"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?"

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: \[ \boxed{2\sqrt{2}} \]