Among the right-angled triangles with an 8 unit perimeter, which one has the smallest hypotenuse?
I. solution. Let the hypotenuse of the right triangle be denoted by $c$, and one of its acute angles by $\alpha$. Then the perimeter is $c+c \cdot \sin \alpha+c \cdot \cos \alpha=8$, and thus

$$
c=\frac{8}{1+\sin \alpha+\cos \alpha}=\frac{8}{1+\sin \alpha+\sin \left(90^{\circ}-\alpha\right)}
$$

Based on the identity $\sin \alpha+\sin \beta=2 \cdot \sin \frac{\alpha+\beta}{2} \cdot \cos \frac{\alpha-\beta}{2}$:

$$
c=\frac{8}{1+2 \cdot \sin 45^{\circ} \cdot \cos \left(\alpha-45^{\circ}\right)}
$$

From this form, it is clear that $c$ is minimal precisely when $0^{\circ}<\alpha<90^{\circ}$, and $\alpha=45^{\circ}$. In this case, $c=8-4 \sqrt{2}$, and the other leg is also $8-4 \sqrt{2}$.

II. solution. Let the legs of the right triangle be $a$ and $b$, and the hypotenuse be $c$. Then $a+b+c=8$, and $a^2+b^2=c^2$. Substituting $b=8-a-c$ into the second equation, we get $c^2=(8-a-c)^2+a^2$, which simplifies to $c=8-a+\frac{32}{8-a}$. We can estimate the sum $8-a+\frac{32}{8-a}$ using the arithmetic and geometric mean inequality:

$$
8-a+\frac{32}{8-a} \geq 2 \cdot \sqrt{(8-a) \cdot \frac{32}{8-a}}=2 \sqrt{32}
$$

Therefore, $c$ is minimal precisely when $8-a=\frac{32}{8-a}$, i.e., when $a=8-4 \sqrt{2}$. In this case, $c=8 \sqrt{2}-8$, and with some calculation, it follows that the other leg is also $8-4 \sqrt{2}$.

III. solution. We prove the following generalization of the problem: If the perimeter of a triangle is $K$ and one of its angles is $\gamma$, then the side $c$ opposite $\gamma$ is minimal precisely when the other two sides are of equal length.

Using the usual notation and the sine rule:

$$
K=\frac{c \cdot \sin \alpha}{\sin \gamma}+\frac{c \cdot \sin \beta}{\sin \gamma}+c, \quad \text { i.e., } \quad K=\frac{c}{\sin \gamma}(\sin \alpha+\sin \beta+\sin \gamma)
$$

Thus,

$$
c=\frac{K \cdot \sin \gamma}{\sin \alpha+\sin \beta+\sin \gamma}
$$

Similarly to what was seen in the first solution,

$$
\sin \alpha+\sin \beta=2 \cdot \sin \frac{180^{\circ}-\gamma}{2} \cdot \cos \frac{\alpha-\beta}{2}
$$

Therefore, considering that $\sin \frac{180^{\circ}-\gamma}{2}>0, c$ is minimal precisely when $\cos \frac{\alpha-\beta}{2}$ is maximal, i.e., when $\alpha=\beta$.