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Question-72294

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Question Number 72294 by mr W last updated on 27/Oct/19
Commented by mr W last updated on 27/Oct/19
the distances from a point to three vertices of a square are known. find the side length of the square.
$${the}\:{distances}\:{from}\:{a}\:{point}\:{to}\:{three} \\ $$$${vertices}\:{of}\:{a}\:{square}\:{are}\:{known}. \\ $$$${find}\:{the}\:{side}\:{length}\:{of}\:{the}\:{square}. \\ $$
Commented by Prithwish sen last updated on 27/Oct/19
s_3 ^2 +s_2 ^2 = a^2 .....(i) s_3 ^2 +s_1 ^2 = b^2 .....(ii) s_1 ^2 +s_4 ^2 = c^2 ......(iii) and s_1 +s_2 = s_3 +s_4 = s .....(iv) solving all these s_2 = ((s^2 −b^2 +a^2 )/(2s)) ....(v) s_3 = ((s^2 +b^2 −c^2 )/(2s)) putting these on (i) [((s^2 −b^2 +a^2 )/(2s))]^2 +[((s^2 +b^2 −c^2 )/(2s))]^2 = a^2 2s^4 −2s^2 (a^2 +c^2 )+[(a^2 −b^2 )^2 +(c^2 −b^2 )^2 ]=0 ∴s^2 = ((a^2 +c^2 )/2)±(√(b^2 (a^2 +c^2 −b^2 )−[((a^2 −c^2 )/2)]^2 ))
$$\boldsymbol{\mathrm{s}}_{\mathrm{3}} ^{\mathrm{2}} +\boldsymbol{\mathrm{s}}_{\mathrm{2}} ^{\mathrm{2}} \:=\:\boldsymbol{\mathrm{a}}^{\mathrm{2}} …..\left(\boldsymbol{\mathrm{i}}\right) \\ $$$$\boldsymbol{\mathrm{s}}_{\mathrm{3}} ^{\mathrm{2}} +\boldsymbol{\mathrm{s}}_{\mathrm{1}} ^{\mathrm{2}} \:=\:\boldsymbol{\mathrm{b}}^{\mathrm{2}} …..\left(\boldsymbol{\mathrm{ii}}\right) \\ $$$$\boldsymbol{\mathrm{s}}_{\mathrm{1}} ^{\mathrm{2}} +\boldsymbol{\mathrm{s}}_{\mathrm{4}} ^{\mathrm{2}} =\:\boldsymbol{\mathrm{c}}^{\mathrm{2}} ……\left(\boldsymbol{\mathrm{iii}}\right) \\ $$$$\boldsymbol{\mathrm{and}}\:\boldsymbol{\mathrm{s}}_{\mathrm{1}} +\boldsymbol{\mathrm{s}}_{\mathrm{2}} \:=\:\boldsymbol{\mathrm{s}}_{\mathrm{3}} +\boldsymbol{\mathrm{s}}_{\mathrm{4}} \:=\:\boldsymbol{\mathrm{s}}\:…..\left(\boldsymbol{\mathrm{iv}}\right) \\ $$$$\boldsymbol{\mathrm{solving}}\:\boldsymbol{\mathrm{all}}\:\boldsymbol{\mathrm{these}} \\ $$$$\boldsymbol{\mathrm{s}}_{\mathrm{2}} =\:\frac{\boldsymbol{\mathrm{s}}^{\mathrm{2}} −\boldsymbol{\mathrm{b}}^{\mathrm{2}} +\boldsymbol{\mathrm{a}}^{\mathrm{2}} }{\mathrm{2}\boldsymbol{\mathrm{s}}}\:….