Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: The figure below depicts a regular $7$ -gon inscribed in a unit circle. [asy]         import geometry; unitsize(3cm); draw(circle((0,0),1),linewidth(1.5)); for (int i = 0; i < 7; ++i) {   for (int j = 0; j < i; ++j) {     draw(dir(i * 360/7) -- dir(j * 360/7),linewidth(1.5));   } } for(int i = 0; i < 7; ++i) {    dot(dir(i * 360/7),5+black); } [/asy] What is the sum of the $4$ th powers of the lengths of all $21$ of its edges and diagonals?
$\textbf{(A) }49 \qquad \textbf{(B) }98 \qquad \textbf{(C) }147 \qquad \textbf{(D) }168 \qquad \textbf{(E) }196$

Solution: There are $7$ segments whose lengths are $2 \sin \frac{\pi}{7}$ $7$ segments whose lengths are $2 \sin \frac{2 \pi}{7}$ $7$ segments whose lengths are $2 \sin \frac{3\pi}{7}$
Therefore, the sum of the $4$ th powers of these lengths is \begin{align*} & 7 \cdot 2^4 \sin^4 \frac{\pi}{7} + 7 \cdot 2^4 \sin^4 \frac{2 \pi}{7} + 7 \cdot 2^4 \sin^4 \frac{3 \pi}{7} \\ & = 7 \cdot 2^4 \left( \frac{1 - \cos \frac{2 \pi}{7}}{2} \right)^2 + 7 \cdot 2^4 \left( \frac{1 - \cos \frac{4 \pi}{7}}{2} \right)^2 + 7 \cdot 2^4 \left( \frac{1 - \cos \frac{6 \pi}{7}}{2} \right)^2 \\ & = 7 \cdot 2^2 \left( 1 - 2 \cos \frac{2 \pi}{7} + \cos^2 \frac{2 \pi}{7} \right) + 7 \cdot 2^2 \left( 1 - 2 \cos \frac{4 \pi}{7} + \cos^2 \frac{4 \pi}{7} \right) + 7 \cdot 2^2 \left( 1 - 2 \cos \frac{6 \pi}{7} + \cos^2 \frac{6 \pi}{7} \right) \\ & = 7 \cdot 2^2 \cdot 3 - 7 \cdot 2^3 \left( \cos \frac{2 \pi}{7} + \cos \frac{4 \pi}{7} + \cos \frac{6 \pi}{7} \right) + 7 \cdot 2^2 \left( \cos^2 \frac{2 \pi}{7} + \cos^2 \frac{4 \pi}{7}  + \cos^2 \frac{6 \pi}{7} \right) \\ & = 7 \cdot 2^2 \cdot 3 - 7 \cdot 2^3 \left( \cos \frac{2 \pi}{7} + \cos \frac{4 \pi}{7} + \cos \frac{6 \pi}{7} \right) + 7 \cdot 2^2 \left( \frac{1 + \cos \frac{4 \pi}{7} }{2} + \frac{1 + \cos \frac{8 \pi}{7} }{2} + \frac{1 + \cos \frac{12 \pi}{7} }{2} \right) \\ & = 7 \cdot 2^2 \cdot 3 - 7 \cdot 2^3 \left( \cos \frac{2 \pi}{7} + \cos \frac{4 \pi}{7} + \cos \frac{6 \pi}{7} \right) + 7 \cdot 2 \cdot 3 + 7 \cdot 2 \left( \cos \frac{4 \pi}{7} + \cos \frac{8 \pi}{7} + \cos \frac{12 \pi}{7} \right) \\ & = 7 \cdot 2^2 \cdot 3 - 7 \cdot 2^3 \left( \cos \frac{2 \pi}{7} + \cos \frac{4 \pi}{7} + \cos \frac{6 \pi}{7} \right) + 7 \cdot 2 \cdot 3 + 7 \cdot 2 \left( \cos \frac{4 \pi}{7} + \cos \frac{6 \pi}{7} + \cos \frac{2 \pi}{7} \right) \\ & = 7 \cdot 2 \cdot 3 \left( 2 + 1 \right) - 7 \cdot 2 \left( 4 - 1 \right) \left( \cos \frac{4 \pi}{7} + \cos \frac{6 \pi}{7} + \cos \frac{2 \pi}{7} \right) \\ & = 7 \cdot 2 \cdot 3 \left( 2 + 1 \right) - 7 \cdot 2 \left( 4 - 1 \right) \cdot \left( - \frac{1}{2} \right) \\ & = \boxed{147} . \]

