Tukukkk

Given a slice of cake, how can it be divided into two equal parts? Ignoring the trivial solution of '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).

One (vertical, straight) cut only.

Place the cake on one side, and then slice down the middle.

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 $frac{2pi}{s} textrm{rad}$. 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 $frac{absin{C}}{2}$. Therefore, the area of our 'inner' piece is $frac{r^2sin{theta}}{2}$.


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; $frac{pi{}R^2}{s} - frac{r^2sin{theta}}{2}$.


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

$frac{r^2sin{theta}}{2} = frac{pi{}R^2}{s} - frac{r^2sin{theta}}{2}$

$2frac{r^2sin{theta}}{2} = frac{pi{}R^2}{s}$

$r^2sin{theta} = frac{pi{}R^2}{s}$

$r^2 = frac{frac{pi{}R^2}{s}}{sin{theta}} = frac{pi{}R^2}{ssin{theta}}$

$r = sqrt{frac{pi{}R^2}{ssin{theta}}}$, 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.

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

$theta = frac{360}{8} = 45^{circ}$

$r = sqrt{frac{pi{}15^2}{8sin{45^{circ}}}} = sqrt{frac{225pi}{8sin{45^{circ}}}} approx 11.1783756707$


Therefore, the cut that divides the pie into two equal slices goes through the two points that are roughly $11.17$ cm from the origin.

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.


$A_{cake} = pi r^2$

$A_{slice} = frac{A_{cake}}{n}$

$theta_{slice} = frac{360^circ}{n}$


$A_{halfSlice} = xy$, 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 $x$:


$y = xtan(theta_{slice})$

$A_{halfSlice} = x^2tan(theta_{slice})$



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


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


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



So for $n=8$ and $r=15cm$, we get $x approx 6.6467cm$.
Still unsure about $n$ below 8 though...

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.