## AMC 12 2019 Test B

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**Instructions**

- This is a 25-question, multiple choice test. Each question is followed by answers marked A, B, C, D and E. Only one of these is correct.
- You will receive 6 points for each correct answer, 2.5 points for each problem left unanswered if the year is before 2006, 1.5 points for each problem left unanswered if the year is after 2006, and 0 points for each incorrect answer.
- No aids are permitted other than scratch paper, graph paper, ruler, compass, protractor and erasers (and calculators that are accepted for use on the test if before 2006. No problems on the test will require the use of a calculator).
- Figures are not necessarily drawn to scale.
- You will have
**75 minutes**working time to complete the test.

Alicia had two containers. The first was $\dfrac{5}{6}$ full of water and the second was empty. She poured all the water from the first container into the second container, at which point the second container was $\dfrac{3}{4}$ full of water. What is the ratio of the volume of the first container to the volume of the second container?

$\textbf{(A) } \dfrac{5}{8} \qquad \textbf{(B) } \dfrac{4}{5} \qquad \textbf{(C) } \dfrac{7}{8} \qquad \textbf{(D) } \dfrac{9}{10} \qquad \textbf{(E) } \dfrac{11}{12}$

$\textbf{D}$

Let the volume of the first container be $x$, and the volume of the second container be $y$. We have $\dfrac34x=\dfrac56y$. So $\dfrac{x}{y}=\dfrac{9}{10}$.

Consider the statement, "If $n$ is not prime, then $n-2$ is prime." Which of the following values of $n$ is a counterexample to this statement?

$\textbf{(A) } 11 \qquad \textbf{(B) } 15 \qquad \textbf{(C) } 19 \qquad \textbf{(D) } 21 \qquad \textbf{(E) } 27$

$\textbf{E}$

Since a counterexample must be a value of $n$ which is not prime, $n$ must be composite, so we eliminate $\text{A}$ and $\text{C}$. Now we subtract $2$ from the remaining answer choices, and we see that the only time $n-2$ is not prime is when $n =27$.

Which one of the following rigid transformations (isometries) maps the line segment $\overline{AB}$ onto the line segment $\overline{A'B'}$ so that the image of $A(-2,1)$ is $A'(2,-1)$ and the image of $B(-1,4)$ is $B'(1,-4)?$

$\textbf{(A) }$ reflection in the $y$-axis

$\textbf{(B) }$ counterclockwise rotation around the origin by $90^{\circ}$

$\textbf{(C) }$ translation by $3$ units to the right and $5$ units down

$\textbf{(D) }$ reflection in the $x$-axis

$\textbf{(E) }$ clockwise rotation about the origin by $180^{\circ}$

$\textbf{E}$

We see that $A(-2,1)$ and $A'(2,-1)$ are symmetric about the origin, as well as $B$ and $B'$. So we can rotate about the origin by $180^\circ$. The answer is E.

A positive integer $n$ satisfies the equation $(n+1)!+(n+2)!=440\cdot n!$. What is the sum of the digits of $n$?

$\textbf{(A) } 2 \qquad \textbf{(B) } 5 \qquad \textbf{(C) } 10\qquad \textbf{(D) } 12 \qquad \textbf{(E) } 15$

$\textbf{C}$

Dividing both sides by $n!$ gives\[(n+1)+(n+2)(n+1)=440 \Rightarrow n^2+4n-437=0 \Rightarrow (n-19)(n+23)=0.\]Since $n$ is non-negative, $n=19$. The answer is $1 + 9 = 10$.

Each piece of candy in a store costs a whole number of cents. Casper has exactly enough money to buy either $12$ pieces of red candy, $14$ pieces of green candy, $15$ pieces of blue candy, or $n$ pieces of purple candy. A piece of purple candy costs $20$ cents. What is the smallest possible value of $n$?

$\textbf{(A) } 18 \qquad \textbf{(B) } 21 \qquad \textbf{(C) } 24\qquad \textbf{(D) } 25 \qquad \textbf{(E) } 28$

$\textbf{B}$

The amount of money is a multiple of 12, 14, and 15. By prime factorization, we have $12=2^2\cdot3$, $14=2\cdot7$, and $15=3\cdot5$. So Casper has at least $2^2\cdot3\cdot5\cdot7=420$ cents. The smallest value of $n$ is $420/20=21$.

In a given plane, points $A$ and $B$ are $10$ units apart. How many points $C$ are there in the plane such that the perimeter of $\triangle ABC$ is $50$ units and the area of $\triangle ABC$ is $100$ square units?

$\textbf{(A) }0\qquad\textbf{(B) }2\qquad\textbf{(C) }4\qquad\textbf{(D) }8\qquad\textbf{(E) }\textrm{infinitely many}$

$\textbf{A}$

Notice that whatever point we pick for $C$, $AB$ will be the base of the triangle. Without loss of generality, let points $A$ and $B$ be $(0,0)$ and $(10,0)$, since for any other combination of points, we can just rotate the plane to make them $(0,0)$ and $(10,0)$ under a new coordinate system. When we pick point $C$, we have to make sure that its $y$-coordinate is $\pm20$, because that's the only way the area of the triangle can be $100$.

