Normat 60:1, 9–14 (2012) 9

Are those Archimedean triplet circles really

triplets?

Hiroshi Okumura

251 Moo 15 Ban Kesorn, Tambol Sila

Amphur Muang Khonkaen 40000, Thailand

hiroshiokmr@gmail.com

1 Introduction

For a point O on the segment AB with |AO| = 2a and |BO| = 2b, let α, β

and γ be the semicircles with diameters AO, BO and AB respectively erected on

the same side. The area surrounded by the three semicircles is called an arbelos

(see Figure 1). The radical axis of the two inner semicircles divides the arbelos

into two curvilinear triangles with congruent incircles. The circles were studied by

Archimedes, and are called the twin circles of Archimedes. Circles congruent to the

twin circles are called Archimedean circles, whose radii is ab/(a + b).

AOB

Figure 1.

AOB

Figure 2. Figure 3.

More than two thousand years after, Leon Bankoﬀ found another Archimedean

circle. If the incircle of the arbelos touches the semicircles α, β at points P and Q,

the circle W

passing through the three points P , Q and O is Archimedean (see

10 Hiroshi Okumura Normat 1/2012

Figure 2) [1]. With this circle he asserted that the twin circles are two members of

the triplet. Later he found one more Archimedean circle W

, which is the maximal

circle touching the external common tangent of α and β and the circular arc of

the semicircle γ cut by the tangent internally [2] (see Figure 3). Since many kinds

of Archimedean circles have been found today, there should be some reason to

designate these circles as a triplet. In this sense, it seems irrelevant that the twin

circles and the circle W

should be regarded as a triplet. In this article we show

that the circle W

forms a real triplet with the twin circles. Also we show that there

are inﬁnite triplet circles (inﬁnite pairs of three congruent circles) in the arbelos.

2 The Archimedean triplet circles

We use the following lemma in the old Japanese geometry [5].

Lemma. A circle C with radius r is divided by a chord t into two arcs and let h be

the distance from the midpoint of one of the arcs to t. If two externally touching

circles C

and C

with radii r

and r

also touch the chord t and the other arc of

the circle C, then h, r, r

and r

are related as

= 2

Proof. We reproduce the proof in [3] for the convenience of the reader (see Figure

4). The centers of C

and C

can be on the opposite sides of the normal dropped on

t from the center of C or on the same side of this normal. From the right triangles

formed by the centers of C and C

(i = 1, 2), the line parallel to t through the

center of C, and the normal dropped on t from the centers of C



(r − r

)

− (h + r

− r)

(r − r

)

− (h + r

− r)



= 2

√

where we used the fact that the segment length of the common external tangent

of C

and C

between the tangency points is equal to 2

√

. The formula of the

lemma follows from this equation.

Figure 4.

Normat 1/2012 Hiroshi Okumura 11

If C

is the circle with radius r

= h/2 touching the chord t and the circle C in the

lemma (see Figure 5), r, r

, r

and r

are related as

= 2

. (1)

Since (1) is symmetric in r

, r

and r

, it also holds when we change the roles of

the circles C

, C

and C

as in Figures 6 and 7.

Figure 5. Figure 6. Figure 7.

Therefore we get:

Theorem. Let C and C

(i = 1, 2, 3) be circles of radii r and r

respectively and let

C be divided by a chord t into two arcs. If two of C

, C

touch externally and

also touch t and one of the arcs, and the remaining is the maximal circle touching

t and the remaining arc, then r, r

, r

and r

are related as (1).

If we regard (1) as a quadratic equation for 1/

√

, it has two positive solutions.

Therefore solving (1) for r

, we always get two positive solutions, one of which is

the radius of C

. The other is equal to the radius of the circle diﬀerent from C

but satisfying the same condition satisﬁed by C

as denoted by the hatched lines

in Figures 8, 9 and 10. We call the circle the conjugate of C

with respect to C

and C

Figure 8. Figure 9. Figure 10.

