arxiv:2008.08950v1 [math.ds] 20 aug 2020 · the reuleaux triangle r,is the (convex) curve formed by...

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A SPECIAL CONIC ASSOCIATED WITH THE

REULEAUX NEGATIVE PEDAL CURVE

LILIANA GABRIELA GHEORGHE AND DAN REZNIK

The Negative Pedal Curve of the Reuleaux Triangle w.r. to a pedal pointM located on its boundary consists of two elliptic arcs and a point P0. In-triguingly, the conic passing through the four arc endpoints and by P0 hasone focus at M . We provide a synthetic proof for this fact using Poncelet’sporism, polar duality and inversive techniques. Additional interesting proper-ties of the Reuleaux negative pedal w.r. to pedal point M are also included.

1. Introduction

Figure 1. The sides of the Reuleaux Triangle R are three circular arcs of circles centeredat each Reuleaux vertex Vi, i = 1, 2, 3. of an equilateral triangle.

The Reuleaux triangle R is the convex curve formed by the arcs of threecircles of equal radii r centered on the vertices V1, V2, V3 of an equilateraltriangle and that mutually intercepts in these vertices; see Figure 1. Thistriangle is mostly known due to its constant width property [4].

Here, we study some properties of the negative pedal curve N of R w.r.to a pedal point M lying on one of its sides. This curve is the envelope oflines passing through points P on R and perpendicular to PM [3, Negative

Date: August, 2020.Keywords and phrases: conic, inversion, pole, polar, dual curve, negative pedal

curve.(2020)Mathematics Subject Classification: 51P99, 60A99.

1

http://arxiv.org/abs/2008.08950v2

2 L. G. GHEORGHE AND D. REZNIK

Pedal Curve]. Trivially, the negative pedal curve of arc V1V2 is a point whichwe call P0. We show that the negative pedal curves of the other two sidesare elliptic arcs with a common focus on M and whose major axis measures2r; see Prop. 1.

Let V3 be the center of the circular arc where pedal point M lies, andlet V1, V2 be the endpoints of said arc. Let arc A1A2 (resp. B1B2) be thenegative pedal image of the Reuleaux side V1V3 (resp. V2V3) where pointA1 is the image of V1, and B1 the image of V2. The endpoints of N whosepreimage is V3 are respectively A2 when V3 is regarded as a point of sideV1V3, and B2 when V3 is regarded as a point of side V2V3 of the Reuleauxtriangle.

Our main result (Theorem 1, Section 2) is an intriguing property of theconic C∗ – called here the endpoint conic – that passes through the endpointsA1, A2, B1, B2 of the negative pedal curve N , and through P0: that oneof its foci is precisely the pedal point M ; see Figure 2. We also give a fullgeometric description of its axes, directrix and vertices, and a criterion foridentifying its type, according to the location of the pedal point M .

In Section 3 we prove other properties of the Reuleaux triangle and itsnegative pedal curve, involving tangencies, collinearities and homotheties.

The proofs combine elementary techniques with inversive arguments andpolar reciprocity. A review of polar reciprocity and other concepts, includingthe description of the negative pedal curve as a locus of points, as well asan alternative description of it as an envelope of lines is postponed to theAppendix.

A SPECIAL CONIC ASSOCIATED W/ THE REULEAUX TRIANGLE 3

Figure 2. The sides of the Reuleaux R are three circular arcs centered at the verticesV1, V2, V3 of an equilateral triangle. Its negative pedal curve N w.r. to a pedal point M onits boundary consists of a point P0 (the antipode of M through V3) and of two elliptic arcsA1A2 and B1B2 (green and blue). The endpoint conic C∗ (purple) passes through P0 andthe four endpoints of the two elliptic arcs of N . It has a focus on M and its focal axis passesthrough the center of the Reuleaux triangle.

2. Main Result: The Endpoint Conic

Referring to Figure 2.

Theorem 1. The conic C∗ which contains P0 and the endpoints A1, A2, B1,B2 of the negative pedal curve of the Reuleaux triangle R with respect to apedal point M located on a side of R, has one focus on M and its axes passthrough the circumcenter of △V1V2V3.

The proof will require some additional steps which steadily use an inversiveapproach and polar duality.

