Problem 1846

Denote by $A_b, A_c$ the reflections of $A$ in $BH$ and $CH$ respectively. $B_c, B_a$ and $C_a, C_b$ are defined similarly. By definition, $l_a = B_cC_b, l_b = C_aA_c, l_c = A_bB_a$. Let $A_1 = l_b \cap l_c, B_1 = l_c \cap l_a, C_1 = l_a \cap l_b$ and let $H_1, O_1$ be the orthocentre and circumcentre of $\triangle A_1B_1C_1$ respectively.

*Claim 1.* $\triangle AA_bA_c \cong \triangle ABC$.
*Proof.* Let $P = BH \cap AC, Q = CH \cap AB$, then it is well known that $\triangle APQ \cong \triangle ABC$. By the dilation with factor 2 centred at $A$, $\triangle APQ$ is sent to $\triangle AA_bA_c$, so we have $\triangle AA_bA_c \cong \triangle ABC$.

*Claim 2.* $\triangle AA_bA_c \cong \triangle AB_aC_a$ and $A_1$ lies on the circumcircle of $\triangle AA_bA_c$, which is centred at $H$.
*Proof.* Since $B_a, C_a$ are reflections of $B, C$ in $AH$, we have $\triangle AB_aC_a \cong \triangle ABC$. Combining this with *Claim 1*, we have $\triangle AA_bA_c \cong \triangle AB_aC_a$ where $A$ is the centre of this similarity. Therefore, $\angle A_c A_1 A_b = \angle A_c AA_b$, meaning $A_1$ lies on $\odot AA_b A_c$. By symmetry, $HA_b = HA = HA_c$, so $H$ is centre of this circle.

*Claim 3.* $\triangle A_1 B_1 C_1 \cong \triangle ABC$.
*Proof.* From *Claim 2* we have
$$\angle C_1 A_1 B_1 = \angle A_c A_1 A_b = \angle A_c AA_b = -\angle CAB$$
and similarly $\angle A_1 B_1 C_1 = -\angle ABC, \angle B_1 C_1 A_1 = -\angle BCA$, which imply $\triangle A_1 B_1 C_1 \cong \triangle ABC$.
Denote the ratio of similitude of $\triangle A_1 B_1 C_1$ and $\triangle ABC$ by $\lambda(= \frac{B_1 C_1}{BC})$, then
$$\lambda = \frac{H_1 A_1}{HA} = \frac{H_1 B_1}{HB} = \frac{H_1 C_1}{HC}.$$
Since $HA = HA_1$ and similarly $HB = HB_1, HC = HC_1$ from *Claim 2*, we get
$$\lambda = \frac{H_1 A_1}{HA_1} = \frac{H_1 B_1}{HB_1} = \frac{H_1 C_1}{HC_1}.$$
Therefore, the circle $A_1 B_1 C_1$ is the Apollonian circle of the segment $HH_1$ with ratio $\lambda$ so the line $HH_1$ passes through $O_1$.