A solution “pulled out of a hat”. An explanation of the trick // Uma solução "tirada da cartola". Uma explicação do truque

 In the last issue of the Exercise Supplement of the Mathematics Gazette, most of the problems seemed trivial. Maybe this S:L 25. 411 is too, but look at the solution I found.

  "S:L25.411.  Consider the sequence  $(x_n)_{n\in\mathbb{N}}$, defined by  $x_{n+1}=\frac{4x_n+1}{2x_n+3}$, for any 

           $n\in\mathbb{N}$  and  $x_0=0$. Show that  $$\lim_{n \to \infty}x_n=1.$$

[Author] Ludovica LAZĂR, Năsăud"

ANSWER  CiP

$x_n=\frac{5^n-2^n}{5^n+2^{n+1}}=\frac{1-\left (\frac{2}{5}\right )^n}{1+2\cdot \left (\frac{2}{5}\right )^n}\;\underset{n \to \infty}{\to} \;1$

   

                           Solution CiP

               Let's consider the sequence

$y_n=\frac{x_n-1}{2\cdot x_n+1}. \tag{1}$

The reverse connection is

$x_n=\frac{1+y_n}{1-2\cdot y_n}. \tag{2}$

The initial conditions are

$x_0=0,\;\;\;y_0=-1. \tag{3}$

          Let's calculate now

$y_{n+1}\;\overset{(1)}{\underset{n\to n+1}{=}}\;\frac{x_{n+1}-1}{2\cdot x_{n+1}+1}\overset{def}{=}\frac{\frac{4x_n+1}{2x_n+3}-1}{2\cdot \frac{4x_n+1}{2x_n+3}+1}=\frac{4x_n+1-2x_n-3}{8x_n+2+2x_n+3}=\frac{2x_n-2}{10x_n+5}=\frac{2}{5}\cdot \frac{x_n-1}{2x_n+1}\overset{(1)}{=}\frac{2}{5}\cdot y_n.$

The equality of the extreme terms tells us that the sequence  $(y_n)_{n\in\mathbb{N}}$  is a geometric progression. So

$y_n=y_0\cdot \left (\frac{2}{5}\right )^n\overset{(3)}{=}-\left (\frac{2}{5}\right )^n,\;\;n\in\mathbb{N} \tag{4}$

          Substituting the value from  (4)  into  (2)  we obtain the formula for  $x_n$  mentioned in the Answer, as well as the value of the limit.

$\blacksquare$

........................................................................................................................................

               Recurrence relations of the form 

$x_{n+1}=\frac{a\cdot x_n+b}{c\cdot x_n+d}\;,\;\;a\cdot d-b\cdot c\neq 0 \tag{H}$

have an unquantified prevalence in Problem Collection books. 

In the book BĂTINEȚU D.M. "Şiruri..." (ALBATROS Publishing House, Bucharest, 1979 - click on the link to download) I came across problem 1.184, page 174-176, together with the solution. With the mention "Bacalaureat, France, 1975".

THAT'S WHERE I TAKEN THE IDEA OF SOLUTION

Luckily, the statement suggested the solution method. In translation

"1.184.  Consider the sequences  $(u_n)_{n\in \mathbb{N}},\;(v_n)_{n\in \mathbb{N}}$  where  $u_0=1$  and

$u_n=\frac{2\cdot u_{n-1}-1}{2\cdot u_{n-1}+5}\;,(\forall)n\in\mathbb{N}^*\;\;;\;\;v_n=\frac{2\cdot u_n+1}{u_n+1}\;,(\forall) n\in\mathbb{N}. \tag{5}$

  $a^{\circ}$)  Show that the sequence  $(v_n)_{n\in\mathbb{N}}$  is a geometric progression,

                         calculate its ratio and  $\lim_{n\to \infty}v_n$.

                                  $b^{\circ}$)  Express  $u_n$  in terms of  $n$  and establish its nature.

(Bacalaureat, Franța, 1975)"

       [The answers are :  $v_n=\frac{3}{2}\cdot \left (\frac{3}{4} \right )^n,\;u_n=\frac{1-v_n}{v_n-2}=\frac{2\cdot 4^n-3^{n+1}}{3^{n+1}-4^{n+1}}\underset{n\to \infty}{\to} -\frac{1}{2}$]

               Fortunately for Romanian-speaking readers, there is the book (I think I'm missing the electronic version) AVADANEI C., AVADANEI N., BORȘ C., CIUREA C.  "De la matematica elementară spre matematica superioară" (Ed. ACADEMIEI [R.S.] România, București, 1987)". Here, Chapter I is dedicated to a systematic study of sequences defined by homographic recurrence relations, pages 7 - 68.

