Problem 1806

1. Define Key Points and Circles:
- Let $H$ and $O$ be the orthocenter and circumcenter of triangle $ABC$, respectively.
- Let $A_1, B_1,$ and $C_1$ be the feet of the altitudes from $A, B,$ and $C$ respectively.
- Let $P$ and $Q$ be the points where the circumcircles of triangles $AB_1C_1$ and $BC_1A_1$ intersect the circumcircle of triangle $ABC$ at points other than $A$ and $B$, respectively.

2. **Claim: $H$ lies on the radical axis of the circles $(AOQ)$ and $(BOP)$.**
- To prove this, consider the power of point $H$ with respect to these circles.
- Let $A'$ and $B'$ be the antipodes of $A$ and $B$ in the circumcircle of $\triangle ABC$.

3. **Prove $H$ lies on the radical axis:**
- Note that $\angle HPA = \angle HC_1A = 90^\circ = \angle A'PA$, so $A' \in PH$.
- Similarly, $B' \in QH$.
- Let $M$ be the second intersection of $HP$ with $(BOP)$ and $N$ be the second intersection of $HQ$ with $(AOQ)$.
- Since $\angle HMO = \angle PMO = \angle PBO = \angle PBB' = \angle PA'B' = \angle HA'B'$, it follows that $MO \parallel A'B'$.
- Similarly, $NO \parallel A'B'$.
- Hence, $MN \parallel A'B'$.

4. Using Thales's Theorem:
- By Thales's theorem in $\triangle HA'B'$, we have:
$$\frac{HM}{HA'} = \frac{HN}{HB'}$$
- This implies:
$$\rho(H, (AOQ)) = HQ \cdot HN = \frac{HN}{HB'} \cdot (HQ \cdot HB') = \frac{HN}{HB'} \cdot \rho(H, (ABC)) = \frac{HM}{HA'} \cdot (HP \cdot HA') = HP \cdot HM = \rho(H, (BOP))$$
- Therefore, $H$ lies on the radical axis of $(AOQ)$ and $(BOP)$.

5. Case Analysis:
- **Case 1: $AQ \nparallel BP$**
- Let $K = AQ \cap BP$.
- Consider the inversion $\Psi$ with respect to the circumcircle of $\triangle ABC$.
- For each $X \in (ABC)$, $\Psi(X) = X$.
- Hence, $\Psi(AQ) = (AOQ)$ and $\Psi(BP) = (BOP)$, implying $K^* = \Psi(K) \in (AOQ) \cap (BOP) \setminus \{O\}$.
- From the claim, $H \in OK^*$.
- Since $K \in OK^*$, it follows that $O, H, K^*,$ and $K$ are collinear.
- Therefore, lines $AQ, BP,$ and the Euler line $HO$ of $\triangle ABC$ are concurrent.

- **Case 2: $AQ \parallel BP$**
- In this case, $AQBP$ forms an isosceles trapezium.
- The circumcenters of $\triangle AOQ$ and $\triangle BOP$ lie on the common perpendicular bisector of segments $AQ$ and $BP$.
- This implies that $(AOQ)$ and $(BOP)$ are tangent at $O$.
- From the claim, $HO$ is tangent to $(BOP)$ at $O$, implying:
$$\angle HOP = \angle OBP = \angle OPB$$
- Hence, $HO \parallel BP \parallel AQ$.

Conclusion:
The lines $AQ, BP,$ and the Euler line of $\triangle ABC$ are either concurrent or parallel to each other.

$\blacksquare$