# Problem Of The Week #367 May 22nd, 2019

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#### anemone

##### MHB POTW Director
Staff member
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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#### anemone

##### MHB POTW Director
Staff member
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! #### anemone

##### MHB POTW Director
Staff member
No one answered last two week's problem, 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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