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
Difficulty Level
Suggested Book
Challenge and Thrill of Pre-college Mathematics

Start with hints

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.

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