Let $O_1O_2O_3$ be an acute angled triangle.Let $\omega_1$, $\omega_2$, $\omega_3$ be the circles with centres $O_1$, $O_2$, $O_3$ respectively such that any of two are tangent to each other. Circumcircle of $O_1O_2O_3$ intersects $\omega_1$ at $A_1$ and $B_1$, $\omega_2$ at $A_2$ and $B_2$, $\omega_3$ at $A_3$ and $B_3$ respectively. Prove that the incenter of triangle which can be constructed by lines $A_1B_1$, $A_2B_2$, $A_3B_3$ and the incenter of $O_1O_2O_3$ are coincide.
Problem
Source: 2023 Turkey Egmo Tst P1
Tags: geometry, incenter, circumcircle
Ege_Saribass
02.09.2024 13:59
Let $A_3B_3 \cap A_2B_2 = X$, $A_3B_3 \cap A_1B_1 = Y$, $A_1B_1 \cap A_2B_2 = Z$.
Let $\omega_1$ and $\omega_2$ touch at $F$, $\omega_2$ and $\omega_3$ touch at $D$, $\omega_3$ and $\omega_1$ touch at $E$. And let $I$ be the incenter of $\triangle O_1O_2O_3$.
Clearly, $D$, $E$ and $F$ are the tangent points of the incircle of $\triangle O_1O_2O_3$. So $ID = IE = IF$.
Hence, $ID$ is radical axis of $\omega_2$ , $\omega_3$. Similar things satisfy for lines $IE$ and $IF$.
We also know that $XA_3.XB_3 = XB_2.XA_2$ thus, $X \in ID$. Similarly, $Y \in IE$, $Z \in IF$.
Let $O$ be the circumcenter of $\triangle O_1O_2O_3$
Now, just consider that $OO_2$ is perpendecular bisector of $A_3B_3$. Similarly, $OO_3$ is perpendecular bisector of $A_2B_2$.
Let $OO_2 \cap XZ = M$ and $OO_3 \cap XY = N$. Obviously, $OO_2 = OO_3$ and $(O_2MDX)$, $(O_3NDX)$.
Conditions above yields $XD$ is bisector of $\angle ZXY$. We can conclude similar things for $Y$ and $Z$.
Thus, $I$ is the incenter of $\triangle XYZ$.
$\blacksquare$
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