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Shape Invariant (Posted on 2008-03-31)

A (not necessarily regular) solid has F faces, each one of which has A sides. It also has V vertices, each of which is the meeting place of B faces.

Show that AF = BV

Submitted by FrankM

Rating: 2.5000 (2 votes)

Solution: (Hide)

AF and BV each give twice the number of edges in the solid.

AF gives the number of edges per face times the number of faces. But each edge occurs in two faces.

BV gives the number of faces (=# of edges) meeting at each vertex times the number of vertices; but each edge appears in two vertices.

Comments: ( You must be logged in to post comments.)

Subject	Author	Date
Tautology	Bractals	2008-03-31 14:05:25
Solution	Robby Goetschalckx	2008-03-31 10:00:00

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