Answer: C 147
Problem: The figure below depicts a regular $7$ -gon inscribed in a unit circle. [asy]         import geometry; unitsize(3cm); draw(circle((0,0),1),linewidth(1.5)); for (int i = 0; i < 7; ++i) {   for (int j = 0; j < i; ++j) {     draw(dir(i * 360/7) -- dir(j * 360/7),linewidth(1.5));   } } for(int i = 0; i < 7; ++i) {    dot(dir(i * 360/7),5+black); } [/asy] What is the sum of the $4$ th powers of the lengths of all $21$ of its edges and diagonals?
$\textbf{(A) }49 \qquad \textbf{(B) }98 \qquad \textbf{(C) }147 \qquad \textbf{(D) }168 \qquad \textbf{(E) }196$

Solution: As explained in Solutions 1 and 2, what we are trying to find is $7 \cdot 2^4 \sin^4 \frac{\pi}{7} + 7 \cdot 2^4 \sin^4 \frac{2 \pi}{7} + 7 \cdot 2^4 \sin^4 \frac{3 \pi}{7}$ . Using trig we get \begin{align*}  & \sin^4 \frac{\pi}{7} + \sin^4 \frac{2 \pi}{7} + \sin^4 \frac{3 \pi}{7} \\ = & \sin^2 \frac{\pi}{7} \left(1 - \cos^2 \frac{\pi}{7} \right) + \sin^2 \frac{2\pi}{7} \left(1 - \cos^2 \frac{2\pi}{7} \right) + \sin^2 \frac{3\pi}{7} \left(1 - \cos^2 \frac{3\pi}{7} \right) \\ = & \sin^2 \frac{\pi}{7} - \left(\frac{1}{2} \sin \frac{2\pi}{7}\right)^2 + \sin^2 \frac{2\pi}{7} - \left(\frac{1}{2} \sin \frac{4\pi}{7}\right)^2 + \sin^2 \frac{3\pi}{7} - \left(\frac{1}{2} \sin \frac{6\pi}{7}\right)^2\\ = & \sin^2 \frac{\pi}{7} - \frac{1}{4} \sin^2 \frac{2\pi}{7} + \sin^2 \frac{2\pi}{7} - \frac{1}{4} \sin^2 \frac{4\pi}{7} + \sin^2 \frac{3\pi}{7} - \frac{1}{4} \sin^2 \frac{6\pi}{7} \\ = & \frac{3}{4} \left(\sin^2 \frac{\pi}{7} + \sin^2 \frac{2\pi}{7} + \sin^2 \frac{3\pi}{7}\right) \\ = & \frac{3}{4} \cdot \frac{1}{2} \left(1 - \cos \frac{2\pi}{7} + 1 - \cos \frac{4\pi}{7} + 1 - \cos \frac{6\pi}{7} \right)\\ = & \frac{3}{4} \cdot \frac{1}{2} \left(3 - \left(-\frac{1}{2}\right)\right) \\ = & \frac{21}{16}. \end{align*} Like in the second solution, we also use the fact that $\cos \frac{2\pi}{7} + \cos \frac{4\pi}{7} + \cos \frac{6\pi}{7} = -\frac{1}{2}$ , which admittedly might need some explanation.
For explanation see supplement
Notice that \begin{align*} \cos \frac{2\pi}{7} + \cos \frac{4\pi}{7} + \cos \frac{6\pi}{7} & = \frac{1}{2}\left(e^\frac{2i\pi}{7}+ e^{-\frac{2i\pi}{7}} \right) + \frac{1}{2}\left( e^\frac{4i\pi}{7}+ e^{-\frac{4i\pi}{7}} \right) + \frac{1}{2}\left( e^\frac{6i\pi}{7}+ e^{-\frac{6i\pi}{7}}\right) \\ & = \frac{1}{2}\left(e^\frac{2i\pi}{7}+ e^{-\frac{2i\pi}{7}}+ e^\frac{4i\pi}{7}+ e^{-\frac{4i\pi}{7}} +e^\frac{6i\pi}{7}+ e^{-\frac{6i\pi}{7}} + 1\right) - \frac{1}{2} \end{align*} In the brackets we have the sum of the roots of the polynomial $x^7 - 1 = 0$ . These sum to $0$ by Vieta’s formulas , and the desired identity follows. See Roots of unity if you have not seen this technique.
Going back to the question: \[7 \cdot 2^4 \sin^4 \frac{\pi}{7} + 7 \cdot 2^4 \sin^4 \frac{2 \pi}{7} + 7 \cdot 2^4 \sin^4 \frac{3 \pi}{7} = 7 \cdot 2^4 \left(\sin^4 \frac{\pi}{7} + \sin^4 \frac{2 \pi}{7} + \sin^4 \frac{3 \pi}{7}\right) = 7 \cdot 2^4 \cdot \frac{21}{16} = \boxed{147}.\] ~obscene_kangaroo