The main idea is to use polar reciprocity: in order to prove that the focusof the C∗ is M , we show that there exists a circle C to which the five polarsof the endpoints A0, A1, B0, B1 and P0 are tangent.

Recall that whenever we perform a polar transform (or a polar duality)of a conic, w.r. to an inversion circle (see Figure 3), points on the conictransform into their polars w.r. to the inversion circle, and those polarsbecome tangents to the dual curve of the (initial) conic [2, Article 306].


Figure 3. The Reuleaux triangle R and its inverse: (i) arcs V ′1V ′3

(blue) and V ′2V ′3

(green)are the inverses of sides V1V3 and V2V3 of R, while line V ′1V

′

2except for segment [V ′

1V ′2] itself

(violet) is the image of arc V1V2; (ii) the polars of A1 and B1 are the tangents in V ′1 and V′

2

to arcs V ′1V ′3

and V ′2V ′3; (iii) the polars of A2 and B2 are the tangents in V ′3 to arcs V

′

1V ′3

andV ′2V ′3; (iv) line V ′

1V ′2

is the polar of P0. (v) all five polars above are tangent to the exinscribedcircle c of △V ′

1O′V ′

2(purple). The angle between circles c0 (green) and C0 (blue) in V ′3 is

120◦ iff V ′3

is on the exinscribed c centered in O (not shown); in this case, the tangents at V ′3

(dashed green and blue) to circles c0 and C0, are also tangent to the exinscribed circle c.

This indirect and inversive approach is appropriate as the polars of A1,A2, B1, B2, and P0 can be readily analyzed.

Classic facts about polar reciprocity guarantee that the dual (curve) of aconic is a circle iff the center of the inversion circle (which, in our case, isM) is the focus of the endpoint conic; see Proposition 9.

The reader not familiar with the topic may find useful the details in Ap-pendix and the references therein.

Referring to Figure 3.


Lemma 1. Let M be a point on the side V1V2 of the Reuleaux triangle andI the inversion circle. Let V ′1 , V

′2 , V

′3 be the inverses of points V1, V2, V3 and

let arcs V ′1V′3 and V

′2V

′3 be, respectively, the inverses of arcs V1V3 and V2V3

of the Reuleaux triangle. Then:

(i) The polars of A1 and B1 are the tangents at V′1 and V

′2 to circular

arcs V ′1V′3 and V

′2V

′3 . The poles of the tangents at V

′1 and V

′2 to arcs

V ′1V′3 and V

′2V

′3 , are the points A1, B1.

(ii) The polars of points A2, B2 are the tangents in V′3 to arcs V

′1V

′3 and

V ′2V′3 , respectively. The poles of the tangents at V

′3 to arcs V

′1V

′3 and

V ′2V′3 , are the points A2, B2.

(iii) The inverse of arc V1V2 is the line V′1V

′2 excluding segment [V

′1 V

′2 ]

and the polar of P0 is the line V′1V

′2 .

Referring to Figure 3:

Proof. Inversion w.r. to a circle maps circles to either circles or lines: thus,the inverses of arcs V1V3 and V2V3 are the two circular arcs V

′1V

′3 and V

′2V

′3 ,

respectively. On the other hand, since arc V1V2 passes through the inversioncenter, its image is the union of two half-lines.

All other statements derive from the description of the negative pedalcurve as the dual of its inverse, as shown in Proposition 10. �

Lemma 2. Using the notation in Lemma 1:

(i) the angles at V ′1, V′2 , and V

′3 between arcs V

′1V

′2, V

′1V

′3 , and V

′1V

′2

respectively (the inverses of the sides of the Reuleaux triangle), are120◦.

(ii) △V ′1O′V ′2 determined by the tangents at V

′1 , V

′2 to said arcs and the

line V ′1V′2 is equilateral.

Proof. These statements derive from the fact that inversion is preserves an-gles between curves. �

Given the above results, Theorem 1 is equivalent to the following Lemma;see Figure 3:

Lemma 3. The five polars of endpoints A1, A2, B1, B2, and P0 are tangentto circle c, the exinscribed circle in △V ′1O

′V ′2, which is externally-tangent toside V ′1V

′2 .

