Question Number 2458 by alib last updated on 20/Nov/15
$${The}\:{medians}\:{of}\:{a}\:{triangle} \\ $$$${are}\:{m}_{\mathrm{1}} ,\:{m}_{\mathrm{2}} ,\:{m}_{\mathrm{3}} . \\ $$$${Find}\:{the}\:{length}\:{of}\:{each}\:{sides}\: \\ $$$${the}\:{triangle}. \\ $$
Answered by prakash jain last updated on 20/Nov/15
![m_1 =(1/2)(√(2c^2 +2b^2 −a^2 ))⇒a^2 =−4m_1 ^2 +2c^2 +2b^2 (1) m_2 =(1/2)(√(2a^2 +2c^2 −b^2 )) ⇒b^2 =−4m_2 ^2 +2a^2 +2c^2 (2) m_3 =(1/2)(√(2a^2 +2b^2 −c^2 ))⇒c^2 =−4m_3 ^2 +2a^2 +2b^2 (3) (2)+(3) b^2 +c^2 =−4(m_2 ^2 +m_3 ^2 )+2(b^2 +c^2 )+4a^2 (b^2 +c^2 )=4[(m_3 ^2 +m_2 ^2 )−a^2 ] subtitute in (1) a^2 =−4m_1 ^2 +8(m_3 ^2 +m_2 ^2 )−8a^2 9a^2 =8m_3 ^2 +8m_2 ^2 −4m_1 ^2 a=(2/3)(√(2m_3 ^2 +2m_2 ^2 −m_1 ^2 )) similarly b=(2/3)(√(2m_3 ^2 +2m_1 ^2 −m_2 ^2 )) c=(2/3)(√(2m_1 ^2 +2m_2 ^2 −m_3 ^2 ))](https://www.tinkutara.com/question/Q2460.png)
$${m}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{2}{c}^{\mathrm{2}} +\mathrm{2}{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }\Rightarrow{a}^{\mathrm{2}} =−\mathrm{4}{m}_{\mathrm{1}} ^{\mathrm{2}} +\mathrm{2}{c}^{\mathrm{2}} +\mathrm{2}{b}^{\mathrm{2}} \:\:\:\:\left(\mathrm{1}\right) \\ $$$${m}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{2}{a}^{\mathrm{2}} +\mathrm{2}{c}^{\mathrm{2}} −{b}^{\mathrm{2}} }\:\Rightarrow{b}^{\mathrm{2}} =−\mathrm{4}{m}_{\mathrm{2}} ^{\mathrm{2}} +\mathrm{2}{a}^{\mathrm{2}} +\mathrm{2}{c}^{\mathrm{2}} \:\:\:\left(\mathrm{2}\right) \\ $$$${m}_{\mathrm{3}} =\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{2}{a}^{\mathrm{2}} +\mathrm{2}{b}^{\mathrm{2}} −{c}^{\mathrm{2}} }\Rightarrow{c}^{\mathrm{2}} =−\mathrm{4}{m}_{\mathrm{3}} ^{\mathrm{2}} +\mathrm{2}{a}^{\mathrm{2}} +\mathrm{2}{b}^{\mathrm{2}} \:\:\:\:\:\left(\mathrm{3}\right) \\ $$$$\left(\mathrm{2}\right)+\left(\mathrm{3}\right) \\ $$$${b}^{\mathrm{2}} +{c}^{\mathrm{2}} =−\mathrm{4}\left({m}_{\mathrm{2}} ^{\mathrm{2}} +{m}_{\mathrm{3}} ^{\mathrm{2}} \right)+\mathrm{2}\left({b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)+\mathrm{4}{a}^{\mathrm{2}} \\ $$$$\left({b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)=\mathrm{4}\left[\left({m}_{\mathrm{3}} ^{\mathrm{2}} +{m}_{\mathrm{2}} ^{\mathrm{2}} \right)−{a}^{\mathrm{2}} \right] \\ $$$${subtitute}\:{in}\:\left(\mathrm{1}\right) \\ $$$${a}^{\mathrm{2}} =−\mathrm{4}{m}_{\mathrm{1}} ^{\mathrm{2}} +\mathrm{8}\left({m}_{\mathrm{3}} ^{\mathrm{2}} +{m}_{\mathrm{2}} ^{\mathrm{2}} \right)−\mathrm{8}{a}^{\mathrm{2}} \\ $$$$\mathrm{9}{a}^{\mathrm{2}} =\mathrm{8}{m}_{\mathrm{3}} ^{\mathrm{2}} +\mathrm{8}{m}_{\mathrm{2}} ^{\mathrm{2}} −\mathrm{4}{m}_{\mathrm{1}} ^{\mathrm{2}} \\ $$$${a}=\frac{\mathrm{2}}{\mathrm{3}}\sqrt{\mathrm{2}{m}_{\mathrm{3}} ^{\mathrm{2}} +\mathrm{2}{m}_{\mathrm{2}} ^{\mathrm{2}} −{m}_{\mathrm{1}} ^{\mathrm{2}} } \\ $$$${similarly} \\ $$$${b}=\frac{\mathrm{2}}{\mathrm{3}}\sqrt{\mathrm{2}{m}_{\mathrm{3}} ^{\mathrm{2}} +\mathrm{2}{m}_{\mathrm{1}} ^{\mathrm{2}} −{m}_{\mathrm{2}} ^{\mathrm{2}} } \\ $$$${c}=\frac{\mathrm{2}}{\mathrm{3}}\sqrt{\mathrm{2}{m}_{\mathrm{1}} ^{\mathrm{2}} +\mathrm{2}{m}_{\mathrm{2}} ^{\mathrm{2}} −{m}_{\mathrm{3}} ^{\mathrm{2}} } \\ $$