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
Практыкаванні для практыкаванняў
Прыклад рашэння
Дамашняя работа
See the book
KUCZMA Marcin Emil
144 Problems of the Austrian-Polish Mathematics Competition 1978-1993
The ACADEMIC DISTRIBUTION CENTER, Freeland, Maryland, 1994
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
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.
$\blacksquare \blacksquare$
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Pag 238 - En
Pag 239 - Fr
ANSWER CiP
If $7 \mid k$ then solutions are $(k,0),\;(k,k),\;(\frac{3k}{7},-\frac{k}{7})$,
and, if any, the pairs $(x_1,y_1),\;(k-x_1+y_1,y_1).$
If $7 \nmid k$ then the solution $(\frac{3k}{7},-\frac{k}{7})$(although rational),
is not made up of integers; the others remain valid.
Examples
If $k=-4$, the equation has solutions
$(-4,0),\;(-4,-4),\;(-1,-3),\;(-6,-3).$
If $k=7$, the equation has solutions
$(7,0),\;(7,7),\;(3,-1),\;(3,6),\;(10,6).$
Solution CiP
There are two pairs $(x,y)$ that obviously check the equation: $(k,0)$ and $(k,k)$.
The given equation is equivalent to
$$x^2-xy+2y^2-kx-ky=0\; ;\; x+y \neq 0. \tag{1} \label{eq1}$$
Let $(x_1,y_1)$ be a solution of equation (1), with the condition $x_1+y_1 \neq 0$. This means that the equation with unknow $x$
$$x^2-(k+y_1)x+2y_1^2-ky_1=0 \tag{2}$$
has the solution $x=x_1$. Being of the second degree for $x$, she has another solution $x_2$ which, according to Viete's Relations, she checks $x_1+x_2=k+y_1$, so $x_2=k-x_1+y_1.$
For the pair $(x_2,y_1)$ to fit the given problem it must be $x_2+y_1 \neq 0$. But $x_2+y_1=0 \; \Leftrightarrow \;k-x_1+2y_1=0 \; \Leftrightarrow \; x_1=k+2y_1$, and writing that $k+2y_1$ check equation (2) we immediately get $y_1=0$, then $x_1=k$.
So for any solution $(x_1,y_1) \neq (k,0)$ of the equation (1) we get another solution by the transformation
$$(x_1,y_1)\; \mapsto \; (k-x_1+y_1,y_1). \tag{3}$$
We have $(k,k) \overset{(3)}{\mapsto} (k-k+k,k)=(k,k)$, so by transforming (3) we do not get a NEW solution. Let's look for all the solutions that remain invariant through this transformation, so $k-x_1+y_1=x_1$, whence $y_1=2x_1-k$. If we write that $(x_1,2x_1-k)$ check equation (1)
$$x_1^2-x_1(2x_1-k)+2(2x_1-k)^2-kx_1-k(2x_1-k)=0$$
$\Leftrightarrow \;7x_1^2-10kx_1+3k^2=0$ we get $x_1=k$ or $x_1=\frac{3k}{7}$. In the secon case $y_1=-\frac{k}{7}$. So the only invariant pairs by transformation (3) are $(k,k)$ and $(\frac{3k}{7}, -\frac{k}{7})$.
So, if $k \;\vdots \; 7$ (means "k is divisible by 7"), then the equation (1) has three solutions $(k,0),\; (k,k),\;(\frac{3k}{7},-\frac{k}{7})$ to which, if there are others $(x_1,y_1)$, an even number of solutions $(x_1,y_1)$ and $(k-x_1+y_1,y_1)$ is added. In total an odd number.
Otherwise ($7 \nmid k$) the solution $(\frac{3k}{7},-\frac{k}{7})$ does not consits of integers, and the number of solutions is even.
$\blacksquare$
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