1,A_2,A_3$
Now, from these three fixed points we take a point in the triangle $ (x,y)$
Let $d(A_i(x,y))$ be distance of our point from the vertex A
Our goal is to minimize:
$$ D(x,y) =  \sum_{k=1}^{3} d(A_i (x,y) )$$
Supposedly we take the gradient of both sides and set the left to zero, we get,
$$ 0 = \nabla \sum_{k=1}^{3} d(A_i (x,y)) $$
or,
$$ 0 = \sum_{k=1}^{3} \nabla d(A_i (x,y) ) $$
And the point where the three unit vectors of $ d(A_i (x,y))$ go to zero is our Torricelli point, But I don't quite get how he define the functions based on distances from vertex. What exactly is the technicalities of this?
Further, I can't find a similar proof online, is this not a well documented proof?
Edit: On some more thought, could I use a similar method for finding the 'Torricelli point' of more complicated shapes? seems like it should easily be doable by the same principles.
For example, finding the 'toricelli point' of the pentagon reduces to the problem of finding a way to arrange the 5 unit vectors such that their sum is zero as shown below. Further speaking, how would one generally find an arrangement such that adds to zero?

A:

There are many questions. Let's try to make a list.

"does different points have different gradients?"

Yes, they do. The gradient of a function is a vector field, meaning the vector varies form point to point.

"But I don't understand direction the gradient should be in?"

"I don't quite get how he define the functions based on distances from vertex. What exactly is the technicalities of this?"
Geometrically we have 2 properties of the gradient:
a) The gradient points in the direction of the fastest increase of the function.
For the function "distance to O" the direction of fastest increase at some P (as per answer to part 1, this will vary as P varies) is the direction of moving along the ray OP, "
1,A_2,A_3 $ 
 Now, from these three fixed points we take a item in the triangulum $ (x, y)$ 
 Let $ d(A_i(x, y))$ be distance of our item from the vertex vitamin a 
 Our goal is to minimize: 
 $ $ D(x, y) =   \sum_{k=1}^{3 } d(A_i (x, y) ) $ $ 
 purportedly we take the gradient of both sides and set the left field to zero, we contract, 
 $ $ 0 = \nabla \sum_{k=1}^{3 } d(A_i (x, y) ) $ $ 
 or, 
 $ $ 0 = \sum_{k=1}^{3 } \nabla d(A_i (x, y) ) $ $ 
 And the compass point where the three unit vectors of $ d(A_i (x, y))$ go to zero is our Torricelli distributor point, But I don't quite get how he define the functions based on distances from vertex. What precisely is the technicalities of this? 
 Further, I can't find a exchangeable proof online, is this not a well documented proof? 
 Edit: On some more idea, could I use a similar method for finding the' Torricelli point' of more complicated supreme headquarters allied powers europe? seems like it should easily be doable by the same principles. 
 For example, finding the' toricelli point' of the pentagon reduces to the problem of finding a way to arrange the 5 unit vectors such that their sum is zero as shown downstairs. Further speaking, how would one broadly witness an arrangement such that adds to zero? 

 A: 

 There constitute many questions. Let's try to make a tilt. 

 " does different points have different gradients? " 

 Yes, they do. The gradient of a function is a vector field, meaning the vector varies form point to target. 

 " But I don't understand focus the gradient should be in? " 

 " I don't quite get how he define the functions based on distances from vertex. What precisely is the technicalities of this? " 
 Geometrically we have 2 properties of the gradient: 
 a) The gradient target in the direction of the fastest increase of the function. 
 For the routine " distance to O " the direction of fast addition at some P (as per answer to part 1, this will vary as phosphorus varies) is the direction of move along the ray OP, "
1,A_2,A_3$
Nlw, from these three fixea points we takg q poinv in ths triangue $ (x,y)$
Let $d(A_i(x,y))$ be distance lf our kjint from the vertdx A
Our glal is ti miiimize:
$$ D(x,y) =  \sum_{k=1}^{3} d(A_i (x,y) )$$
Suppoacdly ce take the gradlent of botv sides and sed ghz left to zero, we get,
$$ 0 = \nabla \sum_{k=1}^{3} d(W_i (x,y)) $$
ot,
$$ 0 = \sum_{k=1}^{3} \nabla d(W_i (x,j) ) $$
And nht point where the three unit vectkrs of $ d(A_i (x,y))$ go to zero is our Torricelli polnt, Hut I don't quite gft how he dgrinq the functiovs based on distances rrom vertex. What exactly is the techuicalities if thlv?
Further, I ran't fpnd a similar proof otline, ix this not a wcll dmcunented proof?
Edit: On smme more thought, cjuld I usa c similar method for dibding the 'Torfuceuli ppiht' of lorx complicatsd shapes? swems like it should ewwily be doabls by tre same principles.
For example, finding the 'todicelli point' of the pebtagon reduces to the problem jf finding a way to arrange the 5 unit vectors sucv thav ghenv rym is zero as shown below. Further speaking, how rkukd one generally find an artajgrient such thaj adds ck aero?

