1681

在直角三角形 ${\displaystyle ABC}$ 中, ${\displaystyle AC=2,BC=1}$, ${\displaystyle D}$ 为斜边 ${\displaystyle AB}$ 上一点, 若 ${\displaystyle \mathrm{\Delta }ACD}$ 与 ${\displaystyle \mathrm{\Delta }BCD}$ 的内切圆面积相等, 则 ${\displaystyle BD=}$ _____.

设 ${\displaystyle \mathrm{\Delta }ABC}$ 的内心为 ${\displaystyle I}$, 内切圆切 ${\displaystyle AB}$ 于 ${\displaystyle E}$, 内切圆半径为 ${\displaystyle r}$. 再设 ${\displaystyle \mathrm{\Delta }ACD,\mathrm{\Delta }BCD}$ 的内心为 ${\displaystyle I_1,I_2}$, 两个内切圆分别切 ${\displaystyle AB}$ 于 ${\displaystyle E_1}$, ${\displaystyle E_2}$, 内切圆半径均为 ${\displaystyle r^{\prime }}$. 设 ${\displaystyle k=\frac{r^{\prime }}{r}}$.
首先容易求出^[1], ${\displaystyle AE=\frac{\sqrt{5}+1}{2}}$, ${\displaystyle BE=\frac{\sqrt{5}-1}{2}}$, ${\displaystyle r=\frac{3-\sqrt{5}}{2}}$.

下面我们正式开始计算. 根据上面的引理, $$D{E}_2=E{E}_1=AE-A{E}_1=(1-k)AE,\text{同理}D{E}_1=(1-k)BE.$$ 又因为 ${\displaystyle \mathrm{\angle }{I}_{1}D{I}_2=\frac{\mathrm{\angle }ADC+\mathrm{\angle }BDC}{2}={90}^{\circ }}$, 所以 ${\displaystyle \mathrm{\Delta }{I}_1{E}_{1}D}$, ${\displaystyle \mathrm{\Delta }D{E}_2{I}_2}$ 相似. 这样 $$(r^{\prime }{)}^2=D{E}_{1}D{E}_2=(1-k{)}^{2}AE\cdot BE=(1-k{)}^2,$$ 也即 ${\displaystyle kr=1-k}$, 推出 ${\displaystyle k=\frac{1}{r+1}=\frac{5+\sqrt{5}}{10}}$. 从而 $$\begin{array}{rl}BD& =B{E}_2+E_{2}D=kBE+(1-k)AE\\ & =\frac{5+\sqrt{5}}{10}\cdot \frac{\sqrt{5}-1}{2}+\frac{5-\sqrt{5}}{10}\cdot \frac{\sqrt{5}+1}{2}=\frac{2\sqrt{5}}{5}.\end{array}$$

设两个内切圆半径为 ${\displaystyle r}$. 则因为 $$S_{\mathrm{\Delta }ACD}=\frac{1}{2}(AC+CD+AD)r,S_{\mathrm{\Delta }BCD}=\frac{1}{2}(BC+CD+BD)r,$$ 所以 $$\frac{S_{\mathrm{\Delta }ACD}}{S_{\mathrm{\Delta }BCD}}=\frac{AD}{BD}=\frac{AC+CD+AD}{BC+CD+BD}=\frac{AC+CD}{BC+CD},$$ (最后一步用了合分比定律), 于是 ${\displaystyle CD=\frac{2BD-AD}{AD-BD}}$. 设 ${\displaystyle \overrightarrow{BD}=\lambda \overrightarrow{BA}}$, 则 ${\displaystyle BD=\sqrt{5}\lambda }$, ${\displaystyle AD=\sqrt{5}(1-\lambda )}$, 于是 $$\begin{array}{}\text{(1)}& CD=\frac{3\lambda -1}{1-2\lambda }.\end{array}$$. 又由 ${\displaystyle \overrightarrow{CD}=\lambda \overrightarrow{CA}+(1-\lambda )\overrightarrow{CB}}$, 于是 $$\begin{array}{}\text{(2)}& C{D}^2=4{\lambda }^2+(1-\lambda {)}^2.\end{array}$$ 联立 (1), (2), 得$$\begin{array}{rl}& {\left(\frac{3\lambda -1}{1-2\lambda }\right)}^2=4{\lambda }^2+(1-\lambda {)}^2\\ \Rightarrow & 20{\lambda }^4-28{\lambda }^3+8{\lambda }^2=4{\lambda }^2(\lambda -1)(5\lambda -2)=0.\end{array}$$解得 ${\displaystyle \lambda =\frac{2}{5}}$ (舍去 ${\displaystyle 0}$ 和 ${\displaystyle 1}$), 于是 ${\displaystyle BD=\frac{2\sqrt{5}}{5}}$.