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Sometimes, interesting clusters of questions may arise in this way.

1. The last problem given at the 1991 Mathematical Olympiad asked to construct an infinite bounded sequence of real numbers having $|x_i-x_j|\cdot |i-j|^{1+\epsilon} \geq 1$ for all $i \neq j$. Every professional mathematician will recognize this as a problem about badly approximable numbers; its solution is to take $x_n$ to be a big enough constant times the fractional part of $n\sqrt{2}$ (and this works with $\epsilon = 0$).

2. A similar looking but rather more difficult problem was given at a 239 St. Petersburg Olympiad: disprove the existence, for all $n \gg 0$ sufficiently large, of a set of $n$ points $P_1,\ldots,P_n$ contained by the disk of radius $10\sqrt{n}$ in $\mathbb{R}^2$ and having pairwise distances $d(P_i,P_j) > \sqrt{|i-j|}$. While it surely must have been motivated by the previous problem's appearance at a previous contest, this is a question about the relationship of transfinite diameter to the capacity of potential theory. Just multiplying the inequalities leads to a sharp estimate with both sides having for logarithms $\frac{1}{2}n^2\log{n} + O(n^2)$, and there is no immediate contradiction. If the estimate held with an error term of only $o(n^2)$ then, scaling the picture by $\sqrt{n}$ and letting $n \to \infty$, the points would attain the transfinite diameter of the disk, which would only be possible if they charged the equilibrium measure - the unform measure supported on the boundary circle, - forcing them to determine distances smaller than $o(1/\sqrt{n})$. We only have an estimate with $O(n^2)$, but we may still retain and justify the intuition that the limit measure $\nu$ in the scaled picture, obtained from the Dirac mass of the points after extraction of weak limit in $n \to \infty$, is very far from the two-dimensional Lebesque measure $\mu$. Since the original disk has positive $\nu$-measure, we may expect to find a nested chain of disks $D$ with an increasing concentration of points $\nu(D)/\mu(D)$. Making this precise will eventually produce a disk $D$ having $\nu(D)/\mu(D)$ arbitrarily large, and again, this would force $D$ to contain points of distance only $o(1/\sqrt{n})$. This means arbitrarily small distances in the unscaled picture, a contradiction.

3. But now, returning to diophantine approximations, one may bring elements from both questions and ask, with the optimal exponent $1/2$, to produce a bounded set of points $P_n$ in the plane having $d(P_i,P_j) \cdot |i-j|^{\frac{1}{2}+\epsilon} \geq 1$. Indeed the sequence $\big( \{n\sqrt{2} \}, \{n\sqrt{3}\} \big)$ has the desired property (where $\sqrt{2}$ and $\sqrt{3}$ can be replaced by generic real numbers, which will be the more direct route to proving the statement). But the proof is considerably harder (the diophantine variant would require some form of Schmidt's theorem), and I do not know if the statement continues to hold with $\epsilon = 0$.

Sometimes, interesting clusters of questions may arise in this way.

1. The last problem given at the 1991 Mathematical Olympiad asked to construct an infinite bounded sequence of real numbers having $|x_i-x_j|\cdot |i-j|^{1+\epsilon} \geq 1$ for all $i \neq j$. Every professional mathematician will recognize this as a problem about badly approximable numbers; its solution is to take $x_n$ to be a big enough constant times the fractional part of $n\sqrt{2}$ (and this works with $\epsilon = 0$).

