20. Triangle $L M N$ represents a right-angled field with $L M=r$, $L X=p$ and $X N=q$. Jenny and Vicky walk at the same speed in opposite directions along the edge of the field, starting at $X$ at the same time. Their first meeting is at $M$.
Which of these expressions gives $q$ in terms of $p$ and $r$ ?

A $\frac{p}{2}+r$
B $\sqrt{p^{2}+r^{2}}+\frac{p}{2}$
C $\frac{p r}{2 p+r}$
D $\frac{p}{2}$
Solution
C
Because Jenny and Vicky meet after walking at the same speed, they walk the same distance. Therefore
$$
X L+L M=X N+N M .
$$

The question tells us that $X L=p, L M=r$ and $X N=q$, To find $N M$, we apply Pythagoras' Theorem to the right-angled triangle $L M N$. This gives
$$
N M^{2}=L N^{2}+L M^{2}=(p+q)^{2}+r^{2} .
$$

It follows that $N M=\sqrt{(p+q)^{2}+r^{2}}$. Substituting these values in equation (1) gives
$$
p+r=q+\sqrt{(p+q)^{2}+r^{2}} \text {. }
$$

Hence
$$
p+r-q=\sqrt{(p+q)^{2}+r^{2}} .
$$

By squaring both sides of equation (2) we obtain
$$
(p+r-q)^{2}=(p+q)^{2}+r^{2} \text {. }
$$

It follows that
$$
p^{2}+r^{2}+q^{2}+2 p r-2 p q-2 r q=p^{2}+2 p q+q^{2}+r^{2} .
$$

Equation (3) may be rearranged to give
$$
4 p q+2 r q=2 p r
$$
from which it follows that
$$
q(2 p+r)=p r .
$$

Therefore
$$
q=\frac{p r}{2 p+r} .
$$