See also here  !!

[In this book I found Problem 1.184 above, with the mention that it appeared under number 15 896 in GMB no. 9, 1978. Unfortunately the citation is erroneous, it is probably a number from 1976, I have not identified it exactly yet.]

Magic Trick (1) Explanation

For 

$x_{n+1}=\frac{4\cdot x_n+1}{2\cdot x_n+3} \tag{6}$

let's look for a transformation of type 

$y_n=\frac{a\cdot x_n+b}{c\cdot x_n+d} \tag{7}$

such that the latter is a geometric progression. 

A simple calculation gives us  $y_{n+1}\underset{(7)}{=}\frac{a\cdot x_{n+1}+b}{c\cdot x_{n+1}+d}\overset{(6)}{=}\frac{a\cdot \frac{4x_n+1}{2x_n+3}+b}{c\cdot \frac{4x_n+1}{2x_n+3}+d}=\frac{(4a+2b)\cdot x_n+a+3b}{(4c+2d)\cdot x_n+c+3d}.$

          Let's try to find two numbers  $\lambda\;and\;\mu$  such that the previous result is equal to  $\left (\frac{\lambda}{\mu}\right )\cdot y_n$.  We want to have the equality

$\frac{(4a+2b)x_n+a+3b}{(4c+2d)x_n+c+3d}=\frac{\lambda}{\mu}\cdot \frac{ax_n+b}{cx_n+d} \tag{8}$

Then, let's try, maybe we have a chance to succeed, to fulfill the equations

$(4a+2b)\cdot x_n+a+3b=\lambda \cdot (a\cdot x_n+b) \tag {9'}$

$(4c+2d)\cdot x_n+c+3d=\mu \cdot (c\cdot x_n+d) \tag {9"}$

Oh, and if we're not asking too much, it wouldn't hurt to have the equations

$\begin{cases}4a+2b=\lambda \cdot a\\\;a+3b=\lambda \cdot b\\4c+2d=\mu\cdot c\\\;c+3d=\mu\cdot d\end{cases} \tag{10}$

     Let's choose from (10) the first two equations :

$\begin{cases}(4-\lambda)\cdot a+\;\;\;\;\;\;\;\;\;\;\;2b=0\\\;\;\;\;\;\;\;\;\;\;\;\;\;a+(3-\lambda)\cdot b=0.\end{cases} \tag{11}$

For the system of equations with unknowns  $a\; and \;b$  to admit non-trivial solutions, it is necessary and sufficient that its matrix be singular, that is

$\left |\begin{matrix}4-\lambda&2\\1&3-\lambda\end{matrix}\right |=0 \tag{12}$

$\Leftrightarrow \lambda^2-7\cdot\lambda +10=0 \tag{13}$

Equation (13) has the roots  $\lambda_1=2,\;\;\lambda_2=5$. 

Choosing  $\lambda_1=5$ the system (11) is written  $\begin{cases}2\cdot a+2\cdot b=0\\a+b=0\end{cases}$  hence  $b=-a$.

          The last two equations in (10) lead to the same system (11) (of course with different notations), and we have the same equation (12)2. Now we choose  $\mu=5$  and we have the system  $\begin{cases}-c+2\cdot d=0\\c-2\cdot d=0\end{cases}$  hence  $c=2\cdot d$.

          So the possible transformations in (7) are  $y_{n+1}=\frac{a\cdot x_n-a}{2d\cdot x_n+d}=\frac{a}{d}\cdot \frac{x_n-1}{2\cdot x_n+1}$. With  $a=1,\;d=1$  we find what we chose in (1). 

                    REMARK  CiP

                   We see in  (H)  a homographic (or fractionally-linear) transformation

$x\mapsto \frac{a\cdot x+b}{c\cdot x+d},\;\;ad-bc \neq 0$

This can be represented by a matrix

$A=\left (\begin{matrix}a&b\\c&d\end{matrix}\right )$

The connection is not just a convention: there is a deep structural correspondence.

          These constitute Möbius transformation. They form a group, often noted  $\mathbb{PSL}(2)$.

            The composition of transformations corresponds to the multiplication of matrices.

             Since matrices obtained from a particular one and multiplied by an arbitrary nonzero scalar give us the same homographic transformation, it is natural that they act on a projective vector  $\textbf{x}=\left (\begin{matrix}x\\1\end{matrix} \right )$.

