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In-ABC-the-following-relationship-holds-m-b-b-m-c-c-a-2r-n-b-h-b-n-c-h-c-




Question Number 183167 by Shrinava last updated on 21/Dec/22
In   △ABC   the following relationship  holds:  (m_b /b) + (m_c /c) ≤ (a/(2r)) ≤ (n_b /h_b ) + (n_c /h_c )
$$\mathrm{In}\:\:\:\bigtriangleup\mathrm{ABC}\:\:\:\mathrm{the}\:\mathrm{following}\:\mathrm{relationship} \\ $$$$\mathrm{holds}: \\ $$$$\frac{\mathrm{m}_{\boldsymbol{\mathrm{b}}} }{\mathrm{b}}\:+\:\frac{\mathrm{m}_{\boldsymbol{\mathrm{c}}} }{\mathrm{c}}\:\leqslant\:\frac{\mathrm{a}}{\mathrm{2r}}\:\leqslant\:\frac{\mathrm{n}_{\boldsymbol{\mathrm{b}}} }{\mathrm{h}_{\boldsymbol{\mathrm{b}}} }\:+\:\frac{\mathrm{n}_{\boldsymbol{\mathrm{c}}} }{\mathrm{h}_{\boldsymbol{\mathrm{c}}} } \\ $$
Commented by mr W last updated on 22/Dec/22
i think not every body knows what is  what in the equation.
$${i}\:{think}\:{not}\:{every}\:{body}\:{knows}\:{what}\:{is} \\ $$$${what}\:{in}\:{the}\:{equation}. \\ $$
Commented by Frix last updated on 22/Dec/22
That′s plain to see because obviously  ((α/(ρ_b +ρ_c ))+(β/(ρ_a +ρ_c )))(i_c −j_c )>(γ/(ρ_a +ρ_b ))((i_a −j_a )+(i_b −j_b ))
$$\mathrm{That}'\mathrm{s}\:\mathrm{plain}\:\mathrm{to}\:\mathrm{see}\:\mathrm{because}\:\mathrm{obviously} \\ $$$$\left(\frac{\alpha}{\rho_{{b}} +\rho_{{c}} }+\frac{\beta}{\rho_{{a}} +\rho_{{c}} }\right)\left({i}_{{c}} −{j}_{{c}} \right)>\frac{\gamma}{\rho_{{a}} +\rho_{{b}} }\left(\left({i}_{{a}} −{j}_{{a}} \right)+\left({i}_{{b}} −{j}_{{b}} \right)\right) \\ $$

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