Erdős Problem #173

Erdős Problem 173: exact obstructions in an alternating-strip colouring

Statement and outcome

The conjecture asks whether every two-colouring of $\mathbb R^2$ admits a monochromatic congruent copy of every triangle except possibly one congruence class. Congruence, not arbitrary scaling, is required.

We analyse the colouring $c(x,y)=\lfloor y\rfloor\bmod2$ exactly enough to prove its standard equilateral obstruction, and to exhibit a monochromatic copy whenever at least one altitude is not an odd integer. This proves, in particular, that every right triangle occurs in this strip colouring. It does not control arbitrary plane colourings.

A useful two-height criterion

Place one side of a triangle horizontally at height zero, with its third vertex at height $h>0$, the corresponding altitude. These two heights can be translated into strips of one colour if and only if $h$ is not an odd integer. To see sufficiency, reduce $h$ modulo two to $u\in[0,2)$. If $u<1$, translation zero already places both residues in $[0,1)$. If $u>1$, translate by $2-u$, placing the residues at $2-u$ and zero modulo two, again in $[0,1)$. If $u=1$, every vertical translate puts the two heights in opposite-coloured strips.

Consequently any triangle with an altitude not in $\{1,3,5,\ldots\}$ has a monochromatic copy. The minimum-altitude-less-than-one observation from the previous local attempt is a special case, but no smallness assumption is needed here.

The forbidden equilateral triangle

Let $E$ be equilateral of side $s=2/\sqrt3$, hence altitude one. In every orientation its vertical span is at least one. Indeed, its three undirected edge directions are spaced by $60^\circ$; some edge is within $30^\circ$ of vertical and has vertical projection at least $s\cos30^\circ=1$.

On the other hand, if the vertex heights in increasing order are $y_1\le y_2\le y_3$, then \[ y_2-y_1\le1,\qquad y_3-y_2\le1. \tag{1} \] For example, with $A$ the lowest vertex and $B,C$ the other two, the first gap is the minimum of the upward projections of $B-A$ and $C-A$. This is at most the projection of their average $(B+C)/2-A$, a vector of length one. Apply the same argument downward from the highest vertex for the other gap.

If all three vertices were in one strip, their span would be strictly less than one because the strips are half-open, contradicting the span bound. If they were in different strips of the same colour, some consecutive pair of heights would lie in different such strips. The gap between those two heights would be strictly greater than one: the lower height is below the end of its strip, while the next same-coloured strip starts one full unit later. This contradicts $1$. Thus no congruent copy of $E$ is monochromatic, including placements on strip boundaries.

Every right triangle occurs in this colouring

If either leg length is not an odd integer, use the two-height criterion, since the legs are altitudes. Otherwise write the two leg lengths as odd positive integers $a,b$. The third altitude is $h=ab/\sqrt{a^2+b^2}$. Now $a^2+b^2\equiv2\pmod8$, so it is not a square. Its square root is irrational, since a rational square root of an integer is an integer. Therefore $h$ is irrational, in particular not an odd integer, and the same criterion applies.

Assessment and attribution

The earlier gpt_pro_5.2/173.tex established only that sufficiently small triangles fit into one strip. We retain the underlying projection idea with translations stated explicitly, prove the exact equilateral exclusion, and add the two-height criterion. These results leave possible strip-colouring exceptions among triangles whose three altitudes are odd integers; they do not show that only $E$ is exceptional.

The stronger theorem that every right triangle occurs in every two-colouring is Shader's Theorem 2, All Right Triangles Are Ramsey in $E^2$, J. Combin. Theory A 20 $1976$, 385--389, https://www.cs.umd.edu/~gasarch/TOPICS/ERT/ShaderRightTri.pdf. That primary paper was checked, but its full forcing argument is not reproduced or counted as proved here. Problem: https://www.erdosproblems.com/173. The general at-most-one-exception assertion remains unresolved in this attempt. No novelty claim is made.

COMPLETION ESTIMATE

COMPLETION: 30%

PROBLEM

In any $2$-colouring of $\mathbb{R}^2$, for all but at most one triangle $T$, there is a monochromatic congruent copy of $T$.

OUTPUT

1) FORMAL RESTATEMENT

A 2-coloring is a map $c:\mathbb R^2\to\{\text{red},\text{blue}\}$.

Two triangles $T,T'$ in $\mathbb R^2$ are congruent if there is an isometry (composition of translation, rotation, reflection) mapping $T$ to $T'$.

Given a triangle $T$, say that $T$ is Ramsey for the coloring $c$ if there exists a monochromatic congruent copy of $T$, i.e. an isometric image $T'$ of $T$ whose three vertices all have the same color.

The statement to prove can be phrased as:

For every 2-coloring $c$ of $\mathbb R^2$, the set of congruence classes of triangles $T$ that are not Ramsey for $c$ has size at most $1$.

The problem text also notes there are colorings (“alternating strips”) where one equilateral triangle must be excluded, and that Shader proved the statement for a fixed right triangle.

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2) QUICK LITERATURE/CONTEXT CHECK

I will not use or claim any external results beyond what is explicitly written in the problem text.

The problem text reports:

* An alternating-strip coloring can force exclusion of one equilateral triangle. * Shader $1976$ proved the statement for a single right-angled triangle.

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3) ATTACK PLAN

I do not currently have a route to the full theorem. I instead analyze the simplest cited obstruction mechanism (alternating strip colorings) and prove a couple of precise, elementary geometric facts about when a triangle can be made monochromatic in such a coloring.

