Rational numbers- sticks and stones

Question

Moderator Note: At the time that this question was posted, it was from an ongoing contest. The relevant deadline has now passed.

For all rational numbers, we will have a stick of variable length extending along $x=R$ and atop this stick will be a circular "stone" centered at the point where the stick ends. No two such stick-and-stone(consisting of the stick wielding the stone at its centre) constructs for distinct rational numbers can touch or cover any parts of each other(a stick cannot tangent a stone and a stone cannot tangent another stone). Can we construct a set of stick-and-stone figures for all rational numbers ranging from 0 to 1 non-inclusive abiding by these rules? Why or why not? Note again that the heights of these structures can vary and that the radius of the stone must be less than the height of the stick.

Answer 1

You could start by using Thomae's function $f\left(\frac p q\right) = \frac 1 q$, where $\gcd(p,q) = 1$ to define the locations of the stones:

Then you just need to chose the radii of the stones small enough that no two of them will touch. Choosing $r\left(\frac p q\right) = \frac 1 {2q(q+1)}$ ought to be sufficiently small. (Can you see why?)

Edit: Here's a modified version of the graph above, with radii as given above and sticks to go with each stone:

Edit 2: Fixed typo in radius formula ($q+1$, not $1+1$).

Ps. Here's the program I used to generate the modified graph. It's a very slightly modified version of the original code from Wikimedia Commons. The code is provided "as is", with no warranties of correctness, readability, merchantability or fitness for a particular purpose.

Answer 2

If I understand you, for each rational $x_i \in (0,1)$, we are asked to choose a point $(x_i,y_i)$ and a radius $r_i$(given to us? all the same?) so that all the closed disks centered at $(x_i,y_i)$ with radius $r_i$ union the segment $x_i \times [0,y_i]$ are disjoint.

If the $r_i$ are constant, this is not possible. Let $m \gt \frac 1{r_i}$ and consider the sticks and stones for $(1/m, 2/m, 3/m \ldots (m-1)/m)$ whichever one of these has the lowest $y$ will see its stone overlap (both) its neighbor(s) sticks.

If the $r_i$ decrease fast enough, this is possible. If $x_i= \frac ab$ in lowest terms, give it height $\sum_{j=b}^{\infty} j^{-2}$ and radius $1/(2b^2)$. Two rationals, $\frac ab$ and $\frac cd$ in lowest terms with $b \gt d$ will be separated in the $x$ direction by at least $\frac 1{d(d-1)}$ and the stone from $\frac ab$ will be above the stick and stone of $\frac cd$