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# 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?

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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
+238
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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
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#1
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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
+26397
+13

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

Treat this as a quadratic in a:  a2 - (b+c)a +b2+c2-bc = 0

Solve for a using the quadratic formula

a = (b + c ± [(b+c)2 - 4(b2 + c2 - bc)]1/2)/2

a = (b+c ± [b2 + c2 + 2bc - 4b2 - 4c2 +4bc]1/2)/2

a = (b+c ± [6bc - 3b2 - 3c2]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
#3
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Thank you for showing us how to do that Alan.

Melody  Oct 1, 2014
#4
+238
+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
#5
+91434
0

Thanks Quinn.

Melody  Oct 2, 2014

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