Geometrie vectoriala
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Pricope Tidor-Vlad
- Mesaje: 276
- Membru din: Vin Sep 28, 2012 4:04 pm
- Localitate: Botosani
Geometrie vectoriala
Demonstrati ca daca inaltimile $AD; BE; CF$ ale triunghiului $ABC$ satisfac: $4 \overrightarrow{AD}+19 \overrightarrow{BE}+ 25 \overrightarrow{CF}= \overrightarrow{0}$, atunci triunghiul $ABC$ are un unghi cu masura de 60.
“Make things as simple as possible, but not simpler.” - Albert Einstein
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Paul Tirlisan
- Mesaje: 5
- Membru din: Dum Aug 24, 2014 1:45 pm
Re: Geometrie vectoriala
Cum $AD=\frac{2S}{a}$, rotind vectorul $\overrightarrow{AD}$ cu $90^{\circ}$ obtinem $\frac{2S}{a^2}\overrightarrow{BC}$. Analog, rotind vectorii $\overrightarrow{BE}$ si $\overrightarrow{CF}$ cu $90^{\circ}$obtinem $\frac{2S}{b^2}\overrightarrow{CA}$, respectiv $\frac{2S}{c^2}\overrightarrow{AB}$.
Deci rotind vectorul $\overrightarrow{v_1}=4\overrightarrow{AD}+19\overrightarrow{BE}+25\overrightarrow{CF}(=\overrightarrow{0})$ cu $90^{\circ}$ rezulta $\overrightarrow{v_2}=\frac{8S}{a^2}\overrightarrow{BC}+\frac{38S}{b^2}\overrightarrow{CA}+\frac{50S}{c^2}\overrightarrow{AB}$. Cum $\overrightarrow{v_2}=\overrightarrow{0}$, rezulta $\frac{4}{a^2}\overrightarrow{BC}+\frac{19}{b^2}\overrightarrow{CA}+\frac{25}{c^2}\overrightarrow{AB}=\overrightarrow{0}$, sau $\frac{25a^2-4c^2}{a^2c^2}\overrightarrow{AB}+\frac{4b^2-19a^2}{a^2b^2}\overrightarrow{AC}=\overrightarrow{0}$. Vectorii $\overrightarrow{AB}$ si $\overrightarrow{AC}$ fiind necoliniari, rezulta $5a=2c$ si $2b=a\sqrt{19}$.
Deci exista $k>0$ a.i. $a=2k$, $b=k\sqrt{19}$ si $c=5k$. Rezulta $c^2+a^2-b^2=10k^2$ si $ca=10k^2\Rightarrow c^2+a^2-b^2=ca\Leftrightarrow \cos B=\cos 60^{\circ}$. Functia cos fiind strict descrescatoare pe $[0, \pi]$, deci si injectiva, rezulta $B=60^{\circ}$.
Deci rotind vectorul $\overrightarrow{v_1}=4\overrightarrow{AD}+19\overrightarrow{BE}+25\overrightarrow{CF}(=\overrightarrow{0})$ cu $90^{\circ}$ rezulta $\overrightarrow{v_2}=\frac{8S}{a^2}\overrightarrow{BC}+\frac{38S}{b^2}\overrightarrow{CA}+\frac{50S}{c^2}\overrightarrow{AB}$. Cum $\overrightarrow{v_2}=\overrightarrow{0}$, rezulta $\frac{4}{a^2}\overrightarrow{BC}+\frac{19}{b^2}\overrightarrow{CA}+\frac{25}{c^2}\overrightarrow{AB}=\overrightarrow{0}$, sau $\frac{25a^2-4c^2}{a^2c^2}\overrightarrow{AB}+\frac{4b^2-19a^2}{a^2b^2}\overrightarrow{AC}=\overrightarrow{0}$. Vectorii $\overrightarrow{AB}$ si $\overrightarrow{AC}$ fiind necoliniari, rezulta $5a=2c$ si $2b=a\sqrt{19}$.
Deci exista $k>0$ a.i. $a=2k$, $b=k\sqrt{19}$ si $c=5k$. Rezulta $c^2+a^2-b^2=10k^2$ si $ca=10k^2\Rightarrow c^2+a^2-b^2=ca\Leftrightarrow \cos B=\cos 60^{\circ}$. Functia cos fiind strict descrescatoare pe $[0, \pi]$, deci si injectiva, rezulta $B=60^{\circ}$.