[Geometry]
Problem: 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.

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