Now when the perimeter is minimized, by symmetry, we put $C$ in the middle, at $(5, 20)$. We can easily see that $AC$ and $BC$ will both be $\sqrt{20^2+5^2} = \sqrt{425}$. The perimeter of this minimal triangle is $2\sqrt{425} + 10$, which is larger than $50$. Since the minimum perimeter is greater than $50$, there is no triangle that satisfies the condition. The answer is 0.

What is the sum of all real numbers $x$ for which the median of the numbers $4,6,8,17,$ and $x$ is equal to the mean of those five numbers?

$\textbf{(A) } -5 \qquad\textbf{(B) } 0 \qquad\textbf{(C) } 5 \qquad\textbf{(D) } \dfrac{15}{4} \qquad\textbf{(E) } \dfrac{35}{4}$

$\textbf{A}$

The mean is $\dfrac{4+6+8+17+x}{5}=\dfrac{35+x}{5}$.

There are three possibilities for the median: it is either $6$, $8$, or $x$.

$\textbf{Case 1:}$ The median is 6.

$\dfrac{35+x}{5}=6$ has solution $x=-5$, and the sequence is $-5, 4, 6, 8, 17$, which does have median $6$. So this is a valid solution.

$\textbf{Case 2:}$ The median is 8.

$\dfrac{35+x}{5}=8$ gives $x=5$, so the sequence is $4, 5, 6, 8, 17$, which has median $6$. This is not valid.

$\textbf{Case 3:}$ The median is $x$.

$\dfrac{35+x}{5}=x\implies x=\dfrac{35}{4}=8.75$, and the sequence is $4, 6, 8, 8.75, 17$, which has median $8$. This case is therefore again not valid.

Hence, the only possible value of $x$ is $-5.$

Let $f(x) = x^{2}(1-x)^{2}$. What is the value of the sum\[f \left(\frac{1}{2019} \right)-f \left(\frac{2}{2019} \right)+f \left(\frac{3}{2019} \right)-f \left(\frac{4}{2019} \right)+\cdots + f \left(\frac{2017}{2019} \right) - f \left(\frac{2018}{2019} \right)?\]

$\textbf{(A) }0\qquad\textbf{(B) }\dfrac{1}{2019^{4}}\qquad\textbf{(C) }\dfrac{2018^{2}}{2019^{4}}\qquad\textbf{(D) }\dfrac{2020^{2}}{2019^{4}}\qquad\textbf{(E) }1$

$\textbf{A}$

First, note that $f(x) = f(1-x)$. Using this result, we regroup the front term and bottom term together:\[\left( f \left(\frac{1}{2019} \right) - f \left(\frac{2018}{2019} \right) \right) + \left( f \left(\frac{2}{2019} \right) - f \left(\frac{2017}{2019} \right) \right) + \cdots + \left( f \left(\frac{1009}{2019} \right) - f \left(\frac{1010}{2019} \right) \right)\]\[= \left( f \left(\frac{1}{2019} \right) - f \left(\frac{1}{2019} \right) \right) + \left( f \left(\frac{2}{2019} \right) - f \left(\frac{2}{2019} \right) \right) + \cdots + \left( f \left(\frac{1009}{2019} \right) - f \left(\frac{1009}{2019} \right) \right)\]Now it is clear that all the terms will cancel out, so the answer is 0.

For how many integral values of $x$ can a triangle of positive area be formed having side lengths $\log_{2} x, \log_{4} x, 3$?

$\textbf{(A) } 57\qquad \textbf{(B) } 59\qquad \textbf{(C) } 61\qquad \textbf{(D) } 62\qquad \textbf{(E) } 63$

$\textbf{B}$

For these lengths to form a triangle of positive area, the Triangle Inequality tells us that we need\[\log_2{x} + \log_4{x} > 3\]\[\log_2{x} + 3 > \log_4{x}\]\[\log_4{x} + 3 > \log_2{x}.\]The second inequality is redundant, as it's always less restrictive than the last inequality.

Let's raise the first inequality to the power of $4$. This gives $$4^{\log_2{x}} \cdot 4^{\log_4{x}} > 64 \Rightarrow \left(2^2\right)^{\log_2{x}} \cdot x > 64 \Rightarrow x^2 \cdot x > 64$$ Thus, $x > 4$.

Doing the same for the last inequality gives $$4^{\log_4{x}} \cdot 64 > 4^{\log_2{x}} \Rightarrow 64x > x^2 \Rightarrow x < 64$$ Therefore, $x$ is an integer strictly between $4$ and $64$, so the number of possible values of $x$ is $64 - 4 - 1 =59$.

