# Tag Info

49

Here's one solution: I assumed the radius of the hole's curvature matches the curvature radius of the circle, the hole's straight side is equal to the circle's radius, and its curved edges meet the straight edge at right angles. Used Flash Pro to solve this, of all things: it has the useful feature that as you paint while editing an object, it updates all ...

47

Quite an interesting cut I had to make to get it to fit. Shift the bottom piece 2 tiles up and one tile right to produce a 10 x 10 square. You can save all 100 square feet. The end result should look like this

38

Cut along the red lines and move the pieces as indicated by the yellow arrows. As is usual with this kind of dissection, it helps if you look at the area to work out the length of the side of the final square. Given the grid lines, you also know the orientation of the square, so you can try to place a square over the original picture at such a location that ...

36

We will make all of our cuts vertical, so we can treat this as a square which we need to divide into $10$ pieces with equal slices of the area and the perimeter. This is reasonably easy: Choose $10$ points dividing the perimeter into $10$ equal lengths, and then make cuts inwards from each point to the center. The area of a triangular slice is half the ...

32

Yes, it's possible, because one can fit 5 disjoint 1x1 squares in a 2.75x2.75 square: four in the corners, and one in the center rotated 45 degrees. The four cigaret holes can't eliminate all 5 squares. The tilted square fits in a cross whose center's square diagonal has a of length 1. So the cross needs width $1/ \sqrt{2} \approx 0.707 < 0.75$. So, ...

30

Index your rectangle from (0,0) to (10,2). Then cut from These four pieces can be used to make the square. Note that this dissection works without any flipping or even any rotation of the pieces! To show that it's a square is also relatively simple. Easy geometry shows that the angles are right angles, and that the sides are $2\sqrt{5}$.

30

I believe this works as a short proof.

28

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The dimensions are and the tiling looks like this: Working out the dimensions of the rectangle is quite easy. We know its total area is $4209$ (i.e., $2^2 + 5^2 + 7^2 + 9^2 + 16^2 + 25^2 + 28^2 + 33^2 + 36^2$). This factorizes as $3 \times 23 \times 61$, and in order to fit in a square with a side length of 36, the rectangle must be $3 \times 23$ units ...

28

If all cuts are straight cuts and the cake is a rectangular prism or cylinder, it's not possible. From Wikipedia's page on the Cake Number: In mathematics, the cake number, denoted by Cn, is the maximum number of regions into which a 3-dimensional cube can be partitioned by exactly n planes. The cake number is so-called because one may imagine each ...

27

JonTheMon and xnor's solutions assume we have superior equipment and skill, but the question states that we "have a hacksaw". Well with a hacksaw, we must start from an exposed side; we can't start a cut in the middle of a plank! The most efficient "side" would be the current hole: If the hacksaw won't fit in the hole, we "cut an L shaped piece off the ...

26

As pointed out in a previous answer, cutting a hole in the middle of the table may be unfeasible if everything you have is a hacksaw. Using the existing hole as a starting point, the cut can be reduced to the length of a single segment bifurcating on both end to meet the old and the newly cut hole tangentially at a convenient distance. Something like this. ...

24

One of the possible solutions is: Calculations Lets say square length is L and Height H Frosting on top = 0.5 * 0.4L * 0.5L Frosting on bottom = same as top Frosting on side = 0.4L * H Total frosting area = 0.2L2 + 0.4LH Cake volume = top area * height = 0.5 * 0.4L * 0.5L * H = 0.1L2H Validation Total vol = 1L2H is 10 times above ...

22

You cut a square like this: And rotate it 180 degrees. The cut square (or rectangle) simply needs to have its centerpoint be halfway between the hole and the center of the square, and to be large enough to contain the hole. Boy do I feel silly seeing the intended solution...

22

Solutions are as in the following diagram: Addition: My original solution for E has 5 cuts and 9 separate pieces - a simpler solution is shown below, with only four cuts (to the cross) and five pieces. Dimensions are as follows: a) side lengths = 2 and 1 —[2 cuts, 3 pieces] b) side lengths = sqrt(5/2) —[2 cuts, 4 pieces] c) side lengths = 3/sqrt(2) and ...

21

One cut solution for #3.

21

Well, from my viewpoint this is a four-piece dissection, since parts of each piece don't move relatively to each other. They are even connected, to some extent. However, I would completely agree that there are about 24 pieces in this dissection, from a pragmatic viewpoint. At least evaluate an hour-long fiddling with MS paint here.

20

I think for the sake of simplicity this should be part of the answer

20

Look at the cube down a space diagonal. It should appear as a hexagon, which can be divided into three rhombuses by line segments from the center to every other corner. If these three cuts are made, they will divide the cube into three pieces, which must be congruent by symmetry.

20

One (non-straight) cut: (Or a different image by Carl Löndahl: https://nup.pw/YK9cjY.png)

20

Okay, I don't know how people here create images that fast, but I get a more sheet-like solution.

20

The trick to this puzzle is to: (And here are those tilings: the center was already given, and the rest are obtainable from these by rotation.)

19

How's this? I got this by square-izing the icosahedron graph:

19

From the deleted answer from frodoswalker we know the possible ranges of the squares: Our maximum area is $1*2+3*4+5*6+7*8+9*10+11*12+13*14=504$, which means the maximum square width is $\lfloor \sqrt{504} \rfloor = 22$ Our minimum area is $1*14+2*13+3*12+4*11+5*10+6*9+7*8=280$, which means the minimum square width is $\lceil \sqrt{280} \rceil = 17$ ...

19

I think it is possible to do with Which look like

18

Let $A=(-30,60),B=(0,0),C=(150,0),D=(300,300),E=(225,400)$. $F=(-12,84)$ is on $AE$ and $G=(175,50)$ is on $CD$. Then $ABCDE$ is similar to $FABCG$, by a factor of $\sqrt5$. The key point of this solution is that angles $GFA, EAB, ABC, BCD, CDE$ are all equal ($\arctan-2\approx116.57^\circ$) and $FA,AB,BC,CD$ are in geometric sequence. Then, we can rotate ...

18

18

Here is a good way of seeing how this dissection comes about.

17

For part 1: For part 2: For part 3:

17

To divide the cake into n equal pieces, Proof: Here is a drawing to illustrate how it works:

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