MHB Triangle ABC: Prove cot(A/2) + cot(C/2) = 2cot(B/2)

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In triangle ABC, with sides a, b, and c satisfying a + c = 2b, it is required to prove that cot(A/2) + cot(C/2) = 2cot(B/2). The discussion highlights a problem of the week, inviting participants to solve this geometric identity. Members kaliprasad and greg1313 successfully provided correct solutions to the problem. The thread emphasizes the importance of following guidelines for submissions. This mathematical exploration showcases the relationships between the angles and sides of a triangle.
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Here is this week's POTW:

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For a triangle $ABC$ with the sides of $a,\,b,\,$ and $c$ such that $a+c=2b$, prove $$\cot \frac{A}{2}+\cot \frac{C}{2}=2\cot \dfrac{B}{2}$$.

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Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!
 
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Congratulations to the following members for their correct solution: (Smile)
1. kaliprasad
2. greg1313

Solution from kaliprasad:
By applying the law of cotangents to the triangle $ABC$, with the standard notation $s$ to represent semi-perimeter and the area of triangle as $\triangle$, we have
$\cot(\frac{A}{2}) = \dfrac{s(s-a)}{\triangle}\cdots(1)$
$\cot(\frac{B}{2}) = \dfrac{s(s-b)}{\triangle}\cdots(2)$
$\cot(\frac{C}{2}) = \dfrac{s(s-c)}{\triangle}\cdots(3)$

Add (1) and (3) to get

$\begin{align*}\cot(\frac{A}{2}) + \cot(\frac{C}{2})&=\dfrac{s(s-a)}{\triangle} + \dfrac{s(s-c)}{\triangle}\\&=\dfrac{s(s-a)+s(s-c)}{\triangle}\\&=\dfrac{s(s-a+s-c)}{\triangle}\\&= \dfrac{s(2s-(a+c))}{\triangle}\\&= \dfrac{s(2s-2b)}{\triangle} \text{ (Given)}\\&= 2\dfrac{s(s-b)}{\triangle}\\&=2\cot(\frac{B}{2})\text{ from (2)} \end{align*}$

And we are done.
 
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