P.P.T and Areas!

As long as $AB < AC$ , the $MD$ will intersect $AC$ . Therefore, $p^2 - 4 < 4p$ , which solves to $2 - 2\sqrt{2} < p < 2 + 2\sqrt{2}$ or $-0.828 < p < 4.828$ , which makes $p = 3$ the only possible prime number for $p > 2$ .

If $p = 3$ , then $AB = p^2 - 4 = 3^2 - 4 = 5$ , $AC = 4p = 4 \cdot 3 = 12$ , and $BC = p^2 + 4 = 3^2 + 4 = 13$ .

Since $5^2 + 12^2 = 13^2$ , $\triangle ABC$ is a right triangle, and by Thales's Theorem right $\angle A$ will lie on the circle that has $BC$ as its diameter.

Since $M$ is the midpoint of $BC$ , $M$ is the center of that circle, and $MB = MA = MC = \cfrac{13}{2}$ , so $\triangle ACM$ is an isosceles triangle.

As two base angles of an isosceles triangle, $\angle MCA = \angle MAC$ , which means $\triangle AMF \sim \triangle CAB$ by AA similarity.

Since $\triangle AMF \sim \triangle CAB$ , $MF = MA \cdot \cfrac{AB}{AC} = \cfrac{13}{2} \cdot \cfrac{5}{12} = \cfrac{65}{24}$ .

Then $A_{AEDF} = A_{MAED} - A_{\triangle AMF} = AM^2 - \cfrac{1}{2} \cdot MF \cdot AM = \bigg(\cfrac{13}{2} \bigg)^2 - \cfrac{1}{2} \cdot \cfrac{65}{24} \cdot \cfrac{13}{2} = \cfrac{3211}{96} \approx \boxed{33.4479}$ .

Let $p > 2$ .

$(p^2 - 4, 4p, p^2 + 4)$ is a Pythagorean triple $\implies \triangle{ABC}$ is a right triangle with $m\angle{A} = 90^{\circ}$ and $\cos(\theta) = \dfrac{p^2 - 4}{p^2 + 4}.$

Using the law of cosines on $\triangle{BAM}$ with included $\angle{ABC} = \theta \implies$

$\overline{AM}^2 = (p^2 - 4)^2 + \dfrac{p^2 + 4}{4} - 2(p^2 - 4)(\dfrac{p^2 + 4}{2})(\dfrac{p^2 - 4}{p^2 + 4}) \implies \overline{AM} = \dfrac{p^2 + 4}{4}$

$\implies MF = \overline{AF}\cos(\theta) = \dfrac{p^2 - 4}{p^2 + 4}\overline{AF} \implies$ $\dfrac{(p^2 + 4)^2}{4} + \dfrac{(p^2 - 4)^2}{(p^2 + 4)^2}\overline{AF}^2 = \overline{AF}^2$

$\implies \dfrac{(p^2 + 4)^2}{4} = \dfrac{16p^2}{(p^2 + 4)^2}\overline{AF}^2 \implies$ $\overline{AF}^2 = \dfrac{(p^2 + 4)^4}{64p^2} \implies \overline{AF} = \dfrac{(p^2 + 4)^2}{8p}$

$\implies \overline{MF} = \dfrac{(p^2 + 4)(p^2 - 4)}{8p} \implies A_{\triangle{AMF}} =$ $\dfrac{(p^2 + 4)^2(p^2 - 4)}{32p} \implies$

$A_{AEDF} = A_{R} = A_{AEDM} - A_{\triangle{AMF}} = \dfrac{(p^2 + 4)^2(8p - p^2 + 4)}{32p}$ .

To show that there exists a unique prime value of $p$ that satisfies the above construction.

$\lambda = 2\theta - 90^{\circ} \implies m_{MF} = \tan(2\theta - 90^{\circ}) = \dfrac{-\cos(2\theta)}{\sin(2\theta)} = \dfrac{\sin^2(\theta) - \cos^2(\theta)}{2\sin(\theta)\cos(2\theta)} =$

$\dfrac{(p - 2)^2(4 + 4p - p^2)}{8p(p^2 - 4)} > 0$ and we already have $p > 2 \implies$

$p^2 - 4p - 4 < 0 \implies p = \lfloor{2(\sqrt{2} + 1)}\rfloor = 4 \implies (2 < p \leq 4)$

$\implies$ prime $p = 3 \implies A_{R} = \dfrac{3211}{96} = \boxed{33.4479166666666667}$ .