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