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I have a machine that can divide a square pie into 9 equal square pieces using 4 blades:

Square pie in 9 equal pieces

The blades can be moved, but there is only one control - which defines the width of the blades in both dimensions

We simply have to set the width, then place the centre of the pie at the centre of the blades and we can chop any size pie into 9 equal pieces:

Different size pies

Someone has just brought me a round pie.

I still need to locate the centre of the pie in the centre of the blades, so I'm expecting to get 3 different shapes; 1 centre, 4 "corners" and 4 "edges".

If I set the blades to 1/3rd of the diameter of the pie, the "edge" pieces and "corner" pieces will be smaller than the middle one.

But it occurs to me that I can make the blades narrower so that the middle piece is smaller and the outer pieces are bigger.

Circular pies

For a pie of given radius r, is it possible to find a setting that produces 9 equal area pieces? If not, what setting will give 9 pieces of pie with the least variation, i.e. the difference between area of the smallest piece and the largest piece is minimised.

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It is

not possible

to divide the cake into 9 equal pieces in this way,

and

the best you can do (by the criterion given in the question) is to have each parallel pair of blades separated by about 0.558 times the radius of the circle, in which case (taking that radius to be 1, so the "ideal" piece area is about 0.349) the max minus min area is about 0.083.

Details:

Let $d$ be half the separation between the blades. Then the middle piece is of size $4d^2$. If we take our circle to have radius 1, the middle piece is 1/9 of the area iff $4d^2=\pi/9$ or equivalently $d=\sqrt{\pi}/6$. Each "middle-edge" piece, if we add in half of the middle piece, can be divided into a sector of the circle, with angle $2\sin^{-1}d$, and two right-angled triangles with legs $d,\sqrt{1-d^2}$, so the total area of one of these pieces is $\sin^{-1}d+d\sqrt{1-d^2}-2d^2$. The corner piece's area is accordingly $\pi/4+d^2-\sin^{-1}d-d\sqrt{1-d^2}$. At $d=\sqrt{\pi}/6$, we can check that the middle piece's area is about 0.349 while the middle-edge piece's area is about 0.408, a bit bigger. (The corners will be too small.) Taking a look at the graphs of these, the largest-minus-smallest function is largest when corner = middle, at which point d is just over 0.279 and the difference between largest and smallest areas is a little over 0.083.

Some more details:

To see that we don't have middle = corner when $d=\sqrt{\pi}/6$: that would mean $3d^2+\sin^{-1}d+d\sqrt{1-d^2}=\pi/4$ or, since $3d^2=\pi/12$, $\sin^{-1}d+d\sqrt{1-d^2}=\pi/6$. Let's prove that the LHS is too big to equal the RHS. Well, we have $\sin^{-1}d\geq d$ and $\sqrt{1-d^2}\geq 1-d^2$ so it's enough to prove $d+d(1-d^2)>\pi/6$; the RHS equals $6d^2$ so this is equivalent to $2-d^2>6d$ or $d^2+6d-2<0$ or $(d+3)^2-11<0$ or $d+3<\sqrt{11}$. 11 doesn't have a nice square root but $10.89=3.3^2$ does, so it suffices to prove that $d+3<3.3$ or $d<0.3$. Since $d=\sqrt{pi}/6$ this is the same as $\pi<1.8^2=3.24$, and this is both well known and easy to prove.

And

Over the range $d=0\ldots\sqrt{2}/2$ (beyond which the corner piece disappears altogether), the middle piece's area is obviously increasing, and the corner piece's area is obviously decreasing. Initially, of course the middle-edge piece's area is increasing, and in fact it's clear that this holds until the point at which $y=2x$ (think about how much the width and height of the upper middle-edge piece change when you make a small increase in $d$) by which point it's clear that the corners are too small relative to the middle. As you gradually increase $d$, the sequence of events is as follows. Initially middle (inc.) <= edge (inc.) <= corner (dec.); so the first pair to meet is therefore edge and corner, and clearly max-min is decreasing because max is decreasing while min increases. After they cross over we have middle (inc.) <= corner (dec.) <= edge (inc.), so next pair to meet is middle and corner. It's not so obvious what's happening to max-min over this range but it happens to be decreasing. After they cross over we have corner (dec.) <= middle (inc.) <= edge (inc.), and now clearly the max is increasing and the min decreasing so the range is increasing. It turns out that eventually those two increasing things change places and we end up with corner <= edge <= middle, but of course it remains true that max - min = increasing - decreasing = increasing.

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  • $\begingroup$ lots of math, but no final answer? $\endgroup$
    – Ben Barden
    Jan 20 at 14:35
  • $\begingroup$ Looks to me like a final answer is there: look in the first spoiler block and at the end of the second. $\endgroup$
    – Gareth McCaughan
    Jan 22 at 3:52
  • $\begingroup$ Ahh. Yep. I see it. Answer would be improved by putting that up in the initial conclusion. $\endgroup$
    – Ben Barden
    Jan 22 at 15:13
  • $\begingroup$ You might be right. I've reorganized things slightly. $\endgroup$
    – Gareth McCaughan
    Jan 22 at 17:40
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partial answer:

It is not possible to make them all the same size.

