Fie un hexagon $A$$B$$C$$D$$E$$F$ si $M$,$N$,$P$,$Q$,$R$,$S$, mijloacele laturilor $AB$,$BC$,$CD$,$DE$,$EF$,$FA$. Sa se arate ca o conditie necesara si suficienta ca $RN^2=MQ^2+PS^2$ este ca $MQ$ si $PS$ sa fie perpendiculare.
Lauretiu Panaitopol
conditia necesara
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conditia necesara
”God made the integers, all else is the work of man.” Kronecker
- Laurențiu Ploscaru
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Re: conditia necesara
Vom nota afixul unui punct cu litera mică corespunzătoare. Relația $RN^2=MQ^2+PS^2$ se rescrie:
$|\dfrac{b+c-e-f}{2}|^2=|\dfrac{a+b-d-e}{2}|^2+|\dfrac{c+d-a-f}{2}|^2\Leftrightarrow$ $|b+c-e-f|^2=|a+b-d-e|^2+|c+d-a-f|^2\Leftrightarrow$
$|b|^2+|c|^2+|e|^2+|f|^2+b\cdot \overline{c}+c\cdot \overline{b}+e\cdot \overline{f}+f\cdot \overline{e}$ $-b\cdot \overline{e}-e\cdot \overline{b}-b\cdot \overline{f}-f\cdot \overline{b}-c\cdot \overline{e}-e\cdot \overline{c}-c\cdot \overline{f}-f\cdot \overline{c}=$
$|a|^2+|b|^2+|d|^2+|e|^2+a\cdot \overline{b}+b\cdot \overline{a}+d\cdot \overline{e}+e\cdot \overline{d}$ $-a\cdot \overline{d}-d\cdot \overline{a}-a\cdot \overline{e}-e\cdot \overline{a}-b\cdot \overline{d}-d\cdot \overline{b}-b\cdot \overline{e}-e\cdot \overline{b}\ +$
$|c|^2+|d|^2+|a|^2+|f|^2+c\cdot \overline{d}+d\cdot \overline{c}+a\cdot \overline{f}+f\cdot \overline{a}$ $-c\cdot \overline{a}-a\cdot \overline{c}-c\cdot \overline{f}-f\cdot \overline{c}-d\cdot \overline{a}-a\cdot \overline{d}-d\cdot \overline{f}-f\cdot \overline{d}\Leftrightarrow$
$0=2|a|^2+2|d|^2+a\cdot \overline{b}+b\cdot \overline{a}+d\cdot \overline{e}+e\cdot \overline{d}$ $c\cdot \overline{d}+d\cdot \overline{c}+a\cdot \overline{f}+f\cdot \overline{a}-b\cdot \overline{c}-c\cdot \overline{b}-e\cdot \overline{f}-f\cdot \overline{e}\ -$
$a\cdot \overline{e}-e\cdot \overline{a}-b\cdot \overline{d}-d\cdot \overline{b}-c\cdot \overline{a}-a\cdot \overline{c}-d\cdot \overline{f}-f\cdot \overline{d}-$ $2a\cdot \overline{d}-2d\cdot \overline{a}+b\cdot \overline{f}+f\cdot \overline{b}+c\cdot \overline{e}+e\cdot \overline{c}$, relație pe care o notăm cu (R).
De asemenea $MQ\perp PS\Leftrightarrow MS^2+PQ^2=MP^2+SQ^2$, relație care după calcule asemănătoare ca cele de sus este echivalentă cu (R).
Prin urmare cele două condiții sunt echivalente cu (R), deci $MQ\perp SP\Leftrightarrow RN^2=MQ^2+SP^2$.
