Among the inscribed isosceles triangles, which one encloses the solid of the largest volume if the rotation occurs 1) around the base, 2) around a tangent passing through the vertex of the triangle?
(1) Let the base of the triangle be $2 y$, and the distance of the center of the circle from the base be $x$. If the rotation occurs around the base, then the volume of the resulting solid is:

$$
v_{1}=\frac{2}{3}(r+x)^{2} \pi y
$$

the volume is maximized when

$$
u=(r+x)^{2} y
$$

is maximized. But

$$
y=\sqrt{r^{2}-x^{2}}
$$

so

$$
u=(r+x)^{2} \sqrt{r^{2}-x^{2}}
$$

or

$$
u^{2}=(r+x)^{4}\left(r^{2}-x^{2}\right)
$$

or even

$$
u^{2}=(r+x)(r+x)(r+x)(r+x)(r+x)(r-x)
$$

Multiplying by 5

$$
5 u^{2}=(r+x)(r+x)(r+x)(r+x)(r+x)(5 r-5 x)
$$

Since the sum of the factors is a constant number: $10 r$, $5 u^{2}$ is maximized when the factors are equal (K.M.L.V.36. page), so the condition for the maximum is:

$$
r+x=5 r-5 x
$$

from which

$$
x=\frac{2}{3} r
$$

In this case, the height of the triangle is: $r+\frac{2}{3} r=\frac{5}{3} r$; the volume of the cone with the largest volume is:

$$
v_{1}=\frac{50}{81} \sqrt{5} \cdot r^{3} \pi
$$

(2) If the rotation occurs around the tangent passing through the vertex of the triangle, then

$$
v_{2}=2(r+x)^{2} \pi y-\frac{2}{3}(r+x)^{2} \pi y=\frac{4}{3}(r+x)^{2} \pi y
$$

we see that $v_{2}$ is also maximized when $x=\frac{2}{3} r$, that is, when the height of the triangle is: $x=\frac{5}{3} r$; the volume of the resulting solid in this case is:

$$
v_{2}=\frac{100}{81} \sqrt{5} \cdot r^{3} \pi
$$

(Gyula Lukhaub.)

The problem was also solved by: Krausz B., Krisztián Gy., Sasvári G., Weisz J.