AMC12 2010 A

$20-2010+201+2010-201+20=20+20=\boxed{\textbf{(C)}\,40}$.

It is easy to see that the ferry boat takes $6$ trips total. The total number of people taken to the island is \begin{align*}&100+99+98+97+96+95\\ &=6(100)-(1+2+3+4+5)\\ &=600 - 15\\ &=\boxed{585\ \textbf{(A)}}\end{align*}

If we shift $A$ to coincide with $E$, and add new horizontal lines to divide $EFGH$ into five equal parts: This helps us to see that $AD=a/5$ and $AB=2a$, where $a=EF$. Hence $\dfrac{AB}{AD}=\dfrac{2a}{a/5}=10$.

$x$ is negative, so we can just place a negative value into each expression and find the one that is positive. Suppose we use $-1$. $\textbf{(A)} \Rightarrow \frac{-1}{|-1|} = -1$ $\textbf{(B)} \Rightarrow -(-1)^2 = -1$ $\textbf{(C)} \Rightarrow -2^{(-1)} = -\frac{1}{2}$ $\textbf{(D)} \Rightarrow -(-1)^{(-1)} = 1$ $\textbf{(E)} \Rightarrow \sqrt[3]{-1} = -1$ Obviously only $\boxed{\textbf{(D)}}$ is positive.

Let $k$ be the number of points Chelsea currently has. In order to guarantee victory, we must consider the possibility that the opponent scores the maximum amount of points by getting only bullseyes. \begin{align*}k+ 10n + 4(50-n) &> (k-50) + 50\cdot{10}\\ 6n &> 250\end{align*} The lowest integer value that satisfies the inequality is $\boxed{42\ \textbf{(C)}}$.

$x$ is at most $999$, so $x+32$ is at most $1031$. The minimum value of $x+32$ is $1000$. However, the only palindrome between $1000$ and $1032$ is $1001$, which means that $x+32$ must be $1001$. It follows that $x$ is $969$, so the sum of the digits is $\boxed{\textbf{(E)}\ 24}$.

The water tower holds $\frac{100000}{0.1} = 1000000$ times more water than Logan's miniature. The volume of a sphere is: $V=\dfrac{4}{3}\pi r^3$. Since we are comparing the heights (m), we should compare the radii (m) to find the ratio. Since, the radius is cubed, Logan should make his tower $\sqrt[3]{1000000} = 100$ times shorter than the actual tower. This is $\frac{40}{100} = \boxed{0.4}$ meters high, or choice $\textbf{(C)}$. Note: The fact that $1\text{ L}=1000\text{ cm}^3$ doesn't matter since only the ratios are important.

Error creating thumbnail: Unable to save thumbnail to destination Let $\angle BAE = \angle ACD = x$. \begin{align*}\angle BCD &= \angle AEC = 60^\circ\\ \angle EAC + \angle FCA + \angle ECF + \angle AEC &= \angle EAC + x + 60^\circ + 60^\circ = 180^\circ\\ \angle EAC &= 60^\circ - x\\ \angle BAC &= \angle EAC + \angle BAE = 60^\circ - x + x = 60^\circ\end{align*} Since $\frac{AC}{AB} = \frac{1}{2}$ and the angle between the hypotenuse and the shorter side is $60^\circ$, triangle $ABC$ is a $30-60-90$ triangle, so $\angle BCA = \boxed{90^\circ\,\textbf{(C)}}$.

