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

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.

Answer 2

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 3

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)?