Spherical Geometry and the Least Symmetric Triangle
Laney Bowden, Andrea Haynes, Clayton Shonkwiler, Aaron Shukert
SSpherical Geometry and the Least Symmetric Triangle
Laney Bowden, Andrea Haynes, Clayton Shonkwiler, and Aaron Shukert
Colorado State University, Fort Collins, CO
We study the problem of determining the least symmetric triangle, which arises both from puregeometry and from the study of molecular chirality in chemistry. Using the correspondence betweenplanar n -gons and points in the Grassmannian of 2-planes in real n -space introduced by Hausmannand Knutson, this corresponds to finding the point in the fundamental domain of the hyperoctahedralgroup action on the Grassmannian which is furthest from the boundary, which we compute exactly.We also determine the least symmetric obtuse and acute triangles. These calculations provide proto-types for computations on polygon and shape spaces. The equilateral triangle is surely the most symmetric triangle by any reasonable standard, butwhat is the least symmetric triangle? More generally, what is the least symmetric n -gon?As stated, this question is ill-posed, but we can make sense of it using a map defined by Haus-mann and Knutson [8] from the Grassmannian G ( R n ) of 2-dimensional linear subspaces of R n to the collection of ordered planar n -gons up to similarity. The symmetric group S n acts on or-dered n -gons by permuting edges, and this action lifts to an action of the hyperoctahedral group B n (cid:39) ( Z / Z ) n (cid:111) S n on G ( R n ) (see [5]). The fundamental domain of this action is a regionin G ( R n ) whose boundary corresponds to those n -gons which are unchanged by some groupelement: in other words, the boundary corresponds to n -gons with a symmetry.Therefore, we can re-state the question as: which point(s) are furthest from the boundary ofthe fundamental domain of the hyperoctahedral group action on G ( R n ) ? The answer should beimportant in understanding the behavior of permutation-invariant functions on random walks, andshould also be interesting from the perspective of algebraic geometry. We like this question, butdon’t know the general answer. So in this paper, we pose the question precisely and work out theanswer for triangles in as much detail as possible in order to jumpstart the larger problem.In the case of triangles, the Grassmannian G ( R ) is double-covered by the unit sphere S ,which is where we will actually do our calculations. In S , the fundamental domain is simplya spherical triangle that, as we will see, can be interpreted as the space of unordered triangles.Its boundary is precisely the set of isosceles and degenerate triangles and its interior is the setof scalene triangles; note that both isosceles and degenerate triangles have a mirror symmetry.Therefore, the point maximizing the distance to the boundary – which will be our least symmetrictriangle – could just as well be thought of as the most scalene triangle . Our triangle will be differentfrom any of those proposed by Robin [15].The problem of finding the most scalene triangle also arises in chemistry, where the chirality Here ordered means that the order of edges in the n -gon matters: e.g., a triangle with edges ordered from shortest tolongest is distinct from the same triangle with edges ordered from longest to shortest. a r X i v : . [ m a t h . M G ] A ug of molecules plays a key role [13]. A shape is chiral if it is not congruent to its mirror image;otherwise it is achiral .Although three-dimensional notions of chirality are the most physically relevant, substantialeffort has been expended on the study of chirality in two dimensions, which “serves as the firststep for a deeper understanding of three-dimensional chirality in chemistry” [3]. In the case oftriangles, isosceles triangles are achiral since reflection across the median intersecting the oddedge produces a congruent triangle, while scalene triangles are chiral. Consequently, there is anextensive chemistry literature on scalene triangles and in particular the search for the most scalenetriangle [2, 3, 6, 14].In the chemistry literature, the most scalene (or most chiral) triangles are found as maxima ofcertain energy functions defined on various models of triangle space. These models are mostly adhoc , whereas we view the Grassmannian parametrization of polygon space as a principled choiceof model: after all, for any n it yields a model of n -gon space which is a Riemannian symmetricspace, meaning in particular that it has a transitive group of isometries and, since it is compact, acanonical (up to scale) invariant Riemannian metric.Moreover, this model is beginning to gain traction in polymer physics due to its computationaltractability [7, 16, 17], and a version of it for continuous curves has been independently devel-oped for use in shape recognition and classification problems [18]. We see the current paper as aprototype of a geometric approach to finding optimal polygons or shapes.
