## AMC 12 2021 Fall 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.

What is the value of $1234+2341+3412+4123?$

$\textbf{(A)}\: 10{,}000\qquad\textbf{(B)} \: 10{,}010\qquad\textbf{(C)} \: 10{,}110\qquad\textbf{(D)} \: 11{,}000\qquad\textbf{(E)} \: 11{,}110$

$\textbf{E}$

We see that $1, 2, 3,$ and $4$ each appear in the units, tens, hundreds, and thousands digit exactly once. Since $1+2+3+4=10$, we find that the sum is equal to\[10\cdot(1+10+100+1000)= 11{,}110\]

What is the area of the shaded figure shown below?

$\textbf{(A)}\: 4\qquad\textbf{(B)} \: 6\qquad\textbf{(C)} \: 8\qquad\textbf{(D)} \: 10\qquad\textbf{(E)} \: 12$

$\textbf{B}$

To find the area of the shaded figure, we subtract the area of the smaller triangle (base $4$ and height $2$) from the area of the larger triangle (base $4$ and height $5$):\[\frac12\cdot4\cdot5-\frac12\cdot4\cdot2=10-4=6\]

At noon on a certain day, Minneapolis is $N$ degrees warmer than St. Louis. At $4{:}00$ the temperature in Minneapolis has fallen by $5$ degrees while the temperature in St. Louis has risen by $3$ degrees, at which time the temperatures in the two cities differ by $2$ degrees. What is the product of all possible values of $N?$

$\textbf{(A)}\: 10\qquad\textbf{(B)} \: 30\qquad\textbf{(C)} \: 60\qquad\textbf{(D)} \: 100\qquad\textbf{(E)} \: 120$

$\textbf{C}$

At 4:00 the temperature difference is $|N-8|=2$. Hence, we get $N=10\ \text{or}\ 6$. So the product is 60.

Let $n=8^{2022}$. Which of the following is equal to $\dfrac{n}{4}?$

$\textbf{(A)}\: 4^{1010}\qquad\textbf{(B)} \: 2^{2022}\qquad\textbf{(C)} \: 8^{2018}\qquad\textbf{(D)} \: 4^{3031}\qquad\textbf{(E)} \: 4^{3032}$

$\textbf{E}$

We have\[n=8^{2022}= \left(8^\frac{2}{3}\right)^{3033}=4^{3033}.\]Therefore,\[\frac{n}4=4^{3032}\]

Call a fraction $\dfrac{a}{b}$, not necessarily in the simplest form, special if $a$ and $b$ are positive integers whose sum is $15$. How many distinct integers can be written as the sum of two, not necessarily different, special fractions?

$\textbf{(A)}\ 9 \qquad\textbf{(B)}\ 10 \qquad\textbf{(C)}\ 11 \qquad\textbf{(D)}\ 12 \qquad\textbf{(E)}\ 13$

$\textbf{C}$

The special fractions are\[\frac{1}{14},\frac{2}{13},\frac{3}{12},\frac{4}{11},\frac{5}{10},\frac{6}{9},\frac{7}{8},\frac{8}{7},\frac{9}{6},\frac{10}{5},\frac{11}{4},\frac{12}{3},\frac{13}{2},\frac{14}{1}.\]We rewrite them in the simplest form:\[\frac{1}{14},\frac{2}{13},\frac{1}{4},\frac{4}{11},\frac{1}{2},\frac{2}{3},\frac{7}{8},1\frac{1}{7},1\frac{1}{2},2,2\frac{3}{4},4,6\frac{1}{2},14.\]Note that two unlike fractions in the simplest form cannot sum to an integer. So, we only consider the fractions whose denominators appear more than once:\[\frac{1}{4},\frac{1}{2},1\frac{1}{2},2,2\frac{3}{4},4,6\frac{1}{2},14.\]For the set $\{2,4,14\},$ two elements (not necessarily different) can sum to $4,6,8,16,18,28.$

For the set $\left\{\dfrac{1}{2},1\dfrac{1}{2},6\dfrac{1}{2}\right\},$ two elements (not necessarily different) can sum to $1,2,3,7,8,13.$

For the set $\left\{\dfrac{1}{4},2\dfrac{3}{4}\right\},$ two elements (not necessarily different) can sum to $3.$

Together, there are $11$ distinct integers that can be written as the sum of two, not necessarily different, special fractions:\[1,2,3,4,6,7,8,13,16,18,28.\]

The largest prime factor of $16384$ is $2$ because $16384 = 2^{14}$. What is the sum of the digits of the greatest prime number that is a divisor of $16383$?

