Isogonic and Isodynamic Points of a Simplex in a Real Affine Space

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(1) The mirror image of triangle A1A2A3 with respect to a circle with center P is an equilateral triangle. (2) Let Ci denote the intersection of the tripolar of the symmedian K (K deﬁned as the isogonal conjugate of the centroid G) with the sideline opposite the vertex Ai and let Ci denote the circle with center Ci passing through the vertex Ai, i = 1, 2, 3; then P is a common point of the three circles C1, C2, C3. (3) Let C denote the circumcircle of triangle A A A and T C denote the tangent of C 1 2 3 Ai at the point A , let C⋆ denote the intersection of T C with the sideline opposite A and i i Ai i ⋆ ⋆ C i denote the circle with center C i passing through the vertex Ai, i = 1, 2, 3; then P is a C ⋆, C ⋆, C ⋆. common point of the three circles 1 2 3 (4) The pedal triangle of P is equilateral. (5) P is a point that satisﬁes the equation d(P,A1) d23 = d(P,A2) d13 = d(P,A3) d12 .

Let us assume that the triangle A1A2A3 is not equilateral. Then there exist precisely two isodynamic points J1,J2, and the following statements hold:

(6) J1,J2 both lie on the Brocard axis, a line through the symmedian K and the circumcenter O of triangle A1A2A3. (7) One of the two isodynamic points is a point inside the triangle, and the other is the mirror image of this point with respect to the circumcircle. (8) The points O,J1,K,J2 form a harmonic range. C ⋆, C ⋆, C ⋆ A A A The circles 1 2 3 are called the Apollonian circles of the triangle 1 2 3. A ⋆ point Q is a point on C i precisely when d(Q,Aj) dij = , {i, j, k}={1, 2, 3} (⋆⋆) . d(Q,Ak) dik Let T C ⋆ be the tangent of C ⋆ at the point J , k =1, 2 and i =1, 2, 3. Then: Jk i i k (9) The three singular conics T C ⋆ ∪ T C ⋆, T C ⋆ ∪ T C ⋆, T C ⋆ ∪ T C ⋆ are con- Jk 1 Jk 2 Jk 2 Jk 3 Jk 3 Jk 3 gruent in pairs, k = 1,2. In other words, these are unions of two lines with the same angle of intersection.

A point P is an isogonic point (often also called isogonic center) of triangle A1A2A3 if any of the three following (equivalent) statements is true: (10) The three singular conics (P⊔A1)∪(P⊔A2), (P⊔A2)∪(P⊔A3), (P⊔A3)∪(P⊔A1) are congruent in pairs. (11) The inversion of triangle A1A2A3 in a circle with center P leads to an equilateral triangle. (12) The antipedal triangle of P is equilateral.

There exist exactly two isogonic points, F1 and F2. These are the isogonal conjugates of the points J1 and J2, respectively. If F1 is inside the triangle A1A2A3, then it is the Fermat-Torricelli point of the triangle, it ↦ minimizes the function P d(P,A1)+ d(P,A2)+ d(P,A3).

Remarks. Let us look at the situation in elliptic and in hyperbolic planes. In these planes there are two points - let us call them J1 and J2 - which satisfy (2). Both are points on the line K⊔O. One more point on this line is the Lemoine point K̃ , a point whose tripolar is orthogonal to K⊔O and meets the sidelines of triangle A A A in C⋆,C⋆,C⋆. In 1 2 3 1 2 3 ISOGONIC AND ISODYNAMIC POINTS OF A SIMPLEX IN A REAL AFFINE SPACE 3 the affine case, the symmedian agrees with the Lemoine point, but in planes with nonzero Gaussian curvature these are two different triangle centers, see [9]. ̃ The two points J1 and J2 also satisfy (9), and the four points O,J1, K,J2 form a harmonic range. But statements (1), (3), (4), (7) do not apply, in general. The equation given in (5) is responsible for the name of the two centers; it has to be replaced in the hyperbolic case by the equation sinh( 1 d(P,A )) sinh( 1 d ) = sinh( 1 d(P,A )) sinh( 1 d ) = sinh( 1 d(P,A )) sinh( 1 d ) 2 1 2 23 2 2 2 13 2 3 2 12 and in the elliptic case by sin( 1 d(P,A )) = sin( 1 d ) = sin( 1 d(P,A )) sin( 1 d ) = sin( 1 d(P,A )) sin( 1 d ) . 2 1 2 23 2 2 2 13 2 3 2 12 Apollonian circles C ⋆, C ⋆, C ⋆ The name for the circles 1 2 3 is justified also in planes with nonzero Gaussian curvature, because their equations are very similar to equation (⋆⋆), cf.[9].

