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We study some properties of a triad of circles associated with a triangle. Each circle is inside the triangle, tangent to two sides of the triangle, and externally tangent to the circle on the third side as diameter. In particular, we find a nice relation involving the radii of the inner and outer Apollonius circles of the three circles in the triad.
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Sangaku Journal of Mathematics (SJM) ©SJM
ISSN 2534-9562
Volume 7 (2023), pp. 54–70
Received 18 July 2023 Published on-line 3 August 2023
web: http://www.sangaku-journal.eu/
©The Author(s) This article is published with open access.1
A Triad of Circles Associated with a Triangle
Ercole Suppaaand Stanley Rabinowitzb
aVia B. Croce 54, 64100 Teramo, Italia
e-mail: ercolesuppa@gmail.com
web: http://www.esuppa.it/
b545 Elm St Unit 1, Milford, New Hampshire 03055, USA
e-mail: stan.rabinowitz@comcast.net2
web: http://www.StanleyRabinowitz.com/
Abstract. We study some properties of a triad of circles associated with a tri-
angle. Each circle is inside the triangle, tangent to two sides of the triangle, and
externally tangent to the circle on the third side as diameter. In particular, we
find a nice relation involving the radii of the inner and outer Apollonius circles of
the three circles in the triad.
Keywords. Paasche point, Apollonius circle, barycentric coordinates, Mathe-
matica.
Mathematics Subject Classification (2020). 51-02, 51M04.
1. Introduction
Notation. Throughout this paper, we will use the following notation, where
4ABC is a fixed acute triangle in the plane. We let a=BC,b=CA,c=AB,r
is the inradius of 4ABC ,Ris the circumradius of 4ABC ,p=a+b+c
2, = [ABC]
is the area of the triangle, and S= 2∆. We also let Idenote the incenter of
4ABC .
The semicircle erected inwardly on side BC will be named ωaas shown in Figure 1
(left). Semicircles ωband ωcare defined similarly. The circle inside 4ABC ,
tangent to sides AB and AC, and externally tangent to semicircle ωawill be
named γa. Circles γband γcare defined similarly. The radii of circles γa,γb, and
γcare denoted by ρa,ρb, and ρc, respectively. The centers of these circles are
named D,E, and F, respectively, as shown in Figure 1 (right).
For purposes of this paper, these three circles will be called the triad of circles
associated with 4ABC.
1This article is distributed under the terms of the Creative Commons Attribution License
which permits any use, distribution, and reproduction in any medium, provided the original
author(s) and the source are credited.
2Corresponding author
54
Ercole Suppa and Stanley Rabinowitz 55
Figure 1.
This triad of circles appears in a Sangaku described in [3] and reprinted in [6,
problem 6]. The statement in the Sangaku is given as Theorem 1.1.
Theorem 1.1. For the triad of circles associated with 4ABC, we have
r=1
2ρa+ρb+ρc+qρ2
a+ρ2
b+ρ2
c.
A proof of this result can be found in [10]. A variant on this result when the
triangle is not acute can also be found in [10].
It is the purpose of this paper to give other properties of such a triad of circles.
2. Known Results
Before giving new results, we summarize some of the properties already known
about the triad of circles. The following five theorems come from [10].
Theorem 2.1. For the triad of circles associated with 4ABC, we have
ρa=r1tan A
2.
Similar formulas hold for ρband ρc.
Theorem 2.2. Let Pand Qbe the feet of the perpendiculars from Dand Ito
side AC, respectively. Then P Q =IQ. (Figure 2)
P
Q
I
D
A
B C
Figure 2. red lengths are equal
56 A Triad of Circles Associated with a Triangle
Theorem 2.3. The lengths of the common external tangents between any two
circles of the triad are equal. The common length is 2r. (Figure 3)
F
E
D
A
B C
Figure 3. blue lengths are equal
Theorem 2.4. The six points of contact of the triad of circles associated with
4ABC lie on a circle with center Iand radius r2. (Figure 4)
F
E
I
D
A
B C
Figure 4.
This circle will be called the contact circle.
The following corollary follows immediately from Theorem 2.4.
Corollary 2.1. In Figure 5 showing the contact circle and the incircle, the green
area is equal to the blue area.
