Parabolic Triangle

The triangle can be stretched to a unit equilateral triangle $\triangle PQR$ and still preserve the ratio of areas. If the triangle is placed with its center at the origin, the coordinates for $P$ , $Q$ , and $R$ are $P(0, \frac{\sqrt{3}}{3})$ , $Q(\frac{1}{2}, -\frac{\sqrt{3}}{6})$ , and $R(-\frac{1}{2}, -\frac{\sqrt{3}}{6})$ . Also label in $A$ , $B$ , and $C$ as follows:

By symmetry, the red parabola will have an equation of $y = ax^2 + b$ , and its slope will be $y' = 2ax$ . At $Q_x = \frac{1}{2}$ it has a slope of $y' = -\sqrt{3}$ , so $-\sqrt{3} = 2a\frac{1}{2}$ , which solves to $a = -\sqrt{3}$ . Since the red parabola also goes through $Q(\frac{1}{2}, -\frac{\sqrt{3}}{6})$ , we know that $-\frac{\sqrt{3}}{6} = -\sqrt{3} (\frac{1}{2})^2 + b$ which solves to $b = \frac{\sqrt{3}}{12}$ . Therefore, the equation of the red parabola is $y = -\sqrt{3}x + \frac{\sqrt{3}}{12}$ , and the coordinates of $C$ are $C(0, \frac{\sqrt{3}}{12})$ .

Also by symmetry, $AO$ makes a $30°$ angle with the $x$ -axis, so it has a slope of $\tan 30° = \frac{\sqrt{3}}{3}$ which means it is on the line $y = \frac{\sqrt{3}}{3}x$ .

Since $A$ is at the intersection of $y = -\sqrt{3}x + \frac{\sqrt{3}}{12}$ and $y = \frac{\sqrt{3}}{3}x$ , its coordinates are $A(\frac{1}{6}, \frac{\sqrt{3}}{18})$ . The coordinates for $B$ are then $B(0, \frac{\sqrt{3}}{18})$ .

The area of $\triangle ABO$ is $A_{\triangle ABO} = \frac{1}{2} \cdot \frac{1}{6} \cdot \frac{\sqrt{3}}{18} = \frac{\sqrt{3}}{216}$ , and the area of region $ABC$ is $A_{ABC} = \frac{2}{3} \cdot \frac{1}{6} \cdot (\frac{\sqrt{3}}{12} - \frac{\sqrt{3}}{18}) = \frac{\sqrt{3}}{324}$ .

By symmetry, the area of the shaded region is $A_{\text{shaded}} = 6(A_{\triangle ABO} + A_{ABC}) = 6(\frac{\sqrt{3}}{216} + \frac{\sqrt{3}}{324}) = \frac{5\sqrt{3}}{108} = \frac{5}{27} \cdot \frac{\sqrt{3}}{4}$ .

Since the area of a unit equilateral triangle is $\triangle = \frac{\sqrt{3}}{4}$ , by substitution the area of the shaded region is $A_{\text{shaded}} = \boxed{\frac{5}{27}\triangle}$ .

These are known as the Artzt parabolae of the triangle, and a detailed analysis of the areas they create is given here . Amongst other matters, the points of intersection of the parabolae lie on the medians, and divide the medians in the ratio $1:8$ .