Figure 1: Square and Rectangle Puzzle

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-y2)

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

x/y = 1/sqrt(2y-y2), which gives me x2=y/(2-y), and y=2x2/(x2+1)


Figure 2: Since q=sqrt(2y-y2), by similar triangles, y=2x2/(x2+1)

  • At this point I rescaled the whole diagram by a factor of x2+1 to eliminate denominators.

So the segments labeled x are now relabeled x3+x, the segments labeled y and 1-y are now relabeled 2x2 and 1-x2, etc.

To find p, I observe a right triangle with two sides labeled x3+x and 2x2, and so I factor (x3+x)2 - (2x2)2 to get

(x3+x)2 - (2x2)2 = (x3-x)2

So p = x3-x

But wait!  x is smaller than 1, so p=x-x3  (I confess I missed that the first time around!)

So now the bottom edge is labeled (left to right) x3+x, x-x3, and x2-2x+1

I got q=x2-2x+1 simply by subtracting the other two line segments from x2+1, which is the rescaled size of the big square.


Figure 3: the rescaled image. x is replaced by x3+x, y by 2x2, etc.

Now there's another right triangle with two sides labeled x2+1 and 1-x2, so I factor (x2+1)2 - (1-x2)2 to get

(x2+1)2 - (1-x2)2 = (2x)2

So q = 2x

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

0 = q - q = (x2-2x+1) - (2x)
0 = x2 + 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.

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

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

q=x2-2x+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-x3 gives us p=(2-sqrt(3))3-(2-sqrt(3))=14*sqrt(3)-24

And now, we need to make sure (x3+x)+p+q=x2+1 (because we rescaled everything, remember?), and I think that'll prove everything is copascetic.  So for x2+1 we get

x2+1 = (x3+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)

  • Final Answer:  x=2-sqrt(3)

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