# 1996 AHSME Problems/Problem 13

## Problem

Sunny runs at a steady rate, and Moonbeam runs $m$ times as fast, where $m$ is a number greater than 1. If Moonbeam gives Sunny a head start of $h$ meters, how many meters must Moonbeam run to overtake Sunny? $\text{(A)}\ hm\qquad\text{(B)}\ \frac{h}{h+m}\qquad\text{(C)}\ \frac{h}{m-1}\qquad\text{(D)}\ \frac{hm}{m-1}\qquad\text{(E)}\ \frac{h+m}{m-1}$

## Solution

If Sunny runs at a rate of $s$ for $h$. Then the distance covered is $sh$. Now we know that moonbeam runs $m$ times as fast than sunny So moonebeam runs at the rate of $ms$. Now moonbeam gave sunny a headstart of $h$ meters . So he will catch on Sunny at the rate of $s(m-1)$ . At time $\frac{h}{m-1}$ Moon beam will catch on Sunny.Now we are asked how much in meters he have to run to catch on Sunny.That is $\frac{hm}{m-1}$.

### Solution 2

Note that $h$ is a length, while $m$ is a dimensionless constant. Thus, $h$ and $m$ cannot be added, and $B$ and $E$ are not proper answers, since they both contain $h+m$.

Thus, we only concern ourselves with answers $A, C, D$.

If $m$ is a very, very large number, then Moonbeam will have to run just over $h$ meters to reach Sunny. Or, in the language of limits: $\lim_{m\rightarrow \infty} d(m) = h$, where $d(m)$ is the distance Moonbeam needs to catch Sunny at the given rate ratio of $m$.

In option $A$, when $m$ gets large, the distance gets large. Thus, $A$ is not a valid answer.

In option $C$, when $m$ gets large, the distance approaches $0$, not $h$ as desired. This is not a valid answer. (In fact, this is the distance Sunny runs, which does approach $0$ as Moonbeam gets faster and faster.)

In option $D$, when $m$ gets large, the ratio $\frac{m}{m-1}$ gets very close to, but remains just a tiny bit over, the number $1$. Thus, when you multiply it by $h$, the ratio in option $D$ gets very close to, but remains just a tiny bit over, $h$. Thus, the best option out of all the choices is $\boxed{D}$.

### Solution 3

Assume that Sunny originally runs at a unit speed, and thus Moonbeam runs at a rate of $m$.

Choose a new reference frame where Sunny is still, and Moonbeam runs at a rate of $m-1$. In this new reference frame, the distance to be run is still $h$.

Moonbeam runs this distance $h$ in a time of $\frac{h}{m-1}$

Returning to the original reference frame, if Moonbeam runs for $\frac{h}{m-1}$ seconds, Moonbeam will cover a distance of $\frac{hm}{m-1}$, which is option $\boxed{D}$.

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 