Point de Fermat

ABC
2π/3
f:R2→R+
M7→ f(M) = MA +MB +MC
•f P
ABC A, B, C
•P f P
•ABC0BCA0ACB0
ABC AA0BB0CC0P
f(P) = AA0=BB0=CC0
•f
O MA ≥OM −OA
f(M)≥3OM −f(O)
f(M)> f(O)OM > (2/3)f(O)OM ≤(2/3)f(O)
f P
f
P ABC
π/2b Bb C H
A B C
f(H) = HA +HB +HC =HA +BC < BA +BC =f(B)
f(H)< f(C)f B C
Ab A < π/2
π/2≤b A < 2π/3M
Ab A= 2α
MB2=AB2−2AB.AM cos α+AM2
MB =AB −AM cos α+O¡AM2¢
M A MC
f(M) = f(A) + (1 −2 cos α)AM +O¡AM2¢
0<2α < 2π/3 1 −2 cos α < 0f(M)< f(A)M
A f A
f P ABC
P A BC P 0P
P0A < P A P 0B=P B P 0C=P C f(P0)< f(P)
•f
{A, B, C}fC+∞
−−→
gradf(P) = −(−→ u+−→ v+−→ w)
−→
u=−→ P A/P A −→ v=−−→ P B/P C −→ w=−−→ P C/P C
P−→
u+−→ v+−→ w= 0

1 = −→ w2= (−→ u+−→ v)2= 1 + 2−→ u .−→ v+ 1
−→
u .−→ v=−1/2\ AP B = 2π/3\ BP C [ CP A
P AB 2π/3P
A B A C
A P A
f P
ABC
2π/3
•−−→ AB −→ AC
B0C A +π/3
M M0AMM0
f(M) = MA +MB +MC =MM0+MB +M0B0=BM +MM0+M0B0≥BB0
M P M0
P0
³−→
P A, −−→ P B´=2π
3,³−−→
P P 0,−→ P A´=π
3
³−−→
P P 0,−−→ P B´=π B P P 0
³−−−→
P0B0,−−→ P0A´=³−−→ P C, −→ P A´=2π
3,³−−→
P0A, −−→ P0P´=π
3
³−−−→
P0B0,−−→ P0P´=π P, P 0, B0P P 0
BB0
f(P=BP +P P 0+P0B=BB0
AA0CC0
P
ABC0BCA0CAB0f(P) = AA0=BB0=CC0
ABC b A2π/3
f
A