$\textbf{(A)} \: 3\qquad\textbf{(B)} \: 7\qquad\textbf{(C)} \: 10\qquad\textbf{(D)} \: 16\qquad\textbf{(E)} \: 22$

$\textbf{C}$

We have\begin{align*} 16383 & = 2^{14} - 1 \\ & = \left( 2^7 + 1 \right) \left( 2^7 - 1 \right) \\ & = 129 \cdot 127 \\ \end{align*}Since $129$ is composite, $127$ is the largest prime divisible by $16383$. The sum of $127$'s digits is $1+2+7=10$.

Which of the following conditions is sufficient to guarantee that integers $x$, $y$, and $z$ satisfy the equation\[x(x-y)+y(y-z)+z(z-x) = 1?\]

$\textbf{(A)} \: x>y$ and $y=z$

$\textbf{(B)} \: x=y-1$ and $y=z-1$

$\textbf{(C)} \: x=z+1$ and $y=x+1$

$\textbf{(D)} \: x=z$ and $y-1=x$

$\textbf{(E)} \: x+y+z=1$

$\textbf{D}$

According to the given equation, we have $$x^2+y^2+z^2-xy-xz-yz=1$$ $$2x^2+2y^2+2z^2-2xy-2xz-2yz=2$$ $$(x^2-2xy+y^2)+(y^2-2yz+z^2)+(z^2-2zx+x^2)=2$$ $$(x-y)^2+(y-z)^2+(z-x)^2=2$$ It is obvious $x$, $y$, and $z$ are symmetrical. Because $x, y, z$ are integers, $(x-y)^2$, $(y-z)^2$, and $(z-x)^2$ can only equal $0, 1, 1$. So one variable must equal another, and the third variable is $1$ different from those $2$ equal variables. Therefore, the answer is D.

The product of the lengths of the two congruent sides of an obtuse isosceles triangle is equal to the product of the base and twice the triangle's height to the base. What is the measure, in degrees, of the vertex angle of this triangle?

$\textbf{(A)} \: 105 \qquad\textbf{(B)} \: 120 \qquad\textbf{(C)} \: 135 \qquad\textbf{(D)} \: 150 \qquad\textbf{(E)} \: 165$

$\textbf{D}$

Let the lengths of the two congruent sides of the triangle be $a$, the base be $b$, the height be $h$, and the vertex angle be $\theta$. According to the given information, we have $$a^2=2bh$$ The area of the triangle can be wrote in the form of $\dfrac12a^2\sin\theta$ or $\dfrac12bh$. So we have $$\dfrac12a^2\sin\theta=\dfrac12bh$$ Since $a^2=2bh$, we get $$\sin\theta=\dfrac12\rightarrow\theta=150^\circ\text{(it is a obtuse triangle, so $\theta\neq30^\circ$)}$$

Triangle $ABC$ is equilateral with side length $6$. Suppose that $O$ is the center of the inscribed circle of this triangle. What is the area of the circle passing through $A$, $O$, and $C$?

$\textbf{(A)} \: 9\pi \qquad\textbf{(B)} \: 12\pi \qquad\textbf{(C)} \: 18\pi \qquad\textbf{(D)} \: 24\pi \qquad\textbf{(E)} \: 27\pi$

$\textbf{B}$

Apparently $O$ is the center of equilateral triangle $ABC$. So $\angle AOC=120^\circ$, and $AO$ is two thirds of the height of $\triangle ABC$. The height of $\triangle ABC$ is $6\cdot\dfrac{\sqrt3}{2}=3\sqrt3$. Hence, $AO=\dfrac23\cdot3\sqrt3=2\sqrt3$.

Let $O_1$ be the center of the circle passing through $A$, $O$ and $C$. Now $AO_1$ and $OO_1$ are both the radius of the circle, while $\angle AOO_1=60^\circ$. So triangle $AOO_1$ is equilateral. The radius of the circle is $r=AO=2\sqrt3$. Therefore, the area of the circle is $\pi r^2=12\pi$.

What is the sum of all possible values of $t$ between $0$ and $360$ such that the triangle in the coordinate plane whose vertices are\[(\cos 40^\circ,\sin 40^\circ), (\cos 60^\circ,\sin 60^\circ),\textrm{ and } (\cos t^\circ,\sin t^\circ)\]is isosceles?

