A Special Conic Associated with the Reuleaux Negative Pedal Curve
aa r X i v : . [ m a t h . D S ] O c t A SPECIAL CONIC ASSOCIATED WITH THEREULEAUX NEGATIVE PEDAL CURVE
LILIANA GABRIELA GHEORGHE AND DAN REZNIK
The Negative Pedal Curve of the Reuleaux Triangle w.r. to a pedal point M located on its boundary consists of two elliptic arcs and a point P . In-triguingly, the conic passing through the four arc endpoints and by P 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 V i , i = 1 , , . 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 V , V , V 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 P M [3, Negative
Date : August, 2020.
Keywords and phrases: conic, inversion, pole, polar, dual curve, negative pedalcurve. (2020)Mathematics Subject Classification:
Pedal Curve]. Trivially, the negative pedal curve of arc V V is a point whichwe call P . We show that the negative pedal curves of the other two sidesare elliptic arcs with a common focus on M and whose major axis measures r ; see Prop. 1.Let V be the center of the circular arc where pedal point M lies, andlet V , V be the endpoints of said arc. Let arc A A (resp. B B ) be thenegative pedal image of the Reuleaux side V V (resp. V V ) where point A is the image of V , and B the image of V . The endpoints of N whosepreimage is V are respectively A when V is regarded as a point of side V V , and B when V is regarded as a point of side V V 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 endpoints A , A , B , B of the negative pedal curve N , and through P : 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. SPECIAL CONIC ASSOCIATED W/ THE REULEAUX TRIANGLE 3
Figure 2.
The sides of the Reuleaux R are three circular arcs centered at the vertices V , V , V of an equilateral triangle. Its negative pedal curve N w.r. to a pedal point M onits boundary consists of a point P (the antipode of M through V ) and of two elliptic arcs A A and B B (green and blue). The endpoint conic C ∗ (purple) passes through P 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. Main Result: The Endpoint Conic
Referring to Figure 2.
Theorem 1.
The conic C ∗ which contains P and the endpoints A , A , B , B 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 △ V V V . 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 A , A , B , B and P 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]. L. G. GHEORGHE AND D. REZNIK
Figure 3.
The Reuleaux triangle R and its inverse: (i) arcs V ′ V ′ (blue) and V ′ V ′ (green)are the inverses of sides V V and V V of R , while line V ′ V ′ except for segment [ V ′ V ′ ] itself(violet) is the image of arc V V ; (ii) the polars of A and B are the tangents in V ′ and V ′ to arcs V ′ V ′ and V ′ V ′ ; (iii) the polars of A and B are the tangents in V ′ to arcs V ′ V ′ and V ′ V ′ ; (iv) line V ′ V ′ is the polar of P . (v) all five polars above are tangent to the exinscribedcircle c of △ V ′ O ′ V ′ (purple). The angle between circles c (green) and C (blue) in V ′ is ◦ iff V ′ is on the exinscribed c centered in O (not shown); in this case, the tangents at V ′ (dashed green and blue) to circles c and C , are also tangent to the exinscribed circle c . This indirect and inversive approach is appropriate as the polars of A , A , B , B , and P 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, is M ) 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. SPECIAL CONIC ASSOCIATED W/ THE REULEAUX TRIANGLE 5
Lemma 1.
Let M be a point on the side V V of the Reuleaux triangle and I the inversion circle. Let V ′ , V ′ , V ′ be the inverses of points V , V , V andlet arcs V ′ V ′ and V ′ V ′ be, respectively, the inverses of arcs V V and V V of the Reuleaux triangle. Then: (i) The polars of A and B are the tangents at V ′ and V ′ to circulararcs V ′ V ′ and V ′ V ′ . The poles of the tangents at V ′ and V ′ to arcs V ′ V ′ and V ′ V ′ , are the points A , B . (ii) The polars of points A , B are the tangents in V ′ to arcs V ′ V ′ and V ′ V ′ , respectively. The poles of the tangents at V ′ to arcs V ′ V ′ and V ′ V ′ , are the points A , B . (iii) The inverse of arc V V is the line V ′ V ′ excluding segment [ V ′ V ′ ] and the polar of P is the line V ′ V ′ . Referring to Figure 3:
Proof.