I
A
B C
Figure 5. green area = blue area
Ercole Suppa and Stanley Rabinowitz 57
Theorem 2.5. For the triad of circles associated with 4ABC, we have
ρ2
a+ρ2
b+ρ2
c=r2(p4Rr)2
p2.(1)
The following result comes from [4] where it is stated that the result is due to
Tomasz Cie´sla.
Theorem 2.6. Let Ta,Tb, and Tcbe the touch points of the circles in the triad
with their corresponding semicircles as shown in Figure 6. Then ATa,BTb, and
CTcare concurrent.
Figure 6.
The point of concurrence is catalogued as point X1123 in the Encyclopedia of
Triangle Centers [4]. Since reference [4] does not include a proof of this result, we
will give our own proof later in Section 5 of this paper.
The point X1123 is known as the Paasche point of the triangle because Paasche
proved the following result in [7].
Theorem 2.7. Congruent circles with centers A1and A2touch each other exter-
nally at point A0outside 4ABC . Circle (A1)is tangent to AB and BC. Circle
(A2)is tangent to AC and BC . Points B0and C0are defined similarly (Figure 7).
Then AA0,BB0, and CC 0are concurrent.
C'
C2
C1
B'
B2
B1
A' A2
A1
X1123
A
BC
Figure 7.
58 A Triad of Circles Associated with a Triangle
Remark. The same result is true if the pairs of congruent circles are inside the
triangle instead of outside. Figure 8 illustrates this. (Only the two congruent
circles tangent to side BC are shown.) This result comes from [11, Art. 3.5.4,
ex. 4c].
A''
X1123
A
B
C
Figure 8.
The Paasche point can also be characterized as follows according to [2].
Theorem 2.8. In 4ABC, let ωa,ωb, and ωcbe the circles constructed using
sides BC ,C A, and AB, respectively, as diameters. Let be the circle internally
tangent to ωa,ωb, and ωc. Let A0be the touch point between ωaand . Points B0
and C0are defined similarly (Figure 9). Then AA0,BB0, and CC0are concurrent
at X1123, the Paasche point of 4ABC.
𝛺
𝜔c
𝜔b
𝜔a
X1123
C'
A'
B'
A
BC
Figure 9.
A circle that is tangent to three given circles is called an Apollonius circle of those
three circles.
If all three circles lie inside an Apollonius circle, then the Apollonius circle is
called the outer Apollonius circle of the three circles. The outer Apollonius circle
surrounds the three circles and is internally tangent to all three.
Ercole Suppa and Stanley Rabinowitz 59
If all three circles lie outside an Apollonius circle, then the Apollonius circle is
called the inner Apollonius circle of the three circles. The inner Apollonius circle
will either be internally tangent to the three given circles or it will be externally
tangent to all the circles. The following theorem comes from [5].
Theorem 2.9. For the triad of circles associated with 4ABC, the inner Apollo-
nius circle of γa,γb,γc, is internally tangent to the inner Apollonius circle of ωa,
ωb,ωc(Figure 10).
Remark. The inner Apollonius circle of γa,γb,γcis known as the 1st Miyamoto-
Moses-Apollonius circle and the outer Apollonius circle of γa,γb,γcis known as
the 2nd Miyamoto-Moses-Apollonius circle (see [5]).
𝛾a
𝛾b
𝛾c
𝜔c
𝜔b
𝜔a
A
B C
Figure 10.
3. Metric Relationships involving ρa,ρb,ρc
In addition to Theorem 1.1 and Theorem 2.5, the following symmetric relationship
involving ρa,ρb, and ρcholds.
Theorem 3.1. For the triad of circles associated with 4ABC, we have
ρaρb+ρbρc+ρcρa2r(ρa+ρb+ρc)+2r2= 0.
Proof. From Theorem 2.1, we have ρa=r(1 tan A
2), ρb=r(1 tan B
2), and
ρc=r(1 tan C
2). Substituting these values into the expression
ρaρb+ρbρc+ρcρa2r(ρa+ρb+ρc)+2r2
and simplifying (using Mathematica), shows that the expression is equal to
r2cos A+B+C
2sec A
2sec B
2sec C
2.
Since A+B+C=π, this expression is equal to 0.
Theorem 3.2. For the triad of circles associated with 4ABC, we have
p(rρa)(rρb)(rρc) = r4.
60 A Triad of Circles Associated with a Triangle
Proof. This result follows from Theorem 2.1 and the trigonometric identity
tan A
2tan B
2tan C
2=r
p
which comes from [1, p. 358].