Imagine making the cuts one at a time. The first cut removes a box $2\times 2\times 3$. The second cut removes two boxes, each of dimensions $2\times 2\times 0.5$, and the third cut does the same as the second cut, on the last two faces. Hence the total volume of all cuts is $12 + 4 + 4 = 20$. Therefore the volume of the rest of the cube is $3^3 - 20 = 27 - 20 = \boxed{7\ \textbf{(A)}}$. We can visualize it has a $3$ by $3$ cube with a hollow $2$ by $2$ center, along with 6 "windows" cut out too. The cube formula gives the volume of the $3$ by $3$ cube as $3^3 = 27$ and $2^3 = 8.$ Additionally, the area of the "windows" is the 2x2 cut along with a thickness of $\frac{(3-2)}{2} = \frac{1}{2}.$ The volume of all 6 windows is $6(\frac{1}{2})(2)(2) = 12.$ This gives us our answer of $27 - 8 - 12 = \boxed{7\ \textbf{(A)}}.$ We can use Principle of Inclusion-Exclusion (PIE) to find the final volume of the cube. There are 3 "cuts" through the cube that go from one end to the other. Each of these "cuts" has $2 \times 2 \times 3=12$ cubic inches. However, we can not just sum their volumes, as the central $2\times 2\times 2$ cube is included in each of these three cuts. To get the correct result, we can take the sum of the volumes of the three cuts, and subtract the volume of the central cube twice. Hence the total volume of the cuts is $3(2 \times 2 \times 3) - 2(2\times 2\times 2) = 36 - 16 = 20$. Therefore the volume of the rest of the cube is $3^3 - 20 = 27 - 20 = \boxed{7\ \textbf{(A)}}$. We can visualize the final figure and see a cubic frame. We can find the volume of the figure by adding up the volumes of the edges and corners. Each edge can be seen as a $2\times 0.5\times 0.5$ box, and each corner can be seen as a $0.5\times 0.5\times 0.5$ box. $12\cdot{\frac{1}{2}} + 8\cdot{\frac{1}{8}} = 6+1 = \boxed{7\ \textbf{(A)}}$.

$3p-q$ and $3p+q$ are consecutive terms, so the common difference is $(3p+q)-(3p-q) = 2q$. \begin{align*}p+2q &= 9\\ 9+2q &= 3p-q\\ q&=2\\ p&=5\end{align*} The common difference is $4$. The first term is $5$ and the $2010^\text{th}$ term is \[5+4(2009) = \boxed{\textbf{(A) }8041}\]

This problem is quickly solved with knowledge of the laws of exponents and logarithms. \begin{align*} 7^{x+7} &= 8^x \\ 7^x*7^7 &= 8^x \\ \left(\frac{8}{7}\right)^x &= 7^7 \\ x &= \log_{8/7}7^7 \end{align*} Since we are looking for the base of the logarithm, our answer is $\boxed{\textbf{(C)}\ \frac{8}{7}}$.

Start with Brian. If he is a toad, he tells the truth, hence Mike is a frog. If Brian is a frog, he lies, hence Mike is a frog, too. Thus Mike must be a frog. As Mike is a frog, his statement is false, hence there is at most one toad. As there is at most one toad, at least one of Chris and LeRoy is a frog. But then the other one tells the truth, and therefore is a toad. Hence we must have one toad and $\boxed{\textbf{(D)}\ 3}$ frogs.

The image below shows the two curves for $k=4$. The blue curve is $x^2+y^2=k^2$, which is clearly a circle with radius $k$, and the red curve is a part of the curve $xy=k$. In the special case $k=0$ the blue curve is just the point $(0,0)$, and as $0\cdot 0=0$, this point is on the red curve as well, hence they intersect. The case $k<0$ is symmetric to $k>0$: the blue curve remains the same and the red curve is flipped according to the $x$ axis. Hence we just need to focus on $k>0$. Clearly, on the red curve there will always be points arbitrarily far from the origin: for example, as $x$ approaches 0, $y$ approaches $\infty$. Hence the red curve intersects the blue one if and only if it contains a point whose distance from the origin is at most $k$. At this point we can guess that on the red curve the point where $x=y$ is always closest to the origin, and skip the rest of this solution. For an exact solution, fix $k$ and consider any point $(x,y)$ on the red curve. Its distance from the origin is $\sqrt{ x^2 + (k/x)^2 }$. To minimize this distance, it is enough to minimize $x^2 + (k/x)^2$. By the Arithmetic Mean-Geometric Mean Inequality we get that this value is at least $2k$, and that equality holds whenever $x^2 = (k/x)^2$, i.e., $x=\pm\sqrt k$. Now recall that the red curve intersects the blue one if and only if its closest point is at most $k$ from the origin. We just computed that the distance between the origin and the closest point on the red curve is $\sqrt{2k}$. Therefore, we want to find all positive integers $k$ such that $\sqrt{2k} > k$. Clearly the only such integer is $k=1$, hence the two curves are only disjoint for $k=1$ and $k=-1$. This is a total of $\boxed{2\ \textbf{(C)}}$ values.

