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You have three cakes of diameter 20cm, 16cm and 12cm respectively as shown in the figure.

enter image description here

They all have the same height. Now your task is to divide these 3 cakes into 4 equal parts and you are only allowed to cut one cake at a time.

What is the minimum number of cuts required?.

  • The constraints are that you have to cut one cake at a time and the only tool you have at your disposal is a knife.

Hint

You can displace the cakes.

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  • $\begingroup$ I need to read more carefully, I was trying to get 3 equal parts. 4 is much cleaner. $\endgroup$ – Kevin Aug 17 '14 at 18:07
  • $\begingroup$ If you need to cut, let's say 1/16 part of the cake, can you do it in one cut? Or you would need to do 4 cuts to be sure that you got it correctly? $\endgroup$ – klm123 Aug 17 '14 at 18:26
  • $\begingroup$ you cannot measure angles so no you can't cut out 1/16 with just one cut. Remember you just have a knife. $\endgroup$ – Hubble07 Aug 17 '14 at 18:46
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    $\begingroup$ @klm123 Oh you mean like the ticks on the clock face, nice thinking...OK I admit that what you suggest is practically possible but i was looking for a more elegant way of making that cut i mean without drawing any markings on the cake. Think a little outside the box. Also i have added a hint to the question. $\endgroup$ – Hubble07 Aug 17 '14 at 20:09
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    $\begingroup$ From OP's comments, the minimum cut should be 2. And since by cutting the big cake into two $50\pi$ pieces the problem reduces to having one cut on the two smaller cakes to get $50\pi$ if we don't use the cut from the big cake, that means we need to use the cut from the big cake. $64\pi-50\pi = 14\pi$, the required cut from the second cake. That means after the first cut on the big cake, we need to use that cut to cut away $50\pi$ from the second cake. Just need to find out how. $\endgroup$ – justhalf Aug 18 '14 at 1:50
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Two cuts (or four, depending on how you count a cut):

Cut the 20cm cake in half, giving you two pieces that are each one-fourth of the total cake.

Then stack the 12cm on top of the 16cm cake and cut along the 16cm's diameter to the 12cm, then around the 12cm (like a stencil), to the other side of the 16cm's diameter.

I.E., cut along the red line:

Cake Cut Diagram

The formerly 16cm cake will have a 12cm core plus half the annular region, and the 12cm cake plus the other half of the annular region will be of equal size. These two, along with the two halfs of the 20cm cake, form four equal quantities.

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  • $\begingroup$ Yes this is the one i wanted. Elegant $\endgroup$ – Hubble07 Aug 19 '14 at 1:24
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    $\begingroup$ How do you position the 12cm cake properly? $\endgroup$ – Muqo Aug 19 '14 at 2:27
  • $\begingroup$ Argh, I didn't realize that we could just cut that red part. I only considered to cut the 12cm core away, then cut the annular region in half, requiring three cuts. (+1!) $\endgroup$ – justhalf Aug 19 '14 at 8:01
  • $\begingroup$ @Muqo everyone seems to agree that cutting cakes in half, i.e., along a diameter, is fine; thus, we can find the center at the intersection of any two diameters. Or, we use the same "eyeball it" justification for finding the center as well. $\endgroup$ – TheRubberDuck Aug 19 '14 at 16:09
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    $\begingroup$ the solution from @germcd is simpler and easier to do. $\endgroup$ – klm123 Aug 20 '14 at 4:34
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2 cuts

The area of each cake is $100 \pi, 64 \pi, 36 \pi$.

Half of the big cake is $50 \pi$

cut a 4 cm segment from the 16 cm cake by putting the 12 cm cake at one edge and cutting the other end.

using the formula for the area of a segment $Area = 8^2 cos^{-1} \frac{(8-4)}{2} - (8 - 4) \sqrt{2*8*4-4^2} = 12.5 \pi $

Adding it to the 12 cm cake would give a combined area of $48.5\pi$ and the remainder of the 16 cm cake would be $51.5\pi$

Below is an image of the 12cm cake on top of the 16 cm cake and the blue line is where I would cut the cake.