Theorem 3.3. For the triad of circles associated with 4ABC, we have
ρa=(pb)(pc)
p.(2)
Similar formulas hold for ρband ρc.
Proof. Using Theorem 2.1 and the well-known identities
tan A
2=r
paand r=
p,
we get
ρa=r1r
pa=rr2
pa=
p2
p2(pa)(3)
=
pp(pa)(pb)(pc)
p2(pa)=(pb)(pc)
p.(4)
This complete the proof.
4. Barycentric coordinates of centers of γa,γb,γc
Theorem 4.1. The barycentric coordinates of the center of γaare
D=aS + 2(pb)(pc)(b+c) : bS 2b(pb)(pc) : cS 2c(pb)(pc).
Proof. Let ybe the distance between Dand the sideline BC. Summing the areas
of triangles DBC,DC A and DAB we obtain
ay +a+a= 2∆.(5)
Plugging (2) into (5) we get
ay + (b+c)·(pb)(pc)
p=S
ay + (b+c)·S2(pb)(pc)
2p=S
2pay + (b+c)S2(pb)(pc)(b+c)=(a+b+c)S
2pay = 2(pb)(pc)(b+c) + aS
ay =aS + 2(pb)(pc)(b+c)
2p(6)
By using (2) and (6), we obtain
D= DBC : DCA :DAB =ay :a:a
=aS + 2(pb)(pc)(b+c) : bS 2b(pb)(pc) : cS 2c(pb)(pc)
which are the desired barycentric coordinates.
Ercole Suppa and Stanley Rabinowitz 61
Theorem 4.2. The radical center of circles γa,γband γcis the Gergonne point
of 4ABC .
Proof. Using Mathematica and the package baricentricas.m3, it can be proved
that
(a) the radical axis of γaand γbis (pa)x(pb)y= 0;
(b) the radical axis of γband γcis (pb)y(pc)z= 0;
(c) the radical axis of γcand γais (pa)x(pc)z= 0.
An easy verification shows that these radical axes concur at
Ge= (pb)(pc):(pc)(pa) : (pa)(pb),
which is the Gergonne point of 4ABC.
We can also give a purely geometric proof.
Proof. Let the incircle touch the sides of 4ABC at Qa,Qb, and Qcas shown in
Figure 11. From Theorem 2.2, QaEa=QaFa=r. Thus, the tangents from Qa
to γband γcare equal. Since ADc=ADband DcEc=DbFb(Theorem 2.3), this
means AEc=AFb. Hence the tangents from Ato γband γcare equal. The radical
axis of circles γband γcis the locus of points such that the lengths of the tangents
to the two circles from that point are equal. The radical axis of two circles is a
straight line. Therefore, the radical axis of circles γband γcis AQa, the Gergonne
cevian from A.
Similarly, the radical axis of circles γaand γcis the Gergonne cevian from Band
the radical axis of circles γaand γbis the Gergonne cevian from C. Hence, the
radical center of the triad of circles is the intersection point of the three Gergonne
cevians, namely, the Gergonne point of 4ABC .
𝛾a
𝛾b𝛾c
Fa
Fb
Dc
Ea
Ec
Db
Qc
Qa
Qb
A
B C
Figure 11.
3The package baricentricas.m written by F.J.G.Capitan can be freely downloaded from
http://garciacapitan.epizy.com/baricentricas/
62 A Triad of Circles Associated with a Triangle
5. A concurrence at the Paasche point.
In [4] the following result is stated.
Theorem 5.1. Suppose that ABC is an acute triangle. Let γabe the circle
touching CA and AB from inside of ABC and also externally tangent to the
semicircle of diameter BC, in point Ta. Define Tband Tccyclically (Figure 12).
Then ABC is perspective to TaTbTc, and the perspector is X1123, the Paasche point
of 4ABC .
Figure 12.
Proof. We use homogeneous barycentric coordinates with respect to the triangle
ABC . Let Mabe the midpoint of BC. The point Tadivides the segment joining
the centers of the circles γaand ωain the ratio ρa:a
2. Using theorem 4.1 we have
that the sum of coordinates of Dis
aS + 2(pb)(pc)(b+c) + bS 2b(pb)(pc) + cS 2c(pb)(pc)
=(a+b+c)S+ 2(pb)(pc)(b+c)2(b+c)(pb)(pc)
=(a+b+c)S= 2pS.