By the Angle Bisector Theorem, we know that $\frac{AB}{BC} = \frac{3}{8}$. If we use the lowest possible integer values for $AB$ and $BC$ (the lengths of $AD$ and $DC$, respectively), then $AB + BC = AD + DC = AC$, contradicting the Triangle Inequality. If we use the next lowest values ($AB = 6$ and $BC = 16$), the Triangle Inequality is satisfied. Therefore, our answer is $6 + 16 + 3 + 8 = \boxed{33}$, or choice $\textbf{(B)}$.

Let $x$ be the probability of flipping heads. It follows that the probability of flipping tails is $1-x$. The probability of flipping $2$ heads and $2$ tails is equal to the number of ways to flip it times the product of the probability of flipping each coin. \begin{align*}{4 \choose 2}x^2(1-x)^2 &= \frac{1}{6}\\ 6x^2(1-x)^2 &= \frac{1}{6}\\ x^2(1-x)^2 &= \frac{1}{36}\\ x(1-x) &= \pm\frac{1}{6}\end{align*} As for the desired probability $x$ both $x$ and $1-x$ are nonnegative, we only need to consider the positive root, hence \begin{align*}x(1-x) &= \frac{1}{6}\\ 6x^2-6x+1&=0\end{align*} Applying the quadratic formula we get that the roots of this equation are $\frac{3\pm\sqrt{3}}{6}$. As the probability of heads is less than $\frac{1}{2}$, we get that the answer is $\boxed{\textbf{(D)}\ \frac{3-\sqrt{3}}{6}}$.

We can solve this by breaking the problem down into $2$ cases and adding up the probabilities. Case $1$: Bernardo picks $9$. If Bernardo picks a $9$ then it is guaranteed that his number will be larger than Silvia's. The probability that he will pick a $9$ is $\frac{1 \cdot \binom{8}{2}}{\binom{9}{3}} = \frac{\frac{8\cdot7}{2}}{\frac{9\cdot8\cdot7}{3\cdot2\cdot1}}=\frac{1}{3}$. Case $2$: Bernardo does not pick $9$. Since the chance of Bernardo picking $9$ is $\frac{1}{3}$, the probability of not picking $9$ is $\frac{2}{3}$. If Bernardo does not pick $9$, then he can pick any number from $1$ to $8$. Since Bernardo is picking from the same set of numbers as Silvia, the probability that Bernardo's number is larger is equal to the probability that Silvia's number is larger. Ignoring the $9$ for now, the probability that they will pick the same number is the number of ways to pick Bernardo's 3 numbers divided by the number of ways to pick any 3 numbers. We get this probability to be $\frac{3!}{8\cdot{7}\cdot{6}} = \frac{1}{56}$ The probability of Bernardo's number being greater is \[\frac{1-\frac{1}{56}}{2} = \frac{55}{112}\] Factoring the fact that Bernardo could've picked a $9$ but didn't: \[\frac{2}{3}\cdot{\frac{55}{112}} = \frac{55}{168}\] Adding up the two cases we get $\frac{1}{3}+\frac{55}{168} = \boxed{\frac{37}{56}\ \textbf{(B)}}$ We have for case $1$: $\frac{1 \cdot \binom{8}{2}}{\binom{9}{3}}$ since $1$ is the number of ways to pick 9 and $\binom{8}{2}$ is the number of ways to pick the other 2 numbers. $\binom{9}{3}$ means to choose 3 numbers from 9. A common pitfall is saying that the probability of picking the same number is $\frac{8*7*6}{(8*7*6)^2}$. This actually undercounts. Note that picking $3,7,6$ will lead to the same end result as picking $7,3,6$ (order does not matter, since it will be descending no matter what). Thus, we multiply by $3!$ :)

It is clear that $\triangle ACE$ is an equilateral triangle. From the Law of Cosines on $\triangle ABC$, we get that $AC^2 = r^2+1^2-2r\cos{\frac{2\pi}{3}} = r^2+r+1$. Therefore, the area of $\triangle ACE$ is $\frac{\sqrt{3}}{4}(r^2+r+1)$ by area of an equilateral triangle. If we extend $BC$, $DE$ and $FA$ so that $FA$ and $BC$ meet at $X$, $BC$ and $DE$ meet at $Y$, and $DE$ and $FA$ meet at $Z$, we find that hexagon $ABCDEF$ is formed by taking equilateral triangle $XYZ$ of side length $r+2$ and removing three equilateral triangles, $ABX$, $CDY$ and $EFZ$, of side length $1$. The area of $ABCDEF$ is therefore $\frac{\sqrt{3}}{4}(r+2)^2-\frac{3\sqrt{3}}{4} = \frac{\sqrt{3}}{4}(r^2+4r+1)$. Based on the initial conditions, \[\frac{\sqrt{3}}{4}(r^2+r+1) = \frac{7}{10}\left(\frac{\sqrt{3}}{4}\right)(r^2+4r+1)\] Simplifying this gives us $r^2-6r+1 = 0$. By Vieta's Formulas we know that the sum of the possible value of $r$ is $\boxed{\textbf{(E)}\ 6}$.

