Water Droplet Rings

I thank @Morteza Sepehrinia for helping me in solving this problem.

Let $A$ , $r$ and $t$ be area of rings, radius of rings and time respectively.

Growth rate of area for rings is $\pi$ , hence for each ring $A=\pi t$ and $r=\sqrt{t}$ . Let radius of smallest ring be $r$ , we get $r+a_k=\sqrt{t_k}; \ \ 1\le k \le 6 ; \ \ a_k \in \text{{0, 5, 8, 10, 11, 12}}$ Now the time difference between every two rings is a positive integer, that is if $m>n$ then $t_m-t_n=s_{m,n} \in \mathbb{N}$ or $(r+a_m)^2-(r+a_n)^2=s_{m,n} \\ 2r= \frac{s_{m,n}-(a_m^2-a_n^2)}{a_m-a_n} \ \ \star$ To reach the desired condition of problem, RHS of the equation $\star$ must be minimum, which results in minimum $r$ and consequently minimum time where first Fibonacci rings occur.

Now note that $\frac{s_{m,n}-(a_m^2-a_n^2)}{a_m-a_n} \ge \frac{\min(s_{m,n}-(a_m^2-a_n^2))}{\max(a_m-a_n)}=\frac{1}{\max(a_m-a_n)}; \ \ m>n$ This is not sufficient, because when equality occurs we have: $\frac{s_{m,n}-(a_m^2-a_n^2)}{a_m-a_n} =\frac{1}{\max(a_m-a_n)} \\ s_{m,n}=\frac{a_m-a_n}{\max(a_m-a_n)}+a_m^2-a_n^2$ It follows that $\max(a_m-a_n) | a_m-a_n$ for all $m>n$ .

So for any given sequence of natural numbers $a_k$ first rings occur for $r_{min}=\frac{1}{2 (\max(a_m-a_n)) }$ such that $\max(a_m-a_n)$ divides $a_m-a_n$ for all $m>n$ .

For this problem maximum of $a_m-a_n$ that divides all of other $a_m-a_n$ 's is $1$ . So $r=\frac{1}{2}$ and $r+a_6=\sqrt{t_6} \\ \frac{1}{2}+12=\sqrt{t_6} \\ t_6=\frac{625}{4}$

Let $r_0$ be the radius of the inner ring. Then the outer ring has radius $r_0 + 1 + 1 + 2 + 3 + 5 = 12 + r_0$ .

The time difference between the two outer rings of the pattern of six is of the form $(12 + r_0)^2 - (11 + r_0)^2 = 2r_0 + 23,$ which must be an integer. This requires $r_0$ to be a half-integer. Working with other pairs of rings in the pattern yields the same conclusion.

Thus a necessary and sufficient condition is that $r_0 \equiv \tfrac12 \ \text{mod 2}$ . The smallest possible value that satisfies is $r_0 = \tfrac12$ ; the outer ring therefore has radius $12\tfrac12$ , and $t = (12\tfrac12)^2 = \frac{625}{4}.$ Thus the solution is $\boxed{629}$ .

A lenient/intuitive proof

Let the radius of rings from the inner one to outer one be $r_1$ to $r_6$ . (unit: $cm$ )

$\left\{\begin{matrix}r_1=x \\ r_2=x+5y \\ r_3=x+8y \\ r_4=x+10y \\ r_5=x+11y \\ r_6=x+12y \end{matrix}\right.$

where $x$ and $y$ are real numbers

The areas bounded by two consecutive rings are $A_1$ to $A_5$ , in terms of $\pi$ . (unit: $cm^2$ )

$\left\{\begin{matrix}A_1=r_2^2-r_1^2=10xy+25y^2 \\A_2=r_3^2-r_2^2=6xy+39y^2 \\A_3=r_4^2-r_3^2=4xy+36y^2 \\A_4=r_5^2-r_4^2=2xy+42y^2 \\A_5=r_6^2-r_5^2=2xy+46y^2 \end{matrix}\right.$

As ring that grows continuously outward at a rate of $\pi cm^2 / s$ and rings are separated by seconds in integers, the areas are integers.

[Basically next part is matrix reduction using elementary row operations]

Consider $A_2-2\times A_4=9y^2$ and $A_4-A_5=4y^2$ , both expression are still integers. Thus $y_{min}=1$ .

Substitute $y=1$ and you will see that $x_{min}=\frac{1}{2}$ .

Time can be calculated by calculating the area of the first ring = $r_5^2= (\frac{1}{2}+12\times 1)^2=\frac{625}{4}$