To demonstrate, assume that they are the same size. Thus, the side pieces must be the same overall area as the center piece. Supposing an edge size of A, that area must be a^2. By further extension, the distance from the inner square to the outer edge along the cuts must be slightly less than A, jsut as the distance from the inner square to the outer edge along the radius must be slightly greater than A. That's the only way to have the side pieces be the same size as the center piece, and there's jsut not that much play there.

If we then look at the corner pieces, considering where the distance A fromt eh center square is along the cut lines, it becomes clear that they would exist entirely inside of an AxA square drawn using their cuts and those points. They cannot be the same size as the central square, and thus it is not possible for all of the pieces to be the same size.

further insight:

If we start from that position, it is clear that the ideal position is going to have the cuts closer together than that - as we bring the cuts closer together, the corners (which are too small) grow, while the center and sides (which are too large) shrink. The center is shrinking significantly faster than the sides, though. if we push them together by x, the area of the center becomes (roughly) (A-x)(A-x), while the the area of the sides becomes (A-x)(A+x/2)

It is clear, then, that we want to shrink it at least until the point where the center is the same size as the corners. Up until that point, the smallest pieces (the corners) are growing and the largest (the sides) are shrinking. Once you hit that point, however, the smallest becomes the central piece, which is shrinking. If we were optimizing for multiplicative difference we'd be done, but for additive difference it will depend on whether the sides or the center are at that point losing more additively, and I'm not absolutely certain on that. Regardless, the largest is still the sides, which are also shrinking, and that persists until the corners become the same size as the sides, at which point the largest is the corners (growing) and the smallest is the center (shrinking) and it's clear that we should go no further.

Thus the ideal state is the one where the corners are grater than or equal to the center, less than or equal to the sides, and side minus center is as small as possible.

Actually, working that our is pretty close to algebraic. The rounded end on the sides throws it off a touch, but it's close enough that we can largely ignore that and still expect to be basically correct. Finding minimum for (A-x)(A+x/2)-(A-x)(A-x) where x<A. (A-x)*(3x/2) = (3Ax-3xx)/2. We can ignore the scalars, so we're looking at minimum of Ax-xx where A>x. So we want to push that as far as we can... which means we're looking for the case where the sides and corners are the same size. Huh.

Final result: looking for the situation where sides and corners are the same size. Expect the center to be smaller.

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  • $\begingroup$ @xhienne read the last sentence of the question. $\endgroup$
    – Ben Barden
    Jan 19 at 15:05
  • $\begingroup$ Ah OK, I missed that! $\endgroup$
    – xhienne
    Jan 19 at 15:19
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First, we have to find the areas of the pieces (the inner square $I$, the midsections $M$, and the corners $C$) in terms of $r$ and the setting $2w$:

Obviously, $I$ has area $4w^2$.
A sector defined by two corners and a midsection has area $\theta r^2$, where $\theta = cos^{-1}(\frac{w}{r})$, and so $I+2M = (\pi-2\theta)r^2$ and $M = (\frac{\pi}{2}-\theta)r^2 -2w^2$.
Revisiting sector area, we have $M+2C = \theta r^2$, from which a bit of manipulation gives $C = (\theta - \frac{\pi}{4})r^2 + w^2$. Assuming that a perfect division is possible, setting $I=M=C$ and manipulating yields the equations $\pi-2\theta = \frac{12w^2}{r^2}, \frac{3\pi}{4}-2\theta=\frac{3w^2}{r^2}$. Using the proportion of the two differences yields $\theta = \frac{\pi}{3}$, but substituting back into the equations themselves requires $\frac{w}{r}$ to equal $\frac{\sqrt{\pi}}{6}$ when the above $\theta$ forces it to be $\frac{1}{2}$.
Therefore, it doesn't quite work.

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  • $\begingroup$ It's been a bit too long since I've done enough trig to get to the actual solution, but you might find my "further insight" helpful in working out the final answer for yourself. $\endgroup$
    – Ben Barden
    Jan 19 at 18:49
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My answer is:

Yes you can. When you can divide the cake into 3 equal pieces horizontally, you can cut the cake into 9 equal pieces. The side of the middle square will be $\frac {\sqrt\pi r} 3$

Proof?

It's obvious that one could divide the middle part into 3 equal pieces. The edge and the center will have area $\frac {\pi r^2} 9$ Then $\pi r^2-\frac {5\pi r^2} 9=\frac {4\pi r^2} 9$, which can be divided into four.

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    $\begingroup$ Does not necessarily follow. You have three different shapes of piece, and your solution simply assumes that all three can be made the same size. There's no guarantee of that. Or... more precisely, you haven't proved that the side pieces and the corner pieces will be the same size with that size of side. $\endgroup$
    – Ben Barden
    Jan 19 at 14:25

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