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- Feb 14, 2012

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- Thread starter anemone
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- #1

- Feb 14, 2012

- 3,838

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My solution:

\(\displaystyle \angle C=180^{\circ}-2\cdot80^{\circ}=20^{\circ}\)

The Law of Cosines gives us:

\(\displaystyle \overline{CD}^2=2\left(1-\cos\left(20^{\circ} \right) \right)\)

Triangle $CDG$ is similar to the two congruent triangles, hence, we may state:

\(\displaystyle \frac{\overline{CD}}{\overline{DG}}= \frac{1}{\overline{CD}}\implies\overline{DG}= \overline{CD}^2= 2\left(1-\cos\left(20^{\circ} \right) \right)\)

Using a double angle identity for cosine, we may write:

\(\displaystyle \overline{DG}=2\left(1-\left(1-2\sin^2\left(10^{\circ} \right) \right) \right)=4\sin^2\left(10^{\circ} \right)\)

Thus, we see that we require:

\(\displaystyle M+N\sin^2\left(10^{\circ} \right)=\frac{1}{2}\csc^3\left(10^{\circ} \right)\)

With $M=0$, we then have \(\displaystyle M+N=\frac{1}{2}\csc^3\left(10^{\circ} \right)\)

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- Feb 7, 2012

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Remembering that $\sin3\theta = 3\sin\theta - 4\sin^3\theta$ and that $\sin30^\circ = 1/2$, I put $\theta = 10^\circ$, to get $\frac12 = 3\sin10^\circ - 4\sin^310^\circ$. Thus $6\sin10^\circ - 8\sin^310^\circ = 1$. Comparing that with the equation in the previous paragraph, you see that we can take $M=3$, $N=-4$ and hence $M+N = -1$.

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- Feb 14, 2012

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HiMy solution:

\(\displaystyle \angle C=180^{\circ}-2\cdot80^{\circ}=20^{\circ}\)

The Law of Cosines gives us:

\(\displaystyle \overline{CD}^2=2\left(1-\cos\left(20^{\circ} \right) \right)\)

Triangle $CDG$ is similar to the two congruent triangles, hence, we may state:

\(\displaystyle \frac{\overline{CD}}{\overline{DG}}= \frac{1}{\overline{CD}}\implies\overline{DG}= \overline{CD}^2= 2\left(1-\cos\left(20^{\circ} \right) \right)\)

Using a double angle identity for cosine, we may write:

\(\displaystyle \overline{DG}=2\left(1-\left(1-2\sin^2\left(10^{\circ} \right) \right) \right)=4\sin^2\left(10^{\circ} \right)\)

Thus, we see that we require:

\(\displaystyle M+N\sin^2\left(10^{\circ} \right)=\frac{1}{2}\csc^3\left(10^{\circ} \right)\)

With $M=0$, we then have \(\displaystyle M+N=\frac{1}{2}\csc^3\left(10^{\circ} \right)\)

Thanks for participating and since the problem doesn't mention any restriction on the variables $M$ and $N$, you got the full credit for your solution! Bravo!

But even though I am not the question setter, I should have noticed when I solved the problem that I should add the condition on both variables so that their sum is unique. I admit I overlooked this. I'm sorry...

Hi

Remembering that $\sin3\theta = 3\sin\theta - 4\sin^3\theta$ and that $\sin30^\circ = 1/2$, I put $\theta = 10^\circ$, to get $\frac12 = 3\sin10^\circ - 4\sin^310^\circ$. Thus $6\sin10^\circ - 8\sin^310^\circ = 1$. Comparing that with the equation in the previous paragraph, you see that we can take $M=3$, $N=-4$ and hence $M+N = -1$.

Thanks for participating and I must tell you when I arrived at the step where I got $2Q\sin^3 10^{\circ}+2P \sin 10^{\circ}-1=0$, I immediately thought of the triple angle formula for the function of sine and you're recent help to me in one of the simultaneous equation threads was the inspiration to me...so, I wouldn't have solved this problem if not for what you have taught me!