From city $A$, two cars start simultaneously on the same route to $B$, the first at speed $v_{1}$, the second at speed $v_{2}\left(<v_{1}\right)$. $t_{1}$ hours later, a third car starts after them and $t_{2}$ hours after overtaking the first car, it arrives in $B$. After another $t_{3}$ time, the second car also arrives in $B$. What is the distance $AB$ and the speed of the third car? Numerical data: $v_{1}=80, v_{2}=60 \mathrm{~km} /$ hour; $t_{1}=1$ hour, $t_{2}=1$ hour, $t_{3}=2$ hours.
Let the three cars be $K_{1}, K_{2}, K_{3}$, the speed of the latter $v$ (km/h), the driving time until reaching $K_{1}$ be $x$ (hours), and finally the distance $AB$ be $s$ (km). Then $K_{1}$ has driven for $t_{1}+x$ hours until its arrival, the total driving time of $K_{2}$ and $K_{3}$ is $t_{1}+x+t_{2}+t_{3}$, and $x+t_{2}$, respectively. Expressing the corresponding distances with the data of each car and equating them, we get two equations for the unknowns $x, v$:

\[
\begin{gathered}
v_{1}\left(t_{1}+x\right)=v x \\
v_{2}\left(t_{1}+x+t_{2}+t_{3}\right)=v\left(x+t_{2}\right)
\end{gathered}
\]

Multiplying (1) by $x+t_{2}$ and (2) by $x$, the right sides of the equations become equal, so the left sides must also be equal. Rearranging the resulting quadratic equation in $x$, we get:

\[
\left(v_{1}-v_{2}\right) x^{2}+\left\{\left(v_{1}-v_{2}\right)\left(t_{1}+t_{2}\right)-v_{2} t_{3}\right\} x+v_{1} t_{1} t_{2}=0
\]

This is a genuine quadratic equation due to the assumption $v_{2} \neq v_{1}$. The first coefficient and the constant term are positive according to the problem statement, so the only positive root, which fits the waiting time, can only exist if the coefficient of $x$ is negative:

\[
\left(v_{1}-v_{2}\right)\left(t_{1}+t_{2}\right)-v_{2} t_{3}\frac{v_{1}-v_{2}}{v_{2}}\left(t_{1}+t_{2}\right)
\]

The necessary and sufficient condition for the solution to be real is that the discriminant is positive or zero:

\[
\left(v_{1}-v_{2}\right)^{2}\left(t_{1}+t_{2}\right)^{2}-2\left(v_{1}-v_{2}\right)\left[2 v_{1} t_{1} t_{2}+v_{2}\left(t_{1}+t_{2}\right) t_{3}\right]+v_{2}^{2} t_{2}^{2} \geq 0
\]

We refrain from further analysis of (4) due to its complexity, noting only that if (4) holds with the > sign, there are two positive solutions, and if it holds with the = sign, there is one (assuming (3) is satisfied).

Given $x$, from (1) we have:

\[
v=\frac{v_{1}\left(t_{1}+x\right)}{x}
\]

Finally, the distance, from the right side of (2):

\[
s=v\left(x+t_{2}\right)=\frac{v_{1}\left(x+t_{1}\right)\left(x+t_{2}\right)}{x}
\]

II. Now, with the numerical data, the value of the discriminant in (4) is 0 and (3) is satisfied, so there is a single solution:

\[
x=2 \text{ hours, } \quad v=120 \text{ km/h, } \quad s=360 \text{ km. }
\]

Ervin Horváth (Süttő, Secondary School, 7th grade) Ferenc Francia (Veszprém, L. Lovassy Gymnasium, 2nd grade)