The figure below is a map showing $12$ cities and $17$ roads connecting certain pairs of cities. Paula wishes to travel along exactly $13$ of those roads, starting at city $A$ and ending at city $L,$ without traveling along any portion of a road more than once. (Paula is allowed to visit a city more than once.)

How many different routes can Paula take?

$\textbf{(A) } 0 \qquad\textbf{(B) } 1 \qquad\textbf{(C) } 2 \qquad\textbf{(D) } 3\qquad\textbf{(E) } 4$

$\textbf{E}$

Note that of the $12$ cities, $6$ of them ($2$ on the top, $2$ on the bottom, and $1$ on each side) have $3$ edges coming into/out of them (i.e., in graph theory terms, they have degree $3$). Therefore, at least $1$ edge connecting to each of these cities cannot be used. Additionally, the same applies to the start and end points, since we don't want to return to them. Thus there are $6+2=8$ vertices that we know have $1$ unused edge, and we have $17-13=4$ unused edges to work with (since there are $17$ edges in total, and we must use exactly $13$ of them). So the 4 unused edges must connect 8 vertices that each has 1 unused edge. It is not hard to find that there is only one configuration satisfying these conditions:

Note: $\text{X}$s represent unused edges.

Observe that at each of the $2$ cities marked with an $\text{O}$ on a path, there are $2$ possibilities: we can either come back to $O$ clockwise or counterclockwise. This gives us a total of $2\cdot 2 = 4$ valid paths.

How many unordered pairs of edges of a given cube determine a plane?

$\textbf{(A) } 12 \qquad \textbf{(B) } 28 \qquad \textbf{(C) } 36\qquad \textbf{(D) } 42 \qquad \textbf{(E) } 66$

$\textbf{D}$

Without loss of generality, choose one of the $12$ edges of the cube (the red line) to be among the two selected. For two lines in space to determine a common plane, they must either intersect (4 blue lines) or be parallel (3 green lines). So each edge has 7 choices. For each of the $12$ edges, this gives $12\cdot7= 84$ total possible pairings. However, each pair has been counted twice (e.g edge $A$ with $B$, and edge $B$ with $A$). So we divide by 2 to get $42$.

Right triangle $ACD$ with right angle at $C$ is constructed outwards on the hypotenuse $\overline{AC}$ of isosceles right triangle $ABC$ with leg length $1$, as shown, so that the two triangles have equal perimeters. What is $\sin(2\angle BAD)$?

$\textbf{(A) } \dfrac{1}{3} \qquad\textbf{(B) } \dfrac{\sqrt{2}}{2} \qquad\textbf{(C) } \dfrac{3}{4} \qquad\textbf{(D) } \dfrac{7}{9} \qquad\textbf{(E) } \dfrac{\sqrt{3}}{2}$

$\textbf{D}$

We have $CD+AD=AB+BC=2$. Let $CD$ be $x$. By the Pythagorean Theorem, we have $AC=\sqrt2$, $AD=\sqrt{x^2+2}$. So $$x+\sqrt{x^2+2}=2\rightarrow x=\dfrac12$$ Therefore, we get $AD=\dfrac32$, $\cos\angle CAD=\dfrac{2\sqrt2}{3}$. We need to find $$\sin(2\angle BAD)=\sin(90^\circ+2\angle CAD)=\cos(2\angle CAD)=2\cos^2\angle CAD-1=\dfrac79$$

A red ball and a green ball are randomly and independently tossed into bins numbered with positive integers so that for each ball, the probability that it is tossed into bin $k$ is $2^{-k}$ for $k=1,2,3,\ldots.$ What is the probability that the red ball is tossed into a higher-numbered bin than the green ball?

$\textbf{(A) } \dfrac{1}{4} \qquad\textbf{(B) } \dfrac{2}{7} \qquad\textbf{(C) } \dfrac{1}{3} \qquad\textbf{(D) } \dfrac{3}{8} \qquad\textbf{(E) } \dfrac{3}{7}$

$\textbf{C}$

By symmetry, the probability of the red ball landing in a higher-numbered bin is the same as the probability of the green ball landing in a higher-numbered bin. The probability of both landing in the same bin is $$\sum_{k=1}^{\infty}{2^{-k} \cdot 2^{-k}} = \sum_{k=1}^{\infty}2^{-2k} = \frac{1}{3}$$Therefore, since the other two probabilities are the same, they have to be $\dfrac{1-\frac{1}{3}}{2} = \dfrac{1}{3}$.

Let $S$ be the set of all positive integer divisors of $100,000.$ How many numbers are the product of two distinct elements of $S?$

$\textbf{(A) }98\qquad\textbf{(B) }100\qquad\textbf{(C) }117\qquad\textbf{(D) }119\qquad\textbf{(E) }121$

$\textbf{C}$

The prime factorization of $100,000$ is $2^5 \cdot 5^5$. Thus, we choose two numbers $2^a5^b$ and $2^c5^d$ where $0 \le a,b,c,d \le 5$ and $(a,b) \neq (c,d)$, whose product is $2^{a+c}5^{b+d}$, where $0 \le a+c \le 10$ and $0 \le b+d \le 10$.