Let us assume r = a + b, r

= a and r

= b in Theorem. In this case, the centers of

the circles C, C

and C

are collinear. Therefore the ﬁgure consisting of the three

circles is symmetric in this line. This implies that the conjugate of C

with respect

to C

and C

is congruent to C

, i.e., (1) has only one double root (see Figures

11, 12, 13). This implies that the twin circles and W

are congruent. Actually (1)

gives r

= ab/(a + b) in this case. As we have just seen, the congruence of the three

circles is obtained from the same equation at the same time. Hence we may say

that the three circles form a real triplet.

12 Hiroshi Okumura Normat 1/2012

Figure 11. Figure 12. Figure 13.

If two circles C

and C

are ﬁxed, the product of the radii of the circle C

and its

conjugate conjugate with respect to C

and C

is constant for a given circle C by

(1). It equals (ab/(a+ b))

if a and b are the radii of C

and C

. The same assertion

can also be found in Japanese geometry in the case of Figure 8 [4].

Let t be the line lying along the chord of C touching C

in Figure 8. Let us

consider the inversion in the circle with center at the point of tangency of C and C

passing through the points of intersection of C and t. By this inversion t and C are

interchanged, while C

and C

remain unchanged. Therefore they are orthogonal

to the inversion circle. Hence the internal common tangent of the circles C

and

passes through the center of the inversion.

3 Inﬁnite triplets

In this section we show that there are inﬁnite pairs of triplet circles in the arbelos.

We now observe that α, β and γ are not semicircles but circles. Let δ

= δ

β. Let δ

and δ

be the twin circles of Archimedes touching α and β respectively.

To avoid overlapping ﬁgures, let δ

be the reﬂected image of the circles W

in the

line AB (see Figure 14). For i = 1, 2, 3, let us assume that the circles δ

, δ

, ···,

are deﬁned (k ≥ 1), where δ

, δ

are congruent for j = 0, 1, 2, ··· , k. Then

k+1

is the conjugate of δ

k−1

with respect to α and δ

. Now the circles δ

, δ

···, δ

, ··· are deﬁned.

By the deﬁnition, δ

, δ

are congruent for any non-negative integer k. Also

from the deﬁnition, (i) If k is even, δ

is the maximal circle touching a chord t of

γ and the arc of γ cut by t, i.e., it touches t from the side opposite to α. Hence

it does not touch α if k 6= 0, since a chord of γ touches α at its midpoint if and

only if it lies along the radical axis of α and β. While δ

touches α if k is odd. (ii)

, δ

, ···, δ

, ··· are chain of circles touching γ and the radical axis of α and β

from the side opposite to α. (iii) δ

touches α if k is even and does not touch α if

k is odd. The three statements imply that δ

, δ

are diﬀerent for any natural

number k. Therefore we get inﬁnite pairs of three congruent but diﬀerent circles.

Normat 1/2012 Hiroshi Okumura 13

β = δ

= δ

Figure 14.

We now consider each of the triplets as a set, which is denoted by a point on the

two arcs in Figure 15, where D

= {δ

, δ

}. Three points forming vertices

of a triangle describe that they consist of circles satisfying the hypothesis of the

theorem with the circle γ. For two triangles with one side in common in the ﬁgure

(e.g. D

and D

{α}), each of the two opposite vertices is the set consisting

of conjugate of each of the circles belonging to the other vertex with respect to

circles belonging to the common vertices. The ﬁgure shows that the inﬁnte triplets

expressed by D

, D

, ··· do not exhaust all the possible triplets. For example,

triplets expressed by P

, P

, ··· do not appear in the triplets expressed by D

, D

, ···.

14 Hiroshi Okumura Normat 1/2012

{α}D

= {β}

Figure 15.

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. Magazine, 47

(1974) 134–137.

[2] L. Bankoﬀ, The marvelous arbelos, in The lighter side of mathematics 247–253, ed.

R. K. Guy and R. E. Woodrow, Mathematical Association of America 1994.

[3] H. Okumura and M. Watanabe, The twin circles of Archimedes in a skewed

arbelos, Forum Geom., 4 (2004) 229–251.

[4] K. Sato, The sangaku in Tama, Kenseisha, Tokyo, 1969 (in Japanese).

[5] Gazen Yamamoto, Samp¯o Jojutsu, 1841.