Two villages, $A$ and $B$, lie on opposite sides of a straight river in the positions shown. There will be a market in village $B$, and residents of village $A$ wish to attend. The people of village $A$ would like to build a bridge across and perpendicular to the river so that the total route walked by residents of $A$ to the bridge, across the bridge, and onward to $B$ is as short as possible. How can they find the exact location of the bridge, and how long will the route be?
For convenience, we label the point $X$, the foot of the perpendicular from $A$ to the river. We also label $Y$, the point where the perpendicular $B Y$ to the river through $B$ meets the parallel $A Y$ to the river through $A$. The key construction is to translate $B$ southward $1 \mathrm{~km}$ to point $E$. Let $A C D B$ be any route. Then segments $D C$ and $B E$ are congruent and parallel,

![](https://cdn.mathpix.com/cropped/2024_04_17_54f31c14ab607b7a972cg-1.jpg?height=404&width=396&top_left_y=1928&top_left_x=1553)
making a parallelogram $D C E B$. The total route is $A C+C D+D B=A C+C E+E B$. But by the triangle inequality, $A C+C E \geq A E$ so the total route cannot be shorter than $A E+E B=5+1=6 \mathrm{~km}$. (continued)

The residents of $A$ can ensure a 6 -kilometer route by putting the start $C$ of the bridge at the point where $A E$ crosses the lower bank of the river. In this situation, since $A, C$, and $E$ lie on a line, the triangle inequality is an equality, so the total route is $A C+C D+D B=A C+C E+E B=A E+E B=5+1=6$.

Remark. Using the fact that triangles $A C X$ and $E A Y$ are similar if $A, C$, and $E$ are collinear, it is possible to calculate that the optimal location of point $C$ is $0.75 \mathrm{~km}$ east of $X$.