Difference between revisions of "Diagonal"
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The number of diagonals of a polygon with <math>n</math> vertices is given by <math>\frac{n(n-3)}{2}</math>. | The number of diagonals of a polygon with <math>n</math> vertices is given by <math>\frac{n(n-3)}{2}</math>. | ||
====Proofs==== | ====Proofs==== | ||
| − | Proof 1: By direct counting. | + | =====Proof 1: By direct counting.===== |
Pick any vertex <math>P</math> of the <math>n</math>-gon. There are <math>n-3</math> diagonals with this point as an endpoint, one for every vertex except <math>P</math> itself and the two vertices adjacent to <math>P</math> (because the segments connecting <math>P</math> to those two points give edges, not diagonals). There are <math>n</math> vertices and for each vertex there are <math>n-3</math> diagonals so there are <math>n(n-3)</math> diagonals. However, when counting our diagonals, we counted each of them twice, once for each endpoint. So, we simply divide by 2 to get our final formula, <math>\frac{n(n-3)}{2}</math>. | Pick any vertex <math>P</math> of the <math>n</math>-gon. There are <math>n-3</math> diagonals with this point as an endpoint, one for every vertex except <math>P</math> itself and the two vertices adjacent to <math>P</math> (because the segments connecting <math>P</math> to those two points give edges, not diagonals). There are <math>n</math> vertices and for each vertex there are <math>n-3</math> diagonals so there are <math>n(n-3)</math> diagonals. However, when counting our diagonals, we counted each of them twice, once for each endpoint. So, we simply divide by 2 to get our final formula, <math>\frac{n(n-3)}{2}</math>. | ||
| − | Proof 2: By minor trickery. | + | =====Proof 2: By minor trickery.===== |
Every two vertices determine either an edge or a diagonal. Thus, there are a total of <math>n \choose 2</math> edges and diagonals. Since there are exactly <math>n</math> edges, this leaves <math>{n \choose 2} - n = \frac{n(n-1)}2 - n = \frac{n(n-3)}2</math> diagonals. | Every two vertices determine either an edge or a diagonal. Thus, there are a total of <math>n \choose 2</math> edges and diagonals. Since there are exactly <math>n</math> edges, this leaves <math>{n \choose 2} - n = \frac{n(n-1)}2 - n = \frac{n(n-3)}2</math> diagonals. | ||
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Polyhedra have two different kinds of diagonals, face diagonals and space diagonals. A face diagonal of a [[polyhedron]] is a diagonal of one of the [[face]]s of the polyhedron, while a space diagonal is any segment joining two vertices which is neither an edge nor a face diagonal. | Polyhedra have two different kinds of diagonals, face diagonals and space diagonals. A face diagonal of a [[polyhedron]] is a diagonal of one of the [[face]]s of the polyhedron, while a space diagonal is any segment joining two vertices which is neither an edge nor a face diagonal. | ||
| − | For example, [[Tetrahedron|tetrahedra]] have no space or face diagonals. [[Octahedron|Octahedra]] have no face diagonals but have 3 space diagonals. [[Cube]]s have 12 face diagonals (2 on each face) and 4 space diagonals. | + | For example, [[Tetrahedron|tetrahedra]] have no space or face diagonals. [[Octahedron|Octahedra]] have no face diagonals but have 3 space diagonals. [[Cube (geometry) | Cube]]s have 12 face diagonals (2 on each face) and 4 space diagonals. |
Revision as of 11:28, 30 October 2006
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Contents
Polygons
A diagonal of a polygon is any segment joining two vertices other than an edge.
Triangles have no diagonals, while convex quadrilaterals have two interior diagonals, and concave quadrilaterals have one interior and one exterior diagonal.
Number of diagonals
The number of diagonals of a polygon with 
vertices is given by 
.
Proofs
Proof 1: By direct counting.
Pick any vertex 
of the -gon. There are 
diagonals with this point as an endpoint, one for every vertex except itself and the two vertices adjacent to (because the segments connecting to those two points give edges, not diagonals). There are vertices and for each vertex there are diagonals so there are 
diagonals. However, when counting our diagonals, we counted each of them twice, once for each endpoint. So, we simply divide by 2 to get our final formula, .
Proof 2: By minor trickery.
Every two vertices determine either an edge or a diagonal. Thus, there are a total of 
edges and diagonals. Since there are exactly edges, this leaves 
diagonals.
Polyhedra
Polyhedra have two different kinds of diagonals, face diagonals and space diagonals. A face diagonal of a polyhedron is a diagonal of one of the faces of the polyhedron, while a space diagonal is any segment joining two vertices which is neither an edge nor a face diagonal.
For example, tetrahedra have no space or face diagonals. Octahedra have no face diagonals but have 3 space diagonals. Cubes have 12 face diagonals (2 on each face) and 4 space diagonals.
Following the same method as Proof 2 for polygons, we see that the number of edges plus the number of face diagonals plus the number of space diagonals of a polyhedron with vertices is equal to .
2004 AIME I problem 3 is a problem related to polyhedral diagonals.