Notice that this is similar to choosing a divisor of $100,000^2 = 2^{10}5^{10}$, which has $(10+1)(10+1) = 121$ divisors. However, some of the divisors of $2^{10}5^{10}$ cannot be written as a product of two $\textbf{distinct}$ divisors of $2^5 \cdot 5^5$, namely: $1 = 2^05^0$, $2^{10}5^{10}$, $2^{10}$, and $5^{10}$. So the answer is $121-4=117$.

As shown in the figure, line segment $\overline{AD}$ is trisected by points $B$ and $C$ so that $AB=BC=CD=2.$ Three semicircles of radius $1,$ $\stackrel{\huge\frown}{AEB},\stackrel{\huge\frown}{BFC},$ and $\stackrel{\huge\frown}{CGD},$ have their diameters on $\overline{AD},$ and are tangent to line $EG$ at $E,F,$ and $G,$ respectively. A circle of radius $2$ has its center on $F.$ The area of the region inside the circle but outside the three semicircles, shaded in the figure, can be expressed in the form\[\frac{a}{b}\cdot\pi-\sqrt{c}+d,\]where $a,b,c,$ and $d$ are positive integers and $a$ and $b$ are relatively prime. What is $a+b+c+d$?

$\textbf{(A) } 13 \qquad\textbf{(B) } 14 \qquad\textbf{(C) } 15 \qquad\textbf{(D) } 16\qquad\textbf{(E) } 17$

$\textbf{E}$

Let the point $AB$ intersects the large circle be $X$, and the point $CD$ intersects the large circle be $Y$. We see that $FX=FY=2$. The distance between $F$ and line $AB$ is 1, which is half of the length $FX$ or $FY$. So we can see that $\angle XFY=120^\circ$.

We can divide the shaded region into 3 part. The radius of the top semicircle is 2, so its area is $\dfrac12\pi(2)^2=2\pi$. The bottom arc corresponds to a central angle of $120^\circ$, so the bottom area is $\dfrac13\pi(2)^2-\dfrac12\cdot2\cdot2\cdot\sin\angle 120^\circ=\dfrac43\pi-\sqrt3$. The rest part is composed of 4 corners of a circle inscribed in a square, so its area is $2^2-\pi(1)^2=4-\pi$.

Therefore, the area of the shaded region is $2\pi+\dfrac43\pi-\sqrt3+4-\pi=\dfrac73\pi-\sqrt3+4$. The answer is $a+b+c+d=7+3+3+4=17$.

There are lily pads in a row numbered $0$ to $11$, in that order. There are predators on lily pads $3$ and $6$, and a morsel of food on lily pad $10$. Fiona the frog starts on pad $0$, and from any given lily pad, has a $\dfrac{1}{2}$ chance to hop to the next pad, and an equal chance to jump $2$ pads. What is the probability that Fiona reaches pad $10$ without landing on either pad $3$ or pad $6$?

$\textbf{(A) } \dfrac{15}{256} \qquad \textbf{(B) } \dfrac{1}{16} \qquad \textbf{(C) } \dfrac{15}{128}\qquad \textbf{(D) } \dfrac{1}{8} \qquad \textbf{(E) } \dfrac14$

$\textbf{A}$

Since Fiona passes pad 3 and 6, she must land on pad 2, 4, 5, and 7. Now we separate the route into 5 parts.

$\textbf{Part 1: pad 0 to 2}$. It can be achieved by 1-jump and 1-jump with probability $\dfrac14$. or a 2-jump with probability $\dfrac12$. So in total the probability is $\dfrac34$.

$\textbf{Part 2: pad 2 to 4}$. It can be achieved by a 2-jump with probability $\dfrac12$.

$\textbf{Part 3: pad 4 to 5}$. It can be achieved by a 1-jump with probability $\dfrac12$.

$\textbf{Part 4: pad 5 to 7}$. It can be achieved by a 2-jump with probability $\dfrac12$.

$\textbf{Part 5: pad 7 to 10}$. It can be achieved by $1+1+1$, or $1+2$, or $2+1$. So the probability is $\dfrac18+\dfrac14+\dfrac14=\dfrac58$.

Therefore, the final answer is $\dfrac34\cdot\dfrac12\cdot\dfrac12\cdot\dfrac12\cdot\dfrac58=\dfrac{15}{256}$.

How many nonzero complex numbers $z$ have the property that $0, z,$ and $z^3,$ when represented by points in the complex plane, are the three distinct vertices of an equilateral triangle?

