4 Normat 60:1, 4–8 (2012)
Dilations and the arbelos
Hiroshi Okumura
251 Moo 15 Ban Kesorn, Tambol Sila
Amphur Muang Khonkaen 40000, Thailand
hiroshiokmr@gmail.com
1 Introduction
For a point O on a segment AB with |OA| = 2a and |OB| = 2b, let α, β and γ be
semicircles with diameters AO, BO and AB respectively erected on the same side.
The area surrounded by the three semicircles is called an arbelos. The radical axis of
α and β divides the arbelos into two curvilinear triangles with congruent incircles,
which are called the twin circles of Archimedes (see Figure 1). Leon Bankoff found
that the circle orthogonal to α, β and the incircle of the arbelos is congruent to the
twin circles [1] (see Figure 2). Circles congruent to the twin circles are said to be
Archimedean. The common radius of the Archimedean circles is ab/(a + b), which
is denoted by r
A
. In this article we generalize the twin circles, also we get some
other new infinite Archimedean circles whose centers lie on a conic section by using
dilations with center O.
AB O AB O
Figure 1. Figure 2.
We use a rectangular coordinate system with origin O such that the coordinates
of A and B are (2a, 0) and (−2b, 0) respectively. The radical axis of α and β is
denoted by L. Let σ be a dilation with center O and scale factor k > 0. The image
of a point P by σ is denoted by P
σ
. For two points P and Q on the line AB, (P Q)
denotes the semicircle with diameter P Q, where all the semicircles are constructed
in the region y > 0.
Normat 1/2012 Hiroshi Okumura 5
2 The twin circles of Archimedes
In this section, we generalize the twin circles of Archimedes. Floor van Lamoen
has found that for a dilation τ with center A, the circle touching the semicircles
(AO
τ
) externally (AB
τ
) internally and the line L from the side opposite to B is
Archimedean [2]. A similar property also holds for the dilation with center O (see
Figure 3).
B AB
σ
A
σ
O
L
β
α
Figure 3: k = 1.5
Theorem 1. The circle touching the semicircles (OA
σ
) externally (AB
σ
) internally
and the line L from the side opposite to B is Archimedean.
Proof. Let x be the radius of the touching circle. By the Pythagorean theorem
((a + kb) − x)
2
− ((a − kb) − x)
2
= (ka + x)
2
− (ka − x)
2
.
Solving the equation we get x = r
A
.
One of the twin circles touching the semicircle α is obtained when σ is the iden-
tity. By exchanging the roles of the points A and B we get one more Archimedean
circle.
Theorem 2. The circle touching the semicircles (OA
σ
−1
) externally and (A
σ
B
σ
−1
)
internally and the line L from the side opposite to B has radius kr
A
. The point of
tangency of the touching circle and L coincides with the point of tangency of one
of the twin circles touching α and L.
Proof. The first part is proved similarly to Theorem 1 (see Figure 4). The second
part follows from the fact: the segment length of the common external tangent of
two externally touching circles with radii p and q are 2
√
pq.
AB A
σ
B
σ
−1
B
σ
O
L
A
σ
−1
β
α
Figure 4. k = 1.5
6 Hiroshi Okumura Normat 1/2012
Exchanging the roles of the points A and B, we get one more circles of radius kr
A
.
And the twin circles are obtained when σ is the identity. The semicircle (A
σ
B
σ
−1
)
belongs to the pencil of circles determined by the semicircle (AB) and the line L.
3 Two infinite sets of Archimedean circles
From now on, we include the case in which σ has a negative scale factor k, i.e., if
k < 0 then
−−→
OP
σ
= −|k|
−−→
OP for any point P . Let α(k) = (OA
σ
) and β(k) = (OB
σ
)
0
.
Also we denote the line x = 2kr
A
by P
k
. Hence P
0
= L, and P
1
touches the
Archimedean circle touching α, γ and P
0
. The next theorem is also proved similarly
to Theorem 1 (see Figures 5 and 6).