Inversion w.r. to a circle maps circles to either circles or lines: thus,the inverses of arcs V V and V V are the two circular arcs V ′ V ′ and V ′ V ′ ,respectively. On the other hand, since arc V V 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. (cid:3) Lemma 2.
Using the notation in Lemma 1: (i) the angles at V ′ , V ′ , and V ′ between arcs V ′ V ′ , V ′ V ′ , and V ′ V ′ respectively (the inverses of the sides of the Reuleaux triangle), are ◦ . (ii) △ V ′ O ′ V ′ determined by the tangents at V ′ , V ′ to said arcs and theline V ′ V ′ is equilateral.Proof. These statements derive from the fact that inversion is preserves an-gles between curves. (cid:3)
Given the above results, Theorem 1 is equivalent to the following Lemma;see Figure 3:
Lemma 3.
The five polars of endpoints A , A , B , B , and P are tangentto circle c , the exinscribed circle in △ V ′ O ′ V ′ , which is externally-tangent toside V ′ V ′ . 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 ′ toarcs V ′ V ′ and V ′ V ′ respectively are also tangent to the excircle of △ V ′ O ′ V ′ .Referring to Figure 3, this can be restated as follows: Lemma 4.
Let △ V ′ O ′ V ′ be equilateral and let O be the center of its exin-scribed circle c , externally-tangent to side [ V ′ V ′ ] . Let C be another circle,concentric with c , that passes through V ′ (and V ′ ). Finally, let c and C be the two circles tangent to the sides [ OV ′ ] and [ OV ′ ] of the triangle, at V ′ and V ′ , respectively. Then: (i) c and C intersect at an angle of ◦ iff the three circles c , C ,and C pass through one common point; let V ′ be that point. L. G. GHEORGHE AND D. REZNIK (ii) if the condition above is fulfilled, then the two tangents at the com-mon point V ′ to circles c and C are also tangent to the exinscribedcircle c . While assertion (i) is automatic, once we identify circles c and C 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 [ A A . . .A ] be a regular hexagon with inscribed circle c and circumcircle C . Let P be a point on arc A A of C and let P P , P P , . . . , P P be the tangents from P , P , . . . , P to c .Let c be the circle tangent to side [ A A ] of the hexagon at A whosecenter is inside the hexagon; let P be its second intersection point with C .Then: (i) Points P and P coincide and the hexagon [ P P . . . P ] is regularand congruent with [ A A . . . A ] . Both hexagons share the sameincircle and circumcircle. (ii) Let P P be the tangent from P to c ; then it tangents (in P ) thecircle c , as well. Referring to Figure 4:
Proof. i) When we perform the construction of tangent lines P P , . . ., P P ,the process ends in six steps and P = P , thanks to Poncelet’s porism, since c 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 T be the intersection of the perpendicular bisector of segment [ A P ] with line A A . We shall prove that T P is tangent at P to circle c and that T , P , P are collinear.Let c be the center of circle c ; since T is a point on the perpendicularbisector of [ A P ] , and since A and P are the two intersections of circles c and c , then O, c and T are collinear.Next, △ T A c = △ T P c , as they have respectively-congruent sides,hence: ∠ T P c = ∠ T A c = 90 ◦ which proves that line T P is tangent at P to circle c .Furthermore, △ T A O = △ T P O as they have respectively-congruentsides, hence: ∠ T P O = ∠ T A O By hypothesis, T , A , and A are collinear, and △ A A O is equilateral,hence the external angle ∠ T A O = 120 ◦ ; hence ∠ T P O = 120 ◦ as well.Since by (i) △ P OP is equilateral, then ∠ OP P = 60 ◦ and ∠ T P P =180 ◦ , proving that points T , P , P are collinear. (cid:3) SPECIAL CONIC ASSOCIATED W/ THE REULEAUX TRIANGLE 7
Figure 4.
