Can someone help me prove a^2+ab+b^2=c^2 for a 120 degree triangle?

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Discussion Overview

The discussion revolves around proving the conjecture that in a triangle with sides of lengths (a, b, c) where the angle between sides a and b is 120 degrees, the relationship a^2 + ab + b^2 = c^2 holds true. The scope includes mathematical reasoning and proofs related to triangle geometry.

Discussion Character

  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant proposes the conjecture that a^2 + ab + b^2 = c^2 for a triangle with a 120-degree angle between sides a and b.
  • Another participant provides a proof using geometric reasoning involving two triangles, leading to the equation a^2 + 2ag + b^2 = c^2 and substituting values based on the cosine of 60 degrees.
  • A third participant suggests an alternative approach using the cosine law, noting that for C = 120 degrees, cos(C) = -1/2, which leads to a different expression for c^2.
  • A later reply acknowledges the cleverness of the geometric proof and admits to overcomplicating the problem initially.

Areas of Agreement / Disagreement

Participants present different methods to approach the proof, with no consensus on a single method being superior. The discussion remains open to multiple viewpoints and approaches.

Contextual Notes

The discussion does not resolve potential assumptions about the triangle's properties or the implications of using different mathematical approaches. The relationship between the angles and sides is explored but not definitively established.

Ynaught?
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I am hoping someone can help me with a proof for the following conjecture:

If a triangle has sides with lengths (a,b,c) and sides (a,b) enclose an angle of 120 degrees, then:

a^2+ab+b^2=c^2

Mahalo
 
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Hey, Mahalo,
this one is easy.

In the attached picture, the 120-degree angle (on the green triangle) is marked with a black arc, while its 180-complement (on the red triangle), marked in yellow, would be a 60-degree angle.

On the red triangle, g^2 + h^2 = b^2 (eq.1).
On the bigger (red + green) triangle, (a + g)^2 + h^2 = c^2, or a^2 + 2ag + g^2 + h^2 = c^2 (eq.2).
Substituting (eq.1) into (eq.2), we get a^2 + 2ag + b^2 = c^2 (eq.3).
But, on the red triangle, g = b * cos(60) = b * 1/2; thus b = 2g, which turns (eq.3) into your equation.
 

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Dang! That's clever. I would have used the cosine law:
[itex]c^2= a^2+ b^2- 2abcos(C)[/itex]
Since, here, C= 120 degrees, cos(C)= cos(120)= -1/2.
 
Thanks gents. I was trying to make things much too complicated. ;)
 

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