# 5 pointed star

• Oct 15th 2008, 05:04 PM
jmccoid
5 pointed star
what observations can you make about the sum of the angles of five pointed stars - How can you prove your observation?
I know the sum of 5 pointed stars is 180 degrees. Other than saying you just add up the interior angles, is there another way I can offer "proof"?? THANK YOU!!
• Oct 15th 2008, 05:09 PM
skeeter
sketch the circle that contains all 5 "points".

now use the fact that an angle inscribed in a circle equals one-half the measure of its intercepted arc.
• Oct 15th 2008, 05:47 PM
jmccoid
THANKS!!! :)
• Oct 15th 2008, 05:59 PM
Soroban
Hello, jmccoid!

Quote:

What observations can you make about the sum of the angles of five pointed stars?
How can you prove your observation?

Is this a cyclic pentagram? (not necessarily regular)
If so, it can be inscribed in a circle and skeeter's solution is excellent.
Code:

                A               * o *           *          *         *              *     E o                o B       *                  *       *                  *       *                  *       *                *       D o              o C           *          *               * * *

Draw the diagonals: .$\displaystyle AC,\, AD,\, BD,\, BE,\, CE$

An incribed angle is measured by one-half its intercepted arc.

. . $\displaystyle \begin{array}{ccc}\angle A &^m_= &\tfrac{1}{2}\overline{CD} \\ \\[-3mm] \angle B & ^m_= & \tfrac{1}{2}\overline{DE} \\ \\[-3mm] \angle C & ^m_= & \tfrac{1}{2}\overline{EA} \\ \\[-3mm] \angle D & ^m_= & \tfrac{1}{2}\overline{AB} \\ \\[-3mm] \angle E & ^m_= & \tfrac{1}{2}\overline{BC} \end{array}$

Add: .$\displaystyle \angle A + \angle B + \angle C + \angle D + \angle E \;=\;\tfrac{1}{2}\overline{CD} + \tfrac{1}{2}\overline{DE} + \tfrac{1}{2}\overline{EA} + \tfrac{1}{2}\overline{AB} + \tfrac{1}{2}\overline{BC}$

$\displaystyle \text{Sum of the angles} \;=\;\tfrac{1}{2}\left(\overline{AB} + \overline{BC} + \overline{CD} + \overline{DE} + \overline{EA}\right) \;=\;\tfrac{1}{2}(360^o) \;=\;180^o$