1. THE CONSTRUCTION
We introduced the symmetric measure on the space of planar n -gons in [4] by pushing forwardthe uniform measure on G ( R n ) using Hausmann and Knutson’s [8] map from the Grassmannian G ( R n ) whose points correspond to the 2-dimensional linear subspaces of R n . In other words, aplane in R n corresponds to a similarity class of n -gons in the plane, and the measure of a collectionof polygons is the measure of the corresponding subset of the Grassmannian. In fact, we get muchmore than just a probability measure on polygon space: defining the Hausmann–Knutson map tobe a Riemannian submersion yields a Riemannian metric on polygon space.Something special happens when n = 3 : any 2-dimensional subspace of R has a unique linethrough the origin as its orthogonal complement, so we can equivalently think about the space oflines through the origin in R , otherwise known as the projective plane RP . In turn, the projectiveplane is double-covered by the unit sphere S ⊆ R since any unit vector determines a line throughthe origin, and the only way that two distinct unit vectors can lie in the same line is if they areantipodal. Consequently, we can identify the space of planar triangles with the unit sphere inthree-dimensional space. Strictly speaking, scalene triangles are only chiral when viewed as living in a two-dimensional universe. In threedimensions a scalene triangle and its mirror image are related by a rotation.
This construction and identification (and the generalization to Grassmannians) is explainedin [5], but we briefly summarize it here.The space of planar triangles is non-compact since the size of any triangle can be scaled byan arbitrary positive real number. However, since we will primarily be interested in triangles upto similarity, we may as well compactify by fixing a scale for our triangles. As will shortly beapparent, it turns out to be convenient to normalize all triangles to have perimeter 2.Since triangles are determined up to congruence by their three side lengths, we can uniquelyspecify a triangle up to similarity by choosing the side lengths a , b , c such that its perimeter is 2: a + b + c = 2 . This equation determines a plane in abc -space and, since a, b, c ≥ , the intersectionof that plane with the positive orthant contains the parameter space of triangles. This intersectionis a simplex, but not every point in this simplex determines a triangle: the side lengths of a trianglemust also satisfy the three triangle inequalities a ≤ b + c, b ≤ c + a, c ≤ a + b. FIG. 1: The simplex a + b + c = 2 in abc -space. The darker sub-triangle consists of those points for which a, b, c satisfy the triangle inequalities. Rather than incorporate these slightly ungainly inequalities, we change coordinates, defining s a = b + c − a , s b = c + a − b , s c = a + b − c . (1)These quantities have a long history in triangle geometry, most notably as the radii of three mutuallytangent circles centered at the vertices of the triangle.Since we’ve already fixed a + b + c = 2 , these coordinates can also be rewritten as s a = 1 − a , s b = 1 − b , s c = 1 − c . Now we see that s a + s b + s c = 3 − ( a + b + c ) = 1 and the triangle Here the 1 should be thought of as the semiperimeter of the triangle. This is why we fixed the perimeter to be 2. inequalities become the conditions s a ≥ , s b ≥ , s c ≥ , so triangle space is parametrized bythe standard simplex in s a s b s c -space. If we were to perform the analysis from Section 4 on thissimplex, we would get answers similar to those given in the last section of Robin’s paper [15]. However, the simplex only has a finite symmetry group, whereas we would prefer, followingPortnoy [12], a transitive symmetry group, since this induces a canonical probability measure onthe parameter space. Therefore, it is desirable to symmetrize by taking square roots: define newcoordinates x, y, z by x = s a , y = s b , z = s c , (2)so that x + y + z = s a + s b + s c = 1 , and hence the points ( x, y, z ) lie on the unit sphere.The sphere has a transitive group of symmetries, namely the rotation group, and we actually getrather more than just a canonical probability measure: the round metric on the sphere is an invariantRiemannian metric which is unique up to scale.Since each of the eight points ( ± x, ± y, ± z ) map to the same ( s a , s b , s c ) , and hence to the sametriangle, the unit sphere is (generically) an eightfold covering of triangle space. The intersection ofthe sphere with each closed orthant contains points mapping to all possible triangles up to similarity.Combining (1) and (2), we can translate directly between side lengths a, b, c and sphere coordi-nates x, y, z using a = 1 − x , b = 1 − y , c = 1 − z . (3) FIG. 2: On the left we see the partition of the sphere into the 8 fundamental domains of the group ( Z / Z ) which independently flips the signs of the coordinates. On the right is the partition of the sphere into the 48fundamental domains of the hyperoctahedral group B (cid:39) ( Z / Z ) (cid:111) S which permutes the coordinatesand changes their signs. Robin is working in the projection of the simplex to the plane, rather than in the simplex itself. This projection distortsthe Riemannian metric, so computing in the simplex will produce a slightly different answer than Robin’s.
