# A lattice of circles and a line which does not cut any of them.

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?

• Hint: $2\mathbb{Z}+5\mathbb{Z} = \mathbb{Z}$. – 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).

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:

$$\dfrac{-2p+5q}{\sqrt{29}},$$

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:

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

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

• Excellent for providing intuition. – 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 $$\boxed{d_{sup} = {1\over 2\sqrt{29}}}$$.