本帖最后由 青青子衿 于 2018-8-28 14:46 编辑
回复 4# lemondian
另外:期待有没有更简洁的方法,和其它两个问题的解答。
lemondian 发表于 2018-8-24 23:21
第三问的一个例子:
\[ \left(x-\frac{9\left(549-200\sqrt{3}\right)}{3346}\right)^2+\left(y-\frac{30\left(17+43\sqrt{3}\right)}{1673}\right)^2=\left(\frac{10\left(73+30\sqrt{3}\right)}{239}\right)^2 \]
\begin{align*}
L_{a}:y&=\phantom{+}\frac{20\left(147+113\sqrt{3}\right)}{2277}x\\
\\
L_{b}:y&=-\frac{5\left(147+305\sqrt{3}\right)}{3652}x+\frac{105\left(89\sqrt{3}-5\right)}{7304}\\
\\
L_{c}:y&=\phantom{+}\frac{4\left(118\sqrt{3}-147\right)}{715}x+\frac{42\left(4-\sqrt{3}\right)}{65}
\end{align*}
【2018年8月28日补充】
建立平面直角坐标系\(xOy\),点\(A(\frac{39}{14},\frac{20\sqrt{3}}{7})\),点\(B(-\,\frac{7}{2},0)\),点\(C(\,\frac{7}{2},0)\),点\(O_1(0,0)\),点\(O_2(-\frac{5}{14},\frac{10\sqrt{3}}{7})\),点\(O_3(\frac{44}{7},\frac{10\sqrt{3}}{7})\);
以点\(O_1(0,0)\)为圆心作半径为\(\frac{7}{2}\)的圆\(\varGamma_1\colon x^2+y^2=\left(\frac{7}{2}\right)^2\),
以点\(O_2(-\frac{5}{14},\frac{10\sqrt{3}}{7})\)为圆心作半径为\(4\)的圆\(\varGamma_2\colon (x+\frac{5}{14})^2+(y-\frac{10\sqrt{3}}{7})^2=4^2\),
以点\(O_3(\frac{22}{7},\frac{10\sqrt{3}}{7})\)为圆心作半径为\(\frac{5}{2}\)的圆\(\varGamma_3\colon (x-\frac{22}{7})^2+(y-\frac{10\sqrt{3}}{7})^2=\left(\frac{5}{2}\right)^2\);
三圆的外切公切圆\(\varGamma_{ABC}\)的圆心为\(\displaystyle O_{1,2,3}\left(\frac{9\left(549-200\sqrt{3}\right)}{3346},\frac{30\left(17+43\sqrt{3}\right)}{1673}\right)\),半径为\(\displaystyle R_{1,2,3}=\frac{10\left(73+30\sqrt{3}\right)}{239}\)
与圆\(\varGamma_1\)的切点为\(\displaystyle P\left(\frac{3\left(339-400\sqrt{3}\right)}{962},-\frac{10\left(113+27\sqrt{3}\right)}{481}\right)\);
与圆\(\varGamma_2\)的切点为\(\displaystyle Q\left(-\frac{169+6000\sqrt{3}}{3206},\frac{10\left(366+269\sqrt{3}\right)}{1603}\right)\);
与圆\(\varGamma_3\)的切点为\(\displaystyle R\left(\frac{8911+2400\sqrt{3}}{2366},\frac{10\left(177+119\sqrt{3}\right)}{1183}\right)\); |