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Problem Of The Week #447 December 21st 2020

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anemone

MHB POTW Director
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Feb 14, 2012
3,894
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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,894
Congratulations to kaliprasad for his correct solution (Cool), which you can find below! However, I will also include another alternative solution to solve for another set of $(a,\,b)$ and note that those are just two sets of the uncountably many options.

Solution from kaliprasad :
It makes sense to choose a and b to be the multiples of $\frac{\pi}{4}$ excluding odd multiple $\frac{\pi}{2}$ as it goes to infinite.

If we take $a=\frac{\pi}{4}$ then we have $\tan a = 1$

If we choose $\tan b = 0$ then we have $a + b = \frac{2015\pi}{4}$ or $b = \frac{1007\pi}{2}$ or $\tan b$ infinite which is a contradiction.

If we choose $\tan b = 1$ then we have $a + b = \frac{4030\pi}{4}$ or $b = \frac{4029\pi}{4}$ or $\tan b$ 1 which is consistent.

So $a=\frac{\pi}{4},b=\frac{4029\pi}{4}$ is a solution.

Alternative solution:
Let $a=\dfrac{\pi}{4}+n\pi$ and $b=\dfrac{\pi}{4}+m\pi$ for integer values $n$ and $m$.

Substituting these into the original equation and simplifying, we have

$\dfrac{2m+2n+1}{4m+1}=2015$

Observe that $n=5036$ and $m=1$ leading to a possible set of solution, i.e. $(a,\,b)=\left(\dfrac{5\pi}{4},\,\dfrac{20145\pi}{4}\right)$.
 
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