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What is the shortest curve $\gamma$ in $\mathbb{R}^2$ from the origin $o=(0,0)$ to a rational point $p=(a,b)$ that (a) passes through no other rational point, and (b) contains no point a rational distance from both $o$ and $p$?

A rational point is one with rational coordinates. I am wondering if one can reach $p$ via such a highly "irrational route." My guess is that there are curves whose lengths approach the straight-line distance $|p|$. Perhaps the curve should be restricted to a specific class: $C^\infty$, analytic, elliptic, quadratic, circle arc. (Just avoiding rational points [condition (a)] can be accomplished with a circle arc of an appropriately chosen radius.)

This is far from my experience, and it may be that all the variations have trivial, uninteresting answers, in which case I apologize for the distraction.

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What do you mean by (b)? If you start from o and go to p along a curve, the distance from o to your current location changes continuously from 0 to |p|, and you cannot avoid rational numbers. Do you mean “contains no point which is at a rational distance from both o and p at the same time”? –  Tsuyoshi Ito Jul 26 '11 at 12:25
    
@Tsuyoshi: Thanks, you are of course correct! Modified just as you suggest. –  Joseph O'Rourke Jul 26 '11 at 12:41
    
There should also be an uncountable infinity of points (c,d) whose coordinates are irrational, arbitrarily close to the desired line, such that the line segments to either of your points fulfill the desired conditions. Unless you meant a smooth curve? Gerhard "Ask Me About System Design" Paseman, 2011.07.26 –  Gerhard Paseman Jul 26 '11 at 15:12

2 Answers 2

up vote 18 down vote accepted

The simplest smooth curve that avoids all rational points is probably the parabola

$$y = \frac{b}{a}x + \lambda x(a-x)$$

where $\lambda$ is any irrational number.

Now, the set of points in $\mathbb{R}^2$ that are a rational distance from both $o$ and $p$ is countable (because any two rationals determine at most two such points); but the set of irrational $\lambda$ is uncountable. Moreover, as $\lambda$ varies, the resulting parabolas are all disjoint (apart from their end-points $o$ and $p$). So there must exist (uncountably many) $\lambda$ for which the above parabola satisfies your conditions. Also, as you suspect, $\lambda$ can be chosen as close to zero as you like.

Updated In fact we can take any one-parameter family of curves $\{C_{\lambda}\}$ from $o$ to $p$ which are disjoint except at $o$ and $p$. The set of all rational points, together with the set of points that are a rational distance from both $o$ and $p$, is countable, so in any open interval $I$ there exists $\lambda \in I$ such that $\{C_{\lambda}\}$ satisfies OP's conditions (a) and (b).

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Nice! :-) $\mbox{}$ –  Joseph O'Rourke Jul 26 '11 at 13:10

A circle passing throiugh the origin and your other point is determined by just one more point. Each rational point therefore determines a unique circle. Hence there are circles with large radius whose arcs ar arbitrarily close to the minimal distance containing no rational point, because there are uncountably many such.

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