Two particles move along the edges of equilateral $\triangle ABC$ in the direction \[A\Rightarrow B\Rightarrow C\Rightarrow A,\] starting simultaneously and moving at the same speed. One starts at $A$, and the other starts at the midpoint of $\overline{BC}$. The midpoint of the line segment joining the two particles traces out a path that encloses a region $R$. What is the ratio of the area of $R$ to the area of $\triangle ABC$?
$\mathrm {(A)} \frac{1}{16}\qquad \mathrm {(B)} \frac{1}{12}\qquad \mathrm {(C)} \frac{1}{9}\qquad \mathrm {(D)} \frac{1}{6}\qquad \mathrm {(E)} \frac{1}{4}$
First, notice that each of the midpoints of $AB$,$BC$, and $CA$ are on the locus.
Suppose after some time the particles have each been displaced by a short distance $x$, to new positions $A'$ and $M'$ respectively. Consider $\triangle ABM$ and drop a perpendicular from $M'$ to hit $AB$ at $Y$. Then, $BA'=1-x$ and $BM'=1/2+x$. From here, we can use properties of a $30-60-90$ triangle to determine the lengths $YA'$ and $YM'$ as monomials in $x$. Thus, the locus of the midpoint will be linear between each of the three special points mentioned above. It follows that the locus consists of the only triangle with those three points as vertices. (A cheaper way to find the shape of the region is to look at the answer choices: if it were any sort of conic section then the ratio would not generally be rational.)
Comparing inradii between this "midpoint" triangle and the original triangle, the area contained by $R$ must be $\textbf{(A)} \frac{1}{16}$ of the total area.