6. A blacksmith is building a horizontal iron fence consisting of many vertical bars, parallel to each other, each of which is positioned $18 \mathrm{~cm}$ apart from the two adjacent ones. The blacksmith connects the ends of each pair of adjacent bars with a curved bar in the shape of an arc of a circle, placed in the plane of the bars, the highest point of which is $3 \sqrt{3} \mathrm{~cm}$ from the line (dashed in the figure) that passes through the upper ends of all the bars, and is perpendicular to the bars themselves. How long is each of the bars used to build the arcs?

![](https://cdn.mathpix.com/cropped/2024_04_17_fb48e0857bc541793844g-02.jpg?height=354&width=420&top_left_y=2199&top_left_x=1452)
(A) $8 \pi(\sqrt{3}-1) \mathrm{cm}$
(B) $6 \pi \sqrt{3} \mathrm{~cm}$
(C) $12 \pi(\sqrt{3}-1) \mathrm{cm}$
(D) $4 \pi \sqrt{3} \mathrm{~cm}$
(E) $8 \pi \sqrt{3} \mathrm{~cm}$.
6. The answer is (D). Referring to the figure, let $A$ and $B$ be the endpoints of adjacent bars, $V$ the vertex of the arc $\overparen{A B}$, $M$ the midpoint of $A B$, and $O$ the center of the circle to which the arc $\widehat{A B}$ belongs. It must hold that $O B = O V$. Setting $O M = x$, by the Pythagorean Theorem, it must hold that:

$$
\begin{gathered}
O M^{2} + M B^{2} = O V^{2} = (O M + M V)^{2} \\
x^{2} + 81 = (x + 3 \sqrt{3})^{2} = x^{2} + 6 \sqrt{3} x + 27
\end{gathered}
$$

![](https://cdn.mathpix.com/cropped/2024_04_17_fb48e0857bc541793844g-11.jpg?height=389&width=357&top_left_y=131&top_left_x=1523)

It follows that $x = 3 \sqrt{3}$, hence $O M = M V$, $O B = 2 \cdot O M$, and the angle $\widehat{A O B}$ has a measure of $\frac{2 \pi}{3}$. The arc $\overparen{\mathrm{AB}}$ thus has a length of:

$$
\frac{2 \pi}{3} \cdot O B = 4 \pi \sqrt{3}
$$

(Mattei)