Q214369-

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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 (green) PQ=s−2(√3)a red line =(√((s−2(√3)a)^2 −(2a)^2 ))=(√(s^2 +8a^2 −4(√3)sa)) t=red line b=(t/(2(√3)))=(1/2)(√((s^2 +8a^2 −4(√3)sa)/3)) let α=(a/s), β=(b/s) ⇒β=(1/2)(√((1+8α^2 −4(√3)α)/3)) such that b=a, 12α^2 =1+8α^2 −4(√3)α 4α^2 +4(√3)α−1=0 ⇒α=1−((√3)/2)≈0.134

$${s}={side}\:{length}\:{of}\:{large}\:{equilateral} \\ $$$${t}={side}\:{length}\:{of}\:{small}\:{equilateral}\:\left({green}\right) \\ $$$${PQ}={s}−\mathrm{2}\sqrt{\mathrm{3}}{a} \\ $$$${red}\:{line}\:=\sqrt{\left({s}−\mathrm{2}\sqrt{\mathrm{3}}{a}\right)^{\mathrm{2}} −\left(\mathrm{2}{a}\right)^{\mathrm{2}} }=\sqrt{{s}^{\mathrm{2}} +\mathrm{8}{a}^{\mathrm{2}} −\mathrm{4}\sqrt{\mathrm{3}}{sa}} \\ $$$${t}={red}\:{line} \\ $$$${b}=\frac{{t}}{\mathrm{2}\sqrt{\mathrm{3}}}=\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\frac{{s}^{\mathrm{2}} +\mathrm{8}{a}^{\mathrm{2}} −\mathrm{4}\sqrt{\mathrm{3}}{sa}}{\mathrm{3}}} \\ $$$${let}\:\alpha=\frac{{a}}{{s}},\:\beta=\frac{{b}}{{s}} \\ $$$$\Rightarrow\beta=\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\frac{\mathrm{1}+\mathrm{8}\alpha^{\mathrm{2}} −\mathrm{4}\sqrt{\mathrm{3}}\alpha}{\mathrm{3}}} \\ $$$${such}\:{that}\:{b}={a}, \\ $$$$\mathrm{12}\alpha^{\mathrm{2}} =\mathrm{1}+\mathrm{8}\alpha^{\mathrm{2}} −\mathrm{4}\sqrt{\mathrm{3}}\alpha \\ $$$$\mathrm{4}\alpha^{\mathrm{2}} +\mathrm{4}\sqrt{\mathrm{3}}\alpha−\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow\alpha=\mathrm{1}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\approx\mathrm{0}.\mathrm{134} \\ $$

Commented by ajfour last updated on 14/Dec/24