Therefore, by writing the coordinates of Main the form Ma= 0 : pS :pS, we get
Ta=a
2·D+ρa·Ma.
It follows that the first coordinate of Ta=xa:ya:zais given by
xa=a
2(aS + 2(pb)(pc)(b+c)) + ρa·0
=a
2(aS + 2(pb)(pc)(b+c)) .
In a similar way we find that
ya=a
2(bS 2b(pb)(pc)) + ρapS
=1
2(S2(pb)(pc)) (ab +S)
Ercole Suppa and Stanley Rabinowitz 63
and
za=a
2(cS 2c(pb)(pc)) + ρapS
=1
2(S2(pb)(pc)) (ac +S).
Hence
Ta=a2S+ 2a(pb)(pc)(b+c) :
(S2(pb)(pc)) (ab +S) :
(S2(pb)(pc)) (ac +S).
The equation of line ATais zay+yaz= 0, i.e.
ATa: (ac +S)y(ab +S)z= 0.
The cyclic substitution ab,bc,cagives
BTb: (bc +S)x(ab +S)z= 0,
CTc: (bc +S)x(ac +S)y= 0.
A direct verification shows that ATa,BTb, and CTcconcur at the Paasche point
X1123 = (ab +S)(ac +S) : (ab +S)(bc +S):(ac +S)(bc +S).
6. Apollonius circles of γa,γb,γc
In order to find the radii of the inner and outer Apollonius circles tangent to γa,
γb, and γcwe will use the method explained in [9] and some preliminary lemmas.
The more complicated calculations are performed with Mathematica.
Lemma 6.1. If u=E F ,v=DF ,w=DE are the distances between the centers
of the circles γa,γb, and γc, we have
u2=a(b+ca) (a2+ab +ac 2b2+ 4bc 2c2)
(a+b+c)2,
v2=b(ab+c) (2a2+ab + 4ac +b2+bc 2c2)
(a+b+c)2,
w2=c(a+bc) (2a2+ 4ab +ac 2b2+bc +c2)
(a+b+c)2.
Proof. From Theorem 3.3, we have
ρa=(pb)(pc)
p,
and similarly
ρb=(pa)(pc)
p, ρc=(pa)(pb)
p.
Assume, without loss of generality, that ρb< ρc. Let E0be the foot of the
perpendicular from Eto F Fa. Applying the Pythagorean Theorem to triangle
64 A Triad of Circles Associated with a Triangle
Figure 13.
EF E 0(see figure 13), taking into account that EE0=EaFa= 2rand F E0=
ρcρb, we obtain
u2=EF 2=EaF2
a+ (F FaEEa)2= (2r)2+ (ρcρb)2
= 4r2+(pa)(pc)(pa)(pb)
p2
= 4 ·2
p2+(pa)2(bc)2
p2
=4p(pa)(pb)(pc)+(pa)2(bc)2
p2
=a(b+ca) (a2+ab +ac 2b2+ 4bc 2c2)
(a+b+c)2.
The formulas relating to v2and w2can be proved in a similar way.
Lemma 6.2. If θ,ϕ,ψare three positive real numbers such that θ+ϕ+ψ= 360,
then we have
cos2θ+ cos2ϕ+ cos2ψ2 cos θcos ϕcos ψ= 1.
Proof. Using the addition formulas we have
cos2θ+ cos2ϕ+ cos2ψ2 cos θcos ϕcos ψ
= cos2θ+ cos2ϕ+ cos2(360θϕ)2 cos θcos ϕcos(360θϕ)
= cos2θ+ cos2ϕ+ cos2(θ+ϕ)2 cos θcos ϕcos(θ+ϕ)
= cos2θ+ cos2ϕ+ (cos θcos ϕsin θsin ϕ)22 cos θcos ϕ(cos θcos ϕsin θsin ϕ)
= cos2θ+ cos2ϕ+ cos2θcos2ϕ+ sin2θsin2ϕ2 cos2θcos2ϕ
= cos2θ+ cos2ϕ+ sin2θsin2ϕcos2θcos2ϕ
= cos2θ(1 cos2ϕ) + cos2ϕ+ sin2θsin2ϕ
= sin2ϕ(cos2θ+ sin2θ) + cos2ϕ
= sin2ϕ+ cos2ϕ= 1.
