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A Generalization of the Classical Arbelos and the Archimedean Circles
Dimitar Belev
Abstract. On the base of a chord are constructed arbelos and circles with equal radii. The Archimedean circles – twins are obtained as a special case of these circles, when the chord is diameter. It is calculated the radius of this circles and it is found a relation between this radius, the radius of the Archimedean twins and the radii of several circles analogous to the Archimedean circles. There are constructed and considered several constructions analogous to constructions of the classical arbelos. There are found new Archimedean circles.
1. Preliminaries or “Why semicircles?”.
We consider an arbelos, consisted of three semicircles , and having . The three semicircles are mutually tangent in the points A, B and P, CP is a common inner tangent of and (Figure 1).
Figure 1
Archimedes has shown, that the circles and touch respectively to , and PC and to , and PC are equal and have radius . The circles and are called the Archimedean twins.Later[1] are found and are going to be found more circles[2] related with the arbelos (called Archimedean circles) having the same radius.
The questions which are put in this article and which answers we are looking for are:
- Do we can to construct arbelos, for which AB is a chord[3] so as to preserve “maximum” of the characteristics of the classical arbelos?
- Do we can to construct in this arbelos circles so as to preserve “maximum” of the characteristics of the classical Archimedean twins? Which is their radius?
- Do the Archimedean Circles have analogous in this arbelos and what is the relation between theirs radii and the radii of the Archimedean twins and\or the radii of the generalized Archimedean twins?
2. Construction of an arbelos on a chord.
Lemma 1.
Let AB is a chord in the circle and point P lie on the chord AB. If the circles and pass through point P and touch internally the circle respectively at the points A and B, then quadrangle is parallelogram and .
Figure 2
Proof:
Let and (Figure 2). We construct the common tangent t at point A of the circles and , then , and ççOB. Similarly and ççOA. Therefore is quadrangle and .
The constructed arbelos on the chord AB (Figure 2) keep the following important properties of the classical arbelos:
- The contour of the arbelos consists of three arcs with equal measures .
- The circles and touch internally the circle respectively at the points A and B.
- The common point P of the circles and lie on the chord AB.
- independently of the position of point P.
The point P is a point of intersection of the circles and , not a tangent point as in the classical arbelos. This property we „immolate” in our construction.
This arbelos we will call - a-arbelos.
3. Construction of a generalized Archimedean twins.
Because the circles and don’t have common internal tangent at point P, as in the classical arbelos, we construct the two tangents at point P respectively to circles and . Now we consider the circles that touch these tangents and the circles , and as has shown in Figure 3.
Theorem 1.
Let AB is a chord in the circle , and point P lie on the chord AB (Figure 3). The circles and pass through point P and touch internally the circle respectively at the points A and B. Through point P is constructed tangents CD and EF respectively to the circles and . The circles and that touch , and EF and circles and that touch , and CD have equal radii with length .
Figure 3
Proof:
First we will find the radii of the circles and . Let the centers O and are in the other sides (Figure 4) with respect to the line EF. The case when the centers are in one and the same side with respect to the tangent we will consider when we are looking for the radii of the circles and . From Lemma 1 we have that is a parallelogram and therefore . We construct . From the triangles and we calculate in two ways , we have
(1) .
Figure 4
From the triangle we have
and from Lemma 1 - . We replace in (1) and after a calculation we receive
(2) .
The case when K is between and A the calculations are analogous and the result is the same. Because circles and are symmetrical with respect to the line OA, they have equal radii.
Now we will find the radii of the circles and . Let the centers O and are in one and the same side (Figure 5) with respect to the line CD. From Lemma 1 we have that . We construct . From triangles and , calculating in two ways , we have
(3) .
From the triangle
,
and when we replace in (3) we receive the same result
.
Figure 5
The case when K is between O and is analogous. The theorem is proved.
From formula (2) we see that and consequently when . Then , AB is a diameter and the centers , O and lie on AB. Since the angle (Figure 3) between the tangents CD and EF is , then if , the two tangents coincide and circles and become mutually tangent in point P.
So we proved that the classical arbelos and the Archimedean twins with radius are a special case of the constructed in Lemma 1 a-arbelos and in Theorem 1 circles and (respectively circles and ) with radii equal to for angle .
This circles we will call – a-Archimedean twins.
4. Generalized Archimedean circles.
When the chord AB is not a diameter, it divides the circle into two different segments and respectively we obtain two different arbeloses. Considering (Figure 3) Theorem 1 we can see that the radii of the a-Archimedean twins in the both arbeloses are equal. We will analyse the characteristics which circles[4] analogous to some of the Archimedean circles in the classical arbelos have.
We will begin with the circle of Bankoff. In the classical arbelos this is a circle passing through the tangent point of the small semicircles and the tangent points of the incircle in the arbelos with the small semicircles.
