906

求正四棱锥的外接球半径与内切球半径之比的最小值.

不妨设底边边长为 ${\displaystyle 2a}$, 设高 ${\displaystyle h=1}$. 我们找出内切圆、外接圆圆心对应的特征三角形(分别对应图中的 ${\displaystyle \text{蓝色}}$、${\displaystyle \text{红色}}$ 三角形). 如图，${\displaystyle I}$ 是角平分线与 ${\displaystyle y}$ 轴的交点，由角平分线定理得$$\begin{array}{r}\frac{r}{1-r}=\frac{a}{\sqrt{a^2+1}}\Rightarrow r=\frac{a}{a+\sqrt{a^2+1}}.\end{array}$$
${\displaystyle O}$是斜边中垂线${\displaystyle y-{\displaystyle \frac{1}{2}=\sqrt{2}a(x-\frac{\sqrt{2}}{2}a)}}$与${\displaystyle y}$轴的交点，故令${\displaystyle x=0}$, 得$$\begin{array}{r}R=1-y{|}_{x=0}=\frac{1}{2}+a^2.\end{array}$$
设${\displaystyle a={\displaystyle \frac{1}{2}(u-\frac{1}{u}),u>1}}$，有$$\begin{array}{r}\frac{R}{r}=\frac{\frac{1}{4}(u^2+\frac{1}{u^2})}{{\displaystyle \frac{\frac{1}{2}(u-\frac{1}{u})}{\frac{1}{2}(u-\frac{1}{u})+\frac{1}{2}(u+\frac{1}{u})}}}=\frac{1}{2}\cdot \frac{u^4+1}{u^2-1}=\frac{1}{2}((u^2-1)+\frac{2}{u^2-1}+2)\ge \sqrt{2}+1.\end{array}$$