A:

There are manj questyons. Oet's try uo male a list.

"does different poibts have difyerwnt gradients?"

Yes, tkey do. The gxadienj of a function is a vector fneld, msaning the gector vadkes form point tu ppitt.

"But I don't understand diwection tie grcdient sfoulc be ig?"

"I don't qulte gcd how he define thf funetionv based on distances from vertex. What exarvly is the tevhticdlities jf thls?"
Geometrically we have 2 propgrties of the eradient:
a) Nhe gradixnt points ig the directimj of the fasvest incrqase of rhe funzgion.
For the fumction "distance to I" the direction of farfest increase ac wome P (as per amswdr eo pert 1, evis will varf as P xsries) is the dircctkon pf moving along the say KP, "
1,A_2,A_3$ Now, from these three fixed points a in the $ (x,y)$ Let point the vertex A goal is to $$ D(x,y) = \sum_{k=1}^{3} d(A_i (x,y) Supposedly we take the gradient of both sides and set the left to we get, $$ 0 = \nabla \sum_{k=1}^{3} d(A_i (x,y)) $$ or, $$ 0 \sum_{k=1}^{3} d(A_i ) And the point where the three unit vectors of $ d(A_i (x,y))$ go to zero is Torricelli point, But I don't quite get how define the functions based distances from vertex. What exactly the of this? I find similar proof online, this not a well documented proof? Edit: On some more thought, could I use a similar method finding the of more shapes? like should easily be the same principles. For example, finding of the pentagon reduces to the problem of a way arrange the 5 unit vectors such their sum is zero as shown below. Further how would one generally find an arrangement such that adds to zero? A: There are Let's try to make list. "does different have gradients?" they The gradient a function is a vector field, meaning the vector varies form to point. "But I don't understand direction the gradient should "I quite get how define the functions based distances vertex. What exactly is of Geometrically properties the a) The gradient points the direction of the fastest of the function. For the direction of fastest increase at some P per answer to part 1, this will as P varies) is the direction of moving along the ray OP,
1,A_2,A_3$
Now, from these three fixed pOints we takE a poiNt iN thE tRianGle $ (x,Y)$
Let $d(A_i(x,y))$ be disTAnce Of our point from the verteX A
Our GoAL is tO MiNimizE:
$$ D(x,y) =  \sum_{K=1}^{3} D(A_I (X,Y) )$$
SuPpOsEdlY wE TaKe the GraDient of Both sides aNd sEt The left to zerO, We Get,
$$ 0 = \nabla \suM_{k=1}^{3} d(a_i (x,y)) $$
or,
$$ 0 = \sum_{k=1}^{3} \naBla D(A_i (x,y) ) $$
ANd The POint wHerE the tHree unIT vectoRs of $ d(A_i (x,y))$ Go TO zero iS Our TorrICElLi poInt, But I don't quite gET hOW he define the fuNctionS bASeD ON diStaNces from veRtEx. WhaT Exactly IS tHE TEchNIcalities of thIs?
Further, I cAN't fInd a siMiLar PRoof onLine, iS tHIs nOt a well docuMentEd proof?
EdIt: On soME more thOUght, couLd I use A siMilAr meTHoD fOr fInDIng THe 'torRIceLli point' Of MoRe comPlicATED ShapEs? sEems Like iT should easily Be dOablE By tHe samE prinCiplEs.
for exAmple, fIndinG tHe 'toricelli poinT' of tHe pentagoN reDuCes To The prOBlem of FinDinG a way to Arrange THe 5 uNiT VECtOrs such that their suM iS ZErO as shown Below. FURtHeR Speaking, HoW woUld oNE GenerAlly FInD an arranGement SUcH tHat adds To Zero?

A:

THeRe aRe mAny quEStioNs. Let's Try to makE a lisT.

"Does different pOInts have diffeREnT GRaDIentS?"

YeS, they do. The gRadiENt of A funCTiOn iS A vectOr fieLd, MEaNIng the vector varies fOrM point To poiNt.

"But I don't undErstand dirECTIon the grAdieNT sHOuld be in?"

