the-minimum-of-x-y-z-

Question Number 200196 by a.lgnaoui last updated on 15/Nov/23

the minimum of (x + y + z) ?

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 .

Commented by mr W last updated on 15/Nov/23

{(x + y + z)}_{m i n} = \sqrt{\frac{a^{2} + b^{2} + c^{2}}{2} + 2 \sqrt{3} Δ}

Δ = \frac{\sqrt{(a + b + c) (- a + b + c) (a - b + c) (a + b - c)}}{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

{l e t}^{'} s s a y t h e d i s t a n c e o f t h e p o i n t P t o

t h e v e r t e x A i s g i v e n . w e c a n s e e t h a t

t h e p o s i t i o n o f t h e p o i n t P m u s t b e

t h e t o u c h i n g p o i n t f r o m t h e c i r c l e

w i t h c e n t e r a t A a n d t h e e l i p s e w i t h

f o c i i a t p o i n t s B a n d C s u c h t h a t

P B + P C + (P A) i s m i n i m u m . t h a t

m e a n s ∠ A P B = ∠ A P C .

s i m i l a r l y i f t h e d i s t a n c e o f t h e p o i n t

P t o v e r t e x B i s g i v e n, s u c h t h a t t h e

s u m o f d i s t a n c e s A P + C P + (B P) i s

m i n i m u m, ∠ B P A = ∠ B P C .

t h a t m e a n s s u c h t h a t t h e s u m o f t h e

d i s t a n c e s t o a l l v e r t i c e s i s m i m i m u m,

∠ A P B = ∠ B P C = ∠ C P A = \frac{360 °}{3} = 120 °

t h i s p o i n t P i n a t r i a n g l e i s c a l l e d

t h e F e r m a t p o i n t .

Commented by mr W last updated on 16/Nov/23

Commented by mr W last updated on 16/Nov/23

f u r t h e r s e e Q 164174

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