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- Feb 14, 2012

- 3,802

- Thread starter anemone
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- #1

- Feb 14, 2012

- 3,802

- Apr 4, 2014

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To find my triangle I use the equalities: $\dfrac{sin A}{a} = \dfrac{sin B}{b} = \dfrac{sin C}{c}$

I rearrange your equation to: $\sqrt{2}\sin A+\sin C=2\sin B$ and let $b = 1$ then

$\sqrt{2}\sin A+\sin C=2\dfrac {sin B}{b}$. Now since $\dfrac {sin B}{b} = \dfrac{sin C}{c} = \dfrac{sin A}{a}$

we can write

$\sqrt{2}\sin A+\sin C= \dfrac{sin A}{a} + \dfrac{sin C}{c}$

and if we let $a = \dfrac {1}{\sqrt{2}}$ and $ c = 1$ then our equation is satisfied.

Now that we have found the 3 sides of our triangle we can find the 3 angles. From my CRC handbook I find

$\cos A = \dfrac{b^2 + c^2 - a^2}{2bc} = 0.75$

$\cos B = \dfrac{c^2 + a^2 - b^2}{2ca} = 0.3536$

$\cos C = \dfrac{a^2 + b^2 - c^2}{2ab} = 0.3536$

Converting to the angles in degrees I get:

$A = 41.41 , B = 69.30 , C = 69.30 $ (as a crosscheck we can see that they sum to 180)

and then

$\dfrac{3}{sin A} + \dfrac{\sqrt{2}}{sin C} = \dfrac{3}{sin 41.41} + \dfrac{\sqrt{2}}{sin69.30} = \dfrac{3}{0.6614} + \dfrac{\sqrt{2}}{0.9354} = 4.5358 + 1.5119 = 6.0477$ and this is greater than $2(\sqrt{3}+1) = 5.4641$.

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- Feb 14, 2012

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The question says as long as we can prove that the angles in a triangle $ABC$ satisfy $\sqrt{2}\sin A-2\sin B+\sin C=0$, then $\dfrac{3}{\sin A}+\dfrac{\sqrt{2}}{\sin C}\ge 2(\sqrt{3}+1)$ will always hold.I can find _a_ triangle that satisfies $\sqrt{2}\sin A-2\sin B+\sin C=0$ where $\dfrac{3}{sin A}+\dfrac{\sqrt{2}}{\sin C}\ge 2(\sqrt{3}+1)$ but you use $\ge$ , does this mean there are a multitude of triangles that meet these requirements?

The other comment that I have for this unsolved challenge is that the equality holds when $A=\dfrac{\pi}{3}$ and $C=\dfrac{\pi}{4}$.