[math-fun] Four solid angles
How many of you already know (or can figure out) what four solid angles associated with a general tetrahedron add up to 4 pi? I've been told that George Polya, when asked, replied with the correct answer instantly (it was part of his geometric toolkit), but that few of his contemporaries, when polled in an informal survey, were aware of this three-dimensional analogue of the familiar fact about the angles of a triangle summing to pi. I'm guessing that things haven't changed much in the intervening half-century. Jim Propp
On 2015-05-25 11:09, James Propp wrote:
How many of you already know (or can figure out) [w->t?]hat four solid angles associated with a general tetrahedron add up to 4 pi?
I've been told that George Polya, when asked, replied with the correct answer instantly (it was part of his geometric toolkit), but that few of his contemporaries, when polled in an informal survey, were aware of this three-dimensional analogue of the familiar fact about the angles of a triangle summing to pi. I'm guessing that things haven't changed much in the intervening half-century.
Jim Propp I don't get it. Consider the regular case. The four vertex trihedrals obviously sum to far short of 4pi. What other "associated" solid angles are there? Subdividing into four tetrahedra by joining the four vertices to an interior point creates four solid angles which obviously sum to 4pi. But for the sum of the four vertex solid angles I get 4pi/3. I.e., you could squeeze 12 tetrahedra (non-regular, but with a pi/3 solid angle) around a point.
Puzzle: Can a non-regular tetrahedron have equal solid angles? --rwg Now seems a good time to repeat some formulæ for the solid angle at a vertex formed with face angles a, b, c: solidAngle[a_, b_, c_] := {ArcCos[-1 + (1 + Cos[a] + Cos[b] + Cos[c])^2/((1 + Cos[a]) (1 + Cos[b]) (1 + Cos[c]))], 2 ArcCos[1/4 (1 + Cos[a] + Cos[b] + Cos[c]) Sec[a/2] Sec[b/2] Sec[c/2]], 2 ArcSin[1/(4 Sqrt[2]) Sqrt[-1 - Cos[2 a] - Cos[2 b] + 4 Cos[a] Cos[b] Cos[c] - Cos[2 c]] Sec[a/2] Sec[b/2] Sec[c/2]], 4 ArcTan[\[Sqrt](Tan[1/4 (a + b - c)] Tan[1/4 (a - b + c)] Tan[ 1/4 (-a + b + c)] Tan[1/4 (a + b + c)])]} There's some brown belt trig in showing those four equivalent. sphericalPolygonArea[r_, n_, anglesum_] := r^2*(anglesum - (n - 2)*Pi) Largest possible solid angle with vertex angles a,b,c,d: maxSolidAngle[a_, b_, c_, d_] := 2*ArcCos[((Cos[d] + Cos[c] + Cos[b] + Cos[a])/(4*Cos[a/2]*Cos[b/2]*Cos[c/2]* Cos[d/2])) - Tan[a/2]*Tan[b/2]*Tan[c/2]*Tan[d/2]] Dihedral between faces with vertex angles a and b: dihedral[a_, b_, c_] := ArcCos[Csc[a]*Csc[b]*Cos[c] - Cot[a]*Cot[b]] Apex and base solid angles for regular n-gon base, height h pyramidSolidAngle[n_, circumRadius_, h_] := List[2*Pi - 2*n*ArcTan[((h*Tan[Pi/n])/(Sqrt[circumRadius^2 + h^2]))], 2*ArcCos[((h^2*Sin[Pi/n])/(2*(circumRadius^2*(Cos[Pi/n])^2 + h^2))) + Csc[Pi/n]/2]] Apex and base solid angles for regular n-gon base, cos of angle between edge and altitude pyramidSolidAngle[n_, cos_] := {2*Pi - 2*n*ArcTan[cos*Tan[Pi/n]], 2*ArcCos[(Sin[Pi/n]/(2*(1 + Cos[\[Pi]/n]^2*(cos^-2 - 1)))) + Csc[Pi/n]/2]}
For a convex closed surface K in space, there is a possibly-multivalued map f: K -> S^2 to the unit sphere given by each point p of K is sent to the set of points of S^2 representing directions that are outward normal to the set of affine hyperplanes in space that intersect K in the point p (and possibly others), but that intersect no interior points of K. By pulling back the area measure dA from S^2 via f, this defines a measure mu = f*(dA) on K, and the total measure mu(K) = area(S^2) = 4pi. When K is a convex polyhedron, this measure mu is concentrated at the vertices. At each vertex v, mu({v}) is equal to the solid angle subtended by the cone dual to the local cone of K around v. (I.e., dual to the cone that is the union of the faces of K containing v — the set of lines that are "perpendicular" at the points of a small simple closed curve about v to the faces containing v, interpolating in the obvious way across edges.) This whole thing may also be thought of as the limit of the Gauss-Bonnet theorem applied to increasingly good approximations to K by smooth surfaces. ——Dan
On May 25, 2015, at 11:09 AM, James Propp <jamespropp@gmail.com> wrote:
How many of you already know (or can figure out) what four solid angles associated with a general tetrahedron add up to 4 pi?
