# A Euclidean Approach to Eggs and Polycentric Curves

## Abstract

This paper examines the problem of smoothly connecting two arcs of curves with two more arcs, and presents the application of this procedure to the drawing of a class of egg-shaped curves and to the generation of some open and closed polycentric curves. The paper provides a solid mathematical background as well as ‘ruler and compass’ constructions, some of them original techniques. The use of basic, although lengthy, Euclidean proof tools suggests these properties and techniques were within reach—if not known—by architects and scientists long before the invention of analytic geometry. The concepts described in this paper are further developed in its second part (Mazzotti, What Borromini Might Have Known About Ovals. Ruler and Compass Constructions, 2014).

### Keywords

Borromini San Carlo alle Quattro Fontane Geogebra Euclidean geometry Ovals Eggs Polycentric curves## Introduction

Polycentric curves are curves made of circle arcs subsequently smoothly connected. Their use in architecture has been and still is widespread. For centuries it was the easiest way of connecting points with a curve since only the use of a ruler and a compass was needed, and the infinite amount of possibilities made them an interesting tool for decorative purposes [see for example the pictures on pp. 282–285 in Ragazzo’s book (2011)]. The closed convex single-symmetry version—the egg—provides a possible shape for ground plans of buildings, while the closed convex double-symmetry version—the oval—has been used extensively for both horizontal layouts, such as amphitheatres, stadia and church ground plans (see, for example López Moto 2011), as well as vertical layouts, such as domes, arcs and windows (see for example Huerta 2007). Computer Aided Design, when the appropriate background knowledge is used, can today be used to choose among the infinite possible polycentric curves suiting a particular geometric condition.

This work started in 1999 when I was asked by architect Margherita Caputo to help to understand what baroque artist Francesco Borromini (1599–1667) gave himself as a rule when planning the dome of San Carlo alle Quattro Fontane in Rome, back in the first half of the XVIIth century. He appeared to be able to master the oval form even if at the time only the constructions by Sebastiano Serlio (1619) and a few others (López Moto 2011) appear to have been known.

What Felice Ragazzo (1995) had published just a few years earlier—a simple conjecture about ovals, which he also applied to eggs—showed how it is possible to decide, among an infinite number of possibilities, which oval to draw once the measures of the two axes are given. Although the same paper contains a rigorous mathematical proof of this conjecture by Franco Ghione, I thought it interesting to see if it could be proved only by means of Euclidean geometry, the only tool known at the time of Borromini.

More than ten years later I read F. Ragazzo’s extensive book (2011) on polycentric curves (PCCs) where, among many other problems concerning PCCs, he applies an extended version of his conjecture on ovals to a class of constructions of PCCs. The Euclidean proof of it for eggs and PCCs is the subject of the present paper, together with original remarks and original constructions deriving from it. The consequent application to ovals and the possibility to classify the consequent constructions given different choices of parameters is the subject of the second part of this research, published separately in this issue.^{1}

*quarter*ovals inside rectangles. It is the following (author translation from the Italian).

The locus of the connection points for the arcs[of a quarter oval (Ed.)] […]is the circle, which I will callConnection Locus,defined through the following three points[points 1, 2 and 3 on Ragazzo’s original drawing (Fig. 1)]

intersection between the half major axis and the half minor side[of the rectangle inscribing the oval (Ed.)]

endpoint of the half minor side transported compass-wise onto the major side

intersection between the half major side and the half minor axis(Ragazzo 1995)

“Definitions and Properties of Six-Centre Eggs” is devoted to the basics of six-centre egg shapes. “Polycentric Curves: Definitions and Issues” presents PCCs and the class of problems this research deals with. In “The Main Theorems for Eggs: Properties of the Connection Locus” I prove the main theorem for eggs and properties of Ragazzo’s *Connection Locus* (my translation from the Italian). In “Constructing a Class of Six-Centre Eggs” I apply the proven theorem to the construction of a class of six-centre eggs. “The Main Theorem for Polycentric Curves. Constructing a Class of PCCs” is where the main theorem for PCCs is proved and the different constructions deriving from it—both for open and for closed PCCs—are displayed. “Conclusions and Further Developments” is devoted to my further developments on the subject. The “Appendix” contains most of the proofs of the theorems in this paper.

