{2}(w_{2})$ be the
incircle of skewed sector $CDD^{\prime}$. Let $O_{1}(r_{1})$ be the incircle
of $\triangle ADB$ and let $O_{2}(r_{2})$ be the incircle of $\triangle ADC$
(Figure 7). Then

$\frac{1}{r_{1}}+\frac{1}{w_{2}}=\frac{1}{r_{2}}+\frac{1}{w_{1}}.$

Figure 7. $1/r_{1}+1/w_{2}=1/r_{2}+1/w_{1}$

###### Proof.

We give names to the various angles as shown in Figure 8.

Figure 8. angle names

Note that $\angle CDA=\alpha_{2}$ is supplementary to $\angle BDA=\alpha$, so
$\cos(\alpha_{2}/2)=\sin(\alpha/2)$. Four applications of Theorem 2.1 gives

$\displaystyle w_{1}$
$\displaystyle=\frac{4R\sin(\beta/2)\cos(\gamma/2)\sin(\delta/2)\sin(\epsilon/2)}{\sin^{2}(\alpha/2)},$
$\displaystyle w_{2}$
$\displaystyle=\frac{4R\sin(\beta/2)\sin(\gamma/2)\sin(\delta/2)\cos(\epsilon/2)}{\cos^{2}(\alpha/2)},$
$\displaystyle r_{1}$
$\displaystyle=\frac{4R\sin(\beta/2)\sin(\gamma/2)\sin(\epsilon/2)\cos(\epsilon/2)}{\cos(\alpha/2)},$
$\displaystyle r_{2}$
$\displaystyle=\frac{4R\sin(\delta/2)\sin(\epsilon/2)\sin(\gamma/2)\cos(\gamma/2)}{\cos(\alpha/2)}.$

Now form the expression

$S=\frac{1}{w_{1}}+\frac{1}{r_{2}}-\left(\frac{1}{w_{2}}+\frac{1}{r_{1}}\right).$

Then substitute $\alpha=\delta+\epsilon$ and then
$\epsilon=\pi-\beta-\gamma-\delta$. Simplifying the resulting expression
(using a computer algebra system), shows that $S=0$. ∎

The result of Theorem 3.2 is so elegant that it is unlikely that it is true
only because the complicated trigonometric expression, $S$, in the proof just
happens to simplify to 0.

######  Open Question.

Is there a simple proof of Theorem 3.2 that does not involve a large amount of
trigonometric computation requiring computer simplification?

The following theorem appeared on a tablet in 1844 in the Aichi prefecture.
See [3, problem 1.4.7] and [2, p. 22].

######  Theorem 3.3.

Chords $AB$ and $CD$ of a circle meet at $E$. Let $W_{1}(w_{1})$ be the
incircle of skewed sector $BED$ and let $W_{2}(w_{2})$ be the incircle of
skewed sector $AEC$. Let $O_{1}(r_{1})$ be the incircle of $\triangle BED$ and
let $O_{2}(r_{2})$ be the incircle of $\triangle AEC$. See Figure 9. Then

$\frac{1}{r_{1}}+\frac{1}{w_{2}}=\frac{1}{r_{2}}+\frac{1}{w_{1}}.$

Figure 9. $1/r_{1}+1/w_{2}=1/r_{2}+1/w_{1}$

###### Proof.

This proof comes from [11, pp. 26–27]. Let $\alpha$ be the vertex angle of
skewed sector $BED$. Let $\theta_{1}$ be its arc angle an let $\theta_{2}$ be
the arc angle of the opposite skewed sector $AEC$. By Theorem 2.3, we have

$\frac{w_{1}}{r_{1}}=1+\tan(\alpha/2)\tan(\theta_{1}/4)$

which is equivalent to

$\frac{w_{1}}{r_{1}}-1=\tan(\alpha/2)\tan(\theta_{1}/4)$

or

$\frac{1}{r_{1}}-\frac{1}{w_{1}}=\frac{\tan(\alpha/2)\tan(\theta_{1}/4)}{w_{1}}.$

Similarly,

$\frac{1}{r_{2}}-\frac{1}{w_{2}}=\frac{\tan(\alpha/2)\tan(\theta_{2}/4)}{w_{2}}.$

