18. Answer: 25
Solution. We shall show that $-2 \leq \sin 3 A+\sin 3 B+\sin 3 C \leq 3 \sqrt{3} / 2$.
Assume that $A \leq B \leq C$. Then $A \leq 60^{\circ}$. Thus $\sin 3 A \geq 0$. It is clear that $\sin 3 B \geq-1$ and $\sin 3 C \geq-1$. Thus $\sin 3 A+\sin 3 B+\sin 3 C \geq-2$. Let $B=C$. Then $B=C=90^{\circ}-A / 2$. If $A$ is very small, $B$ and $C$ are close to $90^{\circ}$, and thus $\sin 3 A+\sin 3 B+\sin 3 C$ is close to -2 .
Now we show that $\sin 3 A+\sin 3 B+\sin 3 C \leq 3 \sqrt{3} / 2$. First the upper bound can be reached when $A=B=20^{\circ}$ and $C=140^{\circ}$.
Let $X=3 A, Y=3 B$ and $Z=3\left(C-120^{\circ}\right)$. Then $X+Y+Z=180^{\circ}$ and
$$
\sin 3 A+\sin 3 B+\sin 3 C=\sin X+\sin Y+\sin Z \text {. }
$$

Suppose that $X, Y, Z$ satisfy the condition that $X+Y+Z=180^{\circ}$ such that $\sin X+\sin Y+$ $\sin Z$ has the maximum value. We can then show that $X=Y=Z$.
Assume that $X \leq Y \leq Z$. If $X<Z$, then
$$
\sin X+\sin Z=2 \sin \frac{X+Z}{2} \cos \frac{X-Z}{2}<2 \sin \frac{X+Z}{2},
$$
implying that
$$
\sin X+\sin Y+\sin Z<\sin \frac{X+Z}{2}+\sin Y+\sin \frac{X+Z}{2}
$$
which contradicts the assumption that $\sin X+\sin Y+\sin Z$ has the maximum value. Hence $X=Y=Z=60^{\circ}$, implying that $A=20^{\circ}, B=20^{\circ}$ and $C=140^{\circ}$ and
$$
\sin 3 A+\sin 3 B+\sin 3 C=3 \sqrt{3} / 2 \text {. }
$$

Since $\sqrt{3} \approx 1.732$, the answer is then obtained.
18. Let $A, B, C$ be the three angles of a triangle. Let $L$ be the maximum value of
$$
\sin 3 A+\sin 3 B+\sin 3 C \text {. }
$$

Determine $\lfloor 10 L\rfloor$.