Answer: C 147
Problem: The figure below depicts a regular $7$ -gon inscribed in a unit circle. [asy]         import geometry; unitsize(3cm); draw(circle((0,0),1),linewidth(1.5)); for (int i = 0; i < 7; ++i) {   for (int j = 0; j < i; ++j) {     draw(dir(i * 360/7) -- dir(j * 360/7),linewidth(1.5));   } } for(int i = 0; i < 7; ++i) {    dot(dir(i * 360/7),5+black); } [/asy] What is the sum of the $4$ th powers of the lengths of all $21$ of its edges and diagonals?
$\textbf{(A) }49 \qquad \textbf{(B) }98 \qquad \textbf{(C) }147 \qquad \textbf{(D) }168 \qquad \textbf{(E) }196$

Solution: This solution follows the same steps as the trigonometry solutions (Solutions 2 and 3), except it gives an alternate way to prove the statement below true without complex numbers:
\[\cos \frac{2\pi}{7} + \cos \frac{4\pi}{7} + \cos \frac{6\pi}{7} = -\frac{1}{2}\]
\begin{align*} S &= \cos \frac{2\pi}{7} + \cos \frac{4\pi}{7} + \cos \frac{6\pi}{7}, \\ S^2 &= \cos^2 \frac{2\pi}{7} + \cos^2 \frac{4\pi}{7} + \cos^2 \frac{6\pi}{7} + 2\cos \frac{2\pi}{7}\cos \frac{4\pi}{7} + 2\cos \frac{2\pi}{7}\cos \frac{6\pi}{7} + 2\cos \frac{4\pi}{7}\cos \frac{6\pi}{7} \\ &= \left(\frac{1+ \cos \frac{4\pi}{7}}{2}\right) + \left(\frac{1+ \cos \frac{8\pi}{7}}{2}\right) + \left(\frac{1+ \cos \frac{12\pi}{7}}{2}\right) + 2\cos \frac{2\pi}{7}\cos \frac{4\pi}{7} + 2\cos \frac{2\pi}{7}\cos \frac{6\pi}{7} + 2\cos \frac{4\pi}{7}\cos \frac{6\pi}{7} \\ &= \frac{1}{2}(3 + S)  + \left(\cos \frac{6\pi}{7} + \cos \frac{2\pi}{7}\right) + \left(\cos \frac{8\pi}{7} + \cos \frac{4\pi}{7}\right) + \left(\cos \frac{10\pi}{7} + \cos \frac{2\pi}{7}\right) \\ &= \frac{1}{2}(3 + S) + 2\cos \frac{2\pi}{7} + 2\cos \frac{4\pi}{7} + 2\cos \frac{6\pi}{7}\\ &= \frac{1}{2}(3 + S) + 2S. \\ \end{align*} We end up with \[2S^2 - 5S - 3 = 0.\] Using the quadratic formula, we find the solutions for $S$ to be $-\frac{1}{2}$ and $3$ . Because $3$ is impossible, $S = -\frac{1}{2}$ .  
With this result, following similar to steps to Solutions 2 and 3 will get $\boxed{147}$

Answer: C 147
Problem: The figure below depicts a regular $7$ -gon inscribed in a unit circle. [asy]         import geometry; unitsize(3cm); draw(circle((0,0),1),linewidth(1.5)); for (int i = 0; i < 7; ++i) {   for (int j = 0; j < i; ++j) {     draw(dir(i * 360/7) -- dir(j * 360/7),linewidth(1.5));   } } for(int i = 0; i < 7; ++i) {    dot(dir(i * 360/7),5+black); } [/asy] What is the sum of the $4$ th powers of the lengths of all $21$ of its edges and diagonals?
$\textbf{(A) }49 \qquad \textbf{(B) }98 \qquad \textbf{(C) }147 \qquad \textbf{(D) }168 \qquad \textbf{(E) }196$

