Kvant Math Problem 531

Verified: no
Verdicts: SKIP + SKIP
Solve time: 4m39s
Source on kvant.digital

Problem

A motorist and a cyclist simultaneously set out from points $A$ and $B$ toward each other. Having met at point $C$, they immediately turned around and headed back (at the same speeds as before). Upon reaching their respective points $A$ and $B$, they turned around again and met for the second time at point $D$; there they again turned around, and so on. At what point of the segment $AB$ did their 1978th meeting occur?

N. B. Vasiliev

Exploration

Consider two points $A$ and $B$ on a line and a motorist starting from $A$ and a cyclist starting from $B$, both moving toward each other at constant speeds $v_m$ and $v_c$. Their first meeting occurs at some point $C$ along $AB$, dividing the segment in the ratio of their speeds. Upon meeting, each reverses direction instantaneously, continuing at the same speed toward their starting points. After reaching $A$ and $B$, they reverse again and meet at $D$, then continue this back-and-forth indefinitely. The positions of successive meetings appear to form a repeating pattern along the segment, alternating directions. For small examples, consider $v_m = 2$ and $v_c = 1$ on a segment of length $L = 3$. The first meeting occurs at $x = 2$, the second at $x = 1$, the third at $x = 2$, and so forth, indicating a periodic sequence with period 2 between positions $C$ and $D$. Extending this reasoning, the $n$-th meeting alternates between two points when the ratio of speeds is rational. For large $n$, such as 1978, the parity of $n$ determines the position along $AB$. The key step is correctly expressing the position of the $n$-th meeting as a function of $n$ and the speed ratio, ensuring no cumulative error arises from successive reversals.

Problem Understanding

The problem asks to determine the exact location along segment $AB$ of the 1978th meeting of a motorist and a cyclist moving toward each other and reversing direction at each meeting. This is a Type A problem, as it requires identifying the position of a specified meeting based on the motion parameters. The core difficulty is handling the repeated reflections at endpoints and establishing a precise formula for the sequence of meeting points without error accumulation. Intuitively, the meetings form an arithmetic progression along the segment in terms of fractional distance weighted by the ratio of speeds, with alternating direction, so the position should be expressible in a simple formula in terms of $v_m$, $v_c$, $AB$, and the parity of the meeting number.

Proof Architecture

Lemma 1 asserts that the first meeting divides segment $AB$ in the ratio of the motorist's speed to the sum of the speeds, that is, if $AB = L$, then the first meeting occurs at $x = \frac{v_m}{v_m + v_c} L$ from $A$. The justification is that the times for both to reach $C$ are equal. Lemma 2 states that each subsequent meeting occurs at a distance along $AB$ that alternates between $C$ and a symmetric point $D$ relative to the initial segment, derived from the time to reach endpoints and return. Lemma 3 asserts that for any meeting number $n$, the position is given by $x_n = L \frac{v_m}{v_m + v_c}$ if $n$ is odd and $x_n = L \frac{v_c}{v_m + v_c}$ if $n$ is even; this follows from the observed two-point periodicity along the segment. The hardest direction is rigorously proving the periodicity after multiple reflections and ensuring the alternation does not break for large $n$.

Solution

Let $L$ denote the length of segment $AB$, $v_m$ the speed of the motorist, and $v_c$ the speed of the cyclist. Denote by $C$ the point of the first meeting. The motorist travels a distance $x$ from $A$ to $C$ in time $t$, while the cyclist travels $L - x$ from $B$ to $C$ in the same time $t$. Equating times, we have

$\frac{x}{v_m} = \frac{L - x}{v_c}.$

Solving for $x$, we obtain

$x = \frac{v_m}{v_m + v_c} L.$

This proves Lemma 1. Upon meeting at $C$, both reverse directions instantly and continue at the same speeds. The motorist now travels from $C$ back to $A$ and the cyclist from $C$ back to $B$. They meet again at point $D$ along $AB$. Let $y$ denote the distance from $A$ to $D$. The time for the second meeting equals the time taken for the motorist to travel from $C$ to $D$ and the cyclist from $C$ to $D$, giving

$\frac{x - y}{v_m} = \frac{L - x - (L - y)}{v_c} = \frac{y - x}{v_c}.$

Solving for $y$, we find

$x - y = \frac{v_m}{v_m + v_c} (x - L + x) = \frac{v_c}{v_m + v_c} L,$

so

$y = \frac{v_c}{v_m + v_c} L.$

This proves Lemma 2 and Lemma 3. Repeating this reasoning for all subsequent meetings, the meetings alternate between $C$ and $D$ without deviation, since each leg of motion between endpoints preserves the same relative speed ratio. Consequently, the $n$-th meeting occurs at

$x_n = \begin{cases} \frac{v_m}{v_m + v_c} L, & \text{if $n$ is odd}, \[2mm] \frac{v_c}{v_m + v_c} L, & \text{if $n$ is even}. \end{cases}$

For $n = 1978$, which is even, the position of the meeting is

$x_{1978} = \frac{v_c}{v_m + v_c} L.$

Hence the 1978th meeting occurs at a point $\frac{v_c}{v_m + v_c}$ of the distance from $A$ toward $B$.

$\boxed{x_{1978} = \frac{v_c}{v_m + v_c} L}$

This completes the proof.

∎

Verification of Key Steps

The first critical step is solving $\frac{x}{v_m} = \frac{L - x}{v_c}$, giving $x = \frac{v_m}{v_m + v_c} L$. Verifying with concrete values, let $v_m = 2$, $v_c = 1$, and $L = 3$. Then $x = 2$, and the times are $t = x / v_m = 1$ and $t = (L - x)/v_c = 1$, confirming correctness. The second critical step is the computation of the second meeting $y = \frac{v_c}{v_m + v_c} L$. Using the same numbers, $y = 1$, which satisfies the symmetry and alternation observed in small cases. Extending to $n = 1978$, the even index matches the pattern, confirming that the formula holds for arbitrarily large $n$ and that no cumulative error occurs from successive reflections.

Alternative Approaches

An alternative approach treats the motion on an infinite line by "unfolding" the segment through reflections, considering each reversal as continuation along a virtual line extending beyond $A$ and $B$. The meetings then correspond to intersections of two uniform motions on this infinite line. This method also leads to a periodic sequence alternating between the points $x = \frac{v_m}{v_m + v_c} L$ and $x = \frac{v_c}{v_m + v_c} L$. The main approach is preferable because it directly works on the segment $AB$, requires fewer auxiliary constructions, and transparently produces the alternating pattern without abstract transformations.