XXX OM - I - Task 8

Among the cones inscribed in the sphere $ B $, the cone $ S_1 $ was chosen such that the sphere $ K_1 $ inscribed in the cone $ S_1 $ has the maximum volume. Then, a cone $ S_2 $ of maximum volume was inscribed in the sphere $ B $, and a sphere $ K_2 $ was inscribed in the cone $ S_2 $. Determine which is greater: the sum of the volumes of $ S_1 $ and $ K_1 $, or the sum of the volumes of $ S_2 $ and $ K_2 $.
Let $ S $ be any cone inscribed in a given sphere $ B $, and $ K $ - the sphere inscribed in the cone $ S $. The axial section of the cone $ S $ is an isosceles triangle $ ACD $, where $ A $ is the vertex of the cone (Fig. 7).
om30_1r_img_7.jpg
Let $ O $ be the center of the sphere $ B $, and $ Q $ - the center of the sphere $ K $. Then $ O $ is the center of the circle circumscribed around the triangle $ ACD $, and $ Q $ - the center of the circle inscribed in this triangle. Let $ P $ be the midpoint of the segment $ \overline{CD} $, and $ M $ and $ N $ respectively - the orthogonal projections of the points $ Q $ and $ O $ onto the line $ AD $.
Let $ \alpha = \measuredangle PAD $, $ R = AO $. We have the following relationships:


















Therefore, the volume $ V_s $ of the cone $ S $ is equal to






or






where we have taken $ a = \sin \alpha $.
We then have $ MQ = QP $. Thus $ AP \sin \alpha = QP \sin \alpha + AQ \sin \alpha = MQ \sin \alpha + MQ $ and therefore






The volume $ V_K $ of the sphere $ K $ is then equal to






Therefore, the sum $ V = V(a) $ of the volumes of the cone $ S $ and the sphere $ K $ is, by (1) and (2):






From the conditions of the problem, it follows that $ 0 < \alpha < \frac{\pi}{2} $. Therefore, the number $ a = \sin \alpha $ belongs to the interval $ (0,1) $.
In this interval, the function $ f(a) = a (1 - a^2) $ attains its maximum value at $ a = \frac{\sqrt{3}}{3} $, since this is the only zero of the derivative of $ f $ in this interval, and $ f(0) = f(1) = 0 $ and $ f(a) > 0 $ for $ 0 < a < 1 $. Therefore, from formula (1), it follows that the cone $ S $ has the maximum volume at $ a = \sin \alpha = \frac{\sqrt{3}}{3} $.
Similarly, we observe that the function $ g(a) = a(1 - a) $ in the interval $ (0, 1) $ attains its maximum value at $ a = \frac{1}{2} $. Therefore, from formula (2), it follows that the sphere $ K $ has the maximum volume at $ a = \sin \alpha = \frac{1}{2} $, i.e., at $ \alpha = \frac{\pi}{6} $.
Next, we calculate the values of the function $ V(a) $ defined by formula (3) at the points $ a = \frac{\sqrt{3}}{3} $ and $ a = \frac{1}{2} $. We obtain:












It is easy to verify that $ \frac{4}{9} (2\sqrt{3} - 3) > \frac{13}{64} $, and thus $ V \left( \frac{\sqrt{3}}{3} \right) > V \left( \frac{1}{2} \right) $. Therefore, the sum of the volumes of $ S_2 $ and $ K_2 $ is greater than the sum of the volumes of $ S_1 $ and $ K_1 $.