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Mrs. Betty made a squared cake with chocolate frosting for his neighbors to the afternoon tea. However, first she sliced a middle piece for her two grandchildren and cut it in half:

enter image description here

There was no leftovers from the rest of the cake after the afternoon tea.

At the end of the day, Mrs. Betty was anxiously waiting for her two grandchildren to give them the two pieces of cake. However, when they came, they brought a friend with them to also eat a slice of the famous chocolate frosting cake.

Mrs. Betty has now to cut the two slices into several pieces to divide them for the three children. How can she do it by making straight cuts in order to divide equitably the two pieces (with the same amount of sponge cake and chocolate frosting) by the three children?

Please assume that:

a) The two pieces are perfectly equal to each other;
b) The two pieces are perfectly rectangular;
c) The thickness of the frosting is the same on the entire cake.

HINT:

Mrs. Betty can solve the challenge with four straight cuts!

I read a similar puzzle on an old book of mine, but since it was too easy, I decided to create this more elaborated version of it.

EDIT:

Sorry, I forgot! No measurements allowed! Imagine Mrs. Betty has only a straightedge and a compass to help her (like an Euclidian postulate).

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  • 1
    $\begingroup$ Does this require stacking the cakes before cutting? $\endgroup$
    – littlecat
    May 17 at 16:58
  • $\begingroup$ No it doesn't...! $\endgroup$
    – Pspl
    May 18 at 7:12
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    $\begingroup$ A compass isn't my idea of a classical kitchen implement. Mentioning it only after letting this question sit for half a day is a serious omission IMO. Besides, if you allow compass and straight edge then @Lawrence's solution which requires fewer than four cuts no matter how you count them is perfectly valid. $\endgroup$ May 18 at 8:09
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    $\begingroup$ I'm voting to close this puzzle because it has already generated three perfectly good but "unintended" answers which OP invalidated not by closing loopholes but by fundamentally changing the puzzle. Answerers should not be punished for askers incapable of or too lazy to formulate a question in such a way that it matches whatever solution they have in mind. $\endgroup$
    – loopy walt
    May 18 at 17:36
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    $\begingroup$ @Pspl You have accepted an answer that infringes your last edit ("d) The thickness of the pieces is the same as the original slices") so I have reverted that edit. BTW, you say in comment that you have a better solution to your puzzle; would you mind adding it to the answers? $\endgroup$
    – xhienne
    May 19 at 10:03
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1 cut across both slices

She sliced them fillet-style, 1/3 and 2/3 the thickness of each of the original two slices. This maintains the cross-section, frosting and all.

Dividing a line into 3 parts can be done by construction, so no measurement is needed.

As @Jeffrey notes, only one cut per slice of cake is needed, dividing each slice into one thin slice 1/3 the original thickness and one thick slice 2/3 the original thickness. If the original slices are laid side-by-side, this can even be optimised to a single filleting cut across both slices.

One child gets two thin slices, the other two get one thick slice each.

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    $\begingroup$ I interpret the "no measurements allowed" restriction to mean that she cannot determine that 1/3 thickness directly. $\endgroup$ May 17 at 16:54
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    $\begingroup$ You only need to cut 1/3 off each slice, meaning a total of 2 cuts. I like the answer btw :) $\endgroup$
    – Jeffrey
    May 17 at 19:13
  • $\begingroup$ @Jeffrey OP's question clearly states that one isn't allowed to measure the cakes, which is required to execute this strategy. $\endgroup$ May 17 at 19:45
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    $\begingroup$ @DonThousand eeh so? The anwear isn't correct (yet), but it is creative. Most ppl will cut the cake along the transverse/sagittal plane, few will think about the Frontal plane (aka fillet stile). I merely stated I liked it... $\endgroup$
    – Jeffrey
    May 17 at 21:18
  • $\begingroup$ Thanks @Jeffrey! I’ve incorporated your excellent observation into the answer. $\endgroup$
    – Lawrence
    May 17 at 23:52
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Slice as in the picture, give the guest the pieces marked A of both cakes; give one grandchild the rest of slice1 and the other the rest of slice 2.

enter image description here
note: It does not matter which sides have the frosting.

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  • $\begingroup$ The one with all the ‘A’s get more than half the frosting? That’s all the frosting from one of the original slices and some from the other. $\endgroup$
    – Lawrence
    May 17 at 23:51
  • $\begingroup$ @Lawrence note: It does not matter which sides have the frosting. $\endgroup$
    – iBug
    May 18 at 0:47
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    $\begingroup$ @Lawrence The A on top-left covers 1/3 of the left edge and the top edge, and the A on the bottom covers 1/3 of only the bottom edge, and the A on the right covers only 1/3 of the right edge. $\endgroup$
    – iBug
    May 18 at 1:03
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    $\begingroup$ @Lawrence These are the parts that "A" pieces cover on the sides. It doesn't matter which sides have the frosting, even two opposing sides. $\endgroup$
    – iBug
    May 18 at 5:58
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    $\begingroup$ Good work, but the first cut counts as two (one for each slice). Besides that, you should not leave an entire corner of frosting uncut. +1 though $\endgroup$
    – Pspl
    May 18 at 7:27
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Here is a method that allocates fairly, including the iced corner:

We do not require two separate pieces, so we simply stack them.

