4. In triangle $ABC$, medians $BD$ and $CE$ intersect at point $M$. The circle constructed on segment $BM$ as a diameter passes through vertex $C$ and is tangent to line $DE$. It is known that $CM=4$. Find the height $AH$ of triangle $ABC$, the angle $CBD$, and the area of triangle $ABC$.
Answer: $A H=12, \angle C B D=30^{\circ}, S_{A B C}=24 \sqrt{3}$.

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

“Phystech-2016”, mathematics, solutions for ticket 33

$D M=\frac{1}{2} B M=R, C E=\frac{3}{2} C M=6$. Note that in the right triangle $O D T$, the leg $O T=R$, and the hypotenuse $O D=2 R$. Therefore, $\angle T D O=30^{\circ}$. The segment $D E$ is the midline of triangle $A B C$, so $D E \| B C$, $\angle M B C=\angle M D E=30^{\circ}$. From triangle $B M C$, we find that $B C=M C \operatorname{ctg} 30^{\circ}=4 \sqrt{3}$.

The height $A H$ of the original triangle is twice the length of $C E$ (since $C E$ is the midline of triangle $A B H$). Thus, $A H=12$. Then the area of the triangle is $\frac{1}{2} B C \cdot A H=24 \sqrt{3}$.