Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: Suppose that $P(z), Q(z)$ , and $R(z)$ are polynomials with real coefficients, having degrees $2$ $3$ , and $6$ , respectively, and constant terms $1$ $2$ , and $3$ , respectively. Let $N$ be the number of distinct complex numbers $z$ that satisfy the equation $P(z) \cdot Q(z)=R(z)$ . What is the minimum possible value of $N$
$\textbf{(A)}\: 0\qquad\textbf{(B)} \: 1\qquad\textbf{(C)} \: 2\qquad\textbf{(D)} \: 3\qquad\textbf{(E)} \: 5$

Solution: The answer cannot be $0,$ as every nonconstant polynomial has at least $1$ distinct complex root (Fundamental Theorem of Algebra). Since $P(z) \cdot Q(z)$ has degree $2 + 3 = 5,$ we conclude that $R(z) - P(z)\cdot Q(z)$ has degree $6$ and is thus nonconstant.
It now suffices to illustrate an example for which $N = 1$ : Take \begin{align*} P(z)&=z^2+1, \\ Q(z)&=z^3+2, \\ R(z)&=(z+1)^6 + P(z) \cdot Q(z). \end{align*} Note that $R(z)$ has degree $6$ and constant term $3,$ so it satisfies the conditions.
We need to find the solutions to \begin{align*} P(z) \cdot Q(z) &= (z+1)^6 + P(z) \cdot Q(z) \\ 0 &= (z+1)^6. \end{align*} Clearly, the only distinct complex root is $-1,$ so our answer is $N=\boxed{1}.$

Answer: B 1
Problem: Three identical square sheets of paper each with side length $6$ are stacked on top of each other. The middle sheet is rotated clockwise $30^\circ$ about its center and the top sheet is rotated clockwise $60^\circ$ about its center, resulting in the $24$ -sided polygon shown in the figure below. The area of this polygon can be expressed in the form $a-b\sqrt{c}$ , where $a$ $b$ , and $c$ are positive integers, and $c$ is not divisible by the square of any prime. What is $a+b+c$
$(\textbf{A})\: 75\qquad(\textbf{B}) \: 93\qquad(\textbf{C}) \: 96\qquad(\textbf{D}) \: 129\qquad(\textbf{E}) \: 147$

Solution: [asy] defaultpen(fontsize(8)+0.8); size(100); pair A=(0,0); pair B=(1.732,3); pair C=(3,3); pair D=(3,1.732); draw(A--(0,3)--C--(3,0)--A, lightgray+dashed); draw(A--B--C--A); draw(A--D--C, gray); label("$A$",A,W); label("$B$",B,N); label("$C$",C,NE); label("$D$",D,E); label("$E$",(0,3),NW); label("$F$",(3,0),E); [/asy] The $24$ -sided polygon is made out of $24$ shapes like $\triangle ABC$ . Then $\angle BAC=360^\circ/24=15^\circ$ , and $\angle EAC = 45^\circ$ , so $\angle{EAB} = 30^{\circ}$ . Then $EB=AE\tan 30^\circ = \sqrt{3}$ ; therefore $BC=EC-EB=3-\sqrt{3}$ . Thus \[[ABC] = \frac{BC}{EC}\cdot [ACE] = \frac{3-\sqrt{3}}{3}\cdot \frac 92\] and the required area is $24\cdot[ABC] =108-36\sqrt{3}$ . Finally $108+36+3=\boxed{147}$ . 
~lopkiloinm

Answer: E 147
Problem: Three identical square sheets of paper each with side length $6$ are stacked on top of each other. The middle sheet is rotated clockwise $30^\circ$ about its center and the top sheet is rotated clockwise $60^\circ$ about its center, resulting in the $24$ -sided polygon shown in the figure below. The area of this polygon can be expressed in the form $a-b\sqrt{c}$ , where $a$ $b$ , and $c$ are positive integers, and $c$ is not divisible by the square of any prime. What is $a+b+c$
$(\textbf{A})\: 75\qquad(\textbf{B}) \: 93\qquad(\textbf{C}) \: 96\qquad(\textbf{D}) \: 129\qquad(\textbf{E}) \: 147$

