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Question-76346

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Question Number 76346 by Maclaurin Stickker last updated on 26/Dec/19
Commented by Maclaurin Stickker last updated on 26/Dec/19
1. Find x as a function of b and c. 2. Under what conditions does the problem admit a solution?
$$\mathrm{1}.\:{Find}\:{x}\:{as}\:{a}\:{function}\:{of}\:{b}\:{and}\:{c}. \\ $$$$\mathrm{2}.\:{Under}\:{what}\:{conditions}\:{does}\:{the} \\ $$$${problem}\:{admit}\:{a}\:{solution}? \\ $$
Commented by Maclaurin Stickker last updated on 26/Dec/19
I find x=(√((b^2 +c^2 ±2(√(3b^2 c^2 −b^4 −c^4 )))/5)) I didn′t understand the second question, could anyone explain?
$${I}\:{find}\:{x}=\sqrt{\frac{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \pm\mathrm{2}\sqrt{\mathrm{3}{b}^{\mathrm{2}} {c}^{\mathrm{2}} −{b}^{\mathrm{4}} −{c}^{\mathrm{4}} }}{\mathrm{5}}} \\ $$$${I}\:{didn}'{t}\:{understand}\:{the}\:{second} \\ $$$${question},\:{could}\:{anyone}\:{explain}? \\ $$
Commented by Maclaurin Stickker last updated on 26/Dec/19
The answer of the book is: (2/( (√5)+1))≤(b/c)≤(((√5)+1)/2) . How can I get this result?
$${The}\:{answer}\:{of}\:{the}\:{book}\:{is}: \\ $$$$\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}+\mathrm{1}}\leqslant\frac{{b}}{{c}}\leqslant\frac{\sqrt{\mathrm{5}}+\mathrm{1}}{\mathrm{2}}\:.\:{How}\:{can}\:{I}\:{get}\:{this} \\ $$$${result}? \\ $$
Commented by MJS last updated on 26/Dec/19
3b^2 c^2 −b^4 −c^4 ≥0 b^2 +c^2 ∓2(√(3b^2 c^2 −b^4 −c^4 ))≥0 you have to solve these
$$\mathrm{3}{b}^{\mathrm{2}} {c}^{\mathrm{2}} −{b}^{\mathrm{4}} −{c}^{\mathrm{4}} \geqslant\mathrm{0} \\ $$$${b}^{\mathrm{2}} +{c}^{\mathrm{2}} \mp\mathrm{2}\sqrt{\mathrm{3}{b}^{\mathrm{2}} {c}^{\mathrm{2}} −{b}^{\mathrm{4}} −{c}^{\mathrm{4}} }\geqslant\mathrm{0} \\ $$$$\mathrm{you}\:\mathrm{have}\:\mathrm{to}\:\mathrm{solve}\:\mathrm{these} \\ $$
Commented by Maclaurin Stickker last updated on 26/Dec/19
Thank you, sir.
$${Thank}\:{you},\:{sir}. \\ $$
Answered by mr W last updated on 26/Dec/19
Commented by mr W last updated on 26/Dec/19
b sin θ=x b cos θ=(√(c^2 −x^2 ))+x ⇒b^2 =x^2 +((√(c^2 −x^2 ))+x)^2 ⇒b^2 =x^2 +c^2 +2x(√(c^2 −x^2 )) ⇒b^2 −c^2 −x^2 =2x(√(c^2 −x^2 )) ⇒(b^2 −c^2 )^2 −2(b^2 −c^2 )x^2 +x^4 =4x^2 (c^2 −x^2 ) ⇒5x^4 −2(b^2 +c^2 )x^2 +(b^2 −c^2 )^2 =0 ⇒x^2 =((b^2 +c^2 ±2(√(3b^2 c^2 −b^4 −c^4 )))/5) ⇒x=(√((b^2 +c^2 ±2(√(3b^2 c^2 −b^4 −c^4 )))/5)) b cos θ=(√(c^2 −b^2 sin^2 θ))+b sin θ b(cos θ−sin θ)=(√(c^2 −b^2 sin^2 θ)) b^2 (1−sin 2θ)=c^2 −b^2 sin^2 θ ⇒((b/c))^2 =(1/(1+sin^2 θ−sin 2θ)) ⇒((b/c))^2 =(2/(3−(cos 2θ+2sin 2θ))) ⇒((b/c))^2 =(2/(3−(√5) cos (2θ−α))) with α=tan^(−1) 2 ⇒((b/c))_(max) ^2 =(2/(3−(√5)))=((((√5)+1)/2))^2 ⇒((b/c))_(max) =(((√5)+1)/2) at 2θ−α=0 ⇒θ=(α/2)=((tan^(−1) 2)/2)=31.7° ⇒((b/c))_(min) ^2 =(2/(3+(√5)))=((2/( (√5)+1)))^2 ⇒((b/c))_(min) =(2/( (√5)+1)) at 2θ−α=180° ⇒θ=((α+180)/2)=((tan^(−1) 2+180)/2)=121.7° ⇒(2/( (√5)+1))≤(b/c)≤(((√5)+1)/2)