I've been told that George Polya, when asked, replied with the correct answer instantly (it was part of his geometric toolkit), but that few of his contemporaries, when polled in an informal survey, were aware of this three-dimensional analogue of the familiar fact about the angles of a triangle summing to pi. I'm guessing that things haven't changed much in the intervening half-century.
G-B rather than G-P --- not the first time I have failed to spot a special case, bah! So how do these "Gaussian curvature" angles relate to (say) dihedral angles at edges, or to face angles? WFL On 5/25/15, Dan Asimov <dasimov@earthlink.net> wrote:
For a convex closed surface K in space, there is a possibly-multivalued map
f: K -> S^2
to the unit sphere given by each point p of K is sent to the set of points of S^2 representing directions that are outward normal to the set of affine hyperplanes in space that intersect K in the point p (and possibly others), but that intersect no interior points of K.
By pulling back the area measure dA from S^2 via f, this defines a measure
mu = f*(dA) on K,
and the total measure
mu(K) = area(S^2) = 4pi.
When K is a convex polyhedron, this measure mu is concentrated at the vertices. At each vertex v, mu({v}) is equal to the solid angle subtended by the cone dual to the local cone of K around v. (I.e., dual to the cone that is the union of the faces of K containing v — the set of lines that are "perpendicular" at the points of a small simple closed curve about v to the faces containing v, interpolating in the obvious way across edges.)
This whole thing may also be thought of as the limit of the Gauss-Bonnet theorem applied to increasingly good approximations to K by smooth surfaces.
——Dan
On May 25, 2015, at 11:09 AM, James Propp <jamespropp@gmail.com> wrote:
How many of you already know (or can figure out) what four solid angles associated with a general tetrahedron add up to 4 pi?
I've been told that George Polya, when asked, replied with the correct answer instantly (it was part of his geometric toolkit), but that few of his contemporaries, when polled in an informal survey, were aware of this three-dimensional analogue of the familiar fact about the angles of a triangle summing to pi. I'm guessing that things haven't changed much in the intervening half-century.
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
For a convex closed surface K in space, there is a possibly-multivalued map f: K -> S^2 to the unit sphere given by each point p of K is sent to the set of points of S^2 representing directions that are outward normal to the set of affine hyperplanes in space that intersect K in the point p (and possibly others), but that intersect no interior points of K. By pulling back the area measure dA from S^2 via f, this defines a measure mu = f*(dA) on K, and the total measure mu(K) = area(S^2) = 4pi. When K is a convex polyhedron, this measure mu is concentrated at the vertices. At each vertex v, mu({v}) is equal to the solid angle subtended by the cone dual to the local cone of K around v. (I.e., dual to the cone that is the union of the faces of K containing v — the set of lines that are "perpendicular" at the points of a small simple closed curve about v to the faces containing v, interpolating in the obvious way across edges.) This whole thing may also be thought of as the limit of the Gauss-Bonnet theorem applied to increasingly good approximations to K by smooth surfaces. ——Dan
On May 25, 2015, at 11:09 AM, James Propp <jamespropp@gmail.com> wrote:
How many of you already know (or can figure out) what four solid angles associated with a general tetrahedron add up to 4 pi?
I've been told that George Polya, when asked, replied with the correct answer instantly (it was part of his geometric toolkit), but that few of his contemporaries, when polled in an informal survey, were aware of this three-dimensional analogue of the familiar fact about the angles of a triangle summing to pi. I'm guessing that things haven't changed much in the intervening half-century.
