(1) Let $\triangle ABC$ be an isosceles triangle with the base a = BC = 2R and the A-altitude to the base equal to h. The shoulder sides AB = AC are equal to $b = \sqrt{R^2 + h^2}$ and the angle $\alpha$ opposite to the base is given by $\tan \frac \alpha 2 = \frac R h$. The triangle semiperimeter is $s = b + R$ and its inradius r is equal to $r = (s - a) \tan \frac \alpha 2 = (b + R - 2R) \tan \frac \alpha 2 = \left(\sqrt{R^2 + h^2} - R\right) \frac R h$ Let $\square PQRS$ be a square circumscribed around the incircle (I), such that $P, Q \in BC$ and let the square sides SP, QR intersect the triangle sides AB, AC at points K, L and let the square side RS intersect the triangle sides AB, AC at points M, N. The right angle triangles $\triangle MSK \sim \triangle BPK$ are similar, having equal vertical angles at the common vertex K. Since the lines AB, SP intersecting at the point K are tangent to the incircle (I), the line KI bisects the angle $\angle AKP = 180^\circ - \frac \alpha 2$. Therefore, $\angle PKI = \frac{\angle AKP}{2} = 90^\circ - \frac \alpha 4 < 90^\circ$, $KP > r = \frac{SP}{2}$ and since SK + KP = SP, $SK < r = \frac{SP}{2}$. Thus the areas $S(\triangle MSK) < S(\triangle BPK)$ and similarly the areas $S(\triangle NRL) < S(\triangle CQL)$. As a result, the area of the square $\square PQRS$ is less than the area of the triangle $\triangle ABC$: $S(\square PQRS) = S(\triangle MSK) + S(\triangle NRL) + S(PQLNMK) <$ $< S(\triangle BPK) + S(\triangle CQL) + S(PQLNMK) <$ $< S(\triangle BPK) + S(\triangle CQL) + S(PQLNMK) + S(\triangle AMN) = S(\triangle ABC)$ If the isosceles triangle $\triangle ABC$ and the square $\square PQRS$ are rotated around the A-altitude, they result in the defined cone and cylinder. Since all interior points of the triangles $\triangle BPK, \triangle CQL$ are farther away from the axis of rotation than the interior points of the triangles $\triangle MSK, \triangle NRL$ and the 1st pair has larger areas than the 2nd pair, the 1st pair makes a larger contribution to the cone volume $V_1$ than the 2nd pair to the cylinder volume $V_2$. Asa result, $V_1 > V_2$. (2) Denote $x = \tan \frac \alpha 2 = \frac R h$. The cone and cylinder volumes are $V_1 = \frac{\pi hR^2}{3} = \frac{\pi h^3x^2}{3}$ $V_2 = 2 \pi r^3 = 2\pi h^3x^3\left(\sqrt{x^2 + 1} - x\right)^3$ and their ratio is $q = \frac{V_2}{V_1} = 6x\left(\sqrt{x^2 + 1} - x\right)^3$ It is clear that $q \longrightarrow 0$ for both $x \longrightarrow 0$ and for $x \longrightarrow \infty$. The maximum q is given by $\frac{\text dq}{\text dx} = 0$ $6\left(\sqrt{x^2 + 1} - x\right)^3 + 18x\left(\sqrt{x^2 + 1} - x\right)^2 \left(\frac{x}{\sqrt{x^2 + 1}} - 1\right) = 0$ $\sqrt{x^2 + 1} - x + \frac{3x^2}{\sqrt{x^2 + 1}} - 3x = 0$ $4x^2 + 1 = 4x \sqrt{x^2 + 1}$ $16x^4 + 8x^2 + 1 = 16x^4 + 16x^2$ $8x^2 = 1,\ \ \ x = \tan \frac \alpha 2 = \frac{\sqrt 2}{4},\ \ \ \alpha \doteq 38.94^\circ$ The angle $\frac \alpha 2$ can be constructed by constructing a right angle triangle with the legs $\sqrt 2$ and 4. The maximum volume ratio is then: $q_{\text{max}} = \frac{V_2}{V_1} = \frac{6\sqrt 2}{4}\left(\sqrt{\frac 1 8 + 1} - \frac{\sqrt 2}{4}\right)^3 = \frac{3 \sqrt 2}{2 (2 \sqrt 2)^3} (3 - 1)^3 = \frac 3 4$ and the minimum reciprocal volume ratio $k_{\text{min}} = \frac{V_1}{V_2} = \frac{1}{q_{\text{max}}} = \frac 4 3$

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Solution. Let $ O$, $ A$, $ B$, $ \theta$, and $ r$ be the cone's vertex, cone's base's center, a point on the cone's base's circle, $ \angle AOB$, and the radius of the sphere, respectively. Then it follows, from basic trigonometry, that $ OA = \frac {r(1 + \sin\theta)}{\sin\theta}$ and $ AB = \frac {r(1 + \sin\theta)}{\cos\theta}$. Hence, $ V_1 = \frac {\pi AB^2OA}{3} = \frac {\pi\left(\frac {r(1 + \sin\theta)}{\cos\theta}\right)^2\left(\frac {r(1 + \sin\theta)}{\sin\theta}\right)}{3}$ and $ V_2 = 2\pi r^3$. Ergo, \[ k = \frac {\pi\left(\frac {r(1 + \sin\theta)}{\cos\theta}\right)^2\left(\frac {r(1 + \sin\theta)}{\sin\theta}\right)}{6\pi r^3}\implies (1 + 6k)\sin^2\theta + (2 - 6k)\sin\theta + 1 = 0. \] However, the discriminant of the above quadratic equation must be nonnegative, that is, $ (1 - 3k)^2\geq 1 + 6k\implies k\geq\frac {4}{3}$. And the conclusion follows. $ \Box$