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

Question

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.

Answer 1

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.

Answer 2

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.

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