if a,b,care the 3 sides of the triangle ABC,and a^2+b^2+c^2=ab+bc+ac.what is the shape of the triangle?

quinn
Sep 30, 2014

#4**+15 **

this is my method:

2a^2+2b^2+2c^2=2ac+2bc+2ac (both side times two)

a^2+2ab+b^2+b^2+2bc+c^2+a^2+2ac+c^2=0

(a-b)^2+(b-c）^2+(a-c)^2=0

because the square of all real number always more than 0

so (a-b)^2=0 , (b-c)^2=0 ,(a-c)^2=0

so a=b=c

so this is a Equiatreral Triangle

Good job guys!

quinn
Oct 1, 2014

#1**+13 **

This is "non-mathematical," but one possibilty is that the triangle is equilateral.

To see why, let a = b = c = some length ..... call it, "N"

So we have.......

a^2 + b^2 + c^2 = ab + bc + ac ..... then.......

N^2 + N^2 + N^2 = N*N + N*N + N*N .......and......

N^2 + N^2 + N^2 = N^2 + N^2 + N^2 ........ and both sides are equal when a = b = c .........

That's all i've got !!!!!!

CPhill
Sep 30, 2014

#2**+13 **

We can show that Chris's solution is the only one as follows:

Treat this as a quadratic in a: a^{2} - (b+c)a +b^{2}+c^{2}-bc = 0

Solve for a using the quadratic formula

a = (b + c ± [(b+c)^{2} - 4(b^{2} + c^{2} - bc)]^{1/2})/2

a = (b+c ± [b^{2} + c^{2} + 2bc - 4b^{2} - 4c^{2} +4bc]^{1/2})/2

a = (b+c ± [6bc - 3b^{2} - 3c^{2}]^{1/2})/2

a = (b+c ± [-3(b - c)^{2}]^{1/2})/2

a = (b+c ± i*(b-c)*√3)/2

This can only be real (presumably the triangle has real sides!) if b = c so that the imaginary part vanishes.

When b = c then a = (b + c)/2 = (b + b)/2 = b

So all the sides are the same length.

Alan
Sep 30, 2014

#4**+15 **

Best Answer

this is my method:

2a^2+2b^2+2c^2=2ac+2bc+2ac (both side times two)

a^2+2ab+b^2+b^2+2bc+c^2+a^2+2ac+c^2=0

(a-b)^2+(b-c）^2+(a-c)^2=0

because the square of all real number always more than 0

so (a-b)^2=0 , (b-c)^2=0 ,(a-c)^2=0

so a=b=c

so this is a Equiatreral Triangle

Good job guys!

quinn
Oct 1, 2014