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).
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
$$ \tan a\cot a\tan c\cot c = 1 $$