But

$\frac{\tan(\theta_{1}/4)}{w_{1}}=\frac{\tan(\theta_{2}/4)}{w_{2}}$

by Theorem 4.2 (which will be proved in the next section). Thus,

$\frac{1}{r_{1}}-\frac{1}{w_{1}}=\frac{1}{r_{2}}-\frac{1}{w_{2}}$

and the result follows. ∎

######  Theorem 3.4.

Chords $AB$ and $CD$ of a circle meet at $E$, with $\angle AEC=\alpha$. Let
$W_{1}(w_{1})$ be the incircle of skewed sector $BED$ and let $W_{2}(w_{2})$
be the incircle of skewed sector $AEC$. Let $O_{1}(r_{1})$ be the incircle of
$\triangle BED$ and let $O_{2}(r_{2})$ be the incircle of $\triangle AEC$. See
Figure 9. Then

$r_{1}r_{2}=w_{1}w_{2}\cos^{2}\frac{\alpha}{2}.$

###### Proof.

Let $m(\stackrel{{\scriptstyle\mbox{\large$\frown$}}}{{DB}})=\theta_{1}$ and
$m(\stackrel{{\scriptstyle\mbox{\large$\frown$}}}{{CA}})=\theta_{2}$. Applying
Theorem 2.2 to skewed sector $BED$ gives

$\frac{w_{1}}{r_{1}}=\frac{\cos(\theta_{2}/4)}{\cos(\alpha/2)\cos(\theta_{1}/4)}.$

Applying Theorem 2.2 to skewed sector $AEC$ gives

$\frac{w_{2}}{r_{2}}=\frac{\cos(\theta_{1}/4)}{\cos(\alpha/2)\cos(\theta_{2}/4)}.$

Multiplying these two equations gives

$\frac{w_{1}w_{2}}{r_{1}r_{2}}=\frac{1}{\cos^{2}(\alpha/2)}$

and the result follows by cross-multiplying. ∎

######  Theorem 3.5.

Cevians $AD$ and $CF$ in $\triangle ABC$ meet at $P$ and $\angle BFC=\angle
BDA$. The cevians are extended to meet the circumcircle of $\triangle ABC$ at
points $D^{\prime}$ and $F^{\prime}$, respectively, as shown in Figure 10. Let
$W_{1}(w_{1})$ be the incircle of skewed sector $BDD^{\prime}$ and let
$W_{2}(w_{2})$ be the incircle of skewed sector $BFF^{\prime}$. Let
$O_{1}(r_{1})$ be the incircle of $\triangle BDD^{\prime}$ and let
$O_{2}(r_{2})$ be the incircle of $\triangle BFF^{\prime}$. Then

$\frac{w_{1}}{r_{1}}=\frac{w_{2}}{r_{2}}.$

Figure 10. $w_{1}/r_{1}=w_{2}/r_{2}$

###### Proof.

Let
$m(\stackrel{{\scriptstyle\mbox{\large$\frown$}}}{{BD^{\prime}}})=\theta_{1}$,
$m(\stackrel{{\scriptstyle\mbox{\large$\frown$}}}{{F^{\prime}B}})=\theta_{2}$,
and $m(\stackrel{{\scriptstyle\mbox{\large$\frown$}}}{{CA}})=\phi$. Applying
Theorem 2.2 to skewed sector $BDD^{\prime}$ using $\alpha_{1}=\angle
BDD^{\prime}$ gives

$\frac{w_{1}}{r_{1}}=\frac{\cos(\phi/4)}{\cos(\alpha_{1}/2)\cos(\theta_{1}/4)}.$

Applying Theorem 2.2 to skewed sector $BFF^{\prime}$ using $\alpha_{2}=\angle
BFF^{\prime}$ gives

$\frac{w_{2}}{r_{2}}=\frac{\cos(\phi/4)}{\cos(\alpha_{2}/2)\cos(\theta_{2}/4)}.$

But $\alpha_{1}=\alpha_{2}$ because they are supplementary to the two given
angles. Chords $AD^{\prime}$ and $BC$ intercept arcs of measures $\theta_{1}$
and $\phi$, so $\alpha_{1}=(\theta_{1}+\phi)/2$. Similarly,
$\alpha_{2}=(\theta_{2}+\phi)/2$. Thus, $\theta_{1}=\theta_{2}$ because
$\alpha_{1}=\alpha_{2}$. Therefore, $w_{1}/r_{1}=w_{2}/r_{2}$. ∎

