How 9 Cheenta students ranked in top 100 in ISI and CMI Entrances?

# Understand the problem

[/et_pb_text][et_pb_text _builder_version="3.22.4" text_font="Raleway||||||||" background_color="#f4f4f4" box_shadow_style="preset2" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px"]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 $\triangle A_1B_1C_1$.

[/et_pb_text][et_pb_tabs active_tab_background_color="#0c71c3" inactive_tab_background_color="#000000" _builder_version="3.23.3" tab_text_color="#ffffff" tab_font="||||||||" background_color="#ffffff"][et_pb_tab title="Hint 0" _builder_version="3.22.4"]Do you really need a hint? Try it first!

[/et_pb_tab][et_pb_tab title="Hint 1" _builder_version="3.23.3"]Prove that the reflections of $H$ with respect to the sides $AB,BC,CA$ all lie on the circumcircle of $ABC$.[/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="3.23.3"]Study the reflection of $\odot AHB$ with respect to $AB$.[/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="3.23.3"]Combining the previous hints, show that $A_1,B_1, C_1$ are just reflections of $O$ on the sides of $ABC$.[/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="3.23.3"]

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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