Solution: As shown in Image:2021_AMC_12B_(Nov)_Problem_15,_sol.png , all 12 vertices of three squares form a regular dodecagon (12-gon).
Denote by $O$ the center of this dodecagon.
Hence, $\angle AOB = \frac{360^\circ}{12} = 30^\circ$
Because the length of a side of a square is 6, $AO = 3 \sqrt{2}$
Hence, $AB = 2 AO \sin \frac{\angle AOB}{2} = 3 \left( \sqrt{3} - 1 \right)$
We notice that $\angle MAB = \angle MBA = 30^\circ$ .
Hence, $AM = \frac{AB}{2\cos \angle MAB} = 3 - \sqrt{3}$
Therefore, the area of the region that three squares cover is \begin{align*} & {\rm Area} \ ABCDEFGHIJKL - 12 {\rm Area} \ \triangle MAB \\ & = 12 {\rm Area} \ \triangle OAB - 12 {\rm Area} \ \triangle MAB \\ & = 12 \cdot \frac{1}{2} OA \cdot OB \sin \angle AOB - 12 \cdot \frac{1}{2} MA \cdot MB \sin \angle AMB \\ & = 6 OA^2 \sin \angle AOB - 6 MA^2 \sin \angle AMB \\ & = 108 - 36 \sqrt{3} . \end{align*}
Therefore, the answer is $\boxed{147}$

Answer: E 147
Problem: Suppose $a$ $b$ $c$ are positive integers such that \[a+b+c=23\] and \[\gcd(a,b)+\gcd(b,c)+\gcd(c,a)=9.\] What is the sum of all possible distinct values of $a^2+b^2+c^2$
$\textbf{(A)}\: 259\qquad\textbf{(B)} \: 438\qquad\textbf{(C)} \: 516\qquad\textbf{(D)} \: 625\qquad\textbf{(E)} \: 687$

Solution: Because $a + b + c$ is odd, $a$ $b$ $c$ are either one odd and two evens or three odds.
$\textbf{Case 1}$ $a$ $b$ $c$ have one odd and two evens.
Without loss of generality, we assume $a$ is odd and $b$ and $c$ are even.
Hence, ${\rm gcd} \left( a , b \right)$ and ${\rm gcd} \left( a , c \right)$ are odd, and ${\rm gcd} \left( b , c \right)$ is even.
Hence, ${\rm gcd} \left( a , b \right) + {\rm gcd} \left( b , c \right) + {\rm gcd} \left( c , a \right)$ is even. This violates the condition given in the problem.
Therefore, there is no solution in this case.
$\textbf{Case 2}$ $a$ $b$ $c$ are all odd.
In this case, ${\rm gcd} \left( a , b \right)$ ${\rm gcd} \left( a , c \right)$ ${\rm gcd} \left( b , c \right)$ are all odd.
Without loss of generality, we assume \[ {\rm gcd} \left( a , b \right) \leq {\rm gcd} \left( b , c \right) \leq {\rm gcd} \left( c , a \right) . \] $\textbf{Case 2.1}$ ${\rm gcd} \left( a , b \right) = 1$ ${\rm gcd} \left( b , c \right) = 1$ ${\rm gcd} \left( c , a \right) = 7$
The only solution is $(a, b, c) = (7, 9, 7)$
Hence, $a^2 + b^2 + c^2 = 179$
$\textbf{Case 2.2}$ ${\rm gcd} \left( a , b \right) = 1$ ${\rm gcd} \left( b , c \right) = 3$ ${\rm gcd} \left( c , a \right) = 5$
The only solution is $(a, b, c) = (5, 3, 15)$
Hence, $a^2 + b^2 + c^2 = 259$
$\textbf{Case 2.3}$ ${\rm gcd} \left( a , b \right) = 3$ ${\rm gcd} \left( b , c \right) = 3$ ${\rm gcd} \left( c , a \right) = 3$
There is no solution in this case.
Therefore, putting all cases together, the answer is $179 + 259 = \boxed{438}$

