Prove that a triangle with sides of length, a,b and c is equilateral if & only if (a+b+c)^2 = 3(ab+bc+ac)?

Answer 1

Since a=b=c, so

(a+b+c)^2 = 3(ab+bc+ac)

(3a)^2 = 3(a^2+a^2+a^2)

9a^2 = 3(3a^2)

9a^2 = 9a^2

Answer 2

Proving that triangle is equilateral -> formula holds is easy. just substitute a = b = c
Proving the other way is harder.
Assume the Formula holds.
(a+b+c)^2 = 3(ab+bc+ac)
a^2+b^2+c^2+2ab + 2bc + 2 ac = 3(ab+bc+ac)
a^2+b^2+c^2 = ab + ac + bc
a^2+b^2+c^2 - ab - ac - bc = 0
a^2 + (-b-c)a + b^2 + c^2 - bc = 0

Solving for a, we get
a = [(b+c) +- sqrt((-b-c)^2 - 4*(b^2 + c^2 - bc))]/2
= [b+c +- sqrt(b^2 +2bc + c^2 - 4b^2 - 4c^2 + 4bc)]/2
= [b+c +- sqrt(-3b^2 - 3c^2 + 6bc)]/2
= [b+c +- sqrt(-3(b-c)^2)]/2

Now since there is a -3 inside the square root, the square of anything is always non-negative, and inside the square root we must have a non-negative number,
this is only possible if b-c = 0
Therefore, b = c.
Solve for c, and you prove a = b
So if the equation holds, a = b = c.

Answer 3

(a+b+c)^2 = 3(ab+bc+ca)
<=> (a^2+b^2+c^2+2ab+2bc+2ca) = 3(ab+bc+ca)
<=> a^2+b^2+c^2-ab-bc-ca = 0
<=>2a^2+2b^2+2c^2 - 2ab-2bc-2ca =0
<=>(a^2+b^2-2ab)+(a^2+c^2-2ac) + (b^2+c^2-2bc) =0
<=>(a-b)^2+(a-c)^2+(b-c)^2 = 0
true if only if a=b,a=c b= c or a=b=c equilateal triangle