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# Test of Mathematics Solution Subjective 107 - Perpendiculars from Center

This is a Test of Mathematics Solution Subjective 107 (from ISI Entrance). The book, Test of Mathematics at 10+2 Level is Published by East West Press. This problem book is indispensable for the preparation of I.S.I. B.Stat and B.Math Entrance.

Also visit: I.S.I. & C.M.I. Entrance Course of Cheenta

## 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 )

This is a Test of Mathematics Solution Subjective 107 (from ISI Entrance). The book, Test of Mathematics at 10+2 Level is Published by East West Press. This problem book is indispensable for the preparation of I.S.I. B.Stat and B.Math Entrance.

Also visit: I.S.I. & C.M.I. Entrance Course of Cheenta

## 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 )

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