HMMT-Feb 2013 guts, Problem 17

The lines $y=x, y=2 x$, and $y=3 x$ are the three medians of a triangle with perimeter 1 . Find the length of the longest side of the triangle.

Solution: The three medians of a triangle contain its vertices, so the three vertices of the triangle are $(a, a),(b, 2 b)$ and $(c, 3 c)$ for some $a, b$, and $c$. Then, the midpoint of $(a, a)$ and $(b, 2 b)$, which is $\left(\frac{a+b}{2}, \frac{a+2 b}{2}\right)$, must lie along the line $y=3 x$. Therefore, \[ \begin{aligned} \frac{a+2 b}{2} & =3 \cdot \frac{a+b}{2} \\ a+2 b & =3 a+3 b \\ -2 a & =b \end{aligned} \] Similarly, the midpoint of $(b, 2 b)$ and $(c, 3 c)$, which is $\left(\frac{b+c}{2}, \frac{2 b+3 c}{2}\right)$, must lie along the line $y=x$. Therefore, \[ \begin{aligned} \frac{2 b+3 c}{2} & =\frac{b+c}{2} \\ 2 b+3 c & =b+c \\ b & =-2 c \\ c & =-\frac{1}{2} b=a \end{aligned} \] From this, three points can be represented as $(a, a),(-2 a,-4 a)$, and $(a, 3 a)$. 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 $\frac{\sqrt{58}}{2+\sqrt{34}+\sqrt{58}}$. $\fbox{\frac{\sqrt{58}}{2+\sqrt{34}+\sqrt{58}}}$.

Answer: \frac{\sqrt{58}}{2+\sqrt{34}+\sqrt{58}}