$\textbf{(A)} \: 100 \qquad\textbf{(B)} \: 150 \qquad\textbf{(C)} \: 330 \qquad\textbf{(D)} \: 360 \qquad\textbf{(E)} \: 380$

$\textbf{E}$

Let $A = (\cos 40^{\circ}, \sin 40^{\circ}), B = (\cos 60^{\circ}, \sin 60^{\circ}),$ and $C = (\cos t^{\circ}, \sin t^{\circ}).$ All these points are located on the unit circle.

Case 1: $AB=AC$. Note that $A$ must be the midpoint of $\widehat{BC}.$ It follows that $C = (\cos 20^{\circ}, \sin 20^{\circ}),$ so $t=20.$

Case 2: $BA=BC$. Note that $B$ must be the midpoint of $\widehat{AC}.$ It follows that $C = (\cos 80^{\circ}, \sin 80^{\circ}),$ so $t=80.$

Case 3: $CA=CB$. Note that $C$ must be the midpoint of $\widehat{AB}.$ It follows that $C = (\cos 50^{\circ}, \sin 50^{\circ})$ or $C = (\cos 230^{\circ}, \sin 230^{\circ}),$ so $t=50$ or $t=230.$

Together, the sum of all such possible values of $t$ is $20+80+50+230=380.$

Una rolls $6$ standard $6$-sided dice simultaneously and calculates the product of the $6{ }$ numbers obtained. What is the probability that the product is divisible by $4?$

$\textbf{(A)}\: \dfrac34\qquad\textbf{(B)} \: \dfrac{57}{64}\qquad\textbf{(C)} \: \dfrac{59}{64}\qquad\textbf{(D)} \: \dfrac{187}{192}\qquad\textbf{(E)} \: \dfrac{63}{64}$

$\textbf{C}$

We will use complementary counting to find the probability that the product is not divisible by $4$. Then, we can find the probability that we want by subtracting this from 1. We split this into $2$ cases.

Case 1: The product is not divisible by $2$. We need every number to be odd, and since the chance we roll an odd number is $\dfrac12,$ our probability is $\left(\dfrac12\right)^6=\dfrac1{64}.$

Case 2: The product is divisible by $2$, but not by $4$. We need $5$ numbers to be odd, and one to be divisible by $2$, but not by $4$. There is a $\dfrac12$ chance that an odd number is rolled, a $\dfrac13$ chance that we roll a number satisfying the second condition (only $2$ and $6$ work), and $6$ ways to choose the order in which the even number appears. Our probability is $\left(\dfrac12\right)^5\left(\dfrac13\right)\cdot6=\dfrac1{16}.$

Therefore, the probability the product is not divisible by $4$ is $\dfrac1{64}+\dfrac1{16}=\dfrac{5}{64}$.

Our answer is $1-\dfrac{5}{64}= \dfrac{59}{64}$.

For $n$ a positive integer, let $f(n)$ be the quotient obtained when the sum of all positive divisors of $n$ is divided by $n.$ For example,\[f(14)=(1+2+7+14)\div 14=\frac{12}{7}\]What is $f(768)-f(384)?$

$\textbf{(A)}\ \dfrac{1}{768} \qquad\textbf{(B)}\ \dfrac{1}{192} \qquad\textbf{(C)}\ 1 \qquad\textbf{(D)}\ \dfrac{4}{3} \qquad\textbf{(E)}\ \dfrac{8}{3}$

$\textbf{B}$

The prime factorization of $384$ is $2^7\cdot3,$ so each of its positive divisors is of the form $2^m$ or $2^m\cdot3$ for some integer $m$ such that $0\leq m\leq7.$ We will use this fact to calculate the sum of all its positive divisors. Note that\[2^m + 2^m\cdot3 = 2^m\cdot(1+3) = 2^m\cdot4 = 2^m\cdot2^2 = 2^{m+2}\]is the sum of the two forms of positive divisors for all such $m.$ By geometric series, the sum of all positive divisors of $384$ is\[\sum_{m=0}^{7}2^{m+2}=\sum_{k=2}^{9}2^k = \frac{2^{10}-2^2}{2-1} = 1020,\]from which $f(384) = \dfrac{1020}{384}.$ Similarly, since the prime factorization of $768$ is $2^8 \cdot 3,$ we have $f(768) = \dfrac{2044}{768}.$

