Difference between revisions of "2004 AIME I Problems/Problem 3"
I like pie (talk | contribs) m (AIME box) |
15Pandabears (talk | contribs) (→Solution) |
||
| (2 intermediate revisions by 2 users not shown) | |||
| Line 1: | Line 1: | ||
== Problem == | == Problem == | ||
| − | 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? | + | A [[convex]] polyhedron <math> P </math> has <math>26</math> vertices, <math>60</math> edges, and <math>36</math> faces, <math>24</math> of which are triangular and <math>12</math> 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 [[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. | + | 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, <math>60</math> are edges. Each [[triangle|triangular]] face has <math>0</math> face diagonals and each [[quadrilateral]] face has <math>2</math>, so there are <math>2 \cdot 12 = 24</math> face diagonals. This leaves <math>325 - 60 - 24 = \boxed{241}</math> segments to be the space diagonals. |
== See also == | == See also == | ||
{{AIME box|year=2004|n=I|num-b=2|num-a=4}} | {{AIME box|year=2004|n=I|num-b=2|num-a=4}} | ||
| + | |||
| + | [[Category:Intermediate Combinatorics Problems]] | ||
| + | [[Category:Intermediate Geometry Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 16:56, 1 January 2016
Problem
A convex polyhedron 
has 
vertices, 
edges, and 
faces, 
of which are triangular and 
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 line segments determined by the vertices. Of these, are edges. Each triangular face has 
face diagonals and each quadrilateral face has 
, so there are 
face diagonals. This leaves 
segments to be the space diagonals.
See also
| 2004 AIME I (Problems • Answer Key • Resources) | ||
| Preceded by Problem 2 |
Followed by Problem 4 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