In elliptic and in hyperbolic planes, there are two points, F1 and F2, say, for which statement (10) is true (cf.[9]), and if one of these points F1, F2 lies inside the triangle, this point is the Fermat-Torricelli point of the triangle (cf. [11]). There are strong indications (based on experiments with GeoGebra) that F1 and F2 are exactly the points that satisfy condition (11), see [9]. But (12) does not necessarily apply to them, and F1 and F2 are, in general, not isogonal conjugates of J1 and J2. What is the situation like in a Lorentz-Minkowski plane or in a Galilean plane? Apollo- nian circles exist in a Lorentz-Minkowski plane, however their common points are points on the line at inﬁnity. In a Galilean plane, each Apollonian circle consists of two parallel lines and all these six lines meet at the absolute pole. (A nonsingular circle touches the line at inﬁnity at the absolute pole, and this point is the center of the circle.) So there are no isodynamicpoints in Galilean and Lorentz-Minkowskiplanes, nor are thereisogonicpoints.

3. Generalized Apollonian spheres. Inthis section we generalizeresults andideas publishedby P. Yiu [24]to higherdimensions.

We come back to the general case of an n-simplex Σ with vertices A1, … An+1. Let P =[p1∶ ⋯ ∶pn+1] be a pointnot on any of the (n−1)-sideplanes of Σ, thus p1p2 ⋯ pn+1 ≠ 0. The (n − 1)-plane P ⊔A3 ⊔ ⋯ ⊔An+1 meets the line A1 ⊔A2 at the point P12 ∶= Σ [p1 ∶ p2 ∶ 0 ∶ ⋯ ∶ 0], while the Σ-polar plane of P , this is the (n−1)-plane P ∶= x1 xn+1 ⋆ [x1∶ ⋯ ∶xn+1] + ⋯ + =0 , meetstheline A1⊔A2 atthepoint P ∶= [−p1∶p2∶ p1 pn+1 12 0 ∶ ⋯ ∶ 0]. Theð midpoint of these two points, Q ∶= 1 P + 1 P ⋆ = [−p2∶p2∶0∶ ⋯ ∶0], 12 2 12 2 12 1 2 is a point on the Σ-polar plane of the barycentric square P 2 of P , as can be easily checked. The (n−1)-sphere with diameter [[P , P ⋆ ]] ∶= Q = tP + (1−t)P ⋆ 0 ≤ t ≤ 1 and 12 12 12 12 center Q12 is denoted by 12. ð This sphere 12 meets the circumsphere of Σ orthogonally. Proof A , P ,A , P ⋆ A ,A . The points 1 12 2 12 form a harmonic range. Therefore the points 1 2 are inversive with respect to 12, and every sphere through A1 and A2, especially the circumsphere of Σ, is orthogonal to 12. □ ⋆ The points Pij, Pij ,Qij and (n−1)-spheres ij , 1 ≤ i ≤ j ≤ n+1 are deﬁned likewise. Following Yiu, we will call these spheres generalized Apollonian spheres of Σ with respect to P . The sphere ij is the locus of points R satisfying d(Ai, R) ∶ d(Aj; R) = 1∕ pi ∶ 1∕ pj . From this follows that R is a common point of two spheres ij, jk, 1 ≤ i

FIGURE 1. Three generalized Appollonian circles. All ﬁgures were cre- ated with the software program GeoGebra [25].

