Is it a new discovery on conic section?

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?

• 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.
– zeb
May 8, 2021 at 13:48
• 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.
– zeb
May 8, 2021 at 14:14
• @zeb what is the standard trick you refer to? May 8, 2021 at 16:14
• 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 $A_b = C$.
– zeb
May 8, 2021 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 $$(\heartsuit)$$.

• Seems like I was too slow!
– zeb
May 8, 2021 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:

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