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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: ( Back to comment list | You must be logged in to post comments.)
Tautology Comment 2 of 2 |
Both sides of the equation count the number of face vertices.
  Posted by Bractals on 2008-03-31 14:05:25
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