Geometry Problem 101

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

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 $p^4 + 12p^3 - 76p^2 + 72p + 63 = 0 \implies d = \boxed{37}$ .

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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.