This lemma will be equivalent to the assertion that the focus of C∗ co-incides with pedal point M when we show that the two tangents at V ′3 toarcs V ′1V

′3 and V

′2V

′3 respectively are also tangent to the excircle of △V

′1O

′V ′2 .Referring to Figure 3, this can be restated as follows:

Lemma 4. Let △V ′1O′V ′2 be equilateral and let O be the center of its exin-

scribed circle c, externally-tangent to side [V ′1V′2 ]. Let C be another circle,

concentric with c, that passes through V ′1 (and V′2). Finally, let c0 and C0

be the two circles tangent to the sides [OV ′1 ] and [OV′2 ] of the triangle, at V

′1

and V ′2 , respectively. Then:

(i) c0 and C0 intersect at an angle of 120◦ iff the three circles c0, C0,

and C pass through one common point; let V ′3 be that point.


(ii) if the condition above is fulfilled, then the two tangents at the com-mon point V ′3 to circles c0 and C0 are also tangent to the exinscribedcircle c.

While assertion (i) is automatic, once we identify circles c0 and C0 as theinverses of the circles which define the Reuleaux triangle, assertion (ii) is notobvious and will require additional steps.

Though our construction is in general asymmetric, there are two regularhexagons associated with it, used in the results below.

Lemma 5. Let [A0A1. . .A5] be a regular hexagon with inscribed circle cand circumcircle C. Let P0 be a point on arc A0A1 of C and let P0P1,P1P2, . . . , P5P6 be the tangents from P0, P1, . . . , P5 to c.

Let c0 be the circle tangent to side [A0A5] of the hexagon at A0 whosecenter is inside the hexagon; let P0 be its second intersection point with C.Then:

(i) Points P6 and P0 coincide and the hexagon [P0P1 . . . P5] is regularand congruent with [A0A1 . . . A5]. Both hexagons share the sameincircle and circumcircle.

(ii) Let P0P1 be the tangent from P0 to c; then it tangents (in P0) thecircle c0, as well.

Referring to Figure 4:

Proof. i) When we perform the construction of tangent lines P0P1, . . ., P5P6,the process ends in six steps and P0 = P6, thanks to Poncelet’s porism, sincec and C are the incircle and the circumcircle of a hexagon. Note the latteris regular since its inscribed and circumscribed circles are concentric.

ii) Let T0 be the intersection of the perpendicular bisector of segment[A0P0] with line A0A5. We shall prove that T0P0 is tangent at P0 to circlec0 and that T0, P0, P1 are collinear.

Let c0 be the center of circle c0; since T0 is a point on the perpendicularbisector of [A0P0], and since A0 and P0 are the two intersections of circlesc0 and c, then O, c0 and T0 are collinear.

Next, △T0A0c0 = △T0P0c0, as they have respectively-congruent sides,hence:

∠T0P0c0 = ∠T0A0c0 = 90◦

which proves that line T0P0 is tangent at P0 to circle c0.Furthermore, △T0A0O = △T0P0O as they have respectively-congruent

sides, hence:

∠T0P0O = ∠T0A0O

By hypothesis, T0, A0, and A5 are collinear, and △A0A5O is equilateral,hence the external angle ∠T0A0O = 120

◦; hence ∠T0P0O = 120◦ as well.

Since by (i) △P0OP1 is equilateral, then ∠OP0P1 = 60◦ and ∠T0P0P1 =

180◦, proving that points T0, P0, P1 are collinear.�


Figure 4. The angle between circles c0 (green circle) and C0 (blue circle) in P0 is 120◦ iffP0 is on the circumcircle of the regular hexagon [A0A1 . . . A5] (orange circle).

We now reformulate Lemma 4 as follows:

Lemma 6. Let [A0A1 . . . A5] be a regular hexagon whose incircle is c andcircumcircle is C. Let c0 be the circle tangent to side [A0A5] of the hexagonat A0 and let C0 be the circle tangent to side [A1A2] at A1. Then the anglebetween circles c0 and C0 is 120

◦ iff the three circles c0,C0, and C haveone common point.

Proof. ⇐ First assume circles c0 and C0 intersect at a point P0 on circum-circle C. Referring to Figure 4, if P0 is on arc A0A1 of C then by Lemma (ii)P0P1 is a common tangent to circles c0 and c. In particular, P0P1 is thetangent at P0 to circle c0.

Next, let P0P′5 be the tangent from P0 to the incircle c (distinct from P0P1).

