Triangle Craze

Finding the general area for this particular problem simplifies the work.

Using the law of sines on $\triangle{ABC}$ we obtain:

$\sin(A) = \dfrac{a}{b}\sin(B)$

$\sin(C) = \dfrac{c}{b}\sin(B)$

Using the law of cosines on $\triangle{A^{*}B^{*}C^{*}}$ we obtain:

$\sin^2(B) = \dfrac{a^2}{b^2}\sin^2(B) + \dfrac{c^2}{b^2}\sin^2(B) - \dfrac{2ac}{b^2}\sin^2(B)\cos(A^{*})$

$\implies \cos(A^{*}) = \dfrac{a^2 + c^2 - b^2}{2ac} = \cos(B) \implies \sin(A^{*}) = \sqrt{1 - \cos^2(B))} = \sin(B) \implies h^{*} = \sin(A)\sin(B) \implies$

$A_{\triangle{A^{*}B^{*}C^{*}}} = \dfrac{1}{2}\sin(A)\sin(B)\sin(C) = \boxed{\dfrac{1}{2}(\dfrac{ac}{b^2})\sin^3(B)}$ .

Let prime $p > 2$ .

$(p^2 - 4, 4p, p^2 + 4)$ is a primitive pythagorean triple with $DB = p^2 - 4 \implies AD = p^2 + 6p + 8$ .

For right $\triangle{ACD}$ : $(AC)^2 = (12p + 1)^2 = (p^2 + 6p + 8)^2 + (4p)^2 \implies p^4 + 12p^3 - 76p^2 + 72p + 63 = 0$ .

By inspection $p = 3$ is a root of $p^4 + 12p^3 - 76p^2 + 72p + 63 = 0$ .

Performing long division we obtain: $(p - 3) * (p^3 + 15p^2 - 31p - 21) = 0 \implies p = 3$ and prime $p \geq 3 \implies f(p) = p^3 + 15p^2 - 31p - 2$ 1 is monotonic increasing $\implies f(p) \geq f(3) = 48 \implies p = 3$ is the only prime root satisfying $p^4 + 12p^3 - 76p^2 + 72p + 63 = 0.$

Using the general area derived above we obtain:

$\sin(B) = \dfrac{12}{13} \implies A_{\triangle{A^{*}B^{*}C^{*}}} = \dfrac{1}{2}\dfrac{(13)(40)}{37^2} (\dfrac{12}{13})^3 = \dfrac{(20)(12^3)}{481^2} = \dfrac{34560}{231361} = \dfrac{m}{n} \implies m + n = \boxed{265921}$ .

$$

$$

$$

Note: $(p^2 - 4, 4p, p^2 + 4)$ is a primitive pythagorean triple for any prime $p > 2$ and $(12,35,37)$ is also a primitive pythagorean triple.