Answer: B 438
Problem: Suppose $a$ $b$ $c$ are positive integers such that \[a+b+c=23\] and \[\gcd(a,b)+\gcd(b,c)+\gcd(c,a)=9.\] What is the sum of all possible distinct values of $a^2+b^2+c^2$
$\textbf{(A)}\: 259\qquad\textbf{(B)} \: 438\qquad\textbf{(C)} \: 516\qquad\textbf{(D)} \: 625\qquad\textbf{(E)} \: 687$

Solution: Let $\gcd(a,b)=x$ $\gcd(b,c)=y$ $\gcd(c,a)=z$ . Without the loss of generality, let $x \le y \le z$ . We can split this off into cases:
$x=1,y=1,z=7$ : let $a=7A, c=7C,$ we can try all possibilities of $A$ and $C$ to find that $a=7, b=9, c=7$ is the only solution.
$x=1,y=2,z=6$ : No solutions. By $y$ and $z$ , we know that $a$ $b$ , and $c$ have to all be divisible by $2$ . Therefore, $x$ cannot be equal to $1$
$x=1,y=3,z=5$ : Note that $c$ has to be both a multiple of $3$ and $5$ . Therefore, $c$ has to be a multiple of $15$ . The only solution for this is $a=5, b=3, c=15$
$x=1,y=4,z=4$ : No solutions. By $y$ and $z$ , we know that $a$ $b$ , and $c$ have to all be divisible by $4$ . Therefore, $x$ cannot be equal to $1$
$x=2,y=2,z=5$ : No solutions. By $x$ and $y$ , we know that $a$ $b$ , and $c$ have to all be divisible by $2$ . Therefore, $z$ cannot be equal to $5$
$x=2,y=3,z=4$ : No solutions. By $x$ and $z$ , we know that $a$ $b$ , and $c$ have to all be divisible by $2$ . Therefore, $y$ cannot be equal to $3$
$x=3,y=3,z=3$ : No solutions. As $a$ $b$ , and $c$ have to all be divisible by $3$ $a+b+c$ has to be divisible by $3$ . This contradicts the sum $a+b+c=23$
Putting these solutions together, we have $(7^2+9^2+7^2)+(5^2+3^2+15^2)=179+259=\boxed{438}$

Answer: B 438
Problem: Suppose $a$ $b$ $c$ are positive integers such that \[a+b+c=23\] and \[\gcd(a,b)+\gcd(b,c)+\gcd(c,a)=9.\] What is the sum of all possible distinct values of $a^2+b^2+c^2$
$\textbf{(A)}\: 259\qquad\textbf{(B)} \: 438\qquad\textbf{(C)} \: 516\qquad\textbf{(D)} \: 625\qquad\textbf{(E)} \: 687$

Solution: Since $a+b+c=23$ $\gcd(a,b,c)=23$ or $\gcd(a,b,c)=1$
As $\gcd(a,b)+\gcd(b,c)+\gcd(c,a)=9$ , it is impossible for $\gcd(a,b,c)=23$ , so $\gcd(a,b,c)=1$
This means that $\gcd(a,b)$ $\gcd(b,c)$ , and $\gcd(c,a)$ must all be coprime. The only possible ways for this to be true are $1+1+7=9$ and $1+3+5=9$
Without loss of generality, let $a\le b\le c$ . Since $a+b+c=23$ , then $a=7, b=7, c=9$ or $a=3, b=5, c=15$
$(7^2+7^2+9^2)+(3^2+5^2+15^2)=179+259=\boxed{438}$

Answer: B 438
Problem: Set $u_0 = \frac{1}{4}$ , and for $k \ge 0$ let $u_{k+1}$ be determined by the recurrence \[u_{k+1} = 2u_k - 2u_k^2.\]
This sequence tends to a limit; call it $L$ . What is the least value of $k$ such that \[|u_k-L| \le \frac{1}{2^{1000}}?\]
$\textbf{(A)}\: 10\qquad\textbf{(B)}\: 87\qquad\textbf{(C)}\: 123\qquad\textbf{(D)}\: 329\qquad\textbf{(E)}\: 401$

