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按题意,也就是求斯坦纳椭圆的两个焦点

Posted by Eov_Second at 2017-01-12 14:54:13 on Problem 3793
三角形中面积最大的内切椭圆被称为斯坦纳椭圆,该椭圆与三角形相切于各边的中点,也就是题目中提到的椭圆。

记三角形三个顶点A,B,C是复平面上的三点,则椭圆焦点是函数
F(Z) = (Z-A)(Z-B)(Z-C)导数的两个零点,也就是焦点Z满足方程
F'(Z) = 3Z^2 - 2(A+B+C)*Z + AB + BC + CA = 0
记-2(A+B+C) = a+bi, AB + BC + CA = c+di, Z = x+yi(a,b,c,d,x,y均为实数)
代入上式,整理得到
3(x^2-y^2)+ax-by+c + (6xy+ay+bx+d)i = 0
实部虚部都为0,我们得到了一个二元方程组
3(x^2-y^2) + ax - by + c = 0
6xy + ay + bx + d = 0

利用牛顿迭代法可以求出一组解,利用椭圆的中心恰好是三角形重心的特性,可以很容易根据中点公式得到另一个焦点

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