## AMC 12 2019 Test A

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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.

The area of a pizza with radius $4$ inches is $N$ percent larger than the area of a pizza with radius $3$ inches. What is the integer closest to $N$?

$\textbf{(A) } 25 \qquad\textbf{(B) } 33 \qquad\textbf{(C) } 44\qquad\textbf{(D) } 66 \qquad\textbf{(E) } 78$

$\textbf{E}$

The area of the larger pizza is $16\pi$, while the area of the smaller pizza is $9\pi$. Therefore, the larger pizza is $\dfrac{7\pi}{9\pi} \cdot 100\%$ bigger than the smaller pizza. $\dfrac{7\pi}{9\pi} \cdot 100\% = 77.777....$, which is closest to $78$.

Suppose $a$ is $150\%$ of $b$. What percent of $a$ is $3b$?

$\textbf{(A) } 50 \qquad \textbf{(B) } 66+\dfrac{2}{3} \qquad \textbf{(C) } 150 \qquad \textbf{(D) } 200 \qquad \textbf{(E) } 450$

$\textbf{D}$

Since $a=1.5b$, we have $\dfrac{3b}{a}=2=200\%$.

A box contains $28$ red balls, $20$ green balls, $19$ yellow balls, $13$ blue balls, $11$ white balls, and $9$ black balls. What is the minimum number of balls that must be drawn from the box without replacement to guarantee that at least $15$ balls of a single color will be drawn?

$\textbf{(A) } 75 \qquad\textbf{(B) } 76 \qquad\textbf{(C) } 79 \qquad\textbf{(D) } 84 \qquad\textbf{(E) } 91$

$\textbf{B}$

In the worst case we draw $14$ red balls, $14$ green balls, $14$ yellow balls, $13$ blue balls, $11$ white balls, and $9$ black balls, for a total of $75$ balls, without drawing $15$ balls of any single color. Drawing one more ball guarantees that we will get $15$ balls of a single color — either red, green, or yellow. Thus, the answer is $75 + 1 =76$.

What is the greatest number of consecutive integers whose sum is $45$?

$\textbf{(A) } 9 \qquad\textbf{(B) } 25 \qquad\textbf{(C) } 45 \qquad\textbf{(D) } 90 \qquad\textbf{(E) } 120$

$\textbf{D}$

To get the longest sequence of consecutive integers, we hope the positive integers and negative integers cancel each other, which is $-44, -43, \cdots, 43, 44$. Since the sum is 45, we need to add the number 45 at the end of the sequence. So the number of elements in the sequence $-44, -43, \cdots, 43, 44, 45$ is 90.

Two lines with slopes $\dfrac{1}{2}$ and $2$ intersect at $(2,2)$. What is the area of the triangle enclosed by these two lines and the line $x+y=10$?

$\textbf{(A) } 4 \qquad\textbf{(B) } 4\sqrt{2} \qquad\textbf{(C) } 6 \qquad\textbf{(D) } 8 \qquad\textbf{(E) } 6\sqrt{2}$

$\textbf{C}$

The line of slope $\dfrac12$ passing through $(2,2)$ is $y=\dfrac12x+1$. The line of slope $2$ passing through $(2,2)$ is $y=2x-2$. The intersection points of $y=\dfrac12x+1$, $y=2x-2$ and $y=-x+10$ are $(2,2)$, $(6,4)$ and $(4,6)$. Apparently the triangle enclosed by these lines is isosceles. By the Pythagorean Theorem, the base is $2\sqrt2$, and the altitude is $3\sqrt2$. So the area is 6.

The figure below shows line $\ell$ with a regular, infinite, recurring pattern of squares and line segments.

How many of the following four kinds of rigid motion transformations of the plane in which this figure is drawn, other than the identity transformation, will transform this figure into itself?

$\quad\bullet\quad$some rotation around a point of line $\ell$

$\quad\bullet\quad$some translation in the direction parallel to line $\ell$

$\quad\bullet\quad$the reflection across line $\ell$

$\quad\bullet\quad$some reflection across a line perpendicular to line $\ell$

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

$\textbf{C}$

Statement $1$ is true. A $180^{\circ}$ rotation about the point half way between an up-facing square and a down-facing square will yield the same figure.

Statement $2$ is also true. A translation to the left or right will place the image onto itself when the figures above and below the line realign (the figure goes on infinitely in both directions).

Statement $3$ is false. A reflection across line $\ell$ will change the up-facing squares to down-facing squares and vice versa.

