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Question Number 200196 by a.lgnaoui last updated on 15/Nov/23
the minimum of (x+y+z)?
$$\mathrm{the}\:\mathrm{minimum}\:\mathrm{of}\:\left(\mathrm{x}+\mathrm{y}+\mathrm{z}\right)? \\ $$
Commented by a.lgnaoui last updated on 15/Nov/23
Commented by Frix last updated on 15/Nov/23
You need to find the Fermat Point.
$$\mathrm{You}\:\mathrm{need}\:\mathrm{to}\:\mathrm{find}\:\mathrm{the}\:\mathrm{Fermat}\:\mathrm{Point}. \\ $$
Commented by mr W last updated on 15/Nov/23
(x+y+z)_(min) =(√(((a^2 +b^2 +c^2 )/2)+2(√3)Δ)) Δ=((√((a+b+c)(−a+b+c)(a−b+c)(a+b−c)))/4)
$$\left({x}+{y}+{z}\right)_{{min}} =\sqrt{\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }{\mathrm{2}}+\mathrm{2}\sqrt{\mathrm{3}}\Delta} \\ $$$$\Delta=\frac{\sqrt{\left({a}+{b}+{c}\right)\left(−{a}+{b}+{c}\right)\left({a}−{b}+{c}\right)\left({a}+{b}−{c}\right)}}{\mathrm{4}} \\ $$
Answered by ajfour last updated on 16/Nov/23
Answered by mr W last updated on 16/Nov/23
Commented by mr W last updated on 16/Nov/23
let′s say the distance of the point P to the vertex A is given. we can see that the position of the point P must be the touching point from the circle with center at A and the elipse with focii at points B and C such that PB+PC+(PA) is minimum. that means ∠APB=∠APC. similarly if the distance of the point P to vertex B is given, such that the sum of distances AP+CP+(BP) is minimum, ∠BPA=∠BPC. that means such that the sum of the distances to all vertices is mimimum, ∠APB=∠BPC=∠CPA=((360°)/3)=120° this point P in a triangle is called the Fermat point.
$${let}'{s}\:{say}\:{the}\:{distance}\:{of}\:{the}\:{point}\:{P}\:{to} \\ $$$${the}\:{vertex}\:{A}\:{is}\:{given}.\:{we}\:{can}\:{see}\:{that} \\ $$$${the}\:{position}\:{of}\:{the}\:{point}\:{P}\:{must}\:{be} \\ $$$${the}\:{touching}\:{point}\:{from}\:{the}\:{circle} \\ $$$${with}\:{center}\:{at}\:{A}\:{and}\:{the}\:{elipse}\:{with} \\ $$$${focii}\:{at}\:{points}\:{B}\:{and}\:{C}\:{such}\:{that} \\ $$$${PB}+{PC}+\left({PA}\right)\:{is}\:{minimum}.\:{that}\: \\ $$$${means}\:\angle{APB}=\angle{APC}. \\ $$$${similarly}\:{if}\:{the}\:{distance}\:{of}\:{the}\:{point} \\ $$$${P}\:{to}\:{vertex}\:{B}\:{is}\:{given},\:{such}\:{that}\:{the} \\ $$$${sum}\:{of}\:{distances}\:{AP}+{CP}+\left({BP}\right)\:{is}\: \\ $$$${minimum},\:\angle{BPA}=\angle{BPC}. \\ $$$${that}\:{means}\:{such}\:{that}\:{the}\:{sum}\:{of}\:{the} \\ $$$${distances}\:{to}\:{all}\:{vertices}\:{is}\:{mimimum}, \\ $$$$\angle{APB}=\angle{BPC}=\angle{CPA}=\frac{\mathrm{360}°}{\mathrm{3}}=\mathrm{120}° \\ $$$${this}\:{point}\:{P}\:{in}\:{a}\:{triangle}\:{is}\:{called} \\ $$$${the}\:{Fermat}\:{point}. \\ $$
Commented by mr W last updated on 16/Nov/23
Commented by mr W last updated on 16/Nov/23
further see Q164174
$${further}\:{see}\:{Q}\mathrm{164174} \\ $$

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