######  Theorem 3.6.

Let $H$ be the orthocenter of acute triangle $ABC$. The altitudes $AD$ and
$CF$ are extended to meet the circumcircle of $\triangle ABC$ at points
$D^{\prime}$ and $F^{\prime}$, respectively. Let $W_{1}(w_{1})$ be the
incircle of skewed sector $BDD^{\prime}$ and let $W_{2}(w_{2})$ be the
incircle of skewed sector $BFF^{\prime}$. Let $O_{1}(r_{1})$ be the incircle
of $\triangle BDH$ and let $O_{2}(r_{2})$ be the incircle of $\triangle BFH$.
See Figure 11. Then

$\frac{w_{1}}{r_{1}}=\frac{w_{2}}{r_{2}}.$

Figure 11. $w_{1}/r_{1}=w_{2}/r_{2}$

###### Proof.

Let $r^{\prime}_{1}$ be the inradius of $\triangle BDD^{\prime}$ and let
$r^{\prime}_{2}$ be the inradius of $\triangle BFF^{\prime}$. Since $\angle
BFC=\angle BDA$, by Theorem 3.5, we have
$w_{1}/r^{\prime}_{1}=w_{2}/r^{\prime}_{2}$. Now $\angle CBD^{\prime}=\angle
CAD^{\prime}$ since both angles subtend the same arc. But $\angle
CAD^{\prime}=\angle CBE^{\prime}$ since both angles are complementary to
$\angle ACB$. Thus, $\angle DBD^{\prime}=\angle DBH$. Right triangles
$BDD^{\prime}$ and $BDH$ share a common side. Thus $\triangle
BDD^{\prime}\cong\triangle BDH$. Hence $r_{1}=r^{\prime}_{1}$. Similarly,
$r_{2}=r^{\prime}_{2}$. Therefore, $w_{1}/r_{1}=w_{2}/r_{2}$. ∎

######  Lemma 3.7.

Let $H$ be the orthocenter of acute $\triangle ABC$, and let the altitudes be
$AD$, $BE$, and $CF$ as shown in Figure 12. Circles $O_{1}(r_{1})$,
$O_{2}(r_{2})$, $O_{3}(r_{3})$, and $O_{4}(r_{4})$ are inscribed in triangles
$BHD$, $BHF$, $CAF$, and $ACD$, respectively. Then $r_{1}/r_{2}=r_{4}/r_{3}$.

Figure 12. $r_{1}/r_{2}=r_{4}/r_{3}$

###### Proof.

Note that $\triangle BHF\sim\triangle CAF$. Therefore the figure consisting of
$\triangle BHF$ and its incircle is similar to the figure consisting of
$\triangle CAF$ and its incircle. Corresponding parts of similar figures are
in proportion, so $r_{2}/r_{3}=BH/CA$. In the same manner, $\triangle
BHD\sim\triangle ACD$ which implies that $r_{1}/r_{4}=BH/AC$. Therefore,
$r_{2}/r_{3}=r_{1}/r_{4}$ or $r_{1}/r_{2}=r_{4}/r_{3}$. ∎

######  Corollary 3.8.

Let $H$ be the orthocenter of acute $\triangle ABC$. The altitudes $AD$ and
$CF$ are extended to meet the circumcircle of $\triangle ABC$ at points
$D^{\prime}$ and $F^{\prime}$, respectively. Let $W_{1}(w_{1})$ be the
incircle of skewed sector $BDD^{\prime}$ and let $W_{2}(w_{2})$ be the
incircle of skewed sector $BFF^{\prime}$. Let $O_{3}(r_{3})$ be the incircle
of $\triangle AFC$ and let $O_{4}(r_{4})$ be the incircle of $\triangle ADC$.
See Figure 13. Then $w_{1}/w_{2}=r_{4}/r_{3}$.

