Doctor A wrote: (A.Zaslavsky) The circumradius of triangle $ ABC$ is equal to $ R$. Another circle with the same radius passes through the orthocenter $ H$ of this triangle and intersect its circumcirle in points $ X$, $ Y$. Point $ Z$ is the fourth vertex of parallelogram $ CXZY$. Find the circumradius of triangle $ ABZ$. It should be "parallelogram CXYZ". Then, the circumcircle (ABZ) is equal to R.

mr.danh wrote: Doctor A wrote: (A.Zaslavsky) The circumradius of triangle $ ABC$ is equal to $ R$. Another circle with the same radius passes through the orthocenter $ H$ of this triangle and intersect its circumcirle in points $ X$, $ Y$. Point $ Z$ is the fourth vertex of parallelogram $ CXZY$. Find the circumradius of triangle $ ABZ$. It should be "parallelogram CXYZ". Then, the circumcircle (ABZ) is equal to R. dear mr.danh,I think Doctor A is right X,C,Y,Z lie clockwisely

mr.danh wrote: Doctor A wrote: (A.Zaslavsky) The circumradius of triangle $ ABC$ is equal to $ R$. Another circle with the same radius passes through the orthocenter $ H$ of this triangle and intersect its circumcirle in points $ X$, $ Y$. Point $ Z$ is the fourth vertex of parallelogram $ CXZY$. Find the circumradius of triangle $ ABZ$. It should be "parallelogram CXYZ". Then, the circumcircle (ABZ) is equal to R. how to do it?could you post your idea? thank you in advance!

NO! In the original language of Russian, is CXZY paralelogram!

Denote the intersection of XY and OO1 is N 1.We make the reflection of H w.r.t N is H’ We know ON=O1N so HOH’O1 is a parallelogram => HO1=OH’=R => H’lies on ⊙O => ZHCH’a parallelogram 2. Extend AO meet ⊙O at M HCMB is a parallelogram => MB∥CH∥ZH’ MB=CH=ZH’ => ZBMH’ is a parallelogram we have ZB∥H’M 3.∠MH’C=∠BZH =∠MAC=∠H1AB => B,Z,A,H are concyclic ∠BZA=∠ACB=180-∠AHB 4.using sin law sin∠AZB=AB/2R1 sin∠ACB=AB/2R => R=R1 DONE!

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Remark: the circle HAB is reflection of ABC relatively the line AB ---> (circumcircle radius of $ \triangle {HAB})=$ (circumcircle radius of $ \triangle {ABC})=R.$

Yes. Sorry, I think X,Z are intersections of the circle which passes through H and the circumcircle (ABC). And this problem is not hard. We should denote the reflections of O wrt AB and XY, then make a use of properties of parallelogram. Best regards, mr.danh

This problem can be easily solved by complex number.