An isosceles triangle $ABC$ ($AB = BC$) is given. Circles $\omega_1$ and $\omega_2$ with centres $O_1$ and $O_2$ lie in the angle $ABC$ and touch the sides $AB$ and $CB$ at $A$ and $C$ respectively, and touch each other externally at point $X$. The side $AC$ meets the circles again at points $Y$ and $Z$. $O$ is the circumcenter of the triangle $XYZ$. Lines $O_2 O$ and $O_1 O$ intersect lines $AB$ and $BC$ at points $C_1$ and $A_1$ respectively. Prove that $B$ is the circumcentre of the triangle $A_1 OC_1$.
Problem
Source: Tuymaada 2020 Senior, P6
Tags: geometry, circumcircle, Triangle
06.10.2020 13:53
[asy][asy] size(10cm); pair X = (0,0); pair O1 = (-2,0); pair O2 = (1,0); draw(circle(O1,2)); draw(circle(O2,1)); pair B = (0,3); pair A = 2*foot(X,B,O1)-X; pair C = 2*foot(X,B,O2)-X; draw(A--B--C--A--cycle); pair Y = intersectionpoints(A--C,circle(O1,2))[0]; pair Z = intersectionpoints(A--C,circle(O2,1))[0]; pair O = circumcenter(X,Y,Z); pair C1 = extension(O2,O,A,B); pair A1 = extension(O1,O,B,C); draw(C1--O2); draw(O1--A1); draw(A--X--C); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$X$",X,dir(270)); dot("$Y$",Y,dir(Y)); dot("$Z$",Z,dir(Z)); dot("$O$",O,dir(O)); dot("$O_1$",O1,dir(O1)); dot("$O_2$",O2,dir(O2)); dot("$A_1$",A1,dir(A1)); dot("$C_1$",C1,dir(C1)); draw(Y--X--Z); draw(B--X); [/asy][/asy] Since $AB=BC$, we note that $B$ lies on the radical axis of $(O_1)$ and $(O_2)$ (i.e. the tangent at $X$). Claim: $O$ lies on $BX$. Proof: Noting that $B$ is the circumcenter of $\triangle AXC$, we have $$\angle OXY=90^{\circ}-\frac{1}{2}\angle YOX=90^{\circ}-\angle YZX=90^{\circ}-(180^{\circ}-\angle XZC)=90^{\circ}-\angle XCB=\frac{1}{2}\angle XBC=\angle CAX=\angle YXB$$ Furthermore, $OO_1$ is the perpendicular bisector of $YX$. Also, $$\angle CXY=\angle CXB+\angle BXY=\angle BCX+\angle XAC=\angle BCX+\frac{1}{2}\angle XBC=90^{\circ}$$Hence, $O_1A_1\parallel XC$ and $BO=BA_1$. Similarly, $BC_1=BO$ and we are done.
02.01.2022 11:09
Easy to see B lies on radical axis of circles. step1 : B,O,X are collinear. ∠OXY = 90 - ∠YZX = 90 - ∠BCX = ∠XBC/2 = ∠XAC = ∠BXY step2 : OC1 || XA and OA1 || XC. we know OO2 is perpendicular to ZX, also from last step we have 90 - ∠YZX = ∠XAC so XA is perpendicular to ZX. we have same proof for OA1 || XC. we had BA = BX = BC and know we have OC1 || XA and OA1 || XC so now by Thales theorem we have BC1 = BX = BA1. we're Done.