Geogebra software has been used to draw the different constructions and to get ideas on how to prove the theorems. Animation videos of the different constructions are available on the following website www.mazzottiangelo.eu/en/pcc.asp. This paper only deals with two-dimensional objects.

## Definitions and Properties of Six-Centre Eggs

anegg-shape(or 6-centre eggor simplyegg) is a non-twisted closed convex curve with onesymmetry axis(or simplyaxis) made of six arcs of circle subsequently smoothly connected, i.e. sharing a common tangent (Fig. 2).

Due to symmetry it is enough to study half an egg, say the top part in our picture, and for future purposes, divide the top trapezoid—formed by the symmetry axis, the two tangents orthogonal to it and the tangent through the middle connecting point—into two parts, by drawing the line containing the middle connecting point and the two centres of the corresponding circles.

*A*and*C*as the points of intersection between the half-egg and the axis*K*,*J*,*L*and*N*as the centres of the sequence of four arcs the half egg is made of, with radii*r*_{1},*r*_{2},*r*_{3}and*r*_{4}*H*,*B*and*P*as the connection points of the pairs of arcs

tangents to the circles in

*A*and*C*are orthogonal to*AC**K*and*N*belong to the axisthe triplets

*J*·*K*·*H*,*J*·*L*·*B*and*L*·*N*·*P*are co-linear

Let *O* be the intersection of *AC* with *JB*. If *r*_{1} < *r*_{2}, *O* will lie between *J* and *B*, *K* between *O* and *A*, \( \overline{OA} > \overline{OB} \) and \( \overline{BT} > \overline{AT} \); otherwise (Fig. 3) *J* will lie between *O* and *B* and so on. If *r*_{2} = *r*_{1} the curve becomes a four-centre egg.

A string method for designing some multicentre eggs can be found in Dixon (1987: 76).

Because of it’s similarity to the problem of inscribing an *oval* in a rectangle, in “The Main Theorems for Eggs: Properties of the Connection Locus” i will consider, following closely Ragazzo’s work (1995), the problem of inscribing a half egg into two quadrilaterals, each with two opposite right angles (in Fig. 3: *ATBO* and *COBS*).

## Polycentric Curves: Definitions and Issues

a

polycentric curve—from now onPCC—is a curve made of a finite number of arcs of circle subsequently connected with a common tangent in the common point.

*not*require

*smooth*connections. And since the curve is not closed and convex like before, a connection with common tangents is not necessarily

*smooth*. By this I mean that I am also allowing for so-called cusps and flexes made of arcs sharing a common tangent in their connecting point (the three cases are illustrated in Fig. 4). The reason for this is that the theorem which I will prove in “The Main Theorem for Polycentric Curves. Constructing a Class of PCCs” will not have to deal with the different cases separately. On the other hand artists and craftsmen do enjoy choosing among many different possible PCCs, fitting their conditions or constraints, and cusps and flexes represent a wider choice.

Given two arcs of circle and a point on each, find all the possible pairs of circles sharing a common tangent in a common point, each one having a common tangent with the given arcs at the given point.

## The Main Theorems for Eggs: Properties of the Connection Locus

To be able to inscribe eggs in trapezoids we need to be able to connect arcs of circles.

Given a quadrilateral like *OATB* in Fig. 3, with two opposite right angles, how many pairs of circles such as the ones contained in *OATB* are there? The answer is: an infinite number, all having something in common. But first of all let’s see what happens if I try to impose one of the centres. Anticipating the contents of “Constructing a Class of Six-Centre Eggs”, i will construct such a pair of circles extending the XVIIth century construction of Abraham Bosse for an oval (Bosse 1655: 65–66) (described for example by Dotto 2001) also displayed in the following century by Tosca (see López Moto 2011 for references to both original and electronic facsimile), and then show that they are unique.

Let \( \overline{BT} > \overline{AT} \).

*K*between

*O*and

*A*. We first determine

*F*along the line

*BO*, on the same side of

*O*w.r.t.

*B*, such that \( \overline{BF} = \overline{AK} \) (see Fig. 7), find

*J*as the intersection of the same line with the axis of the segment

*FK*, and then draw the line

*JK*; the intersection with the circle with centre

*K*and radius \( \overline{AK} \) lying inside the quadrilateral is the connection point

*H*for the two arcs, the first being

*AH*with centre

*K*, and the second one being

*HB*with centre

*J*. The drawn arcs enjoy the desired properties.

