fi-le's files
Today, you and your friend are eating cake with raisins on it. Both of you are very polite - so you insist your partner gets half of the cake AND half of the raisins! However, it seems a little bit hard to make the right cut...
Given a square cake with a finite number of raisins - can you always make a single straight cut, so that the area of the cake and the number of raisins is equal on both slices?
(Please take a minute to think about what the answer might be.)
Pondering and trying around, your friend finds a cut that does the job. Hm! Obviously you and your friend are not in the mood for eating cakes any longer! You get out apron, flour and raisins to make more tricky cakes.
As a first step, let's try with a circle shaped cake first. Circles are easier to think about, because the cuts that half the area are pretty obvious: They all go through the center! So here, the cut is described by only one number between 0 and π, namely the angle to the horizontal. Now we can bake a wicked cake, this one:
Truly scary! No matter how you cut it, the raisins are distributed unfairly. So probably, a similar square cake could be found.
Now, nobody wants to throw out their good manners just because of one naughty cake. So in preperation for being confronted with a cake in the future, you ask:
For what cakes can we make a fair cut?
Hm, we first return back to our circle cake. The wickedness of our wicked cake is that multiple raisins are on the same line to the center. So maybe, let's bake it so that no two raisins have this property. For every such cake you give your friend, they can always make a fair cut. So you try your hand at a proof:
As mentioned, on a circle cake, an area-fair cut can be described with a single angle, \(\theta \in [0, \pi]\). The raisins are points on the unit circle in polar coordinates, like this: \( P_i = (r_i, \theta_i) \) with \( \theta_i \neq \theta_j\) for \( i, j \in \{0,..., n\} \). We then define a function \( f : [0, 2\pi] \to \{-n, ..., n \} \) with \(f(\theta)\) as the number of raisins right of the cut minus the raisins left of the cut. By this definition, it's clear that any \(\theta\) with \(f(\theta) = 0\) is a fair cut. Nice!
Next, we see that \(f(\theta) = -f(\theta + \pi)\), because right and left of the cut are flipped by going halfway along the circle with the cut. In particular \(f(0) = - f(\pi) \). Lastly, since \( \theta_i \neq \theta_j\), \( f(\theta)\) only increases or decreases by one. It is constant between raisins. When it hits a raisin, it changes by one - as it leaves it changes one further. Try it! Therefore, \(f\) passes through all natural numbers between \(f(0)\) and \(-f(0)\). It therefore has a root and there exists a fair cut!
Your friend agrees; that does make sense. They add that it looks like you can construct the fair cut, too: Cutting at an angle of \(median \{\theta_1, ... \theta_n \} \) will be a fair cut.
Now you feel ready to answer the original question! Again, you notice a peculiar fact about area-fair cuts through rectangles - they always go through the middle. The middle?! You're in the same situation as with the circle! You simply have to put the coordinate origin on the middle of the cake. Really, the only thing that's different at all is the possible values of \(r_i\). Come to think of it, those were completely irrelevant in the proof! Quickly you pick up the flour again to make a wicked square cake.
Any old (convex) cake can be cut like the circle! Find it's center of "mass"; then all area-fair cuts go through it. Therefore we can describe them by a half rotation. A fair cut exists if and only if the resins have different angles in polar coordinates with the center of mass as their origin!
Bon appétit ~
Some more cake for thought:
- The first two cakes displayed on this page are a little bit random! You can reload to see different variations. I only implemented some more aestetically pleasing and solvable cakes though.
- Would this also work in more than two dimensions? Then, we also want to distribute the raisins in the (hyper)cake.
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Maybe your friend happens to be a measure theorist and suggests thinking about it like this: The cake is a measurable space \((\Omega, \sigma) \subset (\mathbb{R}^2, \mathcal B(\mathbb{R}^2)) \), but with two different measures \(\mu, \nu \). We are looking for \( A \) with special restrictions, so that \( \mu(A) = \mu(A^c) \land \nu(A) = \nu(A^c) \). In the above, we used \(\mu = \lambda, \nu = {1 \over n} \Sigma^n_{k = 1} \delta_{P_k} \).
( By \( \lambda, \delta \) I mean Lebesgue measure and Dirac Delta measure and by \(\mathcal B \) the Borel sigma algebra.)
However, we have additional restrictions on \(A\) - in our case straight cuts - that I wouldn't know to describe well. If that could be resolved somehow, it would be fun to use other measures! For example, we could have a continuously distributed icing as the second measure.