             It is also not difficult to see that  (12) is the characteristic equation associated with the matrix  $\left (\begin{matrix}4&2\\1&3\end{matrix}\right )$. Why it is not exactly the matrix associated with the transformation (6), i.e.  $\left( \begin{matrix}4&1\\2&3\end{matrix}\right)$  but its transpose, I leave to your discretion.

Equation in real numbers // Ligning i reelle tal

 It's problem E : 17 361  in the picture.

 In translation : "E:17361.  Solve the equation  $[x]=\sqrt{\{x\}\cdot (x+1)}$  in the set of

                            real numbers.

 [Author]  Ionuţ ALEXUC Timişoara"

Remark CiP :  No matter how often they are used, the notations  $\{x\},\;[x]$  no longer have any presentation.  : {x} is fractional part and [x] is the integer part of the real number x.

In my answer S means the set of solutions to the given equation. The problem is proposed for 7th grade (students aged 12-14).

ANSWER CiP

$S=\{0,\;\sqrt{2}\}$

               

                               Solution CiP

               The domain on which the equation is defined is the interval  $x\in [-1,\;+\infty)$. (Because  $\{x\}\geqslant 0$)

          But if  $x<0$ , then  $[x]\leqslant -1$,  so strictly negative and then the equation cannot have such solutions. It remains that  $x\geqslant 0$.

           For  $0\leqslant x<1$  we have  $[x]=0$, and how  $x+1\neq 0$ it must  $\{x\}=0$, so  $x=0$.

           For  $x\geqslant 1$ , from  $x-1<[x]\leqslant x\;\Rightarrow\;\;x-1<\sqrt{\{x\}\cdot (x+1)}\leqslant x$

Hence  $(x-1)^2<\{x\}\cdot (x+1)\;\;\Rightarrow\;\{x\}>\frac{(x-1)^2}{x+1}$. But  $\{x\}<1$  so  $1>\frac{(x-1)^2}{x+1}$

which is equivalent to  $x+1>x^2-2x+1\;\Leftrightarrow\;3x>x^2\;\Leftrightarrow\;3>x$.

          So we can have  $[x]=1\;or\;[x]=2$.

          If  $[x]=1$  the equation is written  $1=\sqrt{\{x\}\cdot (1+\{x\}+1)}$

$\Leftrightarrow\;1=\{x\}\cdot(\{x\}+2)\;\Leftrightarrow\;\{x\}^2+2\{x\}=1\;\Leftrightarrow\;(\{x\}+1)^2=2\;\Leftrightarrow\;\{x\}+1=\sqrt{2}.$

Hence  $x=[x]+\{x\}=1+\{x\}=\sqrt{2}$.

          If  $[x]=2$  the equation is written  $2=\sqrt{\{x\}\cdot (2+\{x\}+1)}$

$\Leftrightarrow\;4=\{x\}\cdot (\{x\}+3)\;\Leftrightarrow\;\{x\}^2+3\{x\}=4\;\Rightarrow\;\{x\}=1$  what is not possible.

Both values ​​found for  $x$  verify the equation.

$\blacksquare$

Reading GROSSWALD Emil "Representations of Integers as Sums of Squares" // GROSSWALD Emil "Representations of Integers as Sums of Squares" lesen

 Ein sehr schwer zu findendes Buch. Du kannst es dir hier ansehen.

A very hard-to-find book. You can view it here.

I have some of the books mentioned below in my Electronic Library.

The first encounter with the problem happened while reading SIERPIŃSKI W. Elementary Theory of Numbers (1988). On page 449 we have Problem 2(stated as Exercise) :

   " 2.   Find the complex integers  $x+y\imath$ which

 are representable as sums of the squares of two

 complex integers"

The exercise only has the answer, without any indication, not even bibliographical, of how to obtain it.

Read this solution yourself !!

A necessary condition is:  $y$  is an even number.

Then, for the number  $x+2y\imath$  we have :

$x+2y\imath\;=$  sums of the squares of two complex integers  $\Leftrightarrow$

$\Leftrightarrow$   NOT  both of  $\frac{x}{2}\;and\;y$  are odd integers

This formulation of the necessary and sufficient condition is inspired by NIVEN's article. The discussion is presented in more detail in GROSSWALD, pages 194-195.

        SIERPIŃSKI formulates, as I invited you to see, the necessary and sufficient condition in this way :

in  $x+y\imath\;\;y$  should be even and, in the case  $x=4t+2,\;\;y$  should be divisible by 4

Moreover, the corresponding writing is also presented, which no matter how much I searched, I could not find the way in which it was discovered.

