5) I draw an equilateral triangle and a regular hexagon inscribed in the same circle. What is the ratio of the area of the triangle to that of the hexagon?
(A) $\frac{1}{2}$,
(B) $\frac{1}{3}$
(C) $\frac{\sqrt{3}}{2}$
(D) $\frac{\sqrt{3}}{3}$
(E) $\frac{1}{6}$.
5. The answer is $(\mathbf{A})$.

As indicated in the figure, we can draw the triangle and the hexagon so that they share three vertices.

![](https://cdn.mathpix.com/cropped/2024_04_17_e415f128e14398de33bcg-2.jpg?height=309&width=306&top_left_y=1122&top_left_x=932)

In this way, it is noted that the hexagon is formed by the triangle to which three isosceles triangles $a, b$, and $c$ are added, with bases being the sides of the triangle and a vertex on the circumference. Furthermore, if the vertices of the equilateral triangle are connected to the center of the circle, it is decomposed into three triangles $d$, $e$, and $f$ congruent to $a, b$, and $c$. Therefore, the hexagon has double the area of the triangle, and the required ratio is $1 / 2$.

[Problem proposed by R. Morandin.]