$\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }4\qquad\textbf{(E) }\textrm{infinitely many}$

$\textbf{D}$

Let $z=re^{i\theta}$, then $z^3=r^3e^{i3\theta}$. The distance between 0 and $z=re^{i\theta}$ is $r$. The distance between 0 and $z^3=r^3e^{i3\theta}$ is $r^3$. Since it is an equilateral triangle, we have $r=r^3\rightarrow r=1$.

To get from $z$ to $z^3$, which should be a rotation of $60^{\circ}$ if the triangle is equilateral, we multiply by $z^2 = r^2e^{i2\theta}$. Thus we have $2\theta=\pm\dfrac{\pi}{3} + 2\pi k$ (where $k$ can be any integer). Given that $0\leq\theta\leq2\pi$, we get $\theta=\dfrac{\pi}{6},\dfrac{5\pi}{6},\dfrac{7\pi}{6},\dfrac{11\pi}{6}$. The answer is 4.

Square pyramid $ABCDE$ has base $ABCD,$ which measures $3$ cm on a side, and altitude $\overline{AE}$ perpendicular to the base$,$ which measures $6$ cm. Point $P$ lies on $\overline{BE},$ one third of the way from $B$ to $E;$ point $Q$ lies on $\overline{DE},$ one third of the way from $D$ to $E;$ and point $R$ lies on $\overline{CE},$ two thirds of the way from $C$ to $E.$ What is the area, in square centimeters, of $\triangle PQR?$

$\textbf{(A) } \dfrac{3\sqrt2}{2} \qquad\textbf{(B) } \dfrac{3\sqrt3}{2} \qquad\textbf{(C) } 2\sqrt2 \qquad\textbf{(D) } 2\sqrt3 \qquad\textbf{(E) } 3\sqrt2$

$\textbf{C}$

Using the given data, we can label the points $A(0, 0, 0), B(3, 0, 0), C(3, 3, 0), D(0, 3, 0),$ and $E(0, 0, 6)$. We can also find the points $P = B + \dfrac{1}{3} \overrightarrow{BE} = (3,0,0) + \dfrac{1}{3}(-3, 0, 6) = (3,0,0) + (-1,0,2) = (2, 0, 2)$. Similarly, $Q = (0, 2, 2)$ and $R = (1, 1, 4)$.

Using the distance formula, $PQ = \sqrt{\left(-2\right)^2 + 2^2 + 0^2} = 2\sqrt{2}$, $PR = \sqrt{\left(-1\right)^2 + 1^2 + 2^2} = \sqrt{6}$, and $QR = \sqrt{1^2 + \left(-1\right)^2 + 2^2} = \sqrt{6}$. Using Heron's formula, or by dropping an altitude from $P$ to find the height, we can then find that the area of $\triangle{PQR}$ is $2\sqrt{2}$.

Raashan, Sylvia, and Ted play the following game. Each starts with $\$1$. A bell rings every $15$ seconds, at which time each of the players who currently have money simultaneously chooses one of the other two players independently and at random and gives $\$1$ to that player. What is the probability that after the bell has rung $2019$ times, each player will have $\$1$? (For example, Raashan and Ted may each decide to give $\$1$ to Sylvia, and Sylvia may decide to give her her dollar to Ted, at which point Raashan will have $\$0$, Sylvia will have $\$2$, and Ted will have $\$1$, and that is the end of the first round of play. In the second round Rashaan has no money to give, but Sylvia and Ted might choose each other to give their $\$1$ to, and the holdings will be the same at the end of the second round.)

$\textbf{(A) } \dfrac{1}{7} \qquad\textbf{(B) } \dfrac{1}{4} \qquad\textbf{(C) } \dfrac{1}{3} \qquad\textbf{(D) } \dfrac{1}{2} \qquad\textbf{(E) } \dfrac{2}{3}$

$\textbf{B}$

On the first turn, each player starts off with $\$1$. Each turn after that, there are only two possibilities: either everyone stays at $\$1$, which we will write as $(1-1-1)$, or the distribution of money becomes $\$2-\$1-\$0$ in some order, which we write as $(2-1-0)$. ($(3-0-0)$ cannot be achieved). We will consider these two states separately.

In the $(1-1-1)$ state, each person has two choices for whom to give their dollar to, meaning there are $2^3=8$ possible ways that the money can be rearranged. Note that there are only two ways that we can reach $(1-1-1)$ again:

1. Raashan gives his money to Sylvia, who gives her money to Ted, who gives his money to Raashan.

2. Raashan gives his money to Ted, who gives his money to Sylvia, who gives her money to Raashan.

Thus, the probability of staying in the $(1-1-1)$ state is $\dfrac{1}{4}$, while the probability of going to the $(2-1-0)$ state is $\dfrac{3}{4}$.

