We draw a circle passing through the vertices of triangle $A B C$. What is the surface area of the solid of revolution formed by the rotation of the side $B C$ about the tangent to the circle at point $A$? What is the volume of the solid formed by the rotation of the triangle about the same axis? Apply the results to the case where $A=90^{\circ}$.
The surface generated by the rotation of side $BC$ is the lateral surface of a frustum of a cone, thus:

$$
F=(B E+C F) \pi a
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_c2edcc3ebc2da95a5b4dg-1.jpg?height=420&width=545&top_left_y=370&top_left_x=779)

The diagram shows that

$$
F C A \varangle=A C O \varangle=y
$$

and

$$
O B A \varangle=A B E \varangle=x
$$

From triangles $A F C$ and $E A B$:

$$
F C=b \cos y, \quad B E=c \cos x
$$

but

$$
\cos y=\frac{b}{2 r}, \quad \cos x=\frac{c}{2 r}
$$

or since

$$
r=\frac{a}{2 \sin \alpha}=\frac{a b c}{4 t}
$$

it follows that

$$
\cos y=\frac{2 t}{a c}, \quad \cos x=\frac{2 t}{a b}
$$

and thus

$$
F C=\frac{2 b t}{a c}, \quad B E=\frac{2 c t}{a b}
$$

Substituting these into (1):

$$
F=2 \pi t\left(\frac{b}{c}+\frac{c}{b}\right)
$$

or

$$
F=2 \pi t \frac{b^{2}+c^{2}}{b c}
$$

The volume of the solid of revolution is:

$$
V=F \cdot \frac{m}{3}=2 \pi t \frac{b^{2}+c^{2}}{b c} \cdot \frac{a m}{3 a}=4 \pi t^{2} \frac{b^{2}+c^{2}}{3 a b c}
$$

If $A=90^{\circ}$, then $b^{2}+c^{2}=a^{2} \text{ and } b c=2 t$, so

$$
F=a^{2} \pi \quad \text { and } \quad V=\frac{a b c \pi}{3}
$$

(Freibauer Ede.)

The problem was also solved by: Bella I., Devecis M., Krausz B., Krisztián Gy., Sasvári G., Spitzer Ö.