Problem 1677

1. Define the Medial Triangle and Key Points:
Let $\Delta A_0B_0C_0$ be the medial triangle of $\Delta ABC$. Here, $B_0$ is the midpoint of $AC$, and $C_0$ is the midpoint of $AB$. Let $D$ be the foot of the altitude from $A$, and $G$ be the centroid of $\Delta ABC$. Let $\omega$ be a circle through $B_0$ and $C_0$ that is tangent to the circumcircle $\Omega$ of $\Delta ABC$ at a point $X \neq A$.

2. **Reflect $D$ and Define Parallel Line:**
Let $D_1$ be the reflection of $D$ over $A_0$. Let the parallel to $BC$ through $A$ intersect $\Omega$ at $T$ different from $A$. Since $G$ is also the centroid with respect to $\Delta ADD_1$ and since $ADD_1T$ is a rectangle, it follows that $D$, $G$, and $T$ are collinear.

3. **Redefine $X$ and Establish Cyclic Quadrilaterals:**
Redefine $X$ as the intersection of $DG$ and $\Omega$ different from $T$. Let $DT \cap AD_1 = H$ and $AG \cap \Omega = P$. From known results (e.g., Greece TST 2018 and All-Russian MO 2018 P2), we have that $DA_0PX$, $HB_0CX$, and $XPEF$ are cyclic quadrilaterals.

4. Tangency and Power of a Point:
Let $Q$ be a point on $BC$ such that $QA$ is tangent to $\Omega$ at $A$. Let $C_0B_0 \cap AQ = M$. Then, $M$ is the center of $\odot(QDA)$. Let $\odot(QDA) \cap \Omega$ at $X'$ different from $X$. Let $A'$ be the $A$-antipode of $\Omega$. Simple angle chasing shows $\angle A'X'D = \angle A'XT = \angle A'XD$, implying $A'XX'D$ is cyclic, which is impossible unless $X \equiv X'$. Hence, $X \in \odot(QDA)$, implying $MX = MA$.

5. Using Power of a Point:
Using the power of a point, we can show $MX$ is tangent to $\Omega$. It is also well-known that $MD$ is tangent to $\odot(A_0B_0C_0)$. Hence, $MC_0 \cdot MB_0 = MD^2 = MA^2 = MX^2$, implying $\odot(B_0C_0X)$ is tangent to $\Omega$.

6. Conclusion:
Since $D$, $G$, and $X$ are collinear, we have proven the required result.

$$\boxed{\text{The points } D, G, \text{ and } X \text{ are collinear.}}$$