AMC12 2023 B

The first three glasses each have a full glass. Let's assume that each glass has "1 unit" of juice. It won't matter exactly how much juice everyone has because we're dealing with ratios, and that wouldn't affect our answer. The fourth glass has a glass that is one third. So the total amount of juice will be $1+1+1+\dfrac{1}{3} = \dfrac{10}{3}$. If we divide the total amount of juice by 4, we get $\dfrac{5}{6}$, which should be the amount of juice in each glass. This means that each of the first three glasses will have to contribute $1 - \dfrac{5}{6} = \boxed{\textbf{(C) }\dfrac16}$ to the fourth glass.

Let the price originally be x. Then, after a 20 percent discount, the price is now x−15x=45x. From the discounted price 45x, we now take 7.5100 of 45x and add it to 45x, giving us 45x+(7.5100)(45x)=45x+30500x=45x+350x=4350x. Now we write the inequality 43/50x≤43 and multiply by 5043 on both sides to get x≤50. We want the greatest original price, which would be $50 or option choice B.

Because the triangles are right triangles, we know the hypotenuses are diameters of circles $A$ and $B$. Thus, their radii are 2.5 and 6.5 (respectively). Square the two numbers and multiply $\pi$ to get $6.25\pi$ and $42.25\pi$ as the areas of the circles. Multiply 4 on both numbers to get $25\pi$ and $169\pi$. Cancel out the $\pi$, and lastly, divide, to get your answer $=\boxed{\textbf{(D) }\frac{25}{169}}.$

$6.5$ millimeters is equal to $0.65$ centimeters. $25$ meters is $2500$ centimeters. The answer is $0.65 \times 2500$, so the answer is $\boxed{\textbf{(C) 1,625}}$.

Notice that the $3\times3$ square grid has a total of $12$ possible $2\times1$ rectangles. Suppose you choose the middle square for one of your turns. The middle square is covered by $4$ rectangles, and each of the remaining $8$ squares is covered by a maximum of $2$ uncounted rectangles. This means that the number of turns is at least $1+\frac{12-4}{2}=1+4=5$. Now suppose you don't choose the middle square. The squares on the middle of the sides are covered by at most 3 uncounted rectangles, and the squares on the corners are covered by at most 2 uncounted rectangles. In this case, we see that the least number of turns needed to account for all 12 rectangles is $12\div 3=4.$ To prove that choosing only side squares indeed does cover all 12 rectangles, we need to show that the 3 rectangles per square that cover each side square do not overlap. Drawing the rectangles that cover one square, we see they form a $T$ shape and they do not cover any other side square. Hence, our answer is $4.$

The expressions to the power of even powers are always positive, so we don't need to care about those. We only need to care about $(x-1)^1(x-3)^3(x-5)^5(x-7)^7(x-9)^9$. We need 0, 2, or 4 of the expressions to be negative. The 9 through 10 interval and 10 plus interval make all of the expressions positive. The 5 through 6 and 6 through 7 intervals make two of the expressions negative. The 1 through 2 and 2 through 3 intervals make four of the expressions negative. There are $\boxed{\textbf{(C) 6}}$ intervals.

We have \begin{align*} \sqrt{\frac{\log \left( n^2 \right) - \left( \log n \right)^2}{\log n - 3}} & = \sqrt{\frac{2 \log n - \left( \log n \right)^2}{\log n - 3}} \\ & = \sqrt{\frac{\left( \log n \right) \left( 2 - \log n\right)}{\log n - 3}} . \end{align*} Because $n$ is an integer and $\log n$ is well defined, $n$ must be a positive integer. Case 1: $n = 1$ or $10^2$. The above expression is 0. So these are valid solutions. Case 2: $n \neq 1, 10^2$. Thus, $\log n > 0$ and $2 - \log n \neq 0$. To make the above expression real, we must have $2 < \log n < 3$. Thus, $100 < n < 1000$. Thus, $101 \leq n \leq 999$. Hence, the number of solutions in this case is 899. Putting all cases together, the total number of solutions is $\boxed{\textbf{(E) 901}}$.