"I don't qUite gEt how he defINe the funCtionS based on Distances FROm vertex. whaT exActLy iS THe TechnicalitieS OF thiS?"
GEometriCalLy we havE 2 prOpeRtiEs oF tHe gradienT:
a) The graDiEnT pOiNts In the DIrection Of The FaSteSt incREase of The fuNctiOn.
foR The FunctioN "DiSTAnce To o" tHe diRecTiOn of fAsteST inCrease aT some P (as pEr aNSwer To PaRt 1, this wIll vary as P varIeS) is the direCtIon Of moviNG Along the Ray OP, "
1,A_2,A_3$
Now, from these  three fix ed po int s w etake  a p oint in the tr i angl e $ (x,y)$
Let $d(A_i( x,y)) $b e di s ta nce o f our p o in t fro mth e v er t ex  A
Ou r g oal isto minimiz e:$$  D(x,y) =  \ s um _{k=1}^{3}  d( A_i (x,y) )$ $
S uppose dl y w e  take  th e gra diento f both  sides an ds et the left to z er o, w e get,
$$ 0 = \na b la \sum_{k=1}^{3}  d(A_i  ( x ,y ) )  $$ 
or ,
$$ 0 = \ su m_{k= 1 }^{3} \ n ab l a d(A _ i (x,y) ) $$And the poi n t w here t he  th r ee uni t vec to r s o f $ d(A_i ( x,y) )$ go tozero i s  our To r ricelli  point , B utI do n 't  q uit eg eth ow  he def ine thefu nc tions  bas e d o n di sta nces  from  vertex. What  ex actl y  is  thetechn ical it ies o f this ?
Fur th er, I can't fin d asimilar p roo fonl in e, is this n ota w ell doc umented pro of ? E di t: On some more th ou g h t,  could I  use a si mi l ar metho dfor  fin d i ng th e 'T o rr icelli p oint'o fmo re comp li catedsh ape s?seems like  it sh ould eas ily b e  doable by the same principl e s. F or exam ple , finding t he ' t oric elli po int '  of t he pe nt a go n  reduces to the pro bl em offindi ng a way to a rrange the 5 unit vec tors su c h that their s um is  zero as s h own belo w. Fu rther sp eaking, h o w  would o negen era lly f in d an arrangem e n t su ch  that a dds  to zer o?
A: 

T her eare manyquestion s.  L et 's  tr y tom ake a li st .
"d oes  diff e rent p oints  hav edi f fer ent gra d ie n t s?"
Y es , th eydo . The  gra d ien t of afunctionisa  vec to rfield,meaning the v ec tor varies  f orm  point t o point. 

"But I don't understa n d direc tio n the  gra dient sho uld  be in ?""I don 't qui te ge thow h e def i n ethe  f unctions b a s edon di st ance s fromvertex. What exact l y i s the technic ali ties o fthi s ?" Geo me t ric a l ly we have 2 pr operties o ft he  gradient: a)Th e gradi ent poi nts i n  the di rection o f the fas te st i n c rea se of thefunction .
For the funct i on  "dis tan ce toO"  th e dir ection offaste st inc re ase at  some  P  (as per  answer to part 1, this  willvaryasP varies)  is the  directio n of  moving al ong  th e ray  OP ,  "
1,A_2,A_3$
Now, from_these three_fixed points we take_a point_in_the triangle_$_(x,y)$
Let $d(A_i(x,y))$ be_distance of our_point from the vertex_A
Our goal is_to_minimize:
$$ D(x,y) =  \sum_{k=1}^{3} d(A_i (x,y) )$$
Supposedly we take the gradient of both_sides_and set_the_left_to zero, we get,
$$ 0_= \nabla \sum_{k=1}^{3} d(A_i (x,y))_$$
or,
$$ 0_= \sum_{k=1}^{3} \nabla d(A_i (x,y) ) $$
And the_point_where the three_unit vectors of $ d(A_i (x,y))$ go to zero_is our Torricelli point, But I_don't quite get_how_he_define the functions based_on distances from vertex. What exactly_is the technicalities of this?
Further, I_can't find a similar proof online, is_this not a well documented proof?
Edit:_On some more thought, could_I use_a similar method for finding_the 'Torricelli point'_of more_complicated shapes? seems_like it should easily be doable_by the same_principles.
For example, finding the 'toricelli point'_of_the pentagon reduces_to_the_problem of_finding a way_to_arrange the_5_unit vectors such that their sum_is_zero as shown below. Further speaking, how_would one generally find_an_arrangement such that adds_to zero?

A:

There are many questions._Let's try to make a list.

"does_different points_have different_gradients?"

Yes, they do. The gradient of a function is a vector_field, meaning the vector varies form_point to point.

"But I_don't understand_direction_the gradient should_be_in?"

"I don't_quite get how he define the functions_based on_distances from vertex. What exactly is_the technicalities of this?"
Geometrically_we_have 2 properties of the gradient:
a)_The gradient points in the direction_of the fastest increase of_the_function.
For_the function "distance to O"_the direction of fastest increase at_some P (as_per answer to part 1, this will_vary_as P varies) is the direction_of_moving along the ray OP, "