Therefore, the answer is $f(768)-f(384)=\dfrac{1}{192}.$

Let $c = \dfrac{2\pi}{11}.$ What is the value of\[\frac{\sin 3c \cdot \sin 6c \cdot \sin 9c \cdot \sin 12c \cdot \sin 15c}{\sin c \cdot \sin 2c \cdot \sin 3c \cdot \sin 4c \cdot \sin 5c}?\]

$\textbf{(A)}\ {-}1 \qquad\textbf{(B)}\ {-}\dfrac{\sqrt{11}}{5} \qquad\textbf{(C)}\ \dfrac{\sqrt{11}}{5} \qquad\textbf{(D)}\ \dfrac{10}{11} \qquad\textbf{(E)}\ 1$

$\textbf{E}$

Plugging in $c$, we get\[\frac{\sin 3c \cdot \sin 6c \cdot \sin 9c \cdot \sin 12c \cdot \sin 15c}{\sin c \cdot \sin 2c \cdot \sin 3c \cdot \sin 4c \cdot \sin 5c}=\frac{\sin \frac{6\pi}{11} \cdot \sin \frac{12\pi}{11} \cdot \sin \frac{18\pi}{11} \cdot \sin \frac{24\pi}{11} \cdot \sin \frac{30\pi}{11}}{\sin \frac{2\pi}{11} \cdot \sin \frac{4\pi}{11} \cdot \sin \frac{6\pi}{11} \cdot \sin \frac{8\pi}{11} \cdot \sin \frac{10\pi}{11}}.\]Since $\sin(x+2\pi)=\sin(x),$ $\sin(2\pi-x)=\sin(-x),$ and $\sin(-x)=-\sin(x),$ we get\[\frac{\sin \frac{6\pi}{11} \cdot \sin \frac{12\pi}{11} \cdot \sin \frac{18\pi}{11} \cdot \sin \frac{24\pi}{11} \cdot \sin \frac{30\pi}{11}}{\sin \frac{2\pi}{11} \cdot \sin \frac{4\pi}{11} \cdot \sin \frac{6\pi}{11} \cdot \sin \frac{8\pi}{11} \cdot \sin \frac{10\pi}{11}}=\frac{\sin \frac{6\pi}{11} \cdot \sin \frac{-10\pi}{11} \cdot \sin \frac{-4\pi}{11} \cdot \sin \frac{2\pi}{11} \cdot \sin \frac{8\pi}{11}}{\sin \frac{2\pi}{11} \cdot \sin \frac{4\pi}{11} \cdot \sin \frac{6\pi}{11} \cdot \sin \frac{8\pi}{11} \cdot \sin \frac{10\pi}{11}}=1\]

Suppose that $P(z), Q(z)$, and $R(z)$ are polynomials with real coefficients, having degrees $2$, $3$, and $6$, respectively, and constant terms $1$, $2$, and $3$, respectively. Let $N$ be the number of distinct complex numbers $z$ that satisfy the equation $P(z) \cdot Q(z)=R(z)$. What is the minimum possible value of $N$?

$\textbf{(A)}\: 0\qquad\textbf{(B)} \: 1\qquad\textbf{(C)} \: 2\qquad\textbf{(D)} \: 3\qquad\textbf{(E)} \: 5$

$\textbf{B}$

The answer cannot be $0,$ as every nonconstant polynomial has at least $1$ distinct complex root (Fundamental Theorem of Algebra). Since $P(z) \cdot Q(z)$ has degree $2 + 3 = 5,$ we conclude that $R(z) - P(z)\cdot Q(z)$ has degree $6$ and is thus nonconstant.

It now suffices to illustrate an example for which $N = 1$: Take\begin{align*} P(z)&=z^2+1, \\ Q(z)&=z^3+2, \\ R(z)&=(z+1)^6 + P(z) \cdot Q(z). \end{align*}Note that $R(z)$ has degree $6$ and constant term $3,$ so it satisfies the conditions.

We need to find the solutions to\begin{align*} P(z) \cdot Q(z) &= (z+1)^6 + P(z) \cdot Q(z) \\ 0 &= (z+1)^6. \end{align*}Clearly, the only distinct complex root is $-1,$ so our answer is $N=1.$

Three identical square sheets of paper each with side length $6$ are stacked on top of each other. The middle sheet is rotated clockwise $30^\circ$ about its center and the top sheet is rotated clockwise $60^\circ$ about its center, resulting in the $24$-sided polygon shown in the figure below. The area of this polygon can be expressed in the form $a-b\sqrt{c}$, where $a$, $b$, and $c$ are positive integers, and $c$ is not divisible by the square of any prime. What is $a+b+c$?