Finally, statement $4$ is also false because it will cause the diagonal lines extending from the squares to switch direction.

Thus, only $2$ statements are true.

Melanie computes the mean $\mu$, the median $M$, and the modes of the $365$ values that are the dates in the months of $2019$. Thus her data consist of $12$ $1\textrm{s}$, $12$ $2\textrm{s}$, . . . , $12$ $28\textrm{s}$, $11$ $29\textrm{s}$, $11$ $30\textrm{s}$, and $7$ $31\textrm{s}$. Let $d$ be the median of the modes. Which of the following statements is true?

$\textbf{(A) } \mu < d < M \qquad\textbf{(B) } M < d < \mu \qquad\textbf{(C) } d = M =\mu \qquad\textbf{(D) } d < M < \mu \qquad\textbf{(E) } d < \mu < M$

$\textbf{E}$

In the new sequence of $12$ $1\textrm{s}$, $12$ $2\textrm{s}$, . . . , $12$ $28\textrm{s}$, we have $d=\mu=M$. However, when we extend the sequence to $12$ $1\textrm{s}$, $12$ $2\textrm{s}$, . . . , $12$ $28\textrm{s}$, $11$ $29\textrm{s}$, $11$ $30\textrm{s}$, $7$ $31\textrm{s}$, $d$ keeps unchanged while $\mu$ and $M$ become greater. So $d$ is the smallest.

In a sequence of 365 numbers, $M$ is the $183\textrm{rd}$ one. Since $183=12\cdot15+3$, we get $M=16$. If we extend the sequence to $12$ $1\textrm{s}$, $12$ $2\textrm{s}$, . . . , $12$ $28\textrm{s}$, $12$ $29\textrm{s}$, $12$ $30\textrm{s}$, $12$ $31\textrm{s}$, $\mu$ will be larger. The new mean value $\mu=16$. So in the original sequence $\mu<M$.

Therefore, the answer is $d<\mu<M$.

For a set of four distinct lines in a plane, there are exactly $N$ distinct points that lie on two or more of the lines. What is the sum of all possible values of $N$?

$\textbf{(A) } 14 \qquad \textbf{(B) } 16 \qquad \textbf{(C) } 18 \qquad \textbf{(D) } 19 \qquad \textbf{(E) } 21$

$\textbf{D}$

The maximum number of possible intersections is $ \dbinom42 = 6$, since each pair of lines can intersect at most once. It is possible to obtain $0$, $1$, $3$, $4$, $5$, and $6$ points of intersection, as demonstrated in the following figures:

It is impossible to have 2 intersections because if line 1 intersects only line 2, and line 3 intersects only line 4, we have $1\parallel3\parallel4$ and $3\parallel1\parallel2$. So we get $1\parallel2\parallel3\parallel4$, which means there is no intersection at all.

Therefore, the answer is $0+1+3+4+5+6= 19$.

A sequence of numbers is defined recursively by $a_1 = 1$, $a_2 = \dfrac{3}{7}$, and\[a_n=\frac{a_{n-2} \cdot a_{n-1}}{2a_{n-2} - a_{n-1}}\]for all $n \geq 3$. Then $a_{2019}$ can be written as $\dfrac{p}{q}$, where $p$ and $q$ are relatively prime positive integers. What is $p+q ?$

$\textbf{(A) } 2020 \qquad\textbf{(B) } 4039 \qquad\textbf{(C) } 6057 \qquad\textbf{(D) } 6061 \qquad\textbf{(E) } 8078$

$\textbf{E}$

We have $$\dfrac{1}{a_n} = \dfrac{2a_{n-2}-a_{n-1}}{a_{n-2} \cdot a_{n-1}}=\dfrac{2}{a_{n-1}}-\dfrac{1}{a_{n-2}}$$ in other words, $$\dfrac{1}{a_n}-\dfrac{1}{a_{n-1}} = \dfrac{1}{a_{n-1}}-\dfrac{1}{a_{n-2}}$$ So $\{\dfrac{1}{a_n}\}$ is an arithmetic sequence with step size $\dfrac{7}{3}-1=\dfrac{4}{3}$, which means $$\dfrac{1}{a_{2019}} = 1+2018 \cdot \dfrac{4}{3} = \dfrac{8075}{3}$$ Since the numerator and the denominator are relatively prime, the answer is $8078$.

