Two triangles $ABC$ and $A'B'C'$ are given. The lines $AB$ and $A'B'$ meet at $C_1$ and the lines parallel to them and passing through $C$ and $C'$ meet at $C_2$. The points $A_1,A_2$, $B_1,B_2$ are defined similarly. Prove that $A_1A_2,B_1B_2,C_1C_1$ are either parallel or concurrent.
Problem
Source: Sharygin finals 10.8
Tags: geometry
62861
02.08.2018 03:34
This is IMO Longlist 1976 Problem 15
62861
02.08.2018 05:15
Claim. There is a unique point $P$ (possibly at infinity) with the same barycentric coordinates wrt $\triangle ABC$ and $\triangle A'B'C'$.
We are interested in triples $(x, y, z) \neq (0, 0, 0)$ satisfying
\[x(A - A') + y(B - B') + z(C - C') = 0.\]Defining $A - A' = (p_a, q_a)$ and similar, consider the underdetermined system
\[
\begin{bmatrix}
p_a & p_b & p_c\\
q_a & q_b & q_c
\end{bmatrix}
\cdot
\begin{bmatrix}
x \\ y \\ z
\end{bmatrix}
=
\begin{bmatrix}
0 \\ 0
\end{bmatrix}.
\]If the rows of the $2 \times 3$ matrix are proportional then $\overline{AA'} \parallel \overline{BB'} \parallel \overline{CC'}$, contradiction as $A_1$ doesn't exist. So there exists a unique solution $(x : y : z)$ (up to scaling), which are the desired coordinates.
Lemma. Let $P$ be a point (not at infinity) and $k$, $\ell$ two parallel lines. Consider an affine transform fixing $P$ and let $k'$, $\ell'$ be the images of $k$ and $\ell$. Then $P$, $k \cap k'$, $\ell \cap \ell'$ are collinear.
Note $k' \parallel \ell'$ and $\tfrac{d(P, k)}{d(P, \ell)} = \tfrac{d(P, k')}{d(P, \ell')}$. Then a homothety at $P$ maps $k \to \ell$ and $k' \to \ell'$ or vice versa, implying the collinearity.
We complete the proof by showing $P$ lies on line $A_1A_2$, the other two lines being similar. Suppose $P$ is not at infinity. By the definition of $P$, there exists an affine transform fixing $P$ taking $\triangle ABC \to \triangle A'B'C'$; it also sends lines $BC \to B'C'$ and $AA_2 \to A'A_2$. By the lemma $P$, $A_1$, $A_2$ are collinear, as desired. The case $P$ at infinity can be settled with a continuity argument.
guptaamitu1
26.07.2022 22:45
The second official solution is quite clever. Observe that line $C_1C_2$ is the locus of all points $X$ such that $$ [XAB] \cdot [A'B'C'] = [XA'B'] \cdot [ABC] $$where areas are directed and $[\lambda]$ denotes the area of figure $\lambda$. Then it is not hard to see that lines $A_1A_2,B_1B_2,C_1C_2$ concur (like the intersection of any two lines must lie on the third also). $\blacksquare$