(16) The correct answer is $(A)$.

Triangle $A^{\prime} B^{\prime} C^{\prime}$ is equilateral for evident reasons of symmetry (or, if you prefer, by observing the congruence of triangles $A^{\prime} B^{\prime} A, B^{\prime} C^{\prime} B, C^{\prime} A^{\prime} C$).

Let $D$ be the orthogonal projection of $A^{\prime}$ onto the line $A B$. The right triangle $A D A^{\prime}$ has acute angles of $30^{\circ}$ and $60^{\circ}$: indeed, $\widehat{A^{\prime} A D}=180^{\circ}-\left(\widehat{B A C}+\widehat{C A C^{\prime}}\right)=30^{\circ}$. It follows that $\overline{A^{\prime} D}=\frac{1}{2}$ and that $\overline{A D}=\frac{\sqrt{3}}{2}$, from which we obtain that $\overline{A^{\prime} E}=1+\frac{1}{2}=\frac{3}{2}$ and $\overline{B^{\prime} E}=1+\frac{\sqrt{3}}{2}=\frac{2+\sqrt{3}}{2}$, where $E$ is the orthogonal projection of $A^{\prime}$ onto the line $A^{\prime \prime} B^{\prime}$.

By the Pythagorean theorem, we have $\overline{A^{\prime} B^{\prime}}{ }^{2}=\left(\frac{3}{2}\right)^{2}+\left(\frac{2+\sqrt{3}}{2}\right)^{2}=\frac{16+4 \sqrt{3}}{4}=4+\sqrt{3}$, from which we conclude that $\overline{A^{\prime} B^{\prime}}=\sqrt{4+\sqrt{3}}$.

Question proposed by Carmelo Di Stefano.
16. On the sides of the equilateral triangle $A B C$, which have length 1, three squares are constructed, as shown in the figure. What is the perimeter of the triangle $A^{\prime} B^{\prime} C^{\prime}$?
(A) $3 \sqrt{4+\sqrt{3}}$
(D) $\frac{3}{2} \sqrt{3}$
(E) $\frac{3}{2}+\frac{3}{4} \sqrt{3}$
(C) $3 \sqrt{1+2 \sqrt{3}}$

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