Juca is a scout exploring the surroundings of his campsite. After collecting fruits and wood, he must get water from the river and return to his tent. Let's represent Juca by the letter $J$, the river by the letter $r$, and his tent by the letter $B$. The distance between the feet of the perpendiculars $C$ and $E$, on $r$ from points $J$ and $B$, is $180 m$.

![](https://cdn.mathpix.com/cropped/2024_05_01_96f63abf6bdef97495eag-14.jpg?height=304&width=873&top_left_y=652&top_left_x=457)

What is the shortest distance Juca can travel to return to his tent, passing by the river?

#
Solution

Let's mark point $B^{\prime}$, the symmetric point of $B$ with respect to line $r$. Point $B^{\prime}$ is also $80 \mathrm{~m}$ away from line $r$, and thus we have congruent triangles $\triangle B C D$ and $\triangle B^{\prime} C D$ (side-angle-side), where $D$ is the point where Juca will fetch water from the river. Since the shortest distance between $J$ and $B^{\prime}$ is the length of a straight line segment connecting these two points, $J, D$, and $B^{\prime}$ must be collinear. Let $x$ be the measure of segment $\overline{D E}$. We apply the similarity ratio in triangles $\triangle B C D$ and $\triangle J E D$:

$$
\begin{aligned}
\frac{180-x}{80} & =\frac{x}{100} \\
\frac{180-x}{4} & =\frac{x}{5} \\
4 x & =900-5 x \\
9 x & =900 \\
x & =100 \mathrm{~m}
\end{aligned}
$$

Now, applying the Pythagorean Theorem to triangles $\triangle B C D$ and $\triangle J E D$, we can find the shortest distance between $J$ and $B$, passing through $r$:

$$
\begin{aligned}
J D+D B & =\sqrt{100^{2}+100^{2}}+\sqrt{80^{2}+80^{2}} \\
& =100 \sqrt{2}+80 \sqrt{2} \\
& =180 \sqrt{2} \mathrm{~m}
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_05_01_96f63abf6bdef97495eag-15.jpg?height=577&width=830&top_left_y=293&top_left_x=693)