Solution: Note that terms of the sequence $(u_k)$ lie in the interval $\left(0,\frac12\right),$ strictly increasing.
Since the sequence $(u_k)$ tends to the limit $L,$ we set $u_{k+1}=u_k=L>0.$
The given equation becomes \[L=2L-2L^2,\] from which $L=\frac12.$
The given inequality becomes \[\frac12-\frac{1}{2^{1000}} \leq u_k \leq \frac12+\frac{1}{2^{1000}},\] and we only need to consider $\frac12-\frac{1}{2^{1000}} \leq u_k.$
We have \begin{alignat*}{8} u_0 &= \phantom{1}\frac14 &&= \frac{2^1-1}{2^2}, \\ u_1 &= \phantom{1}\frac38 &&= \frac{2^2-1}{2^3}, \\ u_2 &= \ \frac{15}{32} &&= \frac{2^4-1}{2^5}, \\ u_3 &= \frac{255}{512} &&= \frac{2^8-1}{2^9}, \\ & \phantom{1111} \vdots \end{alignat*} By induction, it can be proven that \[u_k=\frac{2^{2^k}-1}{2^{2^k+1}}=\frac12-\frac{1}{2^{2^k+1}}.\] We substitute this into the inequality, then solve for $k:$ \begin{align*} \frac12-\frac{1}{2^{1000}} &\leq \frac12-\frac{1}{2^{2^k+1}} \\ -\frac{1}{2^{1000}} &\leq -\frac{1}{2^{2^k+1}} \\ 2^{1000} &\leq 2^{2^k+1} \\ 1000 &\leq 2^k+1. \end{align*} Therefore, the least such value of $k$ is $\boxed{10}.$

Answer: A 10
Problem: Set $u_0 = \frac{1}{4}$ , and for $k \ge 0$ let $u_{k+1}$ be determined by the recurrence \[u_{k+1} = 2u_k - 2u_k^2.\]
This sequence tends to a limit; call it $L$ . What is the least value of $k$ such that \[|u_k-L| \le \frac{1}{2^{1000}}?\]
$\textbf{(A)}\: 10\qquad\textbf{(B)}\: 87\qquad\textbf{(C)}\: 123\qquad\textbf{(D)}\: 329\qquad\textbf{(E)}\: 401$

Solution: If we list out the first few values of $k$ , we get the series $\frac{1}{4}, \frac{3}{8}, \frac{15}{32}, \frac{255}{512}$ , which always seems to be a negative power of $2$ away from $\frac{1}{2}$ . We can test this out by setting $u_k=\frac{1}{2}-\frac{1}{2^{n_k}}$ , where $n_0=2$
Now, we get \begin{align*} u_{k+1} &= 2\cdot\left(\frac{1}{2}-\frac{1}{2^{n_{k}}}\right)-2\cdot\left(\frac{1}{2}-\frac{1}{2^{n_{k}}}\right)^2 \\ &= 1-\frac{1}{2^{n_k - 1}}-2\cdot\left(\frac{1}{4}-\frac{1}{2^{n_k}}+\frac{1}{2^{2 \cdot n_k}}\right)\\ &= 1-\frac{1}{2^{n_k - 1}}-\frac{1}{2}+\frac{1}{2^{n_k-1}}-\frac{1}{2^{2 \cdot n_k-1}} \\ &= \frac{1}{2}-\frac{1}{2^{2 \cdot n_k-1}}. \end{align*} This means that this series approaches $\frac{1}{2}$ , as the second term is decreasing. In addition, we find that $n_{k+1}=2 \cdot n_k-1$
We claim that $n_k = 2^k+1$ , which can be proven by induction:
Base Case
We have $n_0=2=2^0+1$
Induction Step
Assuming that the claim is true, we have $n_{k+1}=2 \cdot (2^k+1)-1=2^{k+1}+1$
It follows that $n_{10}=2^{10}+1>1000$ and $n_9=2^9+1<1000$ . Therefore, the least value of $k$ would be $\boxed{10}$