There is no way to have a set with 0. If a set is to have its lowest element as 1, it must have only 1 element: 1. If a set is to have its lowest element as 2, it must have 2, and the other element will be chosen from the natural numbers between 3 and 12, inclusive. To calculate this, we do $\binom{10}{1}$. If the set is the have its lowest element as 3, the other 2 elements will be chosen from the natural numbers between 4 and 12, inclusive. To calculate this, we do $\binom{9}{2}$. We can see a pattern emerge, where the top is decreasing by 1 and the bottom is increasing by 1. In other words, we have to add $1 + \binom{10}{1} + \binom{9}{2} + \binom{8}{3} + \binom{7}{4} + \binom{6}{5}$. This is $1+10+36+56+35+6 = \boxed{\textbf{(D) 144}}$.

First consider, $|x-1|+|y-1| \le 1.$ We can see that it is a square with a side length of $\sqrt{2}$ (diagonal $2$). The area of the square is $\sqrt{2}^2 = 2.$ Next, we insert an absolute value sign into the equation and get $|x-1|+||y|-1| \le 1.$ This will double the square reflecting over x-axis. So now we have $2$ squares. Finally, we add one more absolute value and obtain $||x|-1|+||y|-1| \le 1.$ This will double the squares as we reflect the $2$ squares we already have over the y-axis. Concluding, we have $4$ congruent squares. Thus, the total area is $4\cdot2 =$ $\boxed{\text{(B) 8}}$

The center of the first circle is $(4,0)$. The center of the second circle is $(0,10)$. Thus, the slope of the line that passes through these two centers is $- \frac{10}{4} = - \frac{5}{2}$. Because this line is the perpendicular bisector of the line that passes through two intersecting points of two circles, the slope of the latter line is $\frac{-1}{- \frac{5}{2}} = \boxed{\textbf{(E) } \frac{2}{5}}$.

Let the trapezoid be $ABCD$ with $AD = BC = 1, \; AB = x, CD = 2x$. Extend $AD$ and $BC$ to meet at point $E$. Then, notice $\triangle ABE \sim \triangle DCE$ with side length ratio $1:2$ and $AE = BE = 1$. Thus, $[DCE] = 4 \cdot [ABE]$ and $[ABCD] = [DCE] - [ABE] = \frac{3}{4} \cdot [DCE]$. The problem reduces to maximizing the area of $[DCE]$, an isosceles triangle with legs of length $2$. Analyzing the sine area formula, this is clearly maximized when $\angle DEC = 90^{\circ}$, so $[DCE] = 2$ and $[ABCD] = \frac{3}{4} \cdot 2 = \boxed{\frac{3}{2}}.$

let $z$ = $a+bi$. $z \otimes z = a^{2}+b^{2}i$. This is equal to $z^{2} + 40 = a^{2}-b^{2}+40+2abi$ Since the real values have to be equal to each other, $a^{2}-b^{2}+40 = a^{2}$. Simple algebra shows $b^{2} = 40$, so $b = \pm 2\sqrt{10}$. The imaginary components must also equal each other, meaning $b^{2} = 2ab$, or $b = 2a$. This means $a = \frac{b}{2} = \pm \sqrt{10}$. Thus, the magnitude of $z$ is $\sqrt{a^{2}+b^{2}} = \sqrt{10+40} = \sqrt{50} = 5\sqrt{2}$ $=\text{\boxed{\textbf{(E) }5\sqrt{2}}}$

We can create three equations using the given information. \[4a+4b+4c = 13\] \[2ab+2ac+2bc=\frac{11}{2}\] \[abc=\frac{1}{2}\] We also know that we want $\sqrt{a^2 + b^2 + c^2}$ because that is the length that can be found from using the Pythagorean Theorem. We cleverly notice that $a^2 + b^2 + c^2 = (a+b+c)^2 - 2(ab+ac+bc)$. We know that $a+b+c = \frac{13}{4}$ and $2(ab+ac+bc)=\dfrac{11}2$, so $a^2 + b^2 + c^2 = \left(\frac{13}{4}\right)^2 - \frac{11}{2} = \frac{169-88}{16} = \frac{81}{16}$. So our answer is $\sqrt{\frac{81}{16}} = \boxed{\textbf{(D)}~\tfrac94}$.

