The Circumbilliard: Any Triangle can be a 3-Periodic
aa r X i v : . [ m a t h . D S ] A p r THE CIRCUMBILLIARD: ANY TRIANGLECAN BE A 3-PERIODIC
DAN REZNIK AND RONALDO GARCIA
Abstract.
A Circumconic passes through a triangle’s vertices. Wedefine the Circumbilliard, a circumellipse to a generic triangle for whichthe latter is a 3-periodic. We study its properties and associated loci.
Keywords elliptic billiard, periodic trajectories, triangle center, cir-cumconic, circumellipse, circumhyperbola, conservation, invariance, in-variant.
MSC Introduction
Given a triangle, a circumconic passes through its three vertices and sat-ifies two additional constraints, e.g., center or perspector . We study prop-erties and invariants of such conics derived from a 1d family of triangles:3-periodics in an Elliptic Billiard (EB): these are triangles whose bisectorscoincide with normals to the boundary (bounces are elastic), see Figure 1.Amongst all planar curves, the EB is uniquely integrable [8]. It can beregarded as a special case of Poncelet’s Porism [3]. These two propeties implytwo classic invariances: N -periodics have constant perimeter and envelopa confocal Caustic. The seminal work is [17] and more recent treatmentsinclude [11, 16]. Where reference and polar triangles are perspective [19].
Figure 1.
Video : [13, PL
We have shown 3-periodics also conserve the Inradius-to-Circumradius ratio which implies an invariant sum of cosines, and that their Mittenpunkt is stationary at the EB center [14]. Indeed many such invariants have beeneffectively generalized for N > [1, 2].We have also studied the loci of 3-periodic Triangle Centers over the family:out of the first 100 listed in [9], 29 sweep out ellipses (a remarkable fact onits own) with the remainder sweeping out higher-order curves [6]. Related isthe study of loci described by the Triangle Centers of the Poristic Trianglefamily [12]. Summary of the paper : given a generic triangle T we define its Cir-cumbilliard
CB: a Circumellipse to T for which the latter is a 3-periodic. Wethen analyze the dynamic of geometry of Circumbilliards for triangles de-rived from the 3-periodic family such as the Excentral, Anticomplementary,Medial, and Orthic, as well as the loci swept by their centers. Additionalresults include: • Proposition 5 in Section 3 describes regions of the EB which produceacute, right-triangle, and obtuse 3-periodics. • Theorem 1 in Section 3: The aspect ratio of Circumbilliards of thePoristic Triangle Family [4] is invariant. This is a family of trianglewith fixed Incircle and Circumcircle.A reference table with all Triangle Centers, Lines, and Symbols appearsin Appendix B. Videos of many of the experiments are assembled on Table 2in Section 4. 2.
The Circumbilliard
Let the boundary of the EB satisfy:(1) f ( x, y ) = (cid:16) xa (cid:17) + (cid:16) yb (cid:17) = 1 . Where a > b > denote the EB semi-axes, and c = a − b throughout thepaper. Below we use aspect ratio as the ratio of an ellipse’s semi-axes. Whenreferring to Triangle Centers we adopt Kimberling X i notation [9], e.g., X for the Incenter, X for the Barycenter, etc., see Table 3 in Appendix B.The following five-parameter equation is assumed for all circumconics notpassing through (0 , .(2) c x + c y + c xy + c x + c y = 0 Proposition 1.
Any triangle T = ( P , P , P ) is associated with a uniqueellipse E for which T is a billiard 3-periodic. The center of E is T’s Mit-tenpunkt.Proof. If T is a 3-periodic of E , by Poncelet’s Porism, T is but an elementof a 1d family of 3-periodics, all sharing the same confocal Caustic . Thisfamily will share a common Mittenpunkt X located at the center of E [14]. As does the Poristic Family [4]. Where lines drawn from each Excenter thru sides’ midpoints meet. This turns out to be the Mandart Inellipse I of the family [19]. Appendix A shows how to obtain the parameters for (2) such that it passesthrough P , P , P and is centered on X : this yields a × linear system.Solving it its is obtained a quadratic equation with positive discriminant,hence the conic is an ellipse. (cid:3) E is called the Circumbilliard (CB) of T . Figure 2 shows examples ofCBs for two sample triangles. Figure 2.
