This is also from the ISL. I've only managed to solve (a), which is not at all tough. We just find a triangle with vertices (as complex numbers) of the form $2n,\ 2n\epsilon,\ 2n\epsilon^2$. Since $\epsilon=\frac{-1+\sqrt3}2$, it means that we can find an $n$ s.t. $2n\epsilon$ is inside a disk (by using the "famous Dedekind theorem" , otherwise known as Kronecker's thm). $2n\epsilon^2$ will, of course, be inside a disk as well, because $\epsilon^2$ is the symmetric of $\epsilon$ wrt the $Ox$ axis.

Don't you have any other proof????? A more elementary proof???

Part (a) is a direct application of Jacobi's theorem [ if a is irrational then ( na - floor(na) ) is everywhere dense in the interval (0,1) as n varies over the natural numbers. ] From this the proof of (a) is obvious Anyone for part (b)

This is a very nice problem! proof of (strengthened) part b: Assume the contrary that there exists an equilateral triangle $ ABC$ such that $ A,B,C$ lie in the interior of different disks, where $ |AB|=|BC|=|CA|=s\le124$. Let $ O_1,O_2,O_3$ be the center of the disks which contain $ A,B,C$ respectively. Note that $ s-0.002\le O_1O_2,O_2O_3,O_3O_1\le s+0.002$. Clearly, $ O_1O_2O_3$ is not equilateral, so assume $ O_1O_2>O_2O_3$. So $ O_1O_2^2-O_2O_3^2=(O_1O_2-O_2O_3)(O_1O_2+O_2O_3)\le 0.004\times(2s+0.004)\le 0.004(2\cdot124+0.004)<1$ But $ O_1O_2^2-O_2O_3^2$ must be an integer, contradiction.

Not that it matters much, but the proof I came up with (using complex numbers) shows a bound of > 166, I think.

Part a is pretty easy, you just use that $\{\sqrt{3}k\}$ gets close to $1$ for some integer $k$. For part $b$ we do the following. First we notice that without lose of generality we may assume one point is on the disk centered at the origin and another at the disk centered at $(a,0)$ where a is an integer. Now, we will show that the third point is close to $(\frac{a}{2},\frac{\sqrt{3}}{2}a)$. The idea is that the side of the equilateral triangle must be between $a- 1/500$ and $a+ 1/500$, also the slope between the point near the origin and the point near $(a,0)$ is at most $-\frac{1}{500a}$, hence the third point coming out of the midpoint has slope at least $500a$. Playing around with the equations, it is not hard to bound the difference between the third point and $(\frac{a}{2},\frac{\sqrt{3}}{2}a)$. Once we know this, we have that $\frac{a}{2}$ must be close to an integer, implying it is an integer. So we may write $a = 2k$. Now we need $\sqrt{3}k$ to be close to an integer. Notice that the length of the equilateral would be at least $2(k-\frac{1}{500}) = 2k -\frac{1}{250}$. For $a > 96$, we need only show $k > 50$. So let's assume $k \leq 50$ and reach a contradiction. The continued fraction of $\sqrt{3}$ is $1, [1,2]$, so it is easy to find the convergents. The convergents are $1, 2, \frac{5}{3}, \frac{7}{4}, \frac{19}{11},\frac{26}{15},\frac{71}{41},\frac{97}{56}, \ldots.$ Now, if $\left| \sqrt{3} - \frac{p}{q}\right| < \frac{1}{2q^2}$, then $\frac{p}{q}$ is a convergent. Note this would be equivalent to $\left|\sqrt{3}k - p\right| < \frac{1}{2k}$ implying $\frac{p}{k}$ is a convergent. But we know that for some $k \leq 50$ and an integer $n$, $\left|\sqrt{3}k - n\right| < \frac{1}{1000} < \frac{1}{2k}$, therefore $\frac{n}{k}$ is a convergent with the denominator being at most $50$, so $k$ must be at most $41$. Now, another theorem tells us $\left|\alpha - \frac{p_k}{q_k}\right| > \frac{1}{q_k(q_k+q_{k+1})}$, so we have $\left|41\sqrt{3}-71\right| > \frac{41}{41(41+56)} = \frac{1}{97} > \frac{1}{1000}$, yielding the result we want. Using more convergents one can show a better lowerbound for the sidelength, one could show sidelength $ > 1040$, and using a computer one could show that the sidelength must be bigger than $1559$.