Figure 13. $w_{1}/w_{2}=r_{4}/r_{3}$

###### Proof.

By Theorem 3.6, $w_{1}/w_{2}=r_{1}/r_{2}$. By Lemma 3.7,
$r_{1}/r_{2}=r_{4}/r_{3}$. Therefore, $w_{1}/w_{2}=r_{4}/r_{3}$. ∎

##  4\. Relationships Between the Incircles of Two Skewed Sectors

######  Theorem 4.1.

Chords $AB$ and $CD$ of a circle meet at $E$. Let $W_{1}(w_{1})$ be the circle
inscribed in skewed sector $DEB$ and let $W_{2}(w_{2})$ be circle inscribed in
skewed sector $AEC$, as shown in Figure 14. Then

$\frac{w_{1}}{w_{2}}=\frac{\tan(\theta_{1}/2)}{\tan(\theta_{2}/2)}$

where $\angle BCD=\theta_{1}$ and $\angle ADC=\theta_{2}$.

Figure 14. $w_{1}/w_{2}=\tan(\theta_{1}/2)/\tan(\theta_{2}/2)$

###### Proof.

See [4, pp. 96–97] or [11, p. 26–27]. ∎

Since the measure of an angle inscribed in a circle is half the circular
measure of the intercepted arc, we have the following result.

######  Theorem 4.2.

Chords $AB$ and $CD$ of a circle meet at $E$. Let $W_{1}(w_{1})$ be the circle
inscribed in skewed sector $DEB$ and let $W_{2}(w_{2})$ be the circle
inscribed in skewed sector $AEC$, as shown in Figure 15. Then

$\frac{w_{1}}{w_{2}}=\frac{\tan(\theta_{1}/4)}{\tan(\theta_{2}/4)}$

where $m(\stackrel{{\scriptstyle\mbox{\large$\frown$}}}{{CA}})=\theta_{1}$ and
$m(\stackrel{{\scriptstyle\mbox{\large$\frown$}}}{{DB}})=\theta_{2}$.

Figure 15. $w_{1}/w_{2}=\tan(\theta_{1}/4)/\tan(\theta_{2}/4)$

The reader may wonder if there is any geometric significance to the angle
$\theta_{1}/4$. If $M$ is the midpoint of arc
$\stackrel{{\scriptstyle\mbox{\large$\frown$}}}{{DB}}$, then $\angle
BDM=\frac{1}{2}\stackrel{{\scriptstyle\mbox{\large$\frown$}}}{{MB}}$ and
$\stackrel{{\scriptstyle\mbox{\large$\frown$}}}{{MB}}=\frac{1}{2}\stackrel{{\scriptstyle\mbox{\large$\frown$}}}{{DB}}=\frac{1}{2}\theta_{1}$,
so $\angle BDM=\theta_{1}/4$.

There is a related result involving the incircles of triangles.

######  Theorem 4.3.

Chords $B_{1}B_{2}$ and $C_{1}C_{2}$ of a circle meet at $A$. Let $r_{1}$ and
$r_{2}$ be the inradii of $\triangle B_{1}AC_{1}$ and $\triangle B_{2}AC_{2}$,
respectively, as shown in Figure 16. Let $B_{1}C_{1}=a_{1}$ and let
$B_{2}C_{2}=a_{2}$. Then $=r_{1}/r_{2}=a_{1}/a_{2}$.

Figure 16. $r_{1}/r_{2}=a_{1}/a_{2}$

###### Proof.

This follows from the fact that $\triangle B_{1}AC_{1}\sim\triangle
B_{2}AC_{2}$. ∎

The following theorem comes from [11, Problem 21] and is related to Ajima’s
Theorem.

######  Theorem 4.4.

Chords $B_{1}B_{2}$ and $C_{1}C_{2}$ of a circle meet at $A$. Let
$W_{1}(w_{1})$ be the circle inscribed in skewed sector $B_{1}AC_{1}$ and let
$W_{2}(w_{2})$ be the circle inscribed in skewed sector $B_{2}AC_{2}$. Let
$v_{1}$ and $v_{2}$ be the heights of the segments formed by chords
$B_{1}C_{1}$ and $B_{2}C_{2}$ as shown in Figure 17. Then

$\frac{w_{1}}{w_{2}}=\frac{v_{1}a_{2}}{v_{2}a_{1}}$

where $a_{1}=B_{1}C_{1}$ and $a_{2}=B_{2}C_{2}$.

Figure 17. $w_{1}/w_{2}=v_{1}a_{2}/v_{2}a_{1}$

The following result is due to Pohoatza and Ehrmann, [6].

