Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: What is the degree measure of the acute angle formed by lines with slopes $2$ and $\frac{1}{3}$
$\textbf{(A)} ~30\qquad\textbf{(B)} ~37.5\qquad\textbf{(C)} ~45\qquad\textbf{(D)} ~52.5\qquad\textbf{(E)} ~60$

Solution: We can take any two lines of this form, since the angle between them will always be the same. Let's take $y=2x$ for the line with slope of 2 and $y=\frac{1}{3}x$ for the line with slope of 1/3. Let's take 3 lattice points and create a triangle. Let's use $(0,0)$ $(1,2)$ , and $(3,1)$ . The distance between the origin and $(1,2)$ is $\sqrt{5}$ . The distance between the origin and $(3,1)$ is $\sqrt{10}$ . The distance between $(1,2)$ and $(3,1)$ is $\sqrt{5}$ . We notice that we have a triangle with 3 side lengths: $\sqrt{5}$ $\sqrt{5}$ , and $\sqrt{10}$ . This forms a 45-45-90 triangle, meaning that the angle is $\boxed{45}$

Answer:  45
Problem: What is the degree measure of the acute angle formed by lines with slopes $2$ and $\frac{1}{3}$
$\textbf{(A)} ~30\qquad\textbf{(B)} ~37.5\qquad\textbf{(C)} ~45\qquad\textbf{(D)} ~52.5\qquad\textbf{(E)} ~60$

Solution: Follow Solution 2 up until the lattice points section. Let's use $(0,0)$ $(2,4)$ , and $(9,3)$ . The distance between the origin and $(2,4)$ is $\sqrt{20}$ . The distance between the origin and $(9,3)$ is $\sqrt{90}$ . The distance between $(2,4)$ and $(9,3)$ is $\sqrt{50}$ . Using the Law of Cosines, we see the $50 = 90 + 20 - 2\times\sqrt{20}$ $\times\sqrt{90}$ $\times\cos(\theta)$ , where $\theta$ is the angle we are looking for.
Simplifying, we get $-60 = -2\times(\sqrt{20}) \times(\sqrt{90}) \times\cos(\theta)$
$30 =  \sqrt{1800} \times\cos(\theta)$
$30 =  30\sqrt{2} \times\cos(\theta)$
$\frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}= \cos(\theta)$
Thus, $\theta = \boxed{45}$

Answer: C 45
Problem: What is the degree measure of the acute angle formed by lines with slopes $2$ and $\frac{1}{3}$
$\textbf{(A)} ~30\qquad\textbf{(B)} ~37.5\qquad\textbf{(C)} ~45\qquad\textbf{(D)} ~52.5\qquad\textbf{(E)} ~60$

Solution: We can set up vectors $\vec{a} = <1,2>$ and $\vec{b} = <3,1>$ to represent the two lines. We know that $\frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|} = \cos \theta$ . Plugging the vectors in gives us $\cos \theta = \frac{5}{5\sqrt{2}} = \frac{1}{\sqrt{2}}$ .  From this we get that $\theta = \boxed{45}$

Answer: C 45
Problem: What is the degree measure of the acute angle formed by lines with slopes $2$ and $\frac{1}{3}$
$\textbf{(A)} ~30\qquad\textbf{(B)} ~37.5\qquad\textbf{(C)} ~45\qquad\textbf{(D)} ~52.5\qquad\textbf{(E)} ~60$

Solution: Let $Z_1 = 3 + i$ and $Z_2 = 1 + 2i$ \begin{align*} Z_2 &= Z_1 \cdot re^{i\theta} \\ 1+2i&=(3+i) \cdot re^{i\theta} \\ 1+2i&=(3 + i) \cdot r(\cos\theta + i\sin\theta) \\ 1+2i&=3r\cos\theta - r\sin\theta  + 3ri\sin\theta + ri\cos\theta \\ \end{align*}
From this we have: \begin{align} 1 &= 3r\cos\theta - r\sin\theta \\ 2 &= r\cos\theta + 3r\sin\theta  \end{align}
To solve this we must compute $r$ \begin{align*} r &= \frac{|Z_2|}{|Z_1|} \\ &= \frac{\sqrt{5}}{\sqrt{10}} \\ &= \frac{\sqrt{2}}{2} \end{align*}
Using elimination we have: $3\cdot(2) - (1)$ \begin{align*} 5 &= 10r\sin\theta \\ \frac{1}{2r} &= \sin\theta \\ \frac{1}{2\frac{\sqrt{2}}{2}} &= \sin\theta \\ \frac{\sqrt{2}}{2} &= \sin\theta \\ \theta &= \boxed{45}