2. A similar looking but rather more difficult problem was given at a 239 St. Petersburg Olympiad: disprove the existence, for all $n \gg 0$ sufficiently large, of a set of $n$ points $P_1,\ldots,P_n$ contained by the disk of radius $10\sqrt{n}$ in $\mathbb{R}^2$ and having pairwise distances $d(P_i,P_j) > \sqrt{|i-j|}$. While it surely must have been motivated by the previous problem's appearance at a previous contest, this is a question about the relationship of transfinite diameter to the capacity of potential theory. Just multiplying the inequalities leads to a sharp estimate with both sides having for logarithms $\frac{1}{2}n^2\log{n} + O(n^2)$, and there is no immediate contradiction. If the estimate held with an error term of only $o(n^2)$ then, scaling the picture by $\sqrt{n}$ and letting $n \to \infty$, the points would attain the transfinite diameter of the disk, which would only be possible if they charged the equilibrium measure - the unform measure supported on the boundary circle, - forcing them to determine distances smaller than $o(1/\sqrt{n})$. We only have an estimate with $O(n^2)$, but we may still retain and justify the intuition that the limit measure $\nu$ in the scaled picture, obtained from the Dirac mass of the points after extraction of weak limit in $n \to \infty$, is very far from the two-dimensional Lebesgue measure $\mu$. Since the original disk has positive $\nu$-measure, we may expect to find a nested chain of disks $D$ with an increasing concentration of points $\nu(D)/\mu(D)$. Making this precise will eventually produce a disk $D$ having $\nu(D)/\mu(D)$ arbitrarily large, and again, this would force $D$ to contain points of distance only $o(1/\sqrt{n})$. This means arbitrarily small distances in the unscaled picture, a contradiction.

3. But now, returning to diophantine approximations, one may bring elements from both questions and ask, with the optimal exponent $1/2$, to produce a bounded set of points $P_n$ in the plane having $d(P_i,P_j) \cdot |i-j|^{\frac{1}{2}+\epsilon} \geq 1$. Indeed the sequence $\big( \{n\sqrt{2} \}, \{n\sqrt{3}\} \big)$ has the desired property (where $\sqrt{2}$ and $\sqrt{3}$ can be replaced by generic real numbers, which will be the more direct route to proving the statement). But the proof is considerably harder (the diophantine variant would require some form of Schmidt's theorem), and I do not know if the statement continues to hold with $\epsilon = 0$.

Sometimes, interesting clusters of questions may arise in this way.

1. The last problem given at the 1991 Mathematical Olympiad asked to construct an infinite bounded sequence of real numbers having $|x_i-x_j|\cdot |i-j|^{1+\epsilon} \geq 1$ for all $i \neq j$. Every professional mathematician will recognize this as a problem about badly approximable numbers; its solution is to take $x_n$ to be a big enough constant times the fractional part of $n\sqrt{2}$ (and this works with $\epsilon = 0$).

2. A similar looking but rather more difficult problem was given at a 239 St. Petersburg Olympiad: disprove the existence, for all $n \gg 0$ sufficiently large, of a set of $n$ points $P_1,\ldots,P_n$ contained by the disk of radius $10\sqrt{n}$ in $\mathbb{R}^2$ and having pairwise distances $d(P_i,P_j) > \sqrt{|i-j|}$. While it surely must have been motivated by the previous problem's appearance at a previous contest, this is a question about the relationship of transfinite diameter to the capacity of potential theory. Just multiplying the inequalities leads to a sharp inequality with both sides having for logarithms $\frac{1}{2}n^2\log{n} + O(n^2)$, and there is no immediate contradiction. If the estimate held with an error term of only $o(n^2)$ then, scaling the picture by $\sqrt{n}$ and letting $n \to \infty$, the points would attain the transfinite diameter of the disk, which would only be possible if they charged the equilibrium measure - the unform measure supported on the boundary circle, - forcing them to determine distances smaller than $o(1/\sqrt{n})$. We only have an estimate with $O(n^2)$, but we may still retain and justify the intuition that the limit measure $\nu$ in the scaled picture, obtained from the Dirac mass of the points after extraction of weak limit in $n \to \infty$, is very far from the two-dimensional Lebesque measure $\mu$. Since the original disk has positive $\nu$-measure, we may expect to find a nested chain of disks $D$ with increasing $\nu(D)/\mu(D)$. Making this precise we will eventually have $\nu(D)/\mu(D)$ arbitrarily large, but this would force $D$ to contain points of distance only $o(1/\sqrt{n})$.

3. But now, returning to diophantine approximations, one may bring elements from both questions and ask, with the optimal exponent $1/2$, to produce a bounded set of points $P_n$ in the plane having $d(P_i,P_j) \cdot |i-j|^{\frac{1}{2}+\epsilon} \geq 1$. Indeed the sequence $\big( \{n\sqrt{2} \}, \{n\sqrt{3}\} \big)$ has the desired property (where $\sqrt{2}$ and $\sqrt{3}$ can be replaced by generic real numbers, which will be the more direct route to proving the statement). But the proof is considerably harder (the diophantine variant would require some form of Schmidt's theorem), and I do not know if the statement continues to hold with $\epsilon = 0$.

