There are six neatly-stacked circles of radius r.


➊ Circle with centre at {−1,0}.


➌ Circle with centre at {+1,0}.


➋ Circle between ➊ and ➌ with centre at {x,0}, x ∈ ℝ, r−1 ≤ x−r, x+r ≤ 1−r.


➍ A circle is balanced atop ➊ and ➋, touching both.


➎ A circle is balanced atop ➋ and ➌, touching both.


➏ And a circle is balanced atop the second layer, so touching both ➍ and ➎. 


Obviously if x=0 then by symmetry the horizontal position of the topmost circle (➏) is 0. Prove by geometry that this is so even if x≠0.


(I can prove this, understanding-free, by tiresome algebra in Mathematica. But that’s understanding-free.)



Six stacked circles, not quite symmetrical
I don't know if this is really research-level math, but here is a quick answer. For each $i=1,\dotsc,6$, let $P\_i$ be the center of circle (i) and write $P\_i=(x\_i,y\_i)$. We have $P\_1=(-1,0)$, $P\_3=(1,0)$, and $P\_2=(x,0)$. Since Circle (4) is stacked on top of circles (1) and (2), $x\_4 = \frac{x\_1+x\_2}{2} = \frac{-1+x}{2}$. Similarly $x\_5 = \frac{1+x}{2}$. Now, the points $P\_2,P\_4,P\_5,P\_6$ form a **rhombus**, so the segment $\overline{P\_4 P\_6}$ is parallel to the segment $\overline{P\_2 P\_5}$. And these segments have the same length (namely $2r$) (we already said it's a rhombus...). Therefore the horizontal offsets are equal: $x\_6 - x\_4 = x\_5-x\_2$. Then
$$
 x\_6 = x\_4 + x\_5 - x\_2 = \frac{1+x}{2} + \frac{-1+x}{2} - x = 0.
$$