Denote three roots as $r_1 < r_2 < r_3$. Following from Vieta's formula, $r_1r_2r_3 = -6$. Case 1: All roots are negative. We have the following solution: $\left( -3, -2, -1 \right)$. Case 2: One root is negative and two roots are positive. We have the following solutions: $\left( -3, 1, 2 \right)$, $\left( -2, 1, 3 \right)$, $\left( -1, 2, 3 \right)$, $\left( -1, 1, 6 \right)$. Putting all cases together, the total number of solutions is $\boxed{\textbf{(A) 5}}$.

We examine each of the conditions. The first condition is false. A simple counterexample is $a=3$ and $b=5$. The corresponding value of $c$ is $115$. Since $\gcd(3,14)=1$, condition $I$ would imply that $\gcd(c,210)=1.$ However, $\gcd(115,210)$ is clearly not $1$ (they share a common factor of $5$). Condition $I$ is false so that we can rule out choices $A,B,$ and $C$. We now decide between the two answer choices $D$ and $E$. What differs between them is the validity of condition $II$, so it suffices to check $II$ simply. We look at statement $II$'s contrapositive to prove it. The contrapositive states that if $\gcd(a,14)\neq1$ and $\gcd(b,15)\neq1$, then $\gcd(c,210)\neq1.$ In other words, if $a$ shares some common factor that is not $1$ with $14$ and $b$ shares some common factor that is not $1$ with $15$, then $c$ also shares a common factor that is not $1$ with $210$. Let's say that $a=a'\cdot n$, where $a'$ is a factor of $14$ not equal to $1$. (So $a'$ is the common factor.) We can rewrite the given equation as $15a+14b=c\implies15(a'n)+14b=c.$ We can express $14$ as $a'\cdot n'$, for some positive integer $n'$ (this $n'$ can be $1$). We can factor $a'$ out to get $a'(15n+bn')=c.$ Since all values in this equation are integers, $c$ must be divisible by $a'$. Since $a'$ is a factor of $14$, $a'$ must also be a factor of $210$, a multiple of $14$. Therefore, we know that $c$ shares a common factor with $210$ (which is $a'$), so $\gcd(c,210)\neq1$. This is what $II$ states, so therefore $II$ is true. Thus, our answer is $\boxed{\textbf{(E) }\text{II and III only}}.$

This problem asks to find largest $x$ that cannot be written as \[6 a + 10 b + 15 c = x, \hspace{1cm} (1)\] where $a, b, c \in \Bbb Z_+$. Denote by $r \in \left\{ 0, 1 \right\}$ the remainder of $x$ divided by 2. Modulo 2 on Equation (1), we get By using modulus $m \in \left\{ 2, 3, 5 \right\}$ on the equation above, we get $c \equiv r \pmod{2}$. Following from the Chicken McNugget Theorem, we have that any number that is no less than $(3-1)(5-1) = 8$ can be expressed in the form of $3a + 5b$ with $a, b \in \Bbb Z_+$. Therefore, all even numbers that are at least equal to $2 \cdot 8 + 15 \cdot 0 = 16$ can be written in the form of Equation (1) with $a, b, c \in \Bbb Z_+$. All odd numbers that are at least equal to $2 \cdot 8 + 15 \cdot 1 = 31$ can be written in the form of Equation (1) with $a, b, c \in \Bbb Z_+$. The above two cases jointly imply that all numbers that are at least 30 can be written in the form of Equation (1) with $a, b, c \in \Bbb Z_+$. Next, we need to prove that 29 cannot be written in the form of Equation (1) with $a, b, c \in \Bbb Z_+$. Because 29 is odd, we must have $c \equiv 1 \pmod{2}$. Because $a, b, c \in \Bbb Z_+$, we must have $c = 1$. Plugging this into Equation (1), we get $3 a + 5 b = 7$. However, this equation does not have non-negative integer solutions. All analysis above jointly imply that the largest $x$ that has no non-negative integer solution to Equation (1) is 29. Therefore, the answer is $2 + 9 = \boxed{\textbf{(D) 11}}$.

