Let $A',\,B',\,C'$ be points, in which excircles touch corresponding sides of triangle $ABC$. Circumcircles of triangles $A'B'C,\,AB'C',\,A'BC'$ intersect a circumcircle of $ABC$ in points $C_1\ne C,\,A_1\ne A,\,B_1\ne B$ respectively. Prove that a triangle $A_1B_1C_1$ is similar to a triangle, formed by points, in which incircle of $ABC$ touches its sides.
Problem
Source: ARO 2005 - problem 11.3
Tags: geometry, circumcircle, geometry solved
grobber
30.04.2005 15:01
Let $A_{11}$ be the midpoint of the arc $BAC$. We have $BC'=CB',BA_{11}=CA_{11}$ and $\angle A_{11}BC'=\angle A_{11}CB'$. Together these imply that the triangles $A_{11}BC',A_{11}CB'$ are congruent. In particular, we get $\angle A_{11}C'A=\angle A_{11}B'A$, which means that $A_{11}=A_1$, or, in other words, $A_1$ is the midpoint of the arc $BAC$. Similar relations hold for the other letters as well, and a simple angle chase will now show that $B_1C_1\|B''C''$, where $A''B''C''$ is the intouch triangle of $ABC$. This means that $A''B''C''$ and $A_1B_1C_1$ are homothetic, and thus similar.
windrock
28.01.2009 14:47
Lemma: Let $ \delta ABC$ and $ (O)$ is the circumcircle. $ D, E$ are two point in $ AB, AC$ respective at $ D, E$ such that: $ BD = CE$. We have that $ (ADE)$ pass through the midpoint of arc $ BC$ contains $ A$. Use the Lemma we get that: $ A_{1}, B_{1}, C_{1}$ respect are the midpoint of arcs $ BC, CA, AB$ contains$ A, B, C$ Easy to finish the problem
ArbelosYS
06.08.2009 07:44
This is IMO-shortlist 2006 Geometry problem 9 http://www.mathlinks.ro/Forum/viewtopic.php?p=875036#875036 See it ^^ !
plane geometry
06.08.2009 10:09
1.O',D',E',C are concyclic O',D',C1,C are concyclic => O',D',C1,C,E are concyclic C1 is the intersection of ⊙D'E'C and \ ⊙ABC 2. ⊿A1B1C1∽⊿IaIbIc ⊿IaIbIc∽⊿DEF => ⊿A1B1C1∽⊿DEF Done
Attachments:
Night_Witch123
31.01.2019 15:51
Let the contact triangle be $\Delta DEF$, From, First Sparrow Lemma, $\implies$ $A_1,B_1,C_1$ are the midpoints of arc $BAC$, $ABC$, $ACB$, Also, note, $\odot (ABC)$ is the Nine- Point Circle WRT $\Delta I_AI_BI_C$ and $\Delta ABC$ is the orthic triangle WRT $\Delta I_AI_BI_C$, Claim 1: $C_1$ lies on $I_AI_B$ Proof: Since, $C_1$ is the midpoint of arc $ACB$, hence, $\angle C_1AB$ $=$ $\angle C_1CB=90^{\circ}-\frac{C}{2}$ $\implies$ $C_1$ lies on exterior angle bisector of $\angle C$ $\implies$ $C_1 \in I_AI_B$ $\qquad \blacksquare$ Therefore, $\Delta A_1B_1C_1 \longrightarrow $ medial triangle WRT $\Delta I_AI_BI_C$, hence, $A_1B_1 ||I_AI_B$, $A_1C_1 ||I_AI_C$ and $C_1B_1 ||I_CI_B$, some trivial angle chasing, $\implies$ $DE || I_AI_B$ $\implies$ $DE ||A_1B_1$, similarly, $EF ||B_1C_1$ and $FD||A_1C_1$ implies the similarity! $\qquad \blacksquare$