But then R must be a point on ik, as well. The radical (n−1)-plane of the three spheres is perpendicular to the line through their centers and contains the circumcenter O. Therefore, the intersection of all these radical (n−1)-planes is the line through O perpendicular to the 2 Σ-polar plane of P . As a consequence, all (n−1)-spheres ij pass through a point of this line if any of these spheres does. If such a point exists, we call it a generalized isodynamic point of Σ with respect to P . If there is exactly one generalized isodynamic point, then it is a point on the circumcircle. If there are two such points, they are inversives with respect to the circumcircle.

The "classical" case: P = incenter I. In this case, the spheres ij are the (proper) Apol- lonian spheres. Srubǎ r’š paper [21] suggests that the line through O perpendicular to the Σ-polar plane of K = I2 always meets an Apollonian sphere at two points, which are then called isodynamic points. However, this is not the case, not even for n = 3. We present a counter example. There exists a tetrahedron with sidelengths (d12, d13, d14, d23, d24, d34)= (13, 11, 9, 12, 5, 11). We will show that the Apollonian spheres 12, 13, 23 of this tetrahedron do not meet. Let ai denote the area of the facet opposite vertex Ai; then a1 = 6 21,a = 9 403,a = 9 51,a = 6 105. The line A ⊔ K meets the sideplane √ 2 4 √ 3 4 √ 4 √ 4 A ⊔A ⊔A R a 2 a 2 a 2  ,  ,  1 2 3 at the point =[ 1 ∶ 2 ∶ 3 ∶0]. The intersection of the spheres 12 13 23 with the plane A1 ⊔A2 ⊔A3 are the generalized Apollonian circles of the point R and triangle Δ= A1A2A3. P. Yiu [23] gives a criterion to decide whether or not these circles have points in common: They have no common points precisely when the point d 2 d 2 d 2 d 2 d 2 d 2 d 2 d 2 d 2 Q = d 2 23 − 13 − 12 ∶ d 2 13 − 12 − 23 ∶ d 2 12 − 23 − 13 ∶0 23 a 2 a 2 a 2 13 a 2 a 2 a 2 12 a 2 a 2 a 2 1 2 3 2 3 1 3 1 2 lies outside the circumcircle of Δ. ISOGONIC AND ISODYNAMIC POINTS OF A SIMPLEX IN A REAL AFFINE SPACE 5

Since the distance between Q = [ 3326952 , 25180529 , − 18117983 , 0] and the circumcenter of 4504043 27024258 27024258 Δ, O = [ 73 , 121 , 169 , 0] , is bigger than the circumradius of Δ, the Apollonian spheres Δ 210 315 630  ,  ,  12 13 23 of this tetrahedrondo not meet, and isodynamicpoints, when deﬁned as common points of the Apollonian spheres, do not exist for this tetrahedron. Remark. C. Pohoata and V. Zajic [20] presented a generalization of the Apollonian circles diﬀerent from that introduced by P. Yiu.

