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
Find the value of
Solution
Let denote the polynomial. We have
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,
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
~SMO_Team