$19 \cdot 36$ Three non-overlapping regular polygons, of which at least two are congruent, and all have a side length of 1, intersect at point $A$. The sum of the three interior angles of the polygons meeting at $A$ is $360^{\circ}$. This forms a new polygon with $A$ as an interior point. The maximum possible perimeter of this new polygon is
(A) 12.
(B) 14.
(C) 18.
(D) 21.
(E) 24.
(50th American High School Mathematics Examination, 1999)
[Solution] Note that the interior angle of a regular $n$-sided polygon is $(n-2) 180^{\circ} / n$. This gives us the following table:
\begin{tabular}{c|ccccc}
\hline Regular $n$-sided polygon sides $n$ & 3 & 4 & 5 & 6 & 7 \\
\hline Each interior angle & $60^{\circ}$ & $90^{\circ}$ & $108^{\circ}$ & $120^{\circ}$ & $128.6^{\circ}$ \\
\hline
\end{tabular}
\begin{tabular}{cccccc}
\hline 8 & 9 & 10 & 11 & 12 & $\cdots$ \\
\hline $135^{\circ}$ & $140^{\circ}$ & $144^{\circ}$ & $147.2^{\circ}$ & $150^{\circ}$ & $\cdots$ \\
\hline
\end{tabular}

From the problem statement, after a bit of analysis, we can determine that the two congruent regular polygons have 12 sides, and the other polygon is an equilateral triangle. The sum of the interior angles at point $A$ for the three regular polygons is
$$
150^{\circ} \times 2 + 60^{\circ} = 360^{\circ}.
$$

Thus, the perimeter of the new polygon with $A$ as an interior point is
$12 \times 2 + 3 - 3 = 24$ (noting that there are three pairs of overlapping sides). Therefore, the answer is $(E)$.