4. Isogonic points of a simplex. From [4] we adopt the following terminology: A simplex is regular if all its edges are the same length, a simplex is equiareal if all its facets have the same (n−1)-volume, and it is equifacetal if all its facets are congruent. In dimension 3, every equiareal simplex is equifacetal; but in higher dimensions this is not the case. Let us define: A point P is an isogonic point of an n-simplex Σ if its antipedal simplex is equiareal. It follows immediately, that a point P is an isogonic point of an n-simplex Σ if and only if its mirror image Σ⋆ in an (n−1)-sphere with center P is equiareal. The simplex Σ⋆ and the antipedal simplex are similar simplices. Before moving to higher dimensions n, we shall see that our definition is adequate for dimension n =3: If the antipedal simplex is equarial, it is also equifacetal, and the four triads of lines (P⊔Ai)∪(P ⊔Aj)∪(P ⊔Ak), 1 ≤ i < j < k ≤ 4, are congruent in pairs. On the other hand, this congruence also implies that the antipedal simplex is equifacetal. But it is notthe congruence,it is the equiangularitythatjustifies the name isogonic point. Moreover, if P is an isogonic point inside the tetrahedron Σ, then this point is the Fermat-Torricelli point of Σ, see [1]. Let us call the Fermat-Torricelli point F . In 1937 A. Weiszfeld published a method to calculate the barycentric coordinates of F . Starting from a point F0 = [f0,1 ∶ ⋯ ∶ f0,n+1] with f0,1 ⋅ … ⋅ f0,n+1 ≠ 0, F is the limit of an iteration process with a step: Fi+1 =[fi+1,1 ∶ ⋯ ∶ fi+1,n+1], fi+1,k = fi,k∕d(Fi,Ai), see [22,4,14,15]. A geometric description of this process is given in the next section.

5. Z⋆-correspondence.

Let P =[p1∶ ⋯ ∶pn+1] be a pointnoton any (n−1)-sideplane of Σ,  an (n−1)-sphere with P ⋆ A ⋆A ⋆ ⋯ A ⋆  center , and let Σ = 1 2 n+1 be the polar simplex of Σ with respect to . The simplices Σ and Σ⋆ are orthologic, and both orthologic centers coincide at P . Therefore the barycentric coordinates of P with respect to Σ⋆ agree with the barycentric coordinates of P with respect to Σ. Proof of the last statement. We can assume that p1+ ⋯ +pn+1 =1 and that  has radius 1. sgn(p ) ⋆ n i ⋆ It suﬃces to show that i piA i = P . First observethat thevector i ∶= ⋆ (A i −P ) ∑ d(P,A i ) is an outward unit normal vector of the (n−1)-dimensional surface of Σ. When we denote n the (n−1)-volume of the facet opposite vertex Ai by ai, then i ai i is the zero vector. It ∑ 6 MANFRED EVERS

FIGURE 2. The I⋆-transversal and the I⋆-correspondent of a point P . A[P ]A[P ]A[P ] P Triangle 1 2 3 is the antipedal triangle of .

follows:

⋆ ⋆ piA i = piP + pi (A i − P ) Éi Éi Éi a = P + i d(P,A⋆) n ) ⋆ i i Éi n d(P,A ) j aj i ∑ 1 n = P + ai i = P. n i ai É ∑ i Z⋆ z⋆ ⋯ z⋆ Now we take a point having homogeneous coordinates 1∶ ∶ n+1 with respect to Σ⋆. If Z⋆ is diﬀerent from P , its polar (n−1)-plane with respect to  is a hyperplane in  and is called the Z⋆-transversal of P . The Σ-pole of this hyperplane is called the Z⋆-correspondent of P , cf. [6]. It makes sense to choose the centroid G of Σ as the P ⋆- correspondent of P . Let us denote the Z⋆-correspondent of P by P #Z⋆, then p p P #Z⋆ = 1 ∶ ⋯ ∶ n+1 (⋆⋆⋆) . z⋆ z⋆ 1 n+1 Proof. Here only an outline of a proof of this equation is presented; a more detailed proof is given in [6]. We assign to each point R ∈  and each hyperplane ℎ not passing through P a real number ISOGONIC AND ISODYNAMIC POINTS OF A SIMPLEX IN A REAL AFFINE SPACE 7

