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Question-74615




Question Number 74615 by chess1 last updated on 27/Nov/19
Answered by mind is power last updated on 27/Nov/19
a=8−b  (2)⇔(8−b)^2 +b^2 +c^2 =32  x^2 +y^2 ≥(((x+y)^2 )/2)  ⇒(8−b)^2 +b^2 ≥(((8−b+b)^2 )/2)=32  but c^2 +(8−b)^2 +b^2 =32..3  ⇒(8−b)^2 +b^2 =32⇒8−b=b⇒b=4  3⇒c=0  a=8−4=4  (a,b,c)=(4,4,0)
$$\mathrm{a}=\mathrm{8}−\mathrm{b} \\ $$$$\left(\mathrm{2}\right)\Leftrightarrow\left(\mathrm{8}−\mathrm{b}\right)^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} +\mathrm{c}^{\mathrm{2}} =\mathrm{32} \\ $$$$\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} \geqslant\frac{\left(\mathrm{x}+\mathrm{y}\right)^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\Rightarrow\left(\mathrm{8}−\mathrm{b}\right)^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} \geqslant\frac{\left(\mathrm{8}−\mathrm{b}+\mathrm{b}\right)^{\mathrm{2}} }{\mathrm{2}}=\mathrm{32} \\ $$$$\mathrm{but}\:\mathrm{c}^{\mathrm{2}} +\left(\mathrm{8}−\mathrm{b}\right)^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} =\mathrm{32}..\mathrm{3} \\ $$$$\Rightarrow\left(\mathrm{8}−\mathrm{b}\right)^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} =\mathrm{32}\Rightarrow\mathrm{8}−\mathrm{b}=\mathrm{b}\Rightarrow\mathrm{b}=\mathrm{4} \\ $$$$\mathrm{3}\Rightarrow\mathrm{c}=\mathrm{0} \\ $$$$\mathrm{a}=\mathrm{8}−\mathrm{4}=\mathrm{4} \\ $$$$\left(\mathrm{a},\mathrm{b},\mathrm{c}\right)=\left(\mathrm{4},\mathrm{4},\mathrm{0}\right) \\ $$$$ \\ $$
Answered by JDamian last updated on 28/Nov/19
Geometric   approach  Expression (2) stands for a sphere which  radius length is 4(√(2.))  Expression (1) stands for a plane.    The sphere and the plane have contact  in c=0. Therefore, the problem is narrowed  to find the points where a straight line and  a circle intersect:     { ((a+b),=,8),((a^2 +b^2 ),=,(32)) :}       ⇒ a=4,  b=4
$$\boldsymbol{\mathrm{Geometric}}\:\:\:\boldsymbol{\mathrm{approach}} \\ $$$${Expression}\:\left(\mathrm{2}\right)\:{stands}\:{for}\:{a}\:{sphere}\:{which} \\ $$$${radius}\:{length}\:{is}\:\mathrm{4}\sqrt{\mathrm{2}.} \\ $$$${Expression}\:\left(\mathrm{1}\right)\:{stands}\:{for}\:{a}\:{plane}. \\ $$$$ \\ $$$${The}\:{sphere}\:{and}\:{the}\:{plane}\:{have}\:{contact} \\ $$$${in}\:{c}=\mathrm{0}.\:{Therefore},\:{the}\:{problem}\:{is}\:{narrowed} \\ $$$${to}\:{find}\:{the}\:{points}\:{where}\:{a}\:{straight}\:{line}\:{and} \\ $$$${a}\:{circle}\:{intersect}: \\ $$$$ \\ $$$$\begin{cases}{{a}+{b}}&{=}&{\mathrm{8}}\\{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }&{=}&{\mathrm{32}}\end{cases}\:\:\:\:\:\:\:\Rightarrow\:{a}=\mathrm{4},\:\:{b}=\mathrm{4} \\ $$$$ \\ $$

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