Sometimes, interesting clusters of questions may arise in this way.

1. The last problem given at the 1991 Mathematical Olympiad asked to construct an infinite bounded sequence of real numbers having $|x_i-x_j|\cdot |i-j|^{1+\epsilon} \geq 1$ for all $i \neq j$. Every professional mathematician will recognize this as a problem about badly approximable numbers; its solution is to take $x_n$ to be a big enough constant times the fractional part of $n\sqrt{2}$ (and this works with $\epsilon = 0$).

2. A similar looking but rather more difficult problem was given at a 239 St. Petersburg Olympiad: disprove the existence, for all $n \gg 0$ sufficiently large, of a set of $n$ points $P_1,\ldots,P_n$ contained by the disk of radius $10\sqrt{n}$ in $\mathbb{R}^2$ and having pairwise distances $d(P_i,P_j) > \sqrt{|i-j|}$. While it surely must have been motivated by the previous problem's appearance at a previous contest, this is a question about the relationship of transfinite diameter to the capacity of potential theory. Just multiplying the inequalities leads to a sharp estimate with both sides having for logarithms $\frac{1}{2}n^2\log{n} + O(n^2)$, and there is no immediate contradiction. If the estimate held with an error term of only $o(n^2)$ then, scaling the picture by $\sqrt{n}$ and letting $n \to \infty$, the points would attain the transfinite diameter of the disk, which would only be possible if they charged the equilibrium measure - the unform measure supported on the boundary circle, - forcing them to determine distances smaller than $o(1/\sqrt{n})$. We only have an estimate with $O(n^2)$, but we may still retain and justify the intuition that the limit measure $\nu$ in the scaled picture, obtained from the Dirac mass of the points after extraction of weak limit in $n \to \infty$, is very far from the two-dimensional Lebesque measure $\mu$. Since the original disk has positive $\nu$-measure, we may expect to find a nested chain of disks $D$ with an increasing concentration of points $\nu(D)/\mu(D)$. Making this precise will eventually produce a disk $D$ having $\nu(D)/\mu(D)$ arbitrarily large, and again, this would force $D$ to contain points of distance only $o(1/\sqrt{n})$. This means arbitrarily small distances in the unscaled picture, a contradiction.

3. But now, returning to diophantine approximations, one may bring elements from both questions and ask, with the optimal exponent $1/2$, to produce a bounded set of points $P_n$ in the plane having $d(P_i,P_j) \cdot |i-j|^{\frac{1}{2}+\epsilon} \geq 1$. Indeed the sequence $\big( \{n\sqrt{2} \}, \{n\sqrt{3}\} \big)$ has the desired property (where $\sqrt{2}$ and $\sqrt{3}$ can be replaced by generic real numbers, which will be the more direct route to proving the statement). But the proof is considerably harder (the diophantine variant would require some form of Schmidt's theorem), and I do not know if the statement continues to hold with $\epsilon = 0$.

Sometimes, interesting clusters of questions may arise in this way.

1. The last problem given at the 1991 Mathematical Olympiad asked to construct an infinite bounded sequence of real numbers having $|x_i-x_j|\cdot |i-j|^{1+\epsilon} \geq 1$ for all $i \neq j$. Every professional mathematician will recognize this as a problem about badly approximable numbers; its solution is to take $x_n$ to be a big enough constant times the fractional part of $n\sqrt{2}$ (and this works with $\epsilon = 0$).

