CEMC Cayley, 2026

Let $O$ be the point from
which the two boats begin, let $L$ be
the point that The Luna reaches after the first hour, let $P$ be the point that The Tuna reaches
after the first hour, and let $T$ be
the point that The Tuna reaches after it travels north for $90$ minutes from $P$.

The Luna travels from $O$ to $L$ in $1$ hour at a speed of $8\text{ km/h}$, so the length of $OL$ is $8 \text{ km}$.

The Tuna travels from $O$ to $P$ in $1$ hour at a speed of $8\text{ km/h}$, so the length of $OP$ is $8\text{ km}$.

The Tuna travels from $P$ to $T$ in $90$ minutes or $1.5$ hours at a speed of $8$ km/h, so the length of $PT$ is $8\text{ km/h}\times 1.5\text{ h}=12\text{ km}$.

The direction from $O$ to $L$ is northwest, and the direction from
$O$ to $P$ is northeast. Each of these directions
are $45\degree$ from north, and so we
conclude that $\angle LOP = 45\degree + 45\degree = 90\degree$.

In the first hour, The Luna and The Tuna each traveled the same
distance at an angle of $45\degree$
from north, and so the two boats were at exactly the same latitude when
they reached points $L$ and $P$, respectively. Put differently, The
Tuna was directly east of The Luna after the first hour. Since The Tuna
then travels north, we conclude that $\angle LPT = 90\degree$.

We have that $\triangle LOP$ is
right-angled at $O$ and $\triangle LPT$ is right-angled at $P$.

By the Pythagorean theorem, we get $OL^2 + OP^2 = LP^2$ and $LP^2 + PT^2=LT^2$.

Substituting the expression for $LP^2$ from the first of these equations
into the second, and using the lengths calculated above, we have $$\begin{align*} LT^2 &= OL^2+OP^2+PT^2 \\ &= 8^2+8^2+12^2 \\ &= 64+64+144 \\ &= 272\end{align*}$$ Since $\text{LT gt;0}$, we have $LT = \sqrt{272}$. To the nearest
kilometre, the distance between the two boats is $16$.