The angle between circles c (green circle) and C (blue circle) in P is ◦ iff P is on the circumcircle of the regular hexagon [ A A . . . A ] (orange circle). We now reformulate Lemma 4 as follows:
Lemma 6.
Let [ A A . . . A ] be a regular hexagon whose incircle is c andcircumcircle is C . Let c be the circle tangent to side [ A A ] of the hexagonat A and let C be the circle tangent to side [ A A ] at A . Then the anglebetween circles c and C is ◦ iff the three circles c , C , and C haveone common point.Proof. ⇐ First assume circles c and C intersect at a point P on circum-circle C . Referring to Figure 4, if P is on arc A A of C then by Lemma (ii) P P is a common tangent to circles c and c . In particular, P P is thetangent at P to circle c .Next, let P P ′ be the tangent from P to the incircle c (distinct from P P ).Similarly, let P ′ P ′ , P ′ P ′ , . . .P ′ P ′ be the tangents from P ′ , P ′ , . . . , P ′ ∈ C to the incircle c . L. G. GHEORGHE AND D. REZNIK
Then, as above, points P ′ and P coincide, and hexagon [ P ′ P ′ . . . P ′ ] is regular; since hexagons [ P P . . . P ] and [ P ′ P ′ . . . P ′ ] have one commonpoint ( P ), are regular, and are both inscribed in C , they must coincide.Once again, Lemma (ii) guarantees that P P is a common tangent tocircles C and c . In particular, line P P is the tangent at P to circle C .Since hexagon [ P P . . . P ] is regular, ∠ P P P = 120 ◦ . This guaranteesthat the angle between circles c and C , which is the angle between theirtangents at P , is also ◦ . ⇒ By hypothesis, circles c and C intersect at an angle of ◦ . Weshall prove that, necessarily, point P must be on the circumcircle C .Call P ′ the intersection point between circle c and arc A A of circle C .We shall prove that P and P ′ coincide.Let C ′ be the circle tangent at A to line A A that passes through P ′ .Then, by the first part of this proof, circles c and C ′ intersect at an angleof ◦ . So circles C and C ′ would be two circles, both tangent at A toline A A , which intersect circle c at the same angle. Hence circles C and C ′ must coincide, as do points P and P ′ . (cid:3) 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 △ V V V . 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 ′ , V ′ and V ′ are, respectively, the inverses of V , V , V 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 ′ V ′ V ′ ,and the circumcenter of △ V V V . 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 M O will either be parallel or coincide. Since M is the focus of C ∗ , is major axis is line M O and point G is on that line. (cid:3) SPECIAL CONIC ASSOCIATED W/ THE REULEAUX TRIANGLE 9
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, L and L are theinverses of antipodal points N and N , the diameter of c passing through M (point N , theantipode of N 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 V V : 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 ′ V ′ and let D , D be the twointersections between circles c and C ′ . One can check that V ′ is the reflectionof M w.r. to V ′ V ′ ; therefore, M is located both on the reflection of thecircumcircle of △ V ′ V ′ V ′ w.r. to V ′ V ′ , and on arc V V of the Reuleauxtriangle. The location of M with respect to C ′ reveals the which type of conic C ∗ is: it is an ellipses if M is on the arc D D of circle C ′ , a hyperbola, if M ∈ V ′ D or V ′ D and a parabola when M is either D or D . Figure 6.
The endpoint conic C ∗ (dark green) is a parabola iff point M coincides witheither D or D ; its directrix (dark green) is the polar of the circumcenter O of △ V ′ V ′ V ′ (not shown). Some Elementary Properties
Collinearity and Tangencies.
Referring to Figure 7:
Proposition 1.