On the other hand for any such arcs inscribed in *OATB*– part of an egg—with *K* as one of the centres, the only point *F* between *J* and *B* such that \( \overline{FB} = \overline{KA} \), forms with *K* and *B* the same angle as in the above construction, and thus with *K* and *J* the same isosceles triangle. The two drawings have to be the same.

*FK*to meet

*BO*, \( B\hat{F}K \) has to be obtuse, i.e.

*K*

I first wrote the following theorem for ovals, proving Ragazzo’s conjecture (Ragazzo 1995) via Euclidean geometry, and then succeeded in extending it to eggs. The following form allows for an extension to polycentric curves—as will be shown in “The Main Theorem for Polycentric Curves. Constructing a Class of PCCs”—as well as the application to ovals—as shown in the already cited second part of this paper.

**Theorem 1**

*Let OATB be a quadrilateral with*\( \overline{BT} > \overline{AT} \)

*and opposite*\( T\hat{B}O \)

*and*\( O\hat{A}T \)

*right angles*(

*see Fig.*8)

*. Let*\( A\hat{O}B = \alpha \)

*. Let S be the point on BT such that*\( \overline{ST} = \overline{AT} \)

*. Necessary and sufficient condition for H to be the connecting point for two arcs of circle with centres inside*\( B\hat{T}A \)

*and tangent in H, one tangent to the line AT in A and the other tangent to the line TB in B, is for H to belong to the open arc AS of the circle through A, B and S. Moreover, the angle with origin at the centre of such a circle corresponding to the arc AB is equal to α.*

If *H* ≡ *S* or *H* ≡ *A* the arcs in Theorem 1 cannot be built.

*Proof*

See “Appendix” at the end of this paper.

Note that, since \( \overline{AV} = \overline{VS} \), as is easily proved, the line *RV* bisects both concave and convex angles \( A\hat{V}S \), forming a pair of \( \frac{\alpha }{2} \) angles on the convex side, therefore passing through point *T*. Moreover *RK* bisects \( A\hat{K}H \), because triangles *AKR* and *RKH* are equal. But triangles *JRH* and *JRB* are also equal, therefore *JR* bisects \( K\hat{J}O \).

*K*and Theorem 1 have been made and proved in the case \( \overline{BT} > \overline{AT} \). If on the other hand the side of the quadrilateral on the symmetry axis of the egg is smaller than the one on the normal to the middle tangent point, everything works exchanging

*J*with

*K*and

*A*with

*B*. Bosse’s construction can start from a point

*J*between

*O*and

*B*(see Fig. 9), for which the limitation \( \overline{OJ} > \frac{{\overline{OB} - \overline{OA} }}{{1 - \cos A\hat{O}B}} \) can be proved (although for the purpose of drawing we could again start from

*K*on the other side of

*A*w.r.t.

*O*), and the whole of Theorem 1 works in exactly the same manner, since the above exchange of point names—after having mirrored Fig. 8 w.r.t. the bisector of angle

*α*(see Fig. 10)—leaves substance unchanged. I’ll reproduce the changed version, for the sake of completeness.

**Theorem 2**

*Let OATB be a quadrilateral with*\( \overline{AT} > \overline{BT} \)*and opposite*\( T\hat{B}O \)*and*\( O\hat{A}T \)*right angles* (*see Fig.*10)*. Let*\( A\hat{O}B = \alpha \)*. Let S be the point on AT such that*\( \overline{ST} = \overline{BT} \)*. Necessary and sufficient condition for H to be the connecting point for two arcs of circle with centres inside*\( B\hat{T}A \)*and tangent in H, one tangent to the line AT in A and the other tangent to the line TB in B, is for H to belong to the open arc BS of the circle through A, B and S. Moreover, the angle at the centre of such a circle corresponding to the arc AB is equal to α.*

Finally, if \( \overline{BT} = \overline{AT} \), *K* and *J* coincide with *O* and the two arcs become a single one.

**Definition**

I will call the circle through *A*, *B* and *S*, following Ragazzo’s work (1995), from now on *Connection Locus* (my translation), or simply *CL*.

Summing up the properties of point *R* for quarter eggs we have the following.

**Theorem 3**

*The centre R of the Connection Locus for a quarter egg is the incentre of triangle OKJ.*

The fact that *RO* bisects \( J\hat{O}A \) (or its adjacent angle) is another result already conjectured by Ragazzo (1995).