The length of the side opposite to the angle with $120^\circ$ is longest. We denote its value as $x$. Because three side lengths form an arithmetic sequence, the middle-valued side length is $\frac{x + 6}{2}$. Following from the law of cosines, we have \begin{align*} 6^2 + \left( \frac{x + 6}{2} \right)^2 - 2 \cdot 6 \cdot \frac{x + 6}{2} \cdot \cos 120^\circ = x^2 . \end{align*} By solving this equation, we get $x = 14$. Thus, $\frac{x + 6}{2} = 10$. Therefore, the area of the triangle is \begin{align*} \frac{1}{2} 6 \cdot 10 \cdot \sin 120^\circ = \boxed{\textbf{(E) } 15 \sqrt{3}} . \end{align*}

Denote by $A_i$ the average of person with initial $A$ in semester $i \in \left\{1, 2 \right\}$ Thus, $Y_1 = Z_1 + 3$, $Y_2 = Y_1 + 18$, $Y_2 = Z_2 + 3$. Denote by $A_{12}$ the average of person with initial $A$ in the full year. Thus, $Y_{12}$ can be any number in $\left( Y_1 , Y_2 \right)$ and $Z_{12}$ can be any number in $\left( Z_1 , Z_2 \right)$. Therefore, the impossible solution is $\boxed{\textbf{(A)}~\text{Yolanda's quiz average for the academic year was 22 points higher than Zelda's.}}$

Because each bin will have an odd number, they will have at least one ball. So we can put one ball in each bin prematurely. We then can add groups of 2 balls into each bin, meaning we now just have to spread 1010 pairs over 3 bins. This will force every bin to have an odd number of balls. Using stars and bars, we find that this is equal to $\binom{1012}{2}$. This is equal to $\frac{1012\cdot1011}{2}$. The total amount of ways would also be found using stars and bars. That would be $\binom{2023+3-1}{3-1} = \binom{2025}{2}$. Dividing our two quantities, we get $\frac{1012 \cdot 1011 \cdot 2}{2 \cdot 2025 \cdot 2024}$. We can roughly cancel $\frac{1012 \cdot 1011}{2025 \cdot 2024}$ to get $\frac{1}{4}$. The 2 in the numerator and denominator also cancels out, so we're left with $\boxed{\frac{1}{4}}$.

Let Cyrus's starting position be $S$. WLOG, let the place Cyrus lands at for his first jump be $O$. From $O$, Cyrus can reach all the points on $\odot O$. The probability that Cyrus will land less than $1$ unit away from $S$ is $\frac{4 \alpha }{ 2 \pi}$. \[\sin \alpha = \frac{ \frac12 }{2} = \frac14, \quad \alpha = \arcsin \frac14\] Therefore, the answer is \[\frac{4 \arcsin \frac14 }{ 2 \pi} = \boxed{\textbf{(E) } \frac{2 \arcsin \frac{1}{4}}{\pi}}\]

We augment the frustum to a circular cone. Denote by $O$ the apex of the cone. Denote by $A$ the bug and $B$ the honey. By using the numbers given in this problem, the height of the cone is $6 \sqrt{3}$. Thus, $OA = 12$ and $OB = 6$. We unfold the lateral face. So we get a circular sector. The radius is 12 and the length of the arc is $12\pi$. Thus, the central angle of this circular sector is $180^\circ$. Because $A$ and $B$ are opposite in the original frustum, in the unfolded circular cone, $\angle AOB = \frac{180^\circ}{2} = 90^\circ$. Notice that a feasible path between $A$ and $B$ can only fall into the region with the range of radii between $OB = 6$ and $OA = 12$. Therefore, we cannot directly connect $A$ and $B$ and must make a detour. Denote by $AC$ a tangent to the circular sector with radius 6 that meets it at point $C$. Therefore, the shortest path between $A$ and $B$ consists of a segment $AC$ and an arc from $C$ to $B$. Because $OA = 12$, $OC = 6$ and $\angle OCA = 90^\circ$, we have $AC = \sqrt{OA^2 - OC^2} = 6 \sqrt{3}$ and $\angle AOC = 60^\circ$. This implies $\angle COB = \angle AOB - \angle AOC = 30^\circ$. Therefore, the length of the arc between $C$ and $B$ is $OB \cdot \pi \cdot \frac{\angle COB}{180^\circ} = \pi$. Therefore, the shortest distance between $A$ and $B$ is $\boxed{\textbf{(E) } 6 \sqrt{3} + \pi}$.

