The game known in Romania as "Don't Be Angry, Brother!" it's basically LUDO. There is also a German version, mentioned in the title. I would have liked to say "Don't Be Upset, Sister!", because alongside Problem S:L25.296 in Post here, there is also its sister, Problem S:L25.295, which I initially neglected.
"Let the triangle $ABC$ and the points $M\in (AB),\; N\in (AC), \;P\in BC$
be such that $\overrightarrow{BC}=2\overrightarrow{CP},\;\frac{MA}{MB}=\frac{1}{2},\;\;and\;\;\frac{NC}{AC}=\frac{2}{5}.$
a) Express the vectors $\overrightarrow{AM},\;\overrightarrow{AN}$ and $\overrightarrow{AP}$ in terms of the vectors $\overrightarrow{AB}\;and\;\overrightarrow{AC}$.
b) Express the vectors $\overrightarrow{MN}$ and $\overrightarrow{MP}$ in terms of the vectors $\overrightarrow{AB}\;and\;\overrightarrow{AC}$.
c) Show that the points $M,\;N,\;P$ are collinear and determine the ratio $\frac{MN}{MP}.$
No author"
ANSWER CiP
a) $\overrightarrow{AM}=\frac{1}{3}\overrightarrow{AB}$
Solution CiP
a) From $\frac{MA}{MB}=\frac{1}{2}$ and $M\in (AB)$ it follow $\frac{\overline{MA}}{\overline{MB}}=\frac{-1}{2}$ and so
$\frac{\overline{AM}}{\overline{AB}}=-\frac{\overline{MA}}{\overline{MB}-\overline{MA}}=-\frac{-1}{2-(-1)}=\frac{1}{3}$, hence $\overrightarrow{AM}=\frac{1}{3}\overrightarrow{AB}.$
From $\frac{NC}{AC}=\frac{2}{5}$ and $N\in (AC)$ it follow $\frac{\overline{NC}}{\overline{AC}}=\frac{2}{5}$ so
$\frac{\overline{AN}}{\overline{AC}}=\frac{\overline{AC}+\overline{CN}}{\overline{AC}}=\frac{\overline{AC}-\overline{NC}}{\overline{AC}}=\frac{5-2}{5}=\frac{3}{5}$, hence $\overrightarrow{AN}=\frac{3}{5}\overrightarrow{AC}$
Finally $\overrightarrow{AP}=\overrightarrow{AC}+\overrightarrow{CP}=\overrightarrow{AC}+\frac{1}{2}\overrightarrow{BC}=\overrightarrow{AC}+\frac{1}{2}(\overrightarrow{AC}-\overrightarrow{AB})=-\frac{1}{2}\overrightarrow{AB}+\frac{3}{2}\overrightarrow{AC}$.
The same answer can be obtained by observing that $\frac{\overline{PB}}{\overline{PC}}=\frac{3}{1}$ and we apply the formulas (1.1) and (3.2)
$\overrightarrow{AP}=\frac{1}{1-3}\overrightarrow{AB}-\frac{3}{1-3}\overrightarrow{AC}.$
b) $\overrightarrow{MN}=\overrightarrow{AN}-\overrightarrow{AM}=\frac{3}{5}\overrightarrow{AC}-\frac{1}{3}\overrightarrow{AB}$ hence the answer.
$\overrightarrow{MP}=\overrightarrow{AP}-\overrightarrow{AM}=\left (-\frac{1}{2}\overrightarrow{AB}+\frac{3}{2}\overrightarrow{AC} \right )-\frac{1}{3}\overrightarrow{AB}=-\frac{5}{6}\overrightarrow{AB}+\frac{3}{2}\overrightarrow{AC}$
c) With the results from b) we have
$\overrightarrow{MN}=-\frac{1}{3}\overrightarrow{AB}+\frac{3}{5}\overrightarrow{AC}=\frac{2}{5} \cdot \left (-\frac{5}{6}\overrightarrow{AB}+\frac{3}{2}\overrightarrow{AC} \right )=\frac{2}{5}\overrightarrow{MP}$
hence the points $M,\;N,\;P$ are collinear and $\frac{\overline{MN}}{\overline{MP}}=\frac{2}{5}.$
The collinearity of points $M,\;N,\;P$ also results from observing the $\overrightarrow{AP}=-\frac{1}{2}\overrightarrow{AB}+\frac{3}{2}\overrightarrow{AC}=-\frac{1}{2}\cdot 3\overrightarrow{AM}+\frac{3}{2}\cdot \frac{5}{3}\overrightarrow{AN}$ or,
$$\overrightarrow{AP}=\frac{5}{2}\overrightarrow{AN}-\frac{3}{2}\overrightarrow{AM}=\frac{1}{\frac{2}{5}}\overrightarrow{AN}+\left ( 1-\frac{1}{\frac{2}{5}}\right )\overrightarrow{AM}$$
to which we apply formulas (1.1) and (2.2).
$\blacksquare$
