Question Number 139254 by mathdanisur last updated on 25/Apr/21
$${Given}\:{a}\:{convex}\:{hexagon}\:{ABCDEG} \\ $$$${satisfy}:\:{AB}={BC},\:{CD}={DE},\:{EF}={FA}. \\ $$$${Suppose}\:\bigtriangleup{ACE}\:{is}\:{a}\:{right}\:{triangle}. \\ $$$$\left(\frac{{BC}}{{BE}}+\frac{{DE}}{{DA}}+\frac{{FA}}{{FC}}\right)_{\boldsymbol{{min}}} =? \\ $$
Answered by mr W last updated on 27/Apr/21
Commented by mr W last updated on 27/Apr/21
$${BC}=\sqrt{{a}^{\mathrm{2}} +{p}^{\mathrm{2}} } \\ $$$${BE}=\sqrt{{a}^{\mathrm{2}} +\left(\mathrm{2}{b}+{p}\right)^{\mathrm{2}} } \\ $$$${P}=\frac{{BE}}{{BC}}=\frac{\sqrt{{a}^{\mathrm{2}} +\left(\mathrm{2}{b}+{p}\right)^{\mathrm{2}} }}{\:\sqrt{{a}^{\mathrm{2}} +{p}^{\mathrm{2}} }}=\frac{\sqrt{\mathrm{1}+\left(\frac{\mathrm{2}{b}}{{a}}+\frac{{p}}{{a}}\right)^{\mathrm{2}} }}{\:\sqrt{\mathrm{1}+\left(\frac{{p}}{{a}}\right)^{\mathrm{2}} }} \\ $$$${let}\:{x}=\frac{{p}}{{a}},\:\lambda=\frac{{b}}{{a}} \\ $$$${P}=\frac{\sqrt{\mathrm{1}+\left(\mathrm{2}\lambda+{x}\right)^{\mathrm{2}} }}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }} \\ $$$$\frac{{dP}}{{dx}}=\frac{\mathrm{2}\left(\mathrm{2}\lambda+{x}\right)}{\mathrm{2}\sqrt{\mathrm{1}+\left(\mathrm{2}\lambda+{x}\right)^{\mathrm{2}} }\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}−\frac{\mathrm{2}{x}\sqrt{\mathrm{1}+\left(\mathrm{2}\lambda+{x}\right)^{\mathrm{2}} }}{\mathrm{2}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}=\mathrm{0} \\ $$$$\frac{\left(\mathrm{2}\lambda+{x}\right)}{\:\sqrt{\mathrm{1}+\left(\mathrm{2}\lambda+{x}\right)^{\mathrm{2}} }}=\frac{{x}\sqrt{\mathrm{1}+\left(\mathrm{2}\lambda+{x}\right)^{\mathrm{2}} }}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)} \\ $$$$\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\left(\mathrm{2}\lambda+{x}\right)={x}\left[\mathrm{1}+\left(\mathrm{2}\lambda+{x}\right)^{\mathrm{2}} \right] \\ $$$${x}^{\mathrm{2}} +\mathrm{2}\lambda{x}−\mathrm{1}=\mathrm{0} \\ $$$${x}=\sqrt{\lambda^{\mathrm{2}} +\mathrm{1}}−\lambda=\mathrm{1}\:{for}\:\lambda=\frac{{b}}{{a}}=\mathrm{1} \\ $$$${P}_{{max}} =\frac{\sqrt{\mathrm{1}+\left(\mathrm{2}+\mathrm{1}\right)^{\mathrm{2}} }}{\:\sqrt{\mathrm{1}+\mathrm{1}}}=\sqrt{\mathrm{5}} \\ $$$$\left(\frac{{BC}}{{BE}}\right)_{{min}} =\frac{\mathrm{1}}{{P}_{{max}} }=\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}} \\ $$$${similarly} \\ $$$$\left(\frac{{DE}}{{DA}}\right)_{{min}} =\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}} \\ $$$${with}\:{b}={a},\:{c}=\sqrt{\mathrm{2}}{a} \\ $$$${FA}=\sqrt{{c}^{\mathrm{2}} +{r}^{\mathrm{2}} } \\ $$$${FC}={c}+{r} \\ $$$${S}=\frac{{FA}}{{FC}}=\frac{\sqrt{{c}^{\mathrm{2}} +{r}^{\mathrm{2}} }}{{c}+{r}}=\frac{\sqrt{\mathrm{1}+\left(\frac{{r}}{{c}}\right)^{\mathrm{2}} }}{\mathrm{1}+\frac{{r}}{{c}}} \\ $$$${y}=\frac{{r}}{{c}} \\ $$$${S}=\frac{\sqrt{\mathrm{1}+{y}^{\mathrm{2}} }}{\mathrm{1}+{y}} \\ $$$$\frac{{dS}}{{dy}}=\frac{\mathrm{2}{y}}{\mathrm{2}\sqrt{\mathrm{1}+{y}^{\mathrm{2}} }\left(\mathrm{1}+{y}\right)}−\frac{\sqrt{\mathrm{1}+{y}^{\mathrm{2}} }}{\left(\mathrm{1}+{y}\right)^{\mathrm{2}} }=\mathrm{0} \\ $$$$\frac{{y}}{\:\sqrt{\mathrm{1}+{y}^{\mathrm{2}} }}=\frac{\sqrt{\mathrm{1}+{y}^{\mathrm{2}} }}{\mathrm{1}+{y}} \\ $$$${y}^{\mathrm{2}} +{y}=\mathrm{1}+{y}^{\mathrm{2}} \\ $$$${y}=\mathrm{1} \\ $$$${S}_{{min}} =\frac{\sqrt{\mathrm{1}+\mathrm{1}^{\mathrm{2}} }}{\mathrm{1}+\mathrm{1}}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}} \\ $$$$\left(\frac{{BC}}{{BE}}+\frac{{DE}}{{DA}}+\frac{{FA}}{{FC}}\right)_{{min}} =\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}=\frac{\mathrm{2}\sqrt{\mathrm{2}}+\sqrt{\mathrm{5}}}{\:\sqrt{\mathrm{10}}} \\ $$
Commented by mathdanisur last updated on 30/Apr/21
$${thankyou}\:{sir} \\ $$