line $l$ through the point $A$ from triangle $ABC$ . Point $X$ is on line $l$.${\omega}_b$ and ${\omega}_c$ are circles that through points $X,A$ and respectively tanget to $AB$ adn $AC$. tangets from $B,C$ respectively to ${\omega}_b$ and ${\omega}_c$ meet them in $Y,Z$. Prove that by changing $X$, the circumcircle of the circle $XYZ$ passes through two fixed points. Proposed by Ali Zamani
Problem
Source: Iran TST 2023 ; Exam 2 Problem 4
Tags: geometry, Fixed point
a_507_bc
16.03.2023 17:46
Is this exam 2 P5 or the polynomial problem (there are two P5's now)?
AHZOLFAGHARI
16.03.2023 18:51
a_507_bc wrote: Is this exam 2 P5 or the polynomial problem (there are two P5's now)? Sorry. Geometry problem was problem 4 in the second exam. I edited.
VicKmath7
16.03.2023 19:58
The first fixed point is the reflection of $A$ wrt $BC$. Indeed, notice that then $B$ is center of $(AYA')$ and $C$ is center of $(AZA')$, so $\angle ZA'Y=\frac {\angle ACZ} {2}+\frac{\angle ABY} {2}=90^{o}-\angle CAZ+90^{o}-\angle BAY=180^{o}-\angle ZXY$, so $A' \in (XYZ)$.
The second fixed point is the intersection point of the perpendicular from $A'$ to $AA'$ with $\ell$. Redefine this point as $AX \cap (XYZ)=X'$; we will prove that $A'X' \parallel BC$, which is sufficient. Indeed, under inversion at $A$ fixing $(XYZ)$, $X$ goes to $X'$, $Y$ to $AY \cap (XYZ)=Y'$ and $Z$ to $AZ \cap (XYZ)=Z'$, so $X'Z' \parallel AC, X'Y' \parallel AB$ due to the tangencies. Now notice that $\angle Z'X'A' =\angle Z'ZA'=\frac {\angle ACA'} {2}=\angle ACB$, so $A'X' \parallel BC$ and we are done.
sevket12
16.03.2023 20:46
Let $O_b,O_c$ are centers of $\omega_b$ and $\omega_c$ respectively. Obviously $X,Y,Z$ are reflections of $A$ according to $O_bO_c,BO_b,CO_c$. Take an inversion centered $A$ with radius $1$.We want to show that circles $X^* Y^* Z^*$ are coaxial. We can easily see that $X^*$ is circiumcenter of $AO^*_bO^*_c$.$Y^*$ is circiumcenter of $AO^*_bB^*$ and $Z^*$ is circiumcenter of $AO^*_cC^*$ Let $l_b,l_c$ are perpendicular bisector of $AB^*$ and $AC^*$ then $Y^*\in l_b$ and $X^*Y^*$ $\perp l_b$ similarly $Z^* \in l_c$ and $X^*Z^*$ $\perp l_c$ .Let $O$ is circiumcenter of $AB^*C^*$ we can see that circle $X^*Y^*Z^*$ is $(X^*O)$ so these circles has two fixed point which are $O$ and orthogonal projection of $O$ to line $l$ $\blacksquare$
sami1618
05.05.2024 20:37
Consider and inversion about $A$. Let $AX'$ intersect $(AB'C')$ again at $P$ and let $A'$ be the antipode of $A$ w.r.t. $(AB'C')$. These are the desired points.
Claim: Triangles $PB'C'$ and $PY'Z'$ are similar
Claim: The points $Y'$, $X'$, $Z'$ and $P$ are concyclic
Claim: The points $Y'$, $X'$, $Z'$ and $A'$ are concyclic