"Datul pe BRAZDĂ" : condiție NECESARĂ sau SUFICIENTĂ ?? // The condition is N E C E SS A R Y or S U FF I C I E N T ??

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 "... Începi să te dai pe brazdă ..." 

Dar ce urmează  DUPĂ  cele trei  ...  ??

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          I found the following Problem on page 76 :

"C : 9.  Let  $a_1,\; a_2,\,\dots\;,a_{10}$  be real numbers such that $a_{10}\neq 0$  and

$\frac{a_1}{a_2+a_3+\dots +a_{10}}=\frac{a_1+a_2}{a_3+a_4+\dots +a_{10}}=\dots=\frac{a_1+a_2+\dots +a_9}{a_{10}}=K.$

Show that $K\in \left \{-1,\;\frac{a_1}{a_{10}} \right \}$.  Generalization.

{ Author : }  Mihai DICU, teacher, Craiova"

ANSWER CiP

The problem's assumptions are insufficient. In fact, the condition that other denominators appearing in the problem statement must also be non-zero is missing. We address this in the following

                      GENERALIZATION

Let  $m\geqslant 3$  and real numbers  $a_1,\;a_2,\;\dots \;,\;a_m$  sucht that

$a_m \neq 0,\;a_{m-1}+a_m \neq 0, \dots ,\;a_2+\dots +a_m \neq 0. \tag{I}$

If  $K=\frac{a_1}{a_2+\dots +a_m}=\frac{a_1+a_2}{a_3+\dots +a_m}=\dots =\frac{a_1+\dots +a_{m-1}}{a_m}$         (E)

                     then  $K\in \left \{-1,\;\frac{a_1}{a_m} \right \}.$

ANSWER CiP to GENERALIZATION

We have one of the following possibilities :

            EITHER

$a_1+a_2+\dots a_m=0 \tag {$S_1$}$

    when  $K=-1$

            OR 

$a_2=\dots=a_{m-1}=0\;,\;\;a_1,\,a_m\;-\;being\;arbitrary\;, \;a_m\neq 0 \tag{$S_2$}$

    when  $K=\frac{a_1}{a_m}.$

 Solution CiP

SUFFICIENCY

              We fix an arbitrary index  $p\in\{1,\;2,\dots ,m-1\}.$

$(S_1)\;\Rightarrow\; a_1+\dots +a_p=-(a_{p+1}+\dots +a_m)\underset{(I)}{\neq}0\;\Rightarrow\;\frac{a_1+\dots +a_p}{a_{p+1}+\dots +a_m}=-1$

so the equalities  (E)  are satisfied.

$(S_2)\;\Rightarrow\;\frac{a_1+\dots +a_p}{a_{p+1}+\dots +a_m}=\frac{a_1}{a_m}$

and equalities  (E)  are again satisfied.

NECESSITY

               $(E)\Rightarrow K=\frac{(a_1+a_2)-a_1}{(a_3+\dots a_m)-(a_2+a_3+\dots +a_m)}=\frac{a_2}{-a_2}=-1$ , 

this applies when  $a_2 \neq 0$ ; for each of the fractions in  (E)  to have a value of  $-1$ , it is necessary and sufficient that  $\sum a_i=0$.  If  $a_2=0$ then it depends on whether 

$S=a_3+\dots  +a_{m-1}$

 is zero or not. If  $S=0$  then $K=\frac{a_1+0}{a_m}$  and if  $S\neq 0$  then  $K=-1$.

$\blacksquare$