In the $(2-1-0)$ state, we will label the person with $\$2$ as person A, the person with $\$1$ as person B, and the person with $\$0$ as person C. Person A has two options for whom to give money to, and person B has 2 options for whom to give money to, meaning there are total $2\cdot 2 = 4$ ways the money can be redistributed. The only way that the distribution can return to $(1-1-1)$ is if A gives $\$1$ to B, and B gives $\$1$ to C. We check the other possibilities to find that they all lead back to $(2-1-0)$. Thus, the probability of going to the $(1-1-1)$ state is $\dfrac{1}{4}$, while the probability of staying in the $(2-1-0)$ state is $\dfrac{3}{4}$.

No matter which state we are in, the probability of going to the $(1-1-1)$ state is always $\dfrac{1}{4}$. This means that, after the bell rings 2018 times, regardless of what state the money distribution is in, there is a $\dfrac{1}{4}$ probability of going to the $(1-1-1)$ state after the 2019th bell ring. Thus, our answer is $\dfrac{1}{4}$.

Points $A(6,13)$ and $B(12,11)$ lie on circle $\omega$ in the plane. Suppose that the tangent lines to $\omega$ at $A$ and $B$ intersect at a point on the $x$-axis. What is the area of $\omega$?

$\textbf{(A) }\dfrac{83\pi}{8}\qquad\textbf{(B) }\dfrac{21\pi}{2}\qquad\textbf{(C) } \dfrac{85\pi}{8}\qquad\textbf{(D) }\dfrac{43\pi}{4}\qquad\textbf{(E) }\dfrac{87\pi}{8}$

$\textbf{C}$

Let the intersection point on $x$ axis be $C(x,0)$. We have $$AC=BC\rightarrow\sqrt{(x-6)^2 + 13^2} = \sqrt{(x-12)^2 + 11^2}\rightarrow x=5$$ Let the center of circle $\omega$ be $O$. We have $OA\perp AC$ and $OB\perp BC$. The slope of $AC$ is $\dfrac{13-0}{6-5}=13$. So the slope of $OA$ is $-\dfrac1{13}$. Since $A(6,13)$ is a point on $OA$, we get $OA:y=-\dfrac1{13}x+\dfrac{175}{13}$. Similarly, we get $OB:y=-\dfrac{7}{11}x+\dfrac{205}{11}$.

$OA:y=-\dfrac1{13}x+\dfrac{175}{13}$ intersects $OB:y=-\dfrac{7}{11}x+\dfrac{205}{11}$ at point $O(\dfrac{37}{4},\dfrac{51}{4})$. Thus the radius of circle $\omega$ is $OA=\dfrac{\sqrt{170}}4$. Hence, the area is $\dfrac{85}{8}\pi$.

How many quadratic polynomials with real coefficients are there such that the set of roots equals the set of coefficients? (For clarification: If the polynomial is $ax^2+bx+c,a\neq 0,$ and the roots are $r$ and $s,$ then the requirement is that $\{a,b,c\}=\{r,s\}$.)

$\textbf{(A) } 3 \qquad\textbf{(B) } 4 \qquad\textbf{(C) } 5 \qquad\textbf{(D) } 6 \qquad\textbf{(E) } \textrm{infinitely many}$

$\textbf{B}$

Firstly, if $r=s$, then $a=b=c$, so the equation becomes $ax^2 + ax + a = 0 \Rightarrow x^2 + x + 1=0$, which has no real roots.

Hence there are three cases we need to consider:

$\textbf{Case 1:}$ $a=b=r$ and $c=s \neq r$: The equation becomes $ax^2+ax+c=0$, and by Vieta's Formulas, we have $a+c=-1$ and $ac = \dfrac{c}{a}$. This second equation becomes $(a^2-1)c=0$. Hence one possibility is $c=0$, in which case $a=-1$, giving the equation $-x^2 - x = 0$, which has roots $0$ and ${-}1$. This gives one valid solution. On the other hand, if $c\neq 0$, then $a^2-1=0$, so $a = \pm 1$. If $a=1$, we have $c=-2$, and the equation is $x^2 + x - 2 = 0$, which clearly works, giving a second valid solution. If $a=-1$, then we have $c=0$, which has already been considered, so this possibility gives no further valid solutions.

$\textbf{Case 2:}$ $a=c=r$, $b=s \neq r$: The equation becomes $ax^2 + bx + a = 0$, so by Vieta's Formulas, we have $a+b = -\dfrac{b}{a}$ and $ab = 1$. These equations reduce to $a^3 + a + 1 = 0$. By sketching a graph of $f(x)=x^3$ and $f(x)=-x-1$, we see that there is exactly one intersection. This gives a third valid solution.