Answer: C 45
Problem: What is the degree measure of the acute angle formed by lines with slopes $2$ and $\frac{1}{3}$
$\textbf{(A)} ~30\qquad\textbf{(B)} ~37.5\qquad\textbf{(C)} ~45\qquad\textbf{(D)} ~52.5\qquad\textbf{(E)} ~60$

Solution: The lines $y = 2x, y = \frac {1}{3}x$ , and $x = 3$ form a large right triangle and a small right triangle. Call the angle that is formed by the x-axis and the line $y = 2x$ $\alpha$ , and call the angle that is formed by the x-axis and the line $y = \frac {1}{3}x$ $\beta$ . We try to find $\sin (\alpha - \beta)$ first, and then try to see if any of the answer choices match up.
$\sin (\alpha - \beta)$ $\sin \alpha$ $\cos \beta$ $\sin \beta$ $\cos \alpha$
Using soh-cah-toa, we find that $\sin \alpha = \frac {2}{\sqrt 5}, \sin \beta = \frac {1}{\sqrt 10}, \cos \alpha = \frac {1}{\sqrt 5},$ and $\cos \beta = \frac {3}{\sqrt 10}$
Plugging it all in, we find that $\sin (\alpha - \beta) = \frac {5}{\sqrt {50}}$ , which is equivalent to $\frac {\sqrt 2}{2}$ . Since $\sin (\alpha - \beta) = \frac {\sqrt 2}{2}$ , we get that $\alpha - \beta = 45^{\circ}$ . Therefore, the answer is $\boxed{45}$

Answer: C 45
Problem: What is the value of \[2^3 - 1^3 + 4^3 - 3^3 + 6^3 - 5^3 + \dots + 18^3 - 17^3?\]
$\textbf{(A) } 2023 \qquad\textbf{(B) } 2679 \qquad\textbf{(C) } 2941 \qquad\textbf{(D) } 3159 \qquad\textbf{(E) } 3235$

Solution: To solve this problem, we will be using difference of cube, sum of squares and sum of arithmetic sequence formulas.
\[2^3-1^3+4^3-3^3+6^3-5^3+...+18^3-17^3\]
$=(2-1)(2^2+1 \cdot 2+1^2)+(4-3)(4^2+4 \cdot 3+3^2)+(6-5)(6^2+6 \cdot 5+5^2)+...+(18-17)(18^2+18 \cdot 17+17^2)$
$=(2^2+1 \cdot 2+1^2)+(4^2+4 \cdot 3+3^2)+(6^2+6 \cdot 5+5^2)+...+(18^2+18 \cdot 17+17^2)$
$=1^2+2^2+3^2+4^2+5^2+6^2...+17^2+18^2+1 \cdot 2+3 \cdot 4+5 \cdot 6+...+17 \cdot 18$
$=\frac{18(18+1)(36+1)}{6}+1 \cdot 2+3 \cdot 4+5 \cdot 6+...+17 \cdot 18$
we could rewrite the second part as $\sum_{n=1}^{9}(2n-1)(2n)$
$(2n-1)(2n)=4n^2-2n$
$\sum_{n=1}^{9}4n^2=4(\frac{9(9+1)(18+1)}{6})$
$\sum_{n=1}^{9}-2n=-2(\frac{9(9+1)}{2})$
Hence,
$1 \cdot 2+3 \cdot 4+5 \cdot 6+...+17 \cdot 18 = 4(\frac{9(9+1)(18+1)}{6})-2(\frac{9(9+1)}{2})$
Adding everything up:
$2^3-1^3+4^3-3^3+6^3-5^3+...+18^3-17^3$
$=\frac{18(18+1)(36+1)}{6}+4(\frac{9(9+1)(18+1)}{6})-2(\frac{9(9+1)}{2})$
$=3(19)(37)+6(10)(19)-9(10)$
$=2109+1140-90$
$=\boxed{3159}$