The finding of analogous constructions for a-arbelos could be attended with some difficulties since given construction in the classical arbelos could be gained through more than one construction for the a-arbelos. For example the circle of Bankoff could be examined as the incircle in the triangle with vertexes - centers of the small semicircles and center of the incircle of the arbelos. The analogous circles of these circles don’t have “good” characteristics.
Figure 6
Theorem 2.
With the conditions of Lemma 1 is given an arbelos on the chord AB, bounded by the circles , and (Figure 6). Let circles touchs chord AB and circle respectively at the points P and . If the circles are image of the circles under a homothety with center point P and coefficient and the circles intersects the circles and respectively in the points and , then the circles are the incircles of the arbeloses .
Proof:
If the circle touches the chord AB and the circle respectively in the points P and then is a bisector[5] of . Consequently if point is the middle of the opposed arc AB, then point (Figure 6) is found as point of intersection of and circle . The center of the circle is a point of intersection of and the perpendicular through point P to AB. The middle of the segment is a center of the circle . We will show that the circle is the incircle of the upper arbelos[6].
Figure 7
Let EF and AM are tangents (Figure 7) to circle . From Lemma 1 we have that and consequently the triangles APE and AEB are similar, from this
(4) .
But Þ and
(5) circles and are orthogonal.
We consider an inversion (Figure 8) with respect to the circle A(E). This inversion (4) interchanges the points P and B. Then the inversive images of the circles and are the lines[7] and passing respectively through the points P and B and parallels to the tangent t. The circle is invariant with respect to the considered inversion (because of (5)).
Figure 8
For this inversion:
- The images of the points , and are respectively the points , and .
- The circle is inverted into the circle touching to the lines AB and respectively in the points B and .
- The circle is inverted into the circle that pass through the points B, and and touches the line AB in point B (since touches AB in point P).
Then the circles and are images of the circles and under a homothety with center point A and one and the same coefficient . Therefore the circle is image of the circle under homothety with center point B and coefficient . If point is a center of the circle , then because is a midpoint of BL, but and therefore point lie on and
(6) is perpendicular to the lines and .
Let is a midpoint of . Since , then and .
Because and , then the points P, and are collinear[8] and .
From the circle we have , and after calculation we receive that . Therefore the triangles and are similar and from this
(7) and the points , and are collinear.
From (6) and (7) we see that the circle touches the lines and and the circle , consequently the circle touches the circles , and .
For the circle the proof is the same. The theorem is proved.
Corollary 1.
If and are respectively the radii of the circles and , then:
- the arithmetic mean of the radii and is radius of the Archimedean twins;
- the harmonic mean of the radii and is radius r of a-Archimedean twins;
Proof:
Under inversion (Figure 8) with respect to circle A(E) the points and P have images respectively the points and B, that mean , therefore the center of circumcircle[9] of the triangle is the midpoint T of the arc AP. Similarly the midpoint Q of the arc PB is a center of circumcircle[10] of the triangle . Then the line TQ is a midperpendicular of , lie[11] on TQ, is a diameter of circle and can be calculated[12] as a harmonic mean of the bases of the trapezium with vertices the points T, Q and the midpoints of the segments BP and AP. From this
(8) ,
and similarly
(9) .
Now we form the arithmetic mean of the radii and , we replace from (8) and (9), and we have
(10) .
Similarly the harmonic mean of the radii and is
(11) .
From the constructions of Theorem 2 and formulas (8) and (9) we see that the circle of Bankoff is a special case of the circles and for .
Corollary 2.
The circle touches the circles and , and the points of tangency (Figure 6) are the midpoints of the segments and , and the center is a midpoint of the segment .
Proof:
The proof follow immediately from Theorem 2 and Lemma 1.
Note that if , the circle transform into the midway circle[13].
On (Figure 9) we see Archimedean circle number 6 in the online catalogue [2]. Through each of the centers of the small semicircles is constructed a tangent to the other semicircle. The Archimedean circle with a center point P touches this tangents.
We will consider an analogous construction, when AB is a chord.
Figure 9
Theorem 3.
With the conditions of Lemma 1 is given an arbelos on the chord AB, bounded by the circles , and (Figure 10). Through the midpoints of the segments are constructed tangents to the circles . In quadrangle formed from this tangents can be inscribed an Archimedean circle.
Figure 10
Proof:
Let is intersection point of with the line through a point P and perpendicular to AB. From the similar triangles and we have
(12) .
After a similar calculation we see that the point lie on .
Let is perpendicular to the tangent . From the similar triangles and we have
, and from this .
In same way we can see that the distances from a point to the other tangents are equal to . Therefore the circle with a center point and radius
(13)
is incircle of the quadrangle formed from given tangents and it is an Archimedean circle. The theorem is proved.
From formulas (12) and (13) and the mated proof follow that when , the points , and will coincide respectively with the points , P and . Therefore the circle in the classical arbelos is a special case of the constructed in Theorem 3 circle. This circle stays Archimedean independently of the angle a.