We want to get from point $A$ on the highway to the farm marked by $B$ in the shortest possible time using a tractor (see the diagram). The tractor's speed on the highway is $v_{1}=30 \mathrm{~km} / \mathrm{h}$, and on the farmland it is $v_{2}=15 \mathrm{~km} / \mathrm{h}$. Where should we turn off the highway $(a=12 \mathrm{~km}, b=3 \mathrm{~km})$.

![](https://cdn.mathpix.com/cropped/2024_05_02_fb70bba0ee7e4834ca86g-1.jpg?height=177&width=369&top_left_y=318&top_left_x=867)
![](https://cdn.mathpix.com/cropped/2024_05_02_4a6007c8266feff6cf0fg-1.jpg?height=179&width=368&top_left_y=152&top_left_x=865)

I. Solution. The tractor should turn off the road at point $P$. Its speed on the road is $v_{1}=30 \mathrm{~km} / \mathrm{h}$, and on the field $v_{2}=$ $15 \mathrm{~km} / \mathrm{h}$. The time $t_{1}$ spent on the road, based on the first diagram, is:

$$
t_{1}=\frac{a-x}{v_{1}}
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_4a6007c8266feff6cf0fg-1.jpg?height=380&width=679&top_left_y=637&top_left_x=715)

It is clear that the tractor must travel in a straight line on the field. Thus, the distance traveled on the field is:

$$
s=\sqrt{b^{2}+x^{2}}
$$

The time spent on the field is $t_{2}=s / v_{2}$.

We seek the minimum of $t=t_{1}+t_{2}$.

The function $t(x)$ has an extremum where $\frac{\mathrm{d} t(x)}{\mathrm{d} x}=0$. Let's perform the differentiation:

$$
\frac{\mathrm{d} t(x)}{\mathrm{d} x}=-\frac{1}{v_{1}}+\frac{x}{v_{2} \sqrt{b^{2}+x^{2}}}
$$

For the extremum, it must hold that

$$
-\frac{1}{v_{1}}+\frac{x}{v_{2} \sqrt{b^{2}+x^{2}}}=0
$$

From this,

$$
x=x_{0}=\sqrt{\frac{v_{1}^{2}}{v_{1}^{2}-v_{2}^{2}}} b, \quad \text { if } \quad v_{1}>v_{2}
$$

It is not difficult to show that for $x>x_{0}$, $\frac{\mathrm{d} t(x)}{\mathrm{d} x}>0$, and for $0<x<x_{0}$, $\frac{\mathrm{d} t(x)}{\mathrm{d} x}<0$, so $t(x)$ is indeed minimal at $x=x_{0}$.

With the given values, $x_{0}=\sqrt{3} \mathrm{~km}$.

Thus, the distance traveled on the road is

$$
y-(12-\sqrt{3}) \mathrm{km} \approx 10.27 \mathrm{~km}
$$

We note that if $v_{1} \leq v_{2}$, it is obvious that the tractor should immediately turn off the road and travel straight from point $A$ to point $B$.

Tünde Bálint (Dunaújváros, Münnich F. Gimn., III. o. t.)

II. Solution. If we have two light-transmitting media, light travels between them in such a way that the time taken for the light to travel between two points is minimized, according to the principle of Fermat. This leads to Snell's law, which states that

$$
\sin \alpha / \sin \beta=v_{1} / v_{2}
$$

where $v_{1}$ and $v_{2}$ are the speeds of propagation in the two media (2nd diagram).

Considering the above, we can apply the (1) formula to our problem. In our case, $\alpha=90^{\circ}$, so $\sin \alpha=1$. The ratio $v_{1} / v_{2}$ is 2 according to the problem.

Thus,

$$
\begin{gathered}
\sin \beta=1 / 2, \quad \beta=30^{\circ} \\
x=b \operatorname{tg} \beta=\sqrt{3} \mathrm{~km}(\approx 1.73 \mathrm{~km})
\end{gathered}
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_4a6007c8266feff6cf0fg-2.jpg?height=374&width=661&top_left_y=691&top_left_x=729)

Gyula Károlyi (Bp., Fazekas M. Gyak. Gimn., II. o. t.)

Note. Several students used an approximate method to find the minimum. We evaluated the results that were accurate enough as correct solutions.