\left(\boldsymbol{\mathrm{v}}\right) \\ $$$$\boldsymbol{\mathrm{s}}_{\mathrm{3}} \:=\:\frac{\boldsymbol{\mathrm{s}}^{\mathrm{2}} +\boldsymbol{\mathrm{b}}^{\mathrm{2}} −\boldsymbol{\mathrm{c}}^{\mathrm{2}} }{\mathrm{2}\boldsymbol{\mathrm{s}}} \\ $$$$\boldsymbol{\mathrm{putting}}\:\boldsymbol{\mathrm{these}}\:\boldsymbol{\mathrm{on}}\:\left(\boldsymbol{\mathrm{i}}\right) \\ $$$$\left[\frac{\boldsymbol{\mathrm{s}}^{\mathrm{2}} −\boldsymbol{\mathrm{b}}^{\mathrm{2}} +\boldsymbol{\mathrm{a}}^{\mathrm{2}} }{\mathrm{2}\boldsymbol{\mathrm{s}}}\right]^{\mathrm{2}} +\left[\frac{\boldsymbol{\mathrm{s}}^{\mathrm{2}} +\boldsymbol{\mathrm{b}}^{\mathrm{2}} −\boldsymbol{\mathrm{c}}^{\mathrm{2}} }{\mathrm{2}\boldsymbol{\mathrm{s}}}\right]^{\mathrm{2}} =\:\boldsymbol{\mathrm{a}}^{\mathrm{2}} \\ $$$$\mathrm{2}\boldsymbol{\mathrm{s}}^{\mathrm{4}} −\mathrm{2}\boldsymbol{\mathrm{s}}^{\mathrm{2}} \left(\boldsymbol{\mathrm{a}}^{\mathrm{2}} +\boldsymbol{\mathrm{c}}^{\mathrm{2}} \right)+\left[\left(\boldsymbol{\mathrm{a}}^{\mathrm{2}} −\boldsymbol{\mathrm{b}}^{\mathrm{2}} \right)^{\mathrm{2}} +\left(\boldsymbol{\mathrm{c}}^{\mathrm{2}} −\boldsymbol{\mathrm{b}}^{\mathrm{2}} \right)^{\mathrm{2}} \right]=\mathrm{0} \\ $$$$\therefore\boldsymbol{\mathrm{s}}^{\mathrm{2}} \:=\:\frac{\boldsymbol{\mathrm{a}}^{\mathrm{2}} +\boldsymbol{\mathrm{c}}^{\mathrm{2}} }{\mathrm{2}}\pm\sqrt{\boldsymbol{\mathrm{b}}^{\mathrm{2}} \left(\boldsymbol{\mathrm{a}}^{\mathrm{2}} +\boldsymbol{\mathrm{c}}^{\mathrm{2}} −\boldsymbol{\mathrm{b}}^{\mathrm{2}} \right)−\left[\frac{\boldsymbol{\mathrm{a}}^{\mathrm{2}} −\boldsymbol{\mathrm{c}}^{\mathrm{2}} }{\mathrm{2}}\right]^{\mathrm{2}} } \\ $$
Commented by Prithwish sen last updated on 27/Oct/19
Commented by mr W last updated on 27/Oct/19
thanks sir!
$${thanks}\:{sir}! \\ $$
Answered by mr W last updated on 27/Oct/19
a^2 =s^2 +b^2 −2bs cos α ⇒2bs cos α=s^2 +b^2 −a^2 ...(i) c^2 =s^2 +b^2 −2bs cos ((π/2)−α) ⇒2bs sin α=s^2 +b^2 −c^2 ...(ii) (i)^2 +(ii)^2 : 4b^2 s^2 =(s^2 +b^2 −a^2 )^2 +(s^2 +b^2 −c^2 )^2 ⇒s^4 −(a^2 +c^2 )s^2 −b^2 (a^2 +c^2 −b^2 )+((a^4 +c^4 )/2)=0 ⇒s^2 =(1/2){a^2 +c^2 ±(√(a^4 +2a^2 c^2 +c^4 +4b^2 (a^2 +c^2 −b^2 )−2a^4 −2c^4 ))} ⇒s^2 =((a^2 +c^2 )/2)±(√(b^2 (a^2 +c^2 −b^2 )−(((a^2 −c^2 )/2))^2 )) b^2 (a^2 +c^2 −b^2 )−(((a^2 −c^2 )/2))^2 ≥0 b^4 −(a^2 +c^2 )b^2 +(((a^2 −c^2 )/2))^2 ≤0 ⇒b^2 =(1/2)(a^2 +c^2 ±2ac)=(((a±c)^2 )/2) ⇒((∣a−c∣)/( (√2)))≤b≤((a+c)/( (√2))) example: a=1, b=2, b=3 s^2 =((1+9)/2)±(√(4(1+9−4)−(((1−9)/2))^2 )) s^2 =5±2(√2)
$${a}^{\mathrm{2}} ={s}^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{2}{bs}\:\mathrm{cos}\:\alpha \\ $$$$\Rightarrow\mathrm{2}{bs}\:\mathrm{cos}\:\alpha={s}^{\mathrm{2}} +{b}^{\mathrm{2}} −{a}^{\mathrm{2}} \:\:\:…\left({i}\right) \\ $$$${c}^{\mathrm{2}} ={s}^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{2}{bs}\:\mathrm{cos}\:\left(\frac{\pi}{\mathrm{2}}−\alpha\right) \\ $$$$\Rightarrow\mathrm{2}{bs}\:\mathrm{sin}\:\alpha={s}^{\mathrm{2}} +{b}^{\mathrm{2}} −{c}^{\mathrm{2}} \:\:\:…\left({ii}\right) \\ $$$$\left({i}\right)^{\mathrm{2}} +\left({ii}\right)^{\mathrm{2}} : \\ $$$$\mathrm{4}{b}^{\mathrm{2}} {s}^{\mathrm{2}} =\left({s}^{\mathrm{2}} +{b}^{\mathrm{2}} −{a}^{\mathrm{2}} \right)^{\mathrm{2}} +\left({s}^{\mathrm{2}} +{b}^{\mathrm{2}} −{c}^{\mathrm{2}} \right)^{\mathrm{2}} \\ $$$$\Rightarrow{s}^{\mathrm{4}} −\left({a}^{\mathrm{2}} +{c}^{\mathrm{2}} \right){s}^{\mathrm{2}} −{b}^{\mathrm{2}} \left({a}^{\mathrm{2}} +{c}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)+\frac{{a}^{\mathrm{4}} +{c}^{\mathrm{4}} }{\mathrm{2}}=\mathrm{0} \\ $$$$\Rightarrow{s}^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}\left\{{a}^{\mathrm{2}} +{c}^{\mathrm{2}} \pm\sqrt{{a}^{\mathrm{4}} +\mathrm{2}{a}^{\mathrm{2}} {c}^{\mathrm{2}} +{c}^{\mathrm{4}} +\mathrm{4}{b}^{\mathrm{2}} \left({a}^{\mathrm{2}} +{c}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)−\mathrm{2}{a}^{\mathrm{4}} −\mathrm{2}{c}^{\mathrm{4}} }\right\} \\ $$$$\Rightarrow{s}^{\mathrm{2}} =\frac{{a}^{\mathrm{2}} +{c}^{\mathrm{2}} }{\mathrm{2}}\pm\sqrt{{b}^{\mathrm{2}} \left({a}^{\mathrm{2}} +{c}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)−\left(\frac{{a}^{\mathrm{2}} −{c}^{\mathrm{2}} }{\mathrm{2}}\right)^{\mathrm{2}} } \\ $$$${b}^{\mathrm{2}} \left({a}^{\mathrm{2}} +{c}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)−\left(\frac{{a}^{\mathrm{2}} −{c}^{\mathrm{2}} }{\mathrm{2}}\right)^{\mathrm{2}} \geqslant\mathrm{0} \\ $$$${b}^{\mathrm{4}} −\left({a}^{\mathrm{2}} +{c}^{\mathrm{2}} \right){b}^{\mathrm{2}} +\left(\frac{{a}^{\mathrm{2}} −{c}^{\mathrm{2}} }{\mathrm{2}}\right)^{\mathrm{2}} \leqslant\mathrm{0} \\ $$$$\Rightarrow{b}^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}\left({a}^{\mathrm{2}} +{c}^{\mathrm{2}} \pm\mathrm{2}{ac}\right)=\frac{\left({a}\pm{c}\right)^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\Rightarrow\frac{\mid{a}−{c}\mid}{\:\sqrt{\mathrm{2}}}\leqslant{b}\leqslant\frac{{a}+{c}}{\:\sqrt{\mathrm{2}}} \\ $$$${example}: \\ $$$${a}=\mathrm{1},\:{b}=\mathrm{2},\:{b}=\mathrm{3} \\ $$$${s}^{\mathrm{2}} =\frac{\mathrm{1}+\mathrm{9}}{\mathrm{2}}\pm\sqrt{\mathrm{4}\left(\mathrm{1}+\mathrm{9}−\mathrm{4}\right)−\left(\frac{\mathrm{1}−\mathrm{9}}{\mathrm{2}}\right)^{\mathrm{2}} } \\ $$$${s}^{\mathrm{2}} =\mathrm{5}\pm\mathrm{2}\sqrt{\mathrm{2}} \\ $$

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