It is well known that every planar graph has an embedding such that every edge is drawn as a straight line segment (Fáry's Theorem). Kemnitz and Harborth made the following stronger conjecture

Conjecture 1. Every planar graph has a straight line embedding with integer edge lengths.

I was wondering if it is possible to attack this problem with the following approach.

Conjecture 2. Let $X:=\{ x_1, \dots, x_n \}$ be a finite set of points in the plane such that no three points of $X$ are colinear. For any $\epsilon >0$, there exists $X':=\{x_1', \dots, x_n'\}$ such that for all $i, j \in [n]$

1. $d(x_i, x_i') < \epsilon$,
2. $d(x_i', x_j') \in \mathbb{Q}$, and
3. no three points of $X'$ are colinear.

To prove Conjecture 2, it suffices to prove the following conjecture.

Conjecture 3. Let $X:=\{ x_1, \dots, x_n \}$ be a finite set of points in the plane such that no three points of $X$ are colinear and all pairwise distances are rational. Then the set of points which are at rational distance from all points in $X$ is a dense subset of the plane.

Conjecture 3 is trivial for $n=1$ and easy for $n=2$. Almering proved it for $n=3$, and I think it is open for $n>3$. Note that Conjecture 3 is (essentially) a weakening of:

Conjecture 4. There exists a dense subset of the plane with all pairwise distances rational.

This question was posed by Ulam in 1945 (see this mathoverflow question for more background). So, the reason I like Conjecture 3 is that it is still strong enough to prove Conjecture 1, but appears much weaker than Conjecture 4. Unfortunately, Conjecture 3 is beyond my limited area of expertise. Hence:

Question. What are the prospects for proving Conjecture 3? A proof or disproof would be fantastic. However, even arguments suggesting that it is true/false but say beyond current technology would be most welcome.

It is well known that every planar graph has an embedding such that every edge is drawn as a straight line segment (Fáry's Theorem). Kemnitz and Harborth made the following stronger conjecture

Conjecture 1. Every planar graph has a straight line embedding with integer edge lengths.

I was wondering if it is possible to attack this problem with the following approach.

Conjecture 2. Let $X:=\{ x_1, \dots, x_n \}$ be a finite set of points in the plane such that no three points of $X$ are colinear. For any $\epsilon >0$, there exists $X':=\{x_1', \dots, x_n'\}$ such that for all $i, j \in [n]$

1. $d(x_i, x_i') < \epsilon$,
2. $d(x_i', x_j') \in \mathbb{Q}$, and
3. no three points of $X'$ are colinear.

To prove Conjecture 2, it suffices to prove the following conjecture.

Conjecture 3. Let $X:=\{ x_1, \dots, x_n \}$ be a finite set of points in the plane such that no three points of $X$ are colinear and all pairwise distances are rational. Then the set of points which are at rational distance from all points in $X$ is a dense subset of the plane.

Conjecture 3 is trivial for $n=1$ and easy for $n=2$. Almering proved it for $n=3$, and I think it is open for $n>3$. Note that Conjecture 3 is (essentially) a weakening of:

Conjecture 4. There exists a dense subset of the plane with all pairwise distances rational.

This question was posed by Ulam in 1945 (see this mathoverflow question for more background). So, the reason I like Conjecture 3 is that it is still strong enough to prove Conjecture 1, but appears much weaker than Conjecture 4. Unfortunately, Conjecture 3 is beyond my limited area of expertise. Hence:

Question. What are the prospects for proving Conjecture 3? A proof or disproof would be fantastic. However, even arguments suggesting that it is true/false but say beyond current technology would be most welcome.

It is well known that every planar graph has an embedding such that every edge is drawn as a straight line segment (Fáry's Theorem). Kemnitz and Harborth made the following stronger conjecture

Conjecture 1. Every planar graph has a straight line embedding with integer edge lengths.

I was wondering if it is possible to attack this problem with the following approach.

Conjecture 2. Let $X:=\{ x_1, \dots, x_n \}$ be a finite set of points in the plane such that no three points of $X$ are colinear. For any $\epsilon >0$, there exists $x_1', \dots, x_n'$ such that for all $i, j \in [n]$

1. $d(x_i, x_i') < \epsilon$, and
2. $d(x_i', x_j') \in \mathbb{Q}$.

To prove Conjecture 2, it suffices to prove the following conjecture.

Conjecture 3. Let $X:=\{ x_1, \dots, x_n \}$ be a finite set of points in the plane such that no three points of $X$ are colinear and all pairwise distances are rational. Then the set of points which are at rational distance from all points in $X$ is a dense subset of the plane.

Conjecture 3 is trivial for $n=1$ and easy for $n=2$. Almering proved it for $n=3$, and I think it is open for $n>3$. Note that Conjecture 3 is (essentially) a weakening of:

Conjecture 4. There exists a dense subset of the plane with all pairwise distances rational.

This question was posed by Ulam in 1945 (see this mathoverflow question for more background). So, the reason I like Conjecture 3 is that it is still strong enough to prove Conjecture 1, but appears much weaker than Conjecture 4. Unfortunately, Conjecture 3 is beyond my limited area of expertise. Hence:

Question. What are the prospects for proving Conjecture 3? A proof or disproof would be fantastic. However, even arguments suggesting that it is true/false but say beyond current technology would be most welcome.

It is well known that every planar graph has an embedding such that every edge is drawn as a straight line segment (Fáry's Theorem). Kemnitz and Harborth made the following stronger conjecture

Conjecture 1. Every planar graph has a straight line embedding with integer edge lengths.

I was wondering if it is possible to attack this problem with the following approach.

Conjecture 2. Let $X:=\{ x_1, \dots, x_n \}$ be a finite set of points in the plane such that no three points of $X$ are colinear. For any $\epsilon >0$, there exists $X':=\{x_1', \dots, x_n'\}$ such that for all $i, j \in [n]$

1. $d(x_i, x_i') < \epsilon$,
2. $d(x_i', x_j') \in \mathbb{Q}$, and
3. no three points of $X'$ are colinear.

To prove Conjecture 2, it suffices to prove the following conjecture.

Conjecture 3. Let $X:=\{ x_1, \dots, x_n \}$ be a finite set of points in the plane such that no three points of $X$ are colinear and all pairwise distances are rational. Then the set of points which are at rational distance from all points in $X$ is a dense subset of the plane.

Conjecture 3 is trivial for $n=1$ and easy for $n=2$. Almering proved it for $n=3$, and I think it is open for $n>3$. Note that Conjecture 3 is (essentially) a weakening of:

Conjecture 4. There exists a dense subset of the plane with all pairwise distances rational.

This question was posed by Ulam in 1945 (see this mathoverflow question for more background). So, the reason I like Conjecture 3 is that it is still strong enough to prove Conjecture 1, but appears much weaker than Conjecture 4. Unfortunately, Conjecture 3 is beyond my limited area of expertise. Hence:

Question. What are the prospects for proving Conjecture 3? A proof or disproof would be fantastic. However, even arguments suggesting that it is true/false but say beyond current technology would be most welcome.