In a non-isosceles triangle $ABC$, let $I$ be its incenter. The incircle of $ABC$ touches $BC$, $CA$, and $AB$ at $D$, $E$, and $F$, respectively. A line passing through $D$ and perpendicular to $AD$ intersects $IB$ and $IC$ at $A_b$ and $A_c$, respectively. Define the points $B_c$, $B_a$, $C_a$, and $C_b$ similarly. Let $G$ be the intersection of the cevians $AD$, $BE$, and $CF$. The points $O_1$ and $O_2$ are the circumcenter of the triangles $A_bB_cC_a$ and $A_cB_aC_b$, respectively. Prove that $IG$ is the perpendicular bisector of $O_1O_2$.
Problem
Source: Indonesia TST 2016 Round 3
Tags: geometry, incenter, gergonne
25.03.2024 20:28
This is an insane problem, harder than almost every single geo i've done so far, my respect for anyone that has the will to write a complete solution for this since mine is too large that i'm afraid it just won't render. So here are the key steps for the problem in case you were wondering:
This problem also made me wonder if anyone at the TST solved this and also if there is a simplier way than the one i did, maybe using Cayley Bacharach due to the amount of intersections in the graphic.
25.03.2024 22:33
solvedy with Kanav, Choppara, Malay, Kohli, Shyam, Mathur, Ananda, Siddharth, Rushil, Bhaduri, Manas, Mahajan, Narayan, Virat, Ram and Talwar even though we were basically carried by Siddharth and Choppara
25.03.2024 23:02
Anyways another note but by using Desargue's and cyclicity the whole problem statement is equivalent to $A_bB_c \cap A_cC_b - I - G$, which feels bary bashable.
26.03.2024 12:33
Nice problem Sketch: Claim 1 $DA_B = DA_C$
Claim 2 $G$ is the center of spiral similarity sending $EF$ to $A_BB_C$
Claim 3 $DEF \sim A_BB_CC_A$
Claim 4 $A_BB_AC_AA_C$ is cyclic
So we get $I,G$ are corresponding points in triangles, so we’re done.