Ercole Suppa and Stanley Rabinowitz 65
Theorem 6.1. Let ρibe the radius of the inner Apollonius circle externally tan-
gent to γa,γb, and γc(see figure 14). Then
ρ2
i=ρ2
a+ρ2
b+ρ2
c.
Figure 14.
Proof. Let Ube the center of the inner Apollonius circle and let x=ρi. Let us
consider the angles ϕa=EU F ,ϕb=F U D, and ϕc=DU E.
Since ϕa+ϕb+ϕc= 360by Lemma 6.2 we have
cos2ϕa+ cos2ϕb+ cos2ϕc2 cos ϕacos ϕbcos ϕc= 1.(7)
If we substitute
ta= sin2ϕa
2, tb= sin2ϕb
2, tc= sin2ϕc
2(8)
in (7), we obtain
t2
a+t2
b+t2
c2 (tatb+tbtc+tcta)+4tat2tc= 0.(9)
Since UD =x+ρa,U E =x+ρb,U F =x+ρc, the Law of Cosines yields
cos ϕa=(x+ρb)2+ (x+ρc)2u2
2(x+ρb)(x+ρc)
ta=1cos ϕ1
2=u2(ρbρc)2
4(x+ρb)(x+ρc),(10)
and analogously
tb=v2(ρcρa)2
4(x+ρc)(x+ρa), tc=w2(ρaρb)2
4(x+ρa)(x+ρb).(11)
Plugging (10) and (11) in (9) and using Lemma 6.1, after a straightforward cal-
culation, we get an equation of the form f(x)g(x) = 0, where
f(x) = 6(a+b+c)x+ 2ab + 2bc + 2ac a2b2c2+ 12∆ (12)
and
g(x) = 2(a+b+c)x+a2+b2+c22ab 2ac 2bc + 4∆.(13)
The root of (12) is
x=a2+b2+c22ab 2bc 2ca 12∆
6(a+b+c)=2r(4R+r) + 6∆
6p<0.
66 A Triad of Circles Associated with a Triangle
The root of (13) is
x=a2b2c2+ 2ba + 2bc + 2ca 4∆
2(a+b+c)=r(4R+rp)
p>0.
Therefore, discarding the negative root, we have ρi=r(4R+rp)
p. Hence, taking
into account equation (1), we get
ρ2
i=ρ2
a+ρ2
b+ρ2
c.
Corollary 6.1. In Figure 15 showing the triad of circles associated with 4ABC
and the inner Apollonius circle externally tangent to each circle in the triad, we
have that the sum of the yellow areas is equal to the green area.
F
E
D
A
B C
Figure 15. yellow area = green area
Theorem 6.2. Let ρobe the radius of the outer Apollonius circle internally tan-
gent to γa,γb,γc(see Figure 16). We have
ρo=2
3(ρa+ρb+ρc) + qρ2
a+ρ2
b+ρ2
c.
Figure 16.
Proof. The proof is similar to that of Theorem 6.1.
Combining Theorem 1.1, Theorem 6.1, and Theorem 6.2, we get the following
nice results.
Ercole Suppa and Stanley Rabinowitz 67
Corollary 6.2. The inradius rand the radii ρi,ρoof the Apollonius circles satisfy
the relation
3ρo=ρi+ 4r.
Corollary 6.3. The radii ρi,ρoof the Apollonius circles satisfy the relation
3ρo= 2(ρa+ρb+ρc)+3ρi.
Remark. The centers Uand Vof the inner and outer Apollonius circles of γa,
γb, and γcare known ETC centers, namely U=X(52805) and V=X(52806).
Theorem 6.3. Let ωibe the inner Apollonius circle, externally tangent to γa,γb,
and γc. Let Uabe the touch point between ωiand γa. Define Uband Uccyclically.
Let ωobe the outer Apollonius circle, internally tangent to γa,γb, and γc. Let
Vabe the touch point between ωoand γa. Define Vband Vccyclically. Let Gebe
the Gergonne point of 4ABC. Then the points A,Va,Ua, and Geare collinear.
Similarly, the points B,Vb,Ub, and Geare collinear; and the points C,Vc,Uc,
and Geare collinear (Figure 17).
Figure 17.
