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Problem Of The Week #367 May 22nd, 2019

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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,894
Here is this week's POTW:

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Without using a calculator, simplify \(\displaystyle \frac{\displaystyle\sum_{k=1}^{2499}\sqrt{10+{\sqrt{50+\sqrt{k}}}}}{\displaystyle\sum_{k=1}^{2499}\sqrt{10-{\sqrt{50+\sqrt{k}}}}}\).


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

MHB POTW Director
Staff member
Feb 14, 2012
3,894
Hi to all MHB
members!

Despite of the fact that last week's High School's POTW
is more difficult than usual, I am going to give members another week to attempt at a solution!(Smile)
 
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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,894
No one answered last two week's problem,(Sadface) but you can find the suggested solution as below.
Let the numerator and the denominator be $A$ and $B$ respectively.

Letting $a_k=\sqrt{10+\sqrt{50+\sqrt{k}}}$ and $b_k=\sqrt{10-\sqrt{50+\sqrt{k}}}$, we can represent $A$ and $B$ as
\(\displaystyle A=\sum_{1}^{2499}a_k\\B=\sum_{1}^{2499}b_k\),

Letting $p_k=\sqrt{50+\sqrt{k}}$ and $q_k=\sqrt{50-\sqrt{k}}$, since $p_k^2+q_k^2=10^2$ and $p_k>0$ and $q_k>0$, there exists a real number $0<x_k<\dfrac{\pi}{2}$ such that

$p_k=10\cos x_k\\q_k=10\sin x_k$

Then, we get
$a_k=\sqrt{10+10cosx_k}=\sqrt{10+10\left(2\cos^2\dfrac{x_k}{2}-1\right)}=\sqrt{20}\cos\dfrac{x_k}{2}$

$b_k=\sqrt{10-10cosx_k}=\sqrt{10-10\left(1-2\sin^2\dfrac{x_k}{2}\right)}=\sqrt{20}\sin\dfrac{x_k}{2}$

\(\displaystyle
\begin{align}a_{2500-k}&=\sqrt{10+\sqrt{50+\sqrt{(50+\sqrt k)(50-\sqrt k)}}}\\&=\sqrt{10+\sqrt{50+{p_kq_k}}}\\&=\sqrt{10+\sqrt{50+100\cos {x_k}\sin {x_k}}}\\&=\sqrt{10+\sqrt{50(\cos {x_k}+\sin {x_k})^2}}\\&=\sqrt{10+\sqrt{50}\cdot\sqrt2\sin \left(x_k+\frac{\pi}{4}\right)}\\&=\sqrt{10+10\cdot2\cos \left(\frac{x_k}{2}+\frac{\pi}{8}\right)\sin \left(\frac{x_k}{2}+\frac{\pi}{8}\right)}\\&=\sqrt{10\left(\cos \left(\frac{x_k}{2}+\frac{\pi}{8}\right)+\sin \left(\frac{x_k}{2}+\frac{\pi}{8}\right)\right)^2}\\&=\sqrt{10}\left(\cos \left(\frac{x_k}{2}+\frac{\pi}{8}\right)+\sin \left(\frac{x_k}{2}+\frac{\pi}{8}\right)\right)\\&=\frac{\left(\cos \left(\frac{\pi}{8}\right)+\sin \left(\frac{\pi}{8}\right)\right)a_k+\left(\cos \left(\frac{\pi}{8}\right)-\sin \left(\frac{\pi}{8}\right)\right)b_k}{\sqrt2}\\&=\sqrt{\frac{\sqrt2+1}{2\sqrt2}}a_k+\sqrt{\frac{\sqrt2-1}{2\sqrt2}}b_k\end{align}\)

Hence,
$A=\sqrt{\dfrac{\sqrt2+1}{2\sqrt2}}A+\sqrt{\dfrac{\sqrt2-1}{2\sqrt2}}B$

$\dfrac{A}{B}=1+\sqrt{2}+\sqrt{4+2\sqrt{2}}$

or

\(\displaystyle \frac{\displaystyle\sum_{k=1}^{2499}\sqrt{10+{\sqrt{50+\sqrt{k}}}}}{\displaystyle\sum_{k=1}^{2499}\sqrt{10-{\sqrt{50+\sqrt{k}}}}}=1+\sqrt{2}+\sqrt{4+2\sqrt{2}}\)
 
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