Question-219071

Question Number 219071 by Spillover last updated on 19/Apr/25

Answered by mr W last updated on 19/Apr/25

Commented by mr W last updated on 19/Apr/25

l=a+b α=(a/l), β=(b/l), γ=(c/l) ac=(a+b)(a+b−c) ⇒αγ=1−γ ⇒α=((1−γ)/γ) b^2 =a^2 +c^2 −ac ⇒(1−α)^2 =α^2 +γ^2 −αγ ⇒1−2α=γ^2 −αγ ⇒1−((2(1−γ))/γ)=γ^2 −1+γ ⇒γ^3 +γ^2 −4γ+2=0 ⇒(γ−1)(γ^2 +2γ−2)=0 ⇒γ=−1+(√3) ⇒1−γ=2−(√3) fraction of write area =1−3(1−γ) =1−3(2−(√3)) =3(√3)−5≈0.196

$${l}={a}+{b} \\ $$$$\alpha=\frac{{a}}{{l}},\:\beta=\frac{{b}}{{l}},\:\gamma=\frac{{c}}{{l}} \\ $$$${ac}=\left({a}+{b}\right)\left({a}+{b}−{c}\right) \\ $$$$\Rightarrow\alpha\gamma=\mathrm{1}−\gamma\:\Rightarrow\alpha=\frac{\mathrm{1}−\gamma}{\gamma} \\ $$$${b}^{\mathrm{2}} ={a}^{\mathrm{2}} +{c}^{\mathrm{2}} −{ac} \\ $$$$\Rightarrow\left(\mathrm{1}−\alpha\right)^{\mathrm{2}} =\alpha^{\mathrm{2}} +\gamma^{\mathrm{2}} −\alpha\gamma \\ $$$$\Rightarrow\mathrm{1}−\mathrm{2}\alpha=\gamma^{\mathrm{2}} −\alpha\gamma \\ $$$$\Rightarrow\mathrm{1}−\frac{\mathrm{2}\left(\mathrm{1}−\gamma\right)}{\gamma}=\gamma^{\mathrm{2}} −\mathrm{1}+\gamma \\ $$$$\Rightarrow\gamma^{\mathrm{3}} +\gamma^{\mathrm{2}} −\mathrm{4}\gamma+\mathrm{2}=\mathrm{0} \\ $$$$\Rightarrow\left(\gamma−\mathrm{1}\right)\left(\gamma^{\mathrm{2}} +\mathrm{2}\gamma−\mathrm{2}\right)=\mathrm{0} \\ $$$$\Rightarrow\gamma=−\mathrm{1}+\sqrt{\mathrm{3}} \\ $$$$\Rightarrow\mathrm{1}−\gamma=\mathrm{2}−\sqrt{\mathrm{3}} \\ $$$${fraction}\:{of}\:{write}\:{area} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{1}−\mathrm{3}\left(\mathrm{1}−\gamma\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{1}−\mathrm{3}\left(\mathrm{2}−\sqrt{\mathrm{3}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{3}\sqrt{\mathrm{3}}−\mathrm{5}\approx\mathrm{0}.\mathrm{196} \\ $$

Commented by Spillover last updated on 19/Apr/25