Proof. Clearly, by symmetry, it is enough to prove that A,Va,Ua, and Geare
collinear. From Theorem 4.2, we know that Geis the radical center of γa,γb, and
γc. Hence, from the Gergonne construction of Apollonius circles, it follows that
Ua,Va, and Geare collinear. Therefore, it remains to prove that A,Ua, and Geare
collinear. To this end we use barycentric coordinates. We have A=1:0:0and
Ge=1
pa:1
pb:1
pc. The point Uadivides the segment DU joining the centers of
the circles γaand ωiin the ratio ρa:ρi. By using Mathematica, we find that
Ua=a
(pb)(pc):1
pb:1
pc.
The points A,Ua, and Geare collinear because
1 0 0
a
(pb)(pc)
1
pb
1
pc
1
pa
1
pb
1
pc
= 0.
This completes the proof.
68 A Triad of Circles Associated with a Triangle
Remark. We could also show that the lines AUa,BUb, and CUcare concurrent
by using Theorem 2 of [8]. That theorem also shows that the point of concurrence,
Ge, is the internal center of similitude of the incircle of 4ABC and the circle ωi.
Corollary 6.4. Let Uabe the touch point between γaand ωi(Figure 18). Then
the tangents from Uato γband γcare equal.
𝛾a
𝛾b𝛾c
Ua
A
B C
Figure 18. red tangents are equal
Proof. From Theorem 6.3, AUais the Gergonne cevian from vertex A. But from
the proof of Theorem 4.2, this Gergonne cevian is the radical axis of circles γband
γc. Thus the two tangents have the same length.
Theorem 6.4. Let ωi= (U, ρi),ωo= (V, ρo)be the inner and outer Apollonius
circles externally and internally tangent to γa,γb, and γc, respectively. Let Ua,
Vabe the touch point of γawith ωiand ωorespectively. Define Ub,Uc,Vb,Vc
cyclically. Let I=X1,Ge=X7be the incenter and the Gergonne points of
4ABC respectively (Figure 19). Then Uand Vlie on the Soddy line IGeand
UI :I V = 3.
Figure 19.
Proof. This follows directly from the barycentric coordinates for Uand V.
Ercole Suppa and Stanley Rabinowitz 69
7. Other properties
Theorem 7.1. For the triad of circles associated with 4ABC, let x=ρa,y=ρb,
and z=ρc. Let u, v, w be the radii of the greatest circles inscribed in the circular
segments shown in Figure 20. Then
xw +yu +zv =xv +zu +yw.
Figure 20.
Proof. From Theorem 2.1, we have
x=r1tan A
2, y =r1tan B
2, z =r1tan C
2.
Figure 21.
On the other hand, we have (see Figure 21)
u=1
2MN =1
2BM ·tan A
2=a
4tan A
2
and similarly
v=b
4tan B
2, w =c
4tan C
2.
Observe that
tan A
2=r
sa=
s(sa),tan B
2=
s(sb),tan C
2=
s(sc)
70 A Triad of Circles Associated with a Triangle
so, using the Heron formula 2=s(sa)(sb)(sc), we get
u(yz) = a
4tan A
2r1tan B
2r1tan C
2
=ar
4tan A
2tan C
2tan B
2
=ar
4tan A
2tan C
2tan A
2tan B
2
=ar
4
s(sa)
s(sc)
s(sa)
s(sb)
=ar
4sb
ssc
s
=r
4·ac ab
s.
Similarly, we have
v(zx) = r
4·ba bc
s, w(xy) = r
4·cb ca
s.
Therefore
xw +yu +zv (v+zu +yw) =u(yz) + v(zx) + w(xy)
=r
4·ac ab
s+r
4·ba bc
s+r
4·cb ca
s
=r
4·ac ab +ba bc +cb ca
s= 0.
This completes the proof.
References
[1] Titu Andreescu and Oleg Mushkarov, Topics in Geometric Inequalities, XYZ Press, 2019.
[2] Francisco Javier Garcia Capit´an, Problem 2428, Romantics of Geometry Facebook Group,
October 2018.
https://www.facebook.com/groups/parmenides52/posts/1926266464153717/
[3] Honma, ed., Zoku Kanji Samp¯o, Tohoku University Digital Collection, 1849.
