, 2013

Solution:
The three medians of a triangle contain its vertices, so the three vertices of the triangle are $(a, a)$, $(b, 2b)$, and $(c, 3c)$ for some $a$, $b$, and $c$. Then, the midpoint of $(a, a)$ and $(b, 2b)$, which is $\left(\frac{a+b}{2}, \frac{a+2b}{2}\right)$, must lie along the line $y = 3x$. Therefore,
$$\begin{aligned} \frac{a+2b}{2} & = 3 \cdot \frac{a+b}{2} \\ a + 2b & = 3a + 3b \\ -2a & = b \end{aligned}$$
Similarly, the midpoint of $(b, 2b)$ and $(c, 3c)$, which is $\left(\frac{b+c}{2}, \frac{2b+3c}{2}\right)$, must lie along the line $y = x$. Therefore,
$$\begin{aligned} \frac{2b+3c}{2} & = \frac{b+c}{2} \\ 2b + 3c & = b + c \\ b & = -2c \\ c & = -\frac{1}{2}b = a \end{aligned}$$
From this, three points can be represented as $(a, a)$, $(-2a, -4a)$, and $(a, 3a)$. Using the distance formula, the three side lengths of the triangle are $2|a|$, $\sqrt{34}|a|$, and $\sqrt{58}|a|$. Since the perimeter of the triangle is $1$, we find that $|a| = \frac{1}{2 + \sqrt{34} + \sqrt{58}}$ and therefore the longest side length is $\frac{\sqrt{58}}{2 + \sqrt{34} + \sqrt{58}}$.