Answer: A 10
Problem: Set $u_0 = \frac{1}{4}$ , and for $k \ge 0$ let $u_{k+1}$ be determined by the recurrence \[u_{k+1} = 2u_k - 2u_k^2.\]
This sequence tends to a limit; call it $L$ . What is the least value of $k$ such that \[|u_k-L| \le \frac{1}{2^{1000}}?\]
$\textbf{(A)}\: 10\qquad\textbf{(B)}\: 87\qquad\textbf{(C)}\: 123\qquad\textbf{(D)}\: 329\qquad\textbf{(E)}\: 401$

Solution: We are given $u_{k+1} = 2u_k - 2{u_k}^2$ . Multiply this equation by $2$ and subtract $1$ from both sides. The equations can then be written nicely as $2u_{k+1} - 1 = -(2u_k-1)^2$ . Let $v_k = 2u_k - 1$ so that $v_{k+1} = -(v_k)^2$
Clearly, $v_0 = 2u_0 - 1 = -\frac{1}{2}$ . Since the magnitude of $v_0$ is less than $1$ and because our recursive relation for $v_k$ squares the previous term (and negates it), we see that as $k \rightarrow \infty, 2u_k - 1 = v_k \rightarrow 0$ . This means $u_k \rightarrow \frac{1}{2}$ , so $L = \frac{1}{2}$
Isolating $u_k$ in our relation $2u_k - 1 = v_k$ gives us $u_k = \frac{v_k + 1}{2}$ .
Substituting into the inequality, we have $\left|\frac{v_k + 1}{2}-\frac{1}{2}\right| \le \frac{1}{2^{1000}}$ . Rewriting this, we get $|v_k| \le \frac{1}{2^{999}}$
The sequence $\{v_k\}$ is much easier to handle because of its simple recursive relation. Writing out a few terms shows that $|v_k| = \frac{1}{2^{2^k}}$ . Now it just comes down to having $2^k > 999$ , so $k = \boxed{10}$

Answer: A 10
Problem: For real numbers $x$ , let \[P(x)=1+\cos(x)+i\sin(x)-\cos(2x)-i\sin(2x)+\cos(3x)+i\sin(3x)\] where $i = \sqrt{-1}$ . For how many values of $x$ with $0\leq x<2\pi$ does \[P(x)=0?\]
$\textbf{(A)}\ 0 \qquad\textbf{(B)}\  1 \qquad\textbf{(C)}\  2 \qquad\textbf{(D)}\ 3 \qquad\textbf{(E)}\ 4$

Solution: Let $a=\cos(x)+i\sin(x)$ . Now $P(a)=1+a-a^2+a^3$ $P(-1)=-2$ and $P(0)=1$ so there is a real root $a_1$ between $-1$ and $0$ . The other $a$ 's must be complex conjugates since all coefficients of the polynomial are real. The magnitude of those complex $a$ 's squared is $\frac{1}{a_1}$ which is greater than $1$ . If $x$ is real number then $a$ must have magnitude of $1$ , but none of the solutions for $a$ have magnitude of $1$ , so the answer is $\boxed{0}$ ~lopkiloinm

Answer: A 0
Problem: For real numbers $x$ , let \[P(x)=1+\cos(x)+i\sin(x)-\cos(2x)-i\sin(2x)+\cos(3x)+i\sin(3x)\] where $i = \sqrt{-1}$ . For how many values of $x$ with $0\leq x<2\pi$ does \[P(x)=0?\]
$\textbf{(A)}\ 0 \qquad\textbf{(B)}\  1 \qquad\textbf{(C)}\  2 \qquad\textbf{(D)}\ 3 \qquad\textbf{(E)}\ 4$

