Estonian Math Competitions

Let $O$ and $G$ be the circumcenter and centroid of $ABC$ respectively and let $K$ be the midpoint of $BC$ (Fig. 33).

We know that $AG=2GK$. Also we know that $H$, $G$ and $O$ are collinear with $HG=2GO$ (Euler line). So triangles $AHG$ and $KOG$ are similar with scale factor $2$ (by $2$ proportional sides and an equal angle between them). Therefore $AH=2KO$.

Now the Pythagorean theorem in triangle $KOB$ yields $K{O}^2+K{B}^2=O{B}^2=R^2$ and $A{H}^2+B{C}^2=(2KO{)}^2+(2KB{)}^2=4(K{O}^2+K{B}^2)=4R^2$.



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Alternative solution.

Let $B^{\prime }$ be the other end of the diameter to the circumcircle of $ABC$ drawn from $B$ (Fig. 34). Since $A{B}^{\prime }$ and $AB$ are perpendicular and $CH$ and $AB$ are perpendicular, the lines $A{B}^{\prime }$ and $CH$ are parallel. Similarly, we see that $C{B}^{\prime }$ and $AH$ are parallel, meaning that $AHC{B}^{\prime }$ is a parallelogram. Thus $C{B}^{\prime }=AH$.

Now the Pythagorean theorem in triangle $BC{B}^{\prime }$ yields $C{B}^{\prime 2}+B{C}^2=B{B}^{\prime 2}$. As $C{B}^{\prime }=AH$ and $B{B}^{\prime }=2R$, the desired result follows.