51st Ukrainian National Mathematical Olympiad, 4th Round

Answer: No, it cannot.

Suppose, it is possible (fig. 27). Let $O$ be the point of intersection of the lines $XF$, $YB$ and $ZD$. Then by the Ceva's theorem, we have $$\frac{YD}{DX}\cdot \frac{XB}{BZ}\cdot \frac{ZF}{FY}=1.$$

$$\begin{array}{rl}\frac{YD}{DX}& =\frac{YD}{CD}\cdot \frac{CD}{DX}=\frac{YE}{EF}\cdot \frac{CD}{DX},\\ \frac{XB}{BZ}& =\frac{XB}{AB}\cdot \frac{AB}{BZ}=\frac{XC}{CD}\cdot \frac{AB}{BZ},\\ \frac{ZF}{FY}& =\frac{ZF}{FE}\cdot \frac{FE}{FY}=\frac{ZA}{AB}\cdot \frac{FE}{FY}.\end{array}$$ $$Therefore,$$ 1 = \frac{YD}{DX} \cdot \frac{XB}{BZ} \cdot \frac{ZF}{FY} = \left( \frac{YE}{EF} \cdot \frac{CD}{DX} \right) \cdot \left( \frac{XC}{CD} \cdot \frac{AB}{BZ} \right) \cdot \left( \frac{ZA}{AB} \cdot \frac{FE}{FY} \right) = \frac{YE}{YF} \cdot \frac{XC}{XD} \cdot \frac{ZA}{ZB} < 1, $$sinceeachofthefractions$\frac{YF}{XC}$and$\frac{ZA}{XB}$islessthan1.Thiscontradictionprovesthatthelines$XF$,$YB$and$ZD$ cannot be concurrent.