Four circles on a quarter disk: can the circles move?

Question

Four unit circles are on a quarter disk. In the beginning, their centres are the vertices of a square, with two circles each touching a straight edge of the quarter disk, and the other two circles each touching the arc of the quarter disk.

Can the circles move, staying within the perimeter of the quarter disk without overlapping?

My attempt

I tried to use equations (similar to this answer), but the algebra seems to be intractable.

I also tried an intuitive approach. If the circles are arranged like this,

they take up a larger proportion of a quarter disk. And it may seem that, as a general principle, circles can move from a less economical arrangement into a more economical position, but this is not always true, as I demonstrate here.

But I still wonder if there are any general principles that can be used to answer questions like this about whether circles can move.

Answer 1

In fact, the circles can move. While the animation below does not furnish a proof, it is suggestive that the answer is affirmative:

I rotated the figure by $\pi/4$ clockwise and placed the wedge in the first quadrant to make it easier to calculate the coordinates.

Although the animation is not labeled, let us call the tangent circles $C_1, C_2, C_3, C_4$, where $C_1$ is the circle whose center has greatest $x$-coordinate and the others are numbered in clockwise order (so $C_2$ is tangent to the $x$-axis, $C_3$ tangent to the $y$-axis, and $C_4$ is tangent to the arc and has the largest $y$-coordinate).

It is easy to show that for inscribed circles of unit radius, the wedge has radius $R = 1+\sqrt{6(2+\sqrt{2})}$, the proof of which I leave as an exercise. Thus the radius of the blue arc on which the centers of the "outer" two circles lie is $r = R-1$.

From this, we assume that $C_1$ and $C_2$ remain tangent through their movement, and the angle at which the center of $C_1$ makes with the origin and the positive $x$-axis is $\theta \in [\csc^{-1} r, \pi/4 - \csc^{-1} r]$. It is straightforward to show that in order to preserve tangency, $C_2$ must have center at $(x,1)$, where

$$x = r \cos \theta - \sqrt{\frac{6 - r^2(1-\cos 2\theta) + 4r \sin \theta}{2}}. \tag{1}$$

Then $C_3$ will have center at $(1,y)$, where

$$y = 1 + \sqrt{3 + 2x - x^2}. \tag{2}$$

It is intractable to compute the locus of $C_4$ such that it remains tangent to $C_3$ and the arc, so instead, we wish to show that if $C_4$ is kept tangent to $C_1$ as in the animation, the distance between the centers of $C_3$ and $C_4$ will be at least $2$, thus ensuring that they never overlap. To this end, this distance function is expressible in closed form as a function of $\theta$:

$$f(\theta) = \sqrt{(1-r \cos (\theta + 2 \csc^{-1} r))^2 + (y - r \sin (\theta + 2 \csc^{-1} r))^2}, \tag{3}$$ although writing it solely in terms of $\theta$ is too lengthy for this answer. But when plotted, we get

which suggests that indeed, the distance function is strictly monotone and never goes below $2$. While this is not a rigorous proof, it is probably formally provable (although tedious).

Answer 2

As originally packed, four unit circles require a quadrant of radius $R=1+\sqrt {6(2+\sqrt 2)}\approx5.526$, according to @heropup, as can be seen from the corrected figure below.

Since $NQ=1$, and $PQ=OF=ON=SP=\sqrt 2$, and $BS=EO=1+\sqrt 2$, then in right triangle $BPN$ we have$$(1+2\sqrt 2)^2+(1+\sqrt 2)^2=BN^2=6(2+\sqrt 2)^2$$so that $BN=\sqrt{6(2+\sqrt 2)}$, and radius $BR=\sqrt{6(2+\sqrt 2)}+1$

But the same four circles fit into a smaller quadrant if packed as below, where $DE=1$ and radius $BZ=3\sqrt 2+1\approx5.243$, which is less than the first radius $BR$ by $\approx .283$ Starting in the second position, then, but within the larger quadrant, they clearly have room to move so as to touch the circumference. And since any movement is reversible, they can then be moved from that position into the lower one.

A general proposition, perhaps true, might be something like this: Given $n$ equal circles, whose centers form a regular $n$-gon, with each circle tangent to two others and to only one side of a container having $n-1$ sides and at least one rectilineal angle (fig. 1), the circles can be moved, without overlapping, into a position where not all touch a side and one of them touches two sides (fig. 2).

Counterexample(s)?

Answer 3

The answer is YES, we can move the four circles around (at least for a small distance).

Following is my attempt to derive a criterion for ability to move the circles around without bumping into each other. At the end is a modification of this criterion to answer two related questions.

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Derivation of a moveability criterion

Before we start, let us generalize the problem a little bit.

