A shape is created in square ABCD as follows:
A line goes from vertice A to the midpoint of BC and CD
A line goes from vertice B to the midpoint of CD and DA
A line goes from vertice C to the midpoint of DA and AB
A line goes from vertice D to the midpoint of AB and BC
(the resulting figure should look like a star)
What is the ratio of the area of the octogon created in the center of square ABCD to square ABCD?

Let's have some values. The sides of the square be 4 (a multiple of 2, just in case)

From similar triangles, we get the red line as having a length of 1.

From symmetry, we get the central broken blue line as having a length of 2.

Hence, the side of one triangle in the octagon is 1.

As you see, the octagon is a regular octagon and as such, if you divide it into 8 triangles with a vertex at the centre of the shape, you get each triangle with apex angle 45 degrees.

Do you know the area of such a triangle?

Area = \frac12 ab sin \hat{c}

Hence the area of the octagon = area of 8 triangles = 8(\frac12 ab sin \hat{c}) = 8(\frac12(1)(1)sin(45)) = 2\sqrt2

And the area of the square is 4 x 4 = 16.

Ratio of areas = 2\sqrt2 : 16 = 1: 4\sqrt2

http://p1cture.me/images/25518050532797550692.png

EDIT: Sorry, wrong assumption here. See post below.

See here for a nice discussion on the topic:

http://nrich.maths.org/561&part=solution

Okay, the only reason I can find that makes my answer wrong is the assumption that the octagon is a regular octagon... which means that the sides of the triangles are not both 1.

To get length of OP where P is the point passing through the diagonal of the square. And on one point in the octagon, I'll use the coordinate system.

The diagonal has equation y = x.

The line making the upper-right side is y = -2x + 1

we get x = 1/3

And hence, y = 1/3

The length OP becomes \sqrt{\frac19 + \frac19} = \sqrt{\frac29} = \frac{\sqrt2}{3}

Then, area of triangles become:

8\(\frac12(1)\(\frac{\sqrt2}{3}\) sin(45^o)\) = \frac83

And ratio becomes: \frac83 : 16 = 1:6