Rational numbers- sticks and stones

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.

• An example would be useful to clarify what you mean. What is the stick and stone representation for $2/3$? Commented Mar 5, 2012 at 0:09
• Somehow I think the Farey sequence of rationals will be useful here. Or else you want to construct an enumeration of the rationals and assign a value to each rational such that for any two rationals, the sum of the values assigned to them is less than the distance between them. Commented Mar 5, 2012 at 0:17
• A line of some length along x=2/3, topped with a circle centered at the endpoint of that line that is not on the x-axis, or a radius less than the length of the line.
– Ali
Commented Mar 5, 2012 at 0:18
• There is a proof that states that no interval of any size $\epsilon$ is without a rational number. Perhaps that might help?
– Ali
Commented Mar 5, 2012 at 0:19
• Are these lines vertical or horizontal? Commented Mar 5, 2012 at 0:40

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.

• What a neat function! I have not encountered it before. Could you explain why r(p/q) =1/2q(1+1)= 1/4q should work, as well as what the x value of the highest point on the graph is?
– Ali
Commented Mar 5, 2012 at 0:50
• The height of each point depends only on the denominator $q$; greater values of $q$ correspond to lower heights. So, what is the fraction in $(0,1)$ with the smallest denominator? Alternatively, use the symmetry of the picture ... Commented Mar 5, 2012 at 0:59
• Ilmari Karonen- could you please explain why your r(p/q)= 1/2q(1+1) works?
– Ali
Commented Mar 5, 2012 at 1:07
• I did think it was 1/2 but that doesn't exactly make sense- wouldn't f(1/3)> f(1/2)? Or 1/4 for that matter?
– Ali
Commented Mar 5, 2012 at 1:10
• @Aliya: The centers of the stones are located along horizontal rows at height $y_q = \frac1q$ for integer $q$; in particular, the distance between rows $q$ and $q+1$ is $\frac1q - \frac1{q+1} = \frac{q-(q+1)}{q(q+1)} = \frac1{q(q+1)}$. The radius $\frac1{2q(q+1)}$ is half of that. Also, the horizontal distance between stones on the same row is at least as much as the vertical distance, so each stone fits in a nice box that doesn't touch any other stones. Commented Mar 5, 2012 at 1:12

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 $\frac cd$