We will identify the plane $\mathbb{R}^2$ with complex plane $\mathbb{C}$ and let

• $\Gamma$ be the collection of polyline/polygon which allows circular arcs as edges.
• $C(c,r)$ be the circle centered at $c$ with radius $r$.
• $\angle(u,v) = \arg(v\bar{u})$ be the angle one need to rotate something in direction $u$ counterclockwisely into direction $v$.

Start with a convex region $K$ whose boundary $\partial K \in \Gamma$. We place $n$ circles $C_0,\ldots,C_{n-1}$ inside $K$ so that each $C_k$ touches $\partial K$ and two and only two others circles. If one extend indexing of the circles by periodicity and treat $C_{-1}, C_n$ as aliases of $C_{n-1}$ and $C_0$, we can order the circles so that circle $C_k$ is touching circles $C_{k\pm 1}$. The generic problem is:

Can we move all circles simultaneously so that each $C_k$ remains inside $K$, touching $\partial K$ without bumping into other circles.

The initial condition each $C_k$ touches only two circles ensures us for small displacements, $C_k$ will not bump into circles other than $C_{k\pm 1}$. We can concentrate on the relative placement of $C_{k\pm 1}$ with respect to $C_k$.

Let $c_k, r_k$ be center and radius of $C_k$, ie. $C_k = C(c_k,r_k)$. Since $C_k$ is touching $\partial K$, $c_k$ is constrained to lie on some curve $\gamma_k \in \Gamma$. WOLOG, let's assume $C_k$ are ordered along $\partial K$ counterclockwisely and all $\gamma_k$ are parameterized by arc-length with $\gamma_k(0) = c_k$ in positive orientation.

To proceed further, we need to make some technical assumptions:

• [A1] None of $c_k$ is located at corners/endpoints of corresponding $\gamma_k$.

  This means the tangent vectors $t_k = \gamma'_k(0)$ are well defined.
  Let $\alpha_k = \angle(t_{k-1},c_k - c_{k-1})$ and $\beta_k = \angle(c_k - c_{k-1},t_k)$.

• [A2] None of $|\alpha_k|$, $|\beta_k|$ equals to $\frac{\pi}{2}$.

Consider following problem of placement of $n+1$ points $p_0,\ldots,p_n$:

• [C1] $p_0,\ldots,p_n$ are initially located at corresponding $c_k$.
• [C2] $p_0,\ldots,p_n$ are constrained to move on corresponding $\gamma_k$.
• [C3] $|p_k-p_{k-1}| = \ell_k \stackrel{def}{=} r_k + r_{k-1}$ for $k = 1,\ldots, n$.

Under assumption [A1] and [A2], if we move $p_0$ along $\gamma_0$ for a small distance $s_0 = \epsilon$, it is possible to move $p_1$ along $\gamma_1$ for a small distance $s_1$ to maintain condition [C3]. Assumption [A2] ensures $s_1$ is of the order of $\epsilon$. In fact, since $\gamma_1$ is either a straight edge or circular arc at $\epsilon \sim 0$, $s_1$ is locally a $C^\infty$ function in $\epsilon$.

Repeating this sort of argument, one find each $p_k$ will be moved along $\gamma_k$ for a distance $s_k$ of order $\epsilon$. There are $n$ $C^\infty$ functions $s_1(\epsilon),\ldots, s_n(\epsilon)$ defined over some neighborhood of $0$ and $n$ constants $\lambda_1,\ldots,\lambda_n$ such that:

$$p_k(\epsilon) = \gamma_k(s_k(\epsilon))\quad\text{ and }\quad s_k(\epsilon) = \lambda_k \epsilon + O(\epsilon^2)\quad\text{ for }\quad k = 1,\ldots, n$$

Let $C_k(\epsilon) = C(p_k(\epsilon),r_k)$ and focus on dependence of $s_n$ on $\epsilon$.

• If $s_n(\epsilon) > \epsilon$, $p_n(\epsilon)$ will overtake $p_0(\epsilon)$ along $\gamma_0$. $C_{n-1}(\epsilon)$ will bump into $C_0(\epsilon)$ (ie. their interior intersect) and the circles cannot move.

• On the other direction, if $s_n(\epsilon) < \epsilon$, $p_n(\epsilon)$ will lag behind $p_0(\epsilon)$. $C_{n-1}(\epsilon)$ will move away from $C_0(\epsilon)$ and the circles can move.

To see whether one can move the circles, the first thing we need is $\lambda_n$.

If $\lambda_n < 1$, the circles can move. If $\lambda_n > 1$, the circle cannot move.

For the problem at hand, $\lambda_n = 1$ by symmetry. One need the higher order $\epsilon$-dependence of $s_n(\epsilon)$.

