Tough area problem?

Question

Take a square with side length x. Connect the midpoints of each side to their non adjacent verticies so that an octagon forms inside the square. What is the area of the octagon in relation to the original square?

Answer 1

Start with the square whose corners are the points (0,0), (1,0), (1,1), and (0,1). If my calculations are correct, then the vertices of the octagon, starting with the vertex closest to the origin and going counterclockwise, are
( 1/3 , 1/3 )
( 1/2 , 1/4 )
( 2/3 , 1/3 )
( 3/4 , 1/2 )
( 2/3 , 2/3 )
( 1/2 , 3/4 )
( 1/3 , 2/3 )
( 1/4 , 1/2 )
and back to ( 1/3 , 1/3 ).
If these are ok then the area of the octagon is 1/6 (beginning with a unit square).

Answer 2

very clever problem, I don't remember ever seeing this one before...

I solved it by drawing the diagram as you described, then also drawing the two diagonal lines from corner to corner. So your octagon is represented by 8 triangles all connecting at the center. Find the area of one triangle, and multiply by 8 to get the answer.

To get the area of one triangle, you can try one of several methods. For example, you can figure out the angles of the triangle (45, 60, 75), and you know one side b which is actually 1/2 of the square side. The angles 60 and 45 are adjacent to the side, so use a variant of Heron's formula:

A = b^2 / [ 2 (cot 45 + cot 60) ]

cot 45 = 1 and cot 60 = 0.5774

So replace b with 1/2 x where x is your square side, and then solve for A, and then multiply by 8.

Answer 3

I was "need help" but something screwed up and i can't post a solution under that username, so i switched accounts...

Hi. i get 3/14.

My method is kind of messy. let a side of the square be d. Basically, from the area of the square, d^2, I subtracted the sum of all 8 right triangles with one leg being half the side of a square, the other leg being the full side of the adjacent square. this sum is 2d^2. I'm left with the area of the octagon but I also oversubtracted 8 regions by counting them twice (they are the overlap between the 8 triangles). 4 of these overlapping regions are triangles with the base as a side of the square, and legs meeting at a vertex of the octagon. I split each of these regions down the middle, drawing a line segment from the midpoint of the base (also the midpoint of the squae) up to the vertex of the octagon (also the vertex of the triangle) i.e. the altitude. each half of the triangle, when i split it like that, is similar to one of the 8 right triangles we initially summed, but has 1/4 its area, and its area is d^2/4. so half of the area of one of the regions we need to subtract is d^2/16. the area of one of the regions we subtract is d^2/8. we subtract 4 of these regions, so we're subtracting d^2/2.

so far we have:
d^2-(2d^2-d^2/2)

we also need to subtract four triangular regions whose vertices are the corner of a square, the midpoint of a side of the square, and n intersection between two of the lines going from a vertex of teh square to the midpint of a nonadjacent side. finding the area of one of these regions wasn't easy for me.. i decided just to write the equations of the lines and find their intersection point to find the height of the regions, and their base is d/2. i found that the total area of these 4 regions is 5/7 d^2.

so i get d^2-(2d^2-d^2/2-5/7d^2)=3/14 d^2. so the area in relation is 3/14. this answer does look pretty much correct to me, as it seems the area of the octagon should be about 1/4 the area of the square. i think my method is correct but it's not very good, sorry.

TO JEFF
i just looked at your solution for an octagon in a square problem very briefly and decided to calculate the value you described to see what it was .. i found that you get a ratio of about 0.63. it doesn't seem like the area of the octagon should be more than half that of the square, though...