Difference between revisions of "2024 SSMO Team Round Problems/Problem 14"
(Created page with "==Problem== Let <math>a_1, a_2, \dots, a_7</math> be the roots of the polynomial <cmath>x^7+5x^6+9x^5+x^4+x^3+10x^2+5x+1.</cmath> Find the value of <cmath>\left|\prod_{n=1}^7...") |
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==Solution== | ==Solution== | ||
| + | Let <math>f(x)</math> denote the polynomial. We have <cmath>\left(\prod_{n=1}^7 \prod_{m=n+1}^7 (a_na_m-1)\right)^2\prod_{n=1}^{7}(a_n^2-1) = \prod_{n=1}^7\prod_{m=1}^7(a_na_m-1),</cmath> since <math>1 \le n \le 7, n+1 \le m \le 7</math> covers the <math>m>n</math> case of <math>1 \le m,n \le 7,</math> (which is the domain of the RHS product) squaring the nested product doubles it, covering the symmetric <math>n>m</math> case, and the third factor covers the <math>m=n</math> case. Now, note that \begin{align*} | ||
| + | \prod_{m=1}^{7}(a_na_m-1)&=-a_n^7\prod_{m=1}^7\left(\frac{1}{a_n}-a_m\right)\\ | ||
| + | &=-a_n^7f\left(\frac{1}{a_n}\right)\\ | ||
| + | &=-a_n^7-(5a_n^6+10a_n^5+a_n^4+a_n^3+9a_n^2+5a_n+1)\\ | ||
| + | &=(5a_n^6+9a_n^5+a_n^4+a_n^3+10a_n^2+5a_n+1)-(5a_n^6+10a_n^5+a_n^4+a_n^3+9a_n^2+5a_n+1)\\ | ||
| + | &=-a_n^5+a_n^2\\ | ||
| + | &=-(a_n)^2(a_n^3-1). | ||
| + | \end{align*} | ||
| + | So, | ||
| + | \begin{align*} | ||
| + | \prod_{n=1}^7\prod_{n=1}^7(a_na_m-1)&=\prod_{n=1}^7\left(-(a_n)^2(a_n^3-1)\right)\\ | ||
| + | &=-\left(\prod_{n=1}^7a_n\right)^2\prod_{n=1}^7(a_n-1)\prod_{n=1}(a_n-e^{\frac{2\pi i}{3}})\prod_{n=1}(a_n-e^{\frac{4\pi i}{3}})\\ | ||
| + | &=-(-1)^2(-f(1))\left(-f\left(e^{\frac{2\pi i}{3}}\right)\right)\left(-f\left(e^{\frac{4\pi i}{3}}\right)\right)\\ | ||
| + | &=f(1)f\left(e^{\frac{2\pi i}{3}}\right)f\left(e^{\frac{4\pi i}{3}}\right). | ||
| + | \end{align*} | ||
| + | For <math>\omega^3=1</math> and <math>\omega \ne 1,</math> <math>f(\omega) = 19\omega^2+7\omega+7 = 12\omega^2.</math> So, | ||
| + | \begin{align*} | ||
| + | f\left(e^{\frac{2\pi i}{3}}\right) &= 12\left(\frac{-1+i\sqrt{3}}{2}\right)^2 = \frac{-12-12i\sqrt{3}}{2} = -6-6i\sqrt{3}\text{ and }\\ | ||
| + | f\left(e^{\frac{4\pi i}{3}}\right) &= 12\left(\frac{-1-i\sqrt{3}}{2}\right)^2 = \frac{-12+12i\sqrt{3}}{2} = -6+6i\sqrt{3}\implies\\ | ||
| + | f\left(e^{\frac{2\pi i}{3}}\right)f\left(e^{\frac{4\pi i}{3}}\right) &= (-6-6i\sqrt{3})(-6+6i\sqrt{3}) = 144. | ||
| + | \end{align*} | ||
| + | Thus, <cmath>\prod_{n=1}^7\prod_{n=1}^7(a_na_m-1) = f(1)f\left(e^{\frac{2\pi i}{3}}\right)f\left(e^{\frac{4\pi i}{3}}\right) = 33 \cdot 144.</cmath> | ||
| + | Now, we have | ||
| + | \begin{align*} | ||
| + | \prod_{n=1}^7(a_n^2-1)&=\prod_{n=1}^7(a_n-1)\prod_{n-1}^7(a_n+1)\\ | ||
| + | &=(-f(1))(-f(-1))\\ | ||
| + | &=f(1)f(-1)\\ | ||
| + | &=33\cdot1 = 33. | ||
| + | \end{align*} | ||
