Dynamic Geometry: P135

Let the base line of length $4$ be $AB$ , its midpoint $O$ be the origin $(0,0)$ of the $xy$ -plane, the moving point be $C(u,3)$ , and any point or the othocenter of $\triangle ABC$ be $P(x,y)$ . Let $CE$ and $AF$ be the altitudes of $C$ and $A$ respectively. Then $CE$ and $AF$ pass through $P$ .

We note that the gradient of $BC$ is $-\dfrac 3{2-u}$ . Then the gradient of $AF$ , which is perpendicular to $BC$ , is $\dfrac {2-u}3$ and its equation is $\dfrac y{x+2} = \dfrac {2-u}3 \implies y = \dfrac {(2-u)(x+2)}3$ . Since $CE$ intersects $AF$ at $P(x,y)$ , $x = u$ and $y = \dfrac {4-u^2}3 \implies y = \dfrac {4-x^2}3$ , which is a parabola. Then the area under the parabola $A_p$ equals to $\dfrac 23$ the area of the rectangle which inscribes it. That is:

$A_p = \frac 23 \cdot 4 \cdot y(0) = \frac 23 \cdot 4 \cdot \frac 43 = \frac {32}9$

Let the cyan lines meet at $D$ . Since the parabola is symmetrical about the $y$ -axis, the gradients of the two cyan lines are same in magnitude but opposite in direction. $\triangle ABD$ is isosceles and $D$ is on the $y$ -axis. The gradient of $AD$ is given by:

$\frac {dy}{dx} \ \bigg|_{x=-2} = \frac d{dx} \left(\frac {4-x^2}3 \right)\bigg|_{x=-2} = - \frac 23x \ \bigg|_{x=-2} = \frac 43$

Then $OD = 2 \cdot \dfrac 43 = \dfrac 83$ and the area of $\triangle ABD$ , $A_\triangle = \dfrac 12 \cdot 4 \cdot \dfrac 83 = \dfrac {16}3$ . And the area enclosed by the cyan lines and parabola is

$A = A_\triangle - A_p = \frac {16}3 - \frac {32}9 = \frac {16}9$

And the required answer is $p+q = 16+9 = \boxed{25}$ .