Problem 6

A cone of revolution has an inscribed sphere tangent to its base and sloping surface. A cylinder is circumscribed about the sphere, with its base in the base of the cone. If the volumes of the cone and cylinder are

V_1

and

V_2

, respectively: (a) prove that

V_1\ne V_2

; (b) find the smallest possible value of

V_1/V_2

, and in this case construct the half-angle of the cone.

Step 4 of 4: Equality case and construction of the half-angle

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s=\sin\theta=\dfrac13\quad\Longrightarrow\quad\min\dfrac{V_1}{V_2}=\dfrac43

Detailed analysis

Equality holds exactly when $1-3s=0$ , so $s=1/3$ and the minimum ratio is $4/3$ . Construct a right triangle with hypotenuse $3$ and opposite leg $1$ ; the acute angle opposite that leg has $\sin\theta=1/3$ , so it is the required half-angle of the cone.