5
$\begingroup$

Basic question: Can the Euclidean plane be divided into a vertex-to-vertex arrangement of non-overlapping triangles such that every edge has a unique rational length that lies between 1 and some specific rational R greater than 1?

(By edge-to-edge layout, we mean any vertex of the layout is necessarily an end point of all the edges meeting at that vertex. IOW, exactly two triangles meet at each edge in the layout. Also note that the length of each edge has to be different from other edge lengths.)

If this is possible, one can apply further constraints such as "all triangles should have equal area (OR equal perimeter)". Alternatively, one can relax the vertex-to-vertex requirement. One can also replace the requirement that every edge has a unique length with (say) the triangles being pairwise non-congruent.

Note: Requiring the lengths of all edges to be integers rather than rationals would lead to the lengths of the edges being unbounded even if a triangulation with all edges having unique lengths is possible (not sure if this is possible).

$\endgroup$
6
  • 1
    $\begingroup$ Can you explain what is "vertex-to-vertex" arrangement? You can triangulate the plane into equal triangles with sides 3,4,5. $\endgroup$ – Alexandre Eremenko Jan 24 '20 at 10:44
  • 1
    $\begingroup$ Why not equilateral triangles with edge length=1 ? $\endgroup$ – Pietro Majer Jan 24 '20 at 12:08
  • $\begingroup$ thanks. added explanation to vertex-to-vertex and 'uniqueness' of edge lengths. $\endgroup$ – Nandakumar R Jan 24 '20 at 13:30
  • $\begingroup$ If a scalene triangle ABC with rational lengths contains a similar copy DEA, also of rational lengths, and the lengths BD, DC, CE are also rational, then (generically) one should be able to solve the problem by gluing together various rotated and dilated versions of this configuration (the final triangulation resembles a discretised logarithmic spiral). This may be few enough constraints that the relevant Diophantine problem actually admits solutions. $\endgroup$ – Terry Tao Jan 24 '20 at 20:49
  • $\begingroup$ @Terry, will this meet the condition of all side lengths coming from a rational subset of a bounded interval (1+ epsilon, 1+delta)? I'm not seeing how to partition the spiral to do that with small epsilon and delta. Gerhard "I Could Be Missing Something" Paseman, 2020.01.24. $\endgroup$ – Gerhard Paseman Jan 24 '20 at 21:28
1
$\begingroup$

We can at least triangulate an infinite strip with bounded rational side lengths that do not repeat.

We take the strip between $x=0$ and $x=1$, alternating points on each line. The distances will all be within $2\epsilon$ of $2/\sqrt{3}$, and by an appropriate rational dilation we could make them all between $1$ and $R$.

  • Let $(x_0,y_0)=(0,0)$.
  • Let $(x_{n+1},y_{n+1})$=$(1-x_n,y_n+z_n)$ where $z_n$ is of the form $2ab/(a^2-b^2)$, $z_n$ is within $\epsilon$ of $1/\sqrt{3}$, and neither $y_{n+1}-y_{n-1}$ nor $(a^2+b^2)/(a^2-b^2)$ is a distance that has been used before. Then:
  • The diagonal distances between consecutive points are all within $\epsilon$ of $2/\sqrt{3}$.
  • The vertical distances between consecutive points on one line are all within $2\epsilon$ of $2/\sqrt{3}$.

We can find these $a$ and $b$ by choosing $a/b$ sufficiently close to $2+\sqrt{3}$. For example, we might get the following nearly equilateral result for $\epsilon=1/20$, using $(a,b,z)$ of $(4,1,8/15)$, $(7,2,28/45)$, $(11,3,66/112)$, $(15,4,120/209)$.

beginning of triangulation

This triangulates the half-strip, and we can triangulate the whole strip instead by alternating adding new points going up and going down.

With more number theory we might be able to extend this to the left and to the right. The key would be finding enough rational solutions to $$\frac{2ab}{a^2-b^2}+\frac{2cd}{c^2-d^2}=y_{n+1}-y_{n-1},$$ choosing the $y$'s to make that happen, and using the $2ab/(a^2-b^2)$'s as the vertical offsets of points on adjacent vertical lines.

$\endgroup$
0
$\begingroup$

Here is a reduction to a more compact problem. (Actually, it isn't. However, it is closely related to an open problem, and maybe the second step involving "every other" vertex can be salvaged. Oops.)

Given a circle of radius R with R rational and an inscribed regular hexagon, there is K less than R and a concentric circle of radius K, such that this circle contains a countable infinite collection of points x, such that:

a) the distance between x and any vertex of the hexagon is a rational number, and

b) for any two different points x and y in this collection, the twelve distances between one of the two points and one of the hexagon vertices are distinct.

If we have this, take a regular hexagonal tessellation (with sides of length R), and find a point x in side each hexagon and tweak those sides, and then tweak "every other" hexagon vertex to get all distinct edge lengths. (This does not work as stated. We need c) which is a condition that preserves rationality of certain tweaked hexagons.)

Unfortunately, the compact problem resembles the open problem of finding a rational point inside a square, so the quadrilateral version of this problem reduces to a known open problem.

Gerhard "But At Least It's Bounded" Paseman, 2020.01.24.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.