## Constructing a Class of Six-Centre Eggs

Six-centre eggs can be constructed in different ways. They have 9 of freedom, although with some constraints. Take for example four points on a plane (equivalent to eight parameters) and then fix the measure of one or the radii.

*Construction of a (six*-

*centre) egg given four consecutive centres and the radius of the first one*. For example points

*A*,

*B*,

*C*and

*D*and measure

*r*in Fig. 11 (see link to Geogebra animation video on my website www.mazzottiangelo.eu/en/pcc.asp):

draw the line

*AD*draw the circle with centre

*A*and radius*r*, let*M*be the intersection with*AD*find point

*F*as the intersection between this first circle and the line*AB*draw the circle with centre

*B*and radius \( \overline{BF} \) and find its intersection*H*with*BC*draw the circle with centre

*C*and radius \( \overline{CH} \) and find its intersection*I*with*CD*draw the circle with centre

*D*and radius \( \overline{DI} \) and find its intersection*L*with*AD*mirror arcs

*MF*,*FH*,*HI*and*IL*w.r.t.*AD*to get the whole egg.

*right*side. It is also true that the same set of points and radius can yield a different oval if points are chosen

*in a different order*. See Fig. 12, where I have exchanged the roles of

*B*and

*C*and compare it to the egg in Fig. 11. The class of six-centre eggs I am interested in is that which can be built using the CL. That is, I will consider here the problem of inscribing a six-centre egg inside a given isosceles trapezoid, once the tangent point(s) to the oblique side is (are) given. Ragazzo has already solved the problem since the CL is his own invention (see Ragazzo 1995, 2011) although he has not given proof of his conjecture. I will repeat the construction here in a more systematic way referring directly to Theorem 1, which is my own original contribution, dealing also with situations not yet tackled, in order to solve the problem completely.

Given an isosceles trapezoid and two symmetric points on the oblique sides, is it always possible to inscribe a six-centre egg with the same symmetry axis, tangent to the parallel sides with two opposite arcs, and tangent to the oblique sides at the given points exactly where each of the other two pairs of arcs connect? How many are there for any given trapezoid?

*OATB*and

*OBSC*in Fig. 13a) satisfying the conditions of Theorems 1 and 2. That is unless this perpendicular intersects the symmetry axis

*outside*segment \( \overline{AC} \) (see Fig. 13b). We need to find out if the two tangent arcs of circle needed

*inside*the trapezoid with the usual properties can still be drawn in the latter case, and that they are unique.

The answer is *yes*. And the proof is the same as that of Theorem 1, although a few changes need to be made.

**Theorem 4**

*Let OATB be a tangled quadrilateral*(

*see Fig.*14)

*with BT and AO crossing each other,*\( \overline{BT} > \overline{AT} \)

*and opposite*\( T\hat{B}O \)

*and*\( O\hat{A}T \)

*right angles. Let α be the external angle to*\( A\hat{O}B \)

*. Let S be the point on BT such that*\( \overline{ST} = \overline{AT} \)

*. Necessary and sufficient condition for H to be the connecting point for two arcs of circle with centres inside*\( B\hat{T}A \)

*and tangent in H, one tangent to the line AT in A and the other tangent to the line TB in B, is for H to belong to the open arc AS of the circle through A, B and S. Moreover, the angle at the centre of such a circle corresponding to the arc AB is equal to α.*

*Proof*

See “Appendix”.

Thanks to Theorem 4 I can now say that *any* point *B* can be chosen on segment *ST* in Fig. 13 because a CL can be drawn on both sides of *BO* and on each of the two a connection point freely chosen. I can now state that in any given isosceles trapezoid (at least) ∞^{3} six-centre eggs can be inscribed. This is how it is done.

*Construction of a (six*-

*centre) egg inscribed in a given isosceles trapezoid*(see Fig. 15 and link to Geogebra animation video on my website www.mazzottiangelo.eu/en/pcc.asp).

given the isosceles trapezoid

*C*′*D*′*TS*of bases*C*′*S*and*TD*′, draw the symmetry axis*AC*and choose a point*B*on the oblique side*TS*to be the middle connection pointdraw from

*B*the orthogonal line to*ST*and let*O*be the intersection with the*line AC*since \( \overline{BT} < \overline{AT} \) find the point

*S*′ on segment*AT*such that \( \overline{TS'} = \overline{BT} \)since \( \overline{BS} > \overline{SC} \) find the point

