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

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== Problem ==
 
== Problem ==
A [[convex]] [[polyhedron]] <math> P </math> has 26 [[vertex | vertices]], 60 [[edge]]s, and 36 [[face]]s, 24 of which are [[triangle|triangular]], and 12 of which are [[quadrilateral]]s. 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 <math> P </math> have?
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A [[convex]] polyhedron <math> P </math> 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 <math> P </math> have?
  
 
== Solution ==
 
== Solution ==
Every pair of vertices of the polyhedron determines either an edge, a face diagonal or a space diagonal.  We have <math>{26 \choose 2} = \frac{26\cdot25}2 = 325</math> total 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 <math>2 \cdot 12 = 24</math> face diagonals.  This leaves <math>325 - 60 - 24 = 241</math> segments to be the space diagonals.
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Every pair of [[vertex | vertices]] of the [[polyhedron]] determines either an [[edge]], a face [[diagonal]] or a space diagonal.  We have <math>{26 \choose 2} = \frac{26\cdot25}2 = 325</math> total [[line segment]]s determined by the vertices.  Of these, 60 are edges.  Each [[triangle|triangular]] face has 0 face diagonals and each [[quadrilateral]] face has 2, so there are <math>2 \cdot 12 = 24</math> face diagonals.  This leaves <math>325 - 60 - 24 = 241</math> segments to be the space diagonals.
  
 
== See also ==
 
== See also ==
 
* [[2004 AIME I Problems]]
 
* [[2004 AIME I Problems]]

Revision as of 11:42, 29 July 2006

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