Triangle centers from curve shortening The curve-shortening flow transforms curves in the plane by moving each point perpendicularly to the curve at a speed proportional to the curvature at that point. It is usually defined for smooth curves (so you can determine the curvature) but according to a survey by White (Math. Int. 1989) it can be extended to rectifiable varifolds. In particular, according to this extension, it should be well-defined for triangles, which should instantly turn into smooth approximations of themselves and then evolve as normal, shrinking to a point. (This would also follow immediately from a positive answer to a previous question about continuity of the curve-shortening flow, assuming that the continuity is in a sufficiently general space of curves to include triangles and not just smooth curves.) Since the flow isn't affected by Euclidean transformations, this point must be a triangle center. Is it a known triangle center (one of the ones listed at the Encyclopedia of Triangle Centers)? If so, which one? If we could just calculate the flow accurately enough we could look it up in ETC but I don't know how to do the calculation.
If this turns out to be known, a reference would be helpful, so that it could be added to the Wikipedia article.
 A: I think there can be no expression for this "center of mass" that does not involve running the curve shortening flow.  This was proved (along with a lot of other things) for general initial curves by Bryant and Griffiths in this long article  (see the continuation of Example 1 in section 4.4 specifically the top of page 53) and I don't see how restricting to triangles helps.
The only thing you can know is the extinction time which is given by $T=A_0/2\pi$ where $A_0$ is the area of the triangle.
You can estimate where the extinction point is by using Huisken's monotonicity formula, but that is probably the best you can do in general.
A: This seems to be addressed by Ben Chow and Dave Glickenstein in:
MR2281928 (2008b:58023) Reviewed 
Chow, Bennett(1-UCSD); Glickenstein, David(1-AZ)
Semidiscrete geometric flows of polygons. 
Amer. Math. Monthly 114 (2007), no. 4, 316–328. 

They show that that the point is the center of gravity of the triangle.
A: From my understanding, the Brakke flow is not generally unique and so I'm not sure if you can define a unique flow for a polygon like a triangle, which may result in not being able to define a center in this way. Angenent has some work when the curve is in W^2,p, but polygons are not in this space. I'd be interested to know more but I have not been successful at finding a good reference for such small amount of smoothness (help would be appreciated). One could consider the level set formulation, but that may suffer from fattening (which is similar to the nonuniqueness issue). However, most of the literature focuses on nonuniqueness coming from singularities when one starts at a smooth curve, and these would probably not develop the same kind of singularity present in a polygon.
A: I formerly had a handwavy argument here showing that the vanishing point is always near (but not equal) the centroid, but have now replaced it by a more rigorous argument. This doesn't entirely answer the question (which center is the vanishing point), but should at least help eliminate many centers as definitely not being it, because most centers do not have the same property of always being near the centroid.
First, to eliminate the centroid itself: intuitively, the vanishing point shouldn't be the centroid by the idea presented in the answer by foliations: the centroid has a local formula and the vanishing point shouldn't have such a formula. (This should also be true of the Spieker center, the centroid of the triangle's perimeter.) But a little more directly: in a highly-obtuse isosceles triangle, all triangle centers lie on the altitude of the triangle, but the centroid is always exactly 1/3 of the way along the altitude from the base (and the Spieker center is almost on the base), while the vanishing point should be closer to the incenter, near the midpoint of the altitude.
Next, let's show that the vanishing point is near the centroid. The previous handwavy argument used a vague concept of "aspect ratio" (see comments); to make this more concrete let's use the isoperimetric ratio $\rho=L^2/A$ which is known to decreases monotonically through the evolution of a convex curve (Gage 1984).
Consider a smooth convex curve $C$ with diameter $D$, and let $w$ be the width perpendicular to the diameter; then $D=\Theta(L)$ and $A=\Theta(wD)$ so
$w=\Theta(D/\rho)$.
Let $xy$ be a chord of $C$ perpendicular to $D$ through the centroid. Then $xy$ must cross $D$ somewhere within its middle third, and the tangents to $C$ at $x$ and $y$ must form an angle with each other (and with the diameter segment) that is $O(1/\rho)$, else $xy$ would be longer than $w$. We consider all of the symbols $D$, $w$, $x$, and $y$ to vary continuously as the curve evolves, and use $w_0$ (etc.) to refer to their initial values. Note that, because $C$ remains confined to a rectangular box with width $w_0$, it will always be the case that $w=O(w_0)$.
At any point in the evolution of the curve, the same calculation shows that the angle between the two tangents is small, $O(w/D)$. It follows that the rates of area loss on the two sides of $xy$ differ from each other by at most a factor of $1+O(w/D)$, and we know that the total rate of area loss is constant, so the rate at which the area difference between the two sides changes is $O(w/D)$. It follows that the speed at which segment $xy$ can move is $O(1/D)$, because a faster movement would create too much area difference from one side to the other.
Now consider any time period within which the diameter decreases from $D_t$ to $D_t/2$. Surrounding the curve by pair of width-$w$ grim reaper curves (one in each direction), and using the avoidance principle for curve shortening (if a curve is surrounded by another curve, the two curves cannot cross) shows that this happens in time $O(w_tD_t)$. Within this time period, the centroid can only move a distance of $O(w_t)$: the speed of $xy$ controls its motion parallel to the diameter and the length of $xy$ controls its perpendicular motion. After repeating this argument $\Theta(\log\rho)$ times we will have $D\le w_0$, after which the vanishing point is bounded within the $O(w_0)$ remaining diameter of the curve. Therefore, the vanishing point of any smooth curve is within distance $O(D\frac{\log\rho}{\rho})$ of the centroid. 
Triangles aren't smooth but immediately become smooth as soon as you start evolving them, so the same argument applies. I'm not sure whether the logarithmic factor in the bound above is necessary, or whether the vanishing point is always within the closer distance $O(D/\rho)$ of the centroid.
