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# Understand the problem

Let $ABC$ be a triangle, $O$ its circumcenter, $S$ its centroid, and $H$ its orthocenter. Denote by $A_1, B_1$, and $C_1$ the centers of the circles circumscribed about the triangles $CHB, CHA$, and $AHB$, respectively. Prove that the triangle $ABC$ is congruent to the triangle $A_1B_1C_1$ and that the nine-point circle of $\triangle ABC$ is also the nine-point circle of $\triangle A_1B_1C_1$.

##### Source of the problem
IMO longlist 1992
Geometry
Hard
##### Suggested Book
Challenge and Thrill of Pre-college Mathematics

Do you really need a hint? Try it first!

Prove that the reflections of $H$ with respect to the sides $AB,BC,CA$ all lie on the circumcircle of $ABC$.
Study the reflection of $\odot AHB$ with respect to $AB$.
Combining the previous hints, show that $A_1,B_1, C_1$ are just reflections of $O$ on the sides of $ABC$.

From the last hint, we have $BA_1=BO=R=CO=CA_1$ where $R$ is the circumradius of $ABC$. Hence, $BOCA_1$ is a rhombus and $A_1C||BO$. Similarly, $BO||AC_1$ hence $A_1C||AC_1$. As $CAC_1A_1$ is a parallelogram, we also have $A_1C_1=AC$. We can similarly prove that $A_1B_1=AB$ and $B_1C_1=BC$. Thus $ABC\cong A_1B_1C_1$. This implies that it suffices to show that the centres of the two nine-point circles coincide. Remember that the nine-point centre is the midpoint of the line joining the circumcentre and the orthocentre. Claim $H$ is the circumcentre of $A_1B_1C_1$.   Proof Note that $A_1H=R$ (as $A_1$ is the centre of $\odot BHC$ and $\odot BHC$ is a reflection of $\odot ABC$. Similarly, $B_1H=C_1H=R$.     Claim $O$ is the orthocentre of $A_1B_1C_1$.   Proof   As $A_1$ is the reflection of $O$ on $BC$, $A_1O\perp BC$. As $BC\parallel B_1C_1$, $A_1O\perp B_1C_1$. Similarly, $B_1O\perp C_1A_1$ and $C_1O\perp A_1B_1$. Thus $O$ is the orthocentre of $A_1B_1C_1$.

Thus the centres of the nine-point circles of $ABC$ and $A_1B_1C_1$ coincide.