Quique wrote: For part $b$ we do the following. First we notice that without lose of generality we may assume one point is on the disk centered at the origin and another at the disk centered at $(a,0)$ where a is an integer. Why can we do this? Anyway, here's another way to do (b) that can be strengthened quite a bit with some more effort (i.e. by tightening bounds by finding explicit solutions to $N(u+v\sqrt{3})=k$ for small $k$)... By translation and reflection about the $y$-axis if necessary, WLOG let the centers be $(0,0),(a,b),(c,-d)$ (where $a,b,c,d\ge0$), with displacements $(p,q),(x,y),(z,w)$ such that $p^2+q^2,x^2+y^2,z^2+w^2<r^2$ (where $r=1/1000$). By complex numbers, we have \[(a+x-p)+i(b+y-q) = [(c+z-p)+i(-d+w-q)]\left(\frac{1}{2}+i\frac{\sqrt{3}}{2}\right),\]so \[l=(2a-c)-d\sqrt{3} = (p+q\sqrt{3}) + (z-w\sqrt{3}) - 2x \in (-6r,6r)\]and \[m=(2b+d)-c\sqrt{3} = (q-p\sqrt{3}) + (w+z\sqrt{3}) - 2y \in (-6r,6r).\]First note that $2a-c,2b+d\ge0$ or else we have obvious bounding contradictions. Letting $N(u+v\sqrt{3})=u^2-3v^2$, note that $N(l),N(m)\in\{\ldots,-2,1,3,\ldots\}$ since $-1$ is not a quadratic residue modulo $3$. Assume for the sake of contradiction that the side length is at most $96$, whence $\sqrt{c^2+d^2}-2r\le96$ and $c^2+d^2<97^2$. Now we find that \[N(l)\ge3\implies d>\frac{\frac{3}{6r}-6r}{2\sqrt{3}}>144,\]so $N(l)\le1$ and similarly $N(m)\le1$. In addition, \[N(l)=1\implies d>\frac{\frac{1}{6r}-6r}{2\sqrt{3}}>48\]and \[N(l)\le-2\implies d>\frac{\frac{2}{6r}}{2\sqrt{3}}>96,\]so $N(l)=N(m)=1$. But \[N(u+v\sqrt{3})=1\implies (u,v)\in\{(2,1),(7,4),(26,15),(97,56),(362,209),\ldots\},\]so $c=d=56$ and $2a-c=2b+d=97$, a contradiction modulo $2$. Edit: With some computing, we can find that \[l,m\in\{97-56\sqrt{3},265-153\sqrt{3},362-209\sqrt{3},627-362\sqrt{3},724-418\sqrt{3},\ldots\}.\]Also notice that $2a-c\equiv c\pmod{2}$ and $2b+d\equiv d\pmod{2}$, so our smallest candidate (size defined by $c^2+d^2$) is \[\{l,m\}=\{97-56\sqrt{3},362-209\sqrt{3}\}.\]This works when, for example, \[(p,q,x,y,z,w)=\left(\frac{3}{10000},\frac{9}{10000},\frac{r}{20}(7+17\sqrt{3})-\frac{97-56\sqrt{3}}{2},\frac{r}{20}(1+\sqrt{3})-\frac{362-209\sqrt{3}}{2},\frac{4}{10000},\frac{-8}{10000}\right),\]giving us a lower bound of around $\sqrt{56^2+209^2}>216$.

<jgnr> wrote: Clearly, $ O_1O_2O_3$ is not equilateral I don't understand why

1) Consider the triangle $(0,0)$, $(2x,0)$, $\left(x,x\sqrt3\right)$, for integer $x$. Note that $\{k\sqrt3\}$ is dense in $[0,1]$ (we can prove this easily with Pigeonhole) so there must exist $x$ such that $\{x\sqrt3\}<\frac{1}{1000}$. Then, choosing this $x$ gives all three points within $\frac{1}{1000}$ of lattice points. 2) Suppose the contrary, namely that we do have an equilateral triangle with side length at most $96$. For convenience, shift one of the vertices to the origin. As this shift has length at most $\frac{1}{1000}$, we will increase the disks to radius $\frac{1}{500}$ to ensure that the images of the other two vertices still lie within disks. Then, if the centers of the disks which coincide with the other two vertices are $(a,b)$, $(c,d)$ (labeling counterclockwise), then we should have that $a^2+b^2,c^2+d^2\le 96^2$, and that the image of $(a,b)$ after a $\frac{\pi}{3}$ ccw rotation about the origin is at most $\frac{1}{250}$ away from $(c,d)$. Expanding with distance formula, $$\left(\frac{a\sqrt{3}-b}{2}-c\right)^2+\left(\frac{a+b\sqrt{3}}{2}-d\right)^2=(a^2+b^2+c^2+d^2+bc-ad)-(ac+bd)\sqrt{3}\le\frac{1}{250^2}$$Now, note that $$a^2+b^2+c^2+d^2+bc-ad+(ac+bd)\sqrt{3}\le \frac{3+\sqrt 3}{2}(a^2+b^2+c^2+d^2)\le (3+\sqrt 3)96^2$$So, we have that $$(a^2+b^2+c^2+d^2+bc-ad-(ac+bd)\sqrt{3})(a^2+b^2+c^2+d^2+bc-ad+(ac+bd)\sqrt{3})\le \frac{(3+\sqrt3)96^2}{250^2}<1$$which is a contradiction as the LHS is a positive integer.