Each path must go through either the second or the fourth quadrant. Each path that goes through the second quadrant must pass through exactly one of the points $(-4,4)$, $(-3,3)$, and $(-2,2)$. There is $1$ path of the first kind, ${8\choose 1}^2=64$ paths of the second kind, and ${8\choose 2}^2=28^2=784$ paths of the third type. Each path that goes through the fourth quadrant must pass through exactly one of the points $(4,-4)$, $(3,-3)$, and $(2,-2)$. Again, there is $1$ path of the first kind, ${8\choose 1}^2=64$ paths of the second kind, and ${8\choose 2}^2=28^2=784$ paths of the third type. Hence the total number of paths is $2(1+64+784) = \boxed{1698}$.

The probability of drawing a white marble from box $k$ is $\frac{k}{k + 1}$, and the probability of drawing a red marble from box $k$ is $\frac{1}{k+1}$. To stop after drawing $n$ marbles, we must draw a white marble from boxes $1, 2, \ldots, n-1,$ and draw a red marble from box $n.$ Thus, \[P(n) = \left(\frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4} \cdots \frac {n - 1}{n}\right) \cdot \frac{1}{n +1} = \frac{1}{n (n + 1)}.\] So, we must have $\frac{1}{n(n + 1)} < \frac{1}{2010}$ or $n(n+1) > 2010.$ Since $n(n+1)$ increases as $n$ increases, we can simply test values of $n$; after some trial and error, we get that the minimum value of $n$ is $\boxed{\textbf{(A) }45}$, since $45(46) = 2070$ but $44(45) = 1980.$

Since $\left(a_n\right)$ and $\left(b_n\right)$ have integer terms with $a_1=b_1=1$, we can write the terms of each sequence as \begin{align*}&\left(a_n\right) \Rightarrow \{1, x+1, 2x+1, 3x+1, ...\}\\ &\left(b_n\right) \Rightarrow \{1, y+1, 2y+1, 3y+1, ...\}\end{align*} where $x$ and $y$ ($x\leq y$) are the common differences of each, respectively. Since \begin{align*}a_n &= (n-1)x+1\\ b_n &= (n-1)y+1\end{align*} it is easy to see that $a_n \equiv b_n \equiv 1 \mod{(n-1)}$. Hence, we have to find the largest $n$ such that $\frac{a_n-1}{n-1}$ and $\frac{b_n-1}{n-1}$ are both integers; equivalently, we want to maximize $\gcd(a_n-1, b_n-1)$. The prime factorization of $2010$ is $2\cdot{3}\cdot{5}\cdot{67}$. We list out all the possible pairs that have a product of $2010$, noting that these are the possible values of $(a_n, b_n)$ and we need $a_n \leq b_n$: \[(2,1005), (3, 670), (5,402), (6,335), (10,201),(15,134),(30,67)\] and soon find that the largest $n-1$ value is $7$ for the pair $(15, 134)$, and so the largest $n$ value is $\boxed{8\ \textbf{(C)}}$. As above, let $a_n=(n-1)x+1$ and $b_n=(n-1)y+1$ for some $1\leq x\leq y$. Now we get $2010 = a_n b_n = (n-1)^2xy + (n-1)(x+y) + 1$, hence $2009 = (n-1)( (n-1)xy + x + y )$. Therefore $n-1$ divides $2009 = 7^2 \cdot 41$. And as the second term is greater than the first one, we only have to consider the options $n-1\in\{1,7,41\}$. For $n=42$ we easily see that for $x=y=1$ the right side is less than $49$ and for any other $(x,y)$ it is way too large. For $n=8$ we are looking for $(x,y)$ such that $7xy + x + y = 2009/7 = 7\cdot 41$. Note that $x+y$ must be divisible by $7$. We can start looking for the solution by trying the possible values for $x+y$, and we easily discover that for $x+y=21$ we get $xy + 3 = 41$, which has a suitable solution $(x,y)=(2,19)$. Hence $n=8$ is the largest possible $n$. (There is no need to check $n=2$ anymore.) Consider $n=288$, which would imply $b_{288}\ge a_{288}\ge 288$. However then $a_n b_n\ge 288^2>2010$, so we just need to show that $n=8$ is achievable. This is true when $a_n=1+2n$ and $b_n=1+19n$, giving $a_8 b_8=(15)(134)=2010$. Hence the answer is $\boxed{\textbf{(C)}\ 8}$.