Using ( a, b, c ) coordinates, we are implicitly parametrizing ordered triangles, where the orderof the side lengths matters. If we prefer to think of unordered triangles, we can divide by theaction of the permutation group S , which acts by permuting a , b , and c . This action lifts tothe standard permutation action on the sphere, and indeed fits nicely together with the action ofchanging signs of the coordinates above: the hyperoctahedral group B acts on the sphere by signedpermutations, permuting the coordinates ( x, y, z ) and changing their signs. B is the semidirectproduct ( Z / Z ) (cid:111) S of the group ( Z / Z ) of order = 8 which acts by changing signs and thepermutation group S of order
3! = 6 . We can also see B as the finite subgroup of the isometrygroup O (3) consisting of all orthogonal matrices with integer entries.The action of the hyperoctahedral group on polygon space is described in some detail in [5];here we confine ourselves to the observation that B acts freely on points of the sphere with allthree coordinates having distinct, nonzero magnitudes and it has order | B | = 2 ×
3! = 48 . Hence, since elements of B are isometries, the action of B naturally divides the sphere into 48congruent chambers whose boundaries are the great circles where either two coordinates agree(up to sign) or one coordinate is zero. As seen in Figure 2, each chamber is a 45–60–90 sphericaltriangle and we will soon see that we can interpret a chamber as a parameter space for the collectionof unordered triangles.
2. A FIRST SOLUTION
Given our identification of triangles with the unit sphere, the problem of finding the least sym-metric triangle is now simple, at least conceptually: the isosceles triangles are exactly those trian-gles fixed by some permutation of the edge lengths, so we should identify the subset of the spherecorresponding to isosceles triangles, and then determine the point(s) which are furthest from thissubset.In terms of side length coordinates a, b, c , a triangle is isosceles if and only if it satisfies one ofthe equations a = b, b = c, c = a. Using (3) and simplifying slightly, this translates to the equations x = y , y = z , z = x on the sphere. In other words, the subset of the sphere corresponding to isosceles triangles is theintersection of the sphere with the planes x ± y = 0 , y ± z = 0 , z ± x = 0 . Since these are planes through the origin, their intersections with the sphere are great circles, whichare geodesics on the sphere. Note that these great circles are a subset of the circles giving theboundaries of the chambers induced by the action of the hyperoctahedral group.As seen in Figure 3, these great circles determine a tiling of the sphere into 24 spherical trian-gles, each of which is a 60–60–90 triangle. Therefore, there will be exactly 24 points on the spherewhich are furthest from the subset of isosceles triangles: each is the incenter of one of the 24 trian-gles in the tiling. The corresponding most scalene triangles will all be equivalent up to relabelingthe edges, so we can choose any of the 24 triangles and find its incenter.
FIG. 3: Tiling of the sphere determined by the isosceles triangles.
