Posted 12 May 2013 · Report post A regulation golf ball is spherical and has 384 dimples, arranged in a triangular pattern. Most of the dimples are surrounded by six other dimples, but some are surrounded by only five. How many dimples have only five neighbors? 0 Share this post Link to post Share on other sites
0 Posted 12 May 2013 · Report post Twelve. Hexagons infinitely tile the plane, while pentagons (12 of them) tile the fifth platonic solid - the dodecahedron. It takes vertices that join polygons with angles totaling less than 360 to create a "curved" non-planar surface. Since 12 pentagons do it for the dodecahedron, I'm betting 12 instances of a dimple being surrounded by (only) 5 other dimples do the same thing for a golf ball. The remaining "hexagonaly" placed dimples form (distorted planar) regions that cover the ball among these twelve curvature-generating points. 0 Share this post Link to post Share on other sites
0 Posted 12 May 2013 · Report post Twelve. Hexagons infinitely tile the plane, while pentagons (12 of them) tile the fifth platonic solid - the dodecahedron. It takes vertices that join polygons with angles totaling less than 360 to create a "curved" non-planar surface. Since 12 pentagons do it for the dodecahedron, I'm betting 12 instances of a dimple being surrounded by (only) 5 other dimples do the same thing for a golf ball. The remaining "hexagonaly" placed dimples form (distorted planar) regions that cover the ball among these twelve curvature-generating points. but can someone prove it? 0 Share this post Link to post Share on other sites
0 Posted 12 May 2013 · Report post The dodecahedron is a proof by construction: Tile the 12 pentagonal faces with hexagons. Deform the surface into a sphere. What is left to prove? possibly, Show that no other construction (with a number different from twelve) exists. Put the statements of the above proof into equations. 0 Share this post Link to post Share on other sites
0 Posted 12 May 2013 · Report post Hexagons tile the plane. The surface of a sphere closes upon itself. If we wrap the plane into a cylinder, leaving an axis, a hex tiling can absorb the closure. But closing perpendicular to the axis as well, disturbs the tiling in a way that forces the angles at some nodes to sum less than 360. Perhaps there is a consequence regarding nodes and edges (governed by Euler's rule) that would dictate how many of these points have to occur. That could then be a proof. But I don't know how to formulate it. 0 Share this post Link to post Share on other sites
0 Posted 12 May 2013 · Report post Hexagons tile the plane. The surface of a sphere closes upon itself. If we wrap the plane into a cylinder, leaving an axis, a hex tiling can absorb the closure. But closing perpendicular to the axis as well, disturbs the tiling in a way that forces the angles at some nodes to sum less than 360. Perhaps there is a consequence regarding nodes and edges (governed by Euler's rule) that would dictate how many of these points have to occur. That could then be a proof. But I don't know how to formulate it. Yes...using Euler is the way to go in the proof. 0 Share this post Link to post Share on other sites
0 Posted 13 May 2013 · Report post OK I have it. If we draw blue lines to connect dimples with their nearest neighbors and red lines as perpendicular bisectors of the blue lines, the red lines form the edges of a polyhedron. They make hexagons around dimples with six nearest neighbors and pentagons around dimples with five. Each edge is common to two faces, and each vertex is common to three faces. In forming the polyhedron, hexagons and pentagons each contribute 1 face, but they contribute differently to the edge and vertex count: each hexagon associates with (6/2=3) edges and (6/3=2) vertices, while each pentagon pairs with (5/2) edges and (5/3) vertices. Euler says these numbers grow and decrease in lock-step. They are not independent. For any polyhedron,faces + vertices - edges = 2. Let's add things up.f = h + p (total faces = #hexagons + #pentagons)v = 2h + (5/3)p (total vertices)e = 3h + (5/2)p (total edges) 6f = 6h + 6p 6v = 12h + 10p 6e = 18h + 15p6(f+v-e) = 6(2) = (6+12-18)h + (6+10-15)pp = 12. Note that Euler places no restrictions on the number of hexagons: they can tile the infinite plane. But to close that tiled surface into a polyhedron requires us to add somewhere to the mix exactly 12 pentagons. So on that golf ball, populated mainly by dimples having 6 nearest neighbors, somewhere there are 12 dimples that have 5 nearest neighbors. 0 Share this post Link to post Share on other sites
0 Posted 13 May 2013 · Report post Well done Bonanova! 0 Share this post Link to post Share on other sites
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