After cutting square by dotted line, we can end up with 2 rectangles.

If one piece can be placed on top of other one like that it touches all corners of the sides of the big rectangle, what is the short edge, x?

Do you want to solve this yourself? Then stop reading right here! Go back to the puzzle page instead.

**Graeme's solution...**

This is a fun puzzle... I started by labeling the three line segments of the bottom line (from left to right) x, p, and q, and the two line segments of the middle vertical line (from bottom to top) y and 1-y.

I went around the diagram labeling all the other segments of length x, y, p, and q as needed.

Observing right triangle {1, q, 1-y} I got q = sqrt(2y-y^{2})

Then, observing similar right triangles {x, y, p} and {1, q, 1-y} I got

x/y = 1/sqrt(2y-y^{2}), which gives me x^{2}=y/(2-y), and y=2x^{2}/(x^{2}+1)

**At this point I rescaled the whole diagram by a factor of x**^{2}+1 to eliminate denominators.

So the segments labeled x are now relabeled x^{3}+x, the segments labeled y and 1-y are now relabeled 2x^{2} and 1-x^{2}, etc.

To find p, I observe a right triangle with two sides labeled x^{3}+x and 2x^{2}, and so I factor (x^{3}+x)^{2} - (2x^{2})^{2} to get

- (x
^{3}+x)^{2}- (2x^{2})^{2}= (x^{3}-x)^{2}

So p = x^{3}-x

But wait! x is smaller than 1, so p=x-x^{3} (I confess I missed that the first time around!)

So now the bottom edge is labeled (left to right) x^{3}+x, x-x^{3}, and x^{2}-2x+1

I got q=x^{2}-2x+1 simply by subtracting the other two line segments from x^{2}+1, which is the rescaled size of the big square.

Now there's another right triangle with two sides labeled x^{2}+1 and 1-x^{2}, so I factor (x^{2}+1)^{2} - (1-x^{2})^{2} to get

- (x
^{2}+1)^{2}- (1-x^{2})^{2}= (2x)^{2}

So q = 2x

Now I have two different representations of q, so I get

- 0 = q - q = (x
^{2}-2x+1) - (2x)

- 0 = x
^{2}+ 1 - 4x

and then by the quadratic formula I get x = 2-sqrt(3) and one other root, but that one is bigger than 1 so I forget about it.

(which is how I found my mistaken negative value of p earlier), we get**Then, to check that everything fits**

q=2x gives us q=4-2*sqrt(3), and

q=x^{2}-2x+1 gives us q=(2-sqrt(3))^{2}-2(2-sqrt(3))+1 = 4-2*sqrt(3), so they match!

Then, for p, we get

p=x-x^{3} gives us p=(2-sqrt(3))^{3}-(2-sqrt(3))=14*sqrt(3)-24

And now, we need to make sure (x^{3}+x)+p+q=x^{2}+1 (because we rescaled everything, remember?), and I think that'll prove everything is copascetic. So for x^{2}+1 we get

x^{2}+1 = (x^{3}+x)+p+q = (28-16*sqrt(3)) + (14*sqrt(3)-24) + (4-2*sqrt(3)) = 8-4*sqrt(3)

which is indeed one more than the square of 2-sqrt(3)

x=2-sqrt(3)**Final Answer:**