AHSME 1999 Problem 26

Three non-overlapping regular plane polygons, at least two of which are congruent, all have sides of length $1$. The polygons meet at a point $A$ in such a way that the sum of the three interior angles at $A$ is $360^{\circ}$. Thus the three polygons form a new polygon with $A$ as an interior point. What is the largest possible perimeter that this polygon can have? $\mathrm{(A) \ }12 \mathrm{(B) \ }14 \mathrm{(C) \ }18 \mathrm{(D) \ }21 \mathrm{(E) \ } 24$