$\textbf{Case 3:}$ $a=r$, $b=c=s \neq r$: The equation becomes $ax^2 + bx+b=0$, so by Vieta's Formulas, we have $a+b = -\dfrac{b}{a}$ and $ab = \dfrac{b}{a}$. Observe that $b \neq 0$, as if it were $0$, the equation would just have one real root, $0$, so this would not give a valid solution. Thus, taking the second equation and dividing both sides by $b$, we deduce have $a=\dfrac{1}{a}$, so $a=\pm 1$. If $a=1$, we have $1+b=-b$, giving $b=-\dfrac{1}{2}$, so the equation is $x^2 - \dfrac{1}{2}x - \dfrac{1}{2} = 0$, which is a fourth valid solution. If $a=-1$, we have $1+b=b$, which is a contradiction, so this case gives no further valid solutions.

Hence the total number of valid solutions is $4$.

Define a sequence recursively by $x_0=5$ and\[x_{n+1}=\frac{x_n^2+5x_n+4}{x_n+6}\]for all nonnegative integers $n.$ Let $m$ be the least positive integer such that\[x_m\leq 4+\frac{1}{2^{20}}.\]In which of the following intervals does $m$ lie?

$\textbf{(A) } [9,26] \qquad\textbf{(B) } [27,80] \qquad\textbf{(C) } [81,242]\qquad\textbf{(D) } [243,728] \qquad\textbf{(E) } [729,\infty]$

$\textbf{C}$

We first prove that $x_n > 4$ for all $n \ge 0$, by induction. Observe that\[x_{n+1} - 4 = \frac{x_n^2 + 5x_n + 4 - 4(x_n+6)}{x_n+6} = \frac{(x_n - 4)(x_n+5)}{x_n+6}.\]so $x_{n+1} > 4$ if and only if $x_{n} > 4$.

We similarly prove that $x_n$ is decreasing:\[x_{n+1} - x_n = \frac{x_n^2 + 5x_n + 4 - x_n(x_n+6)}{x_n+6} = \frac{4-x_n}{x_n+6} < 0.\]Now we need to estimate the value of $x_{n+1}-4$, which we can do using the rearranged equation:\[x_{n+1} - 4 = (x_n-4)\cdot\frac{x_n + 5}{x_n+6}.\]Since $x_n$ is decreasing, $\dfrac{x_n + 5}{x_n+6}$ is also decreasing, so we have\[\frac{9}{10} < \frac{x_n + 5}{x_n+6} \le \frac{10}{11}\]and\[\frac{9}{10}(x_n-4) < x_{n+1} - 4 \le \frac{10}{11}(x_n-4)\]This becomes\[\left(\frac{9}{10}\right)^n = \left(\frac{9}{10}\right)^n \left(x_0-4\right) < x_{n} - 4 \le \left(\frac{10}{11}\right)^n \left(x_0-4\right) = \left(\frac{10}{11}\right)^n\]The problem thus reduces to finding the least value of $n$ such that\[\left(\frac{9}{10}\right)^n < x_{n} - 4 \le \frac{1}{2^{20}} \text{ and } \left(\frac{10}{11}\right)^{n-1} > x_{n-1} - 4 > \frac{1}{2^{20}}\] Taking logarithms, we get $n \ln \dfrac{9}{10} < -20 \ln 2$ and $(n-1)\ln \dfrac{10}{11} > -20 \ln 2$, i.e.\[n > \frac{20\ln 2}{\ln\frac{10}{9}} \text{ and } n-1 < \frac{20\ln 2}{\ln\frac{11}{10}}\]As approximations, we can use $\ln\dfrac{10}{9} \approx \dfrac{1}{9}$, $\ln\dfrac{11}{10} \approx \dfrac{1}{10}$, and $\ln 2\approx 0.7$. These approximations allow us to estimate\[126 < n < 141,\]which gives C$[81,242]$.

How many sequences of $0$s and $1$s of length $19$ are there that begin with a $0$, end with a $0$, contain no two consecutive $0$s, and contain no three consecutive $1$s?

$\textbf{(A) }55\qquad\textbf{(B) }60\qquad\textbf{(C) }65\qquad\textbf{(D) }70\qquad\textbf{(E) }75$

$\textbf{C}$

Any valid sequence of length $n$ will start with a $0$ followed by either $10$ or $110$. Since $n=19$, we need to add several $10$s and $110$s after the first 0 to make the rest 18 numbers.

$\textbf{Case 1:}$ nine $10$s - there is only $1$ way to arrange them.

$\textbf{Case 2:}$ two $110$s and six $10$s - there are $\dbinom82 = 28$ ways to arrange them.

$\textbf{Case 3:}$ four $110$s and three $10$s - there are $\dbinom74 = 35$ ways to arrange them.

$\textbf{Case 4:}$ six $110$s - there is only $1$ way to arrange them.

Summing the four cases gives $1+28+35+1= 65$.

Let $\omega=-\dfrac{1}{2}+\dfrac{1}{2}i\sqrt3.$ Let $S$ denote all points in the complex plane of the form $a+b\omega+c\omega^2,$ where $0\leq a \leq 1,0\leq b\leq 1,$ and $0\leq c\leq 1.$ What is the area of $S$?