Solution: For $\textrm{Im}(P(x))=0$ , we get \[\sin(2x)=\sin(x)+\sin(3x)=2\sin(2x)\cos(x)\] So either $\sin(2x)=0$ , i.e. $x\in\{0,\pi\}$ or $\cos(x)=\tfrac 12$ , i.e. $x\in \{\pi/3, 5\pi/3\}$
For none of these values do we get $\textrm{Re}(P(x))=0$
Therefore, the answer is $\boxed{0}$

Answer: A 0
Problem: For real numbers $x$ , let \[P(x)=1+\cos(x)+i\sin(x)-\cos(2x)-i\sin(2x)+\cos(3x)+i\sin(3x)\] where $i = \sqrt{-1}$ . For how many values of $x$ with $0\leq x<2\pi$ does \[P(x)=0?\]
$\textbf{(A)}\ 0 \qquad\textbf{(B)}\  1 \qquad\textbf{(C)}\  2 \qquad\textbf{(D)}\ 3 \qquad\textbf{(E)}\ 4$

Solution: We have \begin{align*} P \left( x \right) & = 1 + e^{ix} - e^{i 2x} + e^{i 3x} . \end{align*}
Denote $y = e^{i x}$ . Hence, this problem asks us to find the number of $y$ with $| y| = 1$ that satisfy \[ 1 + y - y^2 + y^3 = 0 . \hspace{1cm} (1) \]
Taking imaginary part of both sides, we have \begin{align*} 0 & = {\rm Im} \ \left( 1 + y - y^2 + y^3 \right) \\ & = \frac{1}{2i} \left( y - \bar y - y^2 + \bar y^2 + y^3 - \bar y^3 \right) \\ & = \frac{y - \bar y}{2i} \left( 1 - y - \bar y + y^2 + y \bar y + \bar y^2 \right) \\ & = {\rm Im} \ y \left( 1 - \left( y + \bar y \right) + \left( y + \bar y \right)^2 - y \bar y \right) \\ & = {\rm Im} \ y \left( 1 - 2 {\rm Re} \ y + 4 \left( {\rm Re} \ y \right)^2 - |y|^2 \right) \\ & = {\rm Im} \ y \left( 1 - 2 {\rm Re} \ y + 4 \left( {\rm Re} \ y \right)^2 - 1 \right) \\ & = 2 {\rm Im} \ y \cdot {\rm Re} \ y \left( 2 {\rm Re} \ y - 1 \right) \\ \end{align*} The sixth equality follows from the property that $|y| = 1$
Therefore, we have either ${\rm Re} \ y = 0$ or ${\rm Im} \ y = 0$ or $2 {\rm Re} \ y - 1 = 0$
Case 1: ${\rm Re} \ y = 0$
Because $|y| = 1$ $y = \pm i$
However, these solutions fail to satisfy Equation (1).
Therefore, there is no solution in this case.
Case 2: ${\rm Im} \ y = 0$
Because $|y| = 1$ $y = \pm 1$
However, these solutions fail to satisfy Equation (1).
Therefore, there is no solution in this case.
Case 3: $2 {\rm Re} \ y - 1 = 0$
Because $|y| = 1$ $y = \frac{1}{2} \pm \frac{\sqrt{3}}{2} i = e^{i \pm \frac{\pi}{3}}$
However, these solutions fail to satisfy Equation (1).
Therefore, there is no solution in this case.
All cases above imply that there is no solution in this problem.
Therefore, the answer is $\boxed{0}$

Answer: A 0
Problem: For real numbers $x$ , let \[P(x)=1+\cos(x)+i\sin(x)-\cos(2x)-i\sin(2x)+\cos(3x)+i\sin(3x)\] where $i = \sqrt{-1}$ . For how many values of $x$ with $0\leq x<2\pi$ does \[P(x)=0?\]
$\textbf{(A)}\ 0 \qquad\textbf{(B)}\  1 \qquad\textbf{(C)}\  2 \qquad\textbf{(D)}\ 3 \qquad\textbf{(E)}\ 4$