The figure below shows $13$ circles of radius $1$ within a larger circle. All the intersections occur at points of tangency. What is the area of the region, shaded in the figure, inside the larger circle but outside all the circles of radius $1$?

$\textbf{(A) } 4 \pi \sqrt{3} \qquad\textbf{(B) } 7 \pi \qquad\textbf{(C) } \pi\left(3\sqrt{3} +2\right) \qquad\textbf{(D) } 10 \pi \left(\sqrt{3} - 1\right) \qquad\textbf{(E) } \pi\left(\sqrt{3} + 6\right)$

$\textbf{A}$

In equilateral triangles $OAB$ and $ABC$, we have $OC=2\sqrt3$. So the radius of the larger circle is $2\sqrt3+1$. The area of the shaded region is $\pi(2\sqrt3+1)^2-13\pi\cdot1^2=4\sqrt3\pi$.

For some positive integer $k$, the repeating base-$k$ representation of the (base-ten) fraction $\dfrac{7}{51}$ is $0.\overline{23}_k = 0.232323..._k$. What is $k$?

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

$\textbf{D}$

We can expand the fraction $0.\overline{23}_k$ as follows:\begin{align*}

0.\overline{23}_k &= 2\cdot k^{-1} + 3 \cdot k^{-2} + 2 \cdot k^{-3} + 3 \cdot k^{-4} + \cdots\\

&=2( k^{-1} + k^{-3} + k^{-5} +\cdots) + 3 (k^{-2} + k^{-4} + k^{-6} + \cdots )\\

&=2\cdot\dfrac{k}{k^2-1}+3\cdot\dfrac{1}{k^2-1}\\

&=\dfrac{2k+3}{k^2-1}\\

&=\dfrac{7}{51}

\end{align*}Hence, we get $k =16$.

Positive real numbers $x \neq 1$ and $y \neq 1$ satisfy $\log_2{x} = \log_y{16}$ and $xy = 64$. What is $(\log_2{\dfrac{x}{y}})^2$?

$\textbf{(A) } \dfrac{25}{2} \qquad\textbf{(B) } 20 \qquad\textbf{(C) } \dfrac{45}{2} \qquad\textbf{(D) } 25 \qquad\textbf{(E) } 32$

$\textbf{B}$

Given that $\log_2{x} = \log_y{16}=4\log_y2=4\cdot\dfrac1{\log_2y}$ and $\log_2{xy}=\log_2x+\log_2y=\log_2{64}=6$, we have \begin{align*}

\log_2x\cdot\log_2y&=4\\

\log_2x+\log_2y&=6

\end{align*} So the required answer is \begin{align*}

(\log_2{\dfrac{x}{y}})^2&=(\log_2x-\log_2y)^2\\

&=(\log_2x+\log_2y)^2-4\log_2x\cdot\log_2y\\

&=6^2-4\cdot4\\

&=20

\end{align*}

How many ways are there to paint each of the integers $2, 3, \dots, 9$ either red, green, or blue so that each number has a different color from each of its proper divisors?

$\textbf{(A)}\ 144\qquad\textbf{(B)}\ 216\qquad\textbf{(C)}\ 256\qquad\textbf{(D)}\ 384\qquad\textbf{(E)}\ 432$

$\textbf{E}$

The $5$ and $7$ can be painted with no restrictions because the set of integers does not contain a multiple or proper factor of $5$ or $7$. There are 3 ways to paint each, giving us $\underline{9}$ ways to paint both. The $2$ is the most restrictive number. There are $\underline{3}$ ways to paint $2$, but without loss of generality, let it be painted red. $4$ cannot be the same color as $2$ or $8$, so there are $\underline{2}$ ways to paint $4$, which automatically determines the color for $8$. $6$ cannot be painted red, so there are $\underline{2}$ ways to paint $6$, but WLOG, let it be painted blue. There are $\underline{2}$ choices for the color for $3$, which is either red or green in this case. Lastly, there are $\underline{2}$ ways to choose the color for $9$.

So the answer is $9 \cdot 3 \cdot 2 \cdot 2 \cdot 2 \cdot 2 =432$.

For a certain complex number $c$, the polynomial\[P(x) = (x^2 - 2x + 2)(x^2 - cx + 4)(x^2 - 4x + 8)\]has exactly 4 distinct roots. What is $|c|$?

