# Problem Of The Week #403 Feb 2nd, 2020

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#### anemone

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

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Monic quadratic polynomials $P(x)$ and $Q(x)$ have the property that $P(Q(x))$ has zeros at $x=-23,\,-21,\,-17$ and $-15$ and $Q(P(x))$ has zeros at $x=-59,\,-57,\,-51$ and $-49$. What is the sum of the minimum values of $P(x)$ and $Q(x)$?

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#### anemone

##### MHB POTW Director
Staff member
Hi MHB,

I was told by castor28 that this week's POTW (High School) was a duplicate of POTW #363, which is true. I am truly sorry for letting this thing happened. I therefore want to thank him for catching the mistake.

Please let me make it up by presenting to you the following problem:

A geometric sequence $(a_n)$ has $a_1=\sin x,\,a_2=\cos x$ and $a_3=\tan x$ for some real number $x$. For what value of $n$ does $a_n=1+\cos x$?

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#### anemone

##### MHB POTW Director
Staff member
Congratulations to the following members for their correct answer!

1. castor28
2. MegaMoh

Solution from castor28 :
Let $q$ be the ratio of the progression. Comparing $a_1$ and $a_2$, we get $q=\cot x$. On the other hand, comparing $a_1$ and $a_3$, we get $q^2 = \dfrac{1}{\cos x}$.

Taken together, these equalities give:
\begin{align*}
\frac{1}{\cos x} &= \frac{\cos^2 x}{\sin^2 x}\\
\cos^3 x &= \sin^2 x = 1 - \cos^2 x\\
(1 + \cos x) &= \frac{1}{\cos^2 x}
\end{align*}

This shows that we must find $n$ such that $a_n=\dfrac{1}{\cos^2 x} = q^4$. As we have $a_4=\tan x \cot x = 1$, we conclude that $a_8$ has the required value.

Note: Solving the equation numerically, we find $q \approx 1.151 > 1$. As the progression is strictly increasing, $a_8$ is the only term with the required value.

Alternate solution from MegaMoh :
$a_1 = \sin{x}$

$a_2 = k \sin{x} = \cos{x}$
$\implies k = \cot{x}$

$a_3 = k^2 \sin{x} = \tan{x}$
$\implies k = \sqrt{\frac1{\cos{x}}} = \cot{x} = \frac{\cos{x}}{\sin{x}}$
$\implies \cos^3{x} = \sin^2{x} = 1 - \cos^2{x}$
$\implies \cos^3{x} + \cos^2{x} - 1 = 0$
$\implies x \approx 0.71532874990708873792278349518063713\text{(rad)}$

$a_n = k^{n-1} \sin{x} = 1 + \cos{x}$
$\implies k^{n-1} = \frac{1+\cos{x}}{\sin{x}} = \frac1{\tan{(\frac{x}{2})}} = \cot{(\frac{x}{2})}$
$\implies \cot^{n-1}{x} = \cot{(\frac{x}{2})}$
$\implies n - 1 = \frac{\ln{(\cot{(\frac{x}{2})})}}{\ln{(\cot{x})}} = 7$
$\therefore$ $n = 8$

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