ANSWER WS

$4t+4u\imath=[(t+1)+u\imath]^2+[u+(1-t)\imath]^2 \tag{1}$

$4t+(4u+2)\imath=[(t+u+1)+(u+1-t)\imath]^2+[(t-u)+(t+u)\imath]^2 \tag{2}$

$(2t+1)+2u\imath=[(t+1)+u\imath]^2+[u-t\imath]^2 \tag{3}$

$(4t+2)+4u\imath=[(t+u+1)+(u-t)\imath]^2+[(t-u+1)+(t+u)\imath]^2 \tag{4}$

                    Examples CiP

          Formulas (1)-(4) cover all possible cases of the numbers  $x+2y\imath$.  Indeed:

     -  when  $x \neq 4t+2$  we can have

                                                 $x=4t$,  when for  $y=2u$  we have (1)

 and for  $y=2u+1$  we have (2)                 

                                                  $x=4t+1\;\;or\;\;x=4t+3$, written together as

                                                            $x=2t+1$, and  $y$  can be any, having (3)

     -  when  $x=4t+2$  then necessarily  $y=2u$, and we have (4)

Squares in the Set of Gauss Integers and the Pythagorean Triple Equation // กำลังสองในเซตของจำนวนเต็มเกาส์และสมการสามเท่าของพีทาโกรัส

 

          Complex numbers of the form  $a+\imath b$  with  $a,b \in \mathbb{Z}$  are called Gaussian integers.

          Here we will consider the problem of finding Gaussian integers that are "perfect squares". For example, we are interested in when the result  $\sqrt{a+\imath b}$ is still a Gaussian integer.

           The problem is solved for example in W. Sierpinski (chapter XIII is dedicated to Gauss integers), page 450, Problem 4. We see that this problem is related to Pythagorean triples. This problem is not completed, unlike Problem 2 about representing as sums of two squares.

          What is said here is only a necessary and sufficient condition that solves our problem :

      "The complex integer  $a+\imath b$  is the square of a complex integer if and only if 

$a^2+b^2=c^2,\;\;\;c+a=2x^2,\;\;\;c-a=2y^2$

   where  $c$ is a natural number and  $x,\;y$  are rational integers. Then

$a+\imath b=(\pm x\pm y\imath)^2$

   the sign should be identical if  $b>0$  and opposite if  $b<0$."

       So the perfect squares in the set of Gauss integers are related to the solutions of the equation of Pythagorean numbers. As shown in another post, for the numbers  $a$  and  $b$  we have the possibilities :

$a=d\cdot 2st,\;\;b=d\cdot (s^2-t^2) \tag{1}$

$a=d\cdot (s^2-t^2),\;\;b=d\cdot 2st \tag{2}$

where  $d\in \mathbb{Z}$  and  $s,\;t$  are coprime integers of opposite parity.

          In the first case  $c+a=d\cdot (s^2+t^2)+d\cdot 2st=d\cdot (s+t)^2$, and $c-a=d\cdot (s^2+t^2)-d\cdot 2st=d\cdot (s-t)^2$. Hence

$\frac{c\pm a}{2}=\frac{d}{2}\cdot (s \pm t)^2 \tag{3}$

          In the second case  $c+a=d\cdot (s^2+t^2)+d\cdot (s^2-t^2)=2ds^2$, and  $c-a=d\cdot (s^2+t^2)-d\cdot (s^2-t^2)=2dt^2$,  hence

$\frac{c+a}{2}=d\cdot s^2,\;\; \frac{c-a}{2}=d\cdot t^2 \tag{4}$

So for  $\frac{c\pm a}{2}$  to be perfect squares,  $d$  must be  $2\alpha^2$ in (3) and  $\alpha^2$ in (4).

Answer CiP: The perfect squares in the set of Gaussian integers are

$4\alpha^2st+2\alpha^2(s^2-t^2)\imath=\left ( \pm\alpha[(s+t)+(s-t)\imath]\right )^2$

$\alpha^2(s^2-t^2)+2\alpha^2st \imath=\left (\pm \alpha [s+t\imath] \right )^2$

where  $\alpha \in \mathbb{Z}$, and  $s,t$ are coprime integers of opposite parity. 