$${b}\:\mathrm{sin}\:\theta={x} \\ $$$${b}\:\mathrm{cos}\:\theta=\sqrt{{c}^{\mathrm{2}} −{x}^{\mathrm{2}} }+{x} \\ $$$$\Rightarrow{b}^{\mathrm{2}} ={x}^{\mathrm{2}} +\left(\sqrt{{c}^{\mathrm{2}} −{x}^{\mathrm{2}} }+{x}\right)^{\mathrm{2}} \\ $$$$\Rightarrow{b}^{\mathrm{2}} ={x}^{\mathrm{2}} +{c}^{\mathrm{2}} +\mathrm{2}{x}\sqrt{{c}^{\mathrm{2}} −{x}^{\mathrm{2}} } \\ $$$$\Rightarrow{b}^{\mathrm{2}} −{c}^{\mathrm{2}} −{x}^{\mathrm{2}} =\mathrm{2}{x}\sqrt{{c}^{\mathrm{2}} −{x}^{\mathrm{2}} } \\ $$$$\Rightarrow\left({b}^{\mathrm{2}} −{c}^{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{2}\left({b}^{\mathrm{2}} −{c}^{\mathrm{2}} \right){x}^{\mathrm{2}} +{x}^{\mathrm{4}} =\mathrm{4}{x}^{\mathrm{2}} \left({c}^{\mathrm{2}} −{x}^{\mathrm{2}} \right) \\ $$$$\Rightarrow\mathrm{5}{x}^{\mathrm{4}} −\mathrm{2}\left({b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right){x}^{\mathrm{2}} +\left({b}^{\mathrm{2}} −{c}^{\mathrm{2}} \right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow{x}^{\mathrm{2}} =\frac{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \pm\mathrm{2}\sqrt{\mathrm{3}{b}^{\mathrm{2}} {c}^{\mathrm{2}} −{b}^{\mathrm{4}} −{c}^{\mathrm{4}} }}{\mathrm{5}} \\ $$$$\Rightarrow{x}=\sqrt{\frac{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \pm\mathrm{2}\sqrt{\mathrm{3}{b}^{\mathrm{2}} {c}^{\mathrm{2}} −{b}^{\mathrm{4}} −{c}^{\mathrm{4}} }}{\mathrm{5}}} \\ $$$$ \\ $$$${b}\:\mathrm{cos}\:\theta=\sqrt{{c}^{\mathrm{2}} −{b}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \:\theta}+{b}\:\mathrm{sin}\:\theta \\ $$$${b}\left(\mathrm{cos}\:\theta−\mathrm{sin}\:\theta\right)=\sqrt{{c}^{\mathrm{2}} −{b}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \:\theta} \\ $$$${b}^{\mathrm{2}} \left(\mathrm{1}−\mathrm{sin}\:\mathrm{2}\theta\right)={c}^{\mathrm{2}} −{b}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \:\theta \\ $$$$\Rightarrow\left(\frac{{b}}{{c}}\right)^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{1}+\mathrm{sin}^{\mathrm{2}} \:\theta−\mathrm{sin}\:\mathrm{2}\theta} \\ $$$$\Rightarrow\left(\frac{{b}}{{c}}\right)^{\mathrm{2}} =\frac{\mathrm{2}}{\mathrm{3}−\left(\mathrm{cos}\:\mathrm{2}\theta+\mathrm{2sin}\:\mathrm{2}\theta\right)} \\ $$$$\Rightarrow\left(\frac{{b}}{{c}}\right)^{\mathrm{2}} =\frac{\mathrm{2}}{\mathrm{3}−\sqrt{\mathrm{5}}\:\mathrm{cos}\:\left(\mathrm{2}\theta−\alpha\right)} \\ $$$${with}\:\alpha=\mathrm{tan}^{−\mathrm{1}} \mathrm{2} \\ $$$$\Rightarrow\left(\frac{{b}}{{c}}\right)_{{max}} ^{\mathrm{2}} =\frac{\mathrm{2}}{\mathrm{3}−\sqrt{\mathrm{5}}}=\left(\frac{\sqrt{\mathrm{5}}+\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\left(\frac{{b}}{{c}}\right)_{{max}} =\frac{\sqrt{\mathrm{5}}+\mathrm{1}}{\mathrm{2}} \\ $$$${at}\:\mathrm{2}\theta−\alpha=\mathrm{0} \\ $$$$\Rightarrow\theta=\frac{\alpha}{\mathrm{2}}=\frac{\mathrm{tan}^{−\mathrm{1}} \mathrm{2}}{\mathrm{2}}=\mathrm{31}.