Similarly, let P ′5P′4, P

′4P

′3, . . .P

′1P

′0 be the tangents from P

′5, P

′4, . . . , P

′1 ∈ C

to the incircle c.


Then, as above, points P ′0 and P0 coincide, and hexagon [P′0P

′1 . . . P

′5]

is regular; since hexagons [P0P1 . . . P5] and [P′0P

′1 . . . P

′5] have one common

point (P0), are regular, and are both inscribed in C, they must coincide.Once again, Lemma (ii) guarantees that P0P5 is a common tangent to

circles C0 and c. In particular, line P0P5 is the tangent at P0 to circle C0.Since hexagon [P0P1 . . . P5] is regular, ∠P1P0P5 = 120

◦. This guaranteesthat the angle between circles c0 and C0, which is the angle between theirtangents at P0, is also 120

◦.⇒ By hypothesis, circles c0 and C0 intersect at an angle of 120

◦. Weshall prove that, necessarily, point P0 must be on the circumcircle C.

Call P ′0 the intersection point between circle c0 and arc A0A1 of circle C.We shall prove that P0 and P

′0 coincide.

Let C′0

be the circle tangent at A1 to line A1A2 that passes through P′0.

Then, by the first part of this proof, circles c0 and C′0

intersect at an angleof 120◦. So circles C0 and C

′0

would be two circles, both tangent at A1 toline A1A2, which intersect circle c0 at the same angle. Hence circles C0 andC

′0

must coincide, as do points P0 and P′0. �

Finally we can prove Theorem 1:

Proof. The above lemmas prove that the focus of C∗ coincides with M . Weend the proof by showing that the axis of C∗ passes through the circumcenterof △V1V2V3. Equivalently, we prove that the directrix of C

∗ is perpendicularto the line that joins points M and G. We shall an inversive argument.

As shown in Proposition 8 in the Appendix, the directrix of a conic whosepolar-dual is some circle, is precisely the polar of the center of that circle(w.r. to the inversion circle). In other words, the directrix of the C∗ is thepolar of O.

Recall V ′1 , V′2 and V

′3 are, respectively, the inverses of V1, V2, V3 w.r.

to the inversion circle centered in M . Recall also that M , O, and G are,respectively, the center of the inversion circle, the circumcenter of △V ′1V

′2V

′3 ,

and the circumcenter of △V1V2V3. Hence M,O,G are collinear.In turn, this implies that the polar of O (that is perpendicular to OM),

is perpendicular to GM , as well.Thus, the axis of C∗ and line MO will either be parallel or coincide. Since

M is the focus of C∗, is major axis is line MO and point G is on that line. �


Figure 5. The endpoint conic C∗ (purple) is the dual of circle c (dashed purple) w.r. tothe inversion circle I centered on M (dashed black); Its directrix is the polar of O, thecircumcenter of c; its major axis passes through G, and its vertices, L1 and L2 are theinverses of antipodal points N1 and N2, the diameter of c passing through M (point N1, theantipode of N2 w.r. to O, not shown).

The above results reveal that the endpoint conic is in fact the polar-dualof a special circle, which depends on the vertices of the Reuleaux triangleand on the location of pedal point M ; see Proposition 8.Using the notation in Lemma 3:

Corollary 1. The endpoint conic C∗ associated with a Reuleaux triangle anda pedal point M is the polar-dual of circle c. Its type depends on the locationof M on arc V1V2: it is an ellipse (resp. hyperbola) if it lies inside (resp.outside) circle c. If it is on said circle, C∗ is a parabola.


Therefore one can (geometrically) construct any of its elements (vertices,other focus) as well as compute its axis and eccentricity.

Observation 1. The type of C∗ can be identified with an additional obser-vation. Referring to Figure 6. With the notation in lemma 6:

Let C′ be reflection of C w.r. to line V ′1V′2 and let D1,D2 be the two

intersections between circles c and C′. One can check that V ′3 is the reflectionof M w.r. to V ′1V

′2; therefore, M is located both on the reflection of the

circumcircle of △V ′1V′2V

′3 w.r. to V

′1V

′2 , and on arc V1V2 of the Reuleaux

triangle. The location of M with respect to C′ reveals the which type of conicC∗ is: it is an ellipses if M is on the arc D1D2 of circle C

′, a hyperbola, ifM ∈ V ′2D2 or V

′1D1 and a parabola when M is either D1 or D2.

