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Question-38775

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Question Number 38775 by ajfour last updated on 29/Jun/18
Commented by ajfour last updated on 29/Jun/18
On a segment of length c , is kept a unit square, overhanging a little; such that (see diagram). Find the length of segment c under the bottom edge of square.
$${On}\:{a}\:{segment}\:{of}\:{length}\:\boldsymbol{{c}}\:,\:{is}\:{kept} \\ $$$${a}\:{unit}\:{square},\:{overhanging}\:{a}\:{little}; \\ $$$${such}\:{that}\:\left({see}\:{diagram}\right). \\ $$$${Find}\:{the}\:{length}\:{of}\:{segment}\:\boldsymbol{{c}} \\ $$$${under}\:{the}\:{bottom}\:{edge}\:{of}\:{square}. \\ $$
Commented by MrW3 last updated on 29/Jun/18
I think you have to give an additional condition, e.g. the radius of the circle, otherwise the segment c is not unique.
$${I}\:{think}\:{you}\:{have}\:{to}\:{give}\:{an}\:{additional} \\ $$$${condition},\:{e}.{g}.\:{the}\:{radius}\:{of}\:{the}\:{circle}, \\ $$$${otherwise}\:{the}\:{segment}\:{c}\:{is}\:{not}\:{unique}. \\ $$
Answered by MrW3 last updated on 30/Jun/18
let R=radius of the circle let a=side of the square (a=1) c=2R−a+x (a/h)=(h/(2R−a))⇒h=(√(a(2R−a))) (x/a)=(a/h)⇒x=(a^2 /h)=(a^2 /( (√(a(2R−a))))) ⇒c=2R−a+(a^2 /( (√(a(2R−a)))))
$${let}\:{R}={radius}\:{of}\:{the}\:{circle} \\ $$$${let}\:{a}={side}\:{of}\:{the}\:{square}\:\left({a}=\mathrm{1}\right) \\ $$$${c}=\mathrm{2}{R}−{a}+{x} \\ $$$$\frac{{a}}{{h}}=\frac{{h}}{\mathrm{2}{R}−{a}}\Rightarrow{h}=\sqrt{{a}\left(\mathrm{2}{R}−{a}\right)} \\ $$$$\frac{{x}}{{a}}=\frac{{a}}{{h}}\Rightarrow{x}=\frac{{a}^{\mathrm{2}} }{{h}}=\frac{{a}^{\mathrm{2}} }{\:\sqrt{{a}\left(\mathrm{2}{R}−{a}\right)}} \\ $$$$\Rightarrow{c}=\mathrm{2}{R}−{a}+\frac{{a}^{\mathrm{2}} }{\:\sqrt{{a}\left(\mathrm{2}{R}−{a}\right)}} \\ $$
Commented by MrW3 last updated on 30/Jun/18
Commented by ajfour last updated on 30/Jun/18
let tan θ = (x/a) tan θ = (x/a) =(a/h) = (h/(c−x)) ⇒ (x^2 /a^2 ) = (a/(c−x)) and as a=1 x^3 −cx^2 +1=0 Let y=(1/x) ⇒ y^3 −cy+1=0 let y=u+v ⇒ u^3 +v^3 +3uv(u+v)−c(u+v)+1=0 or u^3 +v^3 +(u+v)(3uv−c)+1=0 let 3uv = c ⇒ u^3 v^3 = (c^3 /(27)) ⇒ u^3 +v^3 = −1 u^3 , v^3 are roots of z^2 +z+(c^3 /(27)) =0 u^3 , v^3 = ((−1±(√(1−((4c^3 )/(27)))))/2) =((−1)/2)±(√((1/4)−(c^3 /(27)))) y= ((−(1/2)−(√((1/4)−(c^3 /(27))))))^(1/3) +((−(1/2)+(√((1/4)−(c^3 /(27)))) ))^(1/3) x= (1/y) .