*S*″ on segment*SB*such that \( \overline{SS''} = \overline{SC} \)draw the arc

*BS*′ of the CL—with centre*R*—through*A*,*S*′ and*B*, and the arc*CS*″ of the CL—with centre*R*′—through*C*,*S*″ and*B*choose

*H*on arc*BS*′ and*P*on arc*CS*″the axis of segment

*AH*intersects*OA*in point*K*, the axis of*HB*intersects*BO*in point*J*, the axis of*PB*intersects*BO*in point*L*, the axis of segment*PC*intersects*AC*in point*N*arcs

*AH*with centre*K*,*HB*with centre*J*,*BP*with centre*L*and*PC*with centre*N*, and their symmetric arcs, form the desired egg

## The Main Theorem for Polycentric Curves. Constructing a Class of PCCs

As I already pointed out, eggs are an extension of ovals, and both are a very special case of polycentric curves (PCCs). Because of their increased complexity, there are many different approaches to the problem of finding the PCC satisfying this or that condition. Felice Ragazzo’s book tackles and solves five of these problem (see Ragazzo 2011: 17) including the one I talk about in this section—for which he illustrates the whole of the different possibilities of resulting PCCs depending on the mutual position of the given circles (Ragazzo 2011: 146–213). My contribution to the subject is the mathematical proof of the main construction tool and of other properties of the CL involved, as well as some new remarks and the possibility of viewing the constructions I made with Geogebra.

Given two arcs of circle and a point on each, find all the possible pairs of different circles sharing a common tangent in the common point, each one having a common tangent with the given arcs at the given point.

I will tackle the problem by means of extending Theorem 1 to the case of PCCs. Before this, I need to sum up, or prove, all the properties enjoyed by a special quadrilateral and by the *two* circles that one derives from it, forming together the CL. The inspiration for the following approach was Ragazzo’s drawings (Ragazzo 2011: 131–145).

### Systems of Two Tangents

*system of two tangents with base points A and B*(see Fig. 16).

*A*and

*B*on a plane and two non-parallel lines through them

*t*

_{A}and

*t*

_{B}—thought of as tangents in

*A*and

*B*to curves—draw the perpendiculars from

*A*and

*B*,

*n*

_{A}and

*n*

_{B}. Let

*O*=

*n*

_{A}∩

*n*

_{B}and

*T*=

*t*

_{A}∩

*t*

_{B}. Let’s build the CL; let:

*S*_{1}and*S*_{4}be the points on*TB*, the first one on*B*’s side the second on*T*’s side, such that \( \overline{{TS_{1} }} = \overline{{TS_{4} }} = \overline{TA} \)*S*_{2}and*S*_{3}be the points on*TA*, the first one on*A*‘s side the second on*T*’s side, such that \( \overline{{TS_{2} }} = \overline{{TS_{3} }} = \overline{TB} \)*α*the obtuse angle formed by*n*_{A}and*n*_{B}and*β*=*π*−*α*the acute angle formed by*t*_{A}and*t*_{B}

*R*of the circle

*Γ*

_{1}through

*A*,

*B*and

*S*

_{2}just by intersecting the axis of segment

*AB*and the axis of segment

*BS*

_{2}, which is also the bisector of \( A\hat{T}B \). Now note that the two equal triangles

*TRB*and

*TRS*

_{2}are also formed, and so

*S*

_{2}has also to be on

*Γ*

_{1}. The same can be shown for the second circle,

*Γ*

_{2}the circle with centre

*F*through

*A*,

*B*,

*S*

_{3}. Let’s check the angles in the quadrilateral

*AFBR*: since \( \overline{{S_{1} T}} = \overline{AT} \) one has that \( T\hat{S}_{1} A = \frac{\pi - \beta }{2} = \frac{\alpha }{2} \), which implies that \( R\hat{S}_{1} A + R\hat{S}_{1} B = A\hat{S}_{1} B = \pi - \frac{\alpha }{2} \) and so, since \( R\hat{S}_{1} A = R\hat{A}S_{1} \) and \( R\hat{S}_{1} B = R\hat{B}S_{1} \), that \( R\hat{A}S_{1} + R\hat{B}S_{1} = \pi - \frac{\alpha }{2} \); looking at the quadrilateral

*RAS*

_{1}

*B*we can deduce that

It is similarly proved that \( B\hat{F}A = \beta \), and thus that \( R\hat{B}F \) and \( R\hat{A}F \) have to be right angles. We finally notice that *FTS*_{3} and *FTB* are equal therefore *TF* bisects \( B\hat{T}S_{3} \).

