HMMT February

Let $l_{1}$ and $l_{2}$ be the lines as labeled in the above diagram. First, suppose Knot visits $l_{1}$ first, at point $P_{1}$, then $l_{2}$, at point $P_{2}$. Let $K^{\prime}$ be the reflection of $K$ over $l_{1}$, and let $T^{\prime}$ be the reflection of $T$ over $l_{2}$. The length of Knot's path is at least $$K P_{1}+P_{1} P_{2}+P_{2} T=K^{\prime} P_{1}+P_{1} P_{2}+P_{2} T^{\prime} \geq K^{\prime} T^{\prime}$$ by the Triangle Inequality (This bound can be achieved by taking $P_{1}, P_{2}$ to be the intersections of $K^{\prime} T^{\prime}$ with $l_{1}, l_{2}$, respectively.) Also, note that $\measuredangle K^{\prime} O T^{\prime}=90^{\circ}$$, so that$$K^{\prime} T^{\prime}=2 \sqrt{5}$. Now, suppose Knot instead visits l_{2}$first, at point$Q_{2}$, then$l_{1}$, at point$Q_{1}$. Letting$$K^{\prime \prime}$$be the reflection of$K$over$l_{2}$and$$T^{\prime \prime}$$be the reflection of$T$over$l_{1}, by similar logic to before the length of his path is at least the length of $K^{\prime \prime} T^{\prime \prime}$$. However, by inspection$$K^{\prime \prime} T^{\prime \prime}>K^{\prime} T^{\prime}$$, so our answer is$2 $\sqrt{5}\$.$