Two random triangles are shown as well as their Circumbilliards (CBs). Noticetheir axes in general are not horizontal/vertical. An algorithm for computing the CB is givenin Appendix A.
Video: [13, PL Circumbilliards of Derived Triangles
Figure 3 shows CBs for the Excentral, Anticomplementary (ACT), andMedial Triangles, derived from 3-periodics.3.1.
Excentral Triangle.
The locus of the Excenters is shown in Figure 3(left). It is an ellipse similar to the ◦ -rotated locus of X and its axes a e , b e are given by [5, 6]: a e = b + δa , b e = a + δb Where δ = √ a − a b + b . Proposition 2.
The locus of the Excenters the stationary MacBeath Circ-umellipse E ′ [19] of the Excentral Triangles.Proof. The center of E ′ is the Symmedian Point X [19, MacBeath Circum-conic]. The Excentral Triangle’s X coincides with the Mittenpunkt X ofthe reference [9]. Since over the 3-periodics the vertices of the Excentral lieon an ellipse and its center is stationary, the result follows. (cid:3) Proposition 3.
The Excentral CB is centered on X , whose trilinears areirrational, and whose locus is non-elliptic.Proof. X is the Mittenpunkt of the Excentral Triangle [9] and its trilinearsare irrational on the sidelengths. To determine if its locus is an ellipse weuse the algebro-numeric techniques described in [6]. Namely, a least-squaresfit of a zero-centered, axis-aligned ellipse to a sample of X positions of the3-periodic family produces finite error, therefore it cannot be an ellipse. (cid:3) No Triangle Center whose trilinears are irrational on sidelengths has yet been foundwhose locus under the 3-periodic family is an ellipse [6].
DAN REZNIK AND RONALDO GARCIA
Figure 3.
Draw in black in each picture is an a/b = ϕ ≃ . EB and a 3-periodic at t = 7 . ◦ . Left : the CB of the Excentral Triangle (solid green) centered on the latter’s Mittenpunkt is X [9]. Its locus (red) is non-elliptic. Also shown (dashed green) is the elliptic locus of theExcenters (the MacBeath Circumellipse E ′ of the Excentrals [19]), whose center is X [6]; Top Right : the CB of the Anticomplementary Triangle (ACT) (blue), axis-aligned with theEB. Its center is the Gergonne Point X , whose locus (red) is elliptic and similar to the EB[6]. The locus of the ACT vertices is not elliptic (dashed blue); Bottom Right : the CB ofthe Medial Triangle (teal), also axis-aligned with the EB, is centered on X , whose locus(red) is also elliptic and similar to the EB, since it is the midpoint of X X [9]. The locus ofthe medial vertices is a dumb-bell shaped curve (dashed teal). Video: [13, PL
This had been observed in [6] for several irrational centers such as X i , i = X , whose locus is a convexquartic. Other examples include X j , j =
19, 22–27, etc.3.2.
Anticomplementary Triangle (ACT).
The ACT is shown in Fig-ure 3 (top right). The locus of its vertices is clearly not an ellipse.The ACT is perspective with the reference triangle (3-periodic) at X andall of its triangle centers correspond to the anticomplement of correspondingreference ones [19]. The center of the CB of the ACT is therefore X , theanticomplement of X . We have shown the locus of X to be an ellipsesimilar to the EB with axes [6]: ( a , b ) = k ( a, b ) , with: k = 2 δ − a − b c Remark 1.
The axes of the ACT CB are parallel to the EB and of fixedlength.
This stems from the fact the ACT is homothetic to the 3-periodic. Anticomplement: a 1:2 reflection about X . Medial Triangle.
