1981 AMC 12 Problem 25

Let the three equal angles be $\alpha$, and apply the angle-bisector theorem to a suitable subtriangle

Equate law-of-cosines expressions for $\cos\alpha$ in the three adjacent triangles

Let $AB=c,$ $AD=y,$ $AC=b,$ and $AE=z.$ Since $AD$ bisects $\angle BAE,$ and $AE$ bisects $\angle DAC,$ the angle-bisector theorem gives $$\frac{c}{z}=\frac{2}{3},\qquad \frac{y}{b}=\frac{3}{6},$$ so $z=\frac{3c}{2}$ and $b=2y.$ Applying the law of cosines to triangles $ADB,$ $ADE,$ and $AEC$ and equating their common $\cos\alpha$ values gives $$\begin{gathered} 3c^2-2y^2=12,\\ 9c^2-8y^2=-72. \end{gathered}$$ Hence $y^2=54$ and $c^2=40.$ The sides are $AB=2\sqrt{10},$ $BC=11,$ and $AC=2\sqrt{54}=6\sqrt6,$ so the shortest is $2\sqrt{10}.$

Therefore, the correct answer is A.