Let $$ABC$$ be a right-angled triangle with $$\angle B=90^{\circ}$$. Let the length of the altitude $$BD$$ be equal to $$12$$. What is the minimum possible length of $$AC$$, given that $$AC$$ and the perimeter of triangle $$ABC$$ are integers?

Let $$AB=x$$ and $$BC=y$$ be the two perpendicular sides and let the hypotenuse be $$AC=c$$.

1. Right-triangle relation: $$x^{2}+y^{2}=c^{2}$$ $$-(1)$$

2. Altitude formula (altitude from the right angle to the hypotenuse): $$BD=\dfrac{xy}{c}$$. Given $$BD=12$$,

$$\dfrac{xy}{c}=12 \;\Longrightarrow\; xy=12c$$ $$-(2)$$

3. Let the perimeter be $$P=x+y+c$$. Both $$P$$ and $$c$$ are required to be integers, so we must have

$$x+y=S\in\mathbb{Z}$$ $$-(3)$$

4. Relate $$S$$, $$c$$ using (1) and (2):

$$S^{2}=(x+y)^{2}=x^{2}+y^{2}+2xy=c^{2}+24c$$ $$\Longrightarrow\; S^{2}=c(c+24)$$ $$-(4)$$

Because $$S$$ is an integer, the right-hand side must be a perfect square. Hence we need an integer $$c$$ such that $$c(c+24)$$ is a perfect square.

5. Feasibility condition for real, positive $$x,\,y$$: Treat $$x,\,y$$ as the roots of $$t^{2}-St+12c=0$$. Its discriminant must be non-negative:

$$\Delta=S^{2}-48c=(c^{2}+24c)-48c=c^{2}-24c=c(c-24)\ge 0$$ Thus $$c\ge 24$$ $$-(5)$$

6. Search for the smallest integer $$c\ge 24$$ making $$c(c+24)$$ a square.

Try $$c=24$$: $$24\cdot48=1152\neq \text{perfect square}$$

Try $$c=25$$: $$25\cdot49=1225=35^{2}$$ ― perfect square found.

Therefore the minimum possible hypotenuse is $$c=25$$.

7. Verification: For $$c=25$$, equation (4) gives $$S=35$$. With $$S=35$$ and $$xy=12c=300$$, the quadratic $$t^{2}-35t+300=0$$ yields $$t=\dfrac{35\pm\sqrt{35^{2}-4\cdot300}}{2}=\dfrac{35\pm5}{2}$$, giving $$x=20$$ and $$y=15$$.

This is the familiar $$15\text{-}20\text{-}25$$ right triangle, and indeed $$BD=\dfrac{15\cdot20}{25}=12$$ as required, while the perimeter $$15+20+25=60$$ is an integer.

Hence the minimum possible length of $$AC$$ is 25.