Solution: Let $x$ $y$ , and $z$ be the lengths of the chords with arcs $\frac{2\pi}{7}$ $\frac{4\pi}{7}$ and $\frac{6\pi}{7}$ respectively.
Then by the law of cosines we get: \begin{align*} x^2 &= 2\left(1-\cos\frac{2\pi}{7}\right), \\ y^2 &= 2\left(1-\cos\frac{4\pi}{7}\right), \\ z^2 &= 2\left(1-\cos\frac{6\pi}{7}\right). \end{align*} The answer is then just $7(x^4+y^4+z^4)$ (since there's $7$ of each diagonal/side), obtained by summing the squares of the above equations and then multiplying by $7$ \begin{align*} & \quad7\cdot2^2\left( \left(1-\cos\frac{2\pi}{7}\right)^2 + \left(1-\cos\frac{4\pi}{7}\right)^2 + \left(1-\cos\frac{6\pi}{7}\right)^2 \right) \\ &= 7\cdot4\left( \left(1-2\cos\frac{2\pi}{7}+\cos^2\frac{2\pi}{7}\right) + \left(1-2\cos\frac{4\pi}{7}+\cos^2\frac{4\pi}{7}\right) + \left(1-2\cos\frac{6\pi}{7}+\cos^2\frac{6\pi}{7}\right) \right) \\ &= 7\cdot4\left(3-2\left(\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}\right) + \left(\cos^2\frac{2\pi}{7}+\cos^2\frac{4\pi}{7}+\cos^2\frac{6\pi}{7}\right) \right) \\ &= 7\cdot4\left(3-2\left(\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}\right) + \frac12 \left(1+\cos\frac{4\pi}{7}+1+\cos\frac{8\pi}{7}+1+\cos\frac{12\pi}{7}\right) \right) \\ &= 7\cdot4\left(\frac{9}{2}-2\left(\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}\right) + \frac12 \left(\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}+\cos\frac{2\pi}{7}\right) \right) \\ &= 7\cdot4\left(\frac{9}{2}-\frac{3}{2}\left(\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}\right) \right) \\ &= 7\cdot4\left(\frac{9}{2}-\frac{3}{2}\left(-\frac{1}{2}\right)\right) \\ &= \boxed{147}

Answer: C 147
Problem: The figure below depicts a regular $7$ -gon inscribed in a unit circle. [asy]         import geometry; unitsize(3cm); draw(circle((0,0),1),linewidth(1.5)); for (int i = 0; i < 7; ++i) {   for (int j = 0; j < i; ++j) {     draw(dir(i * 360/7) -- dir(j * 360/7),linewidth(1.5));   } } for(int i = 0; i < 7; ++i) {    dot(dir(i * 360/7),5+black); } [/asy] What is the sum of the $4$ th powers of the lengths of all $21$ of its edges and diagonals?
$\textbf{(A) }49 \qquad \textbf{(B) }98 \qquad \textbf{(C) }147 \qquad \textbf{(D) }168 \qquad \textbf{(E) }196$

Solution: Hope you had a ruler handy! This problem can be done with a ruler and basic estimation.
First, measuring the radius of the circle obtains $2.9$ cm (when done on the paper version). Thus, any other measurement we get for the sides/diagonals should be divided by $2.9$
Measuring the sides of the circle gets $2.5$ cm. The shorter diagonals are $4.5$ cm, and the longest diagonals measure $5.6$ cm. Thus, we'd like to estimate \[7\left(\frac{2.5}{2.9}\right)^4 + 7\left(\frac{4.5}{2.9}\right)^4 + 7\left(\frac{5.6}{2.9}\right)^4.\]
We know $\left(\frac{2.5}{2.9}\right)^4$ is slightly less than $1.$ Let's approximate it as 1 for now. Thus, $7\left(\frac{2.5}{2.9}\right)^4 \approx 7.$
Next, $\left(\frac{4.5}{2.9}\right)^4$ is slightly more than $\left(\frac{4.5}{3}\right)^4.$ We know $\left(\frac{4.5}{3}\right)^4 = 1.5^4 = \frac{81}{16},$ slightly more than $5,$ so we can approximate $\left(\frac{4.5}{2.9}\right)^4$ as $5.5.$ Thus, $7\left(\frac{4.5}{2.9}\right)^4 \approx 38.5.$
Finally, $\left(\frac{5.6}{2.9}\right)^4$ is slightly less than $\left(\frac{5.6}{2.8}\right)^4 = 2^4 = 16.$ We say it's around $15,$ so then $7\left(\frac{5.6}{2.9}\right)^4 \approx 105.$
Adding what we have, we get $105 + 38.5 + 1 = 144.5$ as our estimate. We see $\boxed{147}$ is very close to our estimate, so we have successfully finished the problem.