The locus of its vertices is the dumbbell-shapedcurve, which at larger a/b is self-intersecting, and therefore clearly not anellipse, Figure 3 (bottom right).Like the ACT, the Medial is perspective with the reference triangle (3-periodic) at X . All of its triangle centers correspond to the complement of corresponding reference ones [19]. The center of the CB of the Medial istherefore X , the complement of X . This point is known to sit midwaybetween X and X . Remark 2.
The locus of X is an ellipse similar to the EB. This stems from the fact X is stationary and the locus of X is an ellipsesimilar to the EB (above). Therefore its axes will be given by: ( a , b ) = ( a , b ) / Stemming from homothety of 3-periodic and its Medial:
Remark 3.
The axes of the Medial CB are parallel to the EB and of fixedlength.
Superposition of ACT and Medial.Proposition 4.
The Intouchpoints of the ACT (resp. 3-periodic) are on theEB (resp. on the CB of the Medial)Proof.
The first part was proved in [15, Thm. 2]. Because the 3-periodiccan be regarded as the ACT of the Medial, the result follows. (cid:3)
This phenomenon is shown in Figure 4. Also shown is the fact that X i , i =
7, 142, 2, 9, 144 are all collinear and their intermediate intervalsare related as . In [10] this line is known as L ( X , X ) or L . X is the perspector of the ACT and its Intouch Triangle (not shown).3.5. Orthic Triangle.
Let α = p √ − ≃ . . In [15, Thm. 1] weshow that if a/b > α , the 3-periodic family will contain obtuse triangles. Proposition 5. If a/b > α , the 3-periodic is a right triangle when one ofits vertices is at four symmetric points P ⊥ i , i = 1 , , , given by ( ± x ⊥ , ± y ⊥ ) with: (3) x ⊥ = a √ a + 3 b − b δc , y ⊥ = b √− b − a + 4 a δc Proof.
Let the coordinates of the 3-periodic vertices be P = ( x , y ) , P =( x , y ) , P = ( x , y ) as derived in [5].Computing the equation h P − P , P − P i = 0 , after careful algebraicmanipulations, it follows that x satisfies the quartic equation c x − a c ( a + 3 b ) x + a ( a + 2 a b − b ) = 0 . For a/b > p √ − the only positive root in the interval (0 , a ) is given by Complement: a 2:1 reflection about X . DAN REZNIK AND RONALDO GARCIA
Figure 4.
Construction for both ACT and Medial CBs, centered on X and X , respec-tively. The Incircle of the ACT (resp. 3-periodic) is shown blue (resp. green). The formertouches the ACT at the EB and the latter touches the 3-periodic sides at the Medial CB.Also shown is line L (2 ,
7) = L which cointains X i , i = 7 , , , , . Their consecutivedistances are proportional to . X was included since it is the perspector of theACT and its Intouch Triangle (not shown) [19]. Video : [13, PL (4) x ⊥ = a √ a + 3 b − b δc . With y ⊥ obtainable from (1). (cid:3) Equivalently, a 3-periodic will be obtuse iff one of its vertices lies on topor bottom halves of the EB between the P ⊥ i , see Figure 5.Consider the elliptic arc along the EB between ( ± x ⊥ , y ⊥ ) . When a vertexof the 3-periodic lies within (resp. outside) this interval, the 3-periodic isobtuse (resp. acute). Proposition 6.
When a/b > α , the locus of the center of the Orthic CBhas four pieces: 2 for when the 3-periodic is acute (equal to the X locus),and 2 when it is obtuse (equal to the locus of X of T ′′ = P P X .Proof. It is well-known that [9] an acute triangle T has an Orthic whosevertices lie on the sidelines. Furthermore the Orthic’s Mittenpunkt coincideswith the Symmedian X of T . Also known is the fact that: Remark 4.
Let triangle T ′ = P P P be obtuse on P . Its Orthic has onevertex on P P and two others exterior to T ′ . Its Orthocenter X is also Figure 5.
