Putnam 1990: Problem A-4

I think the minimum number of required punches to exhaust the plane is 3. To see that 3 punches are enough, punch at $(0, 0)$, $(0, 1)$, $(e, 0)$ where $e$ is the Euler's constant. Now no point $(x, y)$ can be at rational distance from all of these because, otherwise all of $x^2 + y^2$, $x^2 + (y - 1)^2$, $(x - e)^2 + y^2$ are rational. From first two equations you get that $x, y$ are algebraic hence $e$ satisfies an algebraic equation which is false as $e$ is transcendental. Two punches are insufficient because there is always a point at an integer distance from the center of these punches.

I would think that three points should suffice, but need to invoke measure theory (which is not my favourite topic) to justify it. Clearly any two points will not suffice, since we always find a third point at given rational distances from these that satisfies the triangle inequality. (Take any pair of rational numbers smaller than the distance $d$ between the points but summing to a number$~>d$, and intersect the circles of those radii around the points.)

However having used two points, any remaining point is almost uniquely determined by the rational distances from those two points (at most two points give the same pair of distances). In particular only a countable set of points remain. The question is now whether one can strategically place the third point so that it avoids having a rational distance to any of the remaining points. We must avoid a countable set of circles, and since the union of those has measure zero, there are points that do not lie in the union; choosing such a point as third point should do the job.