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Perpendiculars from Center (Tomato subjective 107)

Problem : If a, b and c are the lengths of the sides of a triangle ABC and  if \( p_1 , p_2 \) and \( p_3 \)   are the lengths of the perpendiculars drawn from the circumcentre onto the sides BC, CA and AB respectively, then show that

\({\displaystyle{\frac{a}{p_1}} + {\frac{b}{p_2}} + {\frac{c}{p_3}} = {\frac{abc}{4{p_1}{p_2}{p_3}}}} \).

Solution : Let  O be the circum centre

\({\displaystyle{\frac{a}{p_1}}}\) = 2 \({\displaystyle \left({\frac{\frac{a}{2}}{p_1}}\right) = 2 \tan \left({\frac{1}{2}}{\angle{BOC}}\right)}\)

Similarly, \({\displaystyle{\frac{b}{p_2}} =  2 \tan \left({\frac{1}{2}}{\angle{AOC}}\right)}\)

& \({\displaystyle{\frac{c}{p_3}} = 2 \tan \left({\frac{1}{2}}{\angle{AOB}}\right)}\)
Now, \({\displaystyle{\frac{1}{2}}{\angle{BOC}} + {\frac{1}{2}}{\angle{AOC}} + {\frac{1}{2}}{\angle{AOB}}}\) = \({\displaystyle{\frac{1}{2}} ({\angle{BOC}} + {\angle{AOC}} + {\angle{AOB}})}\)
= \({\displaystyle{\frac{1}{2}} (360^{\circ})}\) = \({\displaystyle{180^{\circ}}}\)


Lemma
For \({\displaystyle{\alpha},{\beta},{\gamma} > 0}\) & \({\displaystyle{\alpha} + {\beta} + {\gamma} = {\pi}}\)
\({\displaystyle{\tan{\alpha}} + {\tan{\beta}} + {\tan{\gamma}} = {\tan{\alpha}}{\times}{\tan{\beta}}{\times}{\tan{\gamma}}}\)
\({\displaystyle{\alpha} + {\beta} + {\gamma} = {\pi}}\)
\({\Rightarrow}\) \({\displaystyle{\alpha + \beta} = {\pi – \gamma}}\)
\({\Rightarrow}\) \({\displaystyle{\tan (\alpha + \beta)} = {\tan (\pi – \gamma)}}\)
\({\Rightarrow}\) \({\displaystyle{\frac{\tan{\alpha} + \tan{\beta}}{1 – \tan{\alpha} \tan{\beta}}} = -\tan{\gamma}}\)
\({\Rightarrow}\) \({\displaystyle{\tan\alpha + \tan \beta} = {\tan{\alpha}. \tan{\beta} .\tan{\gamma}} – \tan\gamma}\)
\({\Rightarrow}\) \({\displaystyle{\tan{\alpha}} + {\tan{\beta}} + {\tan{\gamma}} = {\tan{\alpha}}{\times}{\tan{\beta}}{\times}{\tan{\gamma}}}\) (lemma proved)


So, \({\displaystyle{\frac{1}{2}}\left({\frac{a}{p_1}} + {\frac{b}{p_2}} + {\frac{c}{p_3}}\right) = \tan({\frac{1}{2}}\angle{AOB}) + \tan({\frac{1}{2}}\angle{BOC}) + \tan({\frac{1}{2}}\angle{COA})}\)
= \({\displaystyle{\tan({\frac{1}{2}}\angle{AOB})\times \tan({\frac{1}{2}}\angle{BOC})\times \tan({\frac{1}{2}}\angle{COA})}}\) [ as \({\displaystyle{\angle{AOB} + \angle{BOC} + \angle{COA} = {360}^{\circ}}}\) ]
= \({\displaystyle{\frac{a}{2{p_1}}}\times{\frac{b}{2{p_2}}}\times{\frac{c}{2{p_3}}}}\)
= \({\displaystyle{\frac{abc}{8{p_1}{p_2}{p_3}}}}\)
\(\Rightarrow \frac{a}{p_1} + \frac{b}{p_2} + \frac{c}{p_3} = \frac{abc}{4{p_1}{p_2}{p_3} } \) ( proved )

© Cheenta 2017

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