Area of octagon constructed in a square The following picture is constructed by connecting each corner of a square with the midpoint of a side from the square that is not adjacent to the corner. These lines create the following red octagon:

The question is, what is the ratio between the area of the octagon and the area of the square. One is supposed to find the solution without a ruler. 
By removing some lines, I find it easy to see that the ratio between the yellow area and the square is 1/4. But I am not sure if this helps.

 A: I think the following image will say more than any text. You can divide the image into smaller squares that will allow you immediately to calculate the ratio.
The ratio of the red area within the whole square is the same as the red area in the big green square to the area of the whole green square. And this is (counting in units of the 9 small squares): $(1+1/4+1/4) : 9 = 1.5 : 9 = 1:6$.

A: A calculated answer. Maybe not the simplest.

A: Let us name all the points on the outer square like this:

So, AB, CD, EF, AD are side of the square of length say $\mathrm{a}$ and E, F, G and H are midpoints of the sides of the square. Now consider line segments AF, GC, BH, ED. AF and GC intersect BH and ED to form another square.

The area of this square is $\frac{a^2}{5}$ as proved from answers to this question. So, side of this square is $\frac{a}{\sqrt{5}}$. The sides of the square act as span of the octagon (S).(See octagon).
 Line HC, AE, GD, FB not drawn for clarity.
Area of octagon in terms of span = $\mathrm{2(\sqrt{2}-1)S^2 = 2(\sqrt{2}-1)\frac{a^2}{5} = \frac{2(\sqrt{2}-1)}{5}{a^2}}$
Therefore, ratio between the area of the octagon and the area of the square = $\mathrm{\frac{2(\sqrt{2}-1)}{5}}$
A: I was frustrated by the solution being 1/6 because that is not the result if one calculates the area of a regular octagon with a radius that is L/4, where L is the length of large square.
The gridded image above makes clear that red shape is close to, but not actually, a regular octagon.
Instead, the horizontal distance from the center of the large square to a vertex is 1.5 times the length of the smallest green square. The angled distance from the center of the large square to a vertex is 'sqrt(2)' times (not 1.5 times) the length of the smallest green square. The vertices do not lie on a circle, so it is not a regular octagon (with equal all the included angles being equal), even though the lengths of all the sides are equal.
