Problem 1751

1. Define the Problem Setup:
- Consider a circle $(O)$ with fixed points $B$ and $C$ such that $BC$ is not the diameter of $(O)$.
- Let $A$ be an arbitrary point on $(O)$, distinct from $B$ and $C$.
- Define $D$, $J$, and $K$ as the midpoints of $BC$, $CA$, and $AB$, respectively.
- Define $E$, $M$, and $N$ as the feet of the perpendiculars from $A$ to $BC$, $B$ to $DJ$, and $C$ to $DK$, respectively.
- The tangents at $M$ and $N$ to the circumcircle of triangle $EMN$ meet at $T$.

2. Identify Key Points and Properties:
- Let $H$ be the orthocenter of $\triangle ABC$. Then $H$ and $D$ lie on the circumcircle of $\triangle EMN$.
- Let $Q$ be a point on the circumcircle of $\triangle EMN$ such that $HEDQ$ forms a rectangle.

3. Use Harmonic Division:
- The harmonic division property states that $-1 = (B, C; D, \infty_{BC})$.
- Projecting through $H$, we get $(M, N; D, Q)$, implying that $T$ lies on $DQ$ and $TD \perp BC$.

4. Lemma Application:
- Consider the medial triangle $\Delta M_AM_BM_C$ of $\Delta ABC$.
- Let $M$ and $N$ be the feet from $B$ and $C$ to $M_AM_C$ and $M_AM_B$, respectively.
- By the sine rule in $\Delta AMM_C$ and $\Delta ANM_B$, we can show that $AM$ and $AN$ are isogonal with respect to $\angle BAC$.

5. Circumcircle Intersections:
- Let $\odot(BMD)$ and $\odot(CND)$ intersect $\odot(ABC)$ at points $F$ and $G$, respectively.
- Let $FD$ and $GD$ intersect $\odot(ABC)$ at points $L$ and $P$.
- Let $DP$ and $DL$ intersect $\odot(BMD)$ and $\odot(CND)$ at points $R$ and $S$, respectively.

6. Cyclic Quadrilaterals and Reflections:
- Since $CP$ and $BL$ are diameters of $\odot(ABC)$, we have $\angle FBC = \angle FMD = \angle FAC$.
- Since $MD \parallel AC$, $M$ lies on $\overline{AF}$ and $N$ lies on $\overline{AG}$.
- $FG \parallel BC \parallel PL$ and $\angle BDF = \angle CDG$.
- Points $R$ and $S$ are reflections of $F$ and $G$ over $BC$, implying $RS \parallel BC$ and $PRSL$ and $BRSC$ are cyclic.

7. Orthocenter Reflection:
- The reflection $H'$ of $H$ over $BC$ lies on $\odot(BFGC)$, implying $H$ lies on $\odot(BRSC)$.
- Let $U$ and $V$ be the midpoints of $BD$ and $CD$, respectively.
- Since $TD$ is the perpendicular bisector of $UV$, $\Delta TMU \cong \Delta TNV$.

8. Cyclic Quadrilateral and Radical Center:
- Since $BUOT$ is cyclic, we have $\angle TBU = \angle UOD = \angle FAC = \angle FBC = \angle RBU$.
- Thus, $R \equiv TB \cap PG \cap \odot(BMD)$ and $S \equiv FL \cap TC \cap \odot(CND)$.
- Therefore, $T$ is the radical center of $\odot(BMD)$, $\odot(CND)$, and $\odot(BRSC)$, which are all fixed circles.

9. Conclusion:
- Since the radical center of fixed circles is itself fixed, $T$ is a fixed point as $A$ moves on $(O)$.

$\blacksquare$