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Inspired by Fitting the 9th piece into the pizza box

2 pizzas with radius r are each cut into 8 identical slices. 6 pieces were eaten so there are 10 pieces left. They need to be put in a square box without cutting or overlapping pieces. What is the minimal side length of the box, expressed in r?

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    $\begingroup$ The $m=4,n=5$ entry here yields a packing with side length $2.081r$: erich-friedman.github.io/mathmagic/0112.html $\endgroup$
    – RobPratt
    Commented Aug 23 at 3:25
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    $\begingroup$ @RobPratt some cool packings on that page. $\endgroup$
    – Dmitry Kamenetsky
    Commented Aug 23 at 5:06
  • $\begingroup$ 2r, because sane people will put the extras in a zip-top bag. 😉 (Or feed them to the dog, as in the preceding question.) $\endgroup$
    – Matthew
    Commented Aug 25 at 4:57

3 Answers 3

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Thanks to 2012rcampion for crunching the numbers to find a minimum side length of

$r\times 2.064\,821\,224\,090\,113\,066\,623\,307\,255\,407\ldots$

For this arrangement of slices:

Two in each corner, two in the center

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  • $\begingroup$ @2012rcampion I believe you are correct. I have no idea how to derive that, but it certainly works. Thank you. $\endgroup$
    – Daniel Mathias
    Commented Aug 23 at 1:06
  • $\begingroup$ I don't know the actual answer, but since yours is the lowest of all answers I accepted it. $\endgroup$
    – Ivo
    Commented Aug 23 at 6:02
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Here is a simple but rather effective packing:

symmetric packing

The sides of the square box are

$\frac {1+\sqrt 2 + \sqrt 3} 2 r \approx 2.073 r$

Explanation:

Let $x$ be the distance between the bottom left corner and the tip $T$ of the lone slice on the left. The distance of $T$ to the top left corner is $\sqrt 3$ because these two points together with the centre of the upper half-pizza form a 30-60-90 triangle. The side of the square is therefore $s = x+\sqrt3$.

Now extend the diagonal side of the lone slice on the left and the base of the square up to their intersection. The extended base has length $1+\sqrt 2$, the extension itself has length $x$. Therefore $s = 1+\sqrt 2-x$. The formula can now be obtained by adding the two expressions for the side length.

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    $\begingroup$ I think you should put a square root on the 1 in your formula. It's totally unnecessary but it would definitely look cooler. $\endgroup$
    – Michael Seifert
    Commented Aug 22 at 19:31
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    $\begingroup$ @MichaelSeifert Why stop there? How would you like $(\sqrt 1 + \sqrt 2 +\sqrt 3) / \sqrt 4$? $\endgroup$
    – Albert.Lang
    Commented Aug 23 at 3:02
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    $\begingroup$ @Albert.Lang I like the way you think $\endgroup$
    – Michael Seifert
    Commented Aug 23 at 11:30
  • $\begingroup$ Understanding the proof is a puzzle in its own right, haha. But I got it now. Needed some time to see which triangles you used to derive the 1+sqrt(2) $\endgroup$
    – justhalf
    Commented Aug 23 at 21:46
  • $\begingroup$ I don't understand how you come to the assumption of a 30-60-90 triangle $\endgroup$
    – kopaka
    Commented Aug 27 at 12:28
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Here's the best setup I could come up with, using unit slices:

Pizza packing optimized

Achieving a length of

$s = 2.137580020435572137..$

Here, the highlighted slice is shifted by $b \approx 0.01626$ in both directions, and the rest of the pieces fit.

This value of $b$ is optimal for this configuration, and the resulting side length turns out to be algebraic :

Let $p_{\pm} = 284+49\sqrt2 + 13\sqrt[3]{164662-2705\sqrt{2} \pm 225 \sqrt{3 \left(62135-39834 \sqrt{2}\right)}}$. Then $${s=\hspace{3px}} \frac{1}{390} \left(474+153 \sqrt{2}+\sqrt{6 (p_{-}+p_{+})}+\sqrt{6 \left(1704+294 \sqrt{2}-p_{-}-p_{+} -8700 \sqrt{\frac{3}{p_{-}+p_{+}}}+9000 \sqrt{\frac{6}{p_{-}+p_{+}}}\right)}\right)$$ and $$b = s - \frac{3 \sqrt{2}}{2}$$

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    $\begingroup$ This is excellent work! I have come to the conclusion that pizza makers should make slightly bigger pizza boxes so we can fit in more pizza slices :) $\endgroup$
    – Dmitry Kamenetsky
    Commented Aug 21 at 12:46
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    $\begingroup$ @DmitryKamenetsky ledopizza.com/why-is-ledo-pizza-square "round pans have become the rage as pizza operators are able to cut costs by giving the appearance of more, while actually offering less." $\endgroup$
    – RobPratt
    Commented Aug 21 at 16:08
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    $\begingroup$ I don't think square pizzas will ever become preferred Italian standard, only if not because classically throwing them turning around to get correct proportions won't result in a square. :-) $\endgroup$
    – FirstName LastName
    Commented Aug 22 at 0:51
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    $\begingroup$ Square pizzas used to be the standard at a pizza place up on Basingstoke Road in Reading opposite Morrisons some few decades ago. I can't remember its name. Best pizzas ever and they were very popular. For the size of box you got (obviously) far more pizza than a circular one. Don't know why they stopped trading because they were the absolute best. $\endgroup$
    – Prime Mover
    Commented Aug 22 at 9:55

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