Substituting $a = b$ we get \[f(2a) + f(0) = 2f(a)^2\] Substituting $a= 0$ we find \[2f(0) = 2f(0)^2 \implies f(0) \in \{0, 1\}.\] This gives \[f(2a) = 2f(a)^2 - f(0) \geq 0-1\] Plugging in $a = \frac{1}{2}$ implies $f(1) \geq -1$, so answer choice $\boxed{\textbf{(E) -2}}$ is impossible.

We start by trying to prove a function of $n$, and then we can apply the function and equate it to $936$ to find the value of $n$. It is helpful to think of this problem in the format $(1+2+3+4+5+6) \cdot (1+2+3+4+5+6) \dots$. Note that if we represent the scenario in this manner, we can think of picking a $1$ for one factor and then a $5$ for another factor to form their product - this is similar thinking to when we have the factorized form of a polynomial. Unfortunately this is not quite accurate to the problem because we can reach the same product in many ways: for example for $n=2$, $4$ can be reached by picking $1$ and $4$ or $2$ and $2$. However, this form gives us insights that will be useful later in the problem. Note that there are only $3$ primes in the set $\{1,2,3,4,5,6\}$: $2,3,$ and $5$. Thus if we're forming the product of possible values of a dice roll, the product has to be written in the form $2^h \cdot 3^i \cdot 5^j$ (the choice of variables will become clear later), for integer nonnegative values $h,i,j$. So now the problem boils down to how many distinct triplets $(h,i,j)$ can be formed by taking the product of $n$ dice values. We start our work on representing $j$: the powers of $5$, because it is the simplest in this scenario because there is only one factor of $5$ in the set. Because of this, having $j$ fives in our prime factorization of the product is equivalent to picking $j$ factors from the polynomial $(1+\dots + 6) \cdots$ and choosing each factor to be a $5$. Now that we've selected $j$ factors, there are $n-j$ factors remaining to choose our powers of $3$ and $2$. Suppose our prime factorization of this product contains $i$ powers of $3$. These powers of $3$ can either come from a $3$ factor or a $6$ factor, but since both $3$ and $6$ contain only one power of $3$, this means that a product with $i$ powers of $3$ corresponds directly to picking $i$ factors from the polynomial, each of which is either $3$ or $6$ (but this distinction doesn't matter when we consider only the powers of $3$. Now we can reframe the problem again. Our method will be as follows: Suppose we choose an arbitrary pair $(i,j)$ that match the requirements, corresponding to the number of $3$'s and the number of $5$'s our product will have. Then how many different $h$ values for the powers of $2$ are possible? In the $i+j$ factors we have already chosen, we obviously can't have any factors of $2$ in the $j$ factors with $5$. However, we can have a factor of $2$ pairing with factors of $3$, if we choose a $6$. The maximal possible power of $2$ in these $i$ factors is thus $2^i$, which occurs when we pick every factor to be $6$. We now have $n-i-j$ factors remaining, and we want to allocate these to solely powers of $2$. For each of these factors, we can choose either a $1,2,$ or $4$. Therefore the maximal power of $2$ achieved in these factors is when