Two 3-periodics are shown: one acute (solid blue) and one obtuse (dashed blue)inscribed into an a/b = 1 . EB. Red arcs along the top and bottom halves of the EBindicate that when a 3-periodic vertex is there, the 3-periodic is obtuse. These only existwhen a/b > α ≃ . . exterior. Furthermore, the Orthic’s Mittenpunkt is the Symmedian Point X of acute triangle T ′′ = P P X . To see this, notice the Orthic of T ′′ is also T ′ . T ′′ must be acute since itsOrthocenter is P . (cid:3) The CB of the orthic is shown in Figures 6 for four 3-periodic configura-tions in an EB whose a/b > α . Proposition 7.
The coordinates ( ± x ∗ , ± y ∗ ) where the locus of the center ofthe Orthic’s CB transitions from one curve to the other are given by: x ∗ = x ⊥ c (cid:0) a + 2 a b − b δ (3 a + b ) + b (cid:1) y ∗ = − y ⊥ c (cid:0) b + 2 a b − a δ (3 b + a ) δ + a (cid:1) Proof.
Let P = ( x , y ) be the right-triangle vertex of a 3-periodic, givenby ( x ⊥ , y ⊥ ) as in (4). Using [5], obtain P = ( p x /q , p y /q ) and P =( p x /q , p y /q ) , with: p x = b c x − a b x y + a c x y − a y p y =2 b x − b c x y + 2 a b x y − a c y q = b ( a + b ) x − a b c x y + a ( a + b ) y p x = b c x + 2 a b x y + a c x y + a y p y = − b x − b c x y − a b x y − a c y q = b ( a + b ) x + 2 a b c x y + a ( a + b ) y It can be shown the Symmedian point X of a right-triangle is the midpointof its right-angle vertex altitude. Computing X using this property leadsto the result. (cid:3) The anti-orthic pre-images of T ′ are both the 3-periodic and T ′′ . DAN REZNIK AND RONALDO GARCIA
Let α eq = p √ − ≃ . be the only positive root of x + 6 x − .It can be shown, see Figure 7: Proposition 8. At a/b = α eq , the locus of the Orthic CB is tangent toEB’s top and bottom vertices. If a 3-periodic vertex is there, the Orthic isequilateral.Proof. Let T be an equilateral with side s eq and center C . Let h be thedistance from any vertex of T to C . It can be easily shown that h/s eq = √ / .Let T ′ be the Excentral Triangle of T : its sides are s eq . Now considerthe upside down equilateral in Figure 7, which is the Orthic of an uprightisosceles 3-periodic. h is clearly the 3-periodic’s height and s eq is its base.The height and width of the upright isosceles are obtained from explicitexpressions for the vertices [5]: s eq = α α − p δ − α − , h = α + δ + 1 α + δ where α = a/b . Setting h/s eq = √ / and solving for α yields the requiredresult for α eq . (cid:3) Summary.
Table 1 summarizes the CBs discussed above, their centers,and their loci. Triangle Center Elliptic Locus3-Periodic X n/aExcentral X NoACT X YesMedial X YesOrthic X ∗ No Table 1.
CBs mentioned in this Section, their Centers and loci types.
Circumbilliard of the Poristic Family.
The Poristic Triangle Fam-ily is a set of triangles (blue) with fixed Incircle and Circumcircle [4]. Itis a cousing of the 3-periodic family in that by definition its Inradius-to-Circumradius r/R ratio is constant.Weaver [18] proved the Antiorthic Axis of this family is stationary. Odehnalshowed the locus of the Excenters is a circle centered on X and of radius R [12]. He also showed that over the family, the locus of the Mittenpunkt X is a circle whose radius is d (4 R + r ) and center is X + ( X − X )(2 R − r ) / (4 R + r ) , where d = | X X | = p R ( R − r ) [12, page 17].Let ρ = r/R and a , b be the semi-axis lengths of the Circumbilliard aporistic triangle. As shown in Figure 8: Theorem 1.
The ratio a /b is invariant over the family and is given by: The line passing through the intersections of reference and Excentral sidelines [19].
Figure 6.