Figure 6. The endpoint conic C∗ (dark green) is a parabola iff point M coincides witheither D1 or D2; its directrix (dark green) is the polar of the circumcenter O of △V ′1V

′

2V ′3

(not shown).

3. Some Elementary Properties

3.1. Collinearity and Tangencies. Referring to Figure 7:

Proposition 1. The negative pedal curve N of the Reuleaux triangle consistsof two elliptic arcs EA and EB and a point P0, the antipode of M w.r. to thecenter of the circle where M is located. The two ellipses EA, EB are centeredon the vertices of the Reuleaux triangle, V1 and V2, have one common focusat M , and their semi-axes are of length r.

Proof. By hypothesis, M belongs to arc V1V2 of the circle centered in V3 thatpasses through V2 and V3. Hence, if P is any point on this arc and we drawthe perpendicular p through P on PM , all these lines will pass through afixed point P0, which is he antipode of M w.r. to center V3.

The second part derives directly from the general construction of the neg-ative pedal curve of a circle. See Proposition 10 in the Appendix. �


Proposition 2. The minor axes of EA and EB pass through P0.

Proof. By the definition of the negative pedal curve, if we regard V1 as apoint on arc V1V2 of the circle centered on V3 on which the pedal point Mlies, then P0V1 will be perpendicular to MV1. Since V1 is the center of EA andsince line MV1 is its major axis, its minor axis will be along P0V1. Similarly,the minor axis of EB is P0V2. �

Proposition 3. Points A2, B2 and V3 are collinear and line A2B2 is acommon tangent to EA and EB.

Proof. By construction, the negative pedal curve of arc V2V3 is the ellipticarc EA, delimited by A1 and A2. This implies that MV3 and A2V3 areperpendicular, as well as MV3 and B2V3. Thus points A2, V3 and B2 arecollinear. Also by construction, the perpendicular to MV3 at V3 is tangent toN at A2 (resp. B2) when V3 is regarded as a point in the V2V3 (resp. V1V3)arc. Hence the points A2, V3 and B2 are collinear (∠A2V3B2 = 180

◦) andA2B2 is the common tangent to EA and EB, at A2 and B2, respectively. �

Proposition 4. Point A1 is on P0V2 and B1 is on P0V1.

Proof. If we regard V1 as a point on arc V1V2 of the circle centered on V3whose negative pedal curve is P0, then, necessarily, V1P0 ⊥ MV1. Similarly,if we regard V1 as a point on arc V1V3 of the circle centered on V2 whosenegative pedal is EB, then by N ’s construction B1V1 ⊥ MV1. Since thisperpendicular must be unique, P0, B1, and V1 are collinear as will be P0, A1,and V2. �

Proposition 5. The line joining the intersection points of EA and EB is theperpendicular bisector of segment [fAfB] and also passes through P0.

Proof. Let U1, U2 denote the points where EA and EB intersect. In orderto prove that P0, U1, and U2 are collinear, we show each one lies on theperpendicular bisector of [fAfB]. Since U1 (resp. U2) is on EA (resp. EB),whose foci are M and fA (resp. M and fB), with major axis of length 2r,then

U1fA + U1M = 2r; U2fB + U1M = 2r.

This implies that U1fA = U1fB and U2fA = U2fB, hence both U1 and U2belong to the perpendicular bisector of [fAfB]. Since we’ve already shownthat P0V1 ⊥ MV1, and since V1 is the center of MfA, this means that P0V1is the perpendicular bisector of [MfA] and this implies that P0fA = P0M .Similarly, P0fB = P0M , and hence P0fA = P0fB. Therefore P0 is also onthe perpendicular bisector of [fAfB], ending the proof. �

3.2. Triangles and Homotheties. Referring to Figure 7:

Proposition 6. The two sides of triangle △fAP0fB, incident on P0, containpoints A2 and B2. The other side contains points A1 and B1.

