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Question Number 182192 by mr W last updated on 05/Dec/22

Commented by mr W last updated on 05/Dec/22

$$findthesmallestareaofinscribed$$$$right-angletriangleinagiven$$$$trianglewithsidesa,b,c.(a⩾b⩾c)$$

Answered by mr W last updated on 06/Dec/22

Commented by mr W last updated on 06/Dec/22

$$BD=\frac{p\sin (\beta +\theta )}{\sin \beta }$$$$DC=\frac{q\sin (\gamma +\frac{\pi }{2}-\theta )}{\sin \gamma }=\frac{q\cos (\gamma -\theta )}{\sin \gamma }$$$$\frac{p\sin (\beta +\theta )}{\sin \beta }+\frac{q\cos (\gamma -\theta )}{\sin \gamma }=a$$$$\frac{p(\sin \beta \cos \theta +\cos \beta \sin \theta )}{\sin \beta }+\frac{q(\cos \gamma \cos \theta +\sin \gamma \sin \theta )}{\sin \gamma }=a$$$$\frac{p}{a}(\cos \theta +\frac{\sin \theta }{\tan \beta })+\frac{q}{a}(\frac{\cos \theta }{\tan \gamma }+\sin \theta )=1$$$$\xi (\cos \theta +\frac{\sin \theta }{\tan \beta })+\eta (\frac{\cos \theta }{\tan \gamma }+\sin \theta )=1$$$$\xi u+\eta v=1$$$$S=\frac{pq}{2}=\frac{\xi \eta a^2}{2}=\frac{\left(\xi u\right)\left(\eta v\right)a^2}{2uv}=\frac{{\left(\sqrt{\left(\xi u\right)\left(\eta v\right)}\right)}^2{a}^2}{2uv}$$$$⩽\frac{{(\xi u+\eta v)}^2{a}^2}{8uv}=\frac{a^2}{8uv}$$$$maximum\frac{a^2}{8uv}at\xi u=\eta v=\frac{1}{2}$$$$S=\frac{a^2}{8uv}=\frac{a^2}{8(\cos \theta +\frac{\sin \theta }{\tan \beta })(\frac{\cos \theta }{\tan \gamma }+\sin \theta )}$$$$\mathrm{\Psi }=(\cos \theta +\frac{\sin \theta }{\tan \beta })(\frac{\cos \theta }{\tan \gamma }+\sin \theta )$$$$\frac{d\mathrm{\Psi }}{d\theta }=(-\sin \theta +\frac{\cos \theta }{\tan \beta })(\frac{\cos \theta }{\tan \gamma }+\sin \theta )+(\cos \theta +\frac{\sin \theta }{\tan \beta })(-\frac{\sin \theta }{\tan \gamma }+\cos \theta )=0$$$$-\frac{\sin \theta \cos \theta }{\tan \gamma }+\frac{{\cos }^2\theta }{\tan \beta \tan \gamma }-{\sin }^2\theta +\frac{\sin \theta \cos \theta }{\tan \beta }-\frac{\sin \theta \cos \theta }{\tan \gamma }-\frac{{\sin }^2\theta }{\tan \beta \tan \gamma }+{\cos }^2\theta +\frac{\sin \theta \cos \theta }{\tan \beta }=0$$$$\tan 2\theta \left(\frac{\tan \beta -\tan \gamma }{1+\tan \beta \tan \gamma }\right)=1$$$$\tan 2\theta \tan (\beta -\gamma )=1$$$$2\theta +\beta -\gamma =\frac{\pi }{2}$$$$\Rightarrow \theta =\frac{\pi }{4}-\frac{\beta -\gamma }{2}$$

Commented by mr W last updated on 06/Dec/22

Commented by mr W last updated on 07/Dec/22

$$\mathrm{\Delta }DEFistherightangledtriangle$$$$withsmallestarea.$$$$AD=\frac{\sqrt{bc(b+c+a)(b+c-a)}}{b+c}$$$$DE=DF=\frac{\sin \frac{\alpha }{2}\sqrt{bc(b+c+a)(b+c-a)}}{(b+c)\sin (\frac{\pi }{4}+\frac{\alpha }{2})}$$$$=\frac{\sin \frac{\alpha }{2}\sqrt{2bc(b+c+a)(b+c-a)}}{(b+c)(\sin \frac{\alpha }{2}+\cos \frac{\alpha }{2})}$$$$=\frac{\sqrt{1-\cos \alpha }\sqrt{bc(b+c+a)(b+c-a)}}{(b+c)\sqrt{1+\sin \alpha }}$$$$S_{min}=\frac{bc(b+c+a)(b+c-a)(1-\cos \alpha )}{2{(b+c)}^2(1+\sin \alpha )}$$$$\Rightarrow S_{min}=\frac{{\left(\frac{2\mathrm{\Delta }}{b+c}\right)}^2}{1+\frac{2\mathrm{\Delta }}{bc}}$$$$with\mathrm{\Delta }=areaof\mathrm{\Delta }ABC.$$$$\mathrm{\Delta }=\frac{\sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}}{4}$$

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