we pick $4$ for all of them, which is equivalent to $2^{2\cdot (n-i-j)}$. Now if we multiply this across the total $n$ factors (or $n$ dice) we have a total of $2^{2n-2i-2j} \cdot 2^i = 2^{2n-i-2j}$, which is the maximal power of $2$ attainable in the product for a pair $(i,j)$. Now note that every power of $2$ below this power is attainable: we can simply just take away a power of $2$ from an existing factor by dividing by $2$. Therefore the powers of $2$, and thus the $h$ value ranges from $h=0$ to $h=2n-i-2j$, so there are a total of $2n+1-i-2j$ distinct values for $h$ for a given pair $(i,j)$. Now to find the total number of distinct triplets, we must sum this across all possible $i$s and $j$s. Lets take note of our restrictions on $i,j$: the only restriction is that $i+j \leq n$, since we're picking factors from $n$ dice. \[\sum_{i+j\leq n}^{} 2n+1-i-2j = \sum_{i+j \leq n}^{} 2n+1 - \sum_{i+j \leq n}^{} i+2j\] We start by calculating the first term. $2n+1$ is constant, so we just need to find out how many pairs there are such that $i+j \leq n$. Set $i$ to $0$: $j$ can range from $0$ to $n$, then set $i$ to $1$: $j$ can range from $0$ to $n-1$, etc. The total number of pairs is thus $n+1+n+n-1+\dots+1 = \frac{(n+1)(n+2)}{2}$. Therefore the left summation evaluates to \[\frac{(2n+1)(n+1)(n+2)}{2}\] Now we calculate $\sum_{i+j \leq n}^{} i+2j$. This simplifies to $\sum_{i+j \leq n}^{} i + 2 \cdot \sum_{i+j \leq n}^{} j$. Note that because $i+j = n$ is symmetric with respect to $i,j$, the sum of $i$ in all of the pairs will be equal to the sum of $j$ in all of the pairs. Thus this is equal to calculating $3 \cdot \sum_{i+j \leq n}^{} i$. In the pairs, $i=1$ appears for $j$ ranging between $0$ and $n-1$ so the sum here is $1 \cdot (n)$. Similarly $i=2$ appears for $j$ ranging from $0$ to $n-2$, so the sum is $2 \cdot (n-1)$. If we continue the pattern, the sum overall is $(n)+2 \cdot (n-1) + 3 \cdot (n-2) + \dots + (n) \cdot 1$. We can rearrange this as $((n)+(n-1)+ \dots + 1) + ((n-1)+(n-2)+ \dots + 1)+ ((n-2)+(n-3)+ \dots + 1) + \dots + 1)$ \[= \frac{(n)(n+1)}{2} + \frac{(n-1)(n)}{2}+ \dots + 1\] We can write this in easier terms as $\sum_{k=0}^{n} \frac{(k)(k+1)}{2} = \frac{1}{2} \cdot \sum_{k=0}^{n} k^2+k$ \[=\frac{1}{2} \cdot( \sum_{k=0}^{n} k^2 + \sum_{k=0}^{n} k)\] \[= \frac{1}{2} \cdot ( \frac{(n)(n+1)(2n+1)}{6} + \frac{(n)(n+1)}{2})\] \[= \frac{1}{2} \cdot ( \frac{(n)(n+1)(2n+1)}{6} + \frac{3n(n+1)}{6}) = \frac{1}{2} \cdot \frac{n(n+1)(2n+4)}{6}\] \[= \frac{n(n+1)(n+2)}{6}\] We multiply this by $3$ to obtain that \[\sum_{i+j \leq n}^{} i+2j = \frac{n(n+1)(n+2)}{2}\] Thus our final answer for the number of distinct triplets $(h,i,j)$ is: \[\sum_{i+j\leq n}^{} 2n+1-i-2j = \frac{(2n+1)(n+1)(n+2)}{2} - \frac{n(n+1)(n+2)}{2}\] \[= \frac{(n+1)(n+2)}{2} \cdot (2n+1-n) = \frac{(n+1)(n+2)}{2} \cdot (n+1)\] \[= \frac{(n+1)^2(n+2)}{2}\] Now most of the work is done. We set this equal to $936$ and prime factorize. $936 = 12 \cdot 78 = 2^3 \cdot 3^2 \cdot 13$, so $(n+1)^2(n+2) = 936 \cdot 2 = 2^4 \cdot 3^2 \cdot 13$. Clearly $13$ cannot be anything squared and $2^4 \cdot 3^2$ is a perfect square, so $n+2 = 13$ and $n = 11 = \boxed{A}$

