Problem S:L25.296 of Vector Calculus in Parallelogram // Parallelogrammda vektorlı esaplavnıñ 296-nci meselesi

 "Gazeta Matematika. Meşğuliyetlernen qoşma" 10/2025 sanlı mecmuasından alındı, 10 saife.

          To further practice vector calculus, consider the problem (in translation):

 "Consider the parallelogram  $ABCD$  and the points  $M\in AB,\; N \in AC$

              such that  $\overrightarrow{AM}=\frac{1}{4}\overrightarrow{AB},\;\; \overrightarrow{AN}=\frac{1}{5}\overrightarrow{AC}$. Express the vectors  $\overrightarrow{DN},\; \overrightarrow{MN}$

  in terms of the vectors  $\overrightarrow{AB}$  and $\overrightarrow{AD}$ , and then show that

                                           the points  $D,\; N,\; M$ are collinear.

 No author."

We will use formulas from the February 20, 2024 Post "Barycentric coordinate on the Straight Line".

Figure according to the statement. The green line  $k$  is a suggestion of the conclusion.

ANSWER  CiP

See  (1)  and  (3)

$$\overrightarrow{DM}=\frac{5}{4}\overrightarrow{DN}=5\overrightarrow{NM}$$

                    Solution CiP

          Let us first consider the point  $M$.  From  $\overrightarrow{AM}=\frac{1}{4}\overrightarrow{AB}\;\;\Leftrightarrow\;\;\frac{\overline{AM}}{\overline{AB}}=\frac{1}{4}$, it 

follows by the Corollary to the LEMMA  $\overrightarrow{DM}=\frac{3}{4}\overrightarrow{DA}+\frac{1}{4}\overrightarrow{DB}$  which is further equal to

$\frac{3}{4}(-\overrightarrow{AD})+\frac{1}{4}(\overrightarrow{DA}+\overrightarrow{AB})=-\frac{3}{4}\overrightarrow{AD}-\frac{1}{4}\overrightarrow{AD}+\frac{1}{4}\overrightarrow{AB}$  so

$$\overrightarrow{DM}=\frac{1}{4}\overrightarrow{AB}-\overrightarrow{AD} \tag{1}$$

From  $\overrightarrow{AN}=\frac{1}{5}\overrightarrow{AC}\;\;\Leftrightarrow\;\;\frac{\overline{AN}}{\overline{AC}}=\frac{1}{5}$  it also results  $\overrightarrow{DN}=\frac{4}{5}\overrightarrow{DA}+\frac{1}{5}\overrightarrow{DC}$  so

$$\overrightarrow{DN}=\frac{1}{5}\overrightarrow{AB}-\frac{4}{5}\overrightarrow{AD} \tag{2}$$

Finally

$\overrightarrow{MN}=\overrightarrow{DN}-\overrightarrow{DM}\;\underset{(1),(2)}{=}\;\left ( \frac{1}{5}\overrightarrow{AB}-\frac{4}{5}\overrightarrow{AD}\right )-\left (\frac{1}{4}\overrightarrow{AB}-\overrightarrow{AD}\right )$  so

$$\overrightarrow{MN}=-\frac{1}{20}\overrightarrow{AB}+\frac{1}{5}\overrightarrow{AD}. \tag{3}$$

          From (1), (2) and (3) we observe the equations  $\overrightarrow{DM}=\frac{5}{4}\overrightarrow{DN}=5\overrightarrow{NM}$,  and it results that the points  $D,\;N,\;M$  are collinear.

$\blacksquare$