In an acute-angled triangle $A B C$, the median $A M$, the bisector $B K$, and the altitude $C H$ are drawn. Can the area of the triangle formed by the intersection points of these segments be greater than $0.499 S_{\mathrm{ABC}}$?
Consider a right triangle $A B C_{1}$ with legs $A B=1$ and $B C_{1}=2 n$. In this triangle, draw the median $A M_{1}$, the angle bisector $B K_{1}$, and the altitude $C_{1} H_{1}$. The area of the triangle formed by these segments is greater than $S_{\mathrm{ABM}_{1}} - S_{\mathrm{ABK}_{1}}$. It is clear that $S_{\mathrm{ABK}_{1}} = (S / 2) - (S / 2 n)$, where $S = S_{\mathrm{ABC}_{1}}$. Therefore, for sufficiently large $n$, the area of the triangle formed by the segments $A M_{1}$, $B K_{1}$, and $C_{1} H_{1}$ will be greater than $0.499 S$. The point $C_{1}$ can be slightly moved so that the right triangle $A B C_{1}$ becomes an acute triangle $A B C$, while the area of the triangle formed by the intersection points of the segments remains greater than $0.499$ of the area of triangle $A B C$.