Denote by $\nu_p (x)$ the number of prime factor $p$ in number $x$. We index Equations given in this problem from (1) to (7). First, we compute $\nu_2 (x)$ for $x \in \left\{ a, b, c, d \right\}$. Equation (5) implies $\max \left\{ \nu_2 (b), \nu_2 (c) \right\} = 1$. Equation (2) implies $\max \left\{ \nu_2 (a), \nu_2 (b) \right\} = 3$. Equation (6) implies $\max \left\{ \nu_2 (b), \nu_2 (d) \right\} = 2$. Equation (1) implies $\nu_2 (a) + \nu_2 (b) + \nu_2 (c) + \nu_2 (d) = 6$. Therefore, all above jointly imply $\nu_2 (a) = 3$, $\nu_2 (d) = 2$, and $\left( \nu_2 (b), \nu_2 (c) \right) = \left( 0 , 1 \right)$ or $\left( 1, 0 \right)$. Second, we compute $\nu_3 (x)$ for $x \in \left\{ a, b, c, d \right\}$. Equation (2) implies $\max \left\{ \nu_3 (a), \nu_3 (b) \right\} = 2$. Equation (3) implies $\max \left\{ \nu_3 (a), \nu_3 (c) \right\} = 3$. Equation (4) implies $\max \left\{ \nu_3 (a), \nu_3 (d) \right\} = 3$. Equation (1) implies $\nu_3 (a) + \nu_3 (b) + \nu_3 (c) + \nu_3 (d) = 9$. Therefore, all above jointly imply $\nu_3 (c) = 3$, $\nu_3 (d) = 3$, and $\left( \nu_3 (a), \nu_3 (b) \right) = \left( 1 , 2 \right)$ or $\left( 2, 1 \right)$. Third, we compute $\nu_5 (x)$ for $x \in \left\{ a, b, c, d \right\}$. Equation (5) implies $\max \left\{ \nu_5 (b), \nu_5 (c) \right\} = 2$. Equation (2) implies $\max \left\{ \nu_5 (a), \nu_5 (b) \right\} = 3$. Thus, $\nu_5 (a) = 3$. From Equations (5)-(7), we have either $\nu_5 (b) \leq 1$ and $\nu_5 (c) = \nu_5 (d) = 2$, or $\nu_5 (b) = 2$ and $\max \left\{ \nu_5 (c), \nu_5 (d) \right\} = 2$. Equation (1) implies $\nu_5 (a) + \nu_5 (b) + \nu_5 (c) + \nu_5 (d) = 7$. Thus, for $\nu_5 (b)$, $\nu_5 (c)$, $\nu_5 (d)$, there must be two 2s and one 0. Therefore, \begin{align*} {\rm gcd} (a,b,c,d) & = \Pi_{p \in \{ 2, 3, 5\}} p^{\min\{ \nu_p (a), \nu_p(b) , \nu_p (c), \nu_p(d) \}} \\ & = 2^0 \cdot 3^1 \cdot 5^0 \\ & = \boxed{\textbf{(C) 3}}. \end{align*}

Since $A$ is folded onto $O$, $AM = MO$ where $M$ is the intersection of $AO$ and the creaseline between $A$ and $O$. Note that the inner pentagon is regular, and therefore similar to the original pentagon, due to symmetry. Because of their similarity, the ratio of the inner pentagon's area to that of the outer pentagon can be represented by $\left(\frac{OM}{ON}\right)^{2} = \left(\frac{\frac{OA}{2}}{OA\sin (\angle OAE)}\right)^{2} = \frac{1}{4\sin^{2}54}$ Remember that $\sin54 = \frac{1+\sqrt5}{4}$. $\cos54 = \sin36$ $4\cos^{3}18-3\cos18 = 2\sin18\cos18$ $4(1-\sin^{2}18)-3-2\sin18=0$ $4\sin^{2}18+2\sin18-1=0$ $\sin18 = \frac{-1+\sqrt5}{4}$ $\sin54 = \cos36 = 1-2\sin^{2}18 = \frac{1+\sqrt5}{4}$ $\sin^{2}54 =\frac{3+\sqrt5}{8}$ Let the inner pentagon be $Z$. $[Z] = \frac{1}{4\sin^{2}54}[ABCDE]$ $= \frac{2(1+\sqrt5)}{3+\sqrt5}$ $= \sqrt5-1$ So the answer is $\boxed{B}$