Let $ABC$ be a triangle with incircle $\gamma$. The circle through $A$ and $B$ tangent to $\gamma$ touches it at $C_2$ and the common tangent at $C_2$ intersects $AB$ at $C_1$. Define the points $A_1$, $B_1$, $A_2$, $B_2$ analogously. Prove that: a) the points $A_1$, $B_1$, $C_1$ are collinear; b) the lines $AA_2$, $BB_2$, $CC_2$ are concurrent.
Problem
Source: Bulgaria EGMO 2023 TST, Day 2, Problem 3
Tags: geometry, incircle, mixtilinear, concurrency, radical axis, Brianchon, Ceva s theorem
EndipifromBMT
08.02.2023 07:45
For part a), we can prove $A_1$, $B_1$, $C_1$ lie on the radial axis of the circumcircle and the incircle.
For part b), Due to Steinbart's theorem, the statement is equivalent to prove $A_1A_2$, $B_1B_2$, $C_1C_2$ are concurrent (See here.)
euclides05
09.02.2023 09:52
This reminds me Romanian Masters of Mathematics 2012 P6...
PROF65
10.02.2023 13:30
Assassino9931 wrote: Let $ABC$ be a triangle with incircle $\gamma$. The circle through $A$ and $B$ tangent to $\gamma$ touches it at $C_2$ and the common tangent at $C_2$ intersects $AB$ at $C_1$. Define the points $A_1$, $B_1$, $A_2$, $B_2$ analogously. Prove that: a) the points $A_1$, $B_1$, $C_1$ are collinear; b) the lines $AA_2$, $BB_2$, $CC_2$ are concurrent. $a)$Let $DEF$ be the touch triangle. $EF\cap BC=D'$ we have $(D'D,BC)=-1 , A_1A_2^2=A_1D^2=AB.AC$ hence $A_1$ is the midpoint of $DD'$ whence $\frac{A_1B}{A_1C}=\frac{DB^2}{DC^2}$ idem for $B_1,C_1$ then using menelaus we conclude. $b)$ Let $ EF\cap A_1A_2=A' , FD\cap B_1B_2=B' , DE\cap C_1C_2=C'$ We have the polars of $A',B',C'$ are $AA_2,BB_2,CC_2$ so it suffices to show that $A',B',C'$ are collinear. Now since $A'A_2 $ is tangent to the incircle then $\frac{A'E}{A'F}=\frac{A_2E^2}{A_2F^2}$ similarly we get $\frac{B'F}{B'D}=\frac{B_2F^2}{B_2D^2},\frac{C'D}{C'E}=\frac{C_2D^2}{C_2E^2}$ thus we needs that $\frac{A_2E }{A_2F } . \frac{B_2F }{B_2D }. \frac{C_2D }{C_2E }=-1$ or $ \frac{\sin \angle A_2DE }{\sin \angle A_2DF } . \frac{\sin \angle B_2EF }{\sin \angle B_2ED }. \frac{\sin \angle C_2FD }{\sin \angle C_2FE }=-1$ $\iff$ $DA_2,EB_2,FC_2$ are concurrent but we have their poles $A_1,B_1,C_1$ are collinear which closes the proof.