For example, the curves x − y = 0 , y + z = 0 , and y − z = 0 determine two of the triangles inthe tiling: one with x and y nonnegative the other its antipodal image. We will focus on the triangle D with x and y both nonnegative, which can also be described as those points ( x, y, z ) on thesphere satisfying | z | ≤ y ≤ x . Translating to side lengths, these inequalities become a ≤ b ≤ c , so D parametrizes triangles with side length written in ascending order.Just as in the plane, the incenter of a spherical triangle is the intersection of the three anglebisectors. The great circles y + z = 0 and y − z = 0 are perpendicular and the angle bisector lieson the equator z = 0 , so the incenter will be a point of the form ( x, y, with ≤ y ≤ x .In order to determine the angle bisector of x − y = 0 and y − z = 0 , we will consider the unitnormal vectors to the planes, namely v = (cid:16) − √ , √ , (cid:17) and v = (cid:16) , √ , − √ (cid:17) . These vectorsform an angle θ = arccos( v · v ) = arccos(1 /
2) = π/ and the vector halfway between them is proportional to the sum v + v = (cid:18) − √ , √ , (cid:19) + (cid:18) , √ , − √ (cid:19) = (cid:18) − √ , √ , − √ (cid:19) . Scaling the above vector by √ , we see that the great circle which bisects x − y = 0 and y − z = 0 is − x + 2 y − z = 0 . The intersection of this great circle with the equator z = 0 is our desired point (cid:16) √ , √ , (cid:17) (seeFigure 4), corresponding to the triangle with side lengths a = 1 − (cid:18) √ (cid:19) = 15 b = 1 − (cid:18) √ (cid:19) = 45 c = 1 − = 1 . FIG. 4: The point (cid:16) √ , √ , (cid:17) corresponding to the least symmetric triangle. It is the incenter of thedisplayed spherical triangle D . The radius (along the sphere) of the inscribed circle is arccos (cid:16) √ (cid:17) ≈ . . As a proposed least symmetric triangle, the fact that the side lengths are in a ratio isgratifying; rather less so is that this is a degenerate triangle with all three sides lying in a line. It is,perhaps, not surprising that the least symmetric triangle would have as much difference as possiblebetween its shortest and longest side lengths subject to the constraints of the triangle inequalities,resulting in a length ratio of r : r + 1 . Both as r → and as r → ∞ , the resulting trianglebecomes isosceles, so the specific value r = 4 is apparently a balance between these two extremessubject to the constraints of the geometry of the sphere.However, degenerate triangles like the one we just found are actually symmetric: reflectingacross the line of degeneracy fixes the degenerate triangle. The issue is that D is, in fact, (almost)a double covering of the space of unordered triangles. The points ( x, y, z ) and ( x, y, − z ) mapto the same triangle, so almost all triangles have two preimages in D ; the exceptions are thosetriangles with c = 1 − z , which come from points of the form ( x, y, . Therefore, theset of triangles we consider symmetric should include not just the isosceles triangles, but also thedegnerate triangles.
3. EXCLUDING DEGENERATE TRIANGLES
The degenerate triangles are those with one side length being half the total perimeter of thetriangle: this forces the other two sides to lie in the same line as the long side. Given our normal-ization that triangles should have perimeter 2, this means that the degenerate triangles are thosewith a = 1 , b = 1 , or c = 1 . In terms of x, y, z coordinates, the degenerate triangles are those with x = 0 or y = 0 or z = 0 . Adding these three great circles to the six corresponding to the isosceles triangles gives the tiling ofthe sphere by 48 congruent 45–60–90 triangles that we saw in Figure 2, and we have now justifiedthe claim that each triangle can be thought of as the space of unordered triangles up to similarity.
FIG. 5: The triangle T bounded by x − y = 0 , y − z = 0 , and z = 0 , the interior of which parametrizesscalene triangles. For specificity, we will focus on the spherical triangle T bounded by x − y = 0 , y − z = 0 ,and z = 0 as our model of the space of triangles. Equivalently, T is the subset of the sphere with ≤ z ≤ y ≤ x , which means that the side lengths a, b, c of the triangles corresponding to pointsin T satisfy ≤ a ≤ b ≤ c ≤ , Note that a triangle can also be doubly-degenerate, with two sides of length 1 and one of length 0. This means thattwo vertices coincide. and the interior of T corresponds to those triangles for which all of the above inequalities are strict.Therefore, a more plausible “least symmetric triangle” will be the incenter of this triangle, asvisualized in Figure 6. Since this will again be the intersection of the angle bisectors, it must be theintersection of the previously determined great circle − x + 2 y − z = 0 (which bisects the angleformed by the sides x − y = 0 and y − z = 0 ) and the great circle halfway between x − y = 0 and z = 0 . FIG. 6: The incenter and incircle of the spherical triangle T which parametrizes the non-degenerate scalenetriangles. The incenter is the point √ √ (cid:0) √ , √ , (cid:1) and the radius of the inscribed circle is arcsin (cid:18)(cid:113) (cid:0) − √ (cid:1)(cid:19) ≈ . . As before, we can find the latter great circle as being perpendicular to the sum of the unit normalvectors (cid:18) − √ , √ , (cid:19) + (0 , ,
1) = (cid:18) − √ , √ , (cid:19) . Multiplying by √ , we see that we’re looking at the great circle − x + y + √ z = 0 , which intersects − x + 2 y − z = 0 at the point ( x, y, z ) = 1 (cid:112)
13 + 6 √ (cid:16) √ , √ , (cid:17) ≈ (0 . , . , . . We have now proved:0
Proposition 1.
