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Qiaochu Yuan
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(Not an answer.)

Let $r = \frac{1}{a}, s = \frac{1}{b}$. Reindex your points to $(x_0, y_0), ... (x_{n-1}, y_{n-1})$ and let $z_k^2 = x_k^2 + y_k^2$. Then $x_k = z_k \cos \frac{2\pi k}{n}, y_k = z_k \sin \frac{2\pi k}{n}$, and the intersection condition becomes

$$z_k^2 \left( r^2 \cos^2 \frac{2\pi k}{n} + s^2 \sin^2 \frac{2\pi k}{n} \right) = 1.$$

Together with the condition that $\prod_{k=0}^{n-1} z_k = 1$, it follows that the desired conditions can be stated as

$$\prod_{k=0}^{n-1} \left( r^2 \cos^2 \frac{2\pi k}{n} + s^2 \sin^2 \frac{2\pi k}{n} \right) = 1.$$

This is likely to be a hard Diophantine equation to solve in general. For $n = 3$, for example, the equation is

$$r^2 (r^2 + 3s^2)^2 = 16.$$

The curves $r(r^2 + 3s^2) = 4$ and $r(r^2 + 3s^2) = -4$ are both elliptic curves, and in general one must use computer algebra to rule out the existence of rational points on such curves. For higher values of $n$ the situation is likely to be even worse.

Qiaochu Yuan
  • 118.2k
  • 40
  • 447
  • 741