$|\dfrac{b+c-e-f}{2}|^2=|\dfrac{a+b-d-e}{2}|^2+|\dfrac{c+d-a-f}{2}|^2\Leftrightarrow$ $|b+c-e-f|^2=|a+b-d-e|^2+|c+d-a-f|^2\Leftrightarrow$
$|b|^2+|c|^2+|e|^2+|f|^2+b\cdot \overline{c}+c\cdot \overline{b}+e\cdot \overline{f}+f\cdot \overline{e}$ $-b\cdot \overline{e}-e\cdot \overline{b}-b\cdot \overline{f}-f\cdot \overline{b}-c\cdot \overline{e}-e\cdot \overline{c}-c\cdot \overline{f}-f\cdot \overline{c}=$
$|a|^2+|b|^2+|d|^2+|e|^2+a\cdot \overline{b}+b\cdot \overline{a}+d\cdot \overline{e}+e\cdot \overline{d}$ $-a\cdot \overline{d}-d\cdot \overline{a}-a\cdot \overline{e}-e\cdot \overline{a}-b\cdot \overline{d}-d\cdot \overline{b}-b\cdot \overline{e}-e\cdot \overline{b}\ +$
$|c|^2+|d|^2+|a|^2+|f|^2+c\cdot \overline{d}+d\cdot \overline{c}+a\cdot \overline{f}+f\cdot \overline{a}$ $-c\cdot \overline{a}-a\cdot \overline{c}-c\cdot \overline{f}-f\cdot \overline{c}-d\cdot \overline{a}-a\cdot \overline{d}-d\cdot \overline{f}-f\cdot \overline{d}\Leftrightarrow$
$0=2|a|^2+2|d|^2+a\cdot \overline{b}+b\cdot \overline{a}+d\cdot \overline{e}+e\cdot \overline{d}$ $c\cdot \overline{d}+d\cdot \overline{c}+a\cdot \overline{f}+f\cdot \overline{a}-b\cdot \overline{c}-c\cdot \overline{b}-e\cdot \overline{f}-f\cdot \overline{e}\ -$
$a\cdot \overline{e}-e\cdot \overline{a}-b\cdot \overline{d}-d\cdot \overline{b}-c\cdot \overline{a}-a\cdot \overline{c}-d\cdot \overline{f}-f\cdot \overline{d}-$ $2a\cdot \overline{d}-2d\cdot \overline{a}+b\cdot \overline{f}+f\cdot \overline{b}+c\cdot \overline{e}+e\cdot \overline{c}$, relație pe care o notăm cu (R).
De asemenea $MQ\perp PS\Leftrightarrow MS^2+PQ^2=MP^2+SQ^2$, relație care după calcule asemănătoare ca cele de sus este echivalentă cu (R).
Prin urmare cele două condiții sunt echivalente cu (R), deci $MQ\perp SP\Leftrightarrow RN^2=MQ^2+SP^2$.
People are strange when you're a stranger,
Faces look ugly when you're alone.
Women seem wicked when you're unwanted,
Streets are uneven when you're down.
Faces look ugly when you're alone.
Women seem wicked when you're unwanted,
Streets are uneven when you're down.
- Laurențiu Ploscaru
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- Membru din: Mie Mai 04, 2011 5:42 pm
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Re: conditia necesara
Să observăm că $\overrightarrow{SP}+\overrightarrow{QM}=\overrightarrow{r_C}/2+\overrightarrow{r_D}/2-\overrightarrow{r_A}/2-\overrightarrow{r_F}/2$ $+\overrightarrow{r_A}/2+\overrightarrow{r_B}/2-\overrightarrow{r_D}/2-\overrightarrow{r_E}/2=$ $\overrightarrow{r_B}/2+\overrightarrow{r_C}/2-\overrightarrow{r_E}-\overrightarrow{r_F}/2=\overrightarrow{RN}$.
Deci $\overrightarrow{SP}+\overrightarrow{QM}=\overrightarrow{RN}\Rightarrow$ $MQ^2+PS^2+2MQ\cdot PS\cdot \cos \widehat{(QM,SP)}=RN^2$.
Prin urmare $RN^2=MQ^2+PS^2\Leftrightarrow \cos \widehat{(QM,SP)}=0\Leftrightarrow MQ\perp PS$.
Deci $\overrightarrow{SP}+\overrightarrow{QM}=\overrightarrow{RN}\Rightarrow$ $MQ^2+PS^2+2MQ\cdot PS\cdot \cos \widehat{(QM,SP)}=RN^2$.
Prin urmare $RN^2=MQ^2+PS^2\Leftrightarrow \cos \widehat{(QM,SP)}=0\Leftrightarrow MQ\perp PS$.
People are strange when you're a stranger,
Faces look ugly when you're alone.
Women seem wicked when you're unwanted,
Streets are uneven when you're down.
Faces look ugly when you're alone.
Women seem wicked when you're unwanted,
Streets are uneven when you're down.
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Re: conditia necesara
Mai exact, deoarece cei trei vectori $\overrightarrow{RN}, \overrightarrow{MQ},\ \overrightarrow{PS}$au suma nula, lungimile lor formeaza un triunghi. Cerinta problemei se reduce acum la teorema lui Pitagora.