Octagon, again

I am splitting my regular octagon into 16 sectors so each one will radiate from a 360/16= 22.5 degree angle at the centre.

$$\\tan22.5=\frac{0.5L_1}{a}\\\\
0.4142\approx \frac{0.5L_1}{a}\\\\
a\approx\frac{0.5L_1}{0.4142}\\\\
a\approx1.2071L_1\\\\
also\\\\
tan22.5\approx\frac{0.5L_2}{a+2}\\\\
0.4142\approx \frac{0.5L_2}{a+2}\\\\
a+2\approx \frac{0.5L_2}{0.4142}\\\\
a\approx \frac{0.5L_2}{0.4142}-2\\\\
a\approx 1.2071L2-2\\\\$$

$$\\so\\\\
1.2071L2-2\approx 1.2071L_1\\\\
1.2071L2\approx 1.2071L_1+2\\\\
L_2\approx L_1+\frac{2}{1.2071}\\\\
L_2\approx L_1+1.6569\\\\$$

so YES for a regular octagon, if you put the 2 unit border around it like you said, the side length will increase by a set amount which is approx   1.6569 units

I am going to "steal" some of Melody's work, here.....

Look at the following pic.....

Let's suppose that the border IS 2 ft wide.  But notice, we can't say WHAT the specific side length of the border might be because, as you see, L₂ in the picture is longer than L_1.

And as Melody has indicated using similar triangles, L₂ = L₁ (r + 2) / r

Since ABQP forms an isosceles trapezoid......we could give an "average" side length = [L₁ + L₂ ]/ 2  = [r*L₁ + L₁(r + 2)] / [2 r] = [r*L₁ + L₁(r + 2)] / d    which is just the length of the median of the trapezoid.

Thus......no general answer as to side length is possible....

Nice pic Chris,  LOL    :))

Hey......I know a good pic when I steal, er.... "borrow"  one .....!!!!!!!

BTW.....how did you make that pic ???   Using GeoGebra ???

Yes I just used GeoGebra.   (This is a free program that anyone can download)

It could have been done more simply I think but this is how I did it

I draw the inside octagon,

then I drew in the diagonals to locate the centre,

then I draw the outer circle

then I extended the diagonals to determine the vertices for the outer octagon

then I drew the outer octogon.

then I did the labelling         :))

It may make a difference if the 2 feet is added perpendicular to the side of the octagon. The drawing shows the 2 feet being added to one of the diagonals.

My question is, when the 2 feet is added perpendicularly, does a right triangle formed with "G" in the depiction bisect the angle G (which originally should have been 67.5)?

If that is the case, then the tangent of that bisected angle should include the opposite side (2), over the adjacent side. This would be half of the length of the exterior length.