Author | Lightnews

Joseph O'Rourke @josephorourke.bsky.social · 6d

These triangles are known to have a periodic billiard path: (1) All acute triangles. (2) All right triangles. (3) All rational triangles. (4) All obtuse triangles with obtuse angle smaller than 5 pi/8 (the 112.4 deg that I quoted). #MathSky #Mathematics #Geometry #Billiards

5

Joseph O'Rourke @josephorourke.bsky.social · 10d

Sharp eyes to notice the two perpendicular bounces. Probably not for all triangles, I agree.

1

Joseph O'Rourke @josephorourke.bsky.social · 10d

Beautiful indeed. And with recent results from the study of translation surfaces.

1

Joseph O'Rourke @josephorourke.bsky.social · 10d

It is *still* unknown whether or not every triangle admits a periodic billiard trajectory. Every triangle with rational angles does. And so does every obtuse triangle of at most 112.4 deg. "112.5 appears to be a natural barrier."
gwtokarsky.github.io. #MathSky #Mathematics #Geometry #Billiards

5 9 54

Joseph O'Rourke @josephorourke.bsky.social · Sep 27

Sure. Have them email me, [email protected].

2

Joseph O'Rourke @josephorourke.bsky.social · Sep 26

Stoker's Conjecture settled by Cho & Kim positively: Every 3D polyhedron is uniquely determined by its dihedral angles and edge lengths, even if nonconvex or self-intersecting (subject to technical restrictions).
doi.org/10.1007/s004...
#MathSky #Mathematics #Geometry #Polyhedra

2 13

Joseph O'Rourke @josephorourke.bsky.social · Sep 25

See also: "Why can't a nonabelian group be 75% abelian?" mathoverflow.net/q/211159/6094

1

Joseph O'Rourke @josephorourke.bsky.social · Sep 22

What is the probability that 4 points chosen uniformly at random on surface of a sphere form a tetrahedron whose four faces are each acute? Asked on MathOverflow (mathoverflow.net/q/498296/6094) with evidence that the answer is 1/12. But not yet resolved.
#MathSky #Mathematics #Geometry #Probability

4 22

Joseph O'Rourke @josephorourke.bsky.social · Sep 17

A monohedral tiling of the plane by "spandrelized" squares.
Each unit square includes a circular arc of a 1/2-radius circle centered at each vertex.
Adams, Colin. "Spandrelized Tilings." Amer. Math. Monthly 132, no. 3 (2025): 199-217.

doi.org/10.1080/0002...
#MathSky #Mathematics #Geometry #Tiling

3 12

Joseph O'Rourke @josephorourke.bsky.social · Sep 14

I wonder in which dimensions is the cylinder/sphere volume ratio rational?
#Mathematics #MathSky #Geometry

1 1

Joseph O'Rourke @josephorourke.bsky.social · Sep 14

Archimedes: "Every cylinder whose base is the greatest circle in a sphere and whose height is equal to the diameter of the sphere has a volume equal to 3/2 the volume of the sphere." Cicero found Archimedes' tomb ~137 yrs later with his famous theorem represented.

#Mathematics #MathSky #Geometry

2 1 10

Joseph O'Rourke @josephorourke.bsky.social · Sep 11

New tiling results on the arXiv, one of which says that determining whether or not two connected polycubes can together tile R^3 is undecidable (Cor. 5.5). A polycube is an object built by gluing cubes face-to-face. (Unrelated fig.)
arxiv.org/abs/2509.07906
#MathSky #Mathematics #Geometry #Tiling

4 20

Joseph O'Rourke @josephorourke.bsky.social · Aug 31

p.4 of their paper details the construction. "the Noperthedron has 3·30=90 vertices." They set three pts C1,C2,C3 and then apply the cyclic group C_30 to each.