######  Theorem 4.5.

Let $D$ be the point on side $BC$ of $\triangle ABC$ such that $AB+BD=AC+CD$.
A circle is circumscribed about $\triangle ABC$. Let $W_{1}(w_{1})$ be the
circle inscribed in skewed sector $ADB$ and let $W_{2}(w_{2})$ be the circle
inscribed in skewed sector $ADC$ (Figure 18). Then $w_{1}=w_{2}$.

Figure 18. $w_{1}=w_{2}$

###### Proof.

Extend $AD$ to meet the circumcircle of $\triangle ABC$ at $D^{\prime}$. Let
$O_{1}(r_{1})$ be the circle inscribed in $\triangle BDD^{\prime}$ and let
$O_{2}(r_{2})$ be the circle inscribed in $\triangle CDD^{\prime}$ (Figure
19). Then $1/r_{1}+1/w_{2}=1/r_{2}+1/w_{1}$ by Theorem 3.2 (with points $A$
and $D^{\prime}$ interchanged). But $r_{1}=r_{2}$ by Theorem 3.4 of [8].
Therefore, $w_{1}=w_{2}$. ∎

See [1] for another proof.

Figure 19. $r_{1}=r_{2}$ and $w_{1}=w_{2}$

##  5\. Relationships Between the Incircles of Six Skewed Sectors

######  Theorem 5.1.

Let $H$ be the orthocenter of acute $\triangle ABC$. The altitudes through $H$
extended to meet the circumcircle of $\triangle ABC$ divide the interior of
that circumcircle into six skewed sectors, each with vertex at $H$, as shown
in Figure 20. Let $W_{i}(w_{i})$ be the circle tangent to two altitudes and
internally tangent to the circumcircle as shown. Then
$w_{1}w_{3}w_{5}=w_{2}w_{4}w_{6}$.

Figure 20. $w_{1}w_{3}w_{5}=w_{2}w_{4}w_{6}$

###### Proof.

Let $\theta_{i}$ be the arc angle of the skewed sector containing circle
$W_{i}$. By Theorem 4.2, $w_{1}/w_{4}=\tan(\theta_{1}/4)/\tan(\theta_{4}/4)$.
Similarly, $w_{2}/w_{5}=\tan(\theta_{2}/4)/\tan(\theta_{5}/4)$ and
$w_{3}/w_{6}=\tan(\theta_{3}/4)/\tan(\theta_{6}/4)$. Consequently,

$\frac{w_{1}w_{3}w_{5}}{w_{2}w_{4}w_{6}}=\frac{w_{1}}{w_{4}}\cdot\frac{w_{3}}{w_{6}}\cdot\frac{w_{5}}{w_{2}}=\frac{\tan(\theta_{1}/4)}{\tan(\theta_{4}/4)}\cdot\frac{\tan(\theta_{3}/4)}{\tan(\theta_{6}/4)}\cdot\frac{\tan(\theta_{5}/4)}{\tan(\theta_{2}/4)}.$

Note that $\angle BAH=\angle BCH$ since both are complementary to $\angle
ABC$. Therefore, $\theta_{1}=\theta_{6}$. Similarly, $\theta_{2}=\theta_{3}$
and $\theta_{4}=\theta_{5}$. Hence

$\frac{w_{1}w_{3}w_{5}}{w_{2}w_{4}w_{6}}=\frac{\tan(\theta_{1}/4)}{\tan(\theta_{5}/4)}\cdot\frac{\tan(\theta_{3}/4)}{\tan(\theta_{1}/4)}\cdot\frac{\tan(\theta_{5}/4)}{\tan(\theta_{3}/4)}=1,$

so $w_{1}w_{3}w_{5}=w_{2}w_{4}w_{6}$. ∎

######  Theorem 5.2.

Let $I$ be the incenter of $\triangle ABC$. The cevians through $I$ extended
to meet the circumcircle of $\triangle ABC$ divide the interior of that
circumcircle into six skewed sectors, each with vertex at $I$, as shown in
Figure 21. Let $W_{i}(w_{i})$ be the circle tangent to two cevians and
internally tangent to the circumcircle as shown. Then
$w_{1}w_{3}w_{5}=w_{2}w_{4}w_{6}$.