Answer: D 3159
Problem: What is the value of \[2^3 - 1^3 + 4^3 - 3^3 + 6^3 - 5^3 + \dots + 18^3 - 17^3?\]
$\textbf{(A) } 2023 \qquad\textbf{(B) } 2679 \qquad\textbf{(C) } 2941 \qquad\textbf{(D) } 3159 \qquad\textbf{(E) } 3235$

Solution: Think about $2^3+4^3+6^3+...+18^3$ . Once we factor out $2^3=8$ , we get $1^3+2^3+...+9^3$ , which can be found using the sum of cubes formula, $(\frac{n(n+1)}{2})^2$ . Now think about $1^3+3^3+...+17^3$ . This is just the previous sum subtracted from the total sum of $18$ cubes. So now we have the two things we need to add. The sum of all the even cubes is $8\cdot (\frac{90}{2})^2\rightarrow 8\cdot 2025 = 16200$ . The sum of all cubes from $1^3$ to $18^3$ is $(\frac{18\cdot 19}{2})^2=29241$ . The sum of the odd cubes is then $29241-16200=13041$ . Thus we get $16200-13041=\boxed{3159}$ ~amcrunner

Answer: D 3159
Problem: What is the value of \[2^3 - 1^3 + 4^3 - 3^3 + 6^3 - 5^3 + \dots + 18^3 - 17^3?\]
$\textbf{(A) } 2023 \qquad\textbf{(B) } 2679 \qquad\textbf{(C) } 2941 \qquad\textbf{(D) } 3159 \qquad\textbf{(E) } 3235$

Solution: Using the same sum of cubes formula, we can rewrite as $2(2^3 + 4^3 + ... + 18^3) - (1^3 + 2^3 + ... + 18^3)$
$= 2(2^3)(1^3 + ... + 9^3) - (1^3 + ... + 18^3)$
$= 16(5 \cdot 9)^2 - (9 \cdot 19)^2 = 9^2(20^2 - 19^2) = 81 \cdot 39 = \boxed{3159}$ ~AoPSuser216

Answer: D 3159
Problem: What is the value of \[2^3 - 1^3 + 4^3 - 3^3 + 6^3 - 5^3 + \dots + 18^3 - 17^3?\]
$\textbf{(A) } 2023 \qquad\textbf{(B) } 2679 \qquad\textbf{(C) } 2941 \qquad\textbf{(D) } 3159 \qquad\textbf{(E) } 3235$

Solution: For any real numbers $x$ and $y$ $x^3-y^3=(x-y)(x^2+xy+y^2)=(x-y)((x-y)^2+3xy)=(x-y)^3+3xy(x-y)$
When $x = y + 1$ , with the above formula, we will get $x^3-y^3=1^3+3xy=1 + 3xy$
Therefore,
$2^3 - 1^3 + 4^3 - 3^3 + \dots + 18^3 - 17^3$
$= (1 + 3\cdot 1\cdot 2) + (1 + 3\cdot 3\cdot 4) + \dots + (1 + 3\cdot 17\cdot 18)$
$= 9 + 3\cdot (1\cdot 2 + 3\cdot 4 + \dots + 17\cdot 18)$
$= 9 + 3 \cdot (2 + 12 + 30 + 56 + 90 + 132 + 182 + 240 + 306)$
$= 9 + 3 \cdot 1050$
$= \boxed{3159}$

Answer: D 3159
Problem: What is the value of \[2^3 - 1^3 + 4^3 - 3^3 + 6^3 - 5^3 + \dots + 18^3 - 17^3?\]
$\textbf{(A) } 2023 \qquad\textbf{(B) } 2679 \qquad\textbf{(C) } 2941 \qquad\textbf{(D) } 3159 \qquad\textbf{(E) } 3235$

Solution: We rewrite the sum as
\[\sum_{k=1}^{9}(2k)^3-(2k-1)^3\] \[=\sum_{k=1}^{9} 12k^2 - 6k + 1\] \[=12\sum_{k=1}^{9}k^2 - 6\sum_{k=1}^{9}k + 9\] \[=12\cdot\frac{9\cdot 10\cdot 19}{6} -6\cdot \frac{9\cdot 10}{2} + 9\] \[=3420 - 270 + 9 = \boxed{3159}\]

