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Driven out of a serious question, when sharing a slice of cake in a coffee shop how can my two friends split it without going down the middle (the cake is likely to crumble if you do this!)

Given a slice of cake, how can it be divided into two equal parts? Without going 'down the middle'.

How far along the already-cut side of the cake do you need to cut? Assume the slice of cake is a standard circle divided into 8 (N), and has a radius of 15cm (r). Bonus points for a more general solution relating to how many pieces it's cut into (N) and how large the cake is (r).

Is there a more practical solution (or 'close enough') without taking a ruler and calculator to the cafe?

One (vertical, straight) cut only.

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  • $\begingroup$ One vertical straight cut? Or following the curve of the circle? $\endgroup$ – Hugh Dec 8 '18 at 22:07
  • $\begingroup$ Vertical straight cut, yes. Should have clarified! $\endgroup$ – user3108295 Dec 8 '18 at 22:30
  • $\begingroup$ Although JohnMark Perry's answer is very funny, this question only has the mathematics tag and does not have the lateral-thinking tag. Based off that, I'm going to say that this belongs in Math.SE, and not here. $\endgroup$ – Hugh Dec 9 '18 at 0:21
  • $\begingroup$ @Hugh well, let's not assume new users know our "tag codes" :-) $\endgroup$ – deep thought Dec 9 '18 at 1:52
  • $\begingroup$ @deepthought technically, yes. I'm making the assumption mainly because it says ignoring the trivial solution in the first line. $\endgroup$ – Hugh Dec 9 '18 at 3:18
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Place the cake on one side, and then slice down the middle.

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    $\begingroup$ Very funny, although I think that this constitutes "trivial"... $\endgroup$ – Hugh Dec 9 '18 at 0:07
  • $\begingroup$ "Note: The "down the middle" solution is just one of the two trivial solutions. Viewn in 3D you can also cut it at half the height." -- that was just given in my answer! But, nice to see that there are other people having the same idea. However, you actually made it a "vertical cut" by turning it! :P $\endgroup$ – SDwarfs Dec 9 '18 at 20:16
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Assuming that we are making a cut that is perpendicular to the line that bisects the angle at the centre of the slice, here is the general solution.

Note that we can simply look at the slice from above, and make a cut such that the areas on both sides of the cut are identical.

First, we can say that the area of the slice is $\frac{\pi{}R^2}{s}$, where $R$ represents the radius of the circle which we are slicing, and $s$ represents the number of slices we are cutting the circle into.

Additionally, the angle subtended by the "sides" of our slice is $\frac{360}{s}^{\circ}$, or $\frac{2\pi}{s} \textrm{rad}$. We can use $\theta$ to represent this angle.

Next, we make our "cut", and we call the distance from the ends of the cut to the origin $r$. Since the 'inner' piece is a triangle where we know the lengths of two sides and an angle, we can use the formula for the area of an SAS triangle $\frac{ab\sin{C}}{2}$. Therefore, the area of our 'inner' piece is $\frac{r^2\sin{\theta}}{2}$.
We can then subtract the area of the 'inner' piece from the area of the whole slice to obtain the area of our 'outer' piece; $\frac{\pi{}R^2}{s} - \frac{r^2\sin{\theta}}{2}$.

Next, we set our two areas equal to each other, and solve for $r$ to obtain the distance from the end of the cut to the origin.

$$\begin{align}\frac{r^2\sin{\theta}}{2} &= \frac{\pi{}R^2}{s} - \frac{r^2\sin{\theta}}{2} \\ 2\frac{r^2\sin{\theta}}{2} &= \frac{\pi{}R^2}{s} \\ r^2\sin{\theta} &= \frac{\pi{}R^2}{s} \\ r^2 &= \frac{\frac{\pi{}R^2}{s}}{\sin{\theta}} \\ &= \frac{\pi{}R^2}{s\sin{\theta}} \\ r &= R\sqrt{\frac{\pi}{s\sin{\theta}}}\end{align}$$
(Where $R$ represents the radius of the circle (or cake), $s$ represents the number of slices into which we cut the circle (or cake), and $\theta$ represents the the angle subtended by the "sides" of our slice.)
Since $s=\frac{2\pi}{\theta}$, we can also write this as $$r=R\sqrt{\frac{\theta}{2\sin\theta}}$$


In the specific case asked in the question, the radius of the slice ($R$) is $15$cm and the number of slices ($s$) is $8$.

$$\begin{align}r &= 15\sqrt{\frac{\pi{}}{8\sin{\theta}}}, \textrm{with } \theta = \frac{360}{8} = 45^{\circ} \\ & = 15\sqrt{\frac{\pi}{8\sin{45^{\circ}}}}\\ r & \approx 11.1783756707 \\ \end{align}$$

Therefore, the cut that divides the pie into two equal slices goes through the two points that are roughly $11.18$ cm from the origin.

