Question Number 204674 by SEKRET last updated on 25/Feb/24
Commented by A5T last updated on 25/Feb/24
$${O}_{\mathrm{1}} {O}_{\mathrm{2}} \:{is}\:{not}\:{well}-{defined},\:{is}\:{it}\:{a}\:{tangent}\:{line}\:{to} \\ $$$${both}\:{quadrants}? \\ $$
Answered by mr W last updated on 25/Feb/24
Commented by mr W last updated on 25/Feb/24
$${there}\:{is}\:{no}\:{unique}\:{solution}! \\ $$$${say}\:{the}\:{side}\:{length}\:{of}\:{square}\:{is}\:{s}. \\ $$$${x}^{\mathrm{2}} +\left({a}+{b}\right)^{\mathrm{2}} =\left(\sqrt{\mathrm{2}}{s}\right)^{\mathrm{2}} \\ $$$$\Rightarrow{x}=\sqrt{\mathrm{2}{s}^{\mathrm{2}} −\left({a}+{b}\right)^{\mathrm{2}} } \\ $$
Commented by SEKRET last updated on 26/Feb/24
$$\:\boldsymbol{\mathrm{thank}}\:\:\boldsymbol{\mathrm{you}}\:\:\boldsymbol{\mathrm{sir}}\: \\ $$
Commented by York12 last updated on 26/Feb/24
$$\mathrm{sir}\:\mathrm{what}\:\mathrm{software}\:\mathrm{is}\:\mathrm{that} \\ $$
Commented by mr W last updated on 26/Feb/24
$${no}\:{special}\:{software}\:{is}\:{used}.\:{each} \\ $$$${smartphone}\:{has}\:{an}\:{app}\:{for}\:{drawing} \\ $$$${images}. \\ $$
Commented by York12 last updated on 26/Feb/24
$$\mathrm{oh}\:\mathrm{thank}\:\mathrm{you} \\ $$
Commented by York12 last updated on 26/Feb/24
$$\mathrm{can}\:\mathrm{not}\:\mathrm{I}\:\mathrm{get}\:\mathrm{contact}\:\mathrm{with}\:\mathrm{you}\:\mathrm{sir}\: \\ $$?
Commented by mr W last updated on 26/Feb/24
$${everybody}\:{can}\:{contact}\:{me}\:{through} \\ $$$${this}\:{forum}. \\ $$