Difference between revisions of "Euler's Polyhedral Formula"

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Let <math>P</math> be any [[convex]] [[polyhedron]], and let <math>V</math>, <math>E</math> and <math>F</math> denote the number of [[vertex|vertices]], [[edge]]s, and [[face]]s, respectively. Then <math>V-E+F=2</math>
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Let <math>P</math> be any [[convex]] [[polyhedron]], and let <math>V</math>, <math>E</math> and <math>F</math> denote the number of [[vertex|vertices]], [[edge]]s, and [[face]]s, respectively. Then <math>V-E+F=2</math>.
  
 
==Observe!==
 
==Observe!==

Revision as of 14:53, 24 April 2020

Let $P$ be any convex polyhedron, and let $V$, $E$ and $F$ denote the number of vertices, edges, and faces, respectively. Then $V-E+F=2$.

Observe!

Apply Euler's Polyhedral Formula on the following polyhedra:

$\begin{tabular}{|c|c|c|c|}\hline Shape & Vertices & Edges & Faces\\ \hline Tetrahedron &4  &6 & 4 \\ \hline Cube/Hexahedron & 8 & 12 & 6\\ \hline Octahedron & 6 & 12 & 8\\ \hline Dodecahedron & 20 & 30 & 12\\ \hline \end{tabular}$

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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