Kvant Math Problem 889
Consider the problem of choosing three points $A$, $B$, $C$ in the plane such that every point $P$ has at least one segment $PA$, $PB$, or $PC$ of irrational length.
Verified: no
Verdicts: SKIP + SKIP
Solve time: 7m03s
Source on kvant.digital
Problem
Do there exist three points $A$, $B$, $C$ in the plane such that for every point of the plane $P$, at least one of the segments $PA$, $PB$, and $PC$ has irrational length?
A. M. Slinko
All-Russian Mathematical Olympiad for School Students (Grade 10, 1984)
Exploration
Consider the problem of choosing three points $A$, $B$, $C$ in the plane such that every point $P$ has at least one segment $PA$, $PB$, or $PC$ of irrational length. Begin with simple examples where points have integer coordinates. For instance, let $A=(0,0)$, $B=(1,0)$, $C=(0,1)$. Then the distance from $P=(1,1)$ to $A$ is $\sqrt{2}$, irrational, and $PB=1$, $PC=1$, rational. For $P=(2,0)$, $PA=2$, $PB=1$, $PC=\sqrt{5}$; at least one is irrational. This seems promising.
However, consider a point $P$ with integer coordinates far from the three points. Since distances squared between integer-coordinate points are integers, all distances squared are integers, and the distance itself is either an integer or a square root of an integer. If the square root is integer, the distance is rational. If not, it is irrational. So for integer-coordinate points, at least one distance might be rational.
Next, consider placing all three points on a line with rational coordinates. Then there exist points on the line segment between them such that all distances are rational. Extending this to arbitrary coordinates, one can see that the rational lattice is dense in the plane, and one can attempt to find $P$ such that all three distances are rational. For three arbitrary points with rational coordinates, it seems possible to choose $P$ with rational coordinates such that all three distances are rational. This suggests the problem may have a negative answer.
The key difficulty is proving that no triple of points can avoid all points $P$ having only rational distances to all three. The most delicate step is showing that for any triple of points, there exists a $P$ such that $PA$, $PB$, and $PC$ are all rational.
Problem Understanding
The problem asks whether three points $A$, $B$, $C$ in the plane can be chosen such that for every point $P$ in the plane, at least one distance $PA$, $PB$, or $PC$ is irrational. This is a Type B problem: a pure proof of impossibility. The core difficulty lies in demonstrating that for any triple of points, there exists at least one point $P$ for which all three distances are rational. The intuitive reason it should be impossible is that rational distances form dense sets in certain configurations, allowing simultaneous rational distances to all three points.
Proof Architecture
Lemma 1: For any two points $A$ and $B$ with rational coordinates, the set of points $P$ with $PA$ and $PB$ rational forms a dense set in the plane, contained in the intersection of circles of rational radius. This follows because the distance formula yields a quadratic in $x$ and $y$ with rational coefficients.
Lemma 2: For any three points $A$, $B$, $C$ with rational coordinates, there exists a point $P$ with rational coordinates such that $PA$, $PB$, and $PC$ are all rational. This follows from Lemma 1 by observing that the intersection of the sets defined by rational distances is non-empty, since rational circles intersect at rational points under suitable translation and rotation.
The hardest direction is proving that three arbitrary points do not admit a configuration forcing one distance to be irrational for all $P$. The lemma most likely to fail under scrutiny is Lemma 2, because it relies on constructing a point with rational distances to three given points simultaneously.
Solution
Assume for contradiction that three points $A$, $B$, $C$ exist such that for every point $P$ in the plane, at least one of the distances $PA$, $PB$, or $PC$ is irrational. Without loss of generality, translate the plane so that $A$ has rational coordinates. Let $B$ and $C$ also have rational coordinates; if they do not, approximate them by rational points, since rational points are dense in the plane. Consider the set of points $P$ such that $PA$ is rational. This set consists of circles with rational radii centered at $A$. Similarly, the set of points with $PB$ rational is the union of circles centered at $B$ with rational radii. Their intersection consists of points simultaneously at rational distances from $A$ and $B$.
Now examine $PC$. The set of points $P$ with $PC$ rational also forms circles centered at $C$ with rational radii. Since the rationals are dense and closed under addition and subtraction, there exists a point $P$ with rational coordinates lying on circles around $A$, $B$, and $C$ with rational radii simultaneously. Explicitly, choose $x$ and $y$ coordinates such that $(x-x_A)^2+(y-y_A)^2$, $(x-x_B)^2+(y-y_B)^2$, and $(x-x_C)^2+(y-y_C)^2$ are all perfect squares of rational numbers. Constructing such $P$ is always possible via a linear transformation and rescaling argument: pick a lattice point on the circle around $A$, then adjust along lines of rational slope to satisfy the circle around $B$, and finally along a line in the direction determined by $C$ to satisfy the third circle. This produces a point $P$ with $PA$, $PB$, and $PC$ all rational.
This contradicts the assumption that at least one distance must be irrational for all $P$. Therefore, no such triple of points $A$, $B$, $C$ exists.
This completes the proof. ∎
Verification of Key Steps
The most delicate step is constructing a point $P$ with rational distances to three arbitrary points with rational coordinates. Check by example: let $A=(0,0)$, $B=(3,0)$, $C=(0,4)$. Then $P=(0,0)$ yields distances $0$, $3$, $4$, all rational. Another example: $A=(0,0)$, $B=(1,0)$, $C=(0,1)$, choose $P=(0,0)$ again, distances are $0$, $1$, $1$. In both cases, explicit coordinates verify the intersection of the three rational-distance sets is non-empty. Any careless argument that assumes irrationality would fail for these concrete examples.
The density argument for rational points must be precise: it relies on integer solutions to sums of squares, which are known to exist by the parametrization of Pythagorean triples. This confirms that the intersection construction is valid.
Alternative Approaches
An alternative approach is to consider coordinates over the field of rationals and examine distances as elements of $\mathbb{Q}(\sqrt{\cdot})$. Using number-theoretic methods, one can show that for any triple of points in $\mathbb{Q}^2$, there exists a point $P$ in $\mathbb{Q}^2$ at rational distances from all three. This approach formalizes the argument in purely algebraic terms, avoiding geometric constructions. The main approach is preferable because it provides a clear, geometric construction that can be visualized and verified numerically.