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Question Number 158622 by ajfour last updated on 06/Nov/21

Commented by ajfour last updated on 06/Nov/21

$$Findmaximumsidelength$$$$ofinscribedsquare\left(s_{max}\right).$$

Answered by mr W last updated on 07/Nov/21

Commented by mr W last updated on 07/Nov/21

$$AB=\frac{s/2}{h}\times \sqrt{a^2-h^2}$$$$CD=\frac{s/2}{h}\times \sqrt{b^2-h^2}$$$$AD=\frac{h}{h-s/2}\times s$$$$AD=AB+BC+CD$$$$\frac{s/2}{h}\times \sqrt{a^2-h^2}+\frac{s/2}{h}\times \sqrt{b^2-h^2}+s=\frac{h}{h-s/2}\times s$$$$\sqrt{a^2-h^2}+\sqrt{b^2-h^2}+2h=\frac{4h^2}{2h-s}$$$$s=2h-\frac{4h^2}{\sqrt{a^2-h^2}+\sqrt{b^2-h^2}+2h}$$$$let\lambda =\frac{s}{a},\mu =\frac{b}{a},\xi =\frac{h}{a}$$$$\lambda =2\xi -\frac{4{\xi }^2}{\sqrt{1-{\xi }^2}+\sqrt{{\mu }^2-{\xi }^2}+2\xi }$$$$\frac{d\lambda }{d\xi }=0for{\lambda }_{max}$$$$...$$

Commented by mr W last updated on 07/Nov/21

Commented by ajfour last updated on 07/Nov/21

$$\mathbb{T}\mathrm{hank}\mathrm{you}\mathrm{sir}.{\mathrm{I}}^{\prime }\mathrm{ll}\mathrm{think}\mathrm{of}$$$$\mathrm{some}\mathrm{better}\mathrm{elliptic}\mathrm{way}.$$

Commented by ajfour last updated on 07/Nov/21

$$s^2=h\sqrt{a^2-h^2}+h\sqrt{b^2-h^2}$$$$-\{\frac{s}{2}p+\frac{s}{2}q+s(h-\frac{s}{2})\}$$$$p+q+s=\sqrt{a^2-h^2}+\sqrt{b^2-h^2}$$$$\Rightarrow hs=(h+\frac{s}{2})(\sqrt{a^2-h^2}+\sqrt{b^2-h^2})$$$$\Rightarrow$$$$\frac{1}{\mathit{s}}=\frac{1}{\sqrt{{\mathit{a}}^2-{\mathit{h}}^2}+\sqrt{{\mathit{b}}^2-{\mathit{h}}^2}}-\frac{1}{2\mathit{h}}$$$$...............$$

Commented by Tawa11 last updated on 07/Nov/21

$$\mathrm{Great}\mathrm{sir}$$

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