Theorem 3. Let k be a real number.
(i) If 0 < k, then the circle touching α(k) externally, α(k + 1) internally and P
k
from the side opposite to the point O is Archimedean.
(ii) If −1 ≤ k < 0, then the circle touching both α(k) and α(k + 1) externally and
P
−k
from the side opposite to the point A is Archimedean.
α(k)
P
k
α
AO
α(k + 1)
B
β
γ
α(k + 1)
α(k)
α
P
−k
AOB
β
γ
Figure 5. 0 < k Figure 6. −1 < k < 0
From (i) in the theorem we get an infinite set consisting of Archimedean circles
touching α(k), α(k + 1) and P
k
for some positive real number k. The line P
1
passes
through the point of intersection of α(2) and γ. This can be proved by using ele-
mentary properties of chords with the Pythagorean theorem. In [3] a more general
aspect is considered. Therefore the circle touching α externally α(2) internally and
the perpendicular to AB through the point of intersection of γ and α(2) from the
side opposite to O is Archimedean from the case k = 1 (see Figure 7).
From (ii) we also get an infinite set consisting of Archimedean circles touching
α(k), α(k + 1) and P
−k
for a real number k satisfying −1 ≤ k < 0.
4 Conic sections
Let (x, y) be the center of the Archimedean circle obtained by (i) in Theorem 3.
Then x = (2k+1)r
A
. While by the Pythagorean theorem, y
2
+(x−ka)
2
= (r
A
+ka)
2
.
0
Those notations are slightly changed from the ones in [3] and [4].
Normat 1/2012 Hiroshi Okumura 7
Eliminating k from the two equations and rearranging, we get
x
2
r
2
A
−
y
2
r
2
A
a/b
= 1.
Therefore (x, y) lies on a part of the hyperbola lying in the quadrant I with focal
points (±
√
ar
A
, 0) and asymptote
y =
r
a
b
x.
Conversely, any point on this curve can be obtained as a center of an Archimedean
circle determined in (i). The asymptote, denoted by the dotted line in Figure 7,
passes through the point of intersection of α(k) and P
k
.
α(3)
α(4)
OB A
β
γ
α
α(2)
P
2
P
3
P
1
P
0
Figure 7.
B O A
P
−k
α(k)
α(k + 1)
α
γ
β
Figure 8. k = −0.25
Let (x, y) be the center of the Archimedean circle determined in (ii), then we can
also get
x
2
r
2
A
+
y
2
r
2
A
(a + 2b)/b
= 1.
Therefore (x, y) lies on a part of the ellipse lying in the region y > 0 together
with the point (r
A
, 0) with minor axis 2r
A
and major axis 2r
A
p
(a + 2b)/b and
8 Hiroshi Okumura Normat 1/2012
focal points
0, ±
√
ar
A
(see Figure 8). Conversely, any point on this curve can be
obtained as a center of an Archimedean circle determined in (ii). The Archimedean
circle touching α, γ and P
0
, touches P
0
at the point (0, 2
√
ar
A
). Therefore the focal
points are obtained as the midpoint of the line segment joining O and the tangent
point and its reflection in the line AB.
Both the conic sections are expressed as follows:
x
2
r
2
A
∓
y
2
r
2
A
(a + b ∓ b)/b
= 1.
Acknowledgments
The author thanks the referee for a number of helpful suggestions.
References
[1] L. Bankoff, Are the twin circles of Archimedes really twins?, Math. Mag., 47
(1974) 214–218.
[2] F. van Lamoen, Archimedean adventures, Forum Geom., 6 (2006) 79–96.
[3] H. Okumura and M. Watanabe, Archimedean circles of Schoch and Woo, Forum
Geom., 4 (2004) 27–34.
[4] H. Okumura and M. Watanabe, Remarks on Woo’s Archimedean circles, Forum
Geom., 7 (2007) 125–128.
