EGMO Problem 3.20 (BAMO 2013/3)

Question

The problem statement follows:

Let $H$ be the orthocenter of an acute angle triangle $ABC$. Consider the circumcenters of triangles $ABH$, $ BCH$, and $CAH$. Prove that they are the vertices of a triangle that is congruent to $ABC$.

So I first showed that the orthocenter of $ABC$ is the circumcenter of the second triangle, $A'B'C'$ and the circumcenter of $ABC$ is the orthocenter of $A'B'C'$. Next if we take a homothety $h$ at $N_9$ of $ABC$ with scale factor $-1$, this will send $H$ to $O$ and vice versa and we'll get a congruent triangle.

But my question is, how do I prove that the triangle formed by taking the homothety is the very triangle in the question, namely $A'B'C'$?

Answer 1

I believe you know the fact that circle $BCH$ is a reflection of circle $ABC$ acros $BC$. Similary is true for the other two circles.

So $$AC' = AO = CO = CA'$$

and by easy angle chase you can see $AC'||CA'$ so $A'CAC'$ is paralelogram, so $A'C' = AC$ and we are done.

Answer 2

$B'C'\;\bot\;AH$ and $BC\;\bot\;AH$, so $BC\;||\; B'C'$.

Similarly, $AB\;||\;A'B'$ and $AC\;||\;A'C'$.

This implies $\triangle ABC \sim \triangle A'B'C'$

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$A'H={BC\over2\sin\angle BHC}={a\over2\sin A}=R$.

Similarly, $B'H=C'H=R$.

So $H$ is the circumcentre of $\triangle A'B'C'$, and its circumradius is $R$, the same as the circumradius of $\triangle ABC$.

Since $\triangle ABC \sim \triangle A'B'C'$, equal circumradii allow us to get $\triangle ABC \cong \triangle A'B'C'$.