Orthic CB for an EB with a/b = 1 . > α , i.e., containing obtuse 3-periodics,which occur when a 3-periodic vertex lies on the top or bottom areas of the EB between the P ⊥ . Top left : 3-periodic is sideways isosceles and acute (vertices outside P ⊥ , so 3 orthicvertices lie on sidelines. The Orthic CB centers is simply the mittenpunkt of the Orthic, i.e, X of the 3-periodic (blue curve: a convex quartic [6]). Top right : The position when avertex is at a P ⊥ and the 3-periodic is a right triangle: its Orthic and CB degenerate to asegment. Here the CB center is at a first (of four) transition points shown in the other insetsas Q i , i = 1 , , , . Bottom left : The 3-periodic is obtuse, the Orthic has two exteriorvertices, and the center of the CB switches to the Symmedian of T ′′ = P P X (red portionof locus). Bottom right: . The 3-periodic is an upright isosceles, still obtuse, the center ofthe Orthic CB reaches its highest point along its locus (red).
Video : [13, PL a b = s ρ + 2( ρ + 1) √ − ρ + 2 ρ ( ρ + 4) where ρ = r/R .Proof. The following expression for r/R has been derived for the 3-periodicfamily of an a, b
EB [7, Equation 7]:(5) ρ = rR = 2( δ − b )( a − δ ) c Solving the above for a/b yields the result. (cid:3) Conclusion
Videos mentioned above have been placed on a playlist [13]. Table 2contains quick-reference links to all videos mentioned, with column “PL
Figure 7. At a/b = α eq ≃ . , the Orthic is an equilateral triangle when a 3-periodicvertex lies on a top or bottom vertex of the EB. Therefore its CB is a circle. Figure 8.
Two configurations (left and right) of the Poristic Triangle Family (blue), whoseIncircle (green) and Circumcircle (purple) are fixed. Here R = 1 , r = 0 . . Over the family,the Circumbilliard (black) has invariant aspect ratio, in this case a /b ≃ . . Also shown isthe circular locus of X [12, page 17]. Video : [13, PL
PL
Table 2.
Videos mentioned in the paper. Column “PL
Acknowledgments
We would like to thank Peter Moses and Clark Kimberling, for theirprompt help with dozens of questions. We would like to thank Boris Odehnalfor his help with some proofs. A warm thanks goes out to Profs. Jair Koillerand Daniel Jaud who provided critical editorial help.The second author is fellow of CNPq and coordinator of Project PRONEX/CNPq/ FAPEG 2017 10 26 7000 508.
References [1] Akopyan, A., Schwartz, R., Tabachnikov, S.: Billiards in ellipses revisited (2020).URL https://arxiv.org/abs/2001.02934 https://arxiv.org/abs/2001.08469 https://books.google.com.br/books?id=QcOmDAEACAAJ (06), 491–504 (2019). URL https://doi.org/10.1080/00029890.2019.1593087
3, 5, 7, 8[6] Garcia, R., Reznik, D., Koiller, J.: Loci of 3-periodics in an elliptic billiard: why somany ellipses? (2020). URL https://arxiv.org/abs/2001.08041
2, 3, 4, 9, 13[7] Garcia, R., Reznik, D., Koiller, J.: New properties of triangular orbits in ellipticbilliards (2020). URL https://arxiv.org/abs/2001.08054 A (376) (2018). DOI https://doi.org/10.1098/rsta.2017.0419 1[9] Kimberling, C.: Encyclopedia of triangle centers (2019). URL https://faculty.evansville.edu/ck6/encyclopedia/ETC.html
2, 3, 4, 6, 12[10] Kimberling, C.: Central lines of triangle centers (2020). URL https://faculty.evansville.edu/ck6/encyclopedia/CentralLines.html (1), 015,005 (2019). DOI 10.1088/1361-6404/ab4748. URL https://arxiv.org/abs/1907.09295 (1), 45–67 (2011) 2,8, 10[13] Reznik, D.: Playlist for “Circumphenomena of 3-Periodics in Elliptic Billiards” (2020).URL https://bit.ly/2WDPeVk