Answer: C 147
Problem: The figure below depicts a regular $7$ -gon inscribed in a unit circle. [asy]         import geometry; unitsize(3cm); draw(circle((0,0),1),linewidth(1.5)); for (int i = 0; i < 7; ++i) {   for (int j = 0; j < i; ++j) {     draw(dir(i * 360/7) -- dir(j * 360/7),linewidth(1.5));   } } for(int i = 0; i < 7; ++i) {    dot(dir(i * 360/7),5+black); } [/asy] What is the sum of the $4$ th powers of the lengths of all $21$ of its edges and diagonals?
$\textbf{(A) }49 \qquad \textbf{(B) }98 \qquad \textbf{(C) }147 \qquad \textbf{(D) }168 \qquad \textbf{(E) }196$

Solution: First, we put the figure in the coordinate plane with the center of the circle at the origin and a vertex on the positive x-axis. Thus, the coordinates of the vertices will be the terminal points of integer multiples of the angle $\frac{2\pi}{7},$ which are \[\left(\cos\dfrac{2\pi n}{7}, \sin\dfrac{2\pi n}{7}\right)\] for integer $n.$ Then, we notice there are three types of diagonals: the ones with chords of arcs $\dfrac{2\pi}{7}, \dfrac{4\pi}{7},$ and $\dfrac{6\pi}{7}.$ We notive there are here are $7$ of each type of diagonal. Then, we use the pythagorean theorem to find the distance from $\left(\cos\frac{2\pi n}{7}, \sin\frac{2\pi n}{7}\right)$ to $(1, 0)$ \[\left(\sqrt{\left(\cos\frac{2\pi n}{7}-1\right)^2+\sin^2\frac{2\pi n}{7}}\right)^4=\left(\sqrt{\cos^2\frac{2\pi n}{7}+\sin^2\frac{2\pi n}{7}-2\cos\frac{2\pi n}{7}+1}\right)^4\] \[=\left(\sqrt{2-2\cos\frac{2\pi n}{7}}\right)^4\] \[=4\cos^2\frac{2\pi n}{7}-8\cos\frac{2\pi n}{7}+4.\] By the cosine double angle identity, $\cos{2\theta}=2\cos^2\theta-1.$ This means that $2\cos^2\theta=\cos{2\theta}+1.$ Substituting this in, \[4\cos^2\frac{2\pi n}{7}-8\cos\frac{2\pi n}{7}+4=2\left(\cos\frac{4\pi n}{7}+1\right)-8\cos\frac{2\pi n}{7}+4=2\cos\frac{4\pi n}{7}-8\cos\frac{2\pi n}{7}+6.\] Summing this up for $n=1,2,3,$ \[2\cos\frac{4\pi}{7}-8\cos\frac{2\pi}{7}+6+2\cos\frac{8\pi}{7}-8\cos\frac{4\pi}{7}+6+2\cos\frac{12\pi}{7}-8\cos\frac{6\pi}{7}+6\] \[=2\left(\cos\frac{4\pi}{7}+\cos\frac{8\pi}{7}+\cos\frac{12\pi}{7}\right)-8\left(\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}\right)+18\] \[=2\left(\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}+\cos\frac{2\pi}{7}\right)-8\left(\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}\right)+18\] \[=2(-0.5)-8(-0.5)+18=21.\] (These equalities are based on $\cos\theta=\cos(2\pi-\theta)$ and $\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}=-\frac{1}{2}.$ ) (For explanation of this see supplement .)
Finally, because there are $7$ of each type of diagonal, the answer is $7\cdot 21=\boxed{147}.$