Answer: D 3159
Problem: What is the value of \[2^3 - 1^3 + 4^3 - 3^3 + 6^3 - 5^3 + \dots + 18^3 - 17^3?\]
$\textbf{(A) } 2023 \qquad\textbf{(B) } 2679 \qquad\textbf{(C) } 2941 \qquad\textbf{(D) } 3159 \qquad\textbf{(E) } 3235$

Solution: We see $=12\cdot\frac{9\cdot 10\cdot 19}{6} -6\cdot \frac{9\cdot 10}{2} + 9 = 9\cdot (12\cdot\frac{10\cdot 19}{6} -6\cdot \frac{10}{2} + 1)$ which is clearly a multiple of 9. The only answer choice which is a multiple of 9 is $\boxed{3159}$ ~Ilaggo2432

Answer: D 3159
Problem: What is the value of \[2^3 - 1^3 + 4^3 - 3^3 + 6^3 - 5^3 + \dots + 18^3 - 17^3?\]
$\textbf{(A) } 2023 \qquad\textbf{(B) } 2679 \qquad\textbf{(C) } 2941 \qquad\textbf{(D) } 3159 \qquad\textbf{(E) } 3235$

Solution: $2^3 - 1^3 + 4^3 - 3^3 + 6^3 - 5^3 + \dots + 18^3 - 17^3$
$=8-1+64-27+216-125+512-343+1000-729+1728-1331+2744-2197+4096-3375+5832-4913$
$= \boxed{3159}$

Answer: D 3159
Problem: What is the value of \[2^3 - 1^3 + 4^3 - 3^3 + 6^3 - 5^3 + \dots + 18^3 - 17^3?\]
$\textbf{(A) } 2023 \qquad\textbf{(B) } 2679 \qquad\textbf{(C) } 2941 \qquad\textbf{(D) } 3159 \qquad\textbf{(E) } 3235$

Solution: Reduce all terms mod 9. This yields:
$2^3 - 1^3 + 4^3 - 3^3 + 6^3 - 5^3 + \dots + 18^3 - 17^3$ $\equiv -1 - 1 + 1 - 0 + 0 - -1 + -1 - 1 + 1 - 0 + 0 - -1 + -1 - 1 + 1 - 0 +0 - -1 (\mod 9)$ $\equiv 0 (\mod 9)$
The only answer choice which is also ≡0 mod 9 is $= \boxed{3159}$

Answer: D 3159
Problem: How many complex numbers satisfy the equation $z^5=\overline{z}$ , where $\overline{z}$ is the conjugate of the complex number $z$
$\textbf{(A)} ~2\qquad\textbf{(B)} ~3\qquad\textbf{(C)} ~5\qquad\textbf{(D)} ~6\qquad\textbf{(E)} ~7$

Solution: When $z^5=\overline{z}$ , there are two conditions: either $z=0$ or $z\neq 0$ . When $z\neq 0$ , since $|z^5|=|\overline{z}|$ $|z|=1$ $z^5\cdot z=z^6=\overline{z}\cdot z=|z|^2=1$ . Consider the $r(\cos \theta +i\sin \theta)$ form, when $z^6=1$ , there are 6 different solutions for $z$ . Therefore, the number of complex numbers satisfying $z^5=\bar{z}$ is $\boxed{7}$

Answer: E 7
Problem: How many complex numbers satisfy the equation $z^5=\overline{z}$ , where $\overline{z}$ is the conjugate of the complex number $z$
$\textbf{(A)} ~2\qquad\textbf{(B)} ~3\qquad\textbf{(C)} ~5\qquad\textbf{(D)} ~6\qquad\textbf{(E)} ~7$

Solution: Let $z = re^{i\theta}.$ We now have $\overline{z} = re^{-i\theta},$ and want to solve
\[r^5e^{5i\theta} = re^{-i\theta}.\]
From this, we have $r = 0$ as a solution, which gives $z = 0$ . If $r\neq 0$ , then we divide by it, yielding
\[r^4e^{5i\theta} = e^{-i\theta}.\]
Dividing both sides by $e^{-i\theta}$ yields $r^4e^{6i\theta} = 1$ . 
Taking the magnitude of both sides tells us that $r^4 = 1$ , so $r^2 = \pm 1$ . However, if $r^2 = -1$ , then $r = \pm i$ , but $r$ must be real. Therefore, $r^2 = 1$
Multiplying both sides by $r^2$
\[r^6\cdot e^{6i\theta} = z^6 = 1.\]
Each of the $6$ th roots of unity is a solution to this, so there are $6 + 1 = \boxed{7}$ solutions.