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Note: The "down the middle" solution is just one of the two trivial solutions. Viewn in 3D you can also cut it at half the height. But, assuming it's a layered cake without special toppings and both partitions shall have the same amount of each layer, it's of course more complicated! :P The area $A$ in top-down view of a full cake of diameter $r$ is: $A = \pi r^2$ and each of $N$ equally sized partitions has therefore an area of $A_{part}=\frac{\pi r²}{N}$. We want to split one of the partitions into two parts, which have the same area. Hence, each part must have an area of $\frac{A_{part}}{2}$. The part which was in the middle of the cake will have the shape of a isosceles triangle, which means two sides and two angles are the same. The third angle (in the middle) of this triangle is $\alpha=\frac{360°}{N}$. It's area is defined by $A = \frac{a^2}{2} sin(\alpha)$ ($a$ being the length of the equal sides). We want this area to be equal to $\frac{A_{part}}{2}$. Therefore we need to solve: $\frac{\pi r^2}{2 N} = \frac{a²}{2}sin(\frac{360°}{N})$ for $a$. The solution to this is: $a = \sqrt{\frac{\pi r^2}{N sin(\frac{360°}{N})}}$. For $N=8$ and $r=15cm$ the solution is: $a = \sqrt{\frac{\pi (15cm)^2}{8 sin(\frac{360°}{8})}} = 11.664 cm$. Hence the cut needs to be made in a distance of a bit more than 11.664 cm measured along the side of the piece starting at the tip.

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  • $\begingroup$ Hi SDwarfs, wonderful answer and well done for working it out! But, I've already given this solution in my answer, below. On Puzzling.SE, we typically don't answer questions with answers that have already been given, since that is redundant. If you want to add your own methods, it's best to edit them into the existing answer. $\endgroup$ – Hugh Dec 9 '18 at 2:28
  • $\begingroup$ Ah, ok... So, how to do that without "sneaking" at other users answers? Also it seems we both have different results, if I compare the numbers. You result is 11.18 cm and mine is 11.664 cm. $\endgroup$ – SDwarfs Dec 9 '18 at 11:54
  • $\begingroup$ In regards to the math, I believe you've missed something somewhere in your calculations: wolframalpha.com/input/… $\endgroup$ – Hugh Dec 9 '18 at 15:38
  • $\begingroup$ In regards to the answer, it's a difficult question. I do not want this to come across as me criticising or attacking you, but... you may as well have not posted it in the first place. This is mainly because our answers are pretty much identical. Not only do we have the same answers, but, the methods we decided to use are pretty much identical. Once again, I do not want this to come across as me criticising or attacking you, but... it kinda feels like you've copied my answer. [2] $\endgroup$ – Hugh Dec 9 '18 at 15:54
  • $\begingroup$ [2] ...Now of course, it's up to the asker to accept the answer of their choosing. If he/she accepts yours, I mean—I was technically first, by thirty or so minutes—but I wouldn't be mad. [3] $\endgroup$ – Hugh Dec 9 '18 at 15:54
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For $n \geq 8:$

Viewing the tip of the slice as the origin and following one edge of the slice, we are trying the find the distance $x$ from the origin in which a perpendicular cut would give two pieces with same area. A perpendicular cut on the slice would result in one piece being a right triangle, so the question is now to find the cut which would give a triangle with area that is half of the slice.

$A_{cake} = \pi r^2$
$A_{slice} = \frac{A_{cake}}{n}$
$\theta_{slice} = \frac{360^\circ}{n}$

$A_{halfSlice} = \frac{xy}{2}$, where $x$ is the distance from the origin, and $y$ is the length of the cut. After finding $y$, we can then put $A_{halfSlice}$ in terms of $x$:

$y = xtan(\theta_{slice})$
$A_{halfSlice} = \frac{x^2tan(\theta_{slice})}{2}$

Finally, we can solve for $x$, given that we know that $A_{halfSlice} = \frac{A_{slice}}{2}$:

$\frac{\pi r^2}{n} = x^2tan(\theta_{slice})$

$x = \frac{\sqrt{\pi}r}{\sqrt{n}\sqrt{tan(\theta_{slice})}}$

So for $n=8$ and $r=15cm$, we get $x \approx 9.399cm$. Still unsure about $n$ below 8 though...

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The new version of the question reads "is there a more practical solution (or 'close enough') without taking a ruler and calculator to the cafe?"

I'd say

cut from an outer corner of the cake to about halfway the opposite side, a bit closer to the outside.

The ratio you are aiming for is

5/9 of the radius

If you could do that, how far off from fair would you be?

- The area of the slice is $$\frac{\pi R^2}{8}$$
- The area of a triangle with one side $R$, one side $\frac{5}{9}R$ and interior angle $45^\circ$ is $$\frac{1}{2}R(\frac{5}{9}R)\sin(45^\circ) = \frac{5R^2}{18\sqrt{2}}$$ - The ratio of those areas is
$$\frac{5}{18\sqrt{2}}\cdot(\frac{\pi}{8})^{-1} = 0.5002...$$

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