Answer:  147
Problem: The figure below depicts a regular $7$ -gon inscribed in a unit circle. [asy]         import geometry; unitsize(3cm); draw(circle((0,0),1),linewidth(1.5)); for (int i = 0; i < 7; ++i) {   for (int j = 0; j < i; ++j) {     draw(dir(i * 360/7) -- dir(j * 360/7),linewidth(1.5));   } } for(int i = 0; i < 7; ++i) {    dot(dir(i * 360/7),5+black); } [/asy] What is the sum of the $4$ th powers of the lengths of all $21$ of its edges and diagonals?
$\textbf{(A) }49 \qquad \textbf{(B) }98 \qquad \textbf{(C) }147 \qquad \textbf{(D) }168 \qquad \textbf{(E) }196$

Solution: Place the figure on the complex plane and let $w = e^{\frac{2\pi}{7}}$ (a $7$ th root of unity). The vertices of the $7$ -gon are $w^0,w^1,w^2,\dots,w^6$ . We wish to find \[\sum_{i=0}^5\sum_{j=i+1}^6\lvert w^i-w^j\rvert^4 = \frac{1}{2}\sum_{i=0}^6\sum_{j=0}^6\lvert w^i-w^j\rvert^4.\] The second expression is more convenient to work with. The factor of $\tfrac{1}{2}$ is because it double-counts each edge (and includes when $i=j$ , but the length is $0$ so it doesn't matter).
Recall the identity $|z|^2 = z\overline{z}$ . Since $|w| = 1$ $\overline{w} = w^{-1}$ , and \begin{align*}     \lvert w^i-w^j\rvert^4 &= ((w^i-w^j)(w^{-i}-w^{-j}))^2 \\     &= (2-w^{i-j}-w^{j-i})^2 \\     &= 4+w^{2i-2j}+w^{2j-2i}+2(-2w^{i-j}-2w^{j-i}+1) \\     &= 6-4w^{i-j}-4w^{j-i}+w^{2i-2j}+w^{2j-2i}. \end{align*}
But notice that, for a fixed value of $i$ , over all values of $j$ from $0$ to $6$ , by properties of modular arithmetic with $w^7 = w^0$ (or expanding),
each of the $w^\bullet$ -terms takes on every value in $w^0,w^1,w^2,\dots,w^6$ exactly once. Since $\sum_{j=0}^6 w^j = 0$ \begin{align*}     \frac{1}{2}\sum_{i=0}^6\sum_{j=0}^6\lvert w^i-w^j\rvert^4 &= \frac{1}{2}\sum_{i=0}^6\sum_{j=0}^6(6-4w^{i-j}-4w^{j-i}+w^{2i-2j}+w^{2j-2i}) \\     &= \frac{1}{2}\sum_{i=0}^6\left(\sum_{j=0}^6 6-4\sum_{j=0}^6 w^{i-j}-4\sum_{j=0}^6 w^{j-i}+\sum_{j=0}^6 w^{2i-2j}+\sum_{j=0}^6 w^{2j-2i}\right) \\     &= \frac{1}{2}\sum_{i=0}^6\left(\sum_{j=0}^6 6-4\sum_{j=0}^6 w^j-4\sum_{j=0}^6 w^j+\sum_{j=0}^6 w^j+\sum_{j=0}^6 w^j\right) \\     &= \frac{1}{2}\sum_{i=0}^6\sum_{j=0}^6 6 \\     &= \frac{1}{2}\cdot7\cdot7\cdot6 \\     &= \boxed{147}

Answer: C 147
Problem: The figure below depicts a regular $7$ -gon inscribed in a unit circle. [asy]         import geometry; unitsize(3cm); draw(circle((0,0),1),linewidth(1.5)); for (int i = 0; i < 7; ++i) {   for (int j = 0; j < i; ++j) {     draw(dir(i * 360/7) -- dir(j * 360/7),linewidth(1.5));   } } for(int i = 0; i < 7; ++i) {    dot(dir(i * 360/7),5+black); } [/asy] What is the sum of the $4$ th powers of the lengths of all $21$ of its edges and diagonals?
$\textbf{(A) }49 \qquad \textbf{(B) }98 \qquad \textbf{(C) }147 \qquad \textbf{(D) }168 \qquad \textbf{(E) }196$

