In the diagram, $A B C D E F$ is a regular hexagon with side length 2. Points $E$ and $F$ are on the $x$-axis and points $A, B, C$, and $D$ lie on a parabola. What is the distance between the two $x$-intercepts of the parabola?

![](https://cdn.mathpix.com/cropped/2024_04_17_d3d67b8c1418cbc5debag-2.jpg?height=428&width=423&top_left_y=2101&top_left_x=1168)
When we translate the parabola and hexagon horizontally, the distance between the $x$-intercepts of the parabola does not change.

This means that we can position the hexagon and parabola in a horizontal position that is more convenient.

Thus, we position the hexagon and parabola so that the $y$-axis passes through the midpoint of $F E$ and so the hexagon and parabola are symmetric across the $y$-axis.

Since $F E=2$, then the midpoint of $F E$ is 1 unit from each of $F$ and $E$.

In other words, the coordinates of $F$ and $E$ are $(-1,0)$ and $(1,0)$, respectively.

Next, we drop a perpendicular from $D$ to $P$ on the $x$-axis.

![](https://cdn.mathpix.com/cropped/2024_04_17_7798dec8e86d239f0636g-03.jpg?height=420&width=417&top_left_y=1389&top_left_x=908)

Since $A B C D E F$ is a regular hexagon, then each of its interior angles measures $120^{\circ}$. This is because the sum of the interior angles is $4 \cdot 180^{\circ}$ or $720^{\circ}$ and so each angle is $\frac{1}{6} \cdot 720^{\circ}$ which equals $120^{\circ}$.

Thus, $\angle D E P=180^{\circ}-120^{\circ}=60^{\circ}$, which means that $\triangle D E P$ is a $30^{\circ}-60^{\circ}-90^{\circ}$ triangle.

Since $D E=2$, then $E P=1$ and $D P=\sqrt{3}$.

Since the coordinates of $E$ are $(1,0)$, then the coordinates of $D$ are $(1+1,0+\sqrt{3})$ or $(2, \sqrt{3})$. Similarly, $C$ is one unit to the left of $D$ and $\sqrt{3}$ units above, which means that the coordinates of $C$ are $(1,2 \sqrt{3}$ ). This is because a right-angled triangle can be drawn with hypotenuse $C D=2$, legs parallel to the axes, and one angle equal to $60^{\circ}$ by reflecting $\triangle E P D$ through the horizontal line through $D$.

Since the hexagon is symmetric across the $y$-axis, then the coordinates of $A$ and $B$ are $(-2, \sqrt{3})$ and $(-1,2 \sqrt{3})$, respectively.

Suppose that the parabola has equation $y=a x^{2}+b x+c$.

Since the points $(-2, \sqrt{3}),(2, \sqrt{3})$ and $(1,2 \sqrt{3})$ lie on the parabola, then

$$
\begin{aligned}
\sqrt{3} & =4 a-2 b+c \\
\sqrt{3} & =4 a+2 b+c \\
2 \sqrt{3} & =a+b+c
\end{aligned}
$$

Subtracting the second equation from the first, we obtain $4 b=0$ and so $b=0$. (We could also have determined this using a symmetry argument.)

Substituting into the second and third equations, we obtain $4 a+c=\sqrt{3}$ and $a+c=2 \sqrt{3}$. Subtracting these equations, we obtain $3 a=-\sqrt{3}$ and so $a=-\frac{\sqrt{3}}{3}$.

Substituting $a=-\frac{\sqrt{3}}{3}$ into $a+c=2 \sqrt{3}$, we obtain $c=2 \sqrt{3}-a=2 \sqrt{3}+\frac{\sqrt{3}}{3}=\frac{7 \sqrt{3}}{3}$.

Therefore, the equation of the parabola is $y=-\frac{\sqrt{3}}{3} x^{2}+\frac{7 \sqrt{3}}{3}$.

We find the $x$-intercepts of the parabola by setting $y=0$ to obtain $0=-\frac{\sqrt{3}}{3} x^{2}+\frac{7 \sqrt{3}}{3}$ or $x^{2}=7$ and so $x= \pm \sqrt{7}$.

Thus, the parabola crosses the $x$-axis at the points $(-\sqrt{7}, 0)$ and $(\sqrt{7}, 0)$.

The distance between these two points is $\sqrt{7}-(-\sqrt{7})=2 \sqrt{7}$.

ANSWER: $2 \sqrt{7}$

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