What is the minimum number of triangle centers sufficient to unambiguously describe a triangle?

Question

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 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?

Answer 1

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).

Answer 2

This is a coda to the given answer. Your problem is a special case of the classical one of reconstructing a triangle from parts, in this case from special points. As suggested by the above, one requires three of these, the simplest example being that of the vertices. You are considerng the case where the points are triangle centres but there are many other interesting ones. To my knowledge, such problems were first considered by Euler and were revived in the last century by Wernick (“A triangle construction from three located points”, Math. Mag. 55 (1982) 227-230). They have received some attention recently in the context of automatic proofs in elementary geometry. There is a systematic approach to such problems and we sketch it briefly. Suppose we are given as data the three special points (in your case three triangle centres) $A_1$, $B_1$, and $C_1$. We now consider an arbitrary triangle $ABC$ and determine when its special points are the above three. Wlog we can assume that the vertices are $(0,0)$, $(c,0)$ and $(p,q)$. We calculate the three special points of this triangle—call them $A_2$, $B_2$ and $C_2$ (they depend on $p$, $q$ and $c$). We then consider the three equations $|B_2C_2|^2=|B_1C_1|^2$, etc, in the variables $p$, $q$ and $c$. Of course, many things can happen, depending on the specific situation, but in general you will have an essentially unique solution and this solves your problem. For simple cases this can be done (and is fun to do) by hand, but in general it is rather tedious or even impossible. It is, however, easy to write a little programme, say with Mathematica, to mechanise the computations. We illustrate this with the case where the data triangle has three triangle centres as vertices. More precisely we suppose that they are determined by three triangle centre functions $f$, $g$ and $h$ of the side lengths $a$, $b$ and $c$ of $ABC$ (we are using the terminology of the Encyclopedia of Triangle Centers—easy to find online). In order to simplify the notation, we assume that these functions are homogeneous and satisfy $f(a,b,c)+f(b,c,a)+f(c,a,b)=1$. Then the required equations are $$[c(g(b,c,a)-h(b,c,a))+p(g(c,a,b)-h(c,a,b))]^2+q^2(g(c.a.b)-h(c,a,b))^2=|B_1C_1|^2$$ and its natural cyclic permutations. (These involve five variables—the side lengths and $p$ and $q$ from the coordinates of the vertices of $ABC$ but they can be reduced to equations in three variables, $(a,b,c)$ or $(p,q,c)$, by using the relationships $p^2+q^2=b^2, (p-c)^2+q^2=a^2$). The case considered by Euler is that of the incentre, circumcentre and orthocentre and the corresponding $f$, $g$ and $h$ can be found in the encyclopedia referred to above. Since there are going on 50,000 triangle centres documented we are talking about a potential of more than $40,000^3$ results.

The seeding article of Euler is in vol. 26 of his Opera Omnia, pp. 139-157 (“A simple solution of some very difficult geometrical problems”—my translation from the latin).