$(\textbf{A})\: 75\qquad(\textbf{B}) \: 93\qquad(\textbf{C}) \: 96\qquad(\textbf{D}) \: 129\qquad(\textbf{E}) \: 147$

$\textbf{E}$

Let $A$ be the center of the figure, and $B$, $C$, $D$ be the vertices on the sides. The $24$-sided polygon is made out of $24$ shapes like $\triangle ABC$. $\angle BAC=360^\circ/24=15^\circ$, $\angle EAC = 45^\circ$, so $\angle{EAB} = 30^{\circ}$. Then $EB=AE\tan 30^\circ = \sqrt{3}$; therefore $BC=EC-EB=3-\sqrt{3}$. So the area of $\triangle ABC$ is\[\dfrac12AE\cdot BC=\dfrac12(3)(3-\sqrt3)=\dfrac32(3-\sqrt3)\]and the required area is $24\cdot\dfrac32(3-\sqrt3) =108-36\sqrt{3}$. Hence, $a+b+c=108+36+3=147$.

Suppose $a$, $b$, $c$ are positive integers such that\[a+b+c=23\]and\[\gcd(a,b)+\gcd(b,c)+\gcd(c,a)=9.\]What is the sum of all possible distinct values of $a^2+b^2+c^2$?

$\textbf{(A)}\: 259\qquad\textbf{(B)} \: 438\qquad\textbf{(C)} \: 516\qquad\textbf{(D)} \: 625\qquad\textbf{(E)} \: 687$

$\textbf{B}$

Because $a + b + c$ is odd, $a$, $b$, $c$ are either one odd and two evens or three odds.

$\textbf{Case 1}$: $a$, $b$, $c$ have one odd and two evens.

Without loss of generality, we assume $a$ is odd and $b$ and $c$ are even.

Hence, ${\rm gcd} \left( a , b \right)$ and ${\rm gcd} \left( a , c \right)$ are odd, and ${\rm gcd} \left( b , c \right)$ is even. Hence, ${\rm gcd} \left( a , b \right) + {\rm gcd} \left( b , c \right) + {\rm gcd} \left( c , a \right)$ is even. This violates the condition given in the problem.

Therefore, there is no solution in this case.

$\textbf{Case 2}$: $a$, $b$, $c$ are all odd.

In this case, ${\rm gcd} \left( a , b \right)$, ${\rm gcd} \left( a , c \right)$, ${\rm gcd} \left( b , c \right)$ are all odd.

Without loss of generality, we assume\[ {\rm gcd} \left( a , b \right) \leq {\rm gcd} \left( b , c \right) \leq {\rm gcd} \left( c , a \right) . \]$\textbf{Case 2.1}$: ${\rm gcd} \left( a , b \right) = 1$, ${\rm gcd} \left( b , c \right) = 1$, ${\rm gcd} \left( c , a \right) = 7$.

The only solution is $(a, b, c) = (7, 9, 7)$.

Hence, $a^2 + b^2 + c^2 = 179$.

$\textbf{Case 2.2}$: ${\rm gcd} \left( a , b \right) = 1$, ${\rm gcd} \left( b , c \right) = 3$, ${\rm gcd} \left( c , a \right) = 5$.

The only solution is $(a, b, c) = (5, 3, 15)$.

Hence, $a^2 + b^2 + c^2 = 259$.

$\textbf{Case 2.3}$: ${\rm gcd} \left( a , b \right) = 3$, ${\rm gcd} \left( b , c \right) = 3$, ${\rm gcd} \left( c , a \right) = 3$.

There is no solution in this case.

Therefore, putting all cases together, the answer is $179 + 259 =438$.

A bug starts at a vertex of a grid made of equilateral triangles of side length $1$. At each step the bug moves in one of the $6$ possible directions along the grid lines randomly and independently with equal probability. What is the probability that after $5$ moves the bug never will have been more than $1$ unit away from the starting position?

