A common 6-in.-radius soccer ball contains 12 pentagons arranged so that every pentagon is separated from the next by the same arc length as one of the spherical (great circle segment) sides of the regular hexagons. As the hexagons are regular, this is the same arc length as one of the sides of the pentagons, as the pentagons also border the hexagons.
Calculate the arc length of a pentagon's side of a new soccer ball using the same radius and instead of one line of separation between pentagons, use two lines of separation between pentagons and consider every new line with a distance equal to a side of a pentagon.
(See picture)
Note: The endpoints of the mentioned lines intersect with the surface of the soccer ball or sphere.
(In reply to
solution by Charlie)
Charlie,
While it's true that "five fit together to completely surround a vertex", each triangle is an equilateral triangle and therefore has three 60-degree interior angles.
If we project the vertex in question down (towards the center of the ball) until it reaches the plane containing the pentagon in question (centered on that vertex), then indeed the triangles are no longer equilateral and they have one angle of 72-degrees (and two angles of 54-degrees).
But now we're not talking about 72 degrees from the center of the ball, but rather from the center of that pentagon.
Am I misunderstanding what you wrote?
- SK
re: solution