picture

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  • $\begingroup$ Yes but that one cut would cut all the 3 cakes simultaneously which is not allowed.you are only allowed to cut one cake at a time. $\endgroup$ – Hubble07 Aug 17 '14 at 19:10
  • $\begingroup$ I missed that part, I edited my answer with a new solution $\endgroup$ – germcd Aug 17 '14 at 20:14
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    $\begingroup$ I appreciate your effort but this is not the cut that i m thinking of, but you are close to it. Think symmetric. $\endgroup$ – Hubble07 Aug 18 '14 at 12:20
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    $\begingroup$ I know its not the solution you're looking for but I'd like to leave it up $\endgroup$ – germcd Aug 18 '14 at 13:05
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    $\begingroup$ I think this solution is better than the accepted answer, since it can actually work with cake and two cuts, while the other solution has sharp turns in its "one" cut, which will scramble the cake with a real knive, you cannot do a sharp 90° turn while cutting, it will mess up the cake $\endgroup$ – Falco Sep 29 '14 at 11:00
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enter image description here

How about put the 12cm cake on top of the 16cm cake and make a cut like this?

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  • $\begingroup$ you are almost there. think symmetric! $\endgroup$ – Hubble07 Aug 18 '14 at 12:21
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    $\begingroup$ You mean like turning a pokeball cake into 3 mushroom head cake and one anti mushroom head cake? $\endgroup$ – hkboy44 Aug 18 '14 at 18:30
  • $\begingroup$ After u have made the cut on the 16cm cake,you can arrange the 12cm and the two pieces of the 16cm cake to form two similarly shaped pieces of 50$pi$h volume. also pokeball cake relates to the cut. $\endgroup$ – Hubble07 Aug 18 '14 at 19:03
  • $\begingroup$ @Muqo (need 9 more rep to post comment), I have given some thought. Try envision the 12cm cake as the "moon cake" and position it on top of the 16cm cake to make a eclipse with its shadow. Then drop the moon cake and you get the pokeball cake!! $\endgroup$ – hkboy44 Aug 19 '14 at 4:18
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  1. Cut the 20cm cake straight in half to get two $50 \pi h$ volumes. Keep the halves together.
  2. Place the 16cm cake flat on top of the cut 20cm cake so that their circle bases are tangent at one end of the cut, called Point A.
  3. Place the 12cm cake flat on top of the 16cm cake so that their circle bases are tangent at a Point B directly above Point A.
  4. Find Point C on the 16cm cake diametrically opposite to Point B using the first cut. Cut the 16cm cake starting at Point C, following the first cut. When you get to the edge of the 12cm cake, turn and cut following the circumference of the 12cm cake until you get to Point B, creating a fifth volume of cake.

You now have three $50 \pi h$ pieces, one $36 \pi h$ piece, and one $14 \pi h$ piece. You made two cuts.

The difference between the 16cm and 12cm cake is $64 \pi h - 36 \pi h = 28 \pi h$, and the method described will allow you to split the difference.

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  • $\begingroup$ Good but the solution that i have does not require you to use the 20cm cake once they are cut. also pictorial solution would be nice. $\endgroup$ – Hubble07 Aug 18 '14 at 7:31
  • $\begingroup$ This is slightly better than EnvisionAndDevelop's solution because here the 20 cm cake serves as a guide for the second cut. $\endgroup$ – Florian F Sep 14 '14 at 16:47
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You can do it in 3 cuts, simply cut each cake in half. Each half of the biggest is a portion, and combining a half from both of the smaller two makes a portion (of which you have 2).

There are $\pi\cdot20^2\cdot h+\pi\cdot16^2\cdot h+\pi\cdot12^2\cdot h = \pi\cdot800\cdot h\space cm^2$ of cake, so each part needs to be $200\cdot\pi\cdot h\space cm^2$. The first two slices are easy, cut the first cake in half. The whole is $20^2\cdot\pi\cdot h\space cm^2$, so each half is the $200\cdot\pi\cdot h\space cm^2$ we need. Now you're left with two differently sized cakes, which we need to partition into two parts. Conveniently, you can simply cut each of those in half, and combine a half from each for $\frac{\pi\cdot16^2\cdot h}{2}+\frac{\pi\cdot12^2\cdot h}{2} = 200\cdot\pi\cdot h\space$, the quantity we need.