The least symmetric triangle has side lengths a = 1 − x = 28 + 2 √ ≈ . b = 1 − y = 82 − √ ≈ . c = 1 − z = 84 + 6 √ ≈ . and is shown in Figure 7. FIG. 7: The least symmetric triangle. Its side lengths fall in the ratio − √ : 3 . Though not quite as simple as for the degenerate triangle from Section 2, the side lengths ofthis triangle still form the pleasing ratio − √ : 3 .
4. REFINEMENTS
Having found the least symmetric triangle, we can hardly resist refining the question and tryingto find the least symmetric obtuse and acute triangles, even though obtuse and acute are not notionsthat naturally generalize to n -gons. To do so, we want to find the point on the sphere furthest fromnot only the isosceles and the degenerate triangles, but also from the right triangles. Hence, weneed to identify the collection of points on the sphere corresponding to the right triangles.In terms of the side lengths a, b, c , the right triangles are uniquely characterized by satisfyingthe Pythagorean theorem, meaning that a + b = c or b + c = a or c + a = b . Translating into x, y, z coordinates, the subset of the sphere corresponding to right triangles is theset of points satisfying one of the following quartic equations: (1 − x ) +(1 − y ) = (1 − z ) , (1 − y ) +(1 − z ) = (1 − x ) , (1 − z ) +(1 − x ) = (1 − y ) . Incidentally, as in Figure 8, the most acute triangle (in the sense of being furthest from the righttriangles) is the equilateral triangle, and the most obtuse is the degenerate triangle with side lengths (1 / , / , . This is reassuringly intuitive: the equilateral triangle is surely the acute triangle leastlike a right triangle, and we would expect the most obtuse triangle to have a ◦ angle.1 FIG. 8: The curves on the left represent the degenerate triangles, the isosceles triangles, and the right trian-gles. On the right we see the points corresponding to the equilateral triangle and to the degenerate isoscelestriangles, which (locally) maximize distance from the right triangles. In both cases the closest right triangleis a 45–45–90 triangle, which is at a distance arccos (cid:18) √ − √ √ − √ (cid:19) ≈ . from the equilateraltriangle and arccos (cid:18)(cid:113) (cid:0) √ − (cid:1)(cid:19) ≈ . from the degenerate isosceles triangles. To find the least symmetric obtuse triangle, we again focus our attention on the spherical triangle T bounded by x − y = 0 , y − z = 0 , and z = 0 . Since the curve of right triangles satisfying theequation (1 − x ) + (1 − y ) = (1 − z ) is the one which intersects T , we focus on it. Since thiscurve is not a geodesic, the region O bounded by it and the great circles x − y = 0 and z = 0 is nota spherical triangle, making it a more substantial challenge to find the point which is maximally farfrom the boundary of O .The point we are after must be equidistant from the great circles x − y = 0 and z = 0 ,so it must lie on the great circle which bisects the angle between them, namely the great circle − x + y + √ z = 0 that we already found in Section 3. This great circle contains the orthonormalvectors u = (cid:18) √ , √ , (cid:19) , u = (cid:18) , − , √ (cid:19) , so it can be parametrized as p ( t ) = cos t u + sin t u = (cid:32) √ t + sin t , √ t − sin t , sin t √ (cid:33) . p ( t ) makes with the z -axis is simply θ ( t ) = arccos( p ( t ) · (0 , , (cid:18) sin t √ (cid:19) , which means that the spherical distance from p ( t ) to the great circle z = 0 (and hence also to thegreat circle x − y = 0 ) is π − θ ( t ) = π − arccos (cid:18) sin t √ (cid:19) = arcsin (cid:18) sin t √ (cid:19) since sin φ = cos( π/ − φ ) for any φ .Therefore, the problem is to determine the value of t for which the spherical distance from p ( t ) to the curve (1 − x ) + (1 − y ) = (1 − z ) is equal to arcsin (cid:16) sin t √ (cid:17) .Using the fact that our points are on the sphere x + y + z = 1 , we can parametrize the curveof right triangles (1 − x ) + (1 − y ) = (1 − z ) by q ( x ) = (cid:32) x, (cid:114) − x x , x (cid:114) − x x (cid:33) , so the spherical distance from p ( t ) to q ( x ) is d ( t, x ) = arccos( p ( t ) · q ( x )) = arccos (cid:32) x √ t + sin t (cid:114) − x x √ t − sin t x (cid:114) − x x sin t √ (cid:33) . Therefore, we are seeking the smallest positive value of t so that d ( t, x ) = arcsin (cid:18) sin t √ (cid:19) or, after eliminating the inverse trig functions, so that x √ t + sin t (cid:114) − x x √ t − sin t x (cid:114) − x x sin t √ (cid:114) − sin t . This equation can be solved for x in, e.g., Mathematica , though the solution is not pleasant: printingout all four solutions would require hundreds of pages. In turn, given x as a function of t , thechallenge is to find the smallest t for which x is real. Since the process of finding it was ad hoc andunenlightening, we simply report that the smallest such t is t = 2 arctan α, α ≈ . is the smallest positive root of the even, palindromic polynomial z − z + 9689 z − z + 100908 z − z + 238166 z − z + 100908 z − z + 9689 z − z + 16 . Complete details can be found in the supplementary materials [1].
FIG. 9: The figure on the left shows the region O together with the point p ( t ) maximally far from theboundary of O and the circle of radius arcsin (cid:16) √ α α (cid:17) ≈ . around the point. The figure on the rightshows the 48 different points and circles under the action of the hyperoctahedral group B which permutesthe coordinates and independently changes their signs.FIG. 10: A comparison between asymmetric triangles. Not surprisingly, the (dashed) least symmetric obtusetriangle has a larger obtuse angle than the (solid) least symmetric triangle. Therefore, the point on the sphere that we’re after is p ( t ) ≈ (0 . , . , . , and converting to side length coordinates yields:4 Proposition 2.
The least symmetric obtuse triangle has side lengths α ) (cid:16) − √ α + 4 α + 2 √ α + α , √ α + 4 α − √ α + α , α (cid:17) ≈ (0 . , . , . . Now, we turn to solving the corresponding problem for acute triangles. The subset
A ⊆ T corresponding to acute triangles is shown on the left in Figure 11; it is bounded by the great circles x − y = 0 and y − z = 0 , as well as the curve of right triangles (1 − x ) + (1 − y ) = (1 − z ) .The point which maximizes distance to the boundary must lie on the great circle halfway between x − y = 0 and y − z = 0 , which has equation x − y + z = 0 and can be parametrized by (cid:101) p ( t ) = cos t (cid:18) √ , √ , √ (cid:19) + sin t (cid:18) √ , , − √ (cid:19) . FIG. 11: The point (cid:101) p ( (cid:101) t ) ≈ (0 . , . , . in A maximally far from the curves corre-sponding to the isosceles and right triangles. The common distance to the boundary of the region A is arcsin (cid:16) (cid:101) α (cid:101) α (cid:17) ≈ . . Permuting the coordinates of (cid:101) p ( (cid:101) t ) yields a ring of six points equidistant fromthe equilateral triangle point (cid:16) √ , √ , √ (cid:17) . The difference is too small to see, but the radius of the circlearound the equilateral point ( (cid:4) ) in the right figure is smaller than that of the inscribed circle around (cid:101) p ( (cid:101) t ) by ≈ . . The distance from (cid:101) p ( t ) to the boundary great circle x − y = 0 is arcsin (cid:0) sin t (cid:1) , and the distanceto the point q ( x ) on the curve of right triangles is (cid:101) d ( t, x ) = arccos (cid:113) − x x +1 cos t √ x (cid:114) − x x + 1 (cid:18) cos t √ − sin t √ (cid:19) + x (cid:18) sin t √ t √ (cid:19) . t for which (cid:101) d ( t, x ) = arcsin (cid:0) sin t (cid:1) .Again, we solve this equation for x , and then find the smallest t which makes x real, which is (cid:101) t = 2 arctan (cid:101) α where (cid:101) α ≈ . is the smallest positive root of the even, palindromic polynomial z − z + 715784192 z − z + 83609604096 z − z + 1410471953408 z − z + 18137673285920 z − z + 56162265469488 z − z + 73382345772378 z − z + 73382345772378 z − z + 56162265469488 z − z + 18137673285920 z − z + 1410471953408 z − z + 83609604096 z − z + 715784192 z − z + 131072 . Again, more details are in the supplementary materials [1].Therefore, the incenter of A is (cid:101) p ( (cid:101) t ) ≈ (0 . , . , . and we can convert to side lengths to conclude: Proposition 3.