$${let}\:\:\mathrm{tan}\:\theta\:=\:\frac{{x}}{{a}} \\ $$$$\:\:\:\:\mathrm{tan}\:\theta\:=\:\frac{{x}}{{a}}\:=\frac{{a}}{{h}}\:=\:\frac{{h}}{{c}−{x}} \\ $$$$\Rightarrow\:\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }\:=\:\frac{{a}}{{c}−{x}}\:\:\:\:\:\:{and}\:{as}\:{a}=\mathrm{1} \\ $$$$\:\:\:\boldsymbol{{x}}^{\mathrm{3}} −\boldsymbol{{cx}}^{\mathrm{2}} +\mathrm{1}=\mathrm{0} \\ $$$$\:{Let}\:\:\:{y}=\frac{\mathrm{1}}{{x}}\:\:\:\:\: \\ $$$$\Rightarrow\:\:\:\:\boldsymbol{{y}}^{\mathrm{3}} −\boldsymbol{{cy}}+\mathrm{1}=\mathrm{0} \\ $$$${let}\:\:\:{y}={u}+{v} \\ $$$$\Rightarrow\:\:{u}^{\mathrm{3}} +{v}^{\mathrm{3}} +\mathrm{3}{uv}\left({u}+{v}\right)−{c}\left({u}+{v}\right)+\mathrm{1}=\mathrm{0} \\ $$$${or}\:\:\:\:{u}^{\mathrm{3}} +{v}^{\mathrm{3}} +\left({u}+{v}\right)\left(\mathrm{3}{uv}−{c}\right)+\mathrm{1}=\mathrm{0} \\ $$$${let}\:\:\:\mathrm{3}{uv}\:=\:{c}\:\:\:\:\Rightarrow\:\:{u}^{\mathrm{3}} {v}^{\mathrm{3}} =\:\frac{{c}^{\mathrm{3}} }{\mathrm{27}} \\ $$$$\Rightarrow\:\:\:\:\:{u}^{\mathrm{3}} +{v}^{\mathrm{3}} \:=\:−\mathrm{1} \\ $$$${u}^{\mathrm{3}} ,\:{v}^{\mathrm{3}} \:{are}\:{roots}\:{of} \\ $$$$\:\:\:\:\:\:\:\:{z}^{\mathrm{2}} +{z}+\frac{{c}^{\mathrm{3}} }{\mathrm{27}}\:=\mathrm{0} \\ $$$$\:\:\:\:\:{u}^{\mathrm{3}} ,\:{v}^{\mathrm{3}} \:\:=\:\frac{−\mathrm{1}\pm\sqrt{\mathrm{1}−\frac{\mathrm{4}{c}^{\mathrm{3}} }{\mathrm{27}}}}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{−\mathrm{1}}{\mathrm{2}}\pm\sqrt{\frac{\mathrm{1}}{\mathrm{4}}−\frac{{c}^{\mathrm{3}} }{\mathrm{27}}} \\ $$$$\:\:\:{y}=\:\sqrt[{\mathrm{3}}]{−\frac{\mathrm{1}}{\mathrm{2}}−\sqrt{\frac{\mathrm{1}}{\mathrm{4}}−\frac{{c}^{\mathrm{3}} }{\mathrm{27}}}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\sqrt[{\mathrm{3}}]{−\frac{\mathrm{1}}{\mathrm{2}}+\sqrt{\frac{\mathrm{1}}{\mathrm{4}}−\frac{{c}^{\mathrm{3}} }{\mathrm{27}}}\:} \\ $$$$\:\:\:\:\:\:\:\:\:{x}=\:\frac{\mathrm{1}}{{y}}\:. \\ $$$$ \\ $$
Commented by MJS last updated on 30/Jun/18
a=1 ⇒ c=((1+(√((2R−1)^3 )))/( (√(2R−1)))); R>(1/2) (dc/dR)=2−(1/( (√((2R−1)^3 )))) ⇒ c is minimal for R=(1/2)+((2)^(1/3) /4); c=((3(2)^(1/3) )/2)
$${a}=\mathrm{1}\:\Rightarrow\:{c}=\frac{\mathrm{1}+\sqrt{\left(\mathrm{2}{R}−\mathrm{1}\right)^{\mathrm{3}} }}{\:\sqrt{\mathrm{2}{R}−\mathrm{1}}};\:{R}>\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\frac{{dc}}{{dR}}=\mathrm{2}−\frac{\mathrm{1}}{\:\sqrt{\left(\mathrm{2}{R}−\mathrm{1}\right)^{\mathrm{3}} }} \\ $$$$\Rightarrow\:{c}\:\mathrm{is}\:\mathrm{minimal}\:\mathrm{for}\:{R}=\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt[{\mathrm{3}}]{\mathrm{2}}}{\mathrm{4}};\:{c}=\frac{\mathrm{3}\sqrt[{\mathrm{3}}]{\mathrm{2}}}{\mathrm{2}} \\ $$
Commented by MJS last updated on 01/Jul/18
x^3 −cx^2 +1=0 x=z+(c/3) (z+(c/3))^3 −c(z+(c/3))^2 +1=0 z^3 −(c^2 /3)z+((27−2c^3 )/(27))=0 u=(((1/(27))c^3 −(1/2)+((√3)/(18))(√(27−4c^3 ))))^(1/3) ; v=(((1/(27))c^3 −(1/2)−((√3)/(18))(√(27−4c^3 ))))^(1/3) z_1 =u+v; x_1 =z_1 +(c/3) z_2 =(−(1/2)−((√3)/2)i)u+(−(1/2)+((√3)/2)i)v; x_2 =z_2 +(c/3) z_3 =(−(1/2)+((√3)/2)i)u+(−(1/2)−((√3)/2)i)v; x_3 =z_3 +(c/3) I think this is easier than to find x=(1/y) with y being a sum of two cubic roots. anyway both methods lead to the same values, if you use a calculator to get approximate values the methods are equal...
