Question Number 214623 by mr W last updated on 13/Dec/24
$${Q}\mathrm{214369} \\ $$
Commented by mr W last updated on 13/Dec/24
Answered by mr W last updated on 13/Dec/24
Commented by mr W last updated on 13/Dec/24
$${s}={side}\:{length}\:{of}\:{large}\:{equilateral} \\ $$$${t}={side}\:{length}\:{of}\:{small}\:{equilateral}\:\left({green}\right) \\ $$$${PQ}={s}−\mathrm{2}\sqrt{\mathrm{3}}{a} \\ $$$${red}\:{line}\:=\sqrt{\left({s}−\mathrm{2}\sqrt{\mathrm{3}}{a}\right)^{\mathrm{2}} −\left(\mathrm{2}{a}\right)^{\mathrm{2}} }=\sqrt{{s}^{\mathrm{2}} +\mathrm{8}{a}^{\mathrm{2}} −\mathrm{4}\sqrt{\mathrm{3}}{sa}} \\ $$$${t}={red}\:{line} \\ $$$${b}=\frac{{t}}{\mathrm{2}\sqrt{\mathrm{3}}}=\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\frac{{s}^{\mathrm{2}} +\mathrm{8}{a}^{\mathrm{2}} −\mathrm{4}\sqrt{\mathrm{3}}{sa}}{\mathrm{3}}} \\ $$$${let}\:\alpha=\frac{{a}}{{s}},\:\beta=\frac{{b}}{{s}} \\ $$$$\Rightarrow\beta=\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\frac{\mathrm{1}+\mathrm{8}\alpha^{\mathrm{2}} −\mathrm{4}\sqrt{\mathrm{3}}\alpha}{\mathrm{3}}} \\ $$$${such}\:{that}\:{b}={a}, \\ $$$$\mathrm{12}\alpha^{\mathrm{2}} =\mathrm{1}+\mathrm{8}\alpha^{\mathrm{2}} −\mathrm{4}\sqrt{\mathrm{3}}\alpha \\ $$$$\mathrm{4}\alpha^{\mathrm{2}} +\mathrm{4}\sqrt{\mathrm{3}}\alpha−\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow\alpha=\mathrm{1}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\approx\mathrm{0}.\mathrm{134} \\ $$
Commented by ajfour last updated on 14/Dec/24
$${I}\:{had}\:{hoped}\:{it}\:{wd}\:{be}\:{difficult}.\:{You} \\ $$$${proved}\:{that}\:{it}\:{isnt}\:{so}.\:{cheers}. \\ $$
Commented by ajfour last updated on 14/Dec/24
https://youtu.be/_nOhjOVtvIQ?si=WlBbeWtccm0GgQxK
Commented by ajfour last updated on 14/Dec/24
Commented by ajfour last updated on 14/Dec/24
$${Above}\:{youtube}\:{link}\:{for}\:{video}. \\ $$