✔ 最佳答案
當 P 為費馬點時, (PA + PB + PC) 為最短, 即 ㄥAPB = ㄥBPC = ㄥCPA = 120°
設 PA = a, PB = b, PC = c, 則
a^2 + b^2 - 2ab cos 120° = (22√3)^2
==> a^2 + b^2 + ab = 1452 . . . . . (i)
b^2 + c^2 + bc = 26^2
==> b^2 + c^2 + bc = 676 . . . . . .(ii)
c^2 + a^2 + ca = 1452 + 676
==> c^2 + a^2 + ca = 2128 . . . . . (iii)
(1/2)ab sin 120° + (1/2)bc sin 120° + (1/2)ca sin 120° = 13*22√3
==> ab + bc + ca = 1144 . . . . . . (iv)
(i) + (ii) + (iii) + 3*(iv), 得
2(a^2 + b^2 + c^2 + 2ab + 2bc + 2ca) = 1452 + 676 + 2128 + 3*1144
==> (a + b + c)^2 = 3844
==> a + b + c = 62
答案 : (PA + PB + PC) 的最短長度是 62