Octo-parallelogram!

Let area of the parallelogram be $A$ then it follows the area of the triangle formed by joining the midpoint of one side with any opposite corner will be $\frac{A}{4}$ . This is the image

Consider the two diagrams, we will attempting to find out the area bounded by regions similar to $MOND$ and that by regions similar to $SPR$ .

Let $x$ be the area of $DNO$ and $y$ be area of $DOM$ , based on the property of median dividing the triangle into two triangles of equal area other areas are labeled.

Now, $MOND = \frac{(2x+y)+(2y+x)}{3} = \frac{A}{6}$

For the other diagram, it can be proved that $XSRZ$ is indeed a parallelogram with area half as $A/2$ , which means area of $SYR$ is $A/8$ .

By inclusion and exclusion of area and some thought it can be said that the area of octagon is $A - (8 \times \frac{A}{4} - 4 \times \frac{A}{6} - 4 \times \frac{A}{8} )$ $= \frac{A}{6}$

Let the parallelogram's area be $P$ and the octagon's area be $O$ . Then, connect the midpoints of the parallelogram, dividing the parallelogram into quadrants. These should also pass through vertices of the octagon. Divide one of the quadrants in a similar fashion. The result will look like this:

Each small quadrant's area is $\frac{P}{16}$ , and the octagon is divided into shapes with area $\frac{O}{4}$ . Now, we will zoom in on the octagon quadrant and divide it into four (we will focus on these smallest quadrants from this point onward).

Part of the octagon is in each of these quadrants of area $\frac{P}{64}$ . Look at the top left quadrant. A triangle is formed in the space between the octagon and the quadrant boundaries. We can see that the triangle has an area equal to one-fourth of the quadrant. The area of the orange shape in this quadrant, then, is $\frac{3}{4}\times\frac{P}{64}=\frac{3P}{256}$ . A similar argument can be made for the bottom right quadrant.

Our final quadrant contains an orange shape labeled $A$ that has all the same angles as the entire section of the octagon in this image, so the shapes are similar. Since the area of the large orange shape is $\frac{O}{4}$ , and $A$ has sides scaled down by $\frac{1}{4}$ , $A = \frac{1}{4}\times\frac{1}{4}\times\frac{O}{4} = \frac{O}{64}$ . Now we have an equation:

$\frac{O}{4} = \frac{P}{64}+\frac{3P}{256}+\frac{3P}{256}+\frac{O}{64}$

$64O = 4P+3P+3P+4O$

$60O = 10P$

$\boxed{\frac{O}{P}=\frac{1}{6}=0.1666}$

I didn't come up with this solution. Credit goes to Presh Talwalkar's video .

By applying a shear, one may assume the parallelogram is a rectangle (whose sides are parallel to the x and y axis). Furthermore, since we are only interested in the ratio of these area, stretching (that is dilating along the x-coordinate but not the y-coordinate) does not change the answer. So one may assume we are dealing with a square (whose center is at the origin). Each quadrant has a 1/4 of the area of the octagon and 1/4 of the area of the square. So we may only consider the ration in the first quadrant. The part of the octagon in the first quadrant is a kite whose corners have coordinate (0,0), (0,1/2), (1/2,0) and (1/3,1/3). [The last coordinate can be computed by intercepting the two lines y = (1-x)/2 and y =1-2x, the other are obvious].

There are many ways to compute the area of this kite. One can split it into two equal triangles (split using the line from (0,0) to (1/3,1/3) ). Each triangle has height 1/3 and basis is 1/2 (this can be read directly from the coordinates). The area of the kite is 1/6. The square has area 1, so the the ratio is 1/6.

你可以把一个平行四边形本质看作正方形，平行四边形只是换了一组基底表示某一向量空间中的正方形。以正方形左下角顶点作为原点建立直角坐标系求解出八边形的八分之一（一个三角形）的面积，以求得结果。

much easier to work with a square ( just a particular type of parallelogram), the solution becomes not too complex