$\textbf{(A)}\ \dfrac{13}{108} \qquad\textbf{(B)}\ \dfrac{7}{54} \qquad\textbf{(C)}\ \dfrac{29}{216} \qquad\textbf{(D)}\ \dfrac{4}{27} \qquad\textbf{(E)}\ \dfrac{1}{16}$

$\textbf{A}$

Let $p(i)$ be the probability that the bug has never been more than 1 unit away from the starting position after $i$ moves. Here we have $p(i)=p_0(i)+p_1(i)$, where 0 represents the bug is now 0 step away from the starting position, and 1 means 1 step away from the starting position. Apparently we have $$p_0(1)=0\quad p_1(1)=1$$ After 2 moves, the probabilities that the bug is 0 or 1 unit away from the starting position are $$p_0(2)=\dfrac16\quad p_1(2)=\dfrac13$$ We can conclude the pattern that $$p_0(i)=\dfrac16p_1(i-1)$$ $$p_1(i)=p_0(i-1)+\dfrac13p_1(i-1)$$ Hence, we can find $$p_0(3)=\dfrac16\cdot\dfrac13=\dfrac{1}{18}\quad p_1(3)=\dfrac16+\dfrac13\cdot\dfrac13=\dfrac{5}{18}$$ $$p_0(4)=\dfrac16\cdot\dfrac{5}{18}=\dfrac{5}{108}\quad p_1(4)=\dfrac{1}{18}+\dfrac13\cdot\dfrac{5}{18}=\dfrac{4}{27}$$ $$p_0(5)=\dfrac16\cdot\dfrac{4}{27}=\dfrac{2}{81}\quad p_1(5)=\dfrac{5}{108}+\dfrac13\cdot\dfrac{4}{27}=\dfrac{31}{324}$$ So the answer is $$p(5)=p_0(5)+p_1(5)=\dfrac{2}{81}+\dfrac{31}{324}=\dfrac{13}{108}$$

Set $u_0 = \dfrac{1}{4}$, and for $k \ge 0$ let $u_{k+1}$ be determined by the recurrence\[u_{k+1} = 2u_k - 2u_k^2.\]

This sequence tends to a limit; call it $L$. What is the least value of $k$ such that\[|u_k-L| \le \frac{1}{2^{1000}}?\]

$\textbf{(A)}\: 10\qquad\textbf{(B)}\: 87\qquad\textbf{(C)}\: 123\qquad\textbf{(D)}\: 329\qquad\textbf{(E)}\: 401$

$\textbf{A}$

Trying the first few values, we get

\begin{align*}

u_0&=\dfrac14=\dfrac12-\dfrac14=\dfrac12-\dfrac{1}{2^2}\\

u_1&=\dfrac38=\dfrac12-\dfrac18=\dfrac12-\dfrac{1}{2^3}\\

u_2&=\dfrac{15}{32}=\dfrac12-\dfrac{1}{32}=\dfrac12-\dfrac{1}{2^5}\\

u_3&=\dfrac{255}{512}=\dfrac12-\dfrac1{512}=\dfrac12-\dfrac{1}{2^9}

\end{align*} We can easily find the pattern that $$u_k=\dfrac12-\dfrac{1}{2^{2^k+1}}$$ We can check it by calculating

\begin{align*}

u_{k+1}&=2u_k-2u_k^2\\

&=2u_k(1-u_k)\\

&=2\left(\dfrac12-\dfrac{1}{2^{2^k+1}}\right)\left(\dfrac12+\dfrac{1}{2^{2^k+1}}\right)\\

&=2\left[\left(\dfrac12\right)^2-\left(\dfrac{1}{2^{2^k+1}}\right)^2\right]\\

&=\dfrac12-\dfrac{1}{2^{2^{k+1}+1}}

\end{align*} Apparently the limit of the sequence is $$L=\dfrac12$$ The given inequality becomes\[\frac12-\frac{1}{2^{1000}} \leq u_k \leq \frac12+\frac{1}{2^{1000}},\] Since $u_k<\dfrac12$, we only need to consider $\dfrac12-\dfrac{1}{2^{1000}} \leq u_k.$ \begin{align*} \frac12-\frac{1}{2^{1000}} &\leq \frac12-\frac{1}{2^{2^k+1}} \\ -\frac{1}{2^{1000}} &\leq -\frac{1}{2^{2^k+1}} \\ 2^{1000} &\leq 2^{2^k+1} \\ 1000 &\leq 2^k+1. \end{align*} Therefore, the least value of $k$ is $10.$

Regular polygons with $5$, $6$, $7$, and $8$ sides are inscribed in the same circle. No two of the polygons share a vertex, and no three of their sides intersect at a common point. At how many points inside the circle do two of their sides intersect?