For a tetrahedron, the construction can be simplified :--- Suppose the tetrahedron scaled to circumradius unity, and project its faces' circumcentres along radial lines from its circumcentre onto its circumsphere. The four spherical triangles with these points as vertices dissect the sphere, and via Dan's observation each has area equal to the (limiting) Gaussian curvature at the tetrahedral vertex within. WFL On 5/25/15, Dan Asimov <dasimov@earthlink.net> wrote:
For a convex closed surface K in space, there is a possibly-multivalued map
f: K -> S^2
to the unit sphere given by each point p of K is sent to the set of points of S^2 representing directions that are outward normal to the set of affine hyperplanes in space that intersect K in the point p (and possibly others), but that intersect no interior points of K.
By pulling back the area measure dA from S^2 via f, this defines a measure
mu = f*(dA) on K,
and the total measure
mu(K) = area(S^2) = 4pi.
When K is a convex polyhedron, this measure mu is concentrated at the vertices. At each vertex v, mu({v}) is equal to the solid angle subtended by the cone dual to the local cone of K around v. (I.e., dual to the cone that is the union of the faces of K containing v — the set of lines that are "perpendicular" at the points of a small simple closed curve about v to the faces containing v, interpolating in the obvious way across edges.)
This whole thing may also be thought of as the limit of the Gauss-Bonnet theorem applied to increasingly good approximations to K by smooth surfaces.
——Dan
On May 25, 2015, at 11:09 AM, James Propp <jamespropp@gmail.com> wrote:
How many of you already know (or can figure out) what four solid angles associated with a general tetrahedron add up to 4 pi?
I've been told that George Polya, when asked, replied with the correct answer instantly (it was part of his geometric toolkit), but that few of his contemporaries, when polled in an informal survey, were aware of this three-dimensional analogue of the familiar fact about the angles of a triangle summing to pi. I'm guessing that things haven't changed much in the intervening half-century.
Dan's got it. (Why am I not surprised?) Putting it succinctly: At each vertex of the tetrahedron, record the solid angle between the (outward-going) vectors perpendicular to the three faces that meet at that vertex. These four solid angles sum to 4 pi. Indeed, if you transport those vectors so that their tails are at the origin, you can see that those solid angles partition the sphere. So now, the next time anyone asks you for "the" 3D analogue of the sum-of-the-angles-of-a-triangle theorem, you, like Polya, can respond instantly. This theorem generalizes to n-dimensional space, for all n. (It's worthwhile taking a few seconds to consider the case n=2, to see how it compares with (a) the familiar fact about the sum of the interior angles of a triangle, and (b) the "baby Gauss-Bonnet" formula that says that as you go around a triangle, the total turning-angle is 2 pi.) Jim Propp On Mon, May 25, 2015 at 5:35 PM, Dan Asimov <dasimov@earthlink.net> wrote:
For a convex closed surface K in space, there is a possibly-multivalued map
f: K -> S^2
to the unit sphere given by each point p of K is sent to the set of points of S^2 representing directions that are outward normal to the set of affine hyperplanes in space that intersect K in the point p (and possibly others), but that intersect no interior points of K.
By pulling back the area measure dA from S^2 via f, this defines a measure
mu = f*(dA) on K,
and the total measure
mu(K) = area(S^2) = 4pi.
When K is a convex polyhedron, this measure mu is concentrated at the vertices. At each vertex v, mu({v}) is equal to the solid angle subtended by the cone dual to the local cone of K around v. (I.e., dual to the cone that is the union of the faces of K containing v — the set of lines that are "perpendicular" at the points of a small simple closed curve about v to the faces containing v, interpolating in the obvious way across edges.)
This whole thing may also be thought of as the limit of the Gauss-Bonnet theorem applied to increasingly good approximations to K by smooth surfaces.
——Dan
On May 25, 2015, at 11:09 AM, James Propp <jamespropp@gmail.com> wrote:
How many of you already know (or can figure out) what four solid angles associated with a general tetrahedron add up to 4 pi?
I've been told that George Polya, when asked, replied with the correct answer instantly (it was part of his geometric toolkit), but that few of his contemporaries, when polled in an informal survey, were aware of this three-dimensional analogue of the familiar fact about the angles of a triangle summing to pi. I'm guessing that things haven't changed much in the intervening half-century.
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participants (6)
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Adam P. Goucher -
Dan Asimov -
Dan Asimov -
Fred Lunnon -
James Propp -
rwg