\mathrm{7}° \\ $$$$\Rightarrow\left(\frac{{b}}{{c}}\right)_{{min}} ^{\mathrm{2}} =\frac{\mathrm{2}}{\mathrm{3}+\sqrt{\mathrm{5}}}=\left(\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}+\mathrm{1}}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\left(\frac{{b}}{{c}}\right)_{{min}} =\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}+\mathrm{1}} \\ $$$${at}\:\mathrm{2}\theta−\alpha=\mathrm{180}° \\ $$$$\Rightarrow\theta=\frac{\alpha+\mathrm{180}}{\mathrm{2}}=\frac{\mathrm{tan}^{−\mathrm{1}} \mathrm{2}+\mathrm{180}}{\mathrm{2}}=\mathrm{121}.\mathrm{7}° \\ $$$$\Rightarrow\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}+\mathrm{1}}\leqslant\frac{{b}}{{c}}\leqslant\frac{\sqrt{\mathrm{5}}+\mathrm{1}}{\mathrm{2}} \\ $$
Commented by mr W last updated on 26/Dec/19
i got this result too, sir.
$${i}\:{got}\:{this}\:{result}\:{too},\:{sir}. \\ $$
Commented by Maclaurin Stickker last updated on 26/Dec/19
thanks for correcting
$${thanks}\:{for}\:{correcting} \\ $$
Commented by mr W last updated on 26/Dec/19
i pushed the ⋗ button too early (unintended) before i checked my working, and you commented too fast :)
$${i}\:{pushed}\:{the}\:\gtrdot\:{button}\:{too}\:{early}\:\left({unintended}\right) \\ $$$${before}\:{i}\:{checked}\:{my}\:{working},\:{and} \\ $$$$\left.{you}\:{commented}\:{too}\:{fast}\::\right) \\ $$
Commented by Maclaurin Stickker last updated on 26/Dec/19
I got x by using algebra: h_(x ) =((2(√(p(p−x)(p−b)(p−c))))/x) p=((b+c+x)/2) (x^2 /2)=(√((((b+c+x)/2))(((b+c−x)/2))(((−b+c+x)/2))(((b−c+x)/2)))) 4x^4 =(b^2 +2bc+c^2 −x^2 )(x^2 −c^2 +2bc−b^2 ) 4x^4 =2b^2 c^2 −b^4 −c^4 −x^4 +2b^2 x^2 +2c^2 x^2 5x^4 =−(b^2 −c^2 )^2 +x^2 (2b^2 +2c^2 ) 5x^4 −x^2 (2b^2 +2c^2 )+(b^2 −c^2 )^2 =0 let y=x^2 5y^2 −y(2b^2 +2c^2 )+(b^2 −c^2 )^2 =0 y=((2b^2 +c^2 ±4(√(3b^2 c^2 −b^4 −c^4 )))/(2.5)) y=((b^2 +c^2 ±2(√(3b^2 c^2 −b^4 −c^4 )))/5) x^2 =((b^2 +c^2 ±2(√(3b^2 c^2 −b^4 −c^4 )))/5) x=(√((b^2 +c^2 ±2(√(3b^2 c^2 −b^4 −c^4 )))/5))
$${I}\:{got}\:\boldsymbol{{x}}\:{by}\:{using}\:{algebra}: \\ $$$${h}_{{x}\:} =\frac{\mathrm{2}\sqrt{{p}\left({p}−{x}\right)\left({p}−{b}\right)\left({p}−{c}\right)}}{{x}} \\ $$$${p}=\frac{{b}+{c}+{x}}{\mathrm{2}} \\ $$$$\frac{{x}^{\mathrm{2}} }{\mathrm{2}}=\sqrt{\left(\frac{{b}+{c}+{x}}{\mathrm{2}}\right)\left(\frac{{b}+{c}−{x}}{\mathrm{2}}\right)\left(\frac{−{b}+{c}+{x}}{\mathrm{2}}\right)\left(\frac{{b}−{c}+{x}}{\mathrm{2}}\right)} \\ $$$$\mathrm{4}{x}^{\mathrm{4}} =\left({b}^{\mathrm{2}} +\mathrm{2}{bc}+{c}^{\mathrm{2}} −{x}^{\mathrm{2}} \right)\left({x}^{\mathrm{2}} −{c}^{\mathrm{2}} +\mathrm{2}{bc}−{b}^{\mathrm{2}} \right) \\ $$$$\mathrm{4}{x}^{\mathrm{4}} =\mathrm{2}{b}^{\mathrm{2}} {c}^{\mathrm{2}} −{b}^{\mathrm{4}} −{c}^{\mathrm{4}} −{x}^{\mathrm{4}} +\mathrm{2}{b}^{\mathrm{2}} {x}^{\mathrm{2}} +\mathrm{2}{c}^{\mathrm{2}} {x}^{\mathrm{2}} \\ $$$$\mathrm{5}{x}^{\mathrm{4}} =−\left({b}^{\mathrm{2}} −{c}^{\mathrm{2}} \right)^{\mathrm{2}} +{x}^{\mathrm{2}} \left(\mathrm{2}{b}^{\mathrm{2}} +\mathrm{2}{c}^{\mathrm{2}} \right) \\ $$$$\mathrm{5}{x}^{\mathrm{4}} −{x}^{\mathrm{2}} \left(\mathrm{2}{b}^{\mathrm{2}} +\mathrm{2}{c}^{\mathrm{2}} \right)+\left({b}^{\mathrm{2}} −{c}^{\mathrm{2}} \right)^{\mathrm{2}} =\mathrm{0} \\ $$$${let}\:{y}={x}^{\mathrm{2}} \\ $$$$\mathrm{5}{y}^{\mathrm{2}} −{y}\left(\mathrm{2}{b}^{\mathrm{2}} +\mathrm{2}{c}^{\mathrm{2}} \right)+\left({b}^{\mathrm{2}} −{c}^{\mathrm{2}} \right)^{\mathrm{2}} =\mathrm{0} \\ $$$${y}=\frac{\mathrm{2}{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \pm\mathrm{4}\sqrt{\mathrm{3}{b}^{\mathrm{2}} {c}^{\mathrm{2}} −{b}^{\mathrm{4}} −{c}^{\mathrm{4}} }}{\mathrm{2}.\mathrm{5}} \\ $$$${y}=\frac{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \pm\mathrm{2}\sqrt{\mathrm{3}{b}^{\mathrm{2}} {c}^{\mathrm{2}} −{b}^{\mathrm{4}} −{c}^{\mathrm{4}} }}{\mathrm{5}} \\ $$$${x}^{\mathrm{2}} =\frac{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \pm\mathrm{2}\sqrt{\mathrm{3}{b}^{\mathrm{2}} {c}^{\mathrm{2}} −{b}^{\mathrm{4}} −{c}^{\mathrm{4}} }}{\mathrm{5}} \\ $$$${x}=\sqrt{\frac{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \pm\mathrm{2}\sqrt{\mathrm{3}{b}^{\mathrm{2}} {c}^{\mathrm{2}} −{b}^{\mathrm{4}} −{c}^{\mathrm{4}} }}{\mathrm{5}}} \\ $$
Commented by Maclaurin Stickker last updated on 26/Dec/19
The result in the book is x=(√((b^2 +c^2 ±2(√(3b^2 c^2 −b^4 −c^4 )))/5)).
$${The}\:{result}\:{in}\:{the}\:{book}\:{is} \\ $$$${x}=\sqrt{\frac{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \pm\mathrm{2}\sqrt{\mathrm{3}{b}^{\mathrm{2}} {c}^{\mathrm{2}} −{b}^{\mathrm{4}} −{c}^{\mathrm{4}} }}{\mathrm{5}}}.\: \\ $$$$ \\ $$
Commented by mr W last updated on 26/Dec/19
very nice sir! this path is the one which i also thought of at first. but i did decide to use an other path by using a parameter (e.g. angle θ) such that the part two of the question also gets answered.
$${very}\:{nice}\:{sir}! \\ $$$${this}\:{path}\:{is}\:{the}\:{one}\:{which}\:{i}\:{also} \\ $$$${thought}\:{of}\:{at}\:{first}.\:{but}\:{i}\:{did}\:{decide} \\ $$$${to}\:{use}\:{an}\:{other}\:{path}\:{by}\:{using}\:{a} \\ $$$${parameter}\:\left({e}.{g}.\:{angle}\:\theta\right)\:{such}\:{that} \\ $$$${the}\:{part}\:{two}\:{of}\:{the}\:{question}\:{also}\:{gets} \\ $$$${answered}. \\ $$

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