2
$\begingroup$

First time, I found a line associated with antipodal points, detail:

Let $ABC$ be a triangle, $(C)$ is circumconic of $ABC$. $P$ and $P'$ are two antipodal points. Construct three lines through $P'$ and parallel to $PA$, $PB$, $PC$ meets $BC$, $CA$, $AB$ respectively at three collinear points, the new line through the center of circumconic.

Question: Is a line associated with antipodal points above known?

enter image description here

Update: But the fact, the result is generalization of the Simson line, I reformulate as follows:

Let $ABC$ be a triangle, $P$ be a point in the plane, let $(C)$ is the Nine point conic of $A$, $B$, $C$, $P$. Let $O$ be arbitrary point on $C$, $P'$ is the reflection of $P$ in $O$. Then three lines through $P'$ and parallel to $PA$, $PB$, $PC$ meet three lines $BC$, $CA$, $AB$ respectively at three collinear point.

When $P$ is the orthorcenter, the line is the Simson line of $P'$

Question again: The generalization of the Simson line above is known?

enter image description here

See also:

$\endgroup$
7
  • 1
    $\begingroup$ This is one those problems that can be solved using the so-called $p,q$ method, after two reductions: 1) by affine invariance, we cann suppose that the circumconic is, in fact, the circumcentre. 2) we can assume that the vertices are $(0,0)$, $(1,0)$ and $(p,q)$ . We can then compute the coordinates of the auxiliary points and so apply the standard collinearity condition that the area of the corresponding triangle be zero. $\endgroup$
    – hordubal
    Oct 10 at 11:59
  • $\begingroup$ Yes, the solution maybe not hard. But my question that is it known? $\endgroup$ Oct 10 at 12:02
  • $\begingroup$ Sorry— misunderstood. No idea if it is known. $\endgroup$
    – hordubal
    Oct 10 at 12:03
  • $\begingroup$ Thank you for your comment. $\endgroup$ Oct 10 at 12:04
  • $\begingroup$ The simson line is very well-known, and have some nice properties, I hope that the line will know and have some nice properties $\endgroup$ Oct 10 at 12:05
1
$\begingroup$

This is not an answer, but I'd like to point out that the concepts in question are projective, although they have a special Euclidean case.

Consider the diagram below. Start with a conic $\gamma$(green), a triangle $ABC$ inscribed in $\gamma$, and a line $\omega$ (black dot-dashed). Let $X$ be the polar of $\omega$ wrt the conic, and draw a line (dotted) through $X$ that meets $\omega$ at $P,P'$. Let the dashed lines through $P'$ meet the respective lines from $P$ to $A,B,C$ at $\omega$. Then the dashed lines meet the triangle sides at collinear points (red), and $X$ lies on this line.

The OP is the special case when $\omega$ is the projective line at infinity.

So, if the line in OP Question 1 is known, and anybody is trying to hunt it down, it may be in the projective geometry literature.

enter image description here

$\endgroup$
1

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.