The negative pedal curve N of the Reuleaux triangle consistsof two elliptic arcs E A and E B and a point P , the antipode of M w.r. to thecenter of the circle where M is located. The two ellipses E A , E B are centeredon the vertices of the Reuleaux triangle, V and V , have one common focusat M , and their semi-axes are of length r .Proof. By hypothesis, M belongs to arc V V of the circle centered in V thatpasses through V and V . Hence, if P is any point on this arc and we drawthe perpendicular p through P on P M , all these lines will pass through afixed point P , which is he antipode of M w.r. to center V .The second part derives directly from the general construction of the neg-ative pedal curve of a circle. See Proposition 10 in the Appendix. (cid:3) SPECIAL CONIC ASSOCIATED W/ THE REULEAUX TRIANGLE 11
Proposition 2.
The minor axes of E A and E B pass through P .Proof. By the definition of the negative pedal curve, if we regard V as apoint on arc V V of the circle centered on V on which the pedal point M lies, then P V will be perpendicular to M V . Since V is the center of E A andsince line M V is its major axis, its minor axis will be along P V . Similarly,the minor axis of E B is P V . (cid:3) Proposition 3.
Points A , B and V are collinear and line A B is acommon tangent to E A and E B .Proof. By construction, the negative pedal curve of arc V V is the ellipticarc E A , delimited by A and A . This implies that M V and A V areperpendicular, as well as M V and B V . Thus points A , V and B arecollinear. Also by construction, the perpendicular to M V at V is tangent to N at A (resp. B ) when V is regarded as a point in the V V (resp. V V )arc. Hence the points A , V and B are collinear ( ∠ A V B = 180 ◦ ) and A B is the common tangent to E A and E B , at A and B , respectively. (cid:3) Proposition 4.
Point A is on P V and B is on P V .Proof. If we regard V as a point on arc V V of the circle centered on V whose negative pedal curve is P , then, necessarily, V P ⊥ M V . Similarly,if we regard V as a point on arc V V of the circle centered on V whosenegative pedal is E B , then by N ’s construction B V ⊥ M V . Since thisperpendicular must be unique, P , B , and V are collinear as will be P , A ,and V . (cid:3) Proposition 5.
The line joining the intersection points of E A and E B is theperpendicular bisector of segment [ f A f B ] and also passes through P .Proof. Let U , U denote the points where E A and E B intersect. In orderto prove that P , U , and U are collinear, we show each one lies on theperpendicular bisector of [ f A f B ] . Since U (resp. U ) is on E A (resp. E B ),whose foci are M and f A (resp. M and f B ), with major axis of length r ,then U f A + U M = 2 r ; U f B + U M = 2 r. This implies that U f A = U f B and U f A = U f B , hence both U and U belong to the perpendicular bisector of [ f A f B ] . Since we’ve already shownthat P V ⊥ M V , and since V is the center of M f A , this means that P V is the perpendicular bisector of [ M f A ] and this implies that P f A = P M .Similarly, P f B = P M , and hence P f A = P f B . Therefore P is also onthe perpendicular bisector of [ f A f B ] , ending the proof. (cid:3) Triangles and Homotheties.
Referring to Figure 7:
Proposition 6.
The two sides of triangle △ f A P f B , incident on P , containpoints A and B . The other side contains points A and B .Proof. The construction of the negative pedal curve of arc V V implies A V ⊥ M V . Since V is the center of the E A , A V is the perpendicular bi-sector of [ M f B ] hence A f B = A M . Since A lies on E A , M A + f A A = 2 r , Figure 7.
