Locus Pocus

While picking random points inside the equilateral triangle with side length 1, you and your colleague notice that some points are “special.” When one of these special points is connected with the triangle’s three vertices, it creates three smaller triangles whose areas form an arithmetic sequence.

After further investigation, you realize these special points all belong to a locus, the total length of which is $L$. What is this length $L$?

As we saw before, given choice of any point $x$ in the triangle, we have three sub-triangles with areas $A_1(x),$ $A_2(x),$ and $A_3(x),$ that are mutually non-overlapping and tile the entirety of the area of the unit equilateral triangle with area $A = \frac{\sqrt{3}}{4}$ square units. If we have chosen our point x in a special way then $A_1$, $A_2$, and $A_3$ are in arithmetic progression and so, without loss of generality we have $A_2(x) = \frac{1}{2} (A_2(x) + A_3(x)),$ or $$2A_2(x) = A_1(x) + A_3(x).$$ Adding $A_2(x)$ to both sides we get $$3A_2(x) = A_1(x)+A_2(x)+A_3(x) = A,$$ so we conclude that $$A_2(x) = \frac{1}{3} A = \frac{\sqrt{3}}{12}.$$

So the locus of special points are those for which at least one of the subtriangles has area $\frac{\sqrt{3}}{12}.$ Without loss of generality, let's assume that the original unit equilateral triangle is located such that its vertices are at $P(-1/2, -\sqrt{3}/6),$ $Q(1/2, -\sqrt{3}/6)$ and $R(0,\sqrt{3}/3).$ Then for instance, for any point $X_t(t,0)$ for $-1/3 \leq t \leq 1/3,$ the area of the triangle $PQX_t$ will be $\sqrt{3}/12.$ So the line segment from $(-1/3,0)$ to $(1/3,0)$ is in the locus of special points and has a length of $2/3.$

However, since the equilateral triangle has trifold rotational symmetry, there will also be a similar sub-locus of length $2/3$ running parallel to the side $PR$ and the side $QR.$ So the locus of special points where the areas of three smaller triangles form an arithmetic sequences has total length of $2$ units.