The $x$ values in which $y=x^6-10x^5+29x^4-4x^3+ax^2$ intersect at $y=bx+c$ are the same as the zeros of $y=x^6-10x^5+29x^4-4x^3+ax^2-bx-c$. Since there are $3$ zeros and the function is never negative, all $3$ zeros must be double roots because the function's degree is $6$. Suppose we let $p$, $q$, and $r$ be the roots of this function, and let $x^3-ux^2+vx-w$ be the cubic polynomial with roots $p$, $q$, and $r$. \begin{align*}(x-p)(x-q)(x-r) &= x^3-ux^2+vx-w\\ (x-p)^2(x-q)^2(x-r)^2 &= x^6-10x^5+29x^4-4x^3+ax^2-bx-c = 0\\ \sqrt{x^6-10x^5+29x^4-4x^3+ax^2-bx-c} &= x^3-ux^2+vx-w = 0\end{align*} In order to find $\sqrt{x^6-10x^5+29x^4-4x^3+ax^2-bx-c}$ we must first expand out the terms of $(x^3-ux^2+vx-w)^2$. \[(x^3-ux^2+vx-w)^2\] \[= x^6-2ux^5+(u^2+2v)x^4-(2uv+2w)x^3+(2uw+v^2)x^2-2vwx+w^2\] [Quick note: Since we don't know $a$, $b$, and $c$, we really don't even need the last 3 terms of the expansion.] \begin{align*}&2u = 10\\ u^2+2v &= 29\\ 2uv+2w &= 4\\ u &= 5\\ v &= 2\\ w &= -8\\ &\sqrt{x^6-10x^5+29x^4-4x^3+ax^2-bx-c} = x^3-5x^2+2x+8\end{align*} All that's left is to find the largest root of $x^3-5x^2+2x+8$. \begin{align*}&x^3-5x^2+2x+8 = (x-4)(x-2)(x+1)\\ &\boxed{\textbf{(A)}\ 4}\end{align*}

If we graph each term separately, we will notice that all of the zeros occur at $\frac{1}{m}$, where $m$ is any integer from $1$ to $119$, inclusive: $|mx-1|=0\implies mx=1\implies x=\frac{1}{m}$. The minimum value of $f(x)$ occurs where the absolute value of the sum of the slopes is at a minimum $\ge 0$, since it is easy to see that the value will be increasing on either side. That means the minimum must happen at some $\frac{1}{m}$. The sum of the slopes at $x = \frac{1}{m}$ is \begin{align*}&\sum_{i=m+1}^{119}i - \sum_{i=1}^{m}i\\ &=\sum_{i=1}^{119}i - 2\sum_{i=1}^{m}i\\ &=-m^2-m+7140\end{align*} Now we want to minimize $-m^2-m+7140$. The zeros occur at $-85$ and $84$, which means the slope is $0$ where $m = 84, 85$. We can now verify that both $x=\frac{1}{84}$ and $x=\frac{1}{85}$ yield $\boxed{49\ \textbf{(A)}}$. You can also think of the slopes playing 'tug of war', where the slope of each absolute function upon passing its $x$-intercept is negated, positively tugging on the remaining negative slopes. The sum of the slopes is $1+2+3+4\dots 119=\sum_{m=1}^{119}m=\frac{119\cdot 120}{2}=60\cdot 119=7140$ So we need to find the least integer $a$ such that $1+2+3+\dots a=\sum_{n=1}^an=\frac{a(a+1)}{2}\ge \frac{7140}{2}=3570:$ \[a(a+1)\ge 7140\implies a^2+a-7140\ge 0\rightarrow a=84\text{ exactly!}\] This "exactly" means that the slope is ZERO between the whole interval $x\in\left(\frac{1}{85},\frac{1}{84}\right)$. We can explicitly evaluate both to check that they are both equal to the desired minimum value of $f(x)$: \[\frac{84+83+\dots+2+1+1+2+\dots+33+34}{85}=\frac{84(85)/2+34(35)/2}{85}=\frac{85(14+84)/2}{85}=49\] \[\frac{83+82+\dots+2+1+1+2+\dots+34+35}{84}=\frac{83(84)/2+35(36)/2}{84}=\frac{84(15+83)/2}{84}=49\] Thus the minimum value of $f(x)$ is $49$.

