Possible Measures of Angle C in Triangle ABC with Given Equation

[utkarshakash]

Homework Statement

If in a triangle ABC, $a^4+b^4+c^4=2c^2(a^2+b^2)$, prove that c=45° or 135°

Homework Equations

The Attempt at a Solution

Rearranging I have

$(a^c-c^2)^2+b^2(b^2-2c^2)=0 \\<br /> <br /> cos C=\dfrac{a^2+b^2-c^2}{2ab} \\<br /> a^2-c^2=2abcosC-b^2$

 

[Mark44]

Homework Statement

If in a triangle ABC, $a^4+b^4+c^4=2c^2(a^2+b^2), prove that c=45° or 135°$

[itex]You have things tangled up here, I believe. The lowercase letters a, b, and c typically represent the lengths of the sides. The uppercase letters A, B, and C typically represent the angle measures. Angle C would be the angle across from the side of length c. It's confusing to me that you use c for what appears to be a side length <u>and</u> an angle.<br /> <br /> <br /> <blockquote data-attributes="" data-quote="utkarshakash" data-source="post: 4234263" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-title"> utkarshakash said: </div> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> <h2>Homework Equations</h2><br /> <br /> <h2>The Attempt at a Solution</h2><br /> Rearranging I have <br /> <br /> $(a^c-c^2)^2+b^2(b^2-2c^2)=0 \\<br /> <br /> cos c=\dfrac{a^2+b^2-c^2}{2ab} \\<br /> a^2-c^2=2abcosC-b^2$ </div> </div> </blockquote>[/itex]

 

[utkarshakash]

You have things tangled up here, I believe. The lowercase letters a, b, and c typically represent the lengths of the sides. The uppercase letters A, B, and C typically represent the angle measures. Angle C would be the angle across from the side of length c. It's confusing to me that you use c for what appears to be a side length and an angle.

I have edited the question. Please see again.

 

[Saitama]

Start by investigating the expansion of $(a^2+b^2-c^2)^2$. Its pretty straightforward after that.

 

[utkarshakash]

Start by investigating the expansion of $(a^2+b^2-c^2)^2$. Its pretty straightforward after that.

Thanks