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Measures of Interior and Exterior Angles of Polygons

Date: 07/07/2005 at 05:34:43
From: Louisa
Subject: Star Polygon

If you arrange a set of points roughly around a circle or an oval, 
and then you connect each point to the next with segments, you should
get a convex polygon. 

Draw five points A through E in a circular path, clockwise.  Connect
every second point with AC, CE, EB, BD, and DA.  Measure the five
angles A through E at the star points, and find the sum of the angle
measures.
 
What happens if you connect every third point to form a star?  What 
would be the sum of the angle measures in this star?  What patterns 
are there for stars with 5, 6, and 7 points?  What then, is the rule 
for n-pointed stars?

I have figured out everything except for the rules for n-pointed 
stars.  Can you help?



Date: 07/07/2005 at 14:16:45
From: Doctor Tom
Subject: Re: Star Polygon

Hi Louisa,

I like to think about all problems like this in terms of something
that a mathematician would call a "winding number".

Let's start with a simple example:  a simple convex polygon.  Suppose
we're trying to find the sum of the interior angles.  You may know
already that for a triangle, it's 180 degrees; for a quadrilateral,
it's 360 degrees, for a pentagon (regular or not) it's 540 degrees,
and generally, for an n-sided polygon, it's 180*(n-2) degrees.

But why?

Imagine a polygon drawn on the ground, and you start on an edge and
walk around it.  You're not turning at all along the lines, but each
time you get to a vertex, you turn some amount.  If you were walking
around a square, for example, each time you got to a vertex, you'd
turn 90 degrees, right?  If you were walking around an equilateral
triangle, you'd turn 120 degrees: in other words, you'd make more than
a 90 degree turn.  For a regular pentagon, you'd only need to turn 72
degrees at each vertex, and so on.

But no matter what the shape, after you've done a complete loop, you
would have turned a total of 360 degees: one complete turn.  For the
equilateral triangle, it would be 3 times that you turned 120 degrees;
for the square, four turns of 90 degress; for the regular pentagon, 5
times 72 degrees, but in all cases, the net turn would be 360 degrees.

The amount that you turn is the EXTERIOR angle at each vertex.  If you
want the interior angle, in every case, it's 180 - exterior.  For the
equilateral triangle, there would be 3 interior angles of 180 - 120 =
60 degrees.  For the square, 4 angles of 180 - 90 = 90 degrees; for
the regular pentagon, 5 interior angles of 180 - 72 = 108 degrees, and
so on.

But note that you'll still turn 360 degrees, even if the figure is not
regular.  For a triangle with interior angles of A, B and C, your
exterior turn is:

  (180 - A) + (180 - B) + (180 - C) = 540 - (A+B+C)

and this is one turn, so it must be 360:

  540 - (A+B+C) = 360, so A+B+C = 540 - 360 = 180 

so the three interior angles of any triangle must add to 180 degress.

Let's do a pentagon (all the others are similar).  Suppose the 
interior angles are A, B, C, D and E.  Your exterior turn is:

  (180 - A) + (180 - B) + (180 - C) + (180 - D) + (180 - E) 
  = 5 x 180 - (A+B+C+D+E) = 900 - (A+B+C+D+E)

and that must be equal to 360 so

  900 - (A+B+D+C+E) = 360, or 900 - 360 = (A+B+C+D+E) = 540

so the interior angles of any pentagon, regular or not, add to 540
degrees.

What's different if we walk around a star in the same way?  The
difference is that (for a 5-pointed star) we make two complete loops,
or 2 x 360 = 720 degrees of exterior turning.

Thus if the interior angles of the star are A, B, C, D and E, the sum
of the exterior angles is:

  (180 - A) + (180 - B) + (180 - C) + (180 - D) + (180 - E) 
  = 5 x 180 - (A+B+C+D+E) = 900 - (A+B+C+D+E)

but this time, instead of adding to 360, it adds to 720, so we obtain
with a little algebra that:

  (A+B+C+D+E) = 900 - 720 = 180.

Look at what happens with a 7-pointed star: depending on how you step
around, you can make either two or three complete loops.  (If you just
skip one point each time around, there are only 2 loops; if you skip 2
points on each step, there are three complete loops.)  With a
9-pointed "star", there are two, three, or four complete loops,
depending on how you step around.

Stars with even numbers of points are a bit odd, since they are made
of two distinct paths, so you have to analyze them as two "stars" with
half the paths.  The six-pointed "Star of David" is actually two
triangles, so you'd look at it as the angles in two triangles,
et cetera.

I hope this helps!


- Doctor Tom, The Math Forum
  http://mathforum.org/dr.math/ 
Associated Topics:
College Triangles and Other Polygons
High School Triangles and Other Polygons

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