Each lattice point is a center of a circle, all with radius $d$. Let line $y={2\over 5}x+n$ doesn't cut or touch any circle, for some $n$. Find the supremum for $d$.

I was trying to maximize $$d={|2x-5y+5n|\over \sqrt{29}}$$ where $x,y$ runs over all integers and we can assume that $n$ is in $(0,1)$, but I don't know how to do it. Any help?

  • 2
    $\begingroup$ Hint: $2\mathbb{Z}+5\mathbb{Z} = \mathbb{Z}$. $\endgroup$ – Mindlack Nov 22 at 11:35

Consider the axis (A) passing through the origin, orthogonal to the family of lines $y=\tfrac25x+p$. All projections of lattice points on (A) are at mutual distances which are integer multiples of $1/\sqrt{29}$ (see explanation below).

enter image description here

If we consider now the projection of the circles on axis (A), we get little segments centered in the projections of lttice points; as they musn't overlap, their maximal radius is therfore $1/(2 \sqrt{29})$

Explanation: If we consider the axis as directed by unit vector $\vec{U}=\binom{-2a}{5a}$ (represented in red) where $a=\dfrac{1}{\sqrt{29}}$, the abscissa of the projection of point $(p;q)$ on axis (A) (in green) is the dot product: $\vec{U}.\binom{p}{q}$ for any $p,q \in \mathbb{Z}$, otherwise said:


showing that all these abscissas are integer multiples of $\dfrac{1}{\sqrt{29}}$.

Remark: In terms of lines, these $\dfrac{1}{\sqrt{29}}$ "gaps" become, by oblique projection, as $\dfrac{1}{\sqrt{29}}\dfrac{1}{\cos \alpha}=1/5$, where $\alpha$ is the angle between axis (A) and $y$ axis.

Henceforth, the lines passing through the centers of the circles have intercepts multiple of $1/5$, i.e., have equations:

$$y=\dfrac{2}{5}x+\dfrac{k}{5}, \ \ \ k \in \mathbb{Z}$$

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  • $\begingroup$ Nice! +1, but is it really maximal? $\endgroup$ – User2020201 Nov 22 at 19:21
  • $\begingroup$ I am going to add a little explanation. $\endgroup$ – Jean Marie Nov 22 at 20:29

This diagram might help you..........................

enter image description here

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  • 1
    $\begingroup$ Excellent for providing intuition. $\endgroup$ – Jean Marie Nov 22 at 11:41

Is this correct? (after sugesstion of @Intelligenti pauca)

So, since everything is ''repeating'' (with period $5$ in direction of $x$ axsis and $2$ in $y$ axsis) we can watch only lattice points in $[0,5)\times [0,2)$. So we are, for fixed $n$, searching for minimum in:

\begin{align}d(0,0) &= {5n\over \sqrt{29}}\\ d(1,0) &= {2+5n\over \sqrt{29}}\\ d(2,0) &= {4+5n\over \sqrt{29}}\\ d(3,0) &= {6+5n\over \sqrt{29}}\\ d(4,0) &= {8+5n\over \sqrt{29}} \end{align}

\begin{align} d(0,1) &= {|5n-5|\over \sqrt{29}}\\ d(1,1) &= {|5n-3|\over \sqrt{29}}\\ d(2,1) &= {|5n-1|\over \sqrt{29}}\\ d(3,1) &= {|5n+1|\over \sqrt{29}}\\ d(4,1) &= {|5n+3|\over \sqrt{29}}\\ \end{align}

The minimum among first 5 is $5n\over \sqrt{29}$ and among other 5 is ${|5n-1|\over \sqrt{29}}$ or ${|5n-3|\over \sqrt{29}}$ (since we are looking for minimum of $f(t)= |2t-a|$, where $a= 5n-1$ and $t\in \{-2,-1,0,1,2\}$ which is clearly at $t=0$ or $t=1$).

  • First case: $5n = |5n-1| \implies n={1\over 10}\implies d={1\over 2\sqrt{29}}$
  • Second case: $5n = |5n-3| \implies n={3\over 10}\implies d={3\over 2\sqrt{29}}$
  • Third case: $|5n-3| = |5n-1| \implies n={4\over 10}\implies d={1\over \sqrt{29}}$

So $\boxed{d_{sup} = {1\over 2\sqrt{29}}}$.

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