1, 3, 4, 6, 9, 10, 11 [14] Reznik, D., Garcia, R., Koiller, J.: Can the elliptic billiard still surpriseus? Math Intelligencer (2019). DOI 10.1007/s00283-019-09951-2. URL https://rdcu.be/b2cg1 https://arxiv.org/abs/2002.00001 Student Mathematical Library , vol. 30.American Mathematical Society, Providence, RI (2005). DOI 10.1090/stml/030. URL . MathematicsAdvanced Study Semesters, University Park, PA 1[18] Weaver, J.H.: Invariants of a poristic system of triangles. Bull. Amer.Math. Soc. (2), 235–240 (1927). DOI 10.1090/S0002-9904-1927-04367-1. URL https://doi.org/10.1090/S0002-9904-1927-04367-1 http://mathworld.wolfram.com
1, 2, 3, 4, 5,6, 8
Appendix A. Computing a Circumconic
Let a Circumconic have center M = ( x m , y m ) Equation (2) is subject tothe following 5 constraints : it must be satisfied for vertices P , P , P , andits gradient must vanish at M : f ( P i ) = 0 , i = 1 , , dgdx ( x m , y m ) = c + c y m + 2 c x m = 0 dgdy ( x m , y m ) = c + c x m + 2 c y m = 0 Written as a linear system: x y x y x y x y x y x y x y x y x y y m x m
00 1 x m y m . c c c c c = − − − Given sidelenghts s , s , s , the coordinates of X = ( x m , y m ) can be ob-tained by converting its Trilinears ( s + s − s :: ... ) to Cartesians [9].Principal axes’ directions are given by the eigenvectors of the Hessianmatrix H (the jacobian of the gradient), whose entries only depend on c , c , and c :(6) H = J ( ∇ g ) = " c c c c The ratio of semiaxes’ lengths is given by the square root of the ratio of H ’s eigenvalues:(7) a/b = p λ /λ If M is set to X one obtains the Circumbilliard. Let U = ( x u , y u ) be an eigenvector of H . The length of the semiaxis along u is given by the distance t which satisfies: g ( M + t U ) = 0 This yields a two-parameter quadratic d + d t , where: d = 1 + c x m + c x m + c y m + c x m y m + c y m d = c x u + c x u y u + c y u The length of the semi-axis associated with U is then t = p − d /d . Theother axis can be computed via (7).The eigenvectors (axes of the conic) of H are given by the zeros of thequadratic form q ( x, y ) = c ( y − x ) + 2( c − c ) xy Appendix B. Table of Symbols
Tables 3 and 4 lists most Triangle Centers and symbols mentioned in thepaper.
Center Meaning Note X Incenter Locus is Ellipse X Barycenter Perspector of Steiner Circum/Inellipses X Circumcenter Locus is Ellipse, Perspector of MX Orthocenter Exterior to EB when 3-periodic is obtuse X Center of the 9-Point Circle X Symmedian Point Locus is Quartic [6] X ∗ X of Orthic Detached from X locus for obtuse triangles X Gergonne Point Perspector of Incircle X Nagel Point Perspector of I , X of ACT Incircle X Mittenpunkt Center of (Circum)billiard X Spieker Point Incenter of Medial X Feuerbach Point on confocal Caustic X Bevan Point X of Excentral X X of the ACT Perspector of I X Anticomplement of X On Circumcircle and EB, J exc center X X of Medial Midpoint of X X , lies on L (2 , X Anticomplement of X Perspector of ACT and its Intouch Triangle X X of the Excentral Triangle Non-elliptic Locus L (2 , ACT-Medial Mittenpunkt Axis Line L [10] Table 3.
Kimberling Centers and Central Lines mentioned in the paper.
Dan Reznik, Data Science Consulting,, Rio de Janeiro, RJ, Brazil
E-mail address : [email protected] Ronaldo Garcia, Inst. de MatemÃątica e EstatÃŋstica,, Univ. Federal deGoiÃąs,, GoiÃćnia, GO, Brazil
E-mail address : [email protected] a, b EB semi-axes a > b > P i , s i Vertices and sidelengths of 3-periodic invariant P s i P ′ i Vertices of the Excentral Triangle a , b Semi-axes of Poristic Circumbilliard r, R, ρ