Cutting four equal segments from the perimiter of a rectangle

Imagine you've got a rectangular cake and you want to cut the edges off to serve to four guests, and you want to cut it into four equal segments. If you cut off the two short edges, then cut the remaining long edges, could you get four equal portions if the width of the slices were just right?

It seems to me that if the cake is Long by Short (L by S), and the segment is sL by sS, then sL = S and 2(sS)+sL = L. Can such a rectangular cake exist? Are there limits where the cake is too square-like or too line-like for this to work?

EDIT: The original intent was to have four congruent (same shape?) pieces, but not necessarily to cut up the whole cake. I tried to draw a picture to represent the idea, though the four pieces in my badly drawn picture aren't actually the same shape. Figuring out how to always make them the same shape is what this is all about, after all :) EDIT 2: Now that I think more about it, it seems that a cake of 2:1 ratio will cut into exactly four equal parts, because the length of the long end of the segment will be exactly twice the length of the short end of the segment. So 2:1 seems to be the most "line-like" rectangle you can cut this way.

On the other hand, it seems you can't cut a square this way, because the length of the "side" pieces are necessarily longer than the length of the "inner" pieces, meaning they're no longer the same shape. So I guess to cut a cake like this, the cake must be greater than 1:1 and less than or equal to 2:1?

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• By "equal", do you mean "congruent" or "equal in area"? – joriki Jan 17 '13 at 9:42

If we let the short side of the end piece be $s$, the end rectangles are $s \times S$. The remaining length of the cake is then $L-2s$. If you want this to be the same as the long side of the end piece you must have $L-2s=S$. To have the ability to cut two more pieces $s \times S$ as in your figure, we must have $S \ge 2s$. Plugging this into the previous, we must have $L \ge 4s$. You can "cut" a square cake this way, you just have to choose $s=0$. Any cake that is not square you can cut this way by choosing $s=\frac 12(L-S)$ until it gets too long so you will more than meet in the middle. This happens when $\frac S2 = \frac 12 (L-S)$ or $L=2S$. For a longer rectangle, the pieces along the long side will either be too long or will overlap in the middle.

Edit in response to clarification. The problem is to cut a cake as in the diagram. The pieces must be equal in size (hence, of course, in area) and you are allowed to have cake left over after cutting.

Let the height of the remaining portion be $m$ and the width of the short side pieces be $s$. As in the question, let the length of the long and short side, respectively, be $L$ and $S$.

The short side pieces will have dimensions $(S, s)$ and the long side pieces will have dimensions $((S-m)/2, L-2s)$. If the areas are equal, we'll require $$S\,s = \left(\frac{S-m}{2}(L-2s)\right)$$ and if the short side and the long side pieces are equal, we'll have either

1. $(S, s)=((S-m)/2, L-2s)$ or
2. $(S, s)=(L-2s, (S-m)/2)$

In the first case, we'll have $S=(S-m)/2$ which implies the impossible conclusion $S\le 0$.

In the second case we'll have \begin{align} (1)\qquad S&=L-2s\\ (2)\qquad s&=\frac{S-m}{2} \end{align} So in the equation above for the areas we have, substituting $s=(S-m)/2$, $$S\left(\frac{S-m}{2}\right)=\left(\frac{S-m}{2}\right)(L-2s)$$ and so $$S=L-2s=L-(S-m)$$ so $2S-L=m$ and since $0\le m\le S$ we have $S\le L$, to no one's surprise..

In the boundary cases, we see that if $m=S$ we have $S=L$ and $s=0$, i.e., the entire (square) cake remains. At the other extreme, if $m=0$ we have $L=2S$ and $s=S/2$, so each piece will have dimensions $(S, S/2)$, in some order. In general, if $m=t\;S$ with $0\le t\le 1$ we'll have equal pieces when \begin{align} L&=(2-t)S\\ s&=\frac{(1-t)S}{2} \end{align}