# Conjecture: If circular coins of any sizes are in a convex polygonal frame, with each coin touching exactly one edge, then all the coins can move

Suppose some circular coins (not necessarily the same size) are in a frame. The coins may be immobile, as in this example:

On the other hand, they may be free to move, as in these examples (in which the coins can move simultaneously):

It is rather tedious to show algebraically that the coins can move, so I tried to find some general principles that allow us to simply look at diagrams like these and know whether the coins can move.

Conjecture: If circular coins (not necessarily the same size) are in a convex polygonal frame, with each coin touching exactly one edge, then all the coins can move.

Is my conjecture true?

Remarks about my conjecture

• The frame must be a polygon, otherwise there would be a counter-example: two coins in the region bounded by $$y=x^2-1$$ and $$y=1-x^2$$, as shown below.
• The frame must be convex, otherwise there would be a counter-example, as shown below.
• Every coin must touch an edge, otherwise there would be a counter-example, as shown below.

EDIT

Zach Teitler has given a counter-example. I have proposed a second conjecture that avoids this counter-example.

EDIT2

My second conjecture also has a counter-example. I have asked another question asking for general principles that are useful in determining whether coins can move.

• How would we formally define “coins can move”? I feel like my intuition can easily betray me. At the moment it's telling me the four circle in the trapezoid can't move. Commented Jan 11, 2023 at 15:03
• We need someone skilled to create animations! Commented Jan 11, 2023 at 15:07
• @AndrejBauer In the state space, the path-connected component of the configuration is larger than a single point. (I'm fairly certain the path-connected components are the same as the connected components, as well.) Commented Jan 11, 2023 at 15:13
• There is a growing literature on configuration spaces of hard disks. In particular, the paper arxiv.org/abs/1108.5719 by Carlsson, Gorham, Kahle and Mason has some pictures of 5 coins/disks in a square which seem to support your conjecture. Matt Kahle's answer to my MO question mathoverflow.net/q/59563/8103 contains further references. I wouldn't be surprised if Matt or one of his coauthors knows the answer to your question. Commented Jan 11, 2023 at 15:59
• @AndrejBauer About the four circles in the trapezoid, here is a demos graph where you can adjust the sliders to move the coins individually.
– Dan
Commented Jan 11, 2023 at 22:15

The following seems like a counterexample to the conjecture as originally stated, allowing different size coins. It doesn't seem like the big coin, with diameter $$1-\epsilon$$, can move right, up, or down. (I apologize for the poor drawing.) (I haven't done "formal" algebra to verify this, but just looking at it, it seems to be so.)

Edit by OP: Here's another look at your idea.

• Look convincing. I think I should edit my conjecture, so that it requires that the coins all be the same size. May I edit this into my question? I don't want to make my question a "moving target", but I don't want the conjecture to just die.
– Dan
Commented Jan 11, 2023 at 14:37
• Sure! I'll edit this answer so that it refers specifically to the different sizes version of the conjecture. Commented Jan 11, 2023 at 14:43
• @Dan, I suggest you make two conjectures, to preserve continuity, with the first one seemingly refuted by Zach's nice example. Commented Jan 11, 2023 at 14:46
• @Dan It is recommended practice (though not an ironclad rule) in such situations to accept the answer to the original conjecture, and then post a separate MO question with the new conjecture. This would give Zach Teitler credit for the correct answer, and also avoid the "moving target" problem. Commented Jan 11, 2023 at 23:15
• @ZachTeitler When I first saw your counter-example, my first thought was, "That's simple, why didn't I think of that?" But upon further reflection, I think it actually required a lot of creativity.
– Dan
Commented Jan 12, 2023 at 5:56

This is to confirm that the construction offered in Zach Teitler's answer works.

Indeed, look at this picture:

The square here is $$[0,1]^2$$. The Black circle is $$C(R,1/2;R)$$, with center $$(R,1/2)$$ and radius $$R$$. The other four circles are $$C(x,y;r)$$ (Red), $$C(u,v;s)$$ (Green), $$C(X,Y;r)$$ (Magenta), and $$C(U,V;s)$$ (Blue), where $$\begin{equation*} X=x=1-r,\quad Y=1-y,\quad U=u,\quad v=1-s,\quad V=s, \end{equation*}$$ $$y\approx0.682$$ is the smallest real root of the polynomial $$9893 - 40154 y + 74218 y^2 - 86208 y^3 + 64832 y^4 - 30720 y^5 + 8192 y^6$$, $$\begin{equation*} R=\frac{1}{8} \left(4 y^2-4 y+\frac{35}{8}\right)\approx0.438,\quad r=\frac5{64}\approx0.078,\quad s=\frac18=0.125, \end{equation*}$$ and $$u\approx0.859$$ is the smallest real root of the polynomial $$256u^2-472 u+403+256 a^2-448 a$$, where in turn $$a$$ is the smallest real root of the polynomial $$8192 a^6-30720 a^5+64832 a^4-86208 a^3+74218 a^2-40154 a+9893.$$

Then each of the five circles touches exactly one edge of the square and the circles are contained in the square. The open discs bounded by the five circles are pairwise disjoint. Also, the Red and Green circles touch each other and the Black one, and the symmetric to them Magenta and Blue circles also touch each other and the Black one.

So, all the required conditions on the five circles and the square hold.

However, if the $$10$$-tuple $$(R,1/2, x, y, u, v, X, Y, U, V)$$ of the centers of the five circles is changed however little, then the condition that all the open discs bounded by the five circles stay pairwise disjoint and inside the square cannot hold.

Note: The verification of the latter statement reduces to solving a linear programming problem with $$10$$ unknowns, albeit with pretty complicated algebraic coefficients.

Here is a pdf image of a Mathematica notebook with detailed calculations.

• Erm.. Something is fishy: the described move diminishes both the $x$ and the $y$ - coordinate difference between the centers of the red and the green circles, so how is it an admissible move? Am I missing anything? Commented Jan 11, 2023 at 21:24
• @fedja : Thank you for your comment. There were two typos in my previous work. With those typos fixed, the answer turns out to be affirmative. Commented Jan 11, 2023 at 22:35
• Great! Now it is believable :-). I also tried to answer an old question of yours about the bound on the $L^p$-norm of the sum of pairwise independent random variables. I'm naturally curious how much you know yourself by now. I showed only that $C_p$ stays bounded for $p\in(1,2)$ but I still have no idea whether it tends to $1$ as $p\to 2$. Do you know the answer to that question? Commented Jan 11, 2023 at 22:59
• This is very nice! Thank you for sharing the computations. Commented Jan 12, 2023 at 2:45
• @ZachTeitler : Thank you for your appreciation. I was just trying to realize your idea. Commented Jan 12, 2023 at 4:03