Nişanlar və Qanunilik

 Ecuson Universitetinin köməkçisi 

 

Nişan orta məktəb müəllimi

 

 

Nəqliyyat kartı, servis  

 

 

POLIGON - canal YouTube

 

 

EPISODUL 2 :"Nobelurile" matematicii (16 OCT 2021)

la link :  https://www.youtube.com/watch?v=2e6TZwOPKgc (YouTube)

         sau https://www.facebook.com/Poligon-cu-Adrian-Manea-????? (Facebook)

Textul prezentarii :https://adrianmanea.xyz/pages/ep2-nobel.php

 

EPISODUL 1 : Matematica naturală... metaforic vorbind (02 OCT 2021)

la link :  https://www.youtube.com/watch?v=jgMdCGs1sXI (YouTube)

         sau https://www.facebook.com/Poligon-cu-Adrian-Manea-109800248130286 (Facebook)

Textul prezentarii : https://adrianmanea.xyz/pages/ep1-metaforic.php

 

EPISODUL 0 : Prezentare (30 SEP 2021)

la link : https://www.youtube.com/watch?v=MHurTIzRxcE

Textul prezentarii : https://adrianmanea.xyz/pages/ep0-prezentare.php

 

نسيت الحروف ... لبعض الوقت ...

 Titlul original: "Scrisori uitate ... de timp ..."

 

 

De la Virgil(~ius ?) IVANESCU, Dr._Tr._Severin 

20.08.1990

19.11.1990

De la Marian DINCA, Bucuresti

01.09.1990

Student Bogdan DUMITRU (de la Institutul Militar de Transmisiuni SIBIU - U.M. 01606)

03.01.1994

Alexandru OANCEA, Timisoara

21.04.1994

 Dan MILITA, Timisoara

11.04.1992

 

 

29.05.1992

Уласцівасці няроўнасцей, якія выкарыстоўваюцца для рашэння няроўнасцей

 

 Практыкаванні для практыкаванняў

 

 

 

 

Прыклад рашэння

 

Дамашняя работа

Punë me shkrim në Matematikë në Semestrin e I -rë, klasën e 7 -të

 

Eine bedingte algebraische Identität : Ein Problem der 1985 Österreich-Polen Mathematischer Wettbewerb // Uwarunkowana tożsamość algebraiczna : Problem austriacko-polskiego konkursu matematycznego 1985

          See the book

KUCZMA Marcin Emil

144 Problems of the Austrian-Polish Mathematics Competition 1978-1993

The ACADEMIC DISTRIBUTION CENTER, Freeland, Maryland, 1994

at Problem 8.1, page 15. Presented also in CRUX_V12n02_Feb(1986), page 19 (The Olympiad Corner:72).

 

          The statement of the problem

          "If $\;a\;,b\;,c\;$ are distinct real numbers whose sum is zero, prove that

$\left [\frac{b-c}{a}+\frac{c-a}{b}+\frac{a-b}{c} \right ]\cdot \left [ \frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}\right ]=9."\tag{1}$

 

           SOLUTION CiP

          Let's make the notations:

$A:=\frac{b-c}{a}+\frac{c-a}{b}+\frac{a-b}{c},$

$B:=\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}.$

When we bring to the same denominator, the numerator of the expression $A$ is

$A_1=bc(b-c)+ca(c-a)+ab(a-b).$ But from the condition $a+b+c=0$ we have

$$c=-a-b,\;\;-c=a+b. \tag{2}$$

Replacing $c$ in $A_1$ we get

 $A_1\overset{(2)}{=}b\cdot (-a-b)(2b+a)+(-a-b)\cdot a(-2a-b)+ab(a-b)=$

$=(-2b^3-3ab^2-a^2b)+(2a^3+3a ^2b+ab^2)+(a^2b-ab^2)=$

$=(2a^3-2b^3)+(3a^2b-3ab^2)=2(a-b)(a^2+ab+b^2)+3ab(a-b)=$

$=(a-b)(2a^2+5ab+2b^2).$

Let's also note that $2a^2+5ab+2b^2=(2a+b)(a+2b)=(a+a+b)(a+b+b)\overset{(2)}{=}(a-c)(b-c).\tag{*}$

