What do you mean "An ugly geo in Cono Sur"? Este problema es cine Let's asumme $P, Q, R, X, Y, Z$ exist Let $T$ be the intersection of the parallels from $B_1$ and $A_2$ to $C_1A_2$ and $C_2A_1$ respectively Let $X'$ = $C_2A_1 \cap B_1A_2$. Analogously, we define $Y'$ and $Z'$ We have $\angle Z'Y'X' = 180^o - \angle A_2B_1C - \angle A_1B_2A = \angle ZYX$, analogously $\angle Z'X'Y' = \angle ZXY$ Then $\triangle XYZ$ is similar to $\triangle X'Y'Z'$ By the definition of T: $\triangle A_2B_1T$ is homothetic to $\triangle X'Y'Z'$, then is also similar to $\triangle XYZ$ Since $CA_2B_1$ is congruent to $RXY$, then $XY = A_2B_1$, so $\triangle XYZ$ and $\triangle A_2B_1T$ are congruent Then $C_1B_2 = ZY = TB_1$. Since C_1B_2 and TB_1 are parallel: $C_1B_2B_1T$ is a paralelogram Analogously, $C_2A_1A_2T$ is a paralelogram too Let's proceed by complex numbers. By the paralelograms we have: $c_2+a_2-a_1 = t = c_1+b_1-b_2 \rightarrow \frac{a_2+b_2+c_2}{3}=\frac{a_1+b_1+c_1}{3}$ Which is equivalent to the fact that the centroids of triangles $A_1B_1C_1$ and $A_2B_2C_2$ coincide. If we start asumming the centroids of triangles $A_1B_1C_1$ and $A_2B_2C_2$ coincide, we can construct $T$ and then construct $P', Q', R'$ by drawing parallels from $X', Y', Z'$ to the sides of $\triangle ABC$ Our new triangle P'Q'R' will follow the properties we want of the triangle PQR with an homotecy (everything is by the same reasons we already had)

Very easy problem. I’ll use vectors, but you really can use any computational method. The condition is trivially equivalent to the existence of translations by vectors $\vec{s}$, $\vec{t}$ and $\vec{u}$ that take $AB_2C_1$, $BA_1C_2$, $CA_2B_1$ to $PYZ$, $QXZ$, $RXY$, respectively. Then, we get: $B_2+\vec{s}=B_1+\vec{u}$ $C_2+\vec{t}=C_1+\vec{s}$ $A_2+\vec{u}=A_1+\vec{t}$ Therefore it is obvious that $A_1+B_1+C_1=A_2+B_2+C_2$, which solves the problem

Gato_combinatorio wrote: What do you mean "An ugly geo in Cono Sur"? Este problema es cine All solutions during the test are bash (no synthetic geo).

As pointed out by the posts above the problem is really easy computationally, but I believe I have a solution that's purely sinthetic. the main ideia is to construct parallelograms. The solution is quite long, but I might write it here if someone is interested. Edit: I'm sorry I havent posted the solution as I said I would but I don't really have time to spare rn. If someone is really interested I have written it on paper but it's in portuguese. (Just DM me and I will send it if possible)

Thelink_20 wrote: As pointed out by the posts above the problem is really easy computationally, but I believe I have a solution that's purely sinthetic. the main ideia is to construct parallelograms. The solution is quite long, but I might write it here if someone is interested. post here