We will use the fact that for any integer $n$, \begin{align*}(5n+1)(5n+2)(5n+3)(5n+4)&=[(5n+4)(5n+1)][(5n+2)(5n+3)]\\ &=(25n^2+25n+4)(25n^2+25n+6)\equiv 4\cdot 6\\ &=24\pmod{25}\equiv -1\pmod{25}.\end{align*} First, we find that the number of factors of $10$ in $90!$ is equal to $\left\lfloor \frac{90}5\right\rfloor+\left\lfloor\frac{90}{25}\right\rfloor=18+3=21$. Let $N=\frac{90!}{10^{21}}$. The $n$ we want is therefore the last two digits of $N$, or $N\pmod{100}$. If instead we find $N\pmod{25}$, we know that $N\pmod{100}$, what we are looking for, could be $N\pmod{25}$, $N\pmod{25}+25$, $N\pmod{25}+50$, or $N\pmod{25}+75$. Only one of these numbers will be a multiple of four, and whichever one that is will be the answer, because $N\pmod{100}$ has to be a multiple of 4. If we divide $90!$ by $5^{21}$ to create $M$ by taking out all the factors of $5$ in $90!$, we can write $N$ as $\frac M{2^{21}}$ where \[M=1\cdot 2\cdot 3\cdot 4\cdot 1\cdot 6\cdot 7\cdot 8\cdot 9\cdot 2\cdots 89\cdot 18,\] where every multiple of 5 is replaced by the number with all its factors of 5 removed. Specifically, every number in the form $5n$ is replaced by $n$, and every number in the form $25n$ is replaced by $n$. The number $M$ can be grouped as follows: \begin{align*}M= &(1\cdot 2\cdot 3\cdot 4)(6\cdot 7\cdot 8\cdot 9)\cdots(86\cdot 87\cdot 88\cdot 89)\\ &\cdot (1\cdot 2\cdot 3\cdot 4)(6\cdot 7\cdot 8\cdot 9)\cdots (16\cdot 17\cdot 18) \\ &\cdot (1\cdot 2\cdot 3).\end{align*} Where the first line is composed of the numbers in $90!$ that aren't multiples of five, the second line is the multiples of five and not 25 after they have been divided by five, and the third line is multiples of 25 after they have been divided by 25. Using the identity at the beginning of the solution, we can reduce $M$ to \begin{align*}M&\equiv(-1)^{18} \cdot (-1)^3(16\cdot 17\cdot 18) \cdot (1\cdot 2\cdot 3) \\ &= 1\cdot -21\cdot 6\\ &= -1\pmod{25} =24\pmod{25}.\end{align*} Using the fact that $2^{10}=1024\equiv -1\pmod{25}$ (or simply the fact that $2^{21}=2097152$ if you have your powers of 2 memorized), we can deduce that $2^{21}\equiv 2\pmod{25}$. Therefore $N=\frac M{2^{21}}\equiv \frac {24}2\pmod{25}=12\pmod{25}$. Finally, combining with the fact that $N\equiv 0\pmod 4$ yields $n=\boxed{\textbf{(A)}\ 12}$.