Solution: This is how I solve this problem:
It's easy to solve for $3$ -gon, $4$ -gon, and $6$ -gon inscribed in a unit circle. (Okay, it's just the weird names for triangle, square, and hexagon)
For $3$ -gon, the sum is equal to $3$ times the $4$ th power of an edge. Thus, \[S_3=3\,\cdot\,\left(\sqrt{3}\right)^4=27.\]
For $4$ -gon, the sum is equal to $4$ times the $4$ th power of an edge, and $2$ times the $4$ th power of the diagonal. Thus, \[S_4=4\,\cdot\,\left(\sqrt{2}\right)^4+2\,\cdot\,\left(2\right)^4=48.\]
For $6$ -gon, the sum is equal to $6$ times the $4$ th power of an edge, $6$ times the $4$ th power of the short diagonal, and $3$ times the $4$ th power of the long diagonal. Thus, \[S_6=6\,\cdot\,\left(1\right)^4+6\,\cdot\,\left(\sqrt{3}\right)^4+3\,\cdot\,\left(2\right)^4=108.\]
Then, I quickly noticed that $27=3\,\cdot\,3^2$ $48=3\,\cdot\,4^2$ , and $108=3\,\cdot\,6^2$ . So reasonably, it will work out this formula, $S_n=3n^2$ . (This step is purely out of guessing, maybe have a look at Solution 8 for more info...)
By inductive reasoning, we got $S_7=3\,\cdot\,7^2=\boxed{147}$

Answer: C 147
Problem: Let $M$ be the least common multiple of all the integers $10$ through $30,$ inclusive. Let $N$ be the least common multiple of $M,32,33,34,35,36,37,38,39,$ and $40.$ What is the value of $\frac{N}{M}?$
$\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 2 \qquad\textbf{(C)}\ 37 \qquad\textbf{(D)}\ 74 \qquad\textbf{(E)}\ 2886$

Solution: By the definition of least common mutiple, we take the greatest powers of the prime numbers of the prime factorization of all the numbers, that we are taking the $\text{lcm}$ of. In this case, \[M = 2^4 \cdot 3^3 \cdot 5^2 \cdot 7 \cdot 11 \cdot 13 \cdot 17 \cdot 19 \cdot 23 \cdot 29.\] Now, using the same logic, we find that \[N = M \cdot 2 \cdot 37,\] because we have an extra power of $2$ and an extra power of $37.$ Thus, $\frac{N}{M} = 2\cdot 37 = \boxed{74}.$

Answer: D 74
Problem: A right rectangular prism whose surface area and volume are numerically equal has edge lengths $\log_{2}x, \log_{3}x,$ and $\log_{4}x.$ What is $x?$
$\textbf{(A)}\ 2\sqrt{6} \qquad\textbf{(B)}\ 6\sqrt{6} \qquad\textbf{(C)}\ 24 \qquad\textbf{(D)}\ 48 \qquad\textbf{(E)}\ 576$

Solution: The surface area of this right rectangular prism is $2(\log_{2}x\log_{3}x+\log_{2}x\log_{4}x+\log_{3}x\log_{4}x).$
The volume of this right rectangular prism is $\log_{2}x\log_{3}x\log_{4}x.$
Equating the numerical values of the surface area and the volume, we have \[2(\log_{2}x\log_{3}x+\log_{2}x\log_{4}x+\log_{3}x\log_{4}x)=\log_{2}x\log_{3}x\log_{4}x.\] Dividing both sides by $\log_{2}x\log_{3}x\log_{4}x,$ we get \[2\left(\frac{1}{\log_{4}x}+\frac{1}{\log_{3}x}+\frac{1}{\log_{2}x}\right)=1. \hspace{15mm} (\bigstar)\] Recall that $\log_{b}a=\frac{1}{\log_{a}b}$ and $\log_{b}\left(a^n\right)=n\log_{b}a,$ so we rewrite $(\bigstar)$ as \begin{align*} 2(\log_{x}4+\log_{x}3+\log_{x}2)&=1 \\ 2\log_{x}24&=1 \\ \log_{x}576&=1 \\ x&=\boxed{576} ~MRENTHUSIASM

Answer: E 576
Problem: What is the number of terms with rational coefficients among the $1001$ terms in the expansion of $\left(x\sqrt[3]{2}+y\sqrt{3}\right)^{1000}?$
$\textbf{(A)}\ 0 \qquad\textbf{(B)}\ 166 \qquad\textbf{(C)}\ 167 \qquad\textbf{(D)}\ 500 \qquad\textbf{(E)}\ 501$