$\textbf{(A) } 2 \qquad \textbf{(B) } \sqrt{6} \qquad \textbf{(C) } 2\sqrt{2} \qquad \textbf{(D) } 3 \qquad \textbf{(E) } \sqrt{10}$

$\textbf{E}$

The polynomial can be factored further broken down into$$P(x) = (x - [1 - i])(x - [1 + i])(x - [2 - 2i])(x - [2 + 2i])(x^2 - cx + 4)$$by using the quadratic formula on each of the quadratic factors. Since the first four roots are all distinct, the term $(x^2 - cx + 4)$ must be a product of any combination of two (not necessarily distinct) factors from the set: $(x - [1 - i]), (x - [1 + i]), (x - [2 - 2i]),$ and $(x - [2 + 2i])$. We need the two factors to yield a constant term of $4$ when multiplied together. The only combinations that work are $(x - [1 - i])$ and $(x - [2 + 2i])$, or $(x - [1+i])$ and $(x - [2-2i])$. When multiplied together, the polynomial is either $(x^2 + [-3 + i]x + 4)$ or $(x^2+[-3-i]x+4)$. Therefore, $c = 3 \pm i$ and $|c| = \sqrt{10}$.

Positive real numbers $a$ and $b$ have the property that\[\sqrt{\log{a}} + \sqrt{\log{b}} + \log \sqrt{a} + \log \sqrt{b} = 100\]and all four terms on the left are positive integers, where $\log$ denotes the base-$10$ logarithm. What is $ab$?

$\textbf{(A) } 10^{52} \qquad \textbf{(B) } 10^{100} \qquad \textbf{(C) } 10^{144} \qquad \textbf{(D) } 10^{164} \qquad \textbf{(E) } 10^{200}$

$\textbf{D}$

Let $\sqrt{\log{a}}$ be $x$ and $\sqrt{\log{b}}$ be $y$. Then $\log \sqrt{a}=\dfrac12\log a=\dfrac12x^2$, $\log \sqrt{b}=\dfrac12\log b=\dfrac12y^2$. Now the equation turns to $$x+y+\dfrac12x^2+\dfrac12y^2=100$$ which can be rearranged as $$(x+1)^2+(y+1)^2=202$$ Since all terms are positive integers, we know that both $x$ and $y$ are positive even numbers. So we find $(x,y)=(10,8)$. Then $a=10^{100}$, $b=10^{64}$. The answer is $ab=10^{164}$.

The numbers $1,2,\dots,9$ are randomly placed into the $9$ squares of a $3 \times 3$ grid. Each square gets one number, and each of the numbers is used once. What is the probability that the sum of the numbers in each row and each column is odd?

$\textbf{(A) }\dfrac{1}{21}\qquad\textbf{(B) }\dfrac{1}{14}\qquad\textbf{(C) }\dfrac{5}{63}\qquad\textbf{(D) }\dfrac{2}{21}\qquad\textbf{(E) } \dfrac17$

$\textbf{B}$

There are 4 even numbers and 5 odd numbers in total. To get an odd sum in a row/column, there must be 2 even numbers or 0 even number in the same row/column. Where there is an even number, there must be an even number in the same row, an even number in the same column, then the location of the 4th even number is confirmed. Connecting the locations of 4 even numbers with lines and we get a rectangle. We have 9 possible rectangles in the $3\times3$ grid. Then we have 4! ways to permute the even numbers in the rectangle, and 5! way to permute the rest 5 odd numbers. The answer is $\dfrac{9\cdot4!\cdot5!}{9!}=\dfrac{1}{14}$.

Let $s_k$ denote the sum of the $\textit{k}$th powers of the roots of the polynomial $x^3-5x^2+8x-13$. In particular, $s_0=3$, $s_1=5$, and $s_2=9$. Let $a$, $b$, and $c$ be real numbers such that $s_{k+1} = a \, s_k + b \, s_{k-1} + c \, s_{k-2}$ for $k = 2$, $3$, $....$ What is $a+b+c$?

