Difference between revisions of "Euler's Polyhedral Formula"
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<math> \begin{tabular}{|c|c|c|c|}\hline Shape & Vertices & Edges & Faces\\ \hline Tetrahedron &4 &6 & 4 \\ \hline Cube/Hexahedron & 8 & 12 & 6\\ \hline Octahedron & 6 & 12 & 8\\ \hline Dodecahedron & 20 & 30 & 12\\ \hline \end{tabular} </math> | <math> \begin{tabular}{|c|c|c|c|}\hline Shape & Vertices & Edges & Faces\\ \hline Tetrahedron &4 &6 & 4 \\ \hline Cube/Hexahedron & 8 & 12 & 6\\ \hline Octahedron & 6 & 12 & 8\\ \hline Dodecahedron & 20 & 30 & 12\\ \hline \end{tabular} </math> | ||
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| + | ==Problem== | ||
| + | A convex polyhedron has for its faces 12 squares, 8 regular hexagons, and 6 regular octagons. At each vertex of the polyhedron one square, one hexagon, and one octagon meet. How many segments joining vertices of the polyhedron lie in the interior of the polyhedron rather than along an edge or a face? (1988 AIME #10) | ||
== See Also == | == See Also == | ||
Revision as of 22:34, 18 March 2017
Let 
be any convex polyhedron, and let 
, 
and 
denote the number of vertices, edges, and faces, respectively. Then 
.
Observe!
Apply Euler's Polyhedral Formula on the following polyhedra:
Problem
A convex polyhedron has for its faces 12 squares, 8 regular hexagons, and 6 regular octagons. At each vertex of the polyhedron one square, one hexagon, and one octagon meet. How many segments joining vertices of the polyhedron lie in the interior of the polyhedron rather than along an edge or a face? (1988 AIME #10)
See Also
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