P = P (R, ℎ) by P (R, ℎ)=0 , if R = P or if the line P ⊔ R does not meet ℎ and (R, ℎ)= 1 , if the point P + t (R − P ) is the intersection of P ⊔ R and ℎ. P t Some properties of P are listed below: ∙ For any points R1, R2, for any real numbers t1,t2 and for an hyperplane ℎ of (A), P (t1R1 + t2R2,ℎ)= t1P (R1,ℎ)+ t2P (R1,ℎ). ∙ A point R lies on a hyperplane ℎ if and only if P (R, ℎ)=1. ∙ Let us denote the (n−1)-sideplane of Σ opposite of the vertex Ai by ℎi,i =1, … , n+1. If P =[p1, … ,pn+1], then P (Ai,ℎi)=1−1∕pi. It can be easily checked that the i-th pi barycentric coordinate of the point P + (Ai − P ) is zero. pi−1 If is the mapping that assigns to each point R its -polar hyperplane, then ⋆ (A i )= ℎi ={R (R, ℎi)=1}, ⋆ ⋆ ⋆ (z i A i )={R zði P (R, ℎi)=1} ⋆ ⋆ ⋆ and ( i z i A i )={ð R i z i P (R, ℎi)=1}. ∑ ∑ ⋆ We calculate the intersectionð of a sideline Ai ⊔Aj of Σ with the hyperplane (Z ). For simplicity, we take (i, j)=(1, 2) and determine the real number x such that ⋆ P (xA1 + (1−x)A2, (Z ))=1. ⋆ 1= P (xA1 + (1−x)A2, (Z )) = x( z⋆ (A ,ℎ )) + (1−x)( z⋆ (A ,ℎ )) É k P 1 k É k P 2 k k k ⋆ ⋆ 1 ⋆ ⋆ 1 = x(1− z 1 + z 1(1− )) + (1−x)(1 − z 2 + z 2(1 − )) p1 p2 z⋆ z⋆ = x(1− 1 )+(1− x)(1 − 2 ) . p1 p2 ⋆ ⋆ p1∕z p2∕z As a result we get x = 1 and 1−x =− 2 . □ p z⋆ p z⋆ p z⋆ p z⋆ 1∕ 1 − 2∕ 2 1∕ 1 − 2∕ 2 In the following we are particularly interested in the Z⋆-correspondents for Z⋆ = G⋆ (centroid of Σ⋆), Z⋆ = I⋆ (incenter of Σ⋆) and Q⋆ = K⋆ (symmedian of Σ⋆): ∙ P #G⋆ = P p p sgn(p ) sgn(pn+1) ∙ P #I⋆ = [ 1 ∶ ⋯ ∶ n+1 ]=[ 1 ∶ ⋯ ∶ ], a⋆ = (n − 1)-volume of a⋆ a⋆ d(P,A ) d(P,A ) i 1 n+1 1 n+1 ⋆ ⋆ the facet of Σ opposite A i . a⋆ a⋆ ⋆ ⋆ i j Explanation: pi ∶ pj = p i ∶ p j = sgn(pi) ∶ sgn(pj) . d(P,Ai) d(P,Aj) p pn 1 1 ∙ P #K⋆ =[ 1 ∶ ⋯ ∶ +1 ]=[ ∶ ⋯ ∶ ] . (a⋆)2 (a⋆ )2 p (d(P ; A ))2 p (d(P ; A ))2 1 n+1 1 1 n+1 n+1

For a point P =[p1, … ,pn+1] not on any sideplane of Σ consider the sequences (Qi)i∈ℕ 1 1 and (Ri)i∈ℕ given by Q0 = R0 = P and Qi+1 = ∶ ⋯ ∶ , d(Qi; A1) d(Qi; An+1) 1 1 R = ∶ ⋯ ∶ . i+1 2 2 p1 (d(Ri; A1)) pn+1 (d(Ri; An+1)) ⋆ ⋆ All points Qi, Ri,i > 0, are points inside ðΣ,ð and Qi+1 = Qi#I , Rð i+1 ð= Ri#K . For 8 MANFRED EVERS

FIGURE 3. The ﬁrst points of the sequences (Qi)i∈ℕ and (Ri)i∈ℕ with ⋆ ⋆ Q0 = R0 = P and Qi+1 = Qi#I , Ri+1 = Ri#K are shown. F1 and F2 are the two isogonic points.