$\textbf{(A)} \; -6 \qquad \textbf{(B)} \; 0 \qquad \textbf{(C)} \; 6 \qquad \textbf{(D)} \; 10 \qquad \textbf{(E)} \; 26$

$\textbf{D}$

Let $p, q$, and $r$ be the roots of the polynomial. We have \begin{align*}

p^3 &= 5p^2 - 8p + 13\\

q^3 &= 5q^2 - 8q + 13\\

r^3 &= 5r^2 - 8r + 13

\end{align*} Rising the power to $k+1$, we get \begin{align*}

p^{k+1} &= 5p^k - 8p^{k-1} + 13p^{k-2}\\

q^{k+1} &= 5q^k - 8q^{k-1} + 13q^{k-2}\\

r^{k+1} &= 5r^k - 8r^{k-1} + 13r^{k-2}

\end{align*} Adding together, we get $$p^{k+1}+q^{k+1}+r^{k+1}= 5\left(p^k+q^k+r^k\right) - 8\left(p^{k-1}+q^{k-1}+r^{k-1}\right) + 13\left(p^{k-2}+q^{k-2}+r^{k-2}\right)$$ which is $$s_{k+1}=5s_k-8s_{k-1}+13s_{k-2}$$ So the answer is $a+b+c=5-8+13=10$.

A sphere with center $O$ has radius $6$. A triangle with sides of length $15, 15,$ and $24$ is situated in space so that each of its sides is tangent to the sphere. What is the distance between $O$ and the plane determined by the triangle?

$\textbf{(A) }2\sqrt{3}\qquad \textbf{(B) }4\qquad \textbf{(C) }3\sqrt{2}\qquad \textbf{(D) }2\sqrt{5}\qquad \textbf{(E) }5\qquad$

$\textbf{D}$

The 3D diagram is shown below:

Here we focus on the plane of the triangle:

The triangle is placed on the sphere so that its three sides are tangent to the sphere. The cross-section of the sphere created by the plane of the triangle is also the incircle of the triangle. To find the inradius, use $\text{area} = \dfrac12\cdot\text{inradius} \cdot \text{perimeter}$. The area of the triangle can be found by drawing an altitude from the vertex between sides with length $15$ to the midpoint of the side with length $24$. The Pythagorean triple $9$ - $12$ - $15$ allows us easily to determine that the base is $24$ and the height is $9$. So the area of the triangle is $\dfrac12\cdot24\cdot9=108$. The perimeter of the triangle is $15 + 15 + 24= 54$. After plugging into the equation, we thus get $108 = \dfrac12\cdot\text{inradius} \cdot 54$, so the inradius is $4$.

Now, let the distance between $O$ and the plane of the triangle be $x$. Choose a point on the incircle and denote it by $A$. The distance $OA$ is $6$, because it is just the radius of the sphere. The distance from point $A$ to the center of the incircle is $4$, because it is the radius of the incircle. By using the Pythagorean Theorem, we thus find $x = \sqrt{6^2-4^2}=\sqrt{20} =2 \sqrt {5}$.

In $\triangle ABC$ with integer side lengths,\[\cos A=\frac{11}{16}, \qquad \cos B= \frac{7}{8}, \qquad \textrm{and} \qquad\cos C=-\frac{1}{4}.\]What is the least possible perimeter for $\triangle ABC$?

$\textbf{(A) } 9 \qquad \textbf{(B) } 12 \qquad \textbf{(C) } 23 \qquad \textbf{(D) } 27 \qquad \textbf{(E) } 44$

$\textbf{A}$

Notice that by the Law of Sines, $a:b:c = \sin{A}:\sin{B}:\sin{C}$, so let's flip all the cosines using $\sin^{2}{x} + \cos^{2}{x} = 1$ ($\sin{x}$ is positive for $0^{\circ} < x < 180^{\circ}$, so we're good there).$$\sin A=\frac{3\sqrt{15}}{16}, \qquad \sin B= \frac{\sqrt{15}}{8}, \qquad \text{and} \qquad\sin C=\frac{\sqrt{15}}{4}$$These are in the ratio $3:2:4$, so our minimal triangle has side lengths $2$, $3$, and $4$. The answer is $2+3+4=9$.

Real numbers between $0$ and $1$, inclusive, are chosen in the following manner. A fair coin is flipped. If it lands heads, then it is flipped again and the chosen number is $0$ if the second flip is heads and $1$ if the second flip is tails. On the other hand, if the first coin flip is tails, then the number is chosen uniformly at random from the closed interval $[0,1]$. Two random numbers $x$ and $y$ are chosen independently in this manner. What is the probability that $|x-y| > \dfrac{1}{2}$?

$\textbf{(A) } \dfrac{1}{3} \qquad \textbf{(B) } \dfrac{7}{16} \qquad \textbf{(C) } \dfrac{1}{2} \qquad \textbf{(D) } \dfrac{9}{16} \qquad \textbf{(E) } \dfrac{2}{3}$

$\textbf{B}$

There are 4 cases depending on what the first coin flip is when determining $x$ and what the first coin flip is when determining $y$.

