Difference between revisions of "Mock AIME 5 2005-2006 Problems/Problem 8"
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== Problem == | == Problem == | ||
| + | Let <math>P</math> be a polyhedron with <math>37</math> faces, all of which are equilateral triangles, squares, or regular pentagons with equal side length. Given there is at least one of each type of face and there are twice as many pentagons as triangles, what is the sum of all the possible number of vertices <math>P</math> can have? | ||
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| + | == Solution == | ||
== Solution == | == Solution == | ||
Latest revision as of 21:18, 8 October 2014
Contents
Problem
Let 
be a polyhedron with 
faces, all of which are equilateral triangles, squares, or regular pentagons with equal side length. Given there is at least one of each type of face and there are twice as many pentagons as triangles, what is the sum of all the possible number of vertices can have?
Solution
Solution
See also
| Mock AIME 5 2005-2006 (Problems, Source) | ||
| Preceded by Problem 7 |
Followed by Problem 9 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
