Comments on BARICENTRIC RELATIONS between THREE COLLINEAR POINTS // ҮС КОЛЛИНЕАРНАЙ ТОЧКА икки ардыгар БАРИЦЕНТРИЧЕСКАЙ СЫҺЫАННАР тустарынан комментарийдар

 We will further specify the relationships between the barycentric coordinates of three points on a line, which appear in the Post here. The notations are slightly modified.

          

          Let  $X,\;Y,\;Z$  be three collinear points. The following relations hold :

$\frac{\overline{XY}}{\overline{XZ}}=r\;\tag{1.1}$

$\overrightarrow{OY}=(1-r)\overrightarrow{OX}+r\overrightarrow{OZ}\;\tag{1.2}$

$Y=(1-r)X+rZ\;\tag{1.3}$

and the writing of (1.2)  $\overrightarrow{OY}=r\overrightarrow{OZ}+(1-r)\overrightarrow{OX}$  also says that

$\frac{\overline{ZY}}{\overline{ZX}}=1-r\;\tag{1.4}$

$\frac{\overline{XZ}}{\overline{XY}}=\frac{1}{r}\;\tag{2.1}$

$\overrightarrow{OZ}=\frac{1}{r}\overrightarrow{OY}+(1-\frac{1}{r})\overrightarrow{OX}\;\tag{2.2}$

$Z=\frac{1}{r}Y+(1-\frac{1}{r})X\;\tag{2.3}$

$\frac{\overline{YZ}}{\overline{YX}}=1-\frac{1}{r}\;\tag{2.4}$

$\frac{\overline{YX}}{\overline{YZ}}=-\frac{r}{1-r}\;\tag{3.1}$

$\overrightarrow{OX}=\frac{1}{1-r}\overrightarrow{OY}-\frac{r}{1-r}\overrightarrow{OZ}\;\tag{3.2}$

$X=\frac{1}{1-r}Y-\frac{r}{1-r}Z\;\tag{3.3}$

$\frac{\overline{ZX}}{\overline{ZY}}=\frac{1}{1-r}\;\tag{3.4}$

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.

$\blacksquare$

How many MISTAKES are allowed in a book ? // Hoeveel fouten zijn toegestaan ​​in een boek ?

 Answer: NOT TOO MUCH

Let's take the first problem from a book.

Page 21, Chapter 1 EQUALITIES, Section 1.1 Equalities of one complex variable, Paragraph 1.A, Problem  #1 (in translation)

               "Let  $z\in\mathbb{C}$ such that  $z^2=-3+4\imath$. Calculate:

                       a)  $A=z^2+\bar z^2;$

                       b)  $B=\left | z-\frac{1}{\bar z } \right |;$

                       c)  $C=z+\frac{1}{\bar z};$

                       d)  $D=Re\;z+Im\;z.$"

Do you see the mistakes ?

ANSWER CiP

A. $-6$  ; B. $\frac{4\sqrt{5}}{5}$  ; C. $\pm\frac{6}{5}(1+2\imath)$  ;  D. $\pm 3$.

In reality  $z=\pm(1+2\imath)$

                         Solution CiP

               Let  $z=x+\imath y$. From  $-3+4\imath=z^2=(x+\imath y)^2=(x^2-y^2)+2xy \cdot \imath$

it follow  $x^2-y^2=-3,\;\;2xy=4$  so  $y=\frac{2}{x}\;\;and\;\;x^2-\frac{4}{x^2}=-3$.  Hence

$x^2=1$  and therefore  $z_{1,2}=\pm(1+2\imath).$

          Regardless of the possible values ​​for  $z$, we have the calculations :

$A=z^2+\overline {z^2}=(-3+4\imath)+\overline {(-3+4\imath)}=(-3+4\imath)+(-3-4\imath)=-6;$

     We then have  $|z^2|=|-3+4\imath|=\sqrt{(-3)^2+4^2}=5=|z|^2=|\bar z|^2$  so on one side  $|\bar z|=\sqrt{5}$, and

$B=\left | \frac{z\cdot \bar z-1}{\bar z} \right |=\left | \frac{|z|^2-1}{\bar z} \right |=\left | \frac{|z^2|-1}{\bar z} \right |=\left | \frac{5-1}{\bar z}\right |=\frac{4}{|\bar z|}=\frac{4}{\sqrt{5}};$

$C=\frac{z \cdot \bar z+1}{\bar z}=\frac{|z|^2+1}{\bar z}=\frac{5+1}{\bar z}=\frac{6\cdot z}{          \bar z \cdot z}=\frac{6z}{|z|^2}=\frac{6}{5}\cdot z\;\tag{1}$

     Finally, we still need

$\bar z=\frac{5}{z}$  and  $\frac{1}{z}=\frac{z}{-3+4\imath}$

to calculate  $D=Re\;z+Im\;z=\frac{z+\bar z}{2}+\frac{z-\bar z}{2\imath}=\frac{z+\frac{5}{z}}{2}+\frac{z-\frac{5}{z}}{2\imath}=\frac{z^2+5}{2z}+\frac{z^2-5}{2\imath z}=$

$\overset{z^2=-3+4\imath}{=}\;\;\frac{2+4\imath}{2z}+\frac{-8+4\imath}{2\imath z}=\frac{1}{z}+\frac{2\imath}{z}+\frac{8\imath ^2}{2\imath z}+\frac{2}{z}=\frac{3+6\imath}{z}=$

$=3(1+2\imath) \cdot \frac{1}{z}=3(1+2\imath) \cdot \frac{z}{-3+4\imath}=3(1+2\imath)\cdot \frac{(-3-4\imath)z}{25}=\frac{3(5-10\imath)}{25}\cdot z=\frac{3}{5}(1-2\imath)\cdot z \tag{2}$

Substituting the values  $z_{1,2}$  ​​into (1) and (2) we obtain the answer.

$\blacksquare$

          Remark CiP  Compare with the official solution :