24. The diagram shows a circle with centre $O$ which lies in a horizontal plane. The diameter $A B$ has length 4 . Point $P$ lies vertically above $O$ and $P O=2 \sqrt{2}$. Point $C$ lies on the semicircular arc $A B$ such that the ratio of the lengths of the arcs $A C$ and $C B$ is $2: 1$.
What is the shortest distance from $A$ to $P C$ ?
A $\sqrt{2}$
B $\sqrt{3}$
C 2
D $2 \sqrt{2}$
E 3
Solution
E

The triangle $P O A$ has a right angle at $O$. Therefore, by Pythagoras's Theorem, $P A^{2}=A O^{2}+P O^{2}=2^{2}+(2 \sqrt{2})^{2}=$ $4+8=12$. It follows that $P A=\sqrt{12}=2 \sqrt{3}$. Similarly, $P C=2 \sqrt{3}$.
Since $\operatorname{arc} A C: \operatorname{arc} C B=2: 1$, it follows that $\angle A O C$ : $\angle C O B=2: 1$. Hence $\angle C O B=60^{\circ}$.

Now $O C$ and $O B$ both have length 2 as they are radii of the circle with centre $O$. Since $\angle C O B=60^{\circ}$, it follows that the triangle $C O B$ is equilateral. Hence $B C=2$.
We now turn our attention to the triangle $A C B$. In this triangle $\angle A C B=90^{\circ}$ because it is the angle in a semicircle, $A B=4$ and $B C=2$. Therefore, by Pythagoras' Theorem, $A B^{2}=A C^{2}+B C^{2}$ and hence $4^{2}=A C^{2}+2^{2}$. It follows that $A C^{2}=4^{2}-2^{2}=16-4=12$. Therefore $A C=\sqrt{12}=2 \sqrt{3}$.
We have therefore shown that $P A C$ is an equilateral triangle in which each side has length $2 \sqrt{3}$.
The shortest distance from $A$ to $P C$ is the length, say $h$, of the perpendicular from $A$ to $P C$. Thus $h$ is the height of the equilateral triangle $P A C$.
We have seen in the solution to Question 22 that the height of an equilateral triangle with side length $x$ is $\frac{\sqrt{3}}{2} x$. It follows that
$$
h=\frac{\sqrt{3}}{2} \times 2 \sqrt{3}=3 \text {. }
$$