geometry
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.
Well, I guess I'll take a stab at this. This is definitely a calculus problem. To take the area between two curves, you want to take the integral of the greater function minus the lesser function. For the yellow area, the greater function is $y=1$. The lesser function will take some manipulation. The formula for the circle is:
$(x-4)^2+(y-4)^2=16$
$(y-4)^2=16-(x-4)^2$
$y-4=-\sqrt{16-(x-4)^2}$
$y=4-\sqrt{16-(x-4)^2}$
So our integral is $\int^3_2[1-(4-\sqrt{16-(x-4)^2}]dx$=$\int^3_2-3dx+\int^3_2\sqrt{16-(x-4)^2}dx$. The first integral is $-3x$ evaluated from 2 to 3, or in other words, $-3(3)-[-3(2)]=-9+6=-3$.
For the second half of that integral, we'll use the info from Andreas's comment. We'll perform a change of variable
$u=x-4,du=dx$
$\int^3_2\sqrt{16-(x-4)^2}dx=\int^{-1}_{-2}\sqrt{16-u^2}du=\frac12[u\sqrt{16-u^2}+16sin^{-1}\frac u4]^{-1}_{-2}=\frac12[(x-4)\sqrt{16-(x-4)^2}+16sin^{-1}\frac{x-4}4]^3_2$
That solves the yellow area. For the other 2, you'll want to know where the 2 functions cross.
$(x-4)^2+(1-4)^2=16$
$(x-4)^2=7$
$x=4-\sqrt7$
For the green area, the 2 functions are the same, but it's evaluated from $4-\sqrt7$ to 2. For the orange area, the greater function is $y=2$, but the lesser function changes. It should be easy to see, though, that the right half is a rectangle. The left half is integrated from 1 to $4-\sqrt7$. Also, the first half of the integral has changed from $\int-3dx$ to $\int-2dx=-2x$. So the total orange area is $2x-\frac12[(x-4)\sqrt{16-(x-4)^2}+16sin^{-1}\frac{x-4}4]$ evaluated from 1 to $4-\sqrt7$ plus $1[2-(4-\sqrt7)]$, or $\sqrt7-2$.
By the "chord-chord" aspect of the Power of a Point Theorem, we have $$|\overline{AX}||\overline{XB}| = |\overline{CX}||\overline{XD}|\quad\to\quad p\cdot p = (p+q)\cdot q \quad\to\quad\frac{p}{q} = \frac{p+q}{p} = \frac{|\overline{CX}|}{|\overline{XE}|}$$
The relation between $p$ and $q$ says exactly that $\frac{p}{q}$ is the golden ratio, $1.618\dots$. $\square$
I can't say that I've seen this construction before, but I don't exactly have a catalog of such things in my head. The power-of-a-point connection seems (almost) "obvious" in retrospect, so it likely has been observed before. Still, it's pretty neat. (It's my favorite of your constructions so far.)
Best Answer
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): $(1+1/4+1/4) : 9 = 1.5 : 9 = 1:6$.