I am looking for a minimal number of properties describing a triangle so that these properties are invariant to the choice of a Cartesian coordinate system as well as to the order in which the triangle points are enumerated. I have a few thousands of data points each of which is a triplet of multidimensional vectors. I thought to treat them as triangles. Not all of them are different. Some are either rotated or flipped versions of others. I wanted to distinguish them by their invariants.


I thought about using distances between certain triangle centers such as the center of the incircle, the circumcenter, the orthocenter, the centroid, etc. However, I found that the number of such centers is very large (around 40,000 listed in the [Encyclopedia of triangle centers](https://faculty.evansville.edu/ck6/) at the date I am writing this question).


Is there a small subset of those triangle centers, distances between which would unambiguously distinguish one triangle from another? What is the minimal amount of such triangle centers? Are there any papers that have information about them?



What is the minimum number of triangle centers sufficient to unambiguously describe a triangle?
If the triangle is equilateral, then all well-defined triangle centres coincide, and it's impossible to determine the size of the equilateral triangle.


Otherwise, three triangle centres suffice: the incentre $I$, circumcentre $O$, and Feuerbach point $F$.


In particular, given these three points we can determine:


* the inradius $r = IF$;
* the circumradius $R$, using Euler's theorem $IO^2 = R(R - 2r)$;


so we can draw the circumcircle and incircle of the triangle. We can also draw the nine-point circle since we know that it has radius $\frac{1}{2}R$ and is internally tangent to the incircle at the Feuerbach point $F$. This gives us the nine-point centre $N$.


Now that we have $N$ and $O$, we can determine the points $G$ and $H$ on the Euler line. As per <https://en.wikipedia.org/wiki/Euler_line#Distances_between_centers>, we can deduce $a^2 + b^2 + c^2$, the sum of the squares of the side-lengths of the triangle.


It's possible to prove that, for any triangle, we have:


$$ a^2 + b^2 + c^2 = 2s^2 - 2r^2 - 8Rr $$


where $R, r$ are aforementioned and $s = \frac{1}{2}(a + b + c)$ is the semiperimeter of the triangle. Consequently, $R, r, s$ are fully determined by the points $I,O,F$. Finally, $abc = 4Rrs$, so the following values are all determined:


* $abc$
* $ab + bc + ca$
* $a + b + c$


These are the coefficients of the cubic polynomial $(x-a)(x-b)(x-c)$ of which the side-lengths $a,b,c$ are the roots. It follows, therefore, that the set $\{a, b, c\}$ of triangle side-lengths can be determined from the points $I, O, F$. Because a triangle is determined (up to congruence) by its side-lengths, it follows that $\{a, b, c\}$ are determined by the distances $IO, OF, FI$, provided they're not all zero (in which case the triangle is equilateral).


