圆锥曲线，两个有6年历史的恐怖题目

amekui

amekui

图片来自网络（只有这一页） 这个引理看得懂但是仍然无法应用在这个问题上
不知道有没有人研究过这类问题

hbghlyj

所求的轨迹在给定的椭圆和圆的二次曲线系中(有四条公切线的二次曲线系)

所以，轨迹方程形如$\frac{x^2}{a_0^2}+\frac{y^2}{b_0^2}=1$
其中$\begin{vmatrix}a_0^2&b_0^2&1\\a^2&b^2&1\\1&1&1\end{vmatrix}=0$
代入一个点就能确定系数

amekui

回复 3# hbghlyj
具体怎么说？还有2楼的那个引理派得上用场吗？

hbghlyj

回复 4# amekui 抱歉刚才网卡没看到回复....
既然已知了曲线系，代入一个点就算出了啊......
第一题,图片中应该是把x,y弄反了,正确是\[\frac{x^2}{\left(\dfrac{a^2 b^2+a^2-b^2}{a^2 b^2-a^2-b^2}\right)^2}+\frac{y^2}{\left(\dfrac{a^2 b^2-a^2+b^2}{a^2 b^2-a^2-b^2}\right)^2}=1\]
第二题,
\[\frac{x^2}{\left(\dfrac{a \left(a^4+b^2\left(a^2-1\right)  \left(a^2 b^2-2 a^2+3 b^2\right)\right)}{a^4-b^2\left(a^2-1\right)  \left(3 a^2 b^2-2 a^2+b^2\right)}\right)^2}+\frac{y^2}{\left(\dfrac{b \left(b^4+a^2 \left(b^2-1\right) \left(a^2 b^2+3 a^2-2 b^2\right)\right)}{b^4-a^2 \left(b^2-1\right) \left(3 a^2 b^2+a^2-2 b^2\right)}\right)^2}=1\]
可以验证它确实在四切线的曲线系里面：\[\left| \begin{array}{ccc} \left(\dfrac{a \left(a^4+\left(a^2-1\right) b^2 \left(b^2 a^2-2 a^2+3 b^2\right)\right)}{a^4-\left(a^2-1\right) b^2 \left(3 b^2 a^2-2 a^2+b^2\right)}\right)^2 & \left(\dfrac{b \left(b^4+a^2 \left(b^2-1\right) \left(b^2 a^2+3 a^2-2 b^2\right)\right)}{b^4-a^2 \left(b^2-1\right) \left(3 b^2 a^2+a^2-2 b^2\right)}\right)^2 & 1 \\ a^2 & b^2 & 1 \\ 1 & 1 & 1 \\\end{array}\right|=0\]

amekui

回复 5# hbghlyj
彭赛列的定理有耳闻。不过对您的这个解释不是很理解，能否提供一些相关的参考资料？
（另外，有无不暴力计算的解析方法？）

amekui

回复 7# hbghlyj
您在确定轨迹的形状时使用了射影几何里的定理。
有无纯解析的方法？我对2楼中的引理如何应用在这个问题中很感兴趣。

kuing

回复 8# amekui

我对2楼中的引理如何应用在这个问题中很感兴趣。
amekui 发表于 2020-7-5 03:24

爆算呗……

交点在椭圆上，即
\[\frac13\left( \frac{1-t_1t_2}{1+t_1t_2} \right)^2+\frac12\left( \frac{t_1+t_2}{1+t_1t_2} \right)^2=1,\]去分母即
\[4-3t_1^2 + 10 t_1 t_2 - 3 t_2^2 + 4 t_1^2 t_2^2=0,\]同理对 `(t_2,t_3)` 和 `(t_3,t_4)` 有类似的式子，需要消掉 `t_2` 和 `t_3`，打开软件如 MMC(A)，输入

1. f[t1_, t2_] = 4 - 3 t1^2 + 10 t1 t2 - 3 t2^2 + 4 t1^2 t2^2
2. Resultant[f[t1, t2], f[t2, t3], t2] // Factor

复制代码

结果为
-(t1 - t3)^2 (1200 - 49 t1^2 + 2402 t1 t3 - 49 t3^2 + 1200 t1^2 t3^2)
因为 `t_1\ne t_3`，即 `1200 - 49 t_1^2 + 2402 t_1 t_3 - 49 t_3^2 + 1200 t_1^2 t_3^2 =0`，再次 Resultant，输入

