Problem 18. In triangle $A B C$, $A C=B C$, $\angle A C B=30^{\circ}$. Point $D$ lies on side $B C$ and $B D=3, D C=6$. Find a point $P$ on side $A C$ of this triangle such that angle $D P C$ is 1.5 times the angle $A P B$.
Solution. \(AB = 2AC \sin 15^{\circ} = 18 \sin 15^{\circ}\). Let \(\angle APB = x\). Then \(\angle DPC = 1.5x\). Applying the Law of Sines to the corresponding triangles, we get:

\[
AP: \sin \left(75^{\circ} + x\right) = 18 \sin 15^{\circ}: \sin x ; AP =
\]

\[
\begin{aligned}
& = 18 \sin 15^{\circ} \sin \left(75^{\circ} + x\right): \sin x ; 6: \sin 1.5x = PC \\
& : \sin \left(30^{\circ} + 1.5x\right) ; PC = 6 \sin \left(30^{\circ} + 1.5x\right): \sin 1.5x .
\end{aligned}
\]

We obtain the equation:

\[
\frac{18 \sin 15^{\circ} \sin \left(75^{\circ} + x\right)}{\sin x} + \frac{6 \sin \left(30^{\circ} + 1.5x\right)}{\sin 1.5x} = 9
\]

From the problem statement, it is clear that the angle \(x\) is acute. Therefore, point \(P\) lies on the segment \(HC\) (where \(BH\) is the height of triangle \(BAC\)).

As point \(P\) moves along the segment \(HC\) (from \(H\) to \(C\)), the angle \(\angle APB\) decreases, and the angle \(\angle DPC\) increases. Therefore, equation (1) has a unique solution.

Solving equation (1) on a calculator, we get \(x = 60^{\circ}\), i.e., \(1.5x = 90^{\circ}\). From the right triangle \(DPC\), we find:

\[
PC = DC \cos 30^{\circ} = 6 \cdot 0.5 \sqrt{3} = 3 \sqrt{3}
\]