In a triangle $$ABC, \angle BAC = 90^\circ$$. Let $$D$$ be the point on $$BC$$ such that $$AB + BD = AC + CD$$. Suppose $$BD : DC = 2 : 1$$. If $$\frac{AC}{AB} = \frac{m+\sqrt{p}}{n}$$, where $$m, n$$ are relatively prime positive integers and $$p$$ is a prime number, determine the value of $$m + n + p$$.

Let the perpendicular sides of the right-angled triangle be $$AB = x$$ and $$AC = y$$. Then $$BC = \sqrt{x^{2}+y^{2}}$$ by the Pythagorean theorem.

The point $$D$$ divides $$BC$$ internally in the ratio $$BD : DC = 2 : 1$$. Write $$BD = 2k$$ and $$DC = k$$, so $$BC = BD + DC = 3k \;\Rightarrow\; k = \dfrac{\sqrt{x^{2}+y^{2}}}{3}$$.

The given condition is $$AB + BD = AC + CD.$$ Substituting the lengths,

$$x + 2k = y + k \;\Longrightarrow\; x - y + k = 0 \;\Longrightarrow\; y - x = k.$$

Replace $$k$$ with its value:

$$y - x = \dfrac{\sqrt{x^{2}+y^{2}}}{3} \quad -(1)$$

Introduce the ratio $$r = \dfrac{y}{x}$$. Then $$y = rx$$ and equation $$-(1)$$ becomes

$$rx - x = \dfrac{\sqrt{x^{2}+r^{2}x^{2}}}{3}.$$

Simplify each side:

$$x(r-1) = \dfrac{x\sqrt{1+r^{2}}}{3}.$$

Because $$x \gt 0$$, divide both sides by $$x$$:

$$3(r-1) = \sqrt{1+r^{2}}.$$

Square both sides to remove the square root:

$$9(r-1)^{2} = 1 + r^{2}.$$

Expand and collect like terms:

$$9(r^{2} - 2r + 1) = 1 + r^{2}$$ $$9r^{2} - 18r + 9 = 1 + r^{2}$$ $$8r^{2} - 18r + 8 = 0.$$

Divide by 2 for simpler coefficients:

$$4r^{2} - 9r + 4 = 0.$$

Solve the quadratic using the discriminant method:

$$r = \dfrac{9 \pm \sqrt{81 - 64}}{8} = \dfrac{9 \pm \sqrt{17}}{8}.$$

We need $$y \gt x$$ (because $$y - x = k \gt 0$$), so $$r \gt 1$$. Choose the positive root:

$$\dfrac{AC}{AB} = r = \dfrac{9 + \sqrt{17}}{8}.$$

Thus $$m = 9$$, $$n = 8$$ and $$p = 17$$ (where $$p$$ is prime). Therefore, $$m + n + p = 9 + 8 + 17 = 34$$.

Final answer: 34