Question Number 18455 by Tinkutara last updated on 21/Jul/17
$$\mathrm{The}\:\mathrm{equation}\:\mathrm{cosec}\:\frac{{x}}{\mathrm{2}}\:+\:\mathrm{cosec}\:\frac{{y}}{\mathrm{2}}\:+ \\ $$ $$\mathrm{cosec}\:\frac{{z}}{\mathrm{2}}\:=\:\mathrm{6},\:\mathrm{where}\:\mathrm{0}\:<\:{x},\:{y},\:{z}\:<\:\frac{\pi}{\mathrm{2}}\:\mathrm{and} \\ $$ $${x}\:+\:{y}\:+\:{z}\:=\:\pi,\:\mathrm{have} \\ $$ $$\left(\mathrm{1}\right)\:\mathrm{Three}\:\mathrm{ordered}\:\mathrm{triplet}\:\left({x},\:{y},\:{z}\right) \\ $$ $$\mathrm{solutions} \\ $$ $$\left(\mathrm{2}\right)\:\mathrm{Two}\:\mathrm{ordered}\:\mathrm{triplet}\:\left({x},\:{y},\:{z}\right) \\ $$ $$\mathrm{solutions} \\ $$ $$\left(\mathrm{3}\right)\:\mathrm{Just}\:\mathrm{one}\:\mathrm{ordered}\:\mathrm{triplet}\:\left({x},\:{y},\:{z}\right) \\ $$ $$\mathrm{solution} \\ $$ $$\left(\mathrm{4}\right)\:\mathrm{No}\:\mathrm{ordered}\:\mathrm{triplet}\:\left({x},\:{y},\:{z}\right)\:\mathrm{solution} \\ $$
Answered by Tinkutara last updated on 02/Aug/17
Commented byTinkutara last updated on 02/Aug/17
![Consider the graph of f(x) = cosec (x/2) in [0, π]. Let A, B and C be three points on this graph. A = (x, cosec (x/2)) B = (y, cosec (y/2)), C = (z, cosec (z/2)) Centroid of this triangle is (((x + y + z)/3), ((cosec (x/2) + cosec (y/2) + cosec (z/2))/3)) Corresponding to this x-coordinate, the y-coordinate on the graph i.e. f(((x + y + z)/3)) = cosec ((x + y + z)/6) = 2 Clearly from the graph it can be seen that ((cosec (x/2) + cosec (y/2) + cosec (z/2))/3) ≥ 2 ⇒ cosec (x/2) + cosec (y/2) + cosec (z/2) ≥ 6 But given that cosec (x/2) + cosec (y/2) + cosec (z/2) = 6 Hence there is only one case of x = y = z = (π/3) .](Q18960.png)
$$\mathrm{Consider}\:\mathrm{the}\:\mathrm{graph}\:\mathrm{of}\:{f}\left({x}\right)\:=\:\mathrm{cosec}\:\frac{{x}}{\mathrm{2}} \\ $$ $$\mathrm{in}\:\left[\mathrm{0},\:\pi\right].\:\mathrm{Let}\:{A},\:{B}\:\mathrm{and}\:{C}\:\mathrm{be}\:\mathrm{three}\:\mathrm{points} \\ $$ $$\mathrm{on}\:\mathrm{this}\:\mathrm{graph}. \\ $$ $${A}\:=\:\left({x},\:\mathrm{cosec}\:\frac{{x}}{\mathrm{2}}\right)\:{B}\:=\:\left({y},\:\mathrm{cosec}\:\frac{{y}}{\mathrm{2}}\right), \\ $$ $${C}\:=\:\left({z},\:\mathrm{cosec}\:\frac{{z}}{\mathrm{2}}\right) \\ $$ $$\mathrm{Centroid}\:\mathrm{of}\:\mathrm{this}\:\mathrm{triangle}\:\mathrm{is} \\ $$ $$\left(\frac{{x}\:+\:{y}\:+\:{z}}{\mathrm{3}},\:\frac{\mathrm{cosec}\:\frac{{x}}{\mathrm{2}}\:+\:\mathrm{cosec}\:\frac{{y}}{\mathrm{2}}\:+\:\mathrm{cosec}\:\frac{{z}}{\mathrm{2}}}{\mathrm{3}}\right) \\ $$ $$\mathrm{Corresponding}\:\mathrm{to}\:\mathrm{this}\:{x}-\mathrm{coordinate}, \\ $$ $$\mathrm{the}\:{y}-\mathrm{coordinate}\:\mathrm{on}\:\mathrm{the}\:\mathrm{graph}\:{i}.{e}. \\ $$ $${f}\left(\frac{{x}\:+\:{y}\:+\:{z}}{\mathrm{3}}\right)\:=\:\mathrm{cosec}\:\frac{{x}\:+\:{y}\:+\:{z}}{\mathrm{6}}\:=\:\mathrm{2} \\ $$ $$\mathrm{Clearly}\:\mathrm{from}\:\mathrm{the}\:\mathrm{graph}\:\mathrm{it}\:\mathrm{can}\:\mathrm{be}\:\mathrm{seen} \\ $$ $$\mathrm{that}\:\frac{\mathrm{cosec}\:\frac{{x}}{\mathrm{2}}\:+\:\mathrm{cosec}\:\frac{{y}}{\mathrm{2}}\:+\:\mathrm{cosec}\:\frac{{z}}{\mathrm{2}}}{\mathrm{3}}\:\geqslant\:\mathrm{2} \\ $$ $$\Rightarrow\:\mathrm{cosec}\:\frac{{x}}{\mathrm{2}}\:+\:\mathrm{cosec}\:\frac{{y}}{\mathrm{2}}\:+\:\mathrm{cosec}\:\frac{{z}}{\mathrm{2}}\:\geqslant\:\mathrm{6} \\ $$ $$\mathrm{But}\:\mathrm{given}\:\mathrm{that}\:\mathrm{cosec}\:\frac{{x}}{\mathrm{2}}\:+\:\mathrm{cosec}\:\frac{{y}}{\mathrm{2}}\:+\:\mathrm{cosec}\:\frac{{z}}{\mathrm{2}}\:=\:\mathrm{6} \\ $$ $$\mathrm{Hence}\:\mathrm{there}\:\mathrm{is}\:\mathrm{only}\:\mathrm{one}\:\mathrm{case}\:\mathrm{of} \\ $$ $${x}\:=\:{y}\:=\:{z}\:=\:\frac{\pi}{\mathrm{3}}\:. \\ $$