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[几何] 配极三角形是透视的

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hbghlyj posted 2022-11-23 07:34 |Read mode
Last edited by hbghlyj 2022-11-23 16:05Elmer G. Rees Notes on Geometry (Universitext) Springer-Verlag 1977 page 78
37. Consider four lines in P2, that meet in six points. Label the six points as a, a1, b, b1, c, c1 in such a way that a and a1 do not lie on any one line and similarly for b, b1 and c, c1. Let C be a non-singular conic such that a, a1 and b, b1 are polar pairs with respect to C. Prove that c, c1 are also polar pairs.
Hence or otherwise show that a triangle and its polar are in perspective, that is the lines aa1, bb1, cc1 are concurrent.

令 C 为非奇异二次曲线,使得 a、a1 和 b、b1 关于 C 是配极点对。证明 c、c1 也是配极点对。
如何证明呢
Chasles's Polar Triangle Theorem

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Ly-lie posted 2022-11-23 11:42
见纯几何吧4444:tieba.baidu.com/p/6820667955?pn=5
屏幕截图 2022-11-23 114115.png

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original poster hbghlyj posted 2022-11-23 21:37
全美经典学习指导系列(Schaum's Outline Series) -  射影几何的理论和问题(Theory and Problems of Projective Geometry)
Chapt. 9 Poles and Polar lines
例题9.5 Prove: If two pairs of opposite vertices of a complete quadrilateral are conjugate points with respect to a conic, so also is the third pair of opposite vertices.
In Fig.9-5, let $A$ and $L$, $B$ and $M$ be the two pairs of opposite vertices of the complete quadrilateral $pqrs$ which are conjugates with respect to a conic $C$ not shown. We are to prove that $D$ and $N$ are also conjugates with respect to the same conic. Denote by $A', B', D'$ the respective intersections of polar lines of $A,B,D$ with respect to $C$ and the side $r$. Now the points $A',B',D'$ are also the harmonic conjugates of the points $A,B,D$ with respect to the two points of $C$ on the line $r$; hence $A,A';B,B';D,D'$ are three reciprocal pairs of an involution of the points on $r$.
By Theorem 6.4, page 66, the lines $A'L,B'M,$ and $D'N$ are on a point $K$. Since both $A'$ and $L$ are on the polar line of $A$ with respect to $C$, that polar line is $A'L$. Similarly, $B'M$ is the polar line of $B$ with respect to $C$ and then, $K$ is the pole of $r$. Also, $D$ is on $r$ and $D'$ is on its polar line with respect to $C$; hence $D'K$ is the polar line of $D$. Finally, since $D'K$ is on $N$, the points $D$ and $N$ are conjugates with respect to $C$.
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