The negative pedal curve N w.r. to pedal point M consists of two arcs of ellipses E A and E B (green and blue), centered on Reuleaux vertices V , V , respectively. They havea common focus at M , and the other foci are f A , f B . Their major axes have length of r ,equal to the diameters of the three Reuleaux circles (dashed). Points P , A , V P , B , V are collinear and along their minor axis. The lines P A and P B are tangent to E A and E B ,respectively. A B is tangent to both ellipses and A , B , V are collinear. The circle (black)passing through M , f A and f B E A , E B (green and blue) is centered on P (antipodal of M w.r. to V ). The distance between the foci f A and f B is constant. Triangle T = △ f A f B P is equilateral and its sides pass through (i) A , (ii) B , (iii) A , B , respectively. Bothintersections U , U of E A with E B lie on the perpendicular bisector of f A f B , hence arecollinear with P . T and △ V V V are homothetic (homothety center M and homothetyratio ). hence f B A + f A A = f A f B . Therefore, triangle inequality implies f B , A ,and f A must be collinear. A similar proof applies to B . In order to provethat P , B , and f A are collinear, we simply show that P f A = P A + A f A .As noted above, A V is perpendicular to P M and V is its midpoint. Hence A V is the perpendicular bisector of [ P M ] ; so P A = M A . Since A lieson E A , we have: P A + A f ′ A = M A + A f ′ A = 2 r The proof for B is similar. (cid:3) Proposition 7.
Triangles △ f A f B P and △ V V V are homothetic at ratio2, and with M the homothety center. Hence, △ f A f B P is equilateral and the SPECIAL CONIC ASSOCIATED W/ THE REULEAUX TRIANGLE 13 distance between f A and f B is the same as r . Furthermore, their barycenters X and X ′ are collinear with M .Proof. Points V , V , V are the midpoints of M f A , M f B , and P M , respec-tively. Thus, V V is a mid-base of △ f A M f B , V V is a mid-base of △ f B P f A and V V is a mid-base of △ P M f A . Hence △ f A f B P and △ V V V are ho-mothetic with ratio 2, and homothety center M . Therefore △ f A f B P isequilateral and the distance between f A and f B is the same as the diameter r of the circles that form the Reuleaux triangle.Thus, △ f A f B P is equilateral with sides twice that of the original triangle: f A f B = 2 V V . This shows that the distance between the pair of foci of E A and E B is constant and equal to the length of their major axes. Note thatlines V f A , V f B , P V 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 X and X ′ denote the barycenters of triangles △ V V V and △ f A f B P , respectively. The barycenter divides the medians in equalproportions, which guarantees △ M X ′ V ∼ △ M X f B . Since M, V , f B arecollinear, so are M, X ′ , X . (cid:3) 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 r , r , r are distinct, and whose centers O i 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 of O , O , O , r , r , r :(i) the distance between foci | f A f B | does not depend on the location of M .(ii) V , A , P as well as V , B , P are collinear.(iii) the line through the two intersections U , U of E A with E B is per-pendicular to f A f B .(iv) line A B is tangent to both E A and E B .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 V V 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 inversionpoint. 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 ∗ the locus of the poles of its tangents, Γ ∗ the envelope of the polars of its points. Then Γ ∗ = Γ ∗ and simply denoteit 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 Γ w.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 passingthrough 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.
SPECIAL CONIC ASSOCIATED W/ THE REULEAUX TRIANGLE 15
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 M is 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 ′ and A ′ , 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 of Γ . 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 M and S ′ ; since S ′ = ( L ′ ) ′ = L, the polar of S = L ′ is the perpendicular in L to line M L . Since inversion is bijective (in fact, it is an involution), if S sweeps Γ ′ , L sweeps Γ , and lines M L 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. (cid:3) 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), then N (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 A , A ,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 ′ A ′ ] where A ′ and A ′ are the inverses of vertices A and A of Γ .(ii) perform the dual of Γ ′ to obtain a conic whose focus is M (theinversion center) and whose vertices are the inverses of A ′ and A ′ ,respectively, i.e., A and A .Then N (Γ) = (cid:2) Γ ′ (cid:3) ∗ . By the above, the conic will be an ellipse (resp.hyperbola), if M is inside (resp. outside) Γ . SPECIAL CONIC ASSOCIATED W/ THE REULEAUX TRIANGLE 17
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 ∈ [ A A ] ; N (Γ) is a hyperbola iff M is not on segment [ A A ] .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. (cid:3) 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
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