## Thirds

Kitt wrote an entry about splitting a pastry in thirds, which has a few different solutions. I hashed out what I thought was a correct solution in the comments but I’d actually made a pretty big mistake that came from me not actually drawing a diagram. So here’s a version with diagrams.

## The problem

Let’s say you have a pastry that’s X units wide and Y units tall, and you want to split it into even thirds.

## Trivial: trisection

One simple solution is to bisect the pastry 1/3 across, then to bisect the larger segment:

This gives you three sections, $$A$$, $$B$$, and $$C$$, which each have an area of $$\frac{XY}{3}$$.

But this has a problem: inequal distribution of crust! $$A$$’s crust length is $$\frac{2X}{3} + Y$$ whereas $$B$$ and $$C$$ have a crust length of $$\frac{2X}{3} + \frac{Y}{2}$$, meaning $$A$$ will always get $$\frac{Y}{2}$$ more crust than $$B$$ and $$C$$. This is clearly unfair!

How else can we solve this problem?

## Trangular cut

Another approach is to make $$A$$ a triangular cut out of the side, and $$B$$ and $$C$$ into trapezoids:

This gives us areas of $$A=\frac{2X}{3}\frac{Y}{2}$$ (i.e. $$A=\frac{XY}{3}$$) and $$B=C=\frac{X+\frac{X}{3}}{2}\frac{Y}{2}$$ which, as you might expect, also simplifies to $$\frac{XY}{3}$$. The crust allotments are now $$A=Y$$ and $$B=C=X+\frac{Y}{2}$$. So we can work out the appropriate pastry size to get equal area and crust:

$Y=X+\frac{Y}{2} \\ \frac{Y}{2}=X$

Or, to check our math a bit differently, we want the $$Y$$ edge to be 1/3 of the total edge length $$2(X+Y)$$, so:

$Y=\frac{2X+2Y}{3} \\ 3Y=2X+2Y \\ Y=2X$

which says exactly the same thing.

So if the pastry is twice as tall as it is wide, we’re done! So in this particular pastry we’d actually want the cuts to look more like this:

Where in the above situation, the areas and the crusts are equal for all three pieces.

However, this isn’t good enough! We want to generalize a solution to all possible pastry aspect ratios!

## Generalized solution

So, first, let’s orient the pastry so that $$X > Y$$, i.e. it’s a “lansdcape” aspect, as above.

Now, we know that if $$Y=\frac{X}{2}$$, both constraints are satisfied. But the crust allocation for $$A$$ is always $$X$$ and $$B=C=Y+\frac{X}{2}$$. Which means that if the aspect of the pastry tends more square, then $$A$$ gets less crust proportionally, and if the rectangular gets more elongated, $$A$$ gets proportionally more crust.

So we need to work out two solutions, one for pastries which are less than a 2:1 aspect, and ones which are more than 2:1.

### When $$X < 2Y$$

We add two offsets, $$a$$ and $$b$$, which affect the pastry division thusly:

In this case, section $$A$$’s crust amount is $$X+2a$$, and so we can solve for $$a$$:

$X+2a=\frac{2X + 2Y}{3} \\ 3X+6a=2X+2Y \\ 6a=2Y-X \\ a=\frac{2Y - X}{6}$

So, from this we can see that $$a$$ is only a sensible (i.e. non-negative) value if $$X \leq 2Y$$, so this setup continues to make sense with our preconditions.

Anyway, now that we’ve solved for $$a$$, we can solve for $$b$$:

$\frac{X}{2} \times \frac{b + Y - a}{2} = \frac{XY}{3} \\ \frac{X(b + Y - a)}{4} = \frac{XY}{3} \\ 3X(b + Y - a) = 4XY \\ b + Y - a = \frac{4Y}{3} \\ b = \frac{Y}{3} + a$

So let’s take a look at Kitt’s pastry conundrum and work out how it could have been split into perfect thirds. We can do an approximate measurement on the image; let’s just use pixels as the size unit here and assume that perspective more or less averages out:

Just to keep life simple let’s say that it’s $$X=270$$ and $$Y=240$$, which happens to be an aspect ratio of 9:8 although that doesn’t really matter right now. If we plug those values into the equation, we get $$a=35$$ and $$b=115$$, so this trisection of the pastry should produce equal area and edge:

We should, of course, double-check our math here.

The total edge of this pastry is $$2(270+240) = 1020$$, so each section should have an edge of 340. The edge of section $$A$$ is $$270 + 2 \times 35 = 340$$, so that checks out. (For that matter, it would be even easier to see that $$B$$ and $$C$$ both have $$135 + 205 = 340$$ crust units.)

The total area of this pastry is $$270 \times 240 = 64800$$, so each section should have an area of 21600. The area of section $$B$$ is $$135 \times \frac{240 - 35 + 115}{2} = 21600$$. So at least for this pastry, the math works out!