$(\textbf{A})\: 52\qquad(\textbf{B}) \: 56\qquad(\textbf{C}) \: 60\qquad(\textbf{D}) \: 64\qquad(\textbf{E}) \: 68$

$\textbf{E}$

Imagine we have $2$ regular polygons with $m$ and $n$ sides and $m>n$ inscribed in a circle without sharing a vertex. We see that each side of the polygon with $n$ sides (the polygon with fewer sides) will be intersected twice. (We can see this because to have a vertex of the $m$-gon on an arc subtended by a side of the $n$-gon, there will be one intersection to “enter” the arc and one to “exit” the arc.)

This means that we will end up with $2$ times the number of sides in the polygon with fewer sides.

If we have polygons with $5,$ $6,$ $7,$ and $8$ sides, we need to consider each possible pair of polygons and count their intersections.

Throughout $6$ of these pairs, the $5$-sided polygon has the least number of sides $3$ times, the $6$-sided polygon has the least number of sides $2$ times, and the $7$-sided polygon has the least number of sides $1$ time.

Therefore the number of intersections is $2\cdot(3\cdot5+2\cdot6+1\cdot7)=68$.

A cube is constructed from $4$ white unit cubes and $4$ blue unit cubes. How many different ways are there to construct the $2 \times 2 \times 2$ cube using these smaller cubes? (Two constructions are considered the same if one can be rotated to match the other.)

$(\textbf{A})\: 7\qquad(\textbf{B}) \: 8\qquad(\textbf{C}) \: 9\qquad(\textbf{D}) \: 10\qquad(\textbf{E}) \: 11$

$\textbf{A}$

This problem is about the relationships between the white unit cubes and the blue unit cubes, which can be solved by Graph Theory. We use a Planar Graph to represent the larger cube. Each vertex of the planar graph represents a unit cube. Each edge of the planar graph represents a shared face between $2$ neighboring unit cubes. Each face of the planar graph represents a face of the larger cube.

Now the problem becomes a Graph Coloring problem of how many ways to assign $4$ vertices blue and $4$ vertices white with Topological Equivalence. For example, in Figure $(1)$, as long as the $4$ blue vertices belong to the same planar graph face, the different planar graphs are considered to be topological equivalent by rotating the larger cube.

Here is how the $4$ blue unit cubes are arranged:

4 blue unit cubes in a plane: Figure (1).

3 blue unit cubes in a plane: Figure (2) to (5). Note that Figure (3) and (4) are not equivalent because one can not be achieved by rotating another one.

2 blue unit cubes in a plane: Figure (6) and (7).

So the answer is $7$.

For real numbers $x$, let\[P(x)=1+\cos(x)+i\sin(x)-\cos(2x)-i\sin(2x)+\cos(3x)+i\sin(3x)\]where $i = \sqrt{-1}$. For how many values of $x$ with $0\leq x<2\pi$ does\[P(x)=0?\]

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

$\textbf{A}$

For $\textrm{Im}(P(x))=0$, we get\[\sin(2x)=\sin(x)+\sin(3x)=2\sin(2x)\cos(x)\]So either $\sin(2x)=0$, i.e. $x\in\{0,\pi\}$ or $\cos(x)=\dfrac 12$, i.e. $x\in \{\pi/3, 5\pi/3\}$.

For none of these values do we get $\textrm{Re}(P(x))=0$.

Therefore, the answer is $0$.

Right triangle $ABC$ has side lengths $BC=6$, $AC=8$, and $AB=10$.

A circle centered at $O$ is tangent to line $BC$ at $B$ and passes through $A$. A circle centered at $P$ is tangent to line $AC$ at $A$ and passes through $B$. What is $OP$?

$\textbf{(A)}\ \dfrac{23}{8} \qquad\textbf{(B)}\ \dfrac{29}{10} \qquad\textbf{(C)}\ \dfrac{35}{12} \qquad\textbf{(D)}\ \dfrac{73}{25} \qquad\textbf{(E)}\ 3$

$\textbf{C}$

We have $OB\perp BC$, $OB=OA$, $PA\perp AC$ and $AP=BP$ in the diagram. Let $M$ be the midpoint of $AB$. Hence, we get $OM\perp AB$ since $OA=OB$. Now we have $\triangle ABC\sim\triangle BOM\sim\triangle POD$. So $$BO=AB\cdot\dfrac{BM}{AC}=10\cdot\dfrac{5}{8}=\dfrac{25}{4}$$ $$OM=BC\cdot\dfrac{BM}{AC}=6\cdot\dfrac{5}{8}=\dfrac{15}{4}$$ $$OD=AC-BO=8-\dfrac{25}{4}=\dfrac74$$ Finally we get $$OP=OB\cdot\dfrac{OD}{OM}=\dfrac{35}{12}$$

What is the average number of pairs of consecutive integers in a randomly selected subset of $5$ distinct integers chosen from the set $\{ 1, 2, 3, …, 30\}$? (For example the set $\{1, 17, 18, 19, 30\}$ has $2$ pairs of consecutive integers.)