Figure 21. $w_{1}w_{3}w_{5}=w_{2}w_{4}w_{6}$

###### Proof.

Let $\theta_{i}$ be the arc angle of the skewed sector containing circle
$W_{i}(w_{i})$. By Theorem 4.2,
$w_{1}/w_{4}=\tan(\theta_{1}/4)/\tan(\theta_{4}/4)$. Similarly,
$w_{2}/w_{5}=\tan(\theta_{2}/4)/\tan(\theta_{5}/4)$ and
$w_{3}/w_{6}=\tan(\theta_{3}/4)/\tan(\theta_{6}/4)$. Consequently,

$\frac{w_{1}w_{3}w_{5}}{w_{2}w_{4}w_{6}}=\frac{w_{1}}{w_{4}}\cdot\frac{w_{3}}{w_{6}}\cdot\frac{w_{5}}{w_{2}}=\frac{\tan(\theta_{1}/4)}{\tan(\theta_{4}/4)}\cdot\frac{\tan(\theta_{3}/4)}{\tan(\theta_{6}/4)}\cdot\frac{\tan(\theta_{5}/4)}{\tan(\theta_{2}/4)}.$

Note that $\angle BAI=\angle CAI$ since $AI$ is an angle bisector. Therefore,
$\theta_{1}=\theta_{2}$. Similarly, $\theta_{3}=\theta_{4}$ and
$\theta_{5}=\theta_{6}$. Hence

$\frac{w_{1}w_{3}w_{5}}{w_{2}w_{4}w_{6}}=\frac{\tan(\theta_{1}/4)}{\tan(\theta_{3}/4)}\cdot\frac{\tan(\theta_{3}/4)}{\tan(\theta_{5}/4)}\cdot\frac{\tan(\theta_{5}/4)}{\tan(\theta_{1}/4)}=1,$

so $w_{1}w_{3}w_{5}=w_{2}w_{4}w_{6}$. ∎

######  Theorem 5.3.

Let $H$ be the orthocenter of acute $\triangle ABC$. The altitudes through $H$
extended to meet the circumcircle of $\triangle ABC$ divide the segments of
the circumcircle bounded by the sides of the triangle into two skewed sectors
each as shown in Figure 22. Let $W_{i}(w_{i})$ be the incircles of the six
skewed sectors formed, situated as shown in Figure 22. Then
$w_{1}w_{3}w_{5}=w_{2}w_{4}w_{6}$.

Figure 22. $w_{1}w_{3}w_{5}=w_{2}w_{4}w_{6}$

###### Proof.

The altitudes of $\triangle ABC$ divide it into six triangles named $T_{1}$
through $T_{6}$ as shown in Figure 22. Let $r_{i}$ be the inradius of triangle
$T_{i}$. By Theorem 3.6, $w_{1}/r_{1}=w_{6}/r_{6}$. Similarly,
$w_{3}/r_{3}=w_{2}/r_{2}$ and $w_{5}/r_{5}=w_{4}/r_{4}$. Therefore,

$\frac{w_{1}w_{3}w_{5}}{r_{1}r_{3}r_{5}}=\frac{w_{6}w_{2}w_{4}}{r_{6}r_{2}r_{4}}.$

But $r_{1}r_{3}r_{5}=r_{2}r_{4}r_{6}$ by Theorem 3.1 of [7]. Therefore,
$w_{1}w_{3}w_{5}=w_{2}w_{4}w_{6}$. ∎

######  Theorem 5.4.

Let $M$ be the centroid of $\triangle ABC$. The medians through $M$ extended
to meet the circumcircle of $\triangle ABC$ divide the segments of the
circumcircle bounded by the sides of the triangle into two skewed sectors each
as shown in Figure 23. Let $W_{i}(w_{i})$ be the incircles of the six skewed
sectors formed, situated as shown in Figure 23. Then

$\frac{1}{w_{1}}+\frac{1}{w_{3}}+\frac{1}{w_{5}}=\frac{1}{w_{2}}+\frac{1}{w_{4}}+\frac{1}{w_{6}}.$

Figure 23. $1/w_{1}+1/w_{3}+1/w_{5}=1/w_{2}+1/w_{4}+1/w_{6}$

###### Proof.

A cevian through a point $P$ inside a triangle $ABC$ divides $\triangle ABC$
into two triangles, known as side triangles. There are six such side
triangles, named $S_{1}$ through $S_{6}$ as shown in Figure 24.

