Five unit circles are in a rectangle. In the beginning, their centres are the vertices of a regular pentagon, and each circle is tangent to two other circles and one edge of the rectangle.
Can the circles move without overlapping?
I will post my answer below. I hope to get a more intuitive answer.
Yes, they can move.
Suppose the top circle moves right a small distance $t$. We will show that none of the circles overlap, by applying Pythagorus' theorem around the ring of circles.
$p=2\left(1+\sin{\frac{\pi}{5}}+\sin{\frac{2\pi}{5}}\right)$
$q=4\left(1+\cos{\frac{2\pi}{5}}\right)$
$a=2\cos{\frac{\pi}{5}}-t$
$b=\sqrt{4-a^2}$
$c=p-2-b$
$d=\sqrt{4-c^2}$
$e=q-4-d$
$f=\sqrt{4-e^2}$
$g=p-2-f$
$h=q-2-a$
$\sqrt{g^2+h^2}$ is an increasing function of $t$, and $\sqrt{g^2+h^2}=2$ when $t=0$.
$\therefore \sqrt{g^2+h^2}>2$ when $t>0$.
This means that when the top circle moves right, it can separate from the circle on its left. So the circles can move without overlapping.
In this desmos animation, you can see that the circles can move, by adjusting the $t$ slider.
(I believe the general case is: If circles of any sizes are each internally tangent to exactly one edge of a convex polygon, then the circles can move without overlapping.)
For unit circle, the length of the rectangle is the pentagon diagonal + $2$, that is $$2\cdot \frac{1+\sqrt 5}{2}+2=3+\sqrt 5=5.2361$$The height of the rectangle is$$2\cdot \sqrt{\left(\frac{1+\sqrt 5}{2}\right)^2-\left(\frac{1}{2}\right)^2}+2=5.0777$$as in the Geogebra image below, since$$\frac{5.2361}{5.0777}=\frac{14.33}{13.90}=1.031$$
I made a wooden frame of this shape, sized just to hold a regular pentagon array of five U.S. quarters, and was easily able to rotate them into the position shown in the next figure.
If the rectangle were changed to a square with side $13.96$, the quarters would fit symmetrically about the diagonal and be more compact than in the original arrangement, since$$13.96^2<14.33\times13.90$$
But even without altering the rectangle, the circles have room to move into the second position. Apart from physical experiment, it is prima facie a more economical use of space—the arrangement is more compact—with three circles crowded into a corner (second position) than with only two along one side and the corners unoccupied (first position).
Here's an attempt at a more intuitive explanation that makes things feel pretty clear to me.
Let the circles be $A,B,C,D,E,$ read from top to bottom and left to right. We're going to move $A$ righwards, $B$ upwards, $C$ downwards, and $D$ and $E$ leftwards, all at a constant rate (though not the same for every circle). We'll argue that this never causes the circles to intersect, so we can continue it until one of them hits a side of the rectangle.
First, let's assert that $D$ and $E$ move left at unit speed. We'll try to work out how fast everything else should move.
Consider the half-plane bounded by the tangent between $B$ and $D$ that contains $B$. If we move this half-plane left at the same rate that $D$ is traveling, this is equivalent to moving it up at some other rate related to the slope of the line. So if $B$ moves upwards at that rate, it'll stay within a half-plane that doesn't overlap $D$ at any point.
Likewise, we can take the half-plane bounded by the $C-E$ tangent which contains $C$, and note that moving it left is equivalent to moving it down, at the same rate as in the previous paragraph by symmetry. So we can also move $C$ downwards at the same speed $B$ is moving upwards.
Now consider the half-plane containing $B$ and bounded by the $A-B$ tangent, and the one containing $C$ and bounded by the $A-C$ tangent. These will move up and down, respectively, at the same rate, which means that they can be thought of as moving left and right, respectively, at some other rate (but still the same one between the two of them, because they meet $A$ at identical slopes). So, by the same logic, the void between these half-planes will keep having space for $A$ as it shifts rightwards at some third constant speed.
Here's a gif of all five circles moving at a constant speed from one end of the square to the other, with a single line for each relevant half-plane. Note that every line touches the corresponding two circles at all times, and all that changes is the relative position of those tangent points.
This is a long comment on @Edward Porcella's answer.
It is not always the case that circles can move from a less economical arrangement to a more economical arrangement, even with the condition that each circle is tangent to two other circles. Here is an example.
Consider packing three equal circles into the region bounded by $y=x^2$ and $y=2-x^2$.
Arrangement $A$:
Arrangement $B$:
In this desmos graph, by sliding $G$ (green circle), $R$ (red circle), and $B$ (blue circle), you can see that:
Imagine we have $p_1 = -d\sin\theta\hat j$ and $p_2 = d\cos\theta\hat i$ and we have $\|p_1-p_2\|^2 = d^2$ As $ \frac{d}{dt}\|p_1-p_2\|^2 = 2(p_1-p_2)\cdot (\dot p_1-\dot p_2) = 0$. Here we have $\dot p_1 = -d\cos\theta\dot\theta\hat j$ and $\dot p_2 = -d\sin\theta\dot\theta\hat i$ and as can be verified, $(p_1-p_2)\cdot (\dot p_1-\dot p_2) = 0$ so a virtual displacement is possible.
Resuming:
Considering the pentagon formed by the circles centers, and with $l$ the regular pentagon side, according to the attached figure we have
$$ \cases{ u = l\cos a\\ v = l\cos b\\ x = l\cos c\\ y = l\cos d } $$
and
$$ \cases{ v = l\sin a\\ x = l\sin b\\ y = l\sin c\\ z = l\sin d} $$
so for virtual displacements we have
$$ \cases{ \delta u = -\tan a\delta v\\ \delta v = -\tan b\delta x\\ \delta x = -\tan c\delta y\\ \delta y = -\tan d\delta z } $$
For compatibility we need
$$ \delta x = \tan a\tan b\tan c\tan d\delta x $$
or as a consequence
$$ \tan a\tan b\tan c\tan d = 1 $$
but in our setup we have $d = \frac{\pi}{2}-a$ and $c = \frac{\pi}{2}-b$ hence in this setup, the displacement is possible because we have
$$ \tan a\cot a\tan c\cot c = 1 $$
