Two national roads intersect at a right angle. On one of them, a car travels at $v_{1}=60 \mathrm{~km} / \mathrm{h}$, and on the other, a car travels at $v_{2}=40 \mathrm{~km} / \mathrm{h}$, both heading towards the intersection. When the distance of the first car from the intersection is $200 \mathrm{~m}$, the distance of the second car is $500 \mathrm{~m}$. When will the two cars be closest to each other, and what is the minimum distance? Where will the first and second car be at that time?
Let's write down the distance between the two cars as a function of time (1st diagram):

![](https://cdn.mathpix.com/cropped/2024_05_02_433f6ccf7c43fec47078g-1.jpg?height=596&width=810&top_left_y=233&top_left_x=641)

1st diagram

$$
d=\sqrt{\left(s_{1}-v_{1} t\right)^{2}+\left(s_{2}-v_{2} t\right)^{2}}
$$

If $d$ is minimal, then $d^{2}$ is also minimal, so it is enough to find the time and value of the minimum of the expression under the square root. We square the expression and complete the square for the resulting quadratic polynomial:

$$
\begin{gathered}
d^{2}=\left(v_{1}^{2}+v_{2}^{2}\right) t^{2}-2\left(s_{1} v_{1}+s_{2} v_{2}\right) t+\left(s_{1}^{2}+s_{2}^{2}\right) \\
=\left(v_{1}^{2}+v_{2}^{2}\right)\left[t-\left(\frac{s_{1} v_{1}+s_{2} v_{2}}{v_{1}^{2}+v_{2}^{2}}\right)\right]^{2}+\left(s_{1}^{2}+s_{2}^{2}-\frac{\left(s_{1} v_{1}+s_{2} v_{2}\right)^{2}}{v_{1}^{2}+v_{2}^{2}}\right)
\end{gathered}
$$

The second term does not depend on time, and the value of the first term is either positive or $0$. $d^{2}$ is minimal when the first term vanishes, i.e.,

$$
t=\frac{s_{1} v_{1}+s_{2} v_{2}}{v_{1}^{2}+v_{2}^{2}}
$$

Then,

$$
\begin{gathered}
d_{\min }^{2}=s_{1}^{2}+s_{2}^{2}-\frac{\left(s_{1} v_{1}+s_{2} v_{2}\right)^{2}}{v_{1}^{2}+v_{2}^{2}}=\frac{\left(s_{1} v_{2}-s_{2} v_{1}\right)^{2}}{v_{1}^{2}+v_{2}^{2}} \\
d_{\min }=\frac{\left(s_{1} v_{1}-s_{2} v_{2}\right)}{\sqrt{v_{1}^{2}+v_{2}^{2}}}
\end{gathered}
$$

With our numerical data:

$$
t=22.15 \mathrm{sec}, \quad d_{\min }=305 \mathrm{~m}
$$

The distance of the cars from the intersection at this time is:

$$
l_{1}=s_{1}-v_{1} t=-169 \mathrm{~m}, \quad l_{2}=s_{2}-v_{2} t=254 \mathrm{~m}
$$

The negative value of $l_{1}$ indicates that the car has passed the intersection.

Samu Péter (Csongrád, Batsányi J. Gym., II. o. t.)

Second solution. Consider the intersection of the roads as a Cartesian coordinate system! We measure distances such that the coordinates of the first car are $\left(s_{1} ; 0\right)=(200 \mathrm{~m} ; 0)$, and those of the second car are $\left(0 ; s_{2}\right)=(0 ; 500 \mathrm{~m})$. The coordinates as functions of time are:

$$
\begin{aligned}
& x_{1}=s_{1}-v_{1} t, \quad y_{1}=0 \\
& x_{2}=0, \quad y_{2}=s_{2}-v_{2} t
\end{aligned}
$$

Using $x_{1}$ and $y_{2}$, we eliminate $t$:

$$
y_{2}=s_{2}-v_{2}\left(\frac{s_{1}-x_{1}}{v_{1}}\right)=x_{1} \frac{v_{2}}{v_{1}}+\frac{s_{2} v_{1}-s_{1} v_{2}}{v_{1}}
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_433f6ccf7c43fec47078g-2.jpg?height=1023&width=1325&top_left_y=131&top_left_x=389)

2nd diagram

The resulting expression describes a straight line (2nd diagram), which gives the position of the second car relative to the first: the $x$-coordinate of any point on the line corresponds to a position of the first car, and the $y$-coordinate corresponds to the position of the second car at the same time. The distance between the two cars is equal to the distance of the point from the origin: the segments connecting the positions of the two cars and the segment $O L$ are the diagonals of the same rectangle. The minimum distance is the distance of the line (3) from the origin. This can be determined using the diagram.

$$
\begin{gathered}
\operatorname{tg} \alpha=v_{2} / v_{1}=2 / 3 \\
y_{0}=s_{2}-\left(v_{2} / v_{1}\right) s_{1} \\
d_{\min }=y_{0} \cos \alpha \\
l_{2}=d_{\min } \cdot \cos \alpha \\
l_{1}=-d_{\min } \cdot \sin \alpha
\end{gathered}
$$

The time at which this occurs can be determined using either equation (1) or (2) by substituting $x_{1}=l_{1}$ or $y_{2}=l_{2}$.

Bagi Barnabás (Barcs, Gym. and Water Engineering Secondary School, II. o. t.)