Figure 24. naming of side triangles

Let $r_{i}$ be the radius of the circle inscribed in triangle $S_{i}$. When
$P$ is the centroid of $\triangle ABC$, Theorem 2.2 from [8] states that

$\frac{1}{r_{1}}+\frac{1}{r_{3}}+\frac{1}{r_{5}}=\frac{1}{r_{2}}+\frac{1}{r_{4}}+\frac{1}{r_{6}}.$

By Theorem 3.2,

$\displaystyle\left(\frac{1}{w_{1}}-\frac{1}{w_{2}}\right)$
$\displaystyle+\left(\frac{1}{w_{3}}-\frac{1}{w_{4}}\right)+\left(\frac{1}{w_{5}}-\frac{1}{w_{6}}\right)$
$\displaystyle=\left(\frac{1}{r_{1}}-\frac{1}{r_{2}}\right)+\left(\frac{1}{r_{3}}-\frac{1}{r_{4}}\right)+\left(\frac{1}{r_{5}}-\frac{1}{r_{6}}\right)=0,$

so $1/w_{1}+1/w_{3}+1/w_{5}=1/w_{2}+1/w_{4}+1/w_{6}$. ∎

######  Theorem 5.5.

Let $H$ be the orthocenter of acute $\triangle ABC$. The altitudes through $H$
divide the triangle into six side triangles, $S_{1}$ through $S_{6}$ as shown
in Figure 24. Let $W_{i}(w_{i})$ be the incircle of the skewed sector
associated with $S_{i}$. Two of these circles are shown in Figure 25. Then
$w_{1}w_{3}w_{5}=w_{2}w_{4}w_{6}$.

Figure 25. $w_{1}w_{3}w_{5}=w_{2}w_{4}w_{6}$

###### Proof.

This follows from Theorem 5.3 by applying Corollary 3.8. ∎

######  Theorem 5.6.

Let $O$ be the circumcenter of $\triangle ABC$. The cevians through $O$
extended to meet the circumcircle of $\triangle ABC$ divide the interior of
that circumcircle into six skewed sectors, each having vertex at $O$, as shown
in Figure 26. Let $W_{i}(w_{i})$ be the circle tangent to two cevians and
internally tangent to the circumcircle as shown. Then

$w_{1}=w_{4},\quad w_{2}=w_{5},\quad w_{3}=w_{6}.$

Figure 26. $w_{1}=w_{4}$, $w_{2}=w_{5}$, $w_{3}=w_{6}$

###### Proof.

It suffices to show that $w_{1}=w_{4}$. Note that $\angle BOD^{\prime}=\angle
AOE^{\prime}$ because they are vertical angles. Also,
$OB=OD^{\prime}=OE^{\prime}=OA$ because they are all radii of circle $O$.
Therefore, skewed sectors $BOD^{\prime}$ and $AOE^{\prime}$ are congruent and
thus their incircles are also congruent. ∎

######  Theorem 5.7.

Let $O$ be the circumcenter of $\triangle ABC$. The cevians through $O$ are
extended to meet the circumcircle of $\triangle ABC$ at points $D^{\prime}$,
$E^{\prime}$, and $F^{\prime}$ as shown in Figure 27. The cevians divide
$\triangle ABC$ into six side triangles named $S_{1}$ through $S_{6}$ as shown
in Figure 24. Six circles, $W_{i}(w_{i})$, are inscribed in the skewed sectors
associated with these side triangles. Two of these circles are shown in Figure
27. Then

$w_{1}=w_{4},\quad w_{2}=w_{5},\quad w_{3}=w_{6}.$

Figure 27. $w_{1}=w_{4}$

###### Proof.

It suffices to show that $w_{1}=w_{4}$. Note that $OA=OB$ because they are
both radii of circle $O$. Thus, $\angle OAB=\angle OBA$ because they are the
base angles of an isosceles triangle. Also, $AD^{\prime}=BE^{\prime}$ because
they are both diameters of circle $O$. The skewed sectors $BAD^{\prime}$ and
$ABE^{\prime}$ have side $AB$ in common. Therefore, skewed sectors
$BAD^{\prime}$ and $ABE^{\prime}$ are congruent and hence their incircles are
also congruent. ∎

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