Problem (Geometry):
Three microphones $A, B$ and $C$ are placed on a line such that $A$ is $1 \mathrm{~km}$ west of $B$ and $C$ is $2 \mathrm{~km}$ east of $B$. A large explosion occurs at a point $P$ not on this line. Each of the three microphones receives the sound. The sound travels at $\frac{1}{3} \mathrm{~km} / \mathrm{s}$. Microphone $B$ receives the sound first, microphone $A$ receives the sound $\frac{1}{2}$ s later, and microphone $C$ receives it $1 \mathrm{~s}$ after microphone $A$. Determine the distance from microphone $B$ to the explosion at $P$.

Solution:
Throughout this solution, we will mostly not include units, but will assume that all lengths are in kilometres, all times are in seconds, and all speeds are in kilometres per second.

We place the points in the coordinate plane with $B$ at $(0,0), A$ on the negative $x$-axis, and $C$ on the positive $x$-axis.

We put $A$ at $(-1,0)$ and $C$ at $(2,0)$.

Suppose that $P$ has coordinates $(x, y)$ and that the distance from $P$ to $B$ is $d \mathrm{~km}$.



Since the sound arrives at $A \frac{1}{2} \mathrm{~s}$ after arriving at $B$ and sound travels at $\frac{1}{3} \mathrm{~km} / \mathrm{s}$, then $A$ is $\left(\frac{1}{2} \mathrm{~s}\right) \cdot\left(\frac{1}{3} \mathrm{~km} / \mathrm{s}\right)=\frac{1}{6} \mathrm{~km}$ farther from $P$ than $B$ is.

Thus, the distance from $P$ to $A$ is $\left(d+\frac{1}{6}\right) \mathrm{km}$.

Since the sound arrives at $C$ an additional 1 second later, then $C$ is an additional $\frac{1}{3} \mathrm{~km}$ farther, and so is $\left(d+\frac{1}{6}\right) \mathrm{km}+\left(\frac{1}{3} \mathrm{~km}\right)=\left(d+\frac{1}{2}\right) \mathrm{km}$ from $P$.

Since the distance from $P$ to $B$ is $d \mathrm{~km}$, then $(x-0)^{2}+(y-0)^{2}=d^{2}$.

Since the distance from $P$ to $A$ is $\left(d+\frac{1}{6}\right) \mathrm{km}$, then $(x+1)^{2}+(y-0)^{2}=\left(d+\frac{1}{6}\right)^{2}$.

Since the distance from $P$ to $C$ is $\left(d+\frac{1}{2}\right) \mathrm{km}$, then $(x-2)^{2}+(y-0)^{2}=\left(d+\frac{1}{2}\right)^{2}$.

When these equations are expanded and simplified, we obtain

$$
\begin{aligned}
x^{2}+y^{2} & =d^{2} \\
x^{2}+2 x+1+y^{2} & =d^{2}+\frac{1}{3} d+\frac{1}{36} \\
x^{2}-4 x+4+y^{2} & =d^{2}+d+\frac{1}{4}
\end{aligned}
$$

Subtracting the first equation from the second, we obtain

$$
2 x+1=\frac{1}{3} d+\frac{1}{36}
$$

Subtracting the first equation from the third, we obtain

$$
-4 x+4=d+\frac{1}{4}
$$

Therefore,

$$
\begin{aligned}
2(2 x+1)+(-4 x+4) & =2\left(\frac{1}{3} d+\frac{1}{36}\right)+\left(d+\frac{1}{4}\right) \\
6 & =\frac{2}{3} d+\frac{1}{18}+d+\frac{1}{4} \\
216 & =24 d+2+36 d+9 \quad \text { (multiplying by } 36) \\
205 & =60 d \\
d & =\frac{41}{12}
\end{aligned}
$$

Therefore, the distance from $B$ to $P$ is $\frac{41}{12} \mathrm{~km}$.

Final answer: $\frac{41}{12}$ km