Let $\rho_k$ be the curvature of $\gamma_k$ at $c_k = \gamma_k(0)$. Choose a coordinate system where $c_{k-1} = 0$ and $c_{k} = \ell_k$, we find

$$p_k - p_{k-1} = \ell_k + \left(s_k + \frac{i}{2}\rho_k s_k^2\right) e^{i\beta_k} - \left(s_{k-1} + \frac{i}{2}\rho_{k-1} s_{k-1}^2\right) e^{-i\alpha_k} + o(\epsilon^2)$$ This leads to $$\begin{align} \Re(p_k - p_{k-1}) &= \ell_k + \left(s_k\cos\beta_k - s_{k-1}\cos\alpha_k\right) - \frac{\epsilon^2}{2}\left(\rho_k\lambda_k^2\sin\beta_k + \rho_{k-1}\lambda_{k-1}^2\sin\alpha_k\right) + o(\epsilon^2)\\ \Im(p_k - p_{k-1}) &= \epsilon(\lambda_k\sin\beta_k + \lambda_{k-1}\sin\alpha_k) + O(\epsilon^2) \end{align} $$ To satisfy [C3], we need

$$s_k\cos\beta_k - s_{k-1}\cos\alpha_k = \frac{\epsilon^2}{2}\left\{ \begin{align} & \left(\rho_k\lambda_k^2\sin\beta_k + \rho_{k-1}\lambda_{k-1}^2\sin\alpha_k\right)\\ - & \frac1{\ell_k}\left[ (\lambda_k\cos\beta_k - \lambda_{k-1}\cos\alpha_k)^2 + (\lambda_k\sin\beta_k + \lambda_{k-s}\sin\alpha_k)^2\right]\end{align}\right\} + o(\epsilon^2)$$ Comparing the $\epsilon$-term on both sides, we find $\frac{\lambda_k}{\lambda_{k-1}} = \frac{\cos\alpha_k}{\cos\beta_k}$. Multiplying these ratios for different $k$, we get:

$$\lambda_k = \prod_{j=1}^k \frac{\cos\alpha_j}{\cos\beta_j}$$

Substitute this back into above expression and simplify, we obtain

$$\frac{s_{k}}{s_{k-1}} = \frac{\lambda_{k}}{\lambda_{k-1}}\left\{ 1 + \frac12\epsilon(\lambda_k\cos\beta_k) \underbrace{\left[\rho_k\frac{\sin\beta_k}{\cos^2\beta_k^2} + \rho_{k-1}\frac{\sin\alpha_k}{\cos^2\alpha_k} - \frac{(\tan\alpha_k + \tan\beta_k)^2}{\ell_k}\right] }_{\text{ call this }\Delta_k}\right\} $$ Multiply these ratios for different $k$ again, we obtain

$$s_n(\epsilon) = \epsilon\lambda_n + \frac{\epsilon^2\lambda_n}{2} \underbrace{\sum_{k=1}^n(\lambda_k \cos\beta_k)\Delta_k}_{\text{ call this }\Delta} + o(\epsilon^2)$$

By a similar argument like the first order case, we arrive at following improved criterion:

If $\lambda_n > 1$ or $\lambda_n = 1$ and $\Delta > 0$, the circles cannot move.
If $\lambda_n < 1$ or $\lambda_n = 1$ and $\Delta < 0$, the circles can move.

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Analysis of original problem

Back to the original problem where all $r_k = 1 \implies \ell_k = 2$. Choose a coordinate systems where the circles are centered at

$$ \begin{cases} c_0 = (1+\sqrt{2},1+2\sqrt{2})\\ c_1 = (1,1+\sqrt{2})\\ c_2 = (1+\sqrt{2},1)\\ c_3 = (1+2\sqrt{2},1+\sqrt{2}) \end{cases} $$

For small $\epsilon$, one can treat $\gamma_1,\gamma_2,\gamma_3$ as straight lines and $\gamma_4$ as a circular arc with radius $$R = \sqrt{(1+\sqrt{2})^2+(1+2\sqrt{2})^2} = \sqrt{6(2+\sqrt{2})}$$ Furthermore, it is easy to see $$\alpha_1 = \theta \stackrel{def}{=} \frac{\pi}{4} + \tan^{-1}\frac{1+\sqrt{2}}{1+2\sqrt{2}} = \tan^{-1}(3+\sqrt{2})$$ In terms of $R,\theta$, the parameters required to compute $\Delta$ is listed below $$\begin{array}{r|cccc} k & \lambda_k & \alpha_k & \beta_k & \rho_k\\ \hline 1 & \sqrt{2}\cos\theta & \theta & \frac{\pi}{4} & 0 \\ 2 & \sqrt{2}\cos\theta & \frac{\pi}{4} & \frac{\pi}{4} & 0 \\ 3 & 1 & \frac{\pi}{4} & \theta & 0\\ 4 & 1 & \frac{\pi}{2} - \theta & \frac{\pi}{2} - \theta & \frac1R \end{array} $$ For the given choice of $R, \theta$, it satisfies an interesting relation $R\cos\theta = 1$. It is not clear whether this is a coincidence or has a deeper meaning. In any event, this simplifies evaluation of $\Delta_k$ a lot.

