PROBLEM 4611 CRUX MATHEMATICORUM , V47n02

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ANSWER CiP

$\frac{1}{\sin^{4}\frac{\pi}{14}}+\frac{1}{\sin^{4}\frac{3\pi}{14}}+\frac{1}{\sin^{4}\frac{5\pi}{14}}=416$

               Solution CiP

          As we showed in a previous post, the numbers

 $x_{1}=sin\frac{\pi}{14},\;x_{2}=-sin\frac{3\pi}{14},\;x_{3}=sin\frac{5\pi}{14}$

 are the roots of the polynomial $8x^{3}-4x^{2}-4x+1$. Then the numbers $y_{k}=\frac{1}{x_{k}}\;,\;k=1,\;2,\;3$ are the roots of the "reverse" equation

$y^{3}-4\cdot y^{2}-4 \cdot y+8=0 \tag{1} \label{eq1}$

 The expression to be calculated is equal to $\sum_{k=1}^{3}y_{k}^{4}$.

    We write Vieta's formulas for the equation (1)

 \begin{cases}y_{1}+y_{2}+y_{3}=4\;\;\; \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(2)\\y_{1}\cdot y_{2}+y_{2}\cdot y_{3}+y_{3} \cdot y_{1}=-4\;\;(3)\\y_{1} \cdot y_{2} \cdot y_{3}=-8 \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(4)\end{cases}

 From \eqref{eq1} we deduce $y_{k}^{3}=4\cdot y_{k}^{2}+4 \cdot y_{k}-8$ $\Rightarrow$

 $y_{k}^{4}=y_{k} \cdot y_{k}^{3}=4\cdot y_{k}^{3}+4 \cdot y_{k}^{2}-8 \cdot y_{k}=$

$=4(4y_{k}^{2}+4y_{k}-8)+4y_{k}^{2}-8y_{k}=20 y_{k}^{2}+8y_{k}-32.$

Then we get

$\sum _{k=1}^{3}y_{k}^{4}=\sum (20 \cdot y_{k}^{2}+8 \cdot y_{k}-32)=20 \cdot \sum_{k=1}^{3} y_{k}^{2}+8 \cdot \sum y_{k} -96  \overset{(2)}{=}$

$=20 \cdot [(\sum y_{k})^{2}-2 \cdot \underset{cycl}{\sum} y_{1}\cdot y_{2}]+8 \cdot 4-96\;\; \underset{(2),(3)}{==}\;\;20 \cdot [4^{2}-2(-4)]-64=416.$

$\blacksquare$

 

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Tracking Number: 11849
Received:        Wed Mar 17 15:14:14 2021

From:  Petre Ciobanu
       Scoala Gimnaziala "Samuil Micu" SADU
       Sibiu, Romania
Email: ptr.ciobanu@gmail.com

Type:  Solve a Crux (Numbered) Problem
       (problem 4611)

Files:
  Pdf_4611.pdf
  Sursa_Latex_4611.doc


Comments:
A close problem that the idea of solving suggested to me is  Problem 28, page 115  from
E.J. BARBEAU - Polynomials (Springer-Verlag, 1989). I dedicated an article to this problem on my blog

https://ogeometrie-cip.blogspot.com/2021/02/an-equation-that-surprising-is-not.html

I also put on the blog the solution of this problem from you

https://ogeometrie-cip.blogspot.com/2021/03/problem-4611-crux-mathematicorum-v47n02.html

Crux Mathematicorum <crux@cms.math.ca>	17:15 (acum 1 minut)
către eu