Solution: By the Binomial Theorem, each term in the expansion is of the form \[\binom{1000}{k}\left(x\sqrt[3]{2}\right)^k\left(y\sqrt{3}\right)^{1000-k}=\binom{1000}{k}2^{\frac k3}3^{\frac{1000-k}{2}}x^k y^{1000-k},\] where $k\in\{0,1,2,\ldots,1000\}.$
This problem is equivalent to counting the values of $k$ such that both $\frac k3$ and $\frac{1000-k}{2}$ are integers. Note that $k$ must be a multiple of $3$ and a multiple of $2,$ so $k$ must be a multiple of $6.$ There are $\boxed{167}$ such values of $k:$ \[6(0), 6(1), 6(2), \ldots, 6(166).\]

Answer: C 167
Problem: Recall that the conjugate of the complex number $w = a + bi$ , where $a$ and $b$ are real numbers and $i = \sqrt{-1}$ , is the complex number $\overline{w} = a - bi$ . For any complex number $z$ , let $f(z) = 4i\hspace{1pt}\overline{z}$ . The polynomial \[P(z) = z^4 + 4z^3 + 3z^2 + 2z + 1\] has four complex roots: $z_1$ $z_2$ $z_3$ , and $z_4$ . Let \[Q(z) = z^4 + Az^3 + Bz^2 + Cz + D\] be the polynomial whose roots are $f(z_1)$ $f(z_2)$ $f(z_3)$ , and $f(z_4)$ , where the coefficients $A,$ $B,$ $C,$ and $D$ are complex numbers. What is $B + D?$
$(\textbf{A})\: {-}304\qquad(\textbf{B}) \: {-}208\qquad(\textbf{C}) \: 12i\qquad(\textbf{D}) \: 208\qquad(\textbf{E}) \: 304$

Solution: By Vieta's formulas, $z_1z_2+z_1z_3+\dots+z_3z_4=3$ , and $B=(4i)^2\left(\overline{z}_1\,\overline{z}_2+\overline{z}_1\,\overline{z}_3+\dots+\overline{z}_3\,\overline{z}_4\right).$
Since $\overline{a}\cdot\overline{b}=\overline{ab},$ \[B=(4i)^2\left(\overline{z_1z_2}+\overline{z_1z_3}+\overline{z_1z_4}+\overline{z_2z_3}+\overline{z_2z_4}+\overline{z_3z_4}\right).\] Since $\overline{a}+\overline{b}=\overline{a+b},$ \[B=(4i)^2\left(\overline{z_1z_2+z_1z_3+\dots+z_3z_4}\right)=-16(\overline{3})=-48\]
Also, $z_1z_2z_3z_4=1,$ and \[D=(4i)^4\left(\overline{z}_1\,\overline{z}_2\,\overline{z}_3\,\overline{z}_4\right)=256\left(\overline{z_1z_2z_3z_4}\right)=256(\overline{1})=256.\]
Our answer is $B+D=256-48=\boxed{208}.$

Answer: D 208
Problem: An organization has $30$ employees, $20$ of whom have a brand A computer while the other $10$ have a brand B computer. For security, the computers can only be connected to each other and only by cables. The cables can only connect a brand A computer to a brand B computer. Employees can communicate with each other if their computers are directly connected by a cable or by relaying messages through a series of connected computers. Initially, no computer is connected to any other. A technician arbitrarily selects one computer of each brand and installs a cable between them, provided there is not already a cable between that pair. The technician stops once every employee can communicate with each other. What is the maximum possible number of cables used?
$\textbf{(A)}\ 190  \qquad\textbf{(B)}\  191 \qquad\textbf{(C)}\  192 \qquad\textbf{(D)}\  195 \qquad\textbf{(E)}\ 196$

Solution: We claim that to maximize the number of cables used, we isolate one computer and connect all cables for the remaining $29$ computers, then connect one more cable for the isolated computer.
If a brand A computer is isolated, then the technician can use at most $19\cdot10+1=191$ cables. If a brand B computer is isolated instead, then the technician can use at most $20\cdot9+1=181$ cables. Therefore, the answer is $\boxed{191}.$

Answer: B 191