A flat ball? (Posted on 2004-04-08)

	Submitted by Federico Kereki
Rating: 3.7500 (4 votes)
Solution:	(Hide)
There must be 12 pentagons, but the number of hexagons may vary, and even be zero. (A quick way of solving this: assuming there IS one answer, if we use no hexagons, we would have a regular polyhedron built out of pentagons: a dodecahedron, with 12 pentagons.) A proof: according to Euler's theorem, E+2=F+V (E=edges, F=faces, V=vertices). In our case, if there are H hexagons and P pentagons, F=H+P, E=(6H+5P)/2 (since every edge belongs to two faces) and V=(6H+5P)/3 (since vertices appear at the junction of three faces). Substituting and equating, you get P=12.

	Subject	Author	Date
	Puzzle Answer	K Sengupta	2022-09-03 22:13:53
	Soccer balls	broll	2016-04-16 02:38:46
	An interesting link (non-regular)	Jer	2004-05-03 09:43:05
	re(2): Johnson solids	Jer	2004-04-27 14:24:22
	re: Johnson solids	Thalamus	2004-04-15 12:32:08
	Johnson solids	Jer	2004-04-15 12:03:24
	re(3): The 720 degree deficit	Tristan	2004-04-13 16:36:27
	re: A related question...	Brian Smith	2004-04-13 14:26:51
	A related question...	SilverKnight	2004-04-13 12:20:41
	re(3): The 720 degree deficit - Euler's formula	SilverKnight	2004-04-12 12:48:21
	re(2): The 720 degree deficit	Tristan	2004-04-12 10:32:31
	re(2): The 720 degree deficit - Euler's formula	Brian Smith	2004-04-12 10:26:10
	re: The 720 degree deficit	Jer	2004-04-12 09:47:25
	Pentagons and Hexagons	Brian Smith	2004-04-12 09:15:46
	The 720 degree deficit	Larry	2004-04-09 19:34:23
	The 720 degree deficit	Larry	2004-04-09 19:34:19
	So far, no cigar...	Federico Kereki	2004-04-09 18:51:00
	re(2): Hint in the title?	Jer	2004-04-09 14:10:40
	re: Hint in the title?	Brian Wainscott	2004-04-09 13:06:20
	Hint in the title?	Jer	2004-04-09 09:37:29
	Bend it like Beckham	Penny	2004-04-08 16:13:08
	answer	Billy Bob	2004-04-08 15:37:57
	re: My thoughts	Thalamus	2004-04-08 15:30:20
	My thoughts	Bryan	2004-04-08 15:28:25
	My thoughts	Bryan	2004-04-08 15:26:09