[4] Clark Kimberling, Encyclopedia of Triangle Centers, entry for X(1123), the Paasche Point.
https://faculty.evansville.edu/ck6/encyclopedia/ETCPart2.html#X1123
[5] Clark Kimberling, Encyclopedia of Triangle Centers, preamble to entry X(52805),
Miyamoto-Moses Points.
https://faculty.evansville.edu/ck6/encyclopedia/ETCPart27.html#X52805
[6] Hiroshi Okumura, Problems 2023–1, Sangaku Journal of Mathematics,7(2023)9–12.
http://www.sangaku-journal.eu/2023/SJM_2023_9-12_problems_2023-1.pdf
[7] Ivan Paasche, Aufgabe P933: Ankreispaare, Praxis der Mathematik 1(1990)40.
[8] Stanley Rabinowitz, Pseudo-Incircles, Forum Geometricorum,6(2006)107–115.
https://forumgeom.fau.edu/FG2006volume6/FG200612.pdf
[9] Milorad R. Stevanovi´c, Predrag B. Petrovi´c, and Marina M. Stevanovi´c, Radii of Circles in
Apollonius’s Problem, Forum Geometricorum,17(2017)359–372.
https://forumgeom.fau.edu/FG2017volume17/FG201735.pdf
[10] Ercole Suppa and Marian Cucoanes, Solution of Problem 2023–1–6, Sangaku Journal of
Mathematics,7(2023)21–28.
http://www.sangaku-journal.eu/2023/SJM_2023_21-28_Suppa,Cucoanes.pdf
[11] Paul Yiu, Introduction to the Geometry of the Triangle, December 2012.
http://math.fau.edu/Yiu/YIUIntroductionToTriangleGeometry121226.pdf
... The sangaku gave a relationship involving the radii of the three circles. Additional properties of this configuration were given in [29] and [30]. For example, in Figure 2 (left), the three blue common tangents are all congruent. ...
... We can generalize many of the results in [30] by replacing the semicircles with arcs having the same angular measure. Let ω a , ω b , and ω c be arcs with the same angular measure θ erected internally on the sides of ABC as shown in Figure 48. ...
... When θ = 180 • , the arcs become semicircles, t = 1, and this result agrees with Theorem 6.1 in [30]. ...
Preprint
Full-text available
We study properties of certain circles associated with a triangle. Each circle is inside the triangle, tangent to two sides of the triangle, and externally tangent to the arc of a circle erected internally on the third side.
Problem 2428, Romantics of Geometry Facebook Group
  • Francisco Javier Garcia Capitán
Francisco Javier Garcia Capitán, Problem 2428, Romantics of Geometry Facebook Group, October 2018. https://www.facebook.com/groups/parmenides52/posts/1926266464153717/
Encyclopedia of Triangle Centers, entry for X(1123), the Paasche Point
  • Clark Kimberling
Clark Kimberling, Encyclopedia of Triangle Centers, entry for X(1123), the Paasche Point. https://faculty.evansville.edu/ck6/encyclopedia/ETCPart2.html#X1123
  • Hiroshi Okumura
Hiroshi Okumura, Problems 2023-1, Sangaku Journal of Mathematics, 7(2023)9-12. http://www.sangaku-journal.eu/2023/SJM_2023_9-12_problems_2023-1.pdf
  • Ivan Paasche
Ivan Paasche, Aufgabe P933: Ankreispaare, Praxis der Mathematik 1(1990)40.
  • Stanley Rabinowitz
  • Pseudo-Incircles
Stanley Rabinowitz, Pseudo-Incircles, Forum Geometricorum, 6(2006)107-115. https://forumgeom.fau.edu/FG2006volume6/FG200612.pdf
  • R Milorad
  • Predrag B Stevanović
  • Marina M Petrović
  • Stevanović
Milorad R. Stevanović, Predrag B. Petrović, and Marina M. Stevanović, Radii of Circles in Apollonius's Problem, Forum Geometricorum, 17(2017)359-372. https://forumgeom.fau.edu/FG2017volume17/FG201735.pdf
Solution of Problem 2023-1-6
  • Ercole Suppa
  • Marian Cucoanes
Ercole Suppa and Marian Cucoanes, Solution of Problem 2023-1-6, Sangaku Journal of Mathematics, 7(2023)21-28. http://www.sangaku-journal.eu/2023/SJM_2023_21-28_Suppa,Cucoanes.pdf
Introduction to the Geometry of the Triangle
  • Paul Yiu
Paul Yiu, Introduction to the Geometry of the Triangle, December 2012. http://math.fau.edu/Yiu/YIUIntroductionToTriangleGeometry121226.pdf