Stuck on Square Root Problem (yep, homework!)

Question

Here's the simple question:

Devon has a piece of poster board 45 cm by 20 cm. His teacher challenges him to cut the board into parts, then rearrange

the parts to form a square. a) What is the side length of the

square? b) What are the fewest cuts

Devon could have made? Explain.

I understand part a (the answer is 30 or the square root of 900) but how many cuts would he have made to make it a square? 45+20 = 65, 65 doesn't divide in 30 and if I try taking 45-15 and moving the 15 to the 20, I now get 35*30 which is not = 900 (1050)? How come then, 45*20 = 90 but when you displace 15, you get a greater answer? I'm sure the error is somewhere in my conversion between the rectangle and the square.

Answer 1

The issue with your calculations is that when you cut the 45 off at the 30/15 mark, you get a piece that is 20*30 and a piece that is 20*15. Putting the 20*15 piece along the 20*30 piece will give you an L-shape, that is 35*30 with a 10*15 rectangle cut out of the corner.

The best way that I can see to do this is in two cuts. First, make the cut that you describe. Now, you have a 20*30 piece and a 20*15 piece. Cut the 20*15 piece into two 10*15 pieces. Now, put those two pieces together along their 10 edges, giving a 10*30 piece. Now, put the 10*30 next to the 20*30 piece along 30 edges, giving a 30*30 square in two cuts.

Answer 2

Part b doesn't have a nice algorithmic solution that I know of. Clearly you can cut the board int $5 \times 5$ squares and rearrange them to make a $30 \times 30$ square. That is a lot of cuts. Many times the answer is a cut (not a single straight line) that is a stairstep, then you move the stairs one notch. It doesn't work here, so I would endorse qalpha's solution.

Answer 3

We describe a nice way to do it, unfortunately in words. It really needs a picture.

Put down your cardboard rectangle, one corner at the origin, the long side along the positive $x$-axis. So the corners of your cardboard rectangle are at $(0,0)$, $(0,45)$, $(45,20)$, and $(0,20)$.

Draw a $30\times 30$ square, with corners $(0,0)$, $(30,0)$, $(30,30)$, and $(0,30)$.

Draw the line that joins $(0,30)$ to $(45,0)$.

This line will meet the top side of your cardboard rectangle at $P=(15,20)$, and will meet the right side of the square at $Q=(30,10)$. Let $R=(30,0)$ and $S=(45,0)$.

All set up! Use a razor knife to cut along the line $PS$. That will slice a substantial triangle from the cardboard. Leave it in place for now.

Use the razor knife to cut straight down along $QR$. This slices off a smallish triangle from the cardboard.

Slide the big triangle upward until its top side agrees with the top line of the square. It will.

Slide the little triangle way up so that it fills in the top left corner of the square. It will.

Done, two cuts.

It is a very pretty construction, works uniformly for all rectangles that are not too skinny. If the rectangle is very skinny, a not too hard adjustment can be made.

You will have to prove that this works. Straight coordinate or similar triangle geometry.

Remark: This construction is one of the steps in the proof of the Bolyai-Gerwien Theorem (which, as is so often the case, was proved a number of years earlier by at least two other people). The result is that if $A$ and $B$ are any polygonal regions with the same area, then $A$ can be cut into a finite number of polygonal pieces that can be reassembled to maske $B$.