$\textbf{Case 1:}$ $x$ is either $0$ or $1$, and $y$ is either $0$ or $1$.

$\textbf{Case 2:}$ $x$ is either $0$ or $1$, and $y$ is chosen from the interval $[0,1]$.

$\textbf{Case 3:}$ $x$ is is chosen from the interval $[0,1]$, and $y$ is either $0$ or $1$.

$\textbf{Case 4:}$ $x$ is is chosen from the interval $[0,1]$, and $y$ is also chosen from the interval $[0,1]$.

Each case has a $\dfrac{1}{4}$ chance of occurring (as it requires two coin flips).

For Case 1, we need $x$ and $y$ to be different. Therefore, the probability for success in Case 1 is $\dfrac{1}{2}$.

For Case 2, if $x$ is 0, we need $y$ to be in the interval $\left(\dfrac{1}{2}, 1\right]$. If $x$ is 1, we need $y$ to be in the interval $\left[0, \dfrac{1}{2}\right)$. Regardless of what $x$ is, the probability for success for Case 2 is $\dfrac{1}{2}$.

By symmetry, Case 3 has the same success rate as Case 2.

For Case 4, we must use geometric probability because there are an infinite number of pairs $(x, y)$ that can be selected, whether they satisfy the inequality or not. Graphing $|x-y| > \dfrac{1}{2}$ gives us the following picture where the shaded area is the set of all the points that fulfill the inequality:

The shaded area is $\dfrac{1}{4}$, which means the probability for success for case 4 is $\dfrac{1}{4}$ (since the total area of the bounding square, containing all possible pairs, is $1$).

Adding up the success rates from each case, we get the probability $\left(\dfrac{1}{4}\right) \cdot \left(\dfrac{1}{2} + \dfrac{1}{2} + \dfrac{1}{2} + \dfrac{1}{4}\right) = \dfrac{7}{16}$.

Let\[z=\frac{1+i}{\sqrt{2}}.\]What is\[\left(z^{1^2}+z^{2^2}+z^{3^2}+\dots+z^{{12}^2}\right) \cdot \left(\frac{1}{z^{1^2}}+\frac{1}{z^{2^2}}+\frac{1}{z^{3^2}}+\dots+\frac{1}{z^{{12}^2}}\right)?\]

$\textbf{(A) } 18 \qquad \textbf{(B) } 72-36\sqrt2 \qquad \textbf{(C) } 36 \qquad \textbf{(D) } 72 \qquad \textbf{(E) } 72+36\sqrt2$

$\textbf{C}$

Note that $z = e^{i\frac{\pi}{4}}$. So $z^{k} = z^{k + 8}$ for all positive integers $k$ because of De Moivre's Theorem. Therefore, we want to look at the exponents of each term modulo $8$.

$1^2, 3^2, 5^2, 7^2, 9^2$ and $11^2$ are all $1 \pmod{8}$

$2^2, 6^2,$ and $10^2$ are all $4 \pmod{8}$

$4^2, 8^2,$ and $12^2$ are all $0 \pmod{8}$

Therefore,

$z^{1^2} =z^{3^2} = z^{5^2} = z^{7^2} =z^{9^2} = z^{11^2} =z$

$z^{2^2} = z^{6^2} = z^{10^2} = z^4= e^{i\pi}= -1$

$z^{4^2} = z^{8^2} = z^{12^2} =z^0= 1$

Then we have\begin{align*} z^{1^2}+z^{2^2}+z^{3^2}+\dots+z^{{12}^2}&=6z\\

\frac{1}{z^{1^2}}+\frac{1}{z^{2^2}}+\frac{1}{z^{3^2}}+\dots+\frac{1}{z^{{12}^2}}&=\dfrac{6}{z}

\end{align*} By multiplication, the answer is 36.

Circles $\omega$ and $\gamma$, both centered at $O$, have radii $20$ and $17$, respectively. Equilateral triangle $ABC$, whose interior lies in the interior of $\omega$ but in the exterior of $\gamma$, has vertex $A$ on $\omega$, and the line containing side $\overline{BC}$ is tangent to $\gamma$. Segments $\overline{AO}$ and $\overline{BC}$ intersect at $P$, and $\frac{BP}{CP} = 3$. Then $AB$ can be written in the form $\dfrac{m}{\sqrt{n}} - \dfrac{p}{\sqrt{q}}$ for positive integers $m$, $n$, $p$, $q$ with $\gcd(m,n) = \gcd(p,q) = 1$. What is $m+n+p+q$? $\phantom{ }$

$\textbf{(A) } 42 \qquad \textbf{(B) }86 \qquad \textbf{(C) } 92 \qquad \textbf{(D) } 114 \qquad \textbf{(E) } 130$

$\textbf{E}$

Let $S$ be the point of tangency between $\overline{BC}$ and $\gamma$, and $M$ be the midpoint of $\overline{BC}$. Note that $AM \perp BS$ and $OS \perp BS$. This implies that $\triangle PMA \sim \triangle PSO$.

