"In the right-angled triangle $$ A B C $$, one of the acute angles is three times the other. If the length of the altitude to the hypotenuse is 2 , what is the length of the hypotenuse?"

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To solve the problem, we start by defining the angles in the right-angled triangle $$ ABC $$.

1. **Define the angles:**
   Let the smaller acute angle be $$ \theta $$. Then, the other acute angle, which is three times the smaller angle, will be $$ 3\theta $$. Since the sum of angles in a triangle is $$ 180^\circ $$ and one angle is $$ 90^\circ $$, we have:
   $$
   \theta + 3\theta + 90^\circ = 180^\circ
   $$
   Simplifying this gives:
   $$
   4\theta = 90^\circ \implies \theta = 22.5^\circ
   $$
   Therefore, the angles in the triangle are $$ 22.5^\circ $$, $$ 67.5^\circ $$, and $$ 90^\circ $$.

2. **Use the altitude to the hypotenuse:**
   The altitude to the hypotenuse is given as $$ h = 2 $$. In a right triangle, the area can be expressed in two ways:
   - Using the base and height (hypotenuse as the base):
     $$
     \text{Area} = \frac{1}{2} \times \text{hypotenuse} \times h
     $$
   - Using the two legs:
     $$
     \text{Area} = \frac{1}{2} \times a \times b
     $$
   where $$ a $$ and $$ b $$ are the lengths of the legs.

3. **Relate the altitude to the hypotenuse:**
   The area can also be expressed in terms of the hypotenuse $$ c $$ and the altitude $$ h $$:
   $$
   \text{Area} = \frac{1}{2} \times c \times h
   $$
   Setting the two area expressions equal gives:
   $$
   \frac{1}{2} \times c \times 2 = \frac{1}{2} \times a \times b
   $$
   Simplifying this, we find:
   $$
   c = a \times b
   $$

4. **Use trigonometric relationships:**
   In triangle $$ ABC $$:
   - The side opposite $$ \theta $$ is $$ a = c \sin(22.5^\circ) $$
   - The side opposite $$ 3\theta $$ is $$ b = c \sin(67.5^\circ) $$

Using the sine values:
   $$
   \sin(22.5^\circ) = \frac{\sqrt{2 - \sqrt{2}}}{2}, \quad \sin(67.5^\circ) = \sin(90^\circ - 22.5^\circ) = \cos(22.5^\circ) = \frac{\sqrt{2 + \sqrt{2}}}{2}
   $$

5. **Substituting into the area equation:**
   We can express $$ c $$ in terms of $$ a $$ and $$ b $$:
   $$
   c = c \sin(22.5^\circ) \cdot c \sin(67.5^\circ
   $$
   This simplifies to:
   $$
   c = c^2 \cdot \frac{\sqrt{2 - \sqrt{2}}}{2} \cdot \frac{\sqrt{2 + \sqrt{2}}}{2}
   $$
   Thus:
   $$
   1 = c \cdot \frac{\sqrt{(2 - \sqrt{2})(2 + \sqrt{2})}}{4}
   $$
   Simplifying the square root:
   $$
   (2 - \sqrt{2})(2 + \sqrt{2}) = 4 - 2 = 2
   $$
   Therefore:
   $$
   1 = c \cdot \frac{\sqrt{2}}{4} \implies c = \frac{4}{\sqrt{2}} = 2\sqrt{2}
   $$

6. **Final answer:**
   The length of the hypotenuse $$ c $$ is:
   $$
   \boxed{2\sqrt{2}}
   $$
The length of the hypotenuse is $$ 2\sqrt{2} $$.

"In the right-angled triangle \( A B C \), one of the acute angles is three times the other. If the length of the altitude to the hypotenuse is 2 , what is the length of the hypotenuse?"

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