*O*≡

*A*or

*O*≡

*B*and

*OATB*becomes a triangle, or even when

*OATB*becomes tangled, as can be shown by checking the above properties on the different cases in Fig. 17.

Summing up, this means that centres *R* and *F* of *Γ*_{1} and *Γ*_{2} can be found simply intersecting the axis of *AB* with the two bisectors of the angles in *T* formed by *t*_{A} and *t*_{B}. Furthermore *RAFB* shares the same angles as *AOBT* when it is not tangled.

### The Main Theorem

We now need a whole new theorem to show that *Γ*_{1} and *Γ*_{2} actually form *the* CL and draw our polycentric curves. The use of analytic geometry to prove this same theorem would probably make things much more straightforward. My choice of using Euclidean geometry is that of remaining consistent with the whole setup of the present work.

**Theorem 5**

*Given a system of tangents with base points A and B, for any H* *∊* *Γ*_{1} *∪* *Γ*_{2}*—{A,* *B,* *S*_{1}, *S*_{2}, *S*_{3}, *S*_{4}*} (see Fig.**16**) there exist two circles Γ*_{3}*and Γ*_{4}*tangent with each other in H, such that t*_{A}*is tangent to Γ*_{3}*in A and t*_{B}*is tangent to Γ*_{4}*in B.*

*Proof*

See “Appendix”.

I now need to prove the reverse of Theorem 5. In order to do so I have to show how *all* pairs of tangent circles, one with tangent *t*_{A} in *A* and the other with tangent *t*_{B} in *B*, can be detected. I will use the following construction, which I found in Ragazzo’s book (2011), and then prove by means of Lemma 7 that such a construction is what one needs to find all possible circles tangent to a given one and tangent in *A* to a straight line through *A*.

*A*be a point,

*t*

_{A}a straight line through it and

*Γ*

_{4}a circle with centre

*J*(see Fig. 18).

let

*n*_{A}be the perpendicular to*t*_{A}in*A*and*E*and*F*the intersections of the parallel to it through*J*with*Γ*_{4}let

*H*_{1}and*H*_{2}be the new intersections of*FA*and*EA*with*Γ*_{4}let

*K*_{1}and*K*_{2}be the intersections of*JH*_{2}and*JH*_{1}with*n*_{A}

Triangles *K*_{1}*H*_{1}*A* and *FJH*_{1} are similar, therefore \( \overline{{K_{1} A}} = \overline{{K_{1} H_{1} }} \) and the circle with centre *K*_{1} and radius \( \overline{{K_{1} A}} \) passes through *H*_{1} and is tangent to *Γ*_{4} since *H*_{1}, *K*_{1} and *J* are co-linear. Moreover triangles *K*_{2}*H*_{2}*A* and *EJH*_{2} are also similar, therefore \( \overline{{K_{2} A}} = \overline{{K_{2} H_{2} }} \) and the circle with centre *K*_{2} and radius \( \overline{{K_{2} A}} \) passes through *H*_{2} and is tangent to *Γ*_{4} since *H*_{2}, *K*_{2} and *J* are co-linear.

If *Γ*_{4} is tangent to *t*_{A} then one of the circles becomes a straight line, while if *A* ∊ *Γ*_{4} no circle can be drawn.

**Lemma 6**

*Any circle tangent to a given circle Γ*_{4}*and to a given straight line t*_{A}*in a given point A can be found by means of the above procedure.*

*Proof*

See “Appendix”.

Knowing how to find all connection points *H* we can prove the property they enjoy.

**Theorem 7**

*Given a system of tangents with base points A and B, for any circles Γ*_{3}*and Γ*_{4}*tangent in H, such that t*_{A}*is tangent to Γ*_{3}*in A and t*_{B}*is tangent to Γ*_{4}*in B, we have that H* *∊* *Γ*_{1} *∪* *Γ*_{2}* —{A,* *B,* *S*_{1}, *S*_{2}, *S*_{3}, *S*_{4}*}.*

*Proof*

See “Appendix”.

Keeping in mind that two points on two smooth curves define a system of tangents as described in “Systems of Two Tangents”, we can finally state the main theorem.

