[bonanova]

This puzzle not mine, but I believe I have found the minimum answer.

If goes like this:

You are given a square pane of glass 1 meter on a side.

You are required to cut the pane into 100 pieces, all of the same area.

There is no requirement what shape the pieces are, nor do they have to all be the same shape.

Just so all 100 pieces have the same area.

The instrument you use is a cutting laser, and you may assume it will cut through glass of arbitrary thickness.

However, the laser consumes vast amounts of energy as it cuts.

For both cost and "green" reasons, you want to accomplish the cutting using as little energy as possible.

If you make multiple cuts - i.e. when the cut reaches the edge of the glass - you may turn the laser off immediately.

So the problem is: minimize the total length of your cuts.

Enjoy

you can stack pieces for cuts after the first.

[Guest]

But of the same size ? How do you do it ? I do not think it is possible

it is possible. Only the area needs to be same. not the shape or size. that can be adjusted.

[Guest]

I would disagree with this. I think it is also interesting to look at upper bounds. I have two ideas for that.

consider cutting 98 parallelograms and 2 triangles. then we require 99 cuts of length sqrt(1+0.4)=1.183m and the total amount of cutting required is 117.14m

Make cuts that are inscribed squares

You could also

cut in a very dense zig zag pattern. This can add a ton of cutting length (theoretically infinite) while adding almost no area to each piece.

[bonanova]

I agree.

My post applies only to rectangles.

[Prime]

With respect to the longest cuts, I suggest we should assume each cut to be one straight line cutting all the way across. (Everything into two pieces.)

But let's return to the lower limit.

I think the theoretical 7 cuts 275cm proven by Bonanova may not be the highest lower limit for dividing 1m square into 100 equal parts. Here is a set of 7 cuts adding up to 280 5/13 cm, which I think is the highest lower limit. (It is not possible to make total cuts shorter.)

First observation: any intermediate fragment must contain a whole number of final fragments. For example, when your resulting pieces must be 100cm² each, you may not have an itermediate fragment of 250cm².

The method is essentially: every consecutive cut with respect to the largest intermediate segment must be made along its shorter side, dividing it as near as possible in half while maintainig whole number of final fragments in each half.

Here is an illustration of 7 cuts required to divide square inot 100 equal area pieces, adding up to 280 5/13 cm.

Let's follow the cuts in the largest remaining segment first. Then show that the cuts in all other segments can be fit in.

1. Cut the square in the middle: 100cm cut, two segments 50x100 each containing 50 end-fragments.

2. Cut 50x100 in half along 50 side: 50cm cut, four segments 50x50, each containing 25 fragments.

3. Cut 50x50 in 13:12 ratio. 50cm cut, 26x50 and 24x50 segment containing 13 and 12 fragments respectively.

4. Cut 26x50 segment along 26cm side in 7:6 ratio. 26cm cut, 26x350/13 and 26x300/13 segments containing 7 and 6.

5. Cut 26x350/13 along 26cm side again in 4:3 ratio. 26cm cut, 26x200/13 and 26x150/13 containing 4 and 3.

6. Cut 26x200/13 along 200/13 cm side exactly in half. 200/13 cm cut. Two 13x200/13 containing 2 fragments each.

7. Cut 13x200/13 along 13cm side in half. 13cm cut. Two final 13x100/13 fragments.

Now let's show how the cuts of smaller remaining fragments could be fit into the above defined cuts.

After the third cut the 24x50 segment could be cut as following:

4. Cut along 24cm side exactly in half. Resulting in two 24x25 segments containing 6 fragments each. The length of the cut 24cm is less than the 4th 26cm cut above, therefore it would fit.

5. Cut along the 24cm side exactly in half. Two 24x12.5 segments containing 3 fragments each. The length of the cut 24cm is less than 26cm in step 5 above.

6. Cut along 12.5cm side in 2:1 ratio. 8x12.5 (final) and 16x12.5 -- 2 fragments. The length 12.5cm is less than 200/13 cm in step 6 above.

7. Cut along 12.5cm side in half. Two 8x12.5 final fragments. The length 12.5cm again is less than 13cm 7th cut above.

In a similar fashion all other remainders can be piled up, alligned, and their cuts will fit into the corresponding cuts of the larger segment. Because we can always find needed proportion cut in the smaller fragment such that the length of the cut would be equal or less the length of the cut in the larger segment.

[Guest]

You could also

cut in a very dense zig zag pattern. This can add a ton of cutting length (theoretically infinite) while adding almost no area to each piece.

For instance, a fractal...

[Prime]

This problem is a much smaller cousin of the unsolved square division on this forum. Interesting how minimization requirement is so much easier than total count problem.

Here are some essential elements for proof of the 280 5/13 as an absolute minimum.

There are several theorems that need a prooves in order to prove the minimum.

1. Prove that dividing into rectangles (at 90-degree angles) can yield shorter total length than cutting at any other angle.

2. Prove that the method of dividing rectangle along shorter side as close to 1/2 of the area as possible while maintaining a whole number of final fragments in each intermediate fragment yeilds smaller sum of cuts than any other method of division.

3. Prove that all minimal cuts in smaller rectangle obtained by the method (2) can fit into the minimal cuts of the larger rectangle obtained in the same step.

I'll start with (2):

As Bonanova showed, the minimal number of cuts to obtain n fragments from one is the nearest power of 2 (p) such that 2^p > n. In the case of 100 it is 7.

When dividing a fragment into two, each of the resulting segments should be less than the nearest (to n) smaller power of 2 in order to lower the number of the remaining cuts. For example, if we divide the 100-unit piece into 65 and 35 unit pieces, then we still have 7 cuts left in order to divide 65-unit piece.

The shortest possible perimeter for a given area rectangle is square. The larger the difference between sides -- the larger is the perimeter. All consecutive cuts are made along the sides of the rectangle, or the resulting fractions thereof. So it stands the reason, that shorter the sides of a rectangle, the shorter total cut length.

That was not a proof -- just an illustration of an idea.

Also note it is more advantageous to make cut along shorter side. Let's say, using the "method" we obtain a rectangle with the sides a and b, where a > b and containing 4m fragments. Let's try two pairs of cuts: along the longer side for the length a and then along shorter for the length b/2 with the total a + b2.

Now let's cut along side b for the length b and then along the resulting 1/2 a for the total b + a/2. If we subtract second exp​ression from the first, we get: a/2 - b/2 which we know is greater than zero. Thus the resulting sum of cuts in the first case was larger.

2m and 2m+1 seggments can be proved in the like fashion.

And that is as much typing as I care to do in one sitting.