Difference between revisions of "2004 AIME I Problems/Problem 3"

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== See also ==
 
== See also ==
* [[2004 AIME I Problems/Problem 2| Previous problem]]
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* [[2004 AIME I Problems/Problem 4| Next problem]]
 
 
 
* [[2004 AIME I Problems]]
 

Revision as of 15:00, 27 April 2008

Problem

A convex polyhedron $P$ has 26 vertices, 60 edges, and 36 faces, 24 of which are triangular and 12 of which are quadrilaterals. A space diagonal is a line segment connecting two non-adjacent vertices that do not belong to the same face. How many space diagonals does $P$ have?

Solution

Every pair of vertices of the polyhedron determines either an edge, a face diagonal or a space diagonal. We have ${26 \choose 2} = \frac{26\cdot25}2 = 325$ total line segments determined by the vertices. Of these, 60 are edges. Each triangular face has 0 face diagonals and each quadrilateral face has 2, so there are $2 \cdot 12 = 24$ face diagonals. This leaves $325 - 60 - 24 = 241$ segments to be the space diagonals.

See also

2004 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 2
Followed by
Problem 4
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