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Question Number 218257 by cherokeesay last updated on 03/Apr/25

Commented by cherokeesay last updated on 03/Apr/25

the 4 points are tangent

$$\mathrm{the}\:\mathrm{4}\:\mathrm{points}\:\mathrm{are}\:\mathrm{tangent} \\ $$

Commented by Nicholas666 last updated on 09/Apr/25

beautifull problem

$${beautifull}\:{problem} \\ $$

Answered by mr W last updated on 05/Apr/25

Commented by mr W last updated on 04/Apr/25

sin α=(1/3) ⇒cos α=((2(√2))/3), tan α=((√2)/4) a=1+(4 tan α−1) sin α=((2+(√2))/3) h=(4/(cos α))−a−2=((4×3)/(2(√2)))−((2+(√2))/3)−2=((8((√2)−1))/3) tan β=((4−2 cos α)/(4−2 sin α))=((2(3−(√2)))/5) b=2 tan ((π/2)−α−β)=(2/(tan (α+β))) =((2(1−((√2)/4)×((2(3−(√2)))/5)))/(((√2)/4)+((2(3−(√2)))/5)))=((4(15−2(√2)))/(31)) S=bh=((4(15−2(√2)))/(31))×((8((√2)−1))/3) =((32(17(√2)−19))/(93))≈1.735

$$\mathrm{sin}\:\alpha=\frac{\mathrm{1}}{\mathrm{3}}\:\Rightarrow\mathrm{cos}\:\alpha=\frac{\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{3}},\:\mathrm{tan}\:\alpha=\frac{\sqrt{\mathrm{2}}}{\mathrm{4}} \\ $$$${a}=\mathrm{1}+\left(\mathrm{4}\:\mathrm{tan}\:\alpha−\mathrm{1}\right)\:\mathrm{sin}\:\alpha=\frac{\mathrm{2}+\sqrt{\mathrm{2}}}{\mathrm{3}} \\ $$$${h}=\frac{\mathrm{4}}{\mathrm{cos}\:\alpha}−{a}−\mathrm{2}=\frac{\mathrm{4}×\mathrm{3}}{\mathrm{2}\sqrt{\mathrm{2}}}−\frac{\mathrm{2}+\sqrt{\mathrm{2}}}{\mathrm{3}}−\mathrm{2}=\frac{\mathrm{8}\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)}{\mathrm{3}} \\ $$$$\mathrm{tan}\:\beta=\frac{\mathrm{4}−\mathrm{2}\:\mathrm{cos}\:\alpha}{\mathrm{4}−\mathrm{2}\:\mathrm{sin}\:\alpha}=\frac{\mathrm{2}\left(\mathrm{3}−\sqrt{\mathrm{2}}\right)}{\mathrm{5}} \\ $$$${b}=\mathrm{2}\:\mathrm{tan}\:\left(\frac{\pi}{\mathrm{2}}−\alpha−\beta\right)=\frac{\mathrm{2}}{\mathrm{tan}\:\left(\alpha+\beta\right)} \\ $$$$\:\:=\frac{\mathrm{2}\left(\mathrm{1}−\frac{\sqrt{\mathrm{2}}}{\mathrm{4}}×\frac{\mathrm{2}\left(\mathrm{3}−\sqrt{\mathrm{2}}\right)}{\mathrm{5}}\right)}{\frac{\sqrt{\mathrm{2}}}{\mathrm{4}}+\frac{\mathrm{2}\left(\mathrm{3}−\sqrt{\mathrm{2}}\right)}{\mathrm{5}}}=\frac{\mathrm{4}\left(\mathrm{15}−\mathrm{2}\sqrt{\mathrm{2}}\right)}{\mathrm{31}} \\ $$$${S}={bh}=\frac{\mathrm{4}\left(\mathrm{15}−\mathrm{2}\sqrt{\mathrm{2}}\right)}{\mathrm{31}}×\frac{\mathrm{8}\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)}{\mathrm{3}} \\ $$$$\:\:\:=\frac{\mathrm{32}\left(\mathrm{17}\sqrt{\mathrm{2}}−\mathrm{19}\right)}{\mathrm{93}}\approx\mathrm{1}.\mathrm{735} \\ $$

Commented by cherokeesay last updated on 04/Apr/25

thank you master !

$${thank}\:{you}\:{master}\:! \\ $$

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