Answer: D 7
Problem: How many complex numbers satisfy the equation $z^5=\overline{z}$ , where $\overline{z}$ is the conjugate of the complex number $z$
$\textbf{(A)} ~2\qquad\textbf{(B)} ~3\qquad\textbf{(C)} ~5\qquad\textbf{(D)} ~6\qquad\textbf{(E)} ~7$

Solution: Let $z = a+bi$
Then, our equation becomes: $(a+bi)^5=a-bi$
Note that since every single term in the expansion contains either an $a$ or $b$ , simply setting $a=0$ and $b=0$ yields a solution.
Now, considering the other case that either $a$ or $b$ does not equal $0$
Multiplying both sides by $a+bi$ (or $z$ ), we get: $(a+bi)^6=a^2+b^2$ (since $i^2=-1$ ).
Substituting $z$ back into the left hand side, we get: $z^6=a^2+b^2$
Note that this will have 6 distinct, non-zero solutions since in this case, we consider that either $a$ or $b$ is not $0$ , and these are simply the sixth roots of a positive real number.
Adding up the solutions, we get $1+6=$ $\boxed{7}$

Answer: E 7
Problem: Usain is walking for exercise by zigzagging across a $100$ -meter by $30$ -meter rectangular field, beginning at point $A$ and ending on the segment $\overline{BC}$ . He wants to increase the distance walked by zigzagging as shown in the figure below $(APQRS)$ . What angle $\theta = \angle PAB=\angle QPC=\angle RQB=\cdots$ will produce in a length that is $120$ meters? (This figure is not drawn to scale. Do not assume that he zigzag path has exactly four segments as shown; there could be more or fewer.)
[asy] import olympiad; draw((-50,15)--(50,15)); draw((50,15)--(50,-15)); draw((50,-15)--(-50,-15)); draw((-50,-15)--(-50,15)); draw((-50,-15)--(-22.5,15)); draw((-22.5,15)--(5,-15)); draw((5,-15)--(32.5,15)); draw((32.5,15)--(50,-4.090909090909)); label("$\theta$", (-41.5,-10.5)); label("$\theta$", (-13,10.5)); label("$\theta$", (15.5,-10.5)); label("$\theta$", (43,10.5)); dot((-50,15)); dot((-50,-15)); dot((50,15)); dot((50,-15)); dot((50,-4.09090909090909)); label("$D$",(-58,15)); label("$A$",(-58,-15)); label("$C$",(58,15)); label("$B$",(58,-15)); label("$S$",(58,-4.0909090909)); dot((-22.5,15)); dot((5,-15)); dot((32.5,15)); label("$P$",(-22.5,23)); label("$Q$",(5,-23)); label("$R$",(32.5,23)); [/asy]
$\textbf{(A)}~\arccos\frac{5}{6}\qquad\textbf{(B)}~\arccos\frac{4}{5}\qquad\textbf{(C)}~\arccos\frac{3}{10}\qquad\textbf{(D)}~\arcsin\frac{4}{5}\qquad\textbf{(E)}~\arcsin\frac{5}{6}$

Solution: Drop an altitude from $P$ to $AB$ and let its base be $x$ . Note that if we repeat this for $Q$ and $R$ , all four right triangles (including $\triangle{RSC}$ ) will have the same trig ratios. By proportion, the hypotenuse $AP$ is $\frac{x}{100}(120) = \frac65 x$ , so $\cos\theta = \frac{x}{(\frac65x)} = \frac56 \Rightarrow \theta = \boxed{56}$

Answer: A 56
Problem: Consider the set of complex numbers $z$ satisfying $|1+z+z^{2}|=4$ . The maximum value of the imaginary part of $z$ can be written in the form $\tfrac{\sqrt{m}}{n}$ , where $m$ and $n$ are relatively prime positive integers. What is $m+n$
$\textbf{(A)}~20\qquad\textbf{(B)}~21\qquad\textbf{(C)}~22\qquad\textbf{(D)}~23\qquad\textbf{(E)}~24$

