Is the set of triangle centers given by polynomial functions dense? Given a function $f$ that satisfies:


*

*Homogeneity: $f(tx,ty,tz) = t^nf(x,y,z)$ for some $n$ and any $x,y,z$.

*Bisymmetric in the second and third variables: $f(x,y,z) = f(x,z,y)$.
We say that the point with trilinear coordinates $f(a,b,c):f(b,c,a):f(c,a,b)$ is a center of the triangle with sides $a$, $b$ and $c$.
Let us restrict ourselves to polynomial functions $f$ and scalene triangles $(a > b > c > 0$). Is the set of triangle centers dense in $\mathbb{R}^2$?
Remark: without the restriction stated before, any point in the plane is a center.
 A: I would say yes.
Suppose two points $P$ and $Q$ are triangle centers. Then the midpoint $R$ between $P$ and $Q$ is a triangle center as well:
$$f_R=(a\,f_P(a,b,c)+b\,f_P(b,c,a)+c\,f_P(c,a,b))\cdot f_Q\\
+(a\,f_Q(a,b,c)+b\,f_Q(b,c,a)+c\,f_Q(c,a,b))\cdot f_P$$
The reflection $S$ of $P$ in $Q$ is a center as well; its trilinears can be computed using
$$f_S=2(a\,f_P(a,b,c)+b\,f_P(b,c,a)+c\,f_P(c,a,b))\cdot f_Q\\
-(a\,f_Q(a,b,c)+b\,f_Q(b,c,a)+c\,f_Q(c,a,b))\cdot f_P$$
Taken together, these two can be used to turn an arbitrary pair of distinct centers into a dense set of centers on the line spanned by these two centers. Let's call one of the points “$0$” and the other “$1$” and consider these as defininig a scale. In any $\varepsilon$ environment on the line there will be a scale mark of the form $\pm\frac{m}{2^k}$. W.l.o.g. the sign is positive (else change the roles of $0$ and $1$). So reflect $i$ in $i+1$ till you reach $m$, then $k$ times take the midpoint between that point and $0$. Other ways to combine the operations of “extend by reflection” and “zoom in using midpoints” are possible.
If you do this not only for two centers, but for three non-collinear ones, you can combine them to get arbitrarily close to any point in the plane. So as long as you can show that a scalene triangle always has three non-collinear centers, you get a dense set of centers.
The way I came up with the above formulas is by considering homogeneous coordinates in a projective plane embedded at $z=1$. The midpoint of points $[x_A:y_A:1]$ and $[x_B:y_B:1]$ is $[x_A+x_B:y_A+y_B:2]$. But this simple addition relies on the representatives having the same $z$ coordinate. So more generally the midpoint of $A=[x_A:y_A:z_A]$ and $B=[x_B:y_B:z_B]$ is $z_B\cdot A+z_A\cdot B$. This relies on the specific embedding at $z=1$, which can also be seen as the position of the line at infinity $l_\infty=[0:0:1]$. If one does not pick specific coordinates for that line, the midpoint could be written as $\langle l_\infty,B\rangle\cdot A+\langle l_\infty,A\rangle\cdot B$. This equation no longer depends on the choice of basis. Trilinear coordinates are just another basis for the projective plane, and the line at infinity in that basis is $[a:b:c]$. This I read from Wikipedia. So the parentheses given above are these scalar products between the line at infinity and the trilinears of one of the points. The same ideas hold for the reflection.