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  • $\begingroup$ good but not the minimum solution. $\endgroup$ – Hubble07 Aug 17 '14 at 18:42
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Just cut all of them in half and combine halves of the cake-16 and cake-12 into two parts.
This would require 3 cuts.

As Duncan said already full volume is $(20^2 + 16^2 + 12^2)*\pi*h = 800 \pi h$. $20^2*\pi*h/2 = 200 \pi h$ will give you two 1/4th parts. And $16^2*\pi*h/2 + 12^2*\pi*h/2= 200 \pi h$ will give you another two 1/4th parts.

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  • $\begingroup$ yes this is a solution but not the minimum one. $\endgroup$ – Hubble07 Aug 17 '14 at 18:41
  • $\begingroup$ @Hubble07, well, if you have an ideal eye you can simple cut required part of cake-14 and add it to cake-12. That will do it in two cuts. But this would not make it a puzzle, would it? Do you have an ideal eye? $\endgroup$ – klm123 Aug 17 '14 at 18:47
  • $\begingroup$ actually you don't need to have an ideal eye to do that second cut. Here symmetry is your friend $\endgroup$ – Hubble07 Aug 17 '14 at 19:18
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Not sure if this is the minimum required, but you can do it in 6 cuts.

Cut the first cake into quarters - two cuts. Cut the second cake into quarters - two cuts. Cut the third cake into quarters - two cuts.

Alternate solution (still 6 cuts):

The total volume of the cakes is $(10^2 + 8^2 + 6^2)*\pi*h = 200 \pi h$. So $4$ equal pieces will have volume of $50 \pi h$ (you can consider just surface area if you like, since $h$ is the same for all). We can get two of those just by cutting the $20$ diameter cake in half ($1$ cut). Next we'd need to add $14 \pi h$ to the $12$ diameter cake to get another $50 \pi h$ group. $14$ out of $64$ is $\frac 7{32}$, which would require $5$ cuts across the middle of the $16$ diameter cake (to cut it into $32$nds) - yielding $6$ total cuts also.

EDIT:

Here's the two cut solution:

1) Cut the $20$ cm cake in half. That gives you two pieces of $50*\pi*h$ volume. The other two cakes are a total of $100*\pi*h$ in volume. 2) Place the $12$ cm cake on top of and tangent to the $16$ cm cake. Start cutting the $16$ cm cake along the edge of the $12$ cm cake. When you reach what would be the end of the diameter for the $12$ cm cake, go perpendicularly out to the edge of the $16$ cm cake. Place the smaller piece of the $16$ cm cake with the $12$ cm cake and you'll have two pieces that are mirror images of each other. Each will be half of $100*\pi*h$.

(Inspiration taken from hkboy's answer - see visual there, but reverse his cut to go from the tangent point out.)

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  • $\begingroup$ Good but far from the minimum possible. Also you took diameter as radius in the volume calculation. You don't need to consider the volume, area of the circle alone will suffice. The answer should be practical i mean you must be able to do the cuts with just the knife. you need to explain how and where exactly are you going to do all the cuts. $\endgroup$ – Hubble07 Aug 17 '14 at 18:08
  • $\begingroup$ Oh, definitely true on the diameter/radius, will update answer. By "you need to explain how and where exactly are you going to do all the cuts" is it enough to say "you cut 1/32 out of the 16 inch cake"? I didn't do that because without some type of measuring device, I figured cutting things in half was the only reasonable measurement you could make. $\endgroup$ – Duncan Aug 18 '14 at 16:44
  • $\begingroup$ Please refer to my comments to others. Remember cutting things in half is not the only reasonable measurement that u can make, its just the first thing that comes to mind. If possible show your cut pictorially. Clue :- Think symmetric. $\endgroup$ – Hubble07 Aug 18 '14 at 17:07
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2 cuts with HORIZONTAL cut. - don't cut the 12cm-diameter cake. it's the first part - cut the 16cm-diameter cake on horizontal way 3/4 and you've got a second part equivalent of the first 12cm-diameter cake - cut the 20cm-diameter cake on horizontal way 6/10 and you've got a third part equivalent of the first 12cm-diameter cake - the final part is composed with 1/4 of the 12cm-diameter cake + 4/10 of the 20cm-diameter cake

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    $\begingroup$ A diagram would really help this answer, $\endgroup$ – Bob Jun 22 '15 at 16:46

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