Given an equilateral triangle $ABC$. Let's draw the three parabolas whose foci are the vertices $A$, $B$, and $C$, and whose directrices are the lines $BC$, $CA$, and $AB$, respectively. What is the area of the part of the triangle that is not covered by the interior of any of the parabolas? ($AB=1$; by the interior of the parabola, we mean the part of the plane that lies on the same side of the parabola as the focus.)
The directrix of the parabola $p$ with focus $A$ is the line $BC$. The tangent at the vertex of $p$ is the perpendicular bisector of the segment $AA_0$, where $A_0$ is the orthogonal projection of the focus $A$ onto the line $BC$. This perpendicular bisector intersects the sides $AB$ and $AC$ at the midpoints $C_0$ and $B_0$ of these sides, respectively, since it bisects every segment connecting the focus to the directrix. The part of the parabola $p$ on the side of the tangent $A A_0$ that contains $A$ is within the triangle $A B_0 C_0$. Therefore, the part of the parabola $p$ within the triangle $H = ABC$ is also within the triangle $A B_0 C_0$.

![](https://cdn.mathpix.com/cropped/2024_05_02_61a21652166913a73492g-1.jpg?height=685&width=656&top_left_y=335&top_left_x=732)

Similarly, we find that the part of the parabola with focus $B$ and directrix $CA$ within $H$ is in the triangle $B C_0 A_0$, and the part of the parabola with focus $C$ and directrix $AB$ within $H$ is in the triangle $C A_0 B_0$. Therefore, the interiors of these three parabolas have no common part within $H$. The area of the region not covered by these parabolas within $H$ can be obtained by subtracting the areas of the regions covered by each parabola from the area of $H$. It is sufficient to determine the area of the region covered by the interior of $p$, because the areas of the regions covered by the other two parabolas are equal to this, as these regions can be obtained by rotating the region covered by $p$ around the centroid $K$ of the triangle $H$ by $120^\circ$ in the appropriate direction.

The parabola $p$ intersects the segment $AB$ at a point $E$ such that the distance $E E_0$ from $E$ to $BC$ is equal to $AE$. In the triangle $A E E_0$, the angle $A E E_0 = 90^\circ + 60^\circ = 150^\circ$ and the angle $E A E_0 = 15^\circ$, so $A E_0$ bisects the angle $B A A_0$. By the angle bisector theorem,

$$
\begin{gathered}
B E_0 : E_0 A_0 = B A : A A_0 = 1 : \frac{\sqrt{3}}{2} \\
B E_0 = 2 - \sqrt{3}, \quad E E_0 = 2 \sqrt{3} - 3
\end{gathered}
$$

By the symmetry of the equilateral triangle and the parabola, the lengths of $C D_0$ and $D D_0$ are the same, where $D_0$ is the projection of $D$ onto $BC$.

Let $O$ be the vertex of $p$, and $F$ the midpoint of $ED$. The area of the region $A D O E$ (which we want to determine) is divided by the line $ED$ into the equilateral triangle $A D E$ and the lens-shaped region $D E O D$. The area of the triangle $A D E$ is:

$$
t_1 = \frac{\sqrt{3}}{4} D E^2 = \frac{\sqrt{3}}{4} (2 \sqrt{3} - 3)^2 = \frac{21 \sqrt{3}}{4} - 9
$$

The area of the lens-shaped region $D E O D$ (which we will prove below) is two-thirds of the product of the base $DE$ and the height $FO$:

$$
t_2 = \frac{2}{3} D E \cdot F O = \frac{2}{3} D E (E E_0 - O A_0) = \frac{2}{3} (2 \sqrt{3} - 3) \left(2 \sqrt{3} - 3 - \frac{\sqrt{3}}{4}\right) = 13 - \frac{15 \sqrt{3}}{2}
$$

Thus, the area of the region $A D O E A$ is:

$$
t = t_1 + t_2 = 4 - \frac{9 \sqrt{3}}{4}
$$

and the area of the region not covered by the three parabolas within $H$ is:

$$
T = \frac{\sqrt{3}}{4} - 3 t = 7 \sqrt{3} - 12 \sim 0.1244
$$

We still need to prove that $t_2 = \frac{2}{3} D E \cdot F O$. Choose the $x$-axis of the coordinate system to be the line $B_0 C_0$ and the $y$-axis to be the line $A A_0$. Then the equation of $p$ is $y = \lambda x^2$, where $\lambda$ is a suitable constant. Let the length of the segment $ED$ be $2 \mu$, then $F O = \lambda \mu^2$, and the area under the arc $D O E$ of the parabola is:

$$
t_3 = \int_{-\mu}^{\mu} \lambda x^2 \, dx = \left[ \frac{\lambda x^3}{3} \right]_{-\mu}^{\mu} = \frac{2}{3} \lambda \mu^3
$$

The area under the segment $ED$ is

$$
t_4 = F O \cdot E D = 2 \lambda \mu^3
$$

$t_3$ is one-third of this, so $t_2$ is two-thirds of it, as we wanted to prove.