**Theorem 8**

*Given two arcs (of circle) and a point on each—A and B—such that the tangents through these points are not parallel, necessary and sufficient condition for H to be the connecting point for a pair of tangent circles, each tangent to one of the arcs in the given point, is for H to belong to Γ*_{1} *∪* *Γ*_{2}*—{A,* *B,* *S*_{1}, *S*_{2}, *S*_{3}, *S*_{4}*}.*

*Proof*

A direct consequence of Theorems 5 and 7.

We just need to prove that, as Ragazzo realised, the CL for two arcs and two points on them such that the tangents are *parallel*, is simply the line through these two points leaving those points out.

**Theorem 9**

*Given two arcs (of circle) and a point on each—A and B—such that the tangents through these points are parallel*(

*see Fig.*19)

*, necessary and sufficient condition for H to be the connecting point for a pair of tangent circles, each tangent to one of the arcs in the given point, is for H to belong to AB—{A,*

*B}.*

*Proof*

See “Appendix”.

### Joining Two Arcs with Two Arcs

*A*and

*B*—where a single tangent is defined, we construct the system of tangents described in “Systems of Two Tangents”, finding

*R*and

*F*as the intersection between the axis of

*AB*and the two bisectors of the angles in

*T*(see Fig. 20). The circles

*Γ*

_{1}and

*Γ*

_{2}through

*A*and

*B*—with centres

*R*and

*F*—are the CL, once we have crossed out points

*S*

_{1},

*S*

_{2},

*S*

_{3},

*S*

_{4}, as well as

*A*and

*B*, as proved by Theorem 8 (see link to Geogebra animation at www.mazzottiangelo.eu/en/pcc.asp).

*H*on

*Γ*

_{1}—{

*A*,

*B*,

*S*

_{1},

*S*

_{2}} and draw the circle through

*A*and

*H*having its centre

*K*on line

*OA*, then find

*J*as the intersection between lines

*HK*and

*OB*and draw the circle with centre

*J*and radius \( \overline{JH} \). We are now free to choose any of the sixteen possible four-arc polycentric curves through

*A*,

*H*and

*B*just by means of deciding which arcs to keep. Figure 21 shows—for a choice of

*H*—four possible choices of arcs (see link to Geogebra animation video on my website www.mazzottiangelo.eu/en/pcc.asp).

*H*on

*Γ*

_{2}—{

*A*,

*B*,

*S*

_{3},

*S*

_{4}}, repeating the same procedure (Fig. 22; see link to Geogebra animation video at www.mazzottiangelo.eu/en/pcc.asp).

Choosing *H* to be *S*_{1}, *S*_{2}, *S*_{3} or *S*_{4} is also possible, when one is ready to accept polycentric curves including segments.

Parallel tangents can also yield four-arc polycentric curves. The construction is a straightforward application of Theorem 9.

### Four-arc Closed PCCs

*A*and

*B*, and defining two lines through

*A*and

*B*, it is also possible to draw four-arc

*closed*polycentric curves with them as first and third connecting points and the given lines as tangents. It is enough to repeat the procedure to draw the four-arc polycentric curve, this time choosing two

*H*points. In this way we will have two sets of two arcs starting and ending in

*A*and

*B*, that is a four-arc

*closed*polycentric curve. Figure 23 shows four possible choices of two points

*H*and

*H*

_{1}and for each a choice (among sixteen) of arcs (see link to Geogebra animation at www.mazzottiangelo.eu/en/pcc.asp).

## Conclusions and Further Developments

Mathematical rigour and practical drawing applications are the topics of this paper. The freeware Geogebra has been crucial to achieve both. Furthermore, only Euclidean geometry has been used throughout this work.

The consequences for the constructions of ovals, and an attempt to classify them, as well as a conjecture on the role they might have played in Baroque architecture, are the topics of the second part of this paper in the present issue of this journal (Mazzotti 2014).

An analytic approach to the study of ovals, eggs and PCCs will be the subject of a future paper.

## Footnotes

- 1.
Angelo Alessandro Mazzotti,

*What Borromini Might Have Known About Ovals: Ruler and Compass Constructions*.

## Notes

### Acknowledgments

I would like to thank Stefano Herzel for his friendship, help and encouragement, Felice Ragazzo for the long talks and the exchange of ideas and Sophie Püschmann for her constant support.

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