Dynamic Geometry: P136

Let the purple triangle be $ABC$ , where $AB=4$ is the horizontal base. Let the midpoint of $AB$ , $M$ be the origin $(0,0)$ of the $xy$ -plane, then $A=(-2,0)$ , $B=(2,0)$ , and $C=(u,3)$ ; and the midpoints of $CA$ and $BC$ be $L\left(\dfrac {u-1}2, \dfrac 32\right)$ and $N\left(\dfrac {u+2}2, \dfrac 32\right)$ respectively.

Let any point on the locus or the center of the nine-point circle of $\triangle ABC$ be $P(x,y)$ . We note that the center of nine-point circle is also the circumcenter of $\triangle LMN$ . Therefore $P$ is the intersection point of the perpendicular side bisectors of $LN$ and $MN$ . Let the midpoints of $LN$ and $MN$ be $E \left(\dfrac u2, \dfrac 32 \right)$ and $F \left(\dfrac {u+2}4, \dfrac 34 \right)$ respectively. Then the equations of

$\begin{cases} EP: & x = \dfrac u2 & ...(1) \\ FP: & \dfrac {y - \frac 34}{x - \frac {u+2}4} = -\dfrac {u+2}3 & ...(2) \end{cases}$

Then the $x$ -coordinate of $P$ , $x = \dfrac u2$ . Substituting $u = 2x$ in equation $(2)$ :

$\begin{aligned} \frac {y-\frac 34}{x - \frac {2x+2}4} & = - \frac {2x+2}3 \\ 12y - 9 & = 4 - 4x^2 \\ y & = \frac {13-4x^2}{12} \end{aligned}$

Therefore the locus is a parabola. The area under a parabola is $\dfrac 23$ of the area of the rectangle that encloses it. The area of under the locus,

$A_p = \frac 23 \cdot y(0) \cdot 2 |x|\bigg|_{y=0} = \frac 23 \cdot \frac {13}{12} \cdot \frac {2\sqrt{13}}2 = \frac {13\sqrt{13}}{18}$

Let the height of the red triangle $DM= h$ . Then $D=(0,h)$ and the equation of $AD$ is $\dfrac y{x+2} = \dfrac h2 \implies y = \dfrac {h(x+2)}2$ . Let the point where $AD$ is tangent to the locus be $Q$ . Then the gradient at $Q$ on the parabola is the same as that of $AD$ or

$\frac {dy}{dx} = \frac d{dx} \left(\frac {13-4x^2}{12} \right) = - \frac 23 x = \frac h2 \implies x = -\frac 34h$

Since $Q$ satisfy the equation of $AD$ and the equation of the locus, we have:

$\begin{aligned} \frac {13-4x^2}{12} & = \frac {h(x+2)}2 \\ 13 - 4x^2 & = 6h(x+2) & \small \blue{\text{Substituting }x = - \frac 34 h} \\ 13 - \frac 94 h^2 & =- \frac 92h^2 + 12h \\ 9h^2 - 12h + 52 & = 0 & \small \blue{\text{The tangent point is below the}} \\ \implies h & = \frac 83 - \frac 2{\sqrt 3} & \small \blue{x\text{-axis for }h = \frac 83+\frac 2{\sqrt 3}.} \end{aligned}$

The area of $\triangle ABD$ , $A_\triangle = \dfrac 12 \cdot 4h = \dfrac {16}3 - \dfrac 4{\sqrt 3}$ and the area enclosed by the locus and the two red segments is

$A = A_\triangle - A_p = \frac {16}3 - \frac {4\sqrt 3}3 - \frac {13\sqrt{13}}{18}$

The required answer is $p+q+m+n+i = 16+3+4+13+18 = \boxed{54}$ .