# 1973 AHSME Problems/Problem 29

## Problem

Two boys start moving from the same point A on a circular track but in opposite directions. Their speeds are 5 ft. per second and 9 ft. per second. If they start at the same time and finish when they first meet at the point A again, then the number of times they meet, excluding the start and finish, is $\textbf{(A)}\ 13 \qquad \textbf{(B)}\ 25 \qquad \textbf{(C)}\ 44 \qquad \textbf{(D)}\ \text{infinity} \qquad \textbf{(E)}\ \text{none of these}$

## Solution

Let $d$ be the length of the track in feet and $x$ be the number of laps that one of the boys did, so time one of the boys traveled before the two finish is $\tfrac{dx}{5}$. Since the time elapsed for both boys is equal, one boy ran $5(\tfrac{dx}{5})$ feet while the other boy ran $9(\tfrac{dx}{5})$ feet. Because both finished at the starting point, both ran an integral number of laps, so $5(\tfrac{dx}{5})$ and $9(\tfrac{dx}{5})$ are multiples of $d$. Because both stopped when both met at the start for the first time, $x = 5$.

Note that between the time a runner finishes a lap and a runner (can be same) finishes a lap, both runners must meet each other. When $0 < x \le 5$ and either $5(\tfrac{dx}{5})$ or $9(\tfrac{dx}{5})$ is a multiple of $d$, one of the runners completed a lap. This is achieved when $x = \tfrac59, 1, \tfrac{10}{9}, \tfrac{15}{9}, 2, \tfrac{20}{9}, \tfrac{25}{9}, 3, \tfrac{30}{9}, \tfrac{35}{9}, 4, \tfrac{40}{9}, 5$, so the two meet each other (excluding start and finish) a total of $\boxed{\textbf{(A)}\ 13}$ times.