Solution: Let $a=\cos(x)+i\sin(x)$ , so by De Moivre $P(x)=a^3-a^2+a+1$ . The problem essentially asks for the number of real roots of $P$ which lie on the complex unit circle.
Let $|r|=1$ be a root of $P$ , and note that we can't have $r^3-r^2+r=0$ , else $P(r)=0$ . Thus, suppose henceforth that $r^3-r^2+r \neq 0$ . We then have $r^3-r^2+r=r^2(r+\tfrac{1}{r}-1)=a^2(2\mathrm{Re}(r)-1)$ , hence the argument of $r^3-r^2+r$ is either the argument of $a^2$ , or the argument of $-a^2$ . Since $r^3-r^2+r=-1$ is real, it follows that $a^2=\pm 1 \implies a \in \{1,i,-1,-i\}$ . Now, we can check all of these values and find that none of them work, yielding an answer of $\boxed{0}$

Answer: A 0
Problem: For real numbers $x$ , let \[P(x)=1+\cos(x)+i\sin(x)-\cos(2x)-i\sin(2x)+\cos(3x)+i\sin(3x)\] where $i = \sqrt{-1}$ . For how many values of $x$ with $0\leq x<2\pi$ does \[P(x)=0?\]
$\textbf{(A)}\ 0 \qquad\textbf{(B)}\  1 \qquad\textbf{(C)}\  2 \qquad\textbf{(D)}\ 3 \qquad\textbf{(E)}\ 4$

Solution: $P(x)$ can be written equivalently as $P(x) = 1 + cis(x) - cis(2x) + cis(3x).$ Thus, we aim to find $x$ such that the sum of the vectors $cis(x)$ $-cis(2x)$ , and $cis(3x)$ is -1. Notice that $cis(x)$ $-cis(2x)$ $cis(3x)$ all lie on the unit circle in the complex plane, and the vector $cis(x) + cis(3x)$ is collinear with $-cis(2x).$ Since $|-cis(2x)| = 1$ and we want the three vectors to sum to -1, we either have $-cis(2x) = 1$ and $cis(x) + cis(3x) = -2$ , or $-cis(2x) = -1$ and $cis(x) + cis(3x) = 0.$ If the first condition is true, $cis(x) = cis(3x)=-1.$ This will imply that $x= \pi.$ But then $-cis(2x) = -1$ , which violates the condition. Similarly, we can show that the second condition cannot be met either. Thus $P(x)$ does not have any solutions on the interval $[0, 2\pi].$ Therefore, the answer is $\boxed{0}$

Answer: A 0
Problem: Triangle $ABC$ has side lengths $AB = 11, BC=24$ , and $CA = 20$ . The bisector of $\angle{BAC}$ intersects $\overline{BC}$ in point $D$ , and intersects the circumcircle of $\triangle{ABC}$ in point $E \ne A$ . The circumcircle of $\triangle{BED}$ intersects the line $AB$ in points $B$ and $F \ne B$ . What is $CF$
$\textbf{(A) } 28 \qquad \textbf{(B) } 20\sqrt{2} \qquad \textbf{(C) } 30 \qquad \textbf{(D) } 32 \qquad \textbf{(E) } 20\sqrt{3}$

Solution: By the Law of Cosine $\cos A = \frac{AC^2 + AB^2 - BC^2}{ 2 \cdot AC \cdot AB} = \frac{20^2 + 11^2 - 24^2}{2\cdot20\cdot11} = -\frac18$
As $ABEC$ is a cyclic quadrilateral, $\angle CEA = \angle CBA$ . As $BDEF$ is a cyclic quadrilateral, $\angle CBA = \angle FEA$
$\because \quad \angle CEA = \angle FEA \quad \text{and} \quad \angle CAE = \angle FAE$
$\therefore \quad \triangle AFE \cong \triangle ACE$ by $ASA$
Hence, $AF = AC = 20$
By the Law of Cosine $CF = \sqrt{20^2 + 20^2 - 2 \cdot 20 \cdot 20 (-\frac18)} = \sqrt{900} = \boxed{30}$

