Diagonals and Axes of Symmetry

Date: 03/31/98 at 03:22:48
From: Paul Dobing
Subject: Polygons - Diagonals and Symmetry

My daughter is in 5th grade and working on diagonals and axes of 
symmetry of polygons. It was presented to her that a regular octagon 
has 8 diagonals and 8 axes of symmetry. Could you explain the concepts 
behind this and confirm that this answer is in fact correct?

Regards,
Paul Dobing

Date: 03/31/98 at 12:13:55
From: Doctor Rob
Subject: Re: Polygons - Diagonals and Symmetry

The number of diagonals is given by n*(n-3)/2 = 8*5/2 = 20, not eight.
This is because if we count at each of the n vertices, each one can be
connected to each of the other n-1 vertices. Two of these connections 
form edges of the polygon, not diagonals, so there are n-3 diagonals 
meeting at each of the n vertices. The total number of ends of 
diagonals is n*(n-3). Each diagonal has two ends. Thus the number of 
diagonals is just n*(n-3)/2. (Can you figure out why this is always a 
whole number, never a fraction?)

Draw a regular octagon and see if you can find all 20 diagonals.

There are eight axes of symmetry. Four of them are perpendicular 
bisectors of the sides of the octagon, and four of them are bisectors 
of the interior angles. (You figure out why there are only four of 
each, not eight!) If an axis of symmetry meets a line segment in the 
figure, the axis must be the perpendicular bisector of the segment, or 
else there must be a second line segment also meeting the axis at the 
same point and making an equal angle with the axis. In the first case, 
you have the perpendicular bisectors of the sides, and in the second 
case, you have the bisectors of the interior angles.

-Doctor Rob,  The Math Forum
Check out our web site! http://mathforum.org/dr.math/