1999 AHSME 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 \qquad \mathrm{(B) \ }14 \qquad \mathrm{(C) \ }18 \qquad \mathrm{(D) \ }21 \qquad \mathrm{(E) \ } 24

Full credit goes to MAA for authoring these problems. These problems were taken on the AOPS website.