tag:blogger.com,1999:blog-3105052.post5638170092321751646..comments2021-04-02T06:45:33.984+01:00Comments on A Welsh Born Icon: Soccer for geeks (Pay attention there at the back Callum!)Nick Brownehttp://www.blogger.com/profile/11805953367604412662noreply@blogger.comBlogger3125tag:blogger.com,1999:blog-3105052.post-63583455764085103092021-02-11T16:44:04.634+00:002021-02-11T16:44:04.634+00:00On reflection you are right. I conflated vertices ...On reflection you are right. I conflated vertices and edges and left out faces in cut and paste. It should have been:<br />...... number of vertices, V = (6*H + 5*P)/3.<br /><br />Number of edges will be:<br />6 edges for each of the hexagons, i.e. 6*H.<br />5 edges for each of the pentagons, i.e. 5*P.<br />But we have counted each edge twice, once for each adjacent polygon, follow the picture<br /><br />Hence, number of edges, E = (6*H + 5*P)/2.<br /><br />Number of faces will be:<br />There are H hexagons and P pentagons, each forming a face. Hence, total number of faces, F = (H + P) .......Nick Brownehttps://www.blogger.com/profile/11805953367604412662noreply@blogger.comtag:blogger.com,1999:blog-3105052.post-34032825143739965152021-02-11T14:20:21.370+00:002021-02-11T14:20:21.370+00:00I think I expressed myself infelicitously rather t...I think I expressed myself infelicitously rather than being wrongNick Brownehttps://www.blogger.com/profile/11805953367604412662noreply@blogger.comtag:blogger.com,1999:blog-3105052.post-64289130907782553402021-02-10T11:47:25.172+00:002021-02-10T11:47:25.172+00:00Each vertex 3 times but each edge 2 times - hence ...Each vertex 3 times but each edge 2 times - hence the division by 3 and by 2.<br /><br />Do be clear Browne!John Brownenoreply@blogger.com