1. ## Halloween

My colleague's daughter wanted to be a stop sign for Halloween, and he needed to cut out such a sign from a piece of cardboard that was 3 feet by 4 feet. What cuts did he make so that the stop sign (a regular octagon) was as big as possible?

2. Intuitively, we want to make the height as large as possible, leaving a little left over material on the sides. (We could alternately think of it as make the sides as large as possible leaving a little left over for the height, it doesn't matter.) Using this way of thinking, obviously, we're thinking of the cardboard as having the 3-foot side running vertically and the 4-foot side running horizontally.

When we see the octagon as triangles and rectangles, we know that the interior rectangle has the same height as the sides of the octagon, which we'll call $a$. Looking at the triangles, then, we want to find out what is the side length of a right triangle with hypotenuse $a$. Whatever that side length is, call it $b$, we want to double it, add it to $a$, and set that equal to 3. When we solve for $a$, we'll know the value of $b$.

Thus he started a cut on the left side (could be on the right, again, not really important how you view this), b units down from the top, at a 45-degree angle from the board (since the interior angle of an octagon is 135) until he cut that piece off. He did a symmetrical cut at the bottom. He also made a cut $a+2b$ units to the right of the upper-left corner (measured from the point before anything had been cut), straight down the cardboard. He then made cuts on the right that were symmetric to those on the left.

I'll see if I can make and post a .pdf showing the picture that I used to think about this.

3. The red line is b and the blue line is a.

4. Originally Posted by matgrl
My colleague's daughter wanted to be a stop sign for Halloween, and he needed to cut out such a sign from a piece of cardboard that was 3 feet by 4 feet. What cuts did he make so that the stop sign (a regular octagon) was as big as possible?
Adding some calculations to ragnar's description:

1. All sides of the octogon have the length a.

2. The width of the board is 3'. It consists of

$2k + a = 3$

Since a is the hypotenuse of an isosceles right triangle with side length k you know:

$k^2+k^2 = a^2~\implies~k = \frac12 a \cdot \sqrt{2}$

Thus the 1st equation becomes:

$a\cdot \sqrt{2}+a=3~\implies~a=\dfrac{3}{1+\sqrt{2}}~\app rox~1.24264'$

5. Elegant solution, earboth!

6. Are there any simplier ways to do this?