1 2

Joseph O'Rourke @josephorourke.bsky.social · Aug 30

Believe it or not, origami stents have been explored: Kuribayashi et al., "Self-deployable origami stent grafts ..."
(doi.org/10.1016/j.ms...)
Here I show a hexagonal design built with origami waterbomb crease patterns.
cs.smith.edu/~jorourke/Ma...
#Mathematics #Geometry #MathSky

1 2 17

Joseph O'Rourke @josephorourke.bsky.social · Aug 28

The Rupert property requires the convex polyhedron P to tunnel by translation through an isometric copy of P. I wonder if twisting while translating would permit any P---even the "Noperthedron"---to pass through itself? #Mathematics #Geometry #MathSky

1 2

Joseph O'Rourke @josephorourke.bsky.social · Aug 27

Yes, the authors clearly had fun! :-)

2

Joseph O'Rourke @josephorourke.bsky.social · Aug 27

The conjecture that every convex polyhedron is Rupert is settled in the negative! The convex body in the image cannot pass straight through a hole inside itself.
arxiv.org/abs/2508.18475
#Mathematics #Geometry #MathSky

1 8 36

Joseph O'Rourke @josephorourke.bsky.social · Jun 25

*Visual Complex Analysis* by Tristan Needham. global.oup.com/academic/pro...

global.oup.com

4

Joseph O'Rourke @josephorourke.bsky.social · Jun 23

A surprising result: 3-space can be filled with disjoint geometric unit-radius circles. So each point of R^3 lies on exactly one circle. The circles may even be chosen to be unlinked. M. Jonsson and J. Wästlund: www.jstor.org/stable/24493....
#MathSky #Geometry #Mathematics

PARTITIONS OF R 3 INTO CURVES on JSTOR

M. JONSSON, J. WÄSTLUND, PARTITIONS OF R 3 INTO CURVES, Mathematica Scandinavica, Vol. 83, No. 2 (1998), pp. 192-204

www.jstor.org

2 8

Joseph O'Rourke @josephorourke.bsky.social · Jun 10

Henry Cohn figured that out: Again, not the cube:
mathoverflow.net/a/73941/6094

Optimal 8-vertex isoperimetric polyhedron?

I know from Marcel Berger's Geometry Revealed: A Jacob's Ladder to Modern Higher Geometry (p.531) that it is not yet established which polyhedron in $\mathbb{R}^3$ on 8 vertices achieves the optimal

mathoverflow.net

3

Joseph O'Rourke @josephorourke.bsky.social · Jun 10

You might guess that the maximal volume 8-vertex polyhedron inscribed in a unit sphere is the cube. But it's not even close : cube 1.54; 8-vertex max 1.82. Proved by Berman and Hanes in 1970. V=8, E=16, F=10. #MathSky #Geometry #Mathematics

Max volume 8-vertex inscribed polyhedron.

3 3 27

Joseph O'Rourke @josephorourke.bsky.social · May 31

Angel-wing net (edge-unfolding) of a nearly flat prismoid, top & bottom two 40-vertex regular polygons. No mathematical significance, just an attractive image. (The two red edges are not cut.) #MathSky #Geometry #Mathematics #MathArt

Net for 40-vertex top & bottom prismoid.

13

Joseph O'Rourke @josephorourke.bsky.social · May 27

Yes, and so works in any dimension. But only in 3D does it look like eyeballs! :-)

3

Joseph O'Rourke @josephorourke.bsky.social · May 26

The Eye-Ball Theorem: Two disjoint spheres S1 and S2. Form cone C1 tangent to S1 with apex at the center of S2, and form cone C2 similarly. Then the radii of the circles of cone/sphere intersections (red) are equal. #MathSky #Geometry

5 11 66

Joseph O'Rourke @josephorourke.bsky.social · May 11

Among every set of six points in 3-space (in general position) are two linked triangles: The Conway-Gordon-Sachs theorem. General position excludes three points collinear and four points coplanar. #MathSky #Geometry

1 4 20