1. Resultant[1200 - 49 t1^2 + 2402 t1 t3 - 49 t3^2 + 1200 t1^2 t3^2, f[t3, t4], t3] // Factor

复制代码

结果为
(4 - 3 t1^2 + 10 t1 t4 - 3 t4^2 + 4 t1^2 t4^2) (2896804 - 7216803 t1^2 + 15552970 t1 t4 - 7216803 t4^2 + 2896804 t1^2 t4^2)
因为最后一个交点不在椭圆上，所以前面的括号不为零，即
\[2896804 - 7216803 t_1^2 + 15552970 t_1 t_4 - 7216803 t_4^2 + 2896804 t_1^2 t_4^2=0,\]而最后交点的坐标
\[x=\frac{1-t_4t_1}{1+t_4t_1}, \, y=\frac{t_4+t_1}{1+t_4t_1},\]解得
\[t_1+t_4=\frac{2y}{x+1}, \, t_1t_4=\frac{1-x}{x+1},\]代入上面的式子中，最终得
\[6048242 x^2+7216803 y^2-8945046=0,\]这就是所求方程。

hbghlyj

回复 9# kuing
在5#的公式中代入$a=\sqrt3,b=\sqrt2$，结果是一样的

amekui

回复 9# kuing
啊这 真是暴力

amekui

再顶一下吧 求一个清晰点的解释

青青子衿

回复 1# amekui
椭圆中内切圆动点轨迹问题（二）
原创 湖南长沙胡晓昊 昊天数学工作室 3月5日
mp.weixin.qq.com/s?__biz=Mzg3NDI5NjA4Nw==&mid=2247483738& ... 8008df0bef325ad97128

amekui

回复 13# 青青子衿
这仍然只是给出了一个答案啊

amekui

唔

hbghlyj

发现《命题人讲座 解析几何》第166页有彭赛列定理的微分法证明

hbghlyj

相关问题

amekui

hbghlyj 发表于 2021-4-27 21:21

链接没了

hbghlyj

https://math.stackexchange.com/q ... rcle-and-an-ellipse
The problem can be generalized so that both curves (the circle and outer ellipse in the diagram) are conics. Then the resulting locus is also a conic. Although the subject matter and concepts are very much that of projective geometry, the methods of proof tend to be more analytic than synthetic.As described in Andrea Del Centina's Poncelet’s porism: a long story of renewed discoveries (also on JSTOR) the problem was studied in the 19th century, and is related to Poncelet's Theorem.Look for Fig. 21 in Del Centina's paper (pg 68) which illustrates the approach in George Salmon's On the problem of the In-and-circumscribed Triangle.
The problem can be generalized so that both curves (the circle and outer ellipse in the diagram) are conics. Then the resulting locus is also a conic. Although the subject matter and concepts are very much that of projective geometry, the methods of proof tend to be more analytic than synthetic.As described in Andrea Del Centina's Poncelet’s porism: a long story of renewed discoveries (also on JSTOR) the problem was studied in the 19th century, and is related to Poncelet's Theorem.Look for Fig. 21 in Del Centina's paper (pg 68) which illustrates the approach in George Salmon's On the problem of the In-and-circumscribed Triangle.

Salmon derives an equation for the locus as a weighted sum of the original conics (($U$ and $V$) and a third conic $F$. ($F$  has the remarkable property that the tangents from any point on $F$ to $U$ and $V$ form a harmonic pencil.)

Further on, look for Fig. 31, which illustrates a theorem by Darboux.



A tangential pencil is the set of all conics tangent to a four given lines (some of which may be complex), and is dual to an ordinary conic pencil: the set of all conics containing four given points (some of which may be complex).

If $K_3=K_2=K_1$,  the quadrilateral circumscribes $K$ (the inner circle/conic in the question) and three of its vertices lie on $K_1 $ (the outer ellipse/conic), as in the original question (see figure below).  Then Darboux's theorem says that the locus of the fourth vertex $Q$ is a conic in the tangential pencil $\mathbf T$ containing $K$ and $K_1$.



To compute the members of $\mathbf T$, let $K'$ and $K'_1$ be the dual conics of $K$ and $K_1$.  Then $\mathbf T$ is the set of duals of members of the pencil containing $K'$ and $K'_1$.

Finally, the paper points out that Darboux observed that his theorem was the dual of Poncelet's General Theorem.  In the comments, @JeanMarie has pointed out that this question has some similarity with another question math.stackexchange.com/q/3509582).  And indeed that one is the dual of this one and is a case of Poncelet's General Theorem.