Solution: First, substitute in $z=a+bi$
\[|1+(a+bi)+(a+bi)^2|=4\] \[|(1+a+a^2-b^2)+ (b+2ab)i|=4\] \[(1+a+a^2-b^2)^2+ (b+2ab)^2=16\] \[(1+a+a^2-b^2)^2+ b^2(1+4a+4a^2)=16\]
Let $p=b^2$ and $q=1+a+a^2$
\[(q-p)^2+ p(4q-3)=16\] \[p^2-2pq+q^2   + 4pq -3p=16\]
We are trying to maximize $b=\sqrt p$ , so we'll turn the equation into a quadratic to solve for $p$ in terms of $q$
\[p^2+(2q-3)p+(q^2-16)=0\] \[p=\frac{(-2q+3)\pm \sqrt{-12q+73}}{2}\]
We want to maximize $p$ . Since $q$ is always negatively contributing to $p$ 's value, we want to minimize $q$
Due to the trivial inequality: $q=1+a+a^2=(a+\frac 12)^2+\frac{3}4 \geq \frac{3}4$
If we plug $q$ 's minimum value in, we get that $p$ 's maximum value is \[p=\frac{(-2(\frac 34)+3)+ \sqrt{-12(\frac 34)+73}}{2}=\frac{\frac 32+ 8}{2}=\frac{19}{4}\]
Then \[b=\frac{\sqrt{19}}{2}\] and \[m+n=\boxed{21}\]

Answer: B 21
Problem: Consider the set of complex numbers $z$ satisfying $|1+z+z^{2}|=4$ . The maximum value of the imaginary part of $z$ can be written in the form $\tfrac{\sqrt{m}}{n}$ , where $m$ and $n$ are relatively prime positive integers. What is $m+n$
$\textbf{(A)}~20\qquad\textbf{(B)}~21\qquad\textbf{(C)}~22\qquad\textbf{(D)}~23\qquad\textbf{(E)}~24$

Solution: We are given that $1+z+z^2=c$ where $c$ is some complex number with magnitude $4$ . Rearranging the quadratic to standard form and applying the quadratic formula, we have \[z=\frac{-1\pm \sqrt{1^2-4(1)(1-c)}}{2}=\frac{-1\pm\sqrt{4c-3}}{2}.\]
The imaginary part of $z$ is maximized when $c=-4$ . (Why? See note below.)
Thus $z=i\sqrt{19}/2$ , and so the answer is $\boxed{21}$

Answer: B 21
Problem: Consider the set of complex numbers $z$ satisfying $|1+z+z^{2}|=4$ . The maximum value of the imaginary part of $z$ can be written in the form $\tfrac{\sqrt{m}}{n}$ , where $m$ and $n$ are relatively prime positive integers. What is $m+n$
$\textbf{(A)}~20\qquad\textbf{(B)}~21\qquad\textbf{(C)}~22\qquad\textbf{(D)}~23\qquad\textbf{(E)}~24$

Solution: [asy] size(250); import TrigMacros; rr_cartesian_axes(-6,5,-5,5,complexplane=true, usegrid = true); Label f; f.p=fontsize(6);  xaxis(-6,5,Ticks(f, 1.0));  yaxis(-5,5,Ticks(f, 1.0)); dot((0,0)); draw(circle((-3/4, 0), 4), red + dashed); dot((-19/4, 0), blue); label("$\phi$", (-19/4, 0), NW); dot((0, 2.18), blue); label("$v'$", (0, 2.18), NE); draw(ellipse((0,0),1.8,2.18), green); [/asy] We can write the given condition as \[\left|\left(z+\frac{1}{2}\right)^2 + \frac{3}{4}\right| = 4.\] Letting $u = \left(z+\frac{1}{2}\right)^2$ , the equation $\left|u + \frac{3}{4}\right| = 4$ equates to the circle centered at $-\frac{3}{4}$ with radius $4$ in the complex plane, call it $\omega$ . Thus the locus of $\left(z+\frac{1}{2}\right)^2$ is $\omega$ . Let $v = z+\frac{1}{2}$ , and since the $+\frac{1}{2}$ does not change $z$ 's imaginary part, we now need to find $v$ with the largest imaginary part such that $v^2$ lies on $\omega$
Note that the point on $\omega$ with largest magnitude is $19/4$ and has argument $\pi$ , call it $\phi$ (The leftmost point on $\omega$ ). The value $v'$ with positive imaginary part such that $(v')^2 = \phi$ has an argument of $\frac{\pi}{2}$ and a magnitude of $\frac{\sqrt{19}}{2}$
Since across all values of $v$ the imaginary part is given by $r\sin{\theta}$ and $v'$ has the largest possible $r$ and the largest possible value of $\sin{\theta},$ it must have the largest imaginary part.
This can non-rigorously be seen by sketching the oval which is the locus of $v$
This gives $19 + 2 \implies \boxed{21}$

