A traditional soccer ball has 20 regular (spherical) hexagons with 12 regular pentagons situated so each is surrounded by five of the hexagons.
Given such a ball that's 22 cm in diameter, what is the arc length of one side where two polygons meet (penta/hexa or hexa/hexa) in centimeters?
(In reply to
re(2): Solution by Jer)
Take for example a regular octahedron which is then puffed out to a sphere:
If you say the "circumference" is across four triangles, alternating apex and midpoint of base, that circumference is 2*sqrt(3) times one of the edges. If the circumference is taken to be the common base of two square pyramids that are back-to-back, this circumference it 4 times one of the edges. One of these two methods must be wrong as they disagree.
When puffed out to a sphere that octahedron traces out octants of the sphere. It's circumference is indeed 4 times the length of one arc of the triangle formed, but that's only because the tracing of the "circumference" of the polyhedron did not go across any of the plane faces in that method.
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Posted by Charlie
on 2024-06-30 07:15:55 |
re(3): Solution
