Let $ABC$ be a non isosceles triangle with circumcircle $(O)$ and incircle $(I)$. Denote $(O_1)$ as the circle that external tangent to $(O)$ at $A'$ and also tangent to the lines $AB,AC$ at $A_b,A_c$ respectively. Define the circles $(O_2), (O_3)$ and the points $B', C', B_c , B_a, C_a, C_b$ similarly. 1. Denote J as the radical center of $(O_1), (O_2), (O_3) $and suppose that $JA'$ intersects $(O_1)$ at the second point $X, JB'$ intersects $(O_2)$ at the second point Y , JC' intersects $(O_3)$ at the second point $Z$. Prove that the circle $(X Y Z)$ is tangent to $(O_1), (O_2), (O_3)$. 2. Prove that $AA', BB', CC'$ are concurrent at the point $M$ and $3$ points $I,M,O$ are collinear.
Problem
Source: 2016 Saudi Arabia Pre-TST Level 4 2.4
Tags: geometry, tangent circles, collinear, concurrent, concurrency
DNCT1
19.09.2020 20:24
Consider the inversion $\mathcal{I}^{JA'.JX}_{J}$ hence $J$ is the radical center of $(O_1),(O_2),(O_3)$ so we have that
$\overline{JA'}.\overline{JX}=\overline{JB'}.\overline{JY}=\overline{JC'}.\overline{JZ}$ then we have
$\mathcal{I}^{JA'.JX}_{J}: X\mapsto A',Y\mapsto B',Z\mapsto C',(XYZ)\mapsto (O)$ hence $(O)$ is tagent to $(O_1),(O_2),(O_3)$ respectively so $(XYZ)$ is tagent to $(O_1),(O_2),(O_3)$ at $X,Y,Z$ as well(complete the proof).
Let $I_a$ be the center of the excircle WRT $\angle BAC$ of $\bigtriangleup ABC$,let the excircle WRT $\angle BAC,\angle ABC,\angle ACB$ tagents to $BC,CA,AB$ at $D,E,F$ respectively,we known that $\overline{AD},\overline{BE},\overline{CF}$ are concurrent(prove this by Ceva's theorem)(1)
Consider the reflection inversion $\mathcal{I}^{AB.AC}_{A}:B\mapsto C, BC\mapsto (O),(O_1)\mapsto (I_a)$ hence the image of the mixtilinear cirlce WRT the vertex $A$ of $\bigtriangleup ABC$ is the cirlcle tangent to $BC,CA,AB$ respectively.Then $\mathcal{I}^{AB.AC}_{A}:A'\mapsto D$ which implies that $\overline{AA'},\overline{AD}$ are isogonal WRT $\angle BAC$ of $\bigtriangleup ABC$.Similarly with $\overline{BB'},\overline{BE}$ are isogonal WRT $\angle ABC$ and $\overline{CC'},\overline{CF}$ are isogonal WRT $\angle ACB$(2).
From (1),(2) we see that $\overline{AA'},\overline{BB'},\overline{CC'}$ are concurrent at $M$(complete the proof).
We known that $\overline{A'I}\cap (O)=L$ is the midpoint of arc$(BAC)$,define $R$ similarly with $R=\overline{B'I}\cap (O)$
Let $ \overline{AI}\cap (O)=K,\overline{BI}\cap (O)=U,\overline{AR}\cap\overline{BL}=S$ hence $\overline{RU}\cap \overline{LK}=O$.
By Pascal's theorem in $(O)$ we have
$$\begin{pmatrix}
A & L &U \\
B & R & K
\end{pmatrix}\Rightarrow \overline{I,S,O}$$and $$\begin{pmatrix}
A & L &B' \\
B & R & A'
\end{pmatrix}\Rightarrow \overline{M,I,S}$$so we have that $\overline{O,I,M}$ (complete the proof).
(My first attachment with the point $T=M$)
Attachments:
paul.thai.1803
20.09.2020 02:53
Nice solution DNCT1! Here is my approach for this problem! a) Inversion $\mathcal{I}_J^k$ with $k = JX.JA' = JY.JB'= JZ.JC'$, by that we have $X \mapsto A', Y \mapsto B', Z \mapsto C'$, so $(O_1) \mapsto (O_1), (O_2) \mapsto (O_2), (O_3) \mapsto (O_3)$ and $(XYZ) \mapsto (O)$. From the problem's requirement, we only need to prove that $(O)$ is tangent to $(O_1), (O_2), (O_3)$, which is obvious from the problem's hypothesis. b) Redefine $M$ as the outside homothety center of $(O)$ and $(I)$, so $\overline{I, M, O}$. We will prove that $AA', BB', CC'$ are concurrent at the point $M$. Using Monge D'Alembert theorem for 3 circles $(O_1), (O), (I)$, we get $\overline{A', A, M}$, so $M \in AA'$. Similarly we also get $M \in BB'$ and $M \in CC'$, hence $AA', BB', CC'$ are concurrent at $M$. $\blacksquare$