# 1998 AHSME Problems/Problem 21

## Problem

In an $h$-meter race, Sunny is exactly $d$ meters ahead of Windy when Sunny finishes the race. The next time they race, Sunny sportingly starts $d$ meters behind Windy, who is at the starting line. Both runners run at the same constant speed as they did in the first race. How many meters ahead is Sunny when Sunny finishes the second race? $\mathrm{(A) \ } \frac dh \qquad \mathrm{(B) \ } 0 \qquad \mathrm{(C) \ } \frac {d^2}h \qquad \mathrm{(D) \ } \frac {h^2}d \qquad \mathrm{(E) \ } \frac{d^2}{h-d}$

## Solution

Let $s$ and $w$ be the speeds of Sunny and Windy. From the first race we know that $\frac sw = \frac h{h-d}$. In the second race, Sunny's track length is $h+d$. She will finish this track in $\frac{h+d}s$. In this time, Windy will run the distance $w\cdot \frac{h+d}s = \frac{(h+d)(h-d)}h$. This is less than $h$, therefore Sunny is ahead. The exact distance between Windy and the finish is $h - \frac{(h+d)(h-d)}h = \frac{ h^2 - (h^2-d^2) }h = \boxed{\frac{d^2}h}$.

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