     In conclusion we have

$$A=\frac{(a-b)(b-c)(a-c)}{abc}.\tag{3}$$

     When we bring to the same denominator, the numerator of the expression $B$ is

$B_1=a(c-a)(a-b)+b(a-b)(b-c)+c(b-c)(c-a)\;\underset{(*)}{\overset{(2)}{=}}\;a(-2a-b)(a-b)+$

$+(ab-b^2)(a+2b)+(-a-b)(-2a^2-5ab-2b^2)=(-2a^3+a^2b+ab^2)+(a^2b+$

$+ab^2-2b^3)+(2a^3+7a^2b+7ab^2+2b^3)=9a^2b+9ab^2=9ab(a+b)\underset{(2)}{=}-9abc.$

     In conclusion we have

$$B=\frac{-9abc}{(b-c)(c-a)(a-b)}.\tag{4}$$ 

          Multiplying the relations $(3)$ and $(4)$ we obtain 

$$[A]\cdot [B]=\frac{(a-b)(b-c)(a-c)}{abc} \cdot \frac{-9abc}{(b-c)(c-a)(a-b)}=9.$$

$\blacksquare$

REMARKS CiP

 

PROBLEM MA127 CRUX MATHEMATICORUM V47 No 6

 Pag 275 - En

Pag 276 - Fr

ANSWER CiP 

$\log_5 12=\frac{2\cdot a+b}{1-a}$

     Solution CiP

          We have $\log_5 12 =\log_5(2^2 \cdot 3)=\log_5 2^2+\log_5 3$, so

$\log_5 12 =2\cdot \log_5 2 + \log_5 3. \tag{1}$

          First, $\frac{1}{a}=\log_2 10=\log_2 (2\cdot 5)=\log_2 2 +\log_2 5=1+\frac{1}{\log_5 2}$ so

$\frac{1}{\log_5 2}=\frac {1}{a}-1$, whence it results

$\log_5 2 =\frac {a}{1-a}. \tag{2}$

           Secondly, $\frac{1}{b}=\log_3 10 =\log_3 2 +\log_3 5=\frac{\log_{10} 2}{\log_{10} 3}+\frac{1}{\log_5 3}$, so $\frac{1}{\log_5 3}=\frac {1}{b}-\frac {a}{b}$, whence it results

$\log_5 3=\frac{b}{1-a}. \tag{3}$

          Replacing formulas (2) and (3) in formula (1) we get the answer.

$\blacksquare$

          

REMARK CiP

          More briefly $\log_5 12=\frac{\log_{10} 12}{\log_{10} 5}=\frac{\log_{10} 2^2 \cdot 3}{\log_{10} \frac{10}{2}}=\frac{\log_{10} 2^2 +\log_{10} 3}{\log_{10} 10-\log_{10} 2}=\frac{2\cdot \log_{10} 2 +\log_{10} 3}{1-\log_{10} 2}$ and we get the same answer.

$\blacksquare \blacksquare$

 

Confirmare trimitere

 

Thanks for sending your material to Crux!

achitat c-da/Corespondenta CRUX

Thank you for sending us your solution for one of the problems found
in Crux Mathematicorum.  Our file server has received your submission
and the appropriate editor will be reviewing it in sequence or as
needed.


Below is a receipt confirming the submission you sent us.



Tracking Number: 12831
Received:        Thu Jul 15 07:10:34 2021

From:  Petre Ciobanu
       Scoala Gimnaziala "Samuil Micu" SADU
       Sibiu, Romania
Email: ptr.ciobanu@gmail.com

Type:  Solve a MathemAttic Problem
       (problem MA127)

Files:
  LATEX_source_of_MA127.doc
  PROBLEM_MA127_CRUX_v47_n6.pdf


Comments:

Crux Mathematicorum <crux@cms.math.ca>	10:12 (acum 22 de minute)
către eu