Answer: B 21
Problem: Consider the set of complex numbers $z$ satisfying $|1+z+z^{2}|=4$ . The maximum value of the imaginary part of $z$ can be written in the form $\tfrac{\sqrt{m}}{n}$ , where $m$ and $n$ are relatively prime positive integers. What is $m+n$
$\textbf{(A)}~20\qquad\textbf{(B)}~21\qquad\textbf{(C)}~22\qquad\textbf{(D)}~23\qquad\textbf{(E)}~24$

Solution: To start, we factor $1+z+z^2$ to get:
\[|(z-\frac{-1+\sqrt{3}i}{2})(z-\frac{-1-\sqrt{3}i}{2})|=4\]
Note that since the magnitude of a product of complex numbers is equal to the product of the magnitudes:
\[|(z+\frac{1-\sqrt{3}i}{2})||(z+\frac{1+\sqrt{3}i}{2})|=4\]
Now, we substitute $z=a+bi$ (Note that we are trying to maximize b):
\[|a+\frac{1}{2}+(b+\frac{\sqrt{3}}{2})i||a+\frac{1}{2}+(b-\frac{\sqrt{3}}{2})i|=4\]
Since we are trying to maximize $b$ , we want the real parts of the components to be as small as possible, which we can do by setting $a=-\frac{1}{2}$ . This leaves us with:
\[|(b+\frac{\sqrt{3}}{2})i||(b-\frac{\sqrt{3}}{2})i|=4\] \[(b+\frac{\sqrt{3}}{2})(b-\frac{\sqrt{3}}{2})=4\] \[b^2-\frac{3}{4}=4\] \[b^2=\frac{19}{4}\] \[b=\frac{\sqrt{19}}{2}\]
This gives us $19 + 2 \implies \boxed{21}$

Answer: B 21
Problem: Flora the frog starts at 0 on the number line and makes a sequence of jumps to the right. In any one jump, independent of previous jumps, Flora leaps a positive integer distance $m$ with probability $\frac{1}{2^m}$
What is the probability that Flora will eventually land at 10?
$\textbf{(A)}~\frac{5}{512}\qquad\textbf{(B)}~\frac{45}{1024}\qquad\textbf{(C)}~\frac{127}{1024}\qquad\textbf{(D)}~\frac{511}{1024}\qquad\textbf{(E)}~\frac{1}{2}$

Solution: Initially, the probability of landing at $10$ and landing past $10$ (summing the infinte series) are exactly the same. Landing before 10 repeats this initial condition, with a different irrelevant scaling factor. Therefore, the probability must be $\boxed{12}$

Answer: E 12
Problem: What is the product of all solutions to the equation \[\log_{7x}2023\cdot \log_{289x}2023=\log_{2023x}2023\]
$\textbf{(A)} ~(\log_{2023}7\cdot \log_{2023}289)^2\qquad\textbf{(B)} ~\log_{2023}7\cdot \log_{2023}289\qquad\textbf{(C)} ~1$ $\textbf{(D)} ~\log_{7}2023\cdot \log_{289}2023\qquad\textbf{(E)} ~(\log_7 2023\cdot\log_{289} 2023)^2$

Solution: For $\log_{7x}2023\cdot \log_{289x}2023=\log_{2023x}2023$ , transform it into $\dfrac{\ln 289+\ln 7}{\ln 7 + \ln x}\cdot \dfrac{\ln 289+\ln 7}{\ln 289 + \ln x}=\dfrac{\ln 289+\ln 7}{\ln 289+\ln 7+\ln x}$ . Replace $\ln x$ with $y$ . Because we want to find the product of all solutions of $x$ , it is equivalent to finding the exponential of the sum of all solutions of $y$ . Change the equation to standard quadratic equation form, the term with 1 power of $y$ is canceled.  By using Vieta, we see that since there does not exist a $by$ term, $\sum y=0$ and $\prod x=e^0=\boxed{1}$

