orl wrote:

Let $H$ be the orthocenter of an acute triangle $ABC$ . The circle centered at the midpoint of $BC$ and passing through $H$ intersects

the line $BC$ at $A_{1}$ , $A_{2}$ . Similarly define the pairs $B_{1}$ , $B_{2}$ and $C_{1}$ , $C_{2}$ . Prove that $A_{1}$ , $A_{2}$ , $B_{1}$ , $B_{2}$ , $C_{1}$ , $C_{2}$ are concyclically.

Proof. Denote the midpoints $D$ , $E$ , $F$ of the sides $[BC]$ , $[CA]$ , $[AB]$ respectively. Prove easily or is well-known that

$\boxed {\ HA^2 + a^2 = HB^2 + b^2 = HC^2 + c^2 = 4R^2\ }\ (*)$ . Thus, $4\cdot\overline {BA_1}\cdot\overline {BA_2} = 4\cdot\left(BD^2 - HD^2\right) =$ $a^2 - 4\cdot HD^2 =$

$a^2 - 2\cdot\left(HB^2 + HC^2\right) + a^2$ $\stackrel {(*)}{\ \ \implies\ \ }$ $\overline {BA_1}\cdot\overline {BA_2} = \frac 12\cdot \left(a^2 + b^2 + c^2\right) - 4R^2$ $\implies$ $\boxed {\ \overline {BA_1}\cdot\overline {BA_2} = 4R^2\prod\cos A\ }$

because prove easily or is well-known that $\boxed {\ \left(a^2 + b^2 + c^2\right) - 8R^2 = 8 R^2\prod\cos A\ } > 0$ (remark that $\{A_1,A_2\}\subset (BC)$ a.s.o.).

Thus, $\overline {BA_1}\cdot \overline {BA_2} = - \overline {A_1B}\cdot\overline {A_1C} = R^2 - OA_1^2\ \implies\ 4R^2\prod\cos A = R^2 - OA_1^2$ $\implies$ $OA_1^2 = R^2\left(1 - 4\prod\cos A\right)$

(symmetrically in $a$ , $b$ , $c$ ). In conclusion, $\boxed {\ \rho = OA_1 = OA_2 = OB_1 = OB_2 = OC_1 = OC_2 = R\sqrt {1 - 4\cos A\cos B\cos C}\ }$ .

Remark. Since $1 - 8\prod\cos A\ge 0$ obtain easily that $\rho\ge 2R\sqrt {\prod\cos A}$ .

Pohoatza wrote:

I guess it would have been a better IMO problem, at least to avoid the "well-known"-type commentaries.

All right !