Problem: In the perspective plane $\mathbb{P}^2$, given a conic $(S)$ and a line $d$ having no intersection with $(S)$. Fixed two points $I$ and $J$ on $d$. Take a variety point $M$ such that the tangents from $M$ to $(S)$ intersect $d$ at two points $R$ and $S$ such that $(IJRS)=-1$. Prove the locus of $M$ is a conic passing through $I$ and $J$.

Here is what I've discovered so far.

  1. Let the tangents from $I$ and $J$ touch $(S)$ at $I_1,I_2$ and $J_1,J_2$ respectively. Then the locus of $M$ is the conic passing through $I,I_1,I_2,J,J_1,J_2$. I also proved those 6 points lie on a conic (let's call it $(S')$).

  2. Take any point on $(S')$, call $A$. Then the intersection $A_1, A_2$ of two tangents from $A$ to $(S)$ and $d$ satisfy $(IJ,A_1A_2)=-1$. Now I'm stucking at proving this claim.

2'. Something that might be useful for proving 2: Let $P$ the polar of $d$ wrt $(S)$. Then $PI$ and $PJ$ tangent to $(S')$.

Can I have some hints for the next steps? Any ideas would be appreciated. Thanks for reading.

  • $\begingroup$ I'm so sorry. Thanks for answering all my problems. I haven't noticed the button until you mention. $\endgroup$
    – RopuToran
    Jan 26 at 15:23

The other answer here uses a proof by Milne that claims "range (c)=range(d)". What does that mean?

On page 29 the notations "range (a)" and "pencil P(a)" are defined. For our purposes, a "range" is any collection of collinear points and a "pencil" is any collection of concurrent lines. Often ranges and pencils contain just four elements so that we can take cross ratios, but in general they contain an arbitrary number of elements.

For shortness we shall often denote a range (abcde ...) by (a), and a pencil P{abcde ...) by P(a)

(Chapter IV is worth a read, or at least a skim)

Unfortunately, I cannot find any explicit definition by Milne of what is meant by "range(a)=range(b)" or "pencil P(a)= pencil P'(a')". From context is seems to be that "range(a)=range(b)" means the two are homographic. In fact, on page 187 pencils and ranges are said to be equal (homographic) as in:

the pencil $\rho(m)$ = the range of poles $(m)$ = the pencil of polars $P(\mu)$

Does this help? Getting back to your question, I interpret

And the range (c) = the range (d), therefore the pencil A(P) = the pencil B(P),


"Because range(c) is homographic to range(d) the pencil A(P) is homographic to B(P)"

In all of the above, ranges and pencils are held to be homographic because they preserve cross ratios. In particular, you should verify that the conjugacy $c \to d$ preserves cross ratio and thus is homographic.

  • $\begingroup$ Now the problem becomes much brighter. I still don't know why they are homographic but as I'm gradually reading the book, I believe I will find out the answer soon. Thanks for helping me and thanks for the awesome book. $\endgroup$
    – RopuToran
    Mar 1 at 8:39
  • 1
    $\begingroup$ Yes, glad you like the book. There's another book by Hatton that I like that also contains the theorem - look at the bottom right of pg 295: archive.org/details/cu31924060184045/page/n309/mode/2up. The proof is on the following page, right hand column. It's a slightly more elegant version than Milne's, but you still have to do some work to understand it. "projective" means "homographic" Let me know if you need more explanation.. $\endgroup$
    – brainjam
    Mar 2 at 6:39

This locus is discussed in Milne, Cross Ratio Geometry, Article 185.

There, $P,A,B$ correspond to OP's $M,I,J$ and the condition that the lines $PA,PB$ are conjugate is equivalent to OP's $R,S$ being harmonic with $I,J$.

The proof relies on the locus conic being defined by the intersections of two homographic pencils centered at $A$ and $B$.

The excerpt is below. But the full reference is given above if background and definitions need to be consulted.

enter image description here

I've added a discussion of the claim "range ($c$) = range ($d$)" in a separate answer.

  • $\begingroup$ I'm trying to approach the problems again and trying to understand the solution clearly. I don't really understand what does range (c) =range (d) refers to? I've been reading the whole definition in the book but still don't quite understand. Does range (c) refers to (cdCD) or range (c) refers to the range containing c and any three other fixed points in the line CD (I have checked this case in GeoGebra and it seems that it's not true ). Sir, if you have read the book, please help me, what does range (c) refer to and why range (c) = range (d) $\endgroup$
    – RopuToran
    Feb 28 at 4:54

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