Q R Q P Z⋆ Z⋆ p a⋆ ⋯ p a⋆ 1 and 1 we have 1 = # with = [sgn( 1) 1 ∶ ∶ sgn( n+1) n+1] and R P Z⋆ Z⋆ p a⋆ 2 ⋯ p a⋆ 2 1 = # with = [sgn( 1)( 1) ∶ ∶ sgn( n+1)( n+1) ]. Both sequences con- verge to the Fermat-Torricelli point F , even if this is a vertex of Σ. It seems that the second sequence in comparison to the ﬁrst converges twice as fast (see Figure 3). But more impor- tantly, the second sequence avoids calculating square roots. If the Fermat-Torricelli point of a simplex lies inside the simplex, then it is an isogonic point.

6. A 3-simplex with five isogonic points. In dimension n> 2, simplices, in general, have more than two isogonic points. We present a 3-simplex with five isogonic points. Consider in the Euclidean space (ℝ4, ⋅) the simplex Σ with the four vertices 0 6 0 2 A1 = ⎛ 0 ⎞ ,A2 = ⎛ 0 ⎞ ,A3 = ⎛ 8 ⎞ ,A4 = ⎛ 2 ⎞ . ⎜ 0 ⎟ ⎜ 0 ⎟ ⎜ 0 ⎟ ⎜ 6 ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ In order to calculate⎝ the isogonic⎠ points,⎝ ⎠ we first⎝ calculate⎠ their⎝ isogonally⎠ conjugated points. These are the points with an equiareal pedal simplex. For a point P which is not a vertex of Σ let ΣP denote the pedal simplex of P and let GP and IP denote the centroid and the incenter of ΣP , respectively. We define a sequence of points (Pi ) i∈ℕ by P0 ∶= P and P ∶= P + G − I . Obviously, this sequence is a constant sequence if and only if P n+1 n Pn Pn is the isogonal conjugate of an isogonic point. Experiments with the interactive geometry tool GeoGebra [25] suggest that each of these sequences converges to one of these limit ISOGONIC AND ISODYNAMIC POINTS OF A SIMPLEX IN A REAL AFFINE SPACE 9 points (rounded to 12 decimal places):

L0 = [0.266996565955, 0.275481800939, 0.217355830792, 0.240165802314], L1 = [−4.180629474014, 2.569387212447, 1.602113038329, 1.009129223238], L2 = [1.193250865914, −1.252645952150, 0.354761022780, 0.704634063455], L3 = [0.713260932730, 0.358215195120, −0.616627271982, 0.545151144132], L4 = [0.657546390333, 0.802131717931, 0.639088262811, −1.098766371077]. The pedal triangles of these points have facets with an area

a0 =2.404772767371,a1 = 122.125536031480,a2 = 19.392997370805, a3 =9.848601171111,a4 = 18.965046082427 . The isogonic points of Σ are

F0 = [0.369979160947, 0.229493293826, 0.163611619856, 0.236915925371], F1 = [−0.297000489955, 0.309278164652, 0.279002561033, 0.708719764270], F2 = [0.388102931405, −0.236608485604, 0.469943106828, 0.378562447371], F3 = [0.382915343108, 0.487963317698, −0.159452369671, 0.288573708865], F4 = [0.645021938255, 0.338403751068, 0.238914519123, −0.222340208446]. The antipedal triangles of these points have facets with an area

̃a0 = 241.637142362610, ̃a1 = 60.087819904352, ̃a2 = 31.387257487815, ̃a3 =5.647726265255, ̃a4 = 31.003305976553 . Finally, we present the two isodynamic points of Σ :

J1 = [0.206439675828, 0.327649375007, 0.263085414624, 0.20282553454], J2 = [2.954833710960, −0.575606610593, −1.403778427224, 0.024551326857].

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