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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.)

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    $\begingroup$ Could you sketch a figure? I cannot see how circle 5 can touch 2 and 3 given 4 is atop 1 & 2. And do you mean circle or disk? And must 4, 5 and 6 have centers along the $x$ axis? $\endgroup$
    – David G. Stork
    Commented Oct 14, 2017 at 0:34
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    $\begingroup$ I think you'd be interested in Burkard Polster, Stacking wine bottles revisited, Math Intelligencer 37.2 (2015) 43-51, repreinted in Mircea Pitici, Ed., The Best Writing on Mathematics 2016, Princeton U Press, 48-65. $\endgroup$
    – Gerry Myerson
    Commented Oct 14, 2017 at 5:51
  • $\begingroup$ Bottom row is ➊ ➋ ➌. Next row is ➍ ➎. Top is just ➏. $\endgroup$
    – jdaw1
    Commented Oct 14, 2017 at 9:26
  • $\begingroup$ Image at jdawiseman.com/2017/20171014_circle_stacking_puzzle.png $\endgroup$
    – jdaw1
    Commented Oct 14, 2017 at 9:44

1 Answer 1

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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 $$ x_6 = x_4 + x_5 - x_2 = \frac{1+x}{2} + \frac{-1+x}{2} - x = 0. $$

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  • $\begingroup$ But the rhombus is not aligned with the flat ‘ground’ surface: it is slightly rotated from that. $\endgroup$
    – jdaw1
    Commented Oct 14, 2017 at 9:27
  • $\begingroup$ Every rhombus is a parallelogram; the opposite sides are parallel. If you prefer, the $P_2P_6$ and $P_4P_5$ diagonals bisect each other, so $\frac{x_2 + x_6}{2} = \frac{x_4 + x_5}{2}$. I hope that clarifies it for you. If you'd like to discuss further then this needs to move to math.stackexchange.com, and we can continue there. Math Overflow is a site for professional mathematicians to discuss their research. $\endgroup$
    – Zach Teitler
    Commented Oct 14, 2017 at 11:02
  • $\begingroup$ Apologies for posting in wrong forum. Answer is elegant and geometric: thank you. $\endgroup$
    – jdaw1
    Commented Oct 14, 2017 at 11:41

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