Proof. The construction of the negative pedal curve of arc V2V3 impliesA1V2 ⊥ MV2. Since V2 is the center of the EA, A1V2 is the perpendicular bi-sector of [MfB ] hence A1fB = A1M . Since A1 lies on EA, MA1+fAA1 = 2r,


Figure 7. The negative pedal curve N w.r. to pedal point M consists of two arcs of ellipsesEA and EB (green and blue), centered on Reuleaux vertices V1, V2, respectively. They havea common focus at M , and the other foci are fA, fB . Their major axes have length of 2r,

equal to the diameters of the three Reuleaux circles (dashed). Points P0, A1, V2 P0, B1, V1are collinear and along their minor axis. The lines P0A1 and P0B1 are tangent to EA and EB ,respectively. A2B2 is tangent to both ellipses and A2, B2, V 3 are collinear. The circle (black)passing through M , fA and fB EA, EB (green and blue) is centered on P0 (antipodal of Mw.r. to V3). The distance between the foci fA and fB is constant. Triangle T = △fAfBP0is equilateral and its sides pass through (i) A2, (ii) B2, (iii) A1, B1, respectively. Bothintersections U1, U2 of EA with EB lie on the perpendicular bisector of fAfB , hence arecollinear with P0. T and △V1V2V3 are homothetic (homothety center M and homothetyratio 2)).

hence fBA1 + fAA1 = fAfB. Therefore, triangle inequality implies fB, A1,and fA must be collinear. A similar proof applies to B1. In order to provethat P0, B2, and fA are collinear, we simply show that P0fA = P0A2+A2fA.As noted above, A2V3 is perpendicular to P0M and V3 is its midpoint. HenceA2V3 is the perpendicular bisector of [P0M ]; so P0A2 = MA2. Since A2 lieson EA, we have:

P0A2 +A2f′A = MA2 +A2f

′A = 2r

The proof for B2 is similar. �

Proposition 7. Triangles △fAfBP0 and △V1V2V3 are homothetic at ratio2, and with M the homothety center. Hence, △fAfBP0 is equilateral and the


distance between fA and fB is the same as 2r. Furthermore, their barycentersX2 and X

′2 are collinear with M .

Proof. Points V1, V2, V3 are the midpoints of MfA, MfB, and P0M , respec-tively. Thus, V1V2 is a mid-base of △fAMfB, V2V3 is a mid-base of △fBP0fAand V3V1 is a mid-base of △P0MfA. Hence △fAfBP0 and △V1V2V3 are ho-mothetic with ratio 2, and homothety center M . Therefore △fAfBP0 isequilateral and the distance between fA and fB is the same as the diameter2r of the circles that form the Reuleaux triangle.

Thus, △fAfBP0 is equilateral with sides twice that of the original triangle:fAfB = 2V1V2. This shows that the distance between the pair of foci of EAand EB is constant and equal to the length of their major axes. Note thatlines V1fA, V2fB , P0V3 intersect at M , hence the two triangles are perspec-tive at M . Due to the parallelism of their sides, their medians will be respec-tively parallel; let X2 and X

′2 denote the barycenters of triangles △V1V2V3

and △fAfBP0, respectively. The barycenter divides the medians in equalproportions, which guarantees △MX ′2V2 ∼ △MX2fB. Since M,V2, fB arecollinear, so are M,X ′2,X2. �

4. Conclusion

We studied some proprieties of the negative pedal curve to an object with aremarkable symmetry: the equilateral Reuleaux triangle. The five endpointsof this curve determine a conic with one focus on pedal point M , for anychoice of M on the third arc of the Reuleaux.

To prove that we adopted an inversive approach, based on the fact thatthe reciprocal of a conic w.r. to a circle is a circle, iff the center of inversionis the focus of said conic.

Since points on the original curve convert to lines tangent to the reciprocalcurve, it sufficed to show that the five lines tangent to the inverted sides ofthe Reuleaux, at their endpoints, are tangent to some circle. Our proof relieson Poncelet’s porism.

One may also consider an asymmetric Reuleaux triangle delimited by threecircles whose radii r1, r2, r3 are distinct, and whose centers Oi are not nec-essarily vertices of an equilateral triangle. Preliminary experiments showthat some properties of the equilateral Reuleaux still in the asymmetriccase. Using the notation in Figure 7, one observes that for any choice ofO1, O2, O3, r1, r2, r3:

(i) the distance between foci |fAfB| does not depend on the location ofM .

(ii) V2, A1, P0 as well as V1, B1, P0 are collinear.(iii) the line through the two intersections U1, U2 of EA with EB is per-

pendicular to fAfB.(iv) line A2B2 is tangent to both EA and EB .

