Geometry Problem 101 Part 2

Let prime $p > 2$ .

For right $\triangle{DEB} \:\ (p^2 - 4, 4p, p^2 + 4)$ is a primitive pythagorean triple $\implies AB = p^2 + 6p + 8$ .

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

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 $p^4 + 12p^3 - 76p^2 + 72p + 63 = 0.$

$p = 3 \implies C:(0,12), D:(30,12)$ and $B:(35,0) \implies m_{BC} = -\dfrac{12}{35} \implies \boxed{y = -\dfrac{12}{35}(x - 35)}$ and $m_{AD} = \dfrac{2}{5} \implies \boxed{y = \dfrac{2}{5}x}$

Solving the two equations above we obtain:

$(x_{0} = \dfrac{210}{13}, y_{0} = \dfrac{84}{13})$ and $x_{0} + y_{0} = \dfrac{294}{13} \implies m + n = \boxed{307}$ .

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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.