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I was told this lovely puzzle recently which I thought people here might enjoy.

Consider a paper punch that can be centered at any point of the plane and that, when operated, removes from the plane precisely those points whose distance from the center is irrational. How many punches are needed to remove every point?

4 Answers4

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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.

hot_queen
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  • Of course, you do not really need the transcendence of a particular number like $e$. A simple cardinality argument shows that there are non algebraic real numbers. – hot_queen Apr 15 '13 at 19:49
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    +1. You even get that $y$, which occurs linearly in the difference between the first two rational numbers, is rational. Therefore I think taking $(0,0)$, $(0,1)$ and $(0,\sqrt2)$ would also do; no transcendentals are needed. – Marc van Leeuwen Apr 15 '13 at 20:01
  • @Marc van Leeuwen: It's not immediately clear that the three points you give work. It's true that y is rational, so x^2 is rational, and everything in x^2 + y^2 - 2\sqrt{2}y + 2 is rational, which would seem to give the contradiction that \sqrt{2} is rational, but what if y=0? – Edward Doolittle Oct 07 '21 at 23:10
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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.

Marc van Leeuwen
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Three punches suffice if they are made at collinear points $A,B,C$ such that $B$ is the midpoint of $AC$ and the square of the length of $AB$ is irrational. For any point $P$, Apollonius's theorem ensures that $$ PA^2+PC^2-2\cdot PB^2=2\cdot AB^2. $$ As the right hand side is irrational, it follows that one of the distances $PA,PB,PC$ is irrational. Therefore the point $P$ has been removed.

Note that Apollonius's theorem applies even when the triangle $\Delta PAC$ is degenerate.

Samrith Ram
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I may not be understanding the question correctly, but it seems like the answer would be three - two points with an irrational distance between themselves, and a third point to eliminate the locus of rational points equidistant between the remaining two points.

Jonathan Rich
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