$$\begin{align} \Delta_1 = \Delta_3 &= \frac{1}{R}\frac{\sin\theta}{\cos^2\theta} - \frac{(1 + \tan\theta)^2}{2} = \tan\theta - \frac{(1+\tan\theta)^2}{2} = -3(2+\sqrt{2})\\ \Delta_2 &= -\frac{(1+1)^2}{2} = -2\\ \Delta_4 &= \frac{2}{R}\frac{\cos\theta}{\sin^2\theta} - \frac{(\cot\theta + \cot\theta)^2}{2} = 2\cot^2\theta - 2\cot^2\theta = 0 \end{align}$$

Since all $\lambda_k\cos\theta_k > 0$ and all $\Delta_k \le 0$ but not all $= 0$, we have

$$\Delta = \sum_{k=1}^4 (\lambda_k \cos\beta_k)\Delta_k < 0$$

Together with $\lambda_4 = 1$, the criterion we derived before tell us we can move the four circles (at least for a small distance).

Whether we can move the circles continuously to the attempted configuration (the one where one unit circle is stuck at the corner of quarter disc) needs more analysis.

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Application to similar problems

Let us review what we have found in original problem.

1. The most surprising result is probably $\Delta_4 = 0$. It turns out it isn't that surprising. It vanishes because $p_3$ and $p_4$ are constrained to move on same circular arc. In general, if $p_{k}$ and $p_{k-1}$ are constrained to move on same straight edge/circular arc with fixed distance apart, they will always travel the same distance. This means $\lambda_k = \lambda_{k-1}$ and $\Delta_k = 0$. In computation of $\lambda_n$ and $\Delta$, we can ignore existence of such a pair of points $p_{k-1}, p_k$.

2. Another interesting result is $\Delta_2 < 0$. It is negative because $\rho_1 = \rho_2 = 0$, ie. $p_1, p_2$ are moving on straight edge portions of $\gamma_1$ and $\gamma_2$.

   In general, if $p_{k}, p_{k-1}$ are moving on straight edge portions of $\gamma_k$ and $\gamma_{k-1}$, we will have $$\Delta_k = -\frac{(\tan\alpha_k + \tan\beta_k)^2}{2}$$ This term is $<0$ except at three cases.

   • $\alpha_k = \beta_k = 0$ : Since the underlying $K$ is convex, this essentially force $\gamma_k$ and $\gamma_{k-1}$ to be same straight edge and reduce to the case discussed before.
   • $\beta_k = -\alpha_k \ne 0$ : This is impossible when underlying $K$ is convex.
   • $\beta_k + \alpha_k = \pi$ : $\gamma_k$ and $\gamma_{k-1}$ is parallel but pointing in opposite directions. This is possible when underlying $K$ contains a pair of parallel non-neighboring edges and $C_k$, $C_{k-1}$ are touch them.

Based on these, we have a alternate criterion.

Under the assumption

• [A0] $K$ is a polygon consists only of straight edges.
• [A1] None of $c_k$ is located at corners/endpoints of corresponding $\gamma_k$.
• [A2'] All $|\alpha_k|$, $|\beta_k| < \frac{\pi}{2}$ and not all of them are $0$.

We will avoid the problematic case $\alpha_k + \beta_k = \pi$. We will find all $\lambda_k\cos\theta_k > 0$, all $\Delta_k \ne 0$ but not all $\Delta_k = 0$. This means $\Delta < 0$ and as a result, arrive at following polygonal version of criterion:

If $\lambda_n > 1$, the circles cannot move. If $\lambda_n \le 1$, the circles can move.

If one look at configurations appear in following questions:

• Five circles in a rectangle: can the circles move?
• Three circles in a triangle: can the circles move?

It is easy to check both configurations satisfy the new set of assumptions and $\lambda_n = 1$ by symmetry. As a result, the circles can move in both cases.

Answer 4

No, none of the circles in the upper diagram can move. If a circle has a tangent contact it can move along or away from that contact, giving a range of $\pi$ radians it can move into. Each of your circles has three points of tangential contact and the restrictions of the three contact points have no direction in common.