Easy proof for $124$: For (1), take $(0,0)$, $(N,0)$, and $(N/2,N\sqrt{3}/2)$ for a positive integer $N$ such that the fractional part of $N\sqrt{3}/2$ is much smaller than $1/1000$ (which exists since $\{k\alpha\}$ is dense in $(0,1)$ as we vary $k$ over the positive integers). For (2), we show that every equilateral triangle must have side-length at least $124$. Suppose the side-length is actually $N$. Work in the complex plane, and let $\omega=e^{2\pi i/6}$. Suppose the three disks the triangle's vertices are centered at $0$, $a$, and $b$ (WLOG we shifted one of them to the origin), and suppose the three vertices are actually $\gamma$, $a+\alpha$, and $b+\beta$, where $|\alpha|,|\beta|,|\gamma|<1/1000$. Note that \[N=|a+\alpha-\gamma|\le |a|+2/1000,\]so $|a|,|b|\ge N-2/1000$. Now, \[\frac{a+\alpha-\gamma}{b+\beta-\gamma}=\omega,\]so \[\frac{a}{b} = \omega+\omega\frac{\beta-\gamma}{b}+\frac{\gamma-\alpha}{b},\]so \[\left|\frac{a}{b}-\omega\right|\le \frac{4/1000}{N-2/1000}.\]We also have \[\left|\frac{a}{b}-\omega\right| = \frac{1}{|b|\cdot|a+b\omega^2|}|(a-b\omega)(a+b\omega^2)| = \frac{|a^2-ab+b^2|}{|b|\cdot|a+b\omega^2|}.\]Since $a,b\in\mathbb{Z}[i]$, we have that $a^2-ab+b^2\in\mathbb{Z}[i]\setminus\{0\}$, so $|a^2-ab+b^2|\ge 1$, so \[\left|\frac{a}{b}-\omega\right|\ge \frac{1}{2(N+2/1000)^2},\]so \[\frac{1}{2(N+2/1000)^2}\le \frac{4/1000}{N-2/1000}.\]It is easy to see that this implies \[N\ge 124.994,\]as desired. Remark: The inequality could be made stronger by using a tighter bound for \[\left|\omega\frac{\beta-\gamma}{b}+\frac{\gamma-\alpha}{b}\right|\]rather than the naive triangle inequality bound we use. Indeed, equality cannot be achieved for that bound, and some annoying work could push this to get around 150 or so (I haven't done all the details myself).

(a) Indeed, consider the equilateral triangle formed by the vertices $(a,0),(-a,0)$ and $(0,\sqrt{3}a)$. By a well-known fact, there exists $a$ such that $\{\sqrt{3}a\}<\frac{1}{1000}$, then for this $a$, the corresponding equilateral triangle will satisfy the condition. (b)Consider such an equilateral triangle $ABC$, lying in the interior of the disk centered at $X,Y,Z$. Now notice that by a well-known fact $XYZ$ cannot formed an equilateral triangle. Suppose $XY> YZ$. Now by triangle inequality, $$|XY-YZ|\leq |XY-AB|+|BC-YZ|\leq \frac{4}{1000}$$Meanwhile, $XY=\sqrt{a}$ and $YZ=\sqrt{b}$ where $a,b$ are integers, so $$XY-YZ=\frac{a-b}{\sqrt{a}+\sqrt{b}}>\frac{1}{2\sqrt{a}}$$which implies $a\leq 125^2$, and so $$AB>\frac{XY}-\frac{1}{1000}>124$$as desired.

Note that for some large enough $x$, $(\sqrt{3}-1)^x<\tfrac{1}{1000}$. Then, if we let $(\sqrt{3}-1)^x=a\sqrt{3}-b$ for positive integers $a$ and $b$ then $a\sqrt{3}$ is within $\frac{1}{1000}$ of an integer. Thus, if we let two of the coordinates be $(0,0)$ and $(2a,0)$, then $(a,\sqrt{3})$ forms an equilateral triangle with the other two and lies inside of a disk. \item Let there be an equilateral triangle with vertices in different discs and has side-length $s$. Let $A$, $B$, $C$ be the centers of those discs. Each side length of $ABC$ is between $s-\tfrac1{500}$ and $s+\tfrac1{500}$. WLOG $PQ>QR$ since $PQR$ is clearly not equilateral. We have \[1\le PQ^2-QR^2=(PQ+QR)(PQ-QR)\le (2s+\tfrac1{250})\cdot (\tfrac{1}{250})\]so $s\ge 125-\tfrac1{500}>96$. We are done.