The question asks for the number of disjoint open intervals, which means we need to find the number of disjoint intervals such that the function is defined within them. We note that since all of the $\sin$ factors are inside a logarithm, the function is undefined where the inside of the logarithm is less than or equal to $0$. First, let us find the number of zeros of the inside of the logarithm. \begin{align*}\sin(\pi x) \cdot \sin(2 \pi x) \cdot \sin (3 \pi x) \cdots \sin(8 \pi x) &= 0\\ \sin(\pi x) &= 0\\ x &= 0, 1\\ \sin(2 \pi x) &= 0\\ x &= 0, \frac{1}{2}, 1\\ \sin(3 \pi x) &= 0\\ x &= 0, \frac{1}{3}, \frac{2}{3}, 1\\ \sin(4 \pi x) &= 0\\ x &= 0, \frac{1}{4}, \frac{2}{4}, \frac{3}{4}, 1\\ &\cdots\end{align*} After counting up the number of zeros for each factor and eliminating the excess cases we get $23$ zeros and $22$ intervals. In order to find which intervals are negative, we must first realize that at every zero of each factor, the sign changes. We also have to be careful, as some zeros are doubled, or even tripled, quadrupled, etc. The first interval $\left(0, \frac{1}{8}\right)$ is obviously positive. This means the next interval $\left(\frac{1}{8}, \frac{1}{7}\right)$ is negative. Continuing the pattern and accounting for doubled roots (which do not flip sign), we realize that there are $5$ negative intervals from $0$ to $\frac{1}{2}$. Since the function is symmetric, we know that there are also $5$ negative intervals from $\frac{1}{2}$ to $1$. And so, the total number of disjoint open intervals is $22 - 2\cdot{5} = \boxed{12\ \textbf{(B)}}$

It should first be noted that given any quadrilateral of fixed side lengths, there is exactly one way to manipulate the angles so that the quadrilateral becomes cyclic. Proof. Given a quadrilateral $ABCD$ where all sides are fixed (in a certain order), we can construct the diagonal $\overline{BD}$. When $BD$ is the minimum allowed by the triangle inequality, one of the angles $\angle DAB$ or $\angle BCD$ will be degenerate and measure $0^\circ$, so opposite angles will sum to less than $180^\circ$. When $BD$ is the maximum allowed, one of the angles will be degenerate and measure $180^\circ$, so opposite angles will sum to more than $180^\circ$. Thus, since the sum of opposite angles increases continuously as $BD$ is lengthened from the minimum to the maximum values, there is a unique value of $BD$ somewhere in the middle such that the sum of opposite angles is exactly $180^\circ$ (due to IVT). Denote $a$, $b$, $c$, and $d$ as the integer side lengths of the quadrilateral. Without loss of generality, let $a\ge b \ge c \ge d$. Since $a+b+c+d = 32$, the Triangle Inequality implies that $a \le 15$. We will now split into $5$ cases. Case $1$: $a = b = c = d$ ($4$ side lengths are equal) Clearly there is only $1$ way to select the side lengths $(8,8,8,8)$, and no matter how the sides are rearranged only $1$ unique quadrilateral can be formed. Case $2$: $a = b = c > d$ or $a > b = c = d$ ($3$ side lengths are equal) If $3$ side lengths are equal, then each of those side lengths can only be integers from $6$ to $10$ except for $8$ (because that is counted in the first case). Obviously there is still only $1$ unique quadrilateral that can be formed from one set of side lengths, resulting in a total of $4$ quadrilaterals. Case $3$: $a = b > c = d$ ($2$ pairs of side lengths are equal) $a$ and $b$ can be any integer from $9$ to $15$, and likewise $c$ and $d$ can be any integer from $1$ to $7$. However, a single set of side lengths can form $2$ different cyclic quadrilaterals (a rectangle and a kite), so the total number of quadrilaterals for this case is $7\cdot{2} = 14$. Case $4$: $a = b > c > d$ or $a > b = c > d$ or $a > b > c = d$ ($2$ side lengths are equal) If the $2$ equal side lengths are each $1$, then the other $2$ sides must each be $15$, which we have already counted in an earlier case. If the equal side lengths are each $2$, there is $1$ possible set of side lengths. Likewise, for side lengths of $3$ there are $2$ sets. Continuing this pattern, we find a total of $1+2+3+4+4+5+7+5+4+4+3+2+1 = 45$ sets of side lengths. (Be VERY careful when adding up the total for this case!) For each set of side lengths, there are $3$ possible quadrilaterals that can be formed, so the total number of quadrilaterals for this case is $3\cdot{45} = 135$. Case $5$: $a > b > c > d$ (no side lengths are equal) Using the same counting principles starting from $a = 15$ and eventually reaching $a = 9$, we find that the total number of possible side lengths is $69$. There are $4!$ ways to arrange the $4$ side lengths, but there is only $1$ unique quadrilateral for $4$ rotations, so the number of quadrilaterals for each set of side lengths is $\frac{4!}{4} = 6$. The total number of quadrilaterals is $6\cdot{69} = 414$. And so, the total number of quadrilaterals that can be made is $414 + 135 + 14 + 4 + 1 = \boxed{568\ \textbf{(C)}}$.