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Question Number 38775 by ajfour last updated on 29/Jun/18

Commented by ajfour last updated on 29/Jun/18

$$Onasegmentoflength\mathit{c},iskept$$$$aunitsquare,overhangingalittle;$$$$suchthat\left(seediagram\right).$$$$Findthelengthofsegment\mathit{c}$$$$underthebottomedgeofsquare.$$

Commented by MrW3 last updated on 29/Jun/18

$$Ithinkyouhavetogiveanadditional$$$$condition,e.g.theradiusofthecircle,$$$$otherwisethesegmentcisnotunique.$$

Answered by MrW3 last updated on 30/Jun/18

$$letR=radiusofthecircle$$$$leta=sideofthesquare(a=1)$$$$c=2R-a+x$$$$\frac{a}{h}=\frac{h}{2R-a}\Rightarrow h=\sqrt{a(2R-a)}$$$$\frac{x}{a}=\frac{a}{h}\Rightarrow x=\frac{a^2}{h}=\frac{a^2}{\sqrt{a(2R-a)}}$$$$\Rightarrow c=2R-a+\frac{a^2}{\sqrt{a(2R-a)}}$$

Commented by MrW3 last updated on 30/Jun/18

Commented by ajfour last updated on 30/Jun/18

$$let\tan \theta =\frac{x}{a}$$$$\tan \theta =\frac{x}{a}=\frac{a}{h}=\frac{h}{c-x}$$$$\Rightarrow \frac{x^2}{a^2}=\frac{a}{c-x}andasa=1$$$${\mathit{x}}^3-{\mathit{c}\mathit{x}}^2+1=0$$$$Lety=\frac{1}{x}$$$$\Rightarrow {\mathit{y}}^3-\mathit{c}\mathit{y}+1=0$$$$lety=u+v$$$$\Rightarrow u^3+v^3+3uv(u+v)-c(u+v)+1=0$$$$or{u}^3+v^3+(u+v)(3uv-c)+1=0$$$$let3uv=c\Rightarrow u^3{v}^3=\frac{c^3}{27}$$$$\Rightarrow u^3+v^3=-1$$$$u^3,v^{3}arerootsof$$$$z^2+z+\frac{c^3}{27}=0$$$$u^3,v^3=\frac{-1\pm \sqrt{1-\frac{4c^3}{27}}}{2}$$$$=\frac{-1}{2}\pm \sqrt{\frac{1}{4}-\frac{c^3}{27}}$$$$y=\sqrt[3]{-\frac{1}{2}-\sqrt{\frac{1}{4}-\frac{c^3}{27}}}$$$$+\sqrt[3]{-\frac{1}{2}+\sqrt{\frac{1}{4}-\frac{c^3}{27}}}$$$$x=\frac{1}{y}.$$$$$$

Commented by MJS last updated on 30/Jun/18

$$a=1\Rightarrow c=\frac{1+\sqrt{{(2R-1)}^3}}{\sqrt{2R-1}};R>\frac{1}{2}$$$$\frac{dc}{dR}=2-\frac{1}{\sqrt{{(2R-1)}^3}}$$$$\Rightarrow c\mathrm{is}\mathrm{minimal}\mathrm{for}R=\frac{1}{2}+\frac{\sqrt[3]{2}}{4};c=\frac{3\sqrt[3]{2}}{2}$$

Commented by MJS last updated on 01/Jul/18

$$x^3-{cx}^2+1=0$$$$x=z+\frac{c}{3}$$$${(z+\frac{c}{3})}^3-c{(z+\frac{c}{3})}^2+1=0$$$$z^3-\frac{c^2}{3}z+\frac{27-2c^3}{27}=0$$$$u=\sqrt[3]{\frac{1}{27}{c}^3-\frac{1}{2}+\frac{\sqrt{3}}{18}\sqrt{27-4c^3}};v=\sqrt[3]{\frac{1}{27}{c}^3-\frac{1}{2}-\frac{\sqrt{3}}{18}\sqrt{27-4c^3}}$$$$z_1=u+v;x_1=z_1+\frac{c}{3}$$$$z_2=(-\frac{1}{2}-\frac{\sqrt{3}}{2}\mathrm{i})u+(-\frac{1}{2}+\frac{\sqrt{3}}{2}\mathrm{i})v;x_2=z_2+\frac{c}{3}$$$$z_3=(-\frac{1}{2}+\frac{\sqrt{3}}{2}\mathrm{i})u+(-\frac{1}{2}-\frac{\sqrt{3}}{2}\mathrm{i})v;x_3=z_3+\frac{c}{3}$$$$\mathrm{I}\mathrm{think}\mathrm{this}\mathrm{is}\mathrm{easier}\mathrm{than}\mathrm{to}\mathrm{find}x=\frac{1}{y}\mathrm{with}$$$$y\mathrm{being}\mathrm{a}\mathrm{sum}\mathrm{of}\mathrm{two}\mathrm{cubic}\mathrm{roots}.\mathrm{anyway}$$$$\mathrm{both}\mathrm{methods}\mathrm{lead}\mathrm{to}\mathrm{the}\mathrm{same}\mathrm{values},\mathrm{if}\mathrm{you}$$$$\mathrm{use}\mathrm{a}\mathrm{calculator}\mathrm{to}\mathrm{get}\mathrm{approximate}\mathrm{values}$$$$\mathrm{the}\mathrm{methods}\mathrm{are}\mathrm{equal}...$$

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