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Question Number 41347 by behi83417@gmail.com last updated on 06/Aug/18

Answered by MJS last updated on 06/Aug/18

$$m_a=\sqrt{\frac{b^2}{2}+\frac{c^2}{2}-\frac{a^2}{4}}$$$$m_b=\sqrt{\frac{a^2}{2}+\frac{c^2}{2}-\frac{b^2}{4}}$$$$m_c=\sqrt{\frac{a^2}{2}+\frac{b^2}{2}-\frac{c^2}{4}}$$$$m_a^2+m_b^2-m_c^2=0$$$$\frac{5c^2}{4}-\frac{a^2}{4}-\frac{b^2}{4}=0$$$$$$$$\mathrm{let}c=1$$$$\Rightarrow b=\sqrt{5-a^2}\Rightarrow 0<a<\sqrt{5}$$$$a+b>c$$$$a+\sqrt{5-a^2}>1\Rightarrow a>-1$$$$a+c>b$$$$a+1>\sqrt{5-a^2}\Rightarrow a>1$$$$b+c>a$$$$\sqrt{5-a^2}+1>a\Rightarrow a<2$$$$$$$$\Rightarrow 1<a<2\wedge b=\sqrt{5-a^2}\wedge c=1$$$$$$$$c^2<\frac{ab}{2}$$$$1<\frac{1}{2}a\sqrt{5-a^2}\Rightarrow 1<a<2$$$$q.e.d.$$

Commented by behi83417@gmail.com last updated on 06/Aug/18

$$thankyoudearMJS.$$$$\text{You can't use 'macro parameter character \#' in math mode}$$$$\Rightarrow a^2+b^2=5c^2$$$$cosC=\frac{a^2+b^2-c^2}{2ab}=\frac{4c^2}{2ab}<1\Rightarrow$$$$\Rightarrow c^2<\frac{ab}{2}\Rightarrow c<\sqrt{\frac{ab}{2}}.◼$$

Commented by MJS last updated on 06/Aug/18

$$\mathrm{typo}...\mathrm{thank}\mathrm{you},\mathrm{I}\mathrm{corrected}\mathrm{it}$$

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