Points $A_1$, $A_2$, $B_1$, $B_2$ lie on the circumcircle of a triangle $ABC$ in such a way that $A_1B_1 \parallel AB$, $A_1A_2 \parallel BC$, $B_1B_2 \parallel AC$. The line $AA_2$ and $CA_1$ meet at point $A'$, and the lines $BB_2$ and $CB_1$ meet at point $B'$. Prove that all lines $A'B'$ concur.
Problem
Source: Sharygin Finals 2023 9.3
Tags: geometry, Sharygin Geometry Olympiad, Sharygin 2023, concurrent lines
NO_SQUARES
02.08.2023 13:04
Hint: use similars triangles and Menelaus theorem This point, in this $A'B'$ concur., lies on line $AB$
NO_SQUARES
20.08.2023 23:55
Also, it is possible to solve it by moving points
sanyalarnab
27.08.2023 18:13
Wooohooo!! Comeback after a looong time with the 670th post!(and did a geo problem after more than half a year) Let $B_2C \cap B_1A \equiv L,A_2C \cap A_1B \equiv P,B_2A_2 \cap B_1A_1 \equiv Q$ and let the tangent to the circumcircle $ABC$ at C intersect $AB$ at $T$.. By Pascal's Theorem on the hexagons $ABB_2CCB_1,BAA_2CCA_1,A_1BB_2A_2CB_1,B_1AA_2B_2CA_1$, we have the following inferences: $$T-B'-L,T-A'-P,P-B'-Q,L-A'-Q$$. Combining these collinearities we have that $T$ lies on $A'B'$. As $T$ is fixed we can conclude that all lines $A'B'$ concur.
Infinityfun
10.11.2023 14:42
Sorry to correct but only these collinearities don't imply all of them lie on a single line. However, there is an easy fix. By angle chasing $A'A_1A_2$ and $B'BC$ are homothetic because their corresponding sides are parallel. So we can say that $LA'B'$ are collinear too. Similiarly, $P$ lies on this line. Combining with your Pascals $T$ also lies on this line. A second solution is below: $CA'\cap AB =F$, parallel from $F$ to $BC$ intersect $AA_2$ at $C'$. By angle chase triangles $A'FC'$ and $B'BC$ are homothetic. $A'B'$, $FB=AB$ and $CC'$ occur. On the other hand, $CFAC'$ are cyclic $\Rightarrow \angle ACC' = \angle AFC' =\angle ABC$ since $FC' \parallel BC$ implying $CC'$ is tangent to $(ABC)$ so $A'B' \cap AB$ is fixed
gnoka
09.01.2024 18:51
This does not hold since you never use the Parallel. sanyalarnab wrote: Wooohooo!! Comeback after a looong time with the 670th post!(and did a geo problem after more than half a year) Let $B_2C \cap B_1A \equiv L,A_2C \cap A_1B \equiv P,B_2A_2 \cap B_1A_1 \equiv Q$ and let the tangent to the circumcircle $ABC$ at C intersect $AB$ at $T$.. By Pascal's Theorem on the hexagons $ABB_2CCB_1,BAA_2CCA_1,A_1BB_2A_2CB_1,B_1AA_2B_2CA_1$, we have the following inferences: $$T-B'-L,T-A'-P,P-B'-Q,L-A'-Q$$. Combining these collinearities we have that $T$ lies on $A'B'$. As $T$ is fixed we can conclude that all lines $A'B'$ concur.