অাই‌সো‌টোমিক অনুবন্ধী

Before reading this post, please read this post.

Given a triangle $ABC$ and $3$ cevians $AA'$, $BB'$, $CC'$, given that the cevians are concurrent at $P$. Reflect $A'$ from the midpoint of $BC$ to get $A''$, and define $B''$ and $C''$ similarly. Then the $3$ new lines $AA''$, $BB''$, $CC''$ are also concurrent at a point $P'$, which we call it the isotomic conjugate of $P$.

Solution: Note that $AC''=BA'$, $C''B=C'A$, $BA''=A'C$, $A''C=A'B$, $CB''=AB'$, and $B''A=B'C$. So, by Ceva theorem, we have $$\frac{AC''}{C''B}\cdot\frac{BA''}{A''C}\cdot \frac{CB''}{B''A}=\frac{BA'}{C'A}\cdot\frac{A'C}{A'B}\cdot \frac{AB'}{B'C}=1$$ and we are done by the converse of the Trig Ceva theorem. Yay.

Interesting Facts: The isotomic conjugate of the centroid of triangle ABC is the centroid itself. The isotomic conjugate of the symmedian point is the third Brocard point, and the isotomic conjugate of the Gergonne point is the Nagel point. Try to prove each of them!

Challenge: Let a triangle $ABC$. Prove that the barycentric coordinates of the isotomic conjugate of the point with barycentric coordinates $(x : y : z)$ is $(\frac{1}{x} : \frac{1}{y} : \frac{1}{z})$. (Note that the coordinates do not involve terms related to the side lengths of the given triangle.)