# Packing pentominoes in a circle

You want to prepare a pizza of 12 flavors. You have 12 oddly-shaped pieces of cheese that you decide to use for the pizza. The shapes happen to be ...

Oh, well, forget it! This isn't going to be even remotely realistic. So here is the problem:

I was playing with pentominoes and figured you can pack them nicely in a circle of radius 5. This immediately cries for the question: Is this optimal? If not, what is the radius of the smallest circle that can accommodate all 12 pentominoes inside without overlap? Show an arrangement that minimizes the radius.

Spoiler alert: the picture above is not optimal.

Scoreboard:
4.84323 loopy wait
4.86594 Florian F
4.88966 Ravi Fernando
4.92443 Daniel Mathias
4.94975 cap
4.98189 Franciszek Remin

• The trivial lower bound is $\sqrt{60/\pi} \sim 4.37$ which is a disk with the same area as the 12 pentominoes but of course you can't reach that because there will always be some waste at the edges. Aug 29, 2022 at 6:25
• My junior's name is Franciszek Remin - if you were so kind to put my son's name on the list of fame. Jan 3 at 12:30
• It is my pleasure. Jan 3 at 15:09

UPDATE 2

A minor improvement. New best radius

4.84323

Arrangement

/UPDATE 2

UPDATE

4.8487

using arrangement

/UPDATE

4.866

using the following scheme

which is obviously heavily indebted to Ravi Fernando's. The improvement is in the left half.

• Congratulations! This matches my result. At last! And your arrangement is even more elegant than mine. Sep 17, 2022 at 19:05
• @FlorianF Just managed to shrink it a bit more. Sep 18, 2022 at 5:06
• Fantastic! Checked and validated. Unfortunately, I cannot accept or upvote a second time. But I updated the scoreboard. Sep 19, 2022 at 19:26

I can get a radius of:

$$\sqrt{\frac{149487}{2} - 975 \sqrt{5873}} \approx 4.88739$$.

the following modification of cap's answer (thanks also to Jaap Scherphuis's comment):

and then

shift the rightmost three pentominoes up by $$c = \frac{77 - \sqrt{5873}}{2} \approx 0.18225$$ units.

The resulting figure has circumcenter located $$\frac{7c-c^2}{14} = \frac{5\sqrt{5873} - 383}{2} \approx 0.08875$$ units left of the center of the middle square; it intersects the circumcircle at the northwest, southwest, and southeast corners, as well as the corner at top of the eastern edge.

EDIT: I found a second solution with the slightly worse radius

$$\sqrt{12110 - 480 \sqrt{634}} \approx 4.88966$$.

the following configuration inspired by Daniel Mathias's answer, with four half-square-unit holes:

and then

shift the four rightmost pentominoes up by $$c = \sqrt{634} - 25 \approx 0.17936$$ units.

The resulting figure has circumcenter $$\frac{4c - c^2}{18} = \frac{6 \sqrt{634} - 151}{2} \approx 0.03807$$ units left of the center of the middle square; it touches its circumcircle at the top and bottom of the left edge, the bottom of the right edge, and the top corner of the X-pentomino. Note that the four pentominoes in the middle don't touch the circumcircle, so they have a little room to wiggle up and down.

I found both of these with the help of

https://cemulate.github.io/polyomino-solver/ to place the pentominoes, and WolframAlpha for coordinate calculations.

• While I congratulate all of you for your efforts, I am a bit embarrassed to announce that this still does not match the best solution I know. But you start to be really close. Aug 28, 2022 at 22:25
• That's great! Really fun problem, I'm looking forward to seeing what further twists it's hiding. Aug 28, 2022 at 23:07
• FYI I found another good arrangement, which is unfortunately very slightly worse than my first. I've added it to my answer in case anyone else can gain some insight from it. Sep 15, 2022 at 6:18

The radius of the smallest pizza that can accommodate all 12 cheeses is

$$\sqrt{3.5^2 + 3.5^2} \approx 4.95$$

The cheeses can be arranged like this: (There is square unit of tomato sauce with no cheese on the right)

• Very nice solution. By the way, the cheese van be placed to put that square of tomato sauce anywhere you like. Aug 27, 2022 at 23:06
• Nice try ...... Aug 28, 2022 at 0:07
• I am new to this puzzle, on what software can I play this, ie arranging and rearranging the pentominoes?? Aug 28, 2022 at 6:16
• @Kutsit I don't know of any software. I made the pentominoes out of legos and used them to find my arrangement. Then I used a spreadsheet to create the graphic. Aug 28, 2022 at 7:26
• @Kutsit I used an on line solver to find the initial pattern and wrote a java program to render it with the circle background Aug 28, 2022 at 7:49

The smallest radius is (apparently less than)

$$\frac12\sqrt{9^2+4^2}=\frac12\sqrt{97}\approx4.9244$$

One such arrangement is shown here:

• You claim it's smallest but don't support that claim. Aug 28, 2022 at 5:36
• On the one side, it is probably difficult to prove optimality, I don't require it. On the other side, I know this one isn't. I know of a better solution. Aug 28, 2022 at 7:55
• Via integer linear programming, I have found that Daniel's radius matches half of the minimum diameter of a set of 60 nonoverlapping unit squares with integer coordinates. If @FlorianF and I are both right, that means every optimal solution will have at least one pentomino with non-integer coordinates. Aug 28, 2022 at 15:34
• Glad to see you looking in the right direction. Who said cheese has to be placed at integer coordinates on a pizza? Aug 28, 2022 at 15:44
• @RobPratt If you can translate that to an actual solution with pentominoes then you clearly beat me. Aug 29, 2022 at 4:47

For reference, here is my solution. • Nice! Now let's see. The distance between the left of the lower edge and the right of the upper edge is $\sqrt{9^2+4^2} = 9.8488$, giving a radius of at least 4.9244. But that line isn't centered. For the actual radius I get 4.9547. Aug 31, 2022 at 11:42