I discovered a problem in plane geometry (there are some nice special cases) as follows:

Let $ABC$ be a triangle and $\Omega$ be arbitrary circumconic. Let two points $A_b, A_c \in BC$, $B_c, B_a \in CA$, $C_a, C_b \in AB$, let the line $AA_b, AA_c$ meet the circumconic again at $A'_b, A'_c$ define $B'_c, B'_a$ and $C'_a, C'_b$ cyclically, let $A'_bA'_c \cap BC = A'$ define $B', C'$ cyclically.

Question: I am looking for a proof that $A', B', C'$ are collinear if and only if $A_b, A_c, B_c, B_a, C_a, C_b$ lie on a conic and Is it a new discovery on conic section?

See also:

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  • $\begingroup$ By a standard trick, it's enough to prove this when $A_b = A_c$, $B_a = B_c$, and $C_a = C_b$. In this special case, you just need to prove that $A', B', C'$ are collinear iff either the three lines $AA_b, BB_c, CC_a$ are concurrent or the three points $A_b, B_c, C_a$ are collinear. $\endgroup$ – zeb May 8 at 13:48
  • $\begingroup$ In this special case, I think the statement is equivalent to the fact that a quadratic Cremona involution where we blow up the points $A,B,C$ takes circumconics of $ABC$ to lines, and vice-versa. I'll try to boil this argument down to something more direct. $\endgroup$ – zeb May 8 at 14:14
  • $\begingroup$ @zeb what is the standard trick you refer to? $\endgroup$ – Fedor Petrov May 8 at 16:14
  • 2
    $\begingroup$ Hold every point in the diagram other than $A_b,A_c,A_b',A_c'$ fixed, and look at what happens when you vary $A_b$. You will see that the map from $A_b$ to $A_c$ is a fractional-linear transformation. Additionally, the map taking $A_b$ to the second intersection of the conic through $A_b,B_a,B_c,C_b,C_a$ with the line $BC$ is also a fractional linear transformation. To check these transformations are the same, you just need to check it at any three points, and it's easy to check it in the degenerate cases $A_b = B$ and $A_b = C$. $\endgroup$ – zeb May 8 at 17:57

It suffices to consider the case when $\Omega$ is a circumcircle, so let it be.

At first, the points $A_b, A_c, B_c, B_a, C_a, C_b$ lie on a conic if and only if $$ \frac{AB_a\cdot AB_c}{AC_a\cdot AC_b}\cdot \frac{BC_a\cdot BC_b}{BA_b\cdot BA_c}\cdot \frac{CA_b\cdot CA_c}{CB_a\cdot CB_c}=1\quad\quad\quad\quad(\heartsuit) $$ (by Pascal theorem, they lie on a conic if and only if $X:=B_aC_a\cap BC$ and two analogous points are collinear. Applying Menelaus theorem we get $XB:XC=(BC_a\cdot AB_a):(AC_a\cdot CB_a)$. Multiplying three such expressions we get $(\heartsuit)$.)

For finding $B'A:B'C$ (we look at collinearity of $A',B',C'$ via Menelaus too) we project the quadruple $(B',A,B_a,C)$ to $\Omega$ from the point $B_a'$. We get $$ \frac{B'A}{B'C}:\frac{B_aA}{B_aC}=\frac{B'_cA}{B'_cC}:\frac{BA}{BC}= \frac{S(B'_cAB)/BA}{S(B'_cCB)/BC}:\frac{BA}{BC}=\frac{BC^2}{BA^2}\cdot \frac{AB_c}{CB_c}. $$ Substituting this to Menelaus we get condition $(\heartsuit)$.

  • $\begingroup$ Seems like I was too slow! $\endgroup$ – zeb May 8 at 14:30

Application of the theorem in post #1. I give a special case and give a proof as follows:

Generalization of conjugate of a point: Let $ABC$ be a triangle, and $\Omega$ is arbitrary circumconic of $ABC$, $P$ be arbitrary point in the plane. Let $AP$, $BP$, $CP$ meet the circumconic at $A'$, $B'$, $C'$. Three lines through $A'$, $B'$, $C'$ and parallel to $BC$, $CA$, $AB$ meets the circumconic again at $A''$, $B''$, $C''$ then $AA'', BB'', CC''$ are concurrent

Proof: Three lines through $A'$, $B'$, $C'$ and parallel to $BC$, $CA$, $AB$ meet $BC$, $CA$, $AB$ at three collinear points at $\infty$.

Let the lines $AA'$, $AA''$ meet $BC$ again at $A_{b}$, $A_{c}$ define $B_{c}$, $B_{a}$ and $C_{a}$, $C_{b}$ are cyclically, apply theorem #1 then six points $A_{b}$, $A_{c}$, $B_{c}$, $B_{a}$, $C_{a}$, $C_{b}$ lie on a conic.

By Carnot theorem we have:

$$ \frac{AB_a\cdot AB_c}{AC_a\cdot AC_b}\cdot \frac{BC_a\cdot BC_b}{BA_b\cdot BA_c}\cdot \frac{CA_b\cdot CA_c}{CB_a\cdot CB_c}=1\quad\quad\quad\quad(1) $$

But $AA'$, $BB'$, $CC'$ are concurrent at $P$, other word $AA_b$, $BB_c$, $CC_a$ are concurrent at $P$, thus:

$$\frac{BA_b}{CA_b}.\frac{CB_b}{AB_b}.\frac{AC_a}{BC_a}=1\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad(2) $$

Since $(1)$ and $(2)$ we get:

$$\frac{BA_c}{CA_c}.\frac{CB_a}{AB_a}.\frac{AC_b}{BC_b}=1\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad(3) $$

Since $(3)$ by converse of the Ceva's theorem we get: Three lines $AA_c$, $BB_a$, $CC_b$ are concurrent, other words $AA'', BB'', CC''$ are concurrent

Professor César Lozada let me know that (via email):

Among all known conjugations, I found your conjugation is:

In general, for the circumconic with center $O=x : y : z$, your transformation for a point $P=u : v : w$ is as simple as:

$Q(O,P) = x*(-x+y+z)*v*w : :$


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