Difference between revisions of "2004 AIME I Problems/Problem 3"
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== Problem == | == Problem == | ||
| − | A convex polyhedron <math> P </math> has 26 vertices, 60 | + | 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? |
== 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. | ||
== See also == | == See also == | ||
* [[2004 AIME I Problems]] | * [[2004 AIME I Problems]] | ||
Revision as of 09:57, 29 July 2006
Problem
A convex polyhedron 
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 have?
Solution
Every pair of vertices of the polyhedron determines either an edge, a face diagonal or a space diagonal. We have 
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 
face diagonals. This leaves 
segments to be the space diagonals.