$\textbf{(A) } \dfrac{1}{2}\sqrt3 \qquad\textbf{(B) } \dfrac{3}{4}\sqrt3 \qquad\textbf{(C) } \dfrac{3}{2}\sqrt3\qquad\textbf{(D) } \dfrac{1}{2}\pi\sqrt3 \qquad\textbf{(E) } \pi$

$\textbf{C}$

Notice that $\omega=e^{\frac{2i\pi}{3}}$, which is one of the cube roots of unity. The other two cube roots are 1 and $\omega^2$. Since $0\leq a \leq 1,0\leq b\leq 1,0\leq c\leq 1$, first we represent all possible value of $c\omega^2$ as blue line in the following diagram:

Then we add $b\omega$ as the red line. $b\omega+c\omega^2$ means the following diagram:

Finally we add $a$ as the green line. Now $a+b\omega+c\omega^2$ turns to:

The diagram is a regular hexagon with side length 1. So the area is $\dfrac32\sqrt3$.

Let $ABCD$ be a convex quadrilateral with $BC=2$ and $CD=6.$ Suppose that the centroids of $\triangle ABC,\triangle BCD,$ and $\triangle ACD$ form the vertices of an equilateral triangle. What is the maximum possible value of the area of $ABCD$?

$\textbf{(A) } 27 \qquad\textbf{(B) } 16\sqrt3 \qquad\textbf{(C) } 12+10\sqrt3 \qquad\textbf{(D) } 9+12\sqrt3 \qquad\textbf{(E) } 30$

$\textbf{C}$

Place an origin at $A$, and assign position vectors of $B = \vec{p}$ and $D = \vec{q}$. Since $AB$ is not parallel to $AD$, vectors $\vec{p}$ and $\vec{q}$ are linearly independent, so we can write $C = m\vec{p} + n\vec{q}$ for some constants $m$ and $n$.

Recall that the centroid of a triangle $\triangle XYZ$ has position vector $\dfrac{1}{3}\left(\vec{x}+\vec{y}+\vec{z}\right)$. Thus the centroid of $\triangle ABC$ is $g_1 = \dfrac{1}{3}(m+1)\vec{p} + \dfrac{1}{3}n\vec{q}$; the centroid of $\triangle BCD$ is $g_2 = \dfrac{1}{3}(m+1)\vec{p} + \dfrac{1}{3}(n+1)\vec{q}$; and the centroid of $\triangle ACD$ is $g_3 = \dfrac{1}{3}m\vec{p} + \dfrac{1}{3}(n+1)\vec{q}$.

Hence $\overrightarrow{G_{1}G_{2}} = \dfrac{1}{3}\vec{q}$, $\overrightarrow{G_{2}G_{3}} = -\dfrac{1}{3}\vec{p}$, and $\overrightarrow{G_{3}G_{1}} = \dfrac{1}{3}\vec{p} - \dfrac{1}{3}\vec{q}$. For $\triangle G_{1}G_{2}G_{3}$ to be equilateral, we need $\left|\overrightarrow{G_{1}G_{2}}\right| = \left|\overrightarrow{G_{2}G_{3}}\right| \Rightarrow \left|\vec{p}\right| = \left|\vec{q}\right| \Rightarrow AB = AD$. Further, $\left|\overrightarrow{G_{1}G_{2}}\right| = \left|\overrightarrow{G_{1}G_{3}}\right| \Rightarrow \left|\vec{p}\right| = \left|\vec{p} - \vec{q}\right| = BD$. Therefore, we have $AB = AD = BD$, so $\triangle ABD$ is equilateral.

Now let the side length of $\triangle ABD$ be $k$, and let $\angle BCD = \theta$. By the Law of Cosines in $\triangle BCD$, we have $k^2 = 2^2 + 6^2 - 2 \cdot 2 \cdot 6 \cdot \cos{\theta} = 40 - 24\cos{\theta}$. Since $\triangle ABD$ is equilateral, its area is $\dfrac{\sqrt{3}}{4}k^2 = 10\sqrt{3} - 6\sqrt{3}\cos{\theta}$, while the area of $\triangle BCD$ is $\dfrac{1}{2} \cdot 2 \cdot 6 \cdot \sin{\theta} = 6 \sin{\theta}$. Thus the total area of $ABCD$ is $10\sqrt{3} + 6\left(\sin{\theta} - \sqrt{3}\cos{\theta}\right) = 10\sqrt{3} + 12\left(\dfrac{1}{2} \sin{\theta} - \dfrac{\sqrt{3}}{2}\cos{\theta}\right) = 10\sqrt{3}+12\sin{\left(\theta-60^{\circ}\right)}$, where in the last step we used the subtraction formula for $\sin$. $\sin{\left(\theta-60^{\circ}\right)}$ has maximum value $1$. So the maximum area is $12+10\sqrt3$.