The least symmetric acute triangle has side lengths (cid:101) α ) (cid:16) − √ (cid:101) α + (cid:101) α + √ (cid:101) α + (cid:101) α , (cid:101) α + (cid:101) α , √ (cid:101) α + (cid:101) α − √ (cid:101) α + (cid:101) α (cid:17) ≈ (0 . , . , . . FIG. 12: The least symmetric acute triangle, with side lengths ≈ (0 . , . , . .
5. CONCLUSION AND OPEN QUESTIONS
The emphasis in the chemistry literature is on finding chirality measures whose maxima areconsidered the most chiral triangles. Of course, our least symmetric triangle from Proposition 1can also be thought of in this way: it maximizes the minimum distance to the 9 great circles rep-resenting isosceles and degenerate triangles. Since this function is continuous but only piecewisesmooth, a natural question to ask is: does there exist a reasonable smooth function on the spherewhich vanishes precisely at the isosceles and degenerate triangles and which is maximized on thehyperoctahedral group orbit of the point √ √ (cid:0) √ , √ , (cid:1) ? And similarly for theother special points we have found? At the very least, we expect, based on numerical experimentsand O’Hara’s results on planar convex bodies [11], that these points are the limits of maxima ofsuitably renormalized Riesz-type potentials as the exponent goes to −∞ .The problem of extending this analysis to n -gons is substantial and interesting. As alluded toin the introduction, the spherical tiling induced by the hyperoctahedral group action on the sphere(Figure 2) generalizes to a decomposition of n -gon space for any n , namely the images of thefundamental domain of the standard hyperoctahedral group action on G ( R n ) . In Section 3 wefound the incenter of T and hence, by taking this point’s hyperoctahedral group orbit, of all theother triangles in the tiling. What are the incenters of the cells in the decomposition of n -gon space,which we can interpret as the least symmetric n -gon?Similarly, what is the least symmetric n -gon in R ? Just as n -gons in the plane are modeledby G ( R n ) , n -gons in space correspond to points in the complex Grassmannian G ( C n ) . Thenatural generalization of the hyperoctahedral group is given by replacing ( Z / n = ( O (1)) n with ( U (1)) n , yielding the group ( U (1)) n (cid:111) S n , which acts on G ( C n ) . The fundamental domain (inthe sense of Hermann [9]) of this action gives the space of unordered n -gons in R , and it seemschallenging to describe this domain and then to find the point furthest from the boundary.Finding the least symmetric triangle can be interpreted as finding the optimal shape satisfyingcertain constraints. As in Section 4, where we added the constraints that triangles should be obtuseor acute, other constraints are also interesting; for example, what is the most knotted trefoil knotin the space of -gons? Since our polygon model has generalizations to continuous curves in theplane [18] and to framed curves in space [10], we are not restricted to asking such questions onlyabout polygons.Finally, while we have presented the triangle from Proposition 1 as the least symmetric triangle,we could also think of it as the median triangle since it is equidistant from the three distinct piecesof the boundary of the region T parametrizing unordered triangles. This framing suggests theobvious question: what is the mean triangle? Or, indeed, the mean n -gon? We intend to addressthis question in future work. Rassat and Fowler [14] showed that any non-isosceles triangle is the most chiral triangle according to some chiralitymeasure in a particular family, but their chirality measures are slightly unnatural to define on the sphere. Meaning furthest from the subset of unknotted polygons. Acknowledgments
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