# Problem Of The Week #364 May 1st, 2019

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#### anemone

##### MHB POTW Director
Staff member
Here is this week's POTW:

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Prove $\sqrt{\sqrt{3}\cos10^{\circ}+1}+\sqrt{\sqrt{3}\cos110^{\circ}+1}+\sqrt{\sqrt{3}\cos130^{\circ}+1}=\sqrt{\dfrac{9}{2}}$.

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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 POTW. Solution from other:
Let $X=\sqrt{a}+\sqrt{b}+\sqrt{c}$ where

\begin{align*}a&=\sqrt{3}\cos 10^{\circ}+1\\&=2\cos 30^{\circ}\cos 10^{\circ}+1\\&=\cos 40^{\circ}+\cos 20^{\circ}+1\\&=2\cos^{2}20^{\circ}+\cos 20^{\circ}\\&=2\cos 20^{\circ}\left(\cos 20^{\circ}+\frac{1}{2}\right)\\&=2\cos 20^{\circ}(\cos 20^{\circ}+\cos 60^{\circ})\\&=2\cos 20^{\circ}2\cos 40^{\circ}\cos 20^{\circ}\\&=4\cos^2 20^{\circ}\cos 40^{\circ}\end{align*}

\begin{align*}b&=\sqrt{3}\cos 110^{\circ}+1\\&=2\sin 60^{\circ}\cos 110^{\circ}+1\\&=\sin 170^{\circ}-\sin 50^{\circ}+1\\&=\sin 10^{\circ}-\sin 50^{\circ}+\sin 90^{\circ}\\&=2\sin 50^{\circ}\cos 40^{\circ}-\sin 50^{\circ}\\&=2\sin 50^{\circ}\left(\cos 40^{\circ}-\frac{1}{2}\right)\\&=2\sin 50^{\circ}2\sin 50^{\circ}\sin 10^{\circ}\\&=4\sin^2 50^{\circ}\sin 10^{\circ}\\&=4\cos^2 40^{\circ}\sin 10^{\circ}\end{align*}

\begin{align*}c&=\sqrt{3}\cos 130^{\circ}+1\\&=2\cos 30^{\circ}\cos 130^{\circ}+1\\&=\cos 160^{\circ}+\cos 100^{\circ}+1\\&=-\cos 20^{\circ}-\sin 10^{\circ}+1\\&=2\sin^2 10^{\circ}-\sin 10^{\circ}\\&=2\sin 10^{\circ}\left(\sin 10^{\circ}-\frac{1}{2}\right)\\&=2\sin 10^{\circ}(\cos 80^{\circ}-\cos 60^{\circ})\\&=2\sin 10^{\circ}(-2\sin 70^{\circ}\sin 10^{\circ})\\&=-4\sin^{2}10^{\circ}\cos 20^{\circ}\end{align*}

\begin{align*}X^{3}&=(\sqrt{a}+\sqrt{b}+\sqrt{c})^{3}\\&=a+b+c+3(\sqrt{a^{2}b}+\sqrt{a^{2}c}+\sqrt{b^{2}a}+\sqrt{b^{2}c}+2\sqrt{abc}+\sqrt{c^{2}a}+\sqrt{c^{2}b})\end{align*}

\begin{align*}a+b+c&=\sqrt{3}\cos 10^{\circ}+1+\sqrt{3}\cos 110^{\circ}+1+\sqrt{3}\cos 130^{\circ}+1\\&=\sqrt{3}(\cos 10^{\circ}+\cos 110^{\circ}+\cos 130^{\circ})+3\\&=\sqrt{3}(2\cos 60^{\circ}\cos 50^{\circ}+\cos 130^{\circ})+3\\&=\sqrt{3}(\cos 50^{\circ}+\cos 130^{\circ})+3\\&=\sqrt{3}(2\cos 90^{\circ}\cos 40^{\circ})+3\\&=\sqrt{3} \cdot 0 + 3\\&=3\end{align*}

\begin{align*}\sqrt{a^{2}b}&=\sqrt{4^3\cos^{4}20^{\circ}\cos^{4}40^{\circ} \cdot \sin 10^{\circ}}\\&=4\cos 20^{\circ}\cos 40^{\circ}\sqrt{\cos 40^{\circ}\cos 20^{\circ}\sin 10^{\circ}}\\&=4\cos 20^{\circ}\cos 40^{\circ}\sqrt{\dfrac{\cos 2(30^{\circ})}{4}}\\&=4\cos 20^{\circ}\cos 40^{\circ}\sqrt{\frac{1}{8}}\\&=2\cos 20^{\circ}\cos 40^{\circ}\end{align*}.

$\sqrt{a^{2}c}=-\cos 20^{\circ}$.
$\sqrt{b^{2}a}=\cos 40^{\circ}$.
$\sqrt{b^{2}c}=-2\cos 40^{\circ}\cos 80^{\circ}=-2\cos 40^{\circ}\sin 10^{\circ}$.

$2\sqrt{abc}=-8\cos 20^{\circ}\cos 40^{\circ}\sin 10^{\circ}=\dfrac{-8}{8}=-1$.
$\sqrt{c^{2}a}=2\sin 10^{\circ}\cos 20^{\circ}$.
$\sqrt{c^{2}b}=\sin 10^{\circ}$.

\begin{align*}X^{3}&=3+3(2\cos 20^{\circ}\cos 40^{\circ}-\cos 20^{\circ}+\cos 40^{\circ}-2\cos 40^{\circ}\sin 10^{\circ}-1+2\sin 10^{\circ}\cos 20^{\circ}+\sin 10^{\circ})\\&=3+3(2(\cos 20^{\circ}\cos 40^{\circ}-\cos 40^{\circ}\sin 10^{\circ}+\sin 10^{\circ}\cos 20^{\circ})-\cos 20^{\circ}+\cos 40^{\circ}-1+\sin 10^{\circ})\\&=3+3(2\left(\dfrac{1}{2}(\cos 60^{\circ}+\cos 20^{\circ})-\dfrac{1}{2}(\sin 50^{\circ}-\sin 30^{\circ})+\dfrac{1}{2}(\sin 30^{\circ}-\sin 10^{\circ})\right)-\sin 10^{\circ}-1+\sin 10^{\circ})\\&=3+3((\dfrac{1}{2}+\cos 20^{\circ}-\sin 50^{\circ}+\dfrac{1}{2}+\dfrac{1}{2}-\sin 10^{\circ})-1)\\&=3+3((\dfrac{3}{2}+\cos 20^{\circ}-(2\sin 30^{\circ}\cos 20^{\circ})-1)\\&=3+3 \cdot \frac{1}{2}\\&=\frac{9}{2}\end{align*}

This implies $X=\sqrt{\frac{9}{2}}$ i.e. $\sqrt{\sqrt{3}\cos10^{\circ}+1}+\sqrt{\sqrt{3}\cos110^{\circ}+1}+\sqrt{\sqrt{3}\cos130^{\circ}+1}=\sqrt{\frac{9}{2}}$.

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