Problem (Geometry):
The square $A R M L$ is contained in the $x y$-plane with $A=(0,0)$ and $M=(1,1)$. Compute the length of the shortest path from the point $(2 / 7,3 / 7)$ to itself that touches three of the four sides of square $A R M L$.

Solution:
Consider repeatedly reflecting square $A R M L$ over its sides so that the entire plane is covered by copies of $A R M L$. A path starting at $(2 / 7,3 / 7)$ that touches one or more sides and returns to $(2 / 7,3 / 7)$ corresponds to a straight line starting at $(2 / 7,3 / 7)$ and ending at the image of $(2 / 7,3 / 7)$ in one of the copies of $A R M L$. To touch three sides, the path must cross three lines, at least one of which must be vertical and at least one of which must be horizontal.



If the path crosses two horizontal lines and the line $x=0$, it will have traveled a distance of 2 units vertically and $4 / 7$ units vertically for a total distance of $\sqrt{2^{2}+(4 / 7)^{2}}$ units. Similarly, the total distance traveled when crossing two horizontal lines and $x=1$ is $\sqrt{2^{2}+(10 / 7)^{2}}$, the total distance traveled when crossing two vertical lines and $y=0$ is $\sqrt{2^{2}+(6 / 7)^{2}}$, and the total distance traveled when crossing two vertical lines and $y=1$ is $\sqrt{2^{2}+(8 / 7)^{2}}$. The least of these is

$$
\sqrt{2^{2}+(4 / 7)^{2}}=\frac{2}{\mathbf{7}} \sqrt{\mathbf{5 3}}
$$

Final answer: $\frac{2}{7} \sqrt{53}$