TITLE: What manifold is produced if all similar triangles in the plane are identified?
QUESTION [8 upvotes]: Consider Euclidean space $\mathbb{R}^6$, we can consider this as the $x$ and $y$ coordinates of the $3$ points of a triangle in $\mathbb{R}^2$. If we identify all points in $\mathbb{R}^6$ whose triangles have the same 3 angles, what manifold do we get under the quotient topology?

REPLY [6 votes]: We can exploit the scale invariance of homogeneous coordinates to capture efficiently the similarity condition and produce a description for which the action of the relevant symmetry groups are especially apparent.
For any oriented triangle $\Delta ABC$, mark one of its vertices, say, $A$, and denote the marked triangle by $(\Delta ABC, A)$. Then, if the lengths of the sides opposite $A, B, C$ are respectively, $a, b, c$, consider the triple $$(t_a, t_b, t_c) := (-a + b + c, -b + c + a, -c + a + b) .$$ We can recover the relative lengths from the triple via $$(a, b, c) = \left(\tfrac{1}{2} (t_b + t_c), \tfrac{1}{2} (t_b + t_c), \tfrac{1}{2} (t_b + t_c)\right) .$$
Geometric construction of $(t_a, t_b, t_c)$ (see this GeoGebra activity for an interactive version)

Now, observe the following:

The sides $a, b, c$ form a triangle iff the inequality $a + b > c$ and its cyclic permutations hold, but these conditions are equivalent to $t_a, t_b, t_c > 0$. Conversely, every triple of positive numbers determine a triangle.
If we scale $(\Delta ABC, A)$ by $\lambda$, the resulting triple scales by the same factor---we get $\lambda t_a, \lambda t_b, \lambda t_c$.
Similar marked triangles give rise to proportional triples.

Thus, by regarding triples as homogeneous coordinates $[t_a, t_b, t_c]$ on a reference triangle $T$, we get an identification between oriented similarity classes of marked triangles and the interior $T^{\circ}$ of $T$.
The reference triangle $T$

Some elementary algebra shows that the equilateral class corresponds to the red point, $[1, 1, 1]$, and the isosceles classes correspond to the green lines, the loci of the points of the form $[t, t, \ast]$ and its cyclic permutations.
Now, we want to disregard the markings, that is, identify $[(\triangle ABC, A)] \sim [(\triangle ABC, B)] \sim [(\triangle ABC, C)]$; we write this equivalence class as $[\triangle ABC]$. These classes are exactly the orbits of the natural $\Bbb Z_3$-action whose generator acts by, say, $[(\triangle ABC, A)] \mapsto [(\triangle ABC, B)]$. Unwinding definitions, via the above identification $\sigma$ just acts on $T^{\circ}$ by cyclic permutation: $$\sigma: [t_a, t_b, t_c] \to [t_b, t_c, t_a] ,$$ or equivalently clockwise rotation by $1/3$ of a turn. Of course, viewed this way $\Bbb Z_3$ is just the oriented symmetry group of $T^{\circ}$. Thus:

We may naturally identify the space of triangles modulo oriented similarity with $T^{\circ} / \Bbb Z_3$.

By the argument in my other answer, this is an orbifold).
If instead we want to consider the space of unoriented triangles modulo similarity, we must also identify $[\triangle ABC, A]$ with $[\triangle ACB, A]$, but this corresponds to the involution $$\tau: (t_a, t_b, t_c) \mapsto (t_a, t_c, t_b) ,$$ that is, reflection across the line through $[1, 0, 0]$ and $[1, 1, 1]$. This reflection and $\sigma$ together generate the full group of symmetries of $T^{\circ}$ and is isomorphic to $S_3$. Thus:

We may naturally identify the space of unoriented triangles modulo similarity with the orbifold $T^{\circ} / S_3$.

Remark In fact, for any triangle $\Delta ABC$, regarded as homogeneous coordinates on $\Delta ABC$, $[t_a, t_b, t_c]$ is the Nagel point of $\Delta ABC$; this is the triangle center complementary to the incenter.