(41C) Area of Triangle (SSS)

[Albert Chan]

Trivia: Area Δ = √((ab-y)*y), max(y/(ab)) = 1/4

Prove:
gap = a-b, y = (c + gap)*(c - gap)/4
dy/dc = 2c > 0, so max(y) when c is also maximize, thus c = b
In other words, Δ is isosceles, maybe equilateral (a = b)

Let k = a/b, thus 2 > k >= 1:
gap = kb - b = b*(k-1)
y = (b + gap)*(b - gap) / 4 = k*(2-k) b² / 4
y/(ab) = y/(kb²) = (2-k)/4
--> max(y/(ab)) = max(0+ to 1/4) = 1/4

Isosceles Δ, b=c: Area Δ = √((ab-y)*y) = ab/4 * √((2+k)*(2-k))
Equilateral Δ, a=b=c, thus k=1: Area Δ = a²/4 * √3 ~ (√3/4) a²