Nevertheless, in this general setting, the focus of the endpoint conic nolonger coincides with the pedal point M . Below, some open questions wecould not yet answer synthetically:

• What is the location of the focus of the endpoint conic if the Reuleauxtriangle is asymmetric? Is it still geometrically meaningful?


• When does the focus of the endpoint conic of an arbitrary Reuleauxtriangle coincides with the pedal point?

• Are there any special locations of M on arc V1V2 for which it stillthe focus of the endpoint conic?

• Is there a poristic family of Reuleaux triangles whose endpoint conichas a focus on M (i.e. for any choice of the pedal point on the thirdarc of the Reuleaux)?

• What are the bounds on the eccentricity of the endpoint conic asso-ciated with some Reuleaux triangle?

Appendix A. Duality and the Negative Pedal Curve

Here we review concepts and results on polar duals [2].Let a circle I be called the inversion circle and its center M the inversion

point. Assume all inversions, the poles and polars below are performed w.r.to I .

The following result provides two equivalent definitions for the dual curve:

Theorem 2. Let Γ be a regular curve, Γ∗1 the locus of the poles of its tangents,Γ∗2 the envelope of the polars of its points. Then Γ

∗1 = Γ

∗2 and simply denote

it Γ∗.Γ∗ is a regular curve and the polars of its points are the tangents to Γ,

while the poles of the tangents of Γ∗ are points of Γ.Further more [Γ∗]∗ = Γ.

This theorem, whose proof is based on the fundamental pole-polar theoremjustifies the dual definition of the curve Γ∗ either as a locus of points or asan envelope of lines, and specifies who the points and the tangents at a dualcurve are.

For more details on poles, polars and polar reciprocity, see e.g. [2].Referring to Figure 8:

Proposition 8. The dual (or the polar dual, or the reciprocal) of a circle Γw.r. to an inversion circle centered at M is a conic Γ∗ whose:

• focus coincides with the inversion center;• vertices are the inverses of the endpoints of the diameter of Γ passing

through the inversion center;• directrix is the polar of the center of Γ.

The dual conic Γ∗ is and ellipse (resp. hyperbola) if the center of I isinside (resp. outside) Γ and a parabola if said center is on said curve.

Proposition 9. The dual of a conic Γ is a circle iff the inversion center isa focus of Γ.

If this is the case, (i) the inverses of the vertices of Γ are a pair of antipodalpoints on the dual circle Γ∗; (ii) the center of Γ∗ is the pole of the directrixof Γ.

These remarkable results are classic; see [1] or [2, Art.309], for a proof andmore details.

There is a natural intertwining between the negative pedal curve, inversion,and polar reciprocity.


Figure 8. The dual of Γ (violet) w.r. to its inversion circle (dashed black) is the curve Γ∗

(dark green), the envelope of polars E′D′ of points D on Γ, as well as loci of E′, the poles ofthe tangents ED to Γ, as D sweeps Γ. The dual of a circle is a conic. Γ∗ is an ellipse iff Mis inside γ, a hyperbola iff M is outside γ, and a parabola, iff M is on γ. Its focus is M , theinversion center (not illustrated), and the directrix is the polar of its center. Its vertices arethe inverses of A′

1and A′

2, the intersection of MO with Γ.

Proposition 10. The negative pedal curve of Γ w.r. to a pedal point isthe reciprocal of its inverse, Γ′ w.r. to a circle centered on that pedal point:N (Γ) = Γ′∗. Therefore (i) the negative pedal curve is the locus of the polesof the tangents to its inverted curve; (ii) the polars of the points of a negativepedal curve N (Γ) are the tangents to its inverted curve Γ′.

Proof. First we prove that the dual of Γ′ is contained in the negative pedal ofΓ. Let S be a point in Γ′; then S is an inverse of some point L in Γ; S = L′;hence, the polar of S is the perpendicular in point S′ to the line joining Mand S′; since S′ = (L′)′ = L, the polar of S = L′ is the perpendicular inL to line ML. Since inversion is bijective (in fact, it is an involution), if Ssweeps Γ′, L sweeps Γ, and lines ML are the set of all tangent lines to thenegative pedal curve of Γ.