Side view (apply 2x -> 4 cuts total):

enter image description here

Kid 1 (1 piece):

enter image description here

Kid 2 (4 pieces):

enter image description here

Kid 3 (4 pieces).

enter image description here

How does it work?

Apply the fact that a cone's volume is 1/3 x base x height to kids 1 and 2.

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  • $\begingroup$ Haha, nice observation. +1 $\endgroup$
    – Lawrence
    May 18 at 3:13
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    $\begingroup$ This is stupidly smart! I haven't even imagined about the cone volume formula. $\endgroup$
    – iBug
    May 18 at 3:19
  • $\begingroup$ Will the icing be shared fairly? I haven't done the proof, but just imagining a limiting case where the non-icing is a skinny column in the bottom-left, the base of the pyramid (kid 1) will get hardly any of it (it's at the point) while the other two will have almost all the non-icing. $\endgroup$
    – Lawrence
    May 18 at 3:23
  • $\begingroup$ @Lawrence the entire method assumes the pieces are lying on their sides, the iced edges are right and back in the pictures. Essentially the same two cut procedure (yielding a skewed pyramid, an upside-down skewed pyramid and two even more skewed three sided pyramids) is applied to four cuboids formed by the sponge portion the icing corner and the two icing edges without the corner. $\endgroup$ May 18 at 3:31
  • $\begingroup$ It's a smart answer indeed. But this is not the solution I was looking for. There is a better solution (which doesn't imply stacking the two slices). $\endgroup$
    – Pspl
    May 18 at 7:24
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I found a solution that divides the cake & frosting into exactly 3 portions without any measuring:

1) Cut each slice diagonally so that we have 2 non-frosted slices and 2 frosted slices.

2) Align the slices to make one non-frosted rectangle over one frosted rectangle.

3) Mark diagonally across both rectangles, and there you'll be able to cut the rectangles into thirds. The non-frosted rectangle is already divided, now to divide the frosted rectangle so that every child gets the same amount of frosting.

4) Cut the frosted rectangle to form a tic-tac-toe grid with an X in each box. Now each child can get 2 of each of the different slices frosted slices and 4 of each of the non-frosted slices of the frosted rectangle.

Demonstration:

enter image description here

With the three portions something fishy is going on...

enter image description here

Poor children are getting served a slice of bread!

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Here's my solution:

Cake-slicing diagram

Start by cutting through the icing at the corner of each slice, from the corner of the icing at a 45-degree angle to the corner of the actual cake. Then, cut from those points through the actual cake to the opposite corner, where there is no icing.

This gives four slices that may not be exactly equal if the cake is rectangular, but they don't have to be exactly equal at this point.

Next, cut each of these three triangular slices into three wedges. Start with the icing: mark points on the outer and inner surfaces of the icing 1/3 and 2/3 of the way from the right-angle corner of the cake triangle. Then, make two straight-line cuts through the icing, dividing the icing into three trapezoids with the same area.

Then, cut through the cake, from the end of the icing cuts to the opposite corner of the triangle.

All three wedges from each one of the original four cake triangles has exactly the same amount of cake and exactly the same amount of icing. So, give one wedge from each of the original four triangles to each kid, and they'll be happy.

However, this solution requires twenty straight-line cuts in total: ten through the cake, and ten through the icing. The image above shows a way to improve this, by only cutting between wedges of different colors. Give the red part to one grandchild, the green part to the second grandchild, and the blue wedges to their friend. This requires only four cake cuts and four icing cuts, making for eight in total.

Furthermore, the two rectangular slices can be stacked on top of each other, so that both can be cut through at the same time. This reduces the total cuts to four.

Finally, if the aspect ratio of the original cake is exactly 2:1, meaning that the two rectangular slices are in fact perfect squares, then the cake and icing do not need to be cut separately, because the cake and icing cuts would already be collinear. By stacking these square slices on top of each other, the puzzle can be solved with only two straight-line cuts.

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  • $\begingroup$ "mark points on the outer and inner surfaces of the icing 1/3 and 2/3 of the way..." But that would require measuring, no? $\endgroup$ May 19 at 0:12
  • $\begingroup$ @riskymysteries It'd be easiest by measuring, sure, but it's possible to divide a line segment into thirds with straightedge and compass. This video shows how: youtube.com/watch?v=2EGO6przl1k $\endgroup$ May 19 at 0:24
  • $\begingroup$ Oh, okay then :) $\endgroup$ May 19 at 18:03
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6 cuts, maintains thickness

This answer relies on the same construction as my other answer to cut a rectangular piece orthogonally into 1/3 and 2/3 portions.

In this answer, if the knife goes through 2 slices, it’s counted as 2 cuts.

Arrange the 2 slices side by side so that the icing runs along the left, top, and right sides of the arrangement.

Cuts 1 & 2: slice off the top rectangle of icing by cutting horizontally where the icing meets the cake. This results in separating the full horizontal length of top icing (in 2 pieces), leaving two pieces of cake with icing along the left and right sides respectively.

Cuts 3 & 4: cut each of the 2 pieces of horizontal icing vertically 1/3 the way across to form 2 sets of 1/3 and 2/3 portions of icing.

Cuts 5 & 6: slice horizontally through the 2 remaining pieces of cake to produce 2 sets of 1/3 and 2/3 pieces. Each piece will have icing on the left or right.

Give one child all the 1/3 pieces, and each of the other 2 children one of the 2/3 pieces of icing and one of the 2/3 pieces of cake.

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