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Shape Invariant (Posted on 2008-03-31) Difficulty: 2 of 5

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

See The Solution Submitted by FrankM    
Rating: 2.5000 (2 votes)

Comments: ( You must be logged in to post comments.)
  Subject Author Date
TautologyBractals2008-03-31 14:05:25
SolutionSolutionRobby Goetschalckx2008-03-31 10:00:00
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