Answer: C 30
Problem: Triangle $ABC$ has side lengths $AB = 11, BC=24$ , and $CA = 20$ . The bisector of $\angle{BAC}$ intersects $\overline{BC}$ in point $D$ , and intersects the circumcircle of $\triangle{ABC}$ in point $E \ne A$ . The circumcircle of $\triangle{BED}$ intersects the line $AB$ in points $B$ and $F \ne B$ . What is $CF$
$\textbf{(A) } 28 \qquad \textbf{(B) } 20\sqrt{2} \qquad \textbf{(C) } 30 \qquad \textbf{(D) } 32 \qquad \textbf{(E) } 20\sqrt{3}$

Solution: Construct the $E$ -antipode, $E^{\prime}\in(ABC)$ . Notice $\triangle CE^{\prime}A\stackrel{+}{\sim}\triangle CBF$ by spiral similarity at $C$ , thus $CF=\dfrac{CB\cdot CA}{CE^{\prime}}=\frac{480}{CE^{\prime}}$ . Let $CE^{\prime}=x$ ; by symmetry $BE^{\prime}=x$ as well and $\cos\angle BE^{\prime}C=\cos\angle A=\tfrac{11^{2}+20^{2}-24^{2}}{2\cdot 11\cdot 20}=-\tfrac{1}{8}$ from Law of Cosines in $\triangle ABC$ , so by Law of Cosines in $\triangle BE^{\prime}C$ we have \[x^{2}+x^{2}+\left(2x^{2}\right)\left(-\dfrac{1}{8}\right)=24^{2}\] from which $x=16$ . Now, $CF=\dfrac{480}{16}=\boxed{30}$

Answer: C 30
Problem: Triangle $ABC$ has side lengths $AB = 11, BC=24$ , and $CA = 20$ . The bisector of $\angle{BAC}$ intersects $\overline{BC}$ in point $D$ , and intersects the circumcircle of $\triangle{ABC}$ in point $E \ne A$ . The circumcircle of $\triangle{BED}$ intersects the line $AB$ in points $B$ and $F \ne B$ . What is $CF$
$\textbf{(A) } 28 \qquad \textbf{(B) } 20\sqrt{2} \qquad \textbf{(C) } 30 \qquad \textbf{(D) } 32 \qquad \textbf{(E) } 20\sqrt{3}$

Solution: Applying Stewart's theorem on $\triangle ABC$ with cevian $\overline{CF}$ using the directed lengths $AF = AC = 20$ and $FB = 11-20 = -9$ , we obtain \begin{align*} (20)(-9)(11) + (CF)(11)(CF) &= (24)(20)(24) + (20)(-9)(20) \\ 11CF^{2} - 1980 &= 11520 - 3600\end{align*} so $CF=\sqrt{\frac{11520 - 3600 + 1980}{11}}=\sqrt{\frac{9900}{11}}=\sqrt{900}=\boxed{30}$

Answer: C 30
Problem: Triangle $ABC$ has side lengths $AB = 11, BC=24$ , and $CA = 20$ . The bisector of $\angle{BAC}$ intersects $\overline{BC}$ in point $D$ , and intersects the circumcircle of $\triangle{ABC}$ in point $E \ne A$ . The circumcircle of $\triangle{BED}$ intersects the line $AB$ in points $B$ and $F \ne B$ . What is $CF$
$\textbf{(A) } 28 \qquad \textbf{(B) } 20\sqrt{2} \qquad \textbf{(C) } 30 \qquad \textbf{(D) } 32 \qquad \textbf{(E) } 20\sqrt{3}$

Solution: Note that $\angle CAF = \angle CAB$ so we may plug into Law of Cosines to find the angle's cosine: \[AB^2+AC^2-2\cdot AB \cdot AC \cdot \cos(\angle CAB) = BC^2 \to \cos(\angle CAB) = -\frac{1}{8}.\]
So, we observe that we can use Law of Cosines again to find $CF$ \[CF^2 = AF^2+AC^2-2 \cdot AF \cdot AC \cdot \cos(\angle CAF) = 900 \to CF=\boxed{30}\] both ways.

Answer: C 30