Let $s$ be the side length of $\triangle ABC$, then it follows that $AM = \dfrac{\sqrt{3}}{2}s$ and $PM = \dfrac{s}{4}$. This means $AP = \dfrac{\sqrt{13}}{4}s$. So $\dfrac{AM}{AP} = \dfrac{2\sqrt{3}}{\sqrt{13}}$. Furthermore, $$\dfrac{AM}{AP}=\dfrac{OS}{OP}=\dfrac{AM+OS}{AP+OP}=\dfrac{\frac{\sqrt3}{2}s+17}{20}$$ Hence, we get\[\frac{\frac{\sqrt{3}}{2}s + 17}{20} = \frac{2\sqrt{3}}{\sqrt{13}}\] The side length of $AB$ is\[s = \frac{80}{\sqrt{13}} - \frac{34}{\sqrt{3}}\]The problem asks for $m + n + p + q = 80 + 13 + 34 + 3 = 130$.

Define binary operations $\diamondsuit$ and $\heartsuit$ by\[a \, \diamondsuit \, b = a^{\log_{7}(b)} \qquad \textrm{and} \qquad a \, \heartsuit \, b = a^{\frac{1}{\log_{7}(b)}}\]for all real numbers $a$ and $b$ for which these expressions are defined. The sequence $(a_n)$ is defined recursively by $a_3 = 3\, \heartsuit\, 2$ and\[a_n = (n\, \heartsuit\, (n-1)) \,\diamondsuit\, a_{n-1}\]for all integers $n \geq 4$. To the nearest integer, what is $\log_{7}(a_{2019})$?

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

$\textbf{D}$

By definition, the recursion becomes $$a_n = \left(n^{\frac1{\log_7(n-1)}}\right)^{\log_7(a_{n-1})}=n^{\frac{\log_7(a_{n-1})}{\log_7(n-1)}}= n^{\log_{n-1}(a_{n-1})}$$ This means $\log_n(a_n) = \log_{n-1}(a_{n-1})$. Thus, we have $$\log_{2019}(a_{2019})= \log_3(a_3) = \log_3\left(3^{\frac1{\log_7(2)}}\right) = \frac1{\log_7(2)} = \log_2(7)$$The problem asks for $\log_7(a_{2019}) = \log_2(2019) \approx11$.

For how many integers $n$ between $1$ and $50$, inclusive, is\[\frac{(n^2-1)!}{(n!)^n}\]an integer? (Recall that $0! = 1$.)

$\textbf{(A) } 31 \qquad \textbf{(B) } 32 \qquad \textbf{(C) } 33 \qquad \textbf{(D) } 34 \qquad \textbf{(E) } 35$

$\textbf{D}$

We extend the original expression as $$\dfrac{1\cdot2\cdots(n^2-1)}{1^n\cdot2^n\cdots n^n}$$ For any $k^n$ in the denominator, if $k<n$, we can find $k, 2k, 3k,\cdots, nk$ in the numerator. So the $k^n$ in the denominator will be canceled when $k<n$. As for the $n^n$ in the denominator, we can find $n, 2n, \cdots, (n-1)n$ in the numerator but not $n^2$. This explains why the original expression is not an integer for some $n$. If $n$ is a prime, the number of prime factor $n$ in the denominator is $n$, and the number of prime factor $n$ in the numerator is $n-1$. So the original is not an integer when $n$ is a prime.