$${x}^{\mathrm{3}} −{cx}^{\mathrm{2}} +\mathrm{1}=\mathrm{0} \\ $$$${x}={z}+\frac{{c}}{\mathrm{3}} \\ $$$$\left({z}+\frac{{c}}{\mathrm{3}}\right)^{\mathrm{3}} −{c}\left({z}+\frac{{c}}{\mathrm{3}}\right)^{\mathrm{2}} +\mathrm{1}=\mathrm{0} \\ $$$${z}^{\mathrm{3}} −\frac{{c}^{\mathrm{2}} }{\mathrm{3}}{z}+\frac{\mathrm{27}−\mathrm{2}{c}^{\mathrm{3}} }{\mathrm{27}}=\mathrm{0} \\ $$$${u}=\sqrt[{\mathrm{3}}]{\frac{\mathrm{1}}{\mathrm{27}}{c}^{\mathrm{3}} −\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{18}}\sqrt{\mathrm{27}−\mathrm{4}{c}^{\mathrm{3}} }};\:{v}=\sqrt[{\mathrm{3}}]{\frac{\mathrm{1}}{\mathrm{27}}{c}^{\mathrm{3}} −\frac{\mathrm{1}}{\mathrm{2}}−\frac{\sqrt{\mathrm{3}}}{\mathrm{18}}\sqrt{\mathrm{27}−\mathrm{4}{c}^{\mathrm{3}} }} \\ $$$${z}_{\mathrm{1}} ={u}+{v};\:{x}_{\mathrm{1}} ={z}_{\mathrm{1}} +\frac{{c}}{\mathrm{3}} \\ $$$${z}_{\mathrm{2}} =\left(−\frac{\mathrm{1}}{\mathrm{2}}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i}\right){u}+\left(−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i}\right){v};\:{x}_{\mathrm{2}} ={z}_{\mathrm{2}} +\frac{{c}}{\mathrm{3}} \\ $$$${z}_{\mathrm{3}} =\left(−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i}\right){u}+\left(−\frac{\mathrm{1}}{\mathrm{2}}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i}\right){v};\:{x}_{\mathrm{3}} ={z}_{\mathrm{3}} +\frac{{c}}{\mathrm{3}} \\ $$$$\mathrm{I}\:\mathrm{think}\:\mathrm{this}\:\mathrm{is}\:\mathrm{easier}\:\mathrm{than}\:\mathrm{to}\:\mathrm{find}\:{x}=\frac{\mathrm{1}}{{y}}\:\mathrm{with} \\ $$$${y}\:\mathrm{being}\:\mathrm{a}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{two}\:\mathrm{cubic}\:\mathrm{roots}.\:\mathrm{anyway} \\ $$$$\mathrm{both}\:\mathrm{methods}\:\mathrm{lead}\:\mathrm{to}\:\mathrm{the}\:\mathrm{same}\:\mathrm{values},\:\mathrm{if}\:\mathrm{you} \\ $$$$\mathrm{use}\:\mathrm{a}\:\mathrm{calculator}\:\mathrm{to}\:\mathrm{get}\:\mathrm{approximate}\:\mathrm{values} \\ $$$$\mathrm{the}\:\mathrm{methods}\:\mathrm{are}\:\mathrm{equal}… \\ $$

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