Two Parabolas and an Equilateral Triangle

Let the equation of the inverted parabola be $y = c - kx^2$ . Then $c$ is the $y$ -intercept we need to find. The intersecting points of the two parabolas or the bottom two vertices of the equilateral triangle $(\pm x_1, y_1)$ satisfy the system of equations of parabola.

$\begin{cases} y = x^2 & ...(1) \\ y = c - kx^2 & ...(2) \end{cases} \implies x^2 = c - kx^2 \implies x_1 = \pm \sqrt{\dfrac c{k+1}}$

Let the altitude of triangle be $h$ , then we have:

$\begin{aligned} c & = y_1 + h = x_1^2 + \sqrt 3 x_1 = \frac c{k+1} + \sqrt{\frac {3c}{k+1}} \\ \frac {kc}{k+1} & = \sqrt{\frac {3c}{k+1}} \end{aligned}$

$\implies c = \dfrac {3(k+1)}{k^2}$ , $x_1 = \dfrac {\sqrt 3}k$ , and area of the triangle $A_\triangle = \dfrac {3\sqrt 3}{k^2}$ .

Since $A_\triangle$ is same as the area between the parabolas and triangle, the area enclosed by the parabolas $A_p = 2A_\triangle$ or $\frac 12 A_p = A_\triangle$ . And we have:

$\begin{aligned} \frac 12 A_p & = \int_0^{x_1} \left({\color{#3D99F6}c}-kx^2 - x^2\right) dx & \small \color{#3D99F6} \text{Note that }c = \frac {3(k+1)}{k^2} \\ & = \int_0^{\frac {\sqrt 3}k} \left({\color{#3D99F6} \frac {3(k+1)}{k^2}}-(k+1)x^2 \right) dx \\ & = \left[\frac {3(k+1)}{k^2}x-\frac {k+1}3x^3 \right]_0^{\frac {\sqrt 3}k} \\ & = \frac {2\sqrt 3(k+1)}{k^3} \end{aligned}$

Therefore, $\dfrac {2\sqrt 3(k+1)}{k^3} = \dfrac {3\sqrt 3}{k^2} \implies 2k+2 = 3k \implies k = 2$ , $c = \dfrac {3(k+1)}{k^2} = \dfrac 94$ , and $a+b = 9+4 = \boxed{13}$ .

Let $P (p, p^2)$ be the rightmost vertex and $Q$ be the topmost vertex of the equilateral triangle, and draw a box outlining the regions between the parabola and the equilateral triangle.

Then the side of the equilateral triangle is $2p$ , so its height is $\sqrt{3}p$ , and its area is $A_{\triangle} = \frac{\sqrt{3}}{4}(2p)^2 = \sqrt{3}p^2$ .

Since $P$ has a $y$ -coordinate of $p^2$ and the height of the equilateral triangle is $\sqrt{3}p$ , the coordinates of $Q$ are $(0, p^2 + \sqrt{3}p)$ . This makes the area of the outlining box $A_{box} = 2p(p^2 + \sqrt{3}p)$ .

Since the area of the equilateral triangle is equal to the area of all the regions between the parabola and the triangle, and the area of the regions between the parabolas (including the equilateral triangle) is $\frac{2}{3}$ the area of the outlined box, $\frac{2}{3}A_{box} = 2A_{\triangle}$ or $\frac{2}{3}2p(p^2 + \sqrt{3}p) = 2\sqrt{3}p^2$ , which solves to $\frac{\sqrt{3}}{2}$ .

Therefore, the $y$ -coordinate of the vertex on the $y$ -axis is $p^2 + \sqrt{3}p = (\frac{\sqrt{3}}{2})^2 + \sqrt{3}(\frac{\sqrt{3}}{2}) = \frac{9}{4}$ , so $a = 9$ , $b = 4$ , and $a + b = \boxed{13}$ .

We want to find $y_0$ .

There are four unknown constants: $\{x_0,y_0,c,m\}$ . To determine $y_0$ , we therefore need to find four restrictions. These will be (1) the $60^\circ$ angle of the equilateral triangle, (2) the intersection of the parabolae at $x=x_0$ , (3) $\text{slope}=\frac{\text{rise}}{\text{run}}$ , and (4) the equivalence of the pink and blue areas.

1. $\frac{y_0-x_0^2}{x_0}=\tan \frac{\pi}{3}=\sqrt{3}\\ x_0^2+\sqrt{3}x_0-y_0=0\qquad (\text{Note that }x_0^2=y_0-\sqrt{3}x_0.)\\ x_0=\frac{-\sqrt{3}\pm \sqrt{3+4y_0}}{2}=\boxed{\frac{\sqrt{3+4y_0}-\sqrt{3}}{2}}\qquad(x_0>0)$
2. $y_0-cx_0^2=x_0^2\\ c=\frac{y_0-x_0^2}{x_0^2}=\frac{\left(\frac{y_0-x_0^2}{x_0}\right)}{x_0}=\boxed{\frac{\sqrt{3}}{x_0}}$
3. $m=\frac{x_0^2-y_0}{x_0}=\boxed{-\sqrt{3}}$
4. $\frac{1}{2}(2x_0)(y_0-x_0^2)=2\int_0^{x_0}y_0-cx^2-(y_0-\sqrt{3}x)+x_0^2-x^2dx\\ x_0(y_0-x_0^2)=2\int_0^{x_0}x_0^2+\sqrt{3}x-(1+c)x^2dx=2\left[x_0^3+\frac{\sqrt{3}}{2}x_0^2-(1+\frac{\sqrt{3}}{x_0})\frac{x_0^3}{3}\right]\\ y_0-x_0^2=(2-\frac{2}{3})x_0^2+2\sqrt{3}(\frac{1}{2}-\frac{1}{3})x_0\\ y_0-\frac{\sqrt{3}}{3}x_0=\frac{7}{3}x_0^2=\frac{7}{3}(y_0-\sqrt{3}x_0)\\ \frac{4}{3}y_0=2\sqrt{3}\left(\frac{\sqrt{3+4y_0}-\sqrt{3}}{2}\right)\\ \frac{4}{3}y_0+3=\sqrt{9+12y_0}\\ \frac{16}{9}y_0^2+8y_0+9=9+12y_0\\ y_0(\frac{16}{9}y_0-4)=0\\ \boxed{\mathbf{y_0=\frac{9}{4}}}\qquad (y_0\neq 0)$

Let the equation of the second parabola be y=k-b(x^2), such that it's vertex is at (0,k). The two paraboli meet at [√{k/(b+1)}, k/(b+1)]. Then the length of the side of the equilateral triangle 'a' is given by a^2=[k/(b+1)]+[bk/(b+1)]^2=4k/(b+1). By the given condition of the problem, area bounded by the two paraboli is twice the area of the triangle. All these yield k=27/4(b+1) and 27(b^2)/4(b+1)=3(b+1) or b=2(since b is positive). Therefore k=27/4(2+1)=9/4.

$\text{p1}=x^2$ and $p2 = y - a x^2$ . Both $a$ and $y$ are arbitary parameters presently and will be computed later. The values of $x$ at the intersections are $\pm\frac{\sqrt{y}}{\sqrt{a+1}}$ . The value of $y$ at the intersections is $\frac{y}{a+1}$ . Equating the area of the triangle and one-half of the area between the parabolas and solving for $a$ gives $2$ . Equating the Euclidean length of a side and the base to make the triangle isosceles and solving for $y$ gives $\frac94$ . Adding the numerator and the denominator gives 13.