Now think about the case when $n$ is not a prime. If we split $n^2$ objects into $n$ groups of size $n$, we have $\dfrac{(n^2)!}{(n!)^n}$ ways. So $\dfrac{(n^2)!}{(n!)^n}$ must be an integer. But we couldn't decide whether $\dfrac{(n^2-1)!}{(n!)^n}=\dfrac{(n^2)!}{(n!)^n}\cdot\dfrac{1}{n^2}$ is an integer or not. If we split $n^2$ objects into $n$ $\textit{unordered}$ groups of size $n$, we have $\dfrac{(n^2)!}{(n!)^n}\cdot\dfrac{1}{n!}=\dfrac{(n^2)!}{(n!)^{n+1}}$ ways. This means $\dfrac{(n^2)!}{(n!)^{n+1}}$ must be an integer. Now the original expression is $$\dfrac{(n^2-1)!}{(n!)^n}=\dfrac{(n^2)!}{(n!)^{n+1}}\cdot\dfrac{n!}{n^2}=\dfrac{(n^2)!}{(n!)^{n+1}}\cdot\dfrac{(n-1)!}{n}$$ If $\dfrac{(n-1)!}{n}$ is an integer, the original expression must be an integer. We only care about the cases when $n$ is not a prime. Since the numerator is much greater than the denominator, when $n$ is not a prime, which means $n$ can be prime factorized into smaller numbers, $\dfrac{(n-1)!}{n}$ should be an integer when $n$ is not extremely small. However, extreme case happened when $n=4$. $\dfrac{(n-1)!}{n}$ is a integer except $n=4$.

Therefore, the original expression $\dfrac{(n^2-1)!}{(n!)^n}$ is not an integer when $n$ is a prime or $n=4$. There are 15 primes between 1 and 50. So the answer is $50-15-1=34$.

Let $\triangle A_0B_0C_0$ be a triangle whose angle measures are exactly $59.999^\circ$, $60^\circ$, and $60.001^\circ$. For each positive integer $n$, define $A_n$ to be the foot of the altitude from $A_{n-1}$ to line $B_{n-1}C_{n-1}$. Likewise, define $B_n$ to be the foot of the altitude from $B_{n-1}$ to line $A_{n-1}C_{n-1}$, and $C_n$ to be the foot of the altitude from $C_{n-1}$ to line $A_{n-1}B_{n-1}$. What is the least positive integer $n$ for which $\triangle A_nB_nC_n$ is obtuse?

$\textbf{(A) } 10 \qquad \textbf{(B) }11 \qquad \textbf{(C) } 13\qquad \textbf{(D) } 14 \qquad \textbf{(E) } 15$

$\textbf{E}$

For all nonnegative integers $n$, let $\angle C_nA_nB_n=x_n$, $\angle A_nB_nC_n=y_n$, and $\angle B_nC_nA_n=z_n$.

Note that quadrilateral $A_0B_0A_1B_1$ is cyclic since $\angle A_0A_1B_0=\angle A_0B_1B_0=90^\circ$; thus, $\angle A_0A_1B_1=\angle A_0B_0B_1=90^\circ-x_0$. By a similar argument, $\angle A_0A_1C_1=\angle A_0C_0C_1=90^\circ-x_0$. Thus, $x_1=\angle A_0A_1B_1+\angle A_0A_1C_1=180^\circ-2x_0$. By a similar argument, $y_1=180^\circ-2y_0$ and $z_1=180^\circ-2z_0$.

Therefore, for any positive integer $n$, we have $x_n=180^\circ-2x_{n-1}$ (identical recurrence relations can be derived for $y_n$ and $z_n$). Iterating this again, we get $x_{n - 1} = 180 - 2x_{n - 2}.$ Subtracting the two, we getting $$x_{n} = -x_{n - 1} + 2x_{n - 2}$$ This recurrence has characteristic equation $x^2 + x - 2 = 0 = (x + 2)(x - 1) = 0 \iff x = -2, 1.$ Now, write\[x_n = p + q \cdot (-2)^n\] When $n=0\ \text{or}\ 1$, we have $$x_0=p+q$$ $$x_1=p-2q=180-2x_0$$ so we get \begin{align*}

p&=60\\

q&=x_0-60

\end{align*} For all positive integers $n$, $x_n=(-2)^n(x_0-60)+60$. Identical equalities hold for $y_n$ and $z_n$.

The problem asks for the smallest $n$ such that either $x_n$, $y_n$, or $z_n$ is greater than $90^\circ$. WLOG, let $x_0=60^\circ$, $y_0=59.999^\circ$, and $z_0=60.001^\circ$. Thus, $x_n=60^\circ$ for all $n$, $y_n=-(-2)^n(0.001)+60$, and $z_n=(-2)^n(0.001)+60$. Solving for the smallest possible value of $n$ in each sequence, we find that $n=15$ gives $y_n>90^\circ$. Therefore, the answer is 15.