8,9 [ Investigation of a quadratic trinomial $\quad$ ]

Authors: Kosukhin O.N., Bednov B.B.

Three athletes started simultaneously from point $A$ and ran in a straight line to point $B$ each at their own constant speed. Upon reaching point $B$, each of them instantly turned around and ran back to the finish line at point $A$ at another constant speed. Their coach ran alongside and was always at the point where the sum of the distances from that point to the participants was the smallest. It is known that the distance from $A$ to $B$ is 60 m and all the athletes finished at the same time. Could the coach have run less than 100 m?
Let's assign numbers to the athletes in descending order of their starting speeds. We will draw graphs of their movements, plotting time on the x-axis and the distance to point $A$ on the y-axis. Let $O$ be the origin, $S$ the point on the y-axis corresponding to point $B \quad (O S=60 \mathrm{~m}), K, L, M$ - points on the graphs of three athletes at the moment they are in point $B, T$ - the point on the x-axis corresponding to the finish time, $P, Q, R$ - points corresponding to the moment of meeting of the first and second, second and third, third and first athletes, respectively, $P^{\prime}, Q_{1}$ and $R^{\prime}$ - projections of these points on the y-axis (see figure).

![](https://cdn.mathpix.com/cropped/2024_05_06_e53884ac984930b8f3cbg-36.jpg?height=981&width=1501&top_left_y=1&top_left_x=273)

Note that for any three given points on a line, there is a unique point for which the sum of the distances to the given points is minimal - this is the middle of the three given points. Therefore, the coach will always be near the athlete who was between the other two. Then the coach's movement graph is described by the broken line $O P R Q T$, and the length $l$ of his path is equal to $O P^{\prime}+P^{\prime} R^{\prime}+R^{\prime} Q^{\prime}+Q^{\prime} O$.

Let the lengths of segments $K L, L M$ and OT be denoted by $a, b$ and $t$ respectively. Since $K L \| O T$, triangles $K P L$ and $Т Р О$ are similar and

$K P: P T=K L: O T=a: t$. Similarly, $L Q: Q T=L M: O T=b: t$ and $K R: R T=K M: O T=(a+b): t$. By the theorem of proportional segments

$S P^{\prime}: P^{\prime} O=K P: P T=a: t$. Therefore, $O P^{\prime}=60 t / a+t$. Similarly, $O Q^{\prime}=60 t / b+t$ and $O R^{\prime}=60 t / a+b+t$. Thus,

$$
\begin{gathered}
\frac{l}{60 t}=\frac{1}{a+t}+\left(\frac{1}{a+t}-\frac{1}{a+b+t}\right)+\left(\frac{1}{b+t}-\frac{1}{a+b+t}\right)+\frac{1}{b+t}= \\
2\left(\frac{1}{a+t}+\frac{1}{b+t}-\frac{1}{a+b+t}\right)
\end{gathered}
$$

Since $b\frac{a}{(2 t-a)(2 t)}$

$\frac{l}{120 t}>\frac{1}{a+t}+\frac{1}{2 t-a}=\frac{3 t}{(a+t)(2 t-a)}-\frac{1}{2 t}$.

The quadratic trinomial $(a+t)(2 t-a)$ (as a function of $a$) takes its maximum value ${ }^{9 t^{2}} / 4$ at $a=t / 2$. Therefore, $l>120 t\left(\frac{4}{3 t}-1 / 2 t\right)=100$.

## Answer

Could not.