Page 97 - En
Page 99 - FrANSWER 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.blogspo
I also put on the blog the solution of this problem from you
https://ogeometrie-cip.blogspo
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