Answer: C 1
Problem: What is the product of all solutions to the equation \[\log_{7x}2023\cdot \log_{289x}2023=\log_{2023x}2023\]
$\textbf{(A)} ~(\log_{2023}7\cdot \log_{2023}289)^2\qquad\textbf{(B)} ~\log_{2023}7\cdot \log_{2023}289\qquad\textbf{(C)} ~1$ $\textbf{(D)} ~\log_{7}2023\cdot \log_{289}2023\qquad\textbf{(E)} ~(\log_7 2023\cdot\log_{289} 2023)^2$

Solution: \[\log_{7x}2023\cdot \log_{289x}2023=\log_{2023x}2023\]
Rearranging it give us:
\[\log_{2023}7x\cdot \log_{2023}289x=\log_{2023}2023x\]
\[(\log_{2023}7+\log_{2023}x)(\log_{2023}289+\log_{2023}x)=(\log_{2023}2023+\log_{2023}x)\]
let $\log_{2023}x$ be $a$ , we get
\[(\log_{2023}7+a)(\log_{2023}289+a)=1+a\]
\[a^2+(\log_{2023}7+\log_{2023}289)a+\log_{2023}7 \cdot \log_{2023}289=1+a\]
\[a^2+\log_{2023}7 \cdot \log_{2023}289-1=0\]
by Vieta's Formulas,
\[a_1+a_2=0\]
\[\log_{2023}{x_1}+\log_{2023}{x_2}=0\]
\[\log_{2023}{x_1x_2}=0\]
\[x_1x_2=\boxed{1}\]

Answer: C 1
Problem: What is the product of all solutions to the equation \[\log_{7x}2023\cdot \log_{289x}2023=\log_{2023x}2023\]
$\textbf{(A)} ~(\log_{2023}7\cdot \log_{2023}289)^2\qquad\textbf{(B)} ~\log_{2023}7\cdot \log_{2023}289\qquad\textbf{(C)} ~1$ $\textbf{(D)} ~\log_{7}2023\cdot \log_{289}2023\qquad\textbf{(E)} ~(\log_7 2023\cdot\log_{289} 2023)^2$

Solution: Similar to solution 1, change the bases first \[\frac{\ln 289+\ln 7}{\ln7 + \ln{x}} \cdot \frac{\ln 289+\ln 7}{2\ln17 + \ln{x}} = \frac{\ln 289+\ln 7}{\ln7 + 2\ln17 + \ln{x}}\] Cancel and cross multiply to get \[(\ln7 + 2\ln17)(\ln7 + 2\ln17 + \ln{x}) = (\ln7 + \ln{x})(2\ln17 + \ln{x})\] Simplify to get \[(\ln{x})^2 = 4(\ln17)^2 + 2\ln17\ln7 + (\ln7)^2\] \[\ln{x} = \pm \sqrt{4(\ln17)^2 + 2\ln17\ln7 + (\ln7)^2}\] The sum of all possible $\ln{x}$ is 0, thus the product of all solutions of $x$ is $\boxed{1}$

Answer: C 1
Problem: What is the product of all solutions to the equation \[\log_{7x}2023\cdot \log_{289x}2023=\log_{2023x}2023\]
$\textbf{(A)} ~(\log_{2023}7\cdot \log_{2023}289)^2\qquad\textbf{(B)} ~\log_{2023}7\cdot \log_{2023}289\qquad\textbf{(C)} ~1$ $\textbf{(D)} ~\log_{7}2023\cdot \log_{289}2023\qquad\textbf{(E)} ~(\log_7 2023\cdot\log_{289} 2023)^2$

Solution: We take the reciprocal of both sides: \[\frac{1}{\log_{7x}2023}\cdot \frac{1}{\log_{289x}2023}=\frac{1}{\log_{2023x}2023}.\] Using logarithm properties, we have \[\log_{2023}7x\cdot \log_{2023}289x=\log_{2023}2023x.\] Simplify to obtain \[2023x^2=2023x,\] from which we have $x=\boxed{1}$

Answer: C 1