The reverse inclusion is similar, if we refer to negative pedal curves as anenvelope of lines. �

Thus, the negative pedal curve, initially defined as an envelope of lines,can also be constructed as a “point curve”, i.e. as the locus of the poles ofthe tangents to its inverse Γ′.

In order to construct the negative pedal of a circle w.r. to a pedal pointthat does not lie on Γ, we first draw its inverse, Γ′ then obtain the latter’sdual. Note that the inverse of a circle can be a circle or a line. The lattercase occurs if the center of inversion is on the inverted circle.


Figure 9. The negative pedal curve N (Γ) (green) of a circle Γ (orange) w.r. to I (dashedblack) is the envelope of lines DE′ (green) where D is a generic point on Γ and DE′ ⊥ MD.Since DE′ is (also) the polar of D′ (green), the inverse of D, on circle γ (dashed violet), thenN (Γ) is the envelope of the polars of its inverse circle γ. Therefore, N (Γ) is the dual of itsinverse circle γ, hence a conic with a focus at M . Its vertices coincide with points A1, A2,the diameter of Γ through M . Here, N (Γ) is an ellipse since M is inside Γ.

Below we describe the negative pedal curve of a circle. Refer to figure 10.

Proposition 11. The negative pedal N (Γ) of a circle Γ, w.r. to a pedalpoint M not located on the boundary of Γ, is a conic, whose (i) focus is M ;center is the center of circle Γ; (ii) vertices are the intersection points of theline that joins the pedal point M and the center of Γ, with the circle Γ; (iii)focal axis is the diameter of Γ.

N (Γ) will be an ellipse (resp. hyperbola), if the pedal point is interior(resp. exterior) to Γ.

The negative pedal curve of a circle, w.r. to a point on the circumference,reduces to a point, the antipode of M .

Proof. First assume that M is not on the circle. Draw the negative pedalcurve of the circle as follows:

(i) let Γ′, be the inverse of Γ; then Γ′ is a circle whose diameter is[A′1A

′2] where A

′1 and A

′2 are the inverses of vertices A1 and A2 of

Γ.(ii) perform the dual of Γ′ to obtain a conic whose focus is M (the

inversion center) and whose vertices are the inverses of A′1 and A′2,

respectively, i.e., A1 and A2.

Then N (Γ) =[

Γ′]∗

. By the above, the conic will be an ellipse (resp.hyperbola), if M is inside (resp. outside) Γ.


Figure 10. N (Γ) (green hyperbola), the negative pedal curve of a circle Γ (orange), beingthe dual circle γ (violet), which is the inverse of Γ, is also the locus of the poles E of tangentsin D′ to γ as D sweeps Γ. The line DE is the tangent to N (Γ) passing through D. N (Γ) isan ellipse (see Fig 9) iff M ∈ [A1A2]; N (Γ) is a hyperbola iff M is not on segment [A1A2].The negative pedal curve of a circle is never a parabola.

If the pedal point M is on the circle, then the inverse is a line, whosereciprocal is a point, its pole. �

References

[1] Akopyan, A., Zaslavsky, A. (2007). Geometry of conics, vol. 26 of Mathematical World.American Mathematical Society, Providence, RI. doi.org/10.1090/mawrld/026.Translated from the 2007 Russian original by Alex Martsinkovsky. 14

[2] Salmon, G. (1869). A treatise on conic sections. London: Longmans, Green, Readerand Dyer. 3, 14

[3] Weisstein, E. (2019). Mathworld. MathWorld–A Wolfram Web Resource.mathworld.wolfram.com. 2

[4] Yaglom, I. M., Boltyanskii, V. G. (1961). Convex Figures. New York: Holt, Rinehart,& Winston. 1

Liliana Gabriela Gheorghe, Universidade Federal de Pernambuco, Dept.

de Matemática, Recife, PE, Brazil

Email address: [email protected]

Dan Reznik, Data Science Consulting, Rio de Janeiro, RJ, Brazil

Email address: [email protected]

doi.org/10.1090/mawrld/026mathworld.wolfram.com

1. Introduction2. Main Result: The Endpoint Conic3. Some Elementary Properties3.1. Collinearity and Tangencies3.2. Triangles and Homotheties

4. ConclusionAppendix A. Duality and the Negative Pedal CurveReferences

arxiv:2008.08950v1 [math.ds] 20 aug 2020 · the reuleaux triangle r,is the (convex) curve formed by...

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