12) In a modern art sculpture, a circle is partially hidden by an equilateral triangle, as shown in the figure: the circle has a diameter equal to the height of the triangle, which measures $\sqrt{6} \mathrm{~m}$. What is the area of the part of the circle not covered by the triangle?
(A) $\left(\frac{3}{2} \pi-\frac{8}{\sqrt{3}}\right) \mathrm{m}^{2}$
(B) $\frac{\pi}{2} \mathrm{~m}^{2}$
(C) $\left(\pi-\frac{3 \sqrt{3}}{4}\right) \mathrm{m}^{2}$
(D) $\left(\frac{3}{2} \pi-\frac{9 \sqrt{3}}{8}\right) \mathrm{m}^{2}$
(E) $\frac{3}{2} \pi \mathrm{m}^{2}$

![](https://cdn.mathpix.com/cropped/2024_04_17_9ee229095b4334063cc2g-2.jpg?height=257&width=302&top_left_y=550&top_left_x=1154)
12. The answer is $(\mathbf{C})$.

Let $A B C$ be the triangle, with $A$ the vertex common to the circle, and $B^{\prime}, C^{\prime}$ the two points of intersection of the sides $A B, A C$ with it. Notice that, by symmetry, the side $B^{\prime} C^{\prime}$ is parallel to $B C$, so $A B^{\prime} C^{\prime}$ is similar to $A B C$ and is equilateral. The center of the circle thus coincides with the centroid of the triangle $A B^{\prime} C^{\prime}$; if $r=\frac{\sqrt{6}}{2}$ is the radius of the circle, the height $h$ of the triangle $A B^{\prime} C^{\prime}$ is then $h=\frac{3}{2} r$ and the side $A B^{\prime}=\frac{2}{\sqrt{3}} h=\sqrt{3} r$. The area of the smaller triangle is therefore $\mathcal{A}\left(A B^{\prime} C^{\prime}\right)=\frac{1}{2} h \cdot A B^{\prime}=\frac{3 \sqrt{3}}{4} r^{2}$, while the area of the part of the circle not covered by the triangle is $\mathcal{A}=\frac{2}{3}\left[\pi r^{2}-\mathcal{A}\left(A B^{\prime} C^{\prime}\right)\right]=\frac{2}{3}\left[\pi-\frac{3 \sqrt{3}}{4}\right] r^{2}=\pi-\frac{3 \sqrt{3}}{4}$.

[Problem proposed by C. Balocchi]