2. A similar looking but much more difficult problem was given at a 239 St. Petersburg Olympiad: disprove the existence, for all $n \gg 0$ sufficiently large, of a set of $n$ points $P_1,\ldots,P_n$ contained by the disk of radius $10\sqrt{n}$ in $\mathbb{R}^2$ and having pairwise distances $d(P_i,P_j) > \sqrt{|i-j|}$. While it surely must have been motivated by the previous problem's appearance at a previous contest, this question is, in disguise, exactly the relationship of transfinite diameter to the capacity of potential theory. Just multiplying the inequalities leads to a sharp estimate in which both sides have for logarithms $\frac{1}{2}n^2\log{n} + O(n^2)$, and there is no immediate contradiction. However, the sharpness of the estimate means precisely that the points fulfil the capacity of the disk; for, rescaling the picture by $\sqrt{n}$ and letting $n \to \infty$, they attain the transfinite diamater. This happens precisely when, in the limit $n \to \infty$, the points charge the equilibrium measure, which is the uniform measure supported on the boundary circle. But then there will be pairs of points with distances only $O(1/\sqrt{n})$ apart ($1/n$ in the scaled picture). The intended solution to the problem, while elementary, was also based on an equidistribution principle.

3. But now, returning to diophantine approximations, one may bring elements from both questions and ask, with the optimal exponent $1/2$, to produce a bounded set of points $P_n$ in the plane having $d(P_i,P_j) \cdot |i-j|^{\frac{1}{2}+\epsilon} \geq 1$. Indeed the sequence $\big( \{n\sqrt{2} \}, \{n\sqrt{3}\} \big)$ has the desired property (where $\sqrt{2}$ and $\sqrt{3}$ can be replaced by generic real numbers, which will be the more direct route to proving the statement). But the proof is considerably harder (the diophantine variant would require some form of Schmidt's theorem), and I do not know if the statement continues to hold with $\epsilon = 0$.

Sometimes, interesting clusters of questions may arise in this way.

1. The last problem given at the 1991 Mathematical Olympiad asked to construct an infinite bounded sequence of real numbers having $|x_i-x_j|\cdot |i-j|^{1+\epsilon} \geq 1$ for all $i \neq j$. Every professional mathematician will recognize this as a problem about badly approximable numbers; its solution is to take $x_n$ to be a big enough constant times the fractional part of $n\sqrt{2}$ (and this works with $\epsilon = 0$).

2. A similar looking but rather more difficult problem was given at a 239 St. Petersburg Olympiad: disprove the existence, for all $n \gg 0$ sufficiently large, of a set of $n$ points $P_1,\ldots,P_n$ contained by the disk of radius $10\sqrt{n}$ in $\mathbb{R}^2$ and having pairwise distances $d(P_i,P_j) > \sqrt{|i-j|}$. While it surely must have been motivated by the previous problem's appearance at a previous contest, this is a question about the relationship of transfinite diameter to the capacity of potential theory. Just multiplying the inequalities leads to a sharp inequality with both sides having for logarithms $\frac{1}{2}n^2\log{n} + O(n^2)$, and there is no immediate contradiction. If the estimate held with an error term of only $o(n^2)$ then, scaling the picture by $\sqrt{n}$ and letting $n \to \infty$, the points would attain the transfinite diameter of the disk, which would only be possible if they charged the equilibrium measure - the unform measure supported on the boundary circle, - forcing them to determine distances smaller than $o(1/\sqrt{n})$. We only have an estimate with $O(n^2)$, but we may still retain and justify the intuition that the limit measure $\nu$ in the scaled picture, obtained from the Dirac mass of the points after extraction of weak limit in $n \to \infty$, is very far from the two-dimensional Lebesque measure $\mu$. Since the original disk has positive $\nu$-measure, we may expect to find a nested chain of disks $D$ with increasing $\nu(D)/\mu(D)$. Making this precise we will eventually have $\nu(D)/\mu(D)$ arbitrarily large, but this would force $D$ to contain points of distance only $o(1/\sqrt{n})$.

3. But now, returning to diophantine approximations, one may bring elements from both questions and ask, with the optimal exponent $1/2$, to produce a bounded set of points $P_n$ in the plane having $d(P_i,P_j) \cdot |i-j|^{\frac{1}{2}+\epsilon} \geq 1$. Indeed the sequence $\big( \{n\sqrt{2} \}, \{n\sqrt{3}\} \big)$ has the desired property (where $\sqrt{2}$ and $\sqrt{3}$ can be replaced by generic real numbers, which will be the more direct route to proving the statement). But the proof is considerably harder (the diophantine variant would require some form of Schmidt's theorem), and I do not know if the statement continues to hold with $\epsilon = 0$.