          Examples CiP

$3+4\imath=(2+\imath)^2=(-2-\imath )^2$ :   $3^2+4^2=5^2,\;\frac{5\pm 3}{2}\in \{4,\;1\}$

$2\imath=(1+\imath)^2=(-1-\imath)^2$  :   $0^2+2^2=2^2,\;\frac{2\pm 0}{2}=1$

$5\pm 12 \imath=(3\pm 2\imath)^2$   :   $5^2+12^2=13^2,\;\frac{13\pm 5}{2}\in \{9,\;4\}$

$12\pm 5\imath \neq perfect\;squares$   :    $12^2+5^2=13^2$  although  $\frac{13\pm 12}{2}\in \left \{\frac{25}{2},\;\frac{1}{2}\right \}$

$24\pm 10\imath=(5\pm \imath)^2$   :   $24^2+10^2=26^2,\;\frac{26\pm 24}{2}\in\{25,\;1\}$

Problem E : 17346

 From GMB 11/2025, page 554. In translation :

  "E:17346.  A natural number  $n$ gives when divided by 7 the remainder 4

                       and when divided by 9 the remainder 5. Find out what remainder is

                       obtained  when dividing  $n$  by 63.

 [Author] Ștefan GOBLEJ, Curtea de Argeș"

ANSWER CiP

32

                            Solution CiP

                    Let  $a$ and  $b$  be the quotients of the divisions of  $n$  by 7 and 9 respectively. Then

$n=7\cdot a+4,\;\;\;n=9\cdot b+5 \;\;\;\;\;a,\;b\in \mathbb{N}\tag{1}$

The equation  $7\cdot a+4=9\cdot b+5$   is equivalent to

$7\cdot a-9\cdot b=1\;\;\;\;\;\;a,b \in \mathbb{N} \tag{2}$

Noting that  $7\cdot 4-9\cdot 3=28-27=1$  , equation  (2)  has a particular solution  $a_0=4,\;b_0=3$.

So the general solution of this [linear Diophantine] equation is

$a=4+9\cdot k,\;\;b=3+7\cdot k,\;\;\;k \in \mathbb{N} \tag{3}$

Substituting the number  $a$  from (3) into (1) we obtain  $n=7\cdot (4+9\cdot k)+4$  so  

$$n=63 \cdot k+32$$

and from this it follows that the remainder of dividing the number  $n$  by 63 is 32.

$\blacksquare$

             Remark CiP  We could have thought like that too, without the theory of Diophantine equations :

$7\cdot a\overset{(2)}{=}1+9\cdot b=1+2\cdot b+7\cdot b\;\Rightarrow\; 7\mid 1+2\cdot b\;\Rightarrow\; 2\cdot b+1=7\cdot k_1,\;k_1 \in \mathbb{N}$. But in the equality  $2\cdot b=7\cdot k_1-1$  we must have  $k_1=odd\;number,\;\; k_1=2\cdot k+1$  so  $2\cdot b=7(2\cdot k+1)-1=14 \cdot k+6$  and from here we obtain  $b=7\cdot k+3$  that is precisely  (3).

          COPILOT's solution : 

We write the numbers that give remainder of 4 when divided by 7 :

4, 11, 18, 25, 32, 39, 46, 53, 60, 67, 74, ...

We check which of them gives a remainder of 5 when divided by 9 :

4  mod 9  = 4

11  mod 9  = 2

18  mod 9  = 0

25  mod 9  =7

32  mod  9  =5  :  BINGO

(CiP note : After at most nine (=9) attempts, the result is definitely found, whether it exists or not.)

So the number that satisfies both conditions is  $n \equiv 32\;\;(mod\;63)$  , hence the remainder when dividing  $n$  by 63 is 32.

LUDO // oder Mensch ärgere dich nich // Another calculation problem with Vectors

           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.

In translation :

"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)  $\overrightarrow{AM}=\frac{1}{3}\overrightarrow{AB}$

$\overrightarrow{AN}=\frac{3}{5}\overrightarrow{AC}$

$\overrightarrow{AP}=-\frac{1}{2}\overrightarrow{AB}+\frac{3}{2}\overrightarrow{AC}$

b)  $\overrightarrow{MN}=-\frac{1}{3}\overrightarrow{AB}+\frac{3}{5}\overrightarrow{AC}$

$\overrightarrow{MP}=-\frac{5}{6}\overrightarrow{AB}+\frac{3}{2}\overrightarrow{AC}$

c)  $\overrightarrow{MN}=\frac{2}{5}\overrightarrow{MP}$

$\frac{MN}{MP}=\frac{2}{5}$

                         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).

$\blacksquare$

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

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

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

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

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

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

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

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

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

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