| + | Substituting this into <cmath>\left(\prod_{n=1}^7 \prod_{m=n+1}^7 (a_na_m-1)\right)^2\prod_{n=1}^{7}(a_n^2-1) = \prod_{n=1}^7\prod_{m=1}^7(a_na_m-1),</cmath> we have <cmath>\left(\prod_{n=1}^7 \prod_{m=n+1}^7 (a_na_m-1)\right)^2 = \frac{33\cdot144}{33} = 144\implies \left| \prod_{n=1}^7 \prod_{m=n+1}^7 (a_na_m-1) \right| = \sqrt{144}\implies\boxed{12}.</cmath> | ||
| + | |||
| + | ~SMO_Team | ||
Latest revision as of 14:42, 10 September 2025
Problem
Let 
be the roots of the polynomial ![\[x^7+5x^6+9x^5+x^4+x^3+10x^2+5x+1.\]](http://latex.artofproblemsolving.com/4/5/d/45d666c0c4fa8f4a151c085a38c03c6f3aa2113b.png)
Find the value of ![\[\left|\prod_{n=1}^7 \prod_{m=n+1}^7 (a_na_m-1)\right|.\]](http://latex.artofproblemsolving.com/8/d/c/8dc382ad11db01ff3d00a8e17eabe97dc149cefe.png)
Solution
Let 
denote the polynomial. We have ![\[\left(\prod_{n=1}^7 \prod_{m=n+1}^7 (a_na_m-1)\right)^2\prod_{n=1}^{7}(a_n^2-1) = \prod_{n=1}^7\prod_{m=1}^7(a_na_m-1),\]](http://latex.artofproblemsolving.com/a/b/a/abaee1a7f919964a93a00594ce3c87e0b0791009.png)
since 
covers the 
case of 
(which is the domain of the RHS product) squaring the nested product doubles it, covering the symmetric 
case, and the third factor covers the 
case. Now, note that \begin{align*}
\prod_{m=1}^{7}(a_na_m-1)&=-a_n^7\prod_{m=1}^7\left(\frac{1}{a_n}-a_m\right)\\
&=-a_n^7f\left(\frac{1}{a_n}\right)\\
&=-a_n^7-(5a_n^6+10a_n^5+a_n^4+a_n^3+9a_n^2+5a_n+1)\\
&=(5a_n^6+9a_n^5+a_n^4+a_n^3+10a_n^2+5a_n+1)-(5a_n^6+10a_n^5+a_n^4+a_n^3+9a_n^2+5a_n+1)\\
&=-a_n^5+a_n^2\\
&=-(a_n)^2(a_n^3-1).
\end{align*}
So,
\begin{align*}
\prod_{n=1}^7\prod_{n=1}^7(a_na_m-1)&=\prod_{n=1}^7\left(-(a_n)^2(a_n^3-1)\right)\\
&=-\left(\prod_{n=1}^7a_n\right)^2\prod_{n=1}^7(a_n-1)\prod_{n=1}(a_n-e^{\frac{2\pi i}{3}})\prod_{n=1}(a_n-e^{\frac{4\pi i}{3}})\\
&=-(-1)^2(-f(1))\left(-f\left(e^{\frac{2\pi i}{3}}\right)\right)\left(-f\left(e^{\frac{4\pi i}{3}}\right)\right)\\
&=f(1)f\left(e^{\frac{2\pi i}{3}}\right)f\left(e^{\frac{4\pi i}{3}}\right).
\end{align*}
For 
and 
So,
\begin{align*}
f\left(e^{\frac{2\pi i}{3}}\right) &= 12\left(\frac{-1+i\sqrt{3}}{2}\right)^2 = \frac{-12-12i\sqrt{3}}{2} = -6-6i\sqrt{3}\text{ and }\\
f\left(e^{\frac{4\pi i}{3}}\right) &= 12\left(\frac{-1-i\sqrt{3}}{2}\right)^2 = \frac{-12+12i\sqrt{3}}{2} = -6+6i\sqrt{3}\implies\\
f\left(e^{\frac{2\pi i}{3}}\right)f\left(e^{\frac{4\pi i}{3}}\right) &= (-6-6i\sqrt{3})(-6+6i\sqrt{3}) = 144.
\end{align*}
Thus, ![\[\prod_{n=1}^7\prod_{n=1}^7(a_na_m-1) = f(1)f\left(e^{\frac{2\pi i}{3}}\right)f\left(e^{\frac{4\pi i}{3}}\right) = 33 \cdot 144.\]](http://latex.artofproblemsolving.com/f/3/6/f3622ca6e41d269a2b0e54c58a090140a3f43674.png)
Now, we have
\begin{align*}
\prod_{n=1}^7(a_n^2-1)&=\prod_{n=1}^7(a_n-1)\prod_{n-1}^7(a_n+1)\\
&=(-f(1))(-f(-1))\\
&=f(1)f(-1)\\
&=33\cdot1 = 33.
\end{align*}
Substituting this into we have ![\[\left(\prod_{n=1}^7 \prod_{m=n+1}^7 (a_na_m-1)\right)^2 = \frac{33\cdot144}{33} = 144\implies \left| \prod_{n=1}^7 \prod_{m=n+1}^7 (a_na_m-1) \right| = \sqrt{144}\implies\boxed{12}.\]](http://latex.artofproblemsolving.com/1/7/9/1794de3fd920ca612c3951ca751b3f2c3daee959.png)
~SMO_Team
