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Question Number 190694 by cortano12 last updated on 09/Apr/23

Commented by nikif99 last updated on 10/Apr/23

Answered by nikif99 last updated on 10/Apr/23

$$$$$$\frac{b}{CD}=\tan 60\Rightarrow CD=\frac{b\sqrt{3}}{3},d=\frac{x\sqrt{3}}{2}\left(1\right)$$$$A_1=bx=b(2-2CD)=2b\left(\frac{3-b\sqrt{3}}{3}\right)\left(2\right),x=\frac{2(3-b\sqrt{3})}{3}\left(3\right)$$$$△BFG:\frac{d-R}{R}=\tan 60\Rightarrow d=R(1+\sqrt{3})\stackrel{\left(1\right)}{\Rightarrow }\frac{x\sqrt{3}}{2}=R(1+\sqrt{3})\stackrel{\left(3\right)}{\Rightarrow }$$$$R=\frac{2(3-b\sqrt{3})\sqrt{3}}{2\times 3(1+\sqrt{3})}\Rightarrow R=\frac{(3-b\sqrt{3})\sqrt{3}}{3(1+\sqrt{3})}\left(4\right)$$$$A_2=\frac{\pi R^2}{2}=\frac{\pi }{2}{\left[\frac{(3-b\sqrt{3})\sqrt{3}}{3(1+\sqrt{3})}\right]}^2\Rightarrow ...\Rightarrow$$$$A_2=\frac{\pi [(2-\sqrt{3})b^2+2(3-2\sqrt{3})b+3(2-\sqrt{3})]}{4}\left(5\right)$$$$a=\frac{2R(d-R)}{2R+(d+R)}{}^{(\ast )seecomment}=\frac{2R\times R\sqrt{3}}{R+d}=\frac{2R^2\sqrt{3}}{R+R(1+\sqrt{3})}\stackrel{\left(4\right)}{\Rightarrow }$$$$a=\frac{2\left[\frac{(3-b\sqrt{3})\sqrt{3}}{3(1+\sqrt{3})}\right]\sqrt{3}}{2+\sqrt{3}}\Rightarrow ...\Rightarrow a=\frac{2(3-b\sqrt{3})}{5+3\sqrt{3}}\left(6\right)$$$$A_3=a^2=\frac{4{(3-b\sqrt{3})}^2}{{(5+3\sqrt{3})}^2}\Rightarrow ...\Rightarrow A_3=\frac{6(b^2-2b\sqrt{3}+3)}{15\sqrt{3}+26}\left(7\right)$$$$A_1+A_2+A_3=f\left(b\right)=...=\frac{6\pi -8\sqrt{3}-3\pi \sqrt{3}}{12}{b}^2+$$$$+\frac{10\sqrt{3}-2\pi \sqrt{3}+3\pi -14}{2}b+\frac{3(12\sqrt{3}-20+2\pi -\pi \sqrt{3}}{4}$$$$f\left(x\right)={kx}^2+mx+nhasmaximumat-\frac{m}{2k}$$$$f\left(b\right)hasmaximumat0.493=b\stackrel{\left(3\right)}{\Rightarrow }x=1.431$$

Commented by nikif99 last updated on 10/Apr/23

$$Asquareinscribedinatriangleof$$$$basebandheighthhassidea=\frac{b\times h}{b+h}$$

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