12. A sphere is placed inside a regular octahedron of side length 6 . Find the greatest possible radius of the sphere.
(2 marks)
一個球體被放進一個邊長為 6 的正八面體內。求該球體半徑的最大可能值。
(2 分)
12. $\sqrt{6}$
12. Each face of the octahedron is an equilateral triangle of side length 6 , the altitude of which is $6 \sin 60^{\circ}=3 \sqrt{3}$. Consequently, half of the octahedron (i.e. the square pyramid in which every edge has length 6) has altitude $\sqrt{(3 \sqrt{3})^{2}-3^{2}}=3 \sqrt{2}$, i.e. the volume of the octahedron is
$$
2 \times \frac{1}{3} \times 6^{2} \times 3 \sqrt{2}=72 \sqrt{2} \text {. }
$$

On the other hand, if we let $r$ be the radius of the inscribed sphere of the regular octahedron and $A$ be the surface area of the octahedron, then the volume of the octahedron would also be equal to $\frac{1}{3} A r$. Now $A=8 \times \frac{1}{2} \times 6 \times 3 \sqrt{3}=72 \sqrt{3}$. Hence we have $\frac{1}{3} \times 72 \sqrt{3} \times r=72 \sqrt{2}$, which gives $r=\sqrt{6}$. This is also the greatest possible radius of a sphere which can be placed inside the tetrahedron.