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4) WORK

Lemma 4.1 (monochromatic copies inside a single strip)

Consider the horizontal strip coloring \[ c(x,y)=\begin{cases} \text{red},& y\in [2k,2k+1)\text{ for some }k\in\mathbb Z,\\ \text{blue},& y\in [2k+1,2k+2)\text{ for some }k\in\mathbb Z. \end{cases} \]

If a triangle $T$ has a congruent copy whose three vertices have $y$-coordinates contained in an interval of length $<1$, then $T$ has a monochromatic copy (in fact, entirely red or entirely blue) in this coloring.

Proof. Let $T'$ be a congruent copy of $T$ whose vertex $y$-coordinates are all in some interval $(\alpha,\alpha+1)$ of length $<1$. Choose an integer $k$ so that $(\alpha,\alpha+1)$ lies entirely inside a single color strip, either inside $[2k,2k+1)$ or inside $[2k+1,2k+2)$. (That is always possible because the strip boundaries occur at integers, and any open interval of length $<1$ can be shifted into the interior of one half of a period-2 partition.) Then all three vertices of $T'$ lie in that strip and hence share the same color. $\square$

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Lemma 4.2 (minimal vertical span of an equilateral triangle)

Let $E_s$ be an equilateral triangle of side length $s>0$. Among all congruent copies of $E_s$ in the plane, the minimal possible difference between the maximum and minimum $y$-coordinates of its vertices equals its altitude \[ h=\frac{\sqrt3}{2}s. \]

Proof. Let $E_1$ be a unit equilateral triangle with vertices $A,B,C$. The three edge directions are separated by $60^\circ$ angles. Fix any placement of $E_1$ and consider the vertical unit direction $u=(0,1)$. For any edge of length $1$ making angle $\phi$ with the horizontal, its vertical projection length is $|\sin\phi|$.

Among the three edge directions of an equilateral triangle, for any given direction $u$, there is an edge whose direction makes angle at most $30^\circ$ from the horizontal-orthogonal direction; equivalently, among the three angles between $u$ and the three edge directions, the minimum is at most $30^\circ$. Therefore for at least one edge $e$ of the triangle, the vertical projection of that edge has length at least $\cos(30^\circ)=\sqrt3/2$.

But the vertical span (max $y$ minus min $y$ among vertices) is at least the vertical projection length of some edge, since that projection is the absolute $y$-difference between its endpoints. Hence the vertical span is always at least $\sqrt3/2$ for the unit triangle.

On the other hand, if one side of the equilateral triangle is placed horizontally, then the third vertex lies exactly $\sqrt3/2$ units above that side, so the vertical span equals $\sqrt3/2$. Thus the minimum possible vertical span for $E_1$ is $\sqrt3/2$. Scaling by factor $s$ multiplies all vertical spans by $s$, so the minimum possible vertical span for $E_s$ is $(\sqrt3/2)s$. $\square$

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Corollary 4.3 (small equilateral triangles always appear in strip coloring)

In the strip coloring of Lemma 4.1, every equilateral triangle of side length $s<\frac{2}{\sqrt3}$ has a monochromatic congruent copy.

Proof. If $s<2/\sqrt3$, then its altitude $(\sqrt3/2)s<1$. By Lemma 4.2 there is a congruent copy with vertical span $<1$, and then Lemma 4.1 yields a monochromatic copy. $\square$

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5) VERIFICATION

* Lemma 4.1 is purely geometric and uses only the definition of the strip coloring. * Lemma 4.2 uses the fact that among three directions separated by $60^\circ$, one is within $30^\circ$ of any chosen direction; this implies a projection lower bound. The inequality is tight by the horizontal-side placement. * Corollary 4.3 shows that strip colorings cannot forbid all equilateral triangles; only sufficiently “large” ones could potentially be obstructed.

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6) FINAL

UNRESOLVED

(i) Strongest fully proved partial result obtained here.

For the standard horizontal alternating-strip 2-coloring, I proved:

* any triangle that can be placed with vertex $y$-coordinates in an interval of length $<1$ has a monochromatic copy (Lemma 4.1); * the minimal possible vertical span of an equilateral triangle of side length $s$ is $(\sqrt3/2)s$ (Lemma 4.2); * consequently, every equilateral triangle with side length $s<2/\sqrt3$ has a monochromatic copy in this strip coloring (Corollary 4.3).

(ii) Exact first gap.

Prove (or disprove) the full statement: in an arbitrary 2-coloring of $\mathbb R^2$, at most one congruence class of triangle fails to have a monochromatic copy. I do not have a mechanism to control arbitrary colorings.

(iii) Top 3 next moves (concrete targets).

1. For strip colorings (and their rotations), determine exactly which triangle shapes/sizes are forbidden and whether exactly one equilateral triangle can be avoided. 2. Prove stability/robustness: show that if two triangle shapes are forbidden then one can derive a contradiction via a geometric configuration forcing both colors. 3. Attempt to extend Shader’s right-triangle argument to general acute/obtuse triangles via decomposition into right triangles or via affine/rotational averaging.

(iv) What a minimal counterexample would likely look like.

A counterexample would be a 2-coloring $c$ and two noncongruent triangles $T_1,T_2$ such that neither has a monochromatic copy. Any such coloring would have to avoid two distinct distance/angle patterns simultaneously; one expects it to be highly structured (e.g., periodic stripes or tilings) rather than random.