4. In triangle $A B C$, medians $B D$ and $C E$ intersect at point $K$. The circle constructed on segment $C K$ as a diameter passes through vertex $B$ and is tangent to line $D E$. It is known that $D K=3$. Find the height $A H$ of triangle $A B C$, the radius of the circle, and the area of triangle $A B C$.
Answer: $A H=18, R=6, S_{A B C}=54 \sqrt{3}$.

Solution. Let the center of the circle be denoted as $O$, the point of tangency of the line $D E$ with the circle as $T$, and the radius of the circle as $R$. Since the medians intersect at a point dividing them in the ratio $2: 1$ from the vertex,

“Phystech-2016", mathematics, solutions for ticket 35 $E K=\frac{1}{2} C K=R, B D=3 D K=9, B K=2 D K=6$. Note that in the right triangle $OET$, the leg $O T=R$, and the hypotenuse $O E=2 R$. Therefore, $\angle T E O=30^{\circ}$. The segment $D E$ is the midline of triangle $A B C$, so $D E \| B C, \angle K C B=\angle K E D=30^{\circ}$. From triangle $B K C$, we find that $B C=B K \operatorname{ctg} 30^{\circ}=6 \sqrt{3}$, $2 R=C K=\frac{B K}{\sin 30^{\circ}}=12$.

The height $A H$ of the original triangle is twice the length of $B D$ (since $B D$ is the midline of triangle $A C H$). Thus, $A H=18$. Then the area of the triangle is $\frac{1}{2} B C \cdot A H=54 \sqrt{3}$.