The theoretical limit for this setup (since we want to restrict ourselves to orienting the pastry such that $$X \geq Y$$) is a width and a height of 1. In that case, we’d have $$a = \frac{1}{6}$$ and $$b = \frac{1}{3} + \frac{1}{6} = \frac{1}{2}$$. Or in other words, it would look like this:

Theoretically the pastry could continue to get narrower as long as $$0 \leq a \leq Y$$ and $$0 \leq b \leq Y$$. We know $$a \geq 0$$ as long as $$X \leq 2Y$$, and we know $$b \geq 0$$ as long as

$\frac{Y}{3} + \frac{2Y - X}{6} \geq 0 \\ \frac{4Y - X}{6} \geq 0 \\$

or, in other words, $$4Y \geq X$$ – which is always true given the above. But what about the other conditions? First, we check that $$a \leq Y$$:

$\frac{2Y - X}{6} \leq Y \\ 2Y - X \leq 6Y \\ {-X} \leq 4Y 0 \leq 4Y + X$

Well, since $$X$$ and $$Y$$ are always positive, it’s safe to say that this condition is true. What about $$b \leq Y$$?

$\frac{Y}{3} + \frac{2Y - X}{6} \leq Y \\ 2Y + 2Y - X \leq 6Y \\ 0 \leq 2Y + X$

Which is, again, always true. So, this setup can always be used for any pastry; if it’s more than twice as wide than it is tall, simply turn it sideways. But this might make for some very slim cuts, so we still want a general solution that works for an extra-wide pastry. Also, it would get weird if we ever have a situation where $$Y - a < b$$. Does that ever happen?

$Y - a < b \\ Y - a < \frac{Y}{3} + a \\ \frac{2Y}{3} < 2a \\ \frac{2Y}{3} < \frac{2Y - X}{3} \\ 2Y < 2Y - X \\ 0 < -X \\ X < 0$

Okay, so we never actually get into that situation, at least.

### When $$X > 2Y$$

So how can we solve for pastries which are wider than a 2:1 aspect? Let’s try this setup:

In this case, section $$A$$’s crust amount is $$X-2a$$, so we again solve for $$a$$:

$X - 2a = \frac{2X + 2Y}{3} \\ 3X - 6a = 2X + 2Y \\ -6a = 2Y - X \\ a = \frac{X - 2Y}{6}$

And now $$a$$ is only sensible (i.e. non-negative) if $$X \geq 2Y$$, so this setup once again makes sense with our preconditions.

So again, now we solve for $$b$$; this time we can use the area of section $$A$$ as our guide, as it’s much more straightforward to compute:

$\frac{(X - 2a)(Y - b)}{2} = \frac{XY}{3} \\ 3(X - 2a)(Y - b) = 2XY \\ 3(XY - 2aY - bX + 2ab) = 2XY \\ 3XY - 6aY - 3bX + 6ab = 2XY \\ 6ab - 3Xb = 6aY - XY \\ b = \frac{6aY - XY}{6a - 3X}\\ b = \frac{(X - 2Y)Y - XY}{X - 2Y - 3X} \\ b = \frac{2Y^2}{2X + 2Y} \\ b = \frac{Y^2}{X + Y}$

So, hey, that simplifies pretty nicely. So, let’s say we have a pastry that’s, say, 3 units wide and 1 unit tall. In this case, $$a=\frac{1}{6}$$ and $$b = \frac{1}{4}$$. Let’s verify that this makes sense!

This theoretical pastry would have a total area of 3, and a total edge of 8. So we’d need section $$A$$ to have 1 unit of area and $$\frac{8}{3}$$ units of edge. How does this work out?

Edge length is $$X-2a = 3 - \frac{2}{6} = \frac{8}{3}$$. Phew!

And the area is $$\frac{(X - 2a)(Y - b)}{2} = \frac{\frac{8}{3} \times \frac{3}{4}}{2} = 1$$. Alright!

Are there any limits to how wide the pastry can be? As always, we want $$a \geq 0$$ and $$b \geq 0$$. We already know this to be true for $$a$$ as long as $$X \geq 2Y$$ (which is the situation we have this setup for anyway), and it’s pretty obvious that $$b \geq 0$$ as long as $$X$$ and $$Y$$ are non-negative real numbers. This will not hold for imaginary pastry heights, though.

We also need $$2a \leq X$$ and $$b \leq Y$$. First thing first:

$2a \leq X \\ \frac{X - 2Y}{3} \leq X \\ X - 2Y \leq 3X \\ {-2Y} \leq 2X \\ 0 \leq X + Y$

Which is always true. And now the other thing:

$\frac{Y^2}{X + Y} \leq Y \\ Y^2 \leq XY + Y^2 \\ 0 \leq XY$

which is also always true.

So, now we know how to always trisect a pastry and keep the same amount of edge and area between all three parts!

## Left as an exercise for the reader

• Figuring out the volume of filling
• Accounting for the width of the crimping