Perspective triangles induce conics.

Question

Let $A, B, C, a, b, c$ be three points and three lines respectively.

Prove (or provide a counterexample) that:

$\triangle ABC$ and $\triangle abc$ are perspective if and only if there exists a conic $\Omega$ (possibly imaginary) such that $(A,a),(B,b),(C,c) $ are pole-polar pairs with respect to $\Omega$.

The reverse proof isn't that important as I already have proof for it using Sondat's Theorem but feel free to post it.

Reverse proof:

Perform a projective transformation taking the conic to a circle. Let its center be $O $. From the definition of the polars $\triangle ABC $ and $\triangle abc$ are orthogonal in both directions from $O $ and thus by Sondat's Theorem they are perspective.$\blacksquare$

Partial forwards proof:

Let $a \cap \overline{CB}, \overline{AC}, \overline{AB}, c, b = F, Y, X, B_1, C_1, b \cap \overline{AB}, \overline{AC} = P, E, c \cap \overline{AB}, \overline{AC} = D, Q$.

Suppose $(X,C_1;Y,B_1) < 0$. Let $O $ be such $\measuredangle XOC_1 = \measuredangle YOB_1 = 90^\circ$($O$ exists because of the assumption).

Perform a projective tansformation that takes $X, Y, B_1, C_1$ to $\overline{OX}_\infty, \overline{OY}_\infty, \overline{OB_1}_\infty, \overline{OC_1}_\infty$.

Now work on the resulting picture.

Since the triangles are perspective we have $\overline{DEF}$. Since $a $ was taken to infinity, $F $ was taken to infinity and thus $\overline{BC} \parallel \overline{DE}$.

From the projection we have $X \perp C_1, Y \perp B_1$ thus $b \perp \overline{AB}, c \perp \overline{AC}$ thus $\overline{PQ}$ is antiparallel to $\overline{DE}$ within $\measuredangle BAC$ thus $\overline{PQ}$ is antiparallel to $\overline{BC}$ within $\measuredangle BAC$ thus $AP \cdot AB = AQ \cdot AC $.

Now if $AP \cdot AB $ is positive draw a circle of radius $\sqrt{AP \cdot AB}$. If it's negative draw an imaginary circle. Thus there exists a circle such that $(A,a), (B,c), (C, c) $ are pole-pairs with respect to it. Reversing the transformation yields a conic in the original picture. $\blacksquare$

Answer 1

Your conjecture is a theorem in Russell, Pure Geometry (1893), pg 153 (Ch. XIV, Art. 3):

Two triangles which are reciprocal for a conic are homologous ; and conversely, if two triangles be homologous they are reciprocal for a conic.

In this book, "homologous" means "in perspective" and "reciprocal" means that the vertices and sidelines of the two triangles are in a pole/polar relationship.

The problem of constructing a conic from two perspective triangles is converted to that of constructing a conic from a self-conjugate (i.e. self-reciprocal) triangle and a pole-polar pair (described in Ch. XXV, Art 12, pg 234-235).

The latter construction can be done with ruler and compass, with the snag that some of the constructed points may be imaginary. Midway through, here's how Russell deals with this caveat:

This completes the theoretical solution of the problem ; and we have shown that one, and only one, conic can be drawn satisfying the given conditions. Practically the above solution is worthless ; for any pair of the points XX', YY', ZZ' may be imaginary. The following is the practical construction when the conic is real.