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Problem of the week #6 - May 7th, 2012

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Jameson

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Jan 26, 2012
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Thank you to Chris L T521 for submitting this problem.

Let $A$, $B$ and $C$ be sides of a triangle, and let $\alpha$ be the angle opposite of side $A$, $\beta$ be the angle opposite of side $B$ and $\gamma$ be the angle opposite of side $C$. Show that

\[\frac{A+B}{C} = \frac{\cos\left(\tfrac{1}{2}(\alpha-\beta)\right)}{\sin\left(\tfrac{1}{2}\gamma\right)}\]

Hint:
Law of sines


Hint 2:
Sum to product formula


Remember to read the POTW submission guidelines to find out how to submit your answers!
 
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Jameson

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Jan 26, 2012
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Congratulations to the following members for their correct solutions:

1) Sudharaka
2) caffeinemachine

Solution:

The Law of Sines tells us that for any triangle $\Delta ABC$, we can relate the angles with their sides:

\[\frac{\sin\alpha}{A}= \frac{\sin\beta}{B}= \frac{\sin\gamma}{C} \]

Using this equality, we can come up with the following relations:

\[\begin{aligned}A\sin\gamma &= C\sin\alpha\\ C\sin\beta &= B\sin\gamma\end{aligned}\]

We start with the LHS and show it gives the RHS. Incorporating the relations from above, we see that

\[\begin{aligned}\frac{A+B}{C} &= \frac{A\sin\gamma + B\sin\gamma}{C\sin\gamma}\\ &= \frac{C\sin\alpha+C\sin\beta}{C\sin\gamma}\\ &= \frac{\sin\alpha + \sin\beta}{\sin\gamma}\end{aligned}\]

By the sum-to-product and double angle identities, we have

\[\begin{aligned}\frac{\sin\alpha + \sin\beta}{\sin\gamma} &= \frac{ 2\sin \left( \frac{\alpha+\beta}{2} \right) \cos \left( \frac{\alpha-\beta}{2} \right)}{\sin\gamma}\\ &= \frac{ 2\sin\left( \frac{\alpha+\beta}{2} \right) \cos\left( \frac{\alpha-\beta}{2} \right)}{2\sin \left( \frac{\gamma}{2}\right) \cos\left( \frac{\gamma}{2} \right)}\end{aligned}\]

Note that $\gamma = \pi - \alpha - \beta$. So it follows that $\cos \left(\dfrac{\gamma}{2}\right) = \cos \left( \dfrac{\pi-\alpha-\beta}{2} \right) = \sin \left( \dfrac{\alpha+\beta}{2} \right)$. Therefore, we're now left with

\[\frac{ 2\sin\left( \frac{\alpha+\beta}{2} \right) \cos\left( \frac{\alpha-\beta}{2} \right)}{2\sin \left( \frac{\gamma}{2}\right) \cos\left( \frac{\gamma}{2} \right)} = \frac{ \cos\left( \tfrac{1}{2}(\alpha-\beta)\right) }{\sin \left( \tfrac{1}{2}\gamma\right)}\]

Therefore, $\dfrac{A+B}{C} = \dfrac{ \cos\left( \tfrac{1}{2}(\alpha-\beta)\right) }{\sin \left( \tfrac{1}{2}\gamma\right)}$. Q.E.D.
 
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