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

Commented by cherokeesay last updated on 03/Apr/25

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

Commented by Nicholas666 last updated on 09/Apr/25

$$beautifullproblem$$

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

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

$$\sin \alpha =\frac{1}{3}\Rightarrow \cos \alpha =\frac{2\sqrt{2}}{3},\tan \alpha =\frac{\sqrt{2}}{4}$$$$a=1+(4\tan \alpha -1)\sin \alpha =\frac{2+\sqrt{2}}{3}$$$$h=\frac{4}{\cos \alpha }-a-2=\frac{4\times 3}{2\sqrt{2}}-\frac{2+\sqrt{2}}{3}-2=\frac{8(\sqrt{2}-1)}{3}$$$$\tan \beta =\frac{4-2\cos \alpha }{4-2\sin \alpha }=\frac{2(3-\sqrt{2})}{5}$$$$b=2\tan (\frac{\pi }{2}-\alpha -\beta )=\frac{2}{\tan (\alpha +\beta )}$$$$=\frac{2(1-\frac{\sqrt{2}}{4}\times \frac{2(3-\sqrt{2})}{5})}{\frac{\sqrt{2}}{4}+\frac{2(3-\sqrt{2})}{5}}=\frac{4(15-2\sqrt{2})}{31}$$$$S=bh=\frac{4(15-2\sqrt{2})}{31}\times \frac{8(\sqrt{2}-1)}{3}$$$$=\frac{32(17\sqrt{2}-19)}{93}\approx 1.735$$

Commented by cherokeesay last updated on 04/Apr/25

$$thankyoumaster!$$

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