$\textbf{(A)}\ \dfrac{2}{3} \qquad\textbf{(B)}\ \dfrac{29}{36} \qquad\textbf{(C)}\ \dfrac{5}{6} \qquad\textbf{(D)}\ \dfrac{29}{30} \qquad\textbf{(E)}\ 1$

$\textbf{A}$

There are $29$ possible pairs of consecutive integers, namely $p_1=\{1,2\}, \cdots, p_{29}=\{29,30\}$. Define a random variable $X_i$, with $X_i=1$, if $p_i$ is part of the 5-element subset, and $X_i=0$ otherwise. Then the number of pairs of consecutive integers in a $5$-element selection is given by the sum $X_1+\cdots + X_{29}$. By linearity of expectation, the expected value is equal to the sum of the $\mathrm{E}[X_i]$:\[\mathrm{E}[X_1+\cdots +X_{29}]=\mathrm{E}[X_1]+\cdots + \mathrm{E}[X_{29}]\]To compute $\mathrm{E}[X_i]$, note that $X_i=1$ for a total of $\dbinom{28}{3}$ out of $\dbinom{30}{5}$ possible selections. Thus\[\mathrm{E}[X_i]=\frac{\dbinom{28}{3}}{\dbinom{30}{5}}= \frac 1{29}\cdot \frac 23\] By symmetry, $\mathrm{E}[X_1]=\mathrm{E}[X_2]=\cdots=\mathrm{E}[X_{29}]$, so \[\mathrm{E}[X_1+\cdots +X_{29}]= \frac{2}{3}\]

Triangle $ABC$ has side lengths $AB = 11, BC=24$, and $CA = 20$. The bisector of $\angle{BAC}$ intersects $\overline{BC}$ in point $D$, and intersects the circumcircle of $\triangle{ABC}$ in point $E \ne A$. The circumcircle of $\triangle{BED}$ intersects the line $AB$ in points $B$ and $F \ne B$. What is $CF$?

$\textbf{(A) } 28 \qquad \textbf{(B) } 20\sqrt{2} \qquad \textbf{(C) } 30 \qquad \textbf{(D) } 32 \qquad \textbf{(E) } 20\sqrt{3}$

$\textbf{C}$

(Ignore the point $E^\prime$ in the diagram.)

Claim: $\triangle ADC \sim \triangle ABE.$

Proof: Note that $\angle CAE = \angle EAB$ and $\angle BCA = \angle BEA$ meaning that our claim is true by AA similarity.

Because of this similarity, we have that\[\frac{AC}{AD} = \frac{AE}{AB} \Longrightarrow AB \cdot AC = AD \cdot AE = AB \cdot AF\]by Power of a Point. Thus, $AC=AF=20.$

We can find $\angle CAB$ by the law of cosines in $\triangle ABC$: \[AB^2+AC^2-2\cdot AB \cdot AC \cdot \cos(\angle CAB) = BC^2 \to \cos(\angle CAB) = -\frac{1}{8}.\] Use law of cosines again to find $CF$ in $\triangle ACF$:\[CF^2 = AF^2+AC^2-2 \cdot AF \cdot AC \cdot \cos(\angle CAF) = 900 \to CF=30\]

For $n$ a positive integer, let $R(n)$ be the sum of the remainders when $n$ is divided by $2$, $3$, $4$, $5$, $6$, $7$, $8$, $9$, and $10$. For example, $R(15) = 1+0+3+0+3+1+7+6+5=26$. How many two-digit positive integers $n$ satisfy $R(n) = R(n+1)\,?$

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

$\textbf{C}$

By adding 1, the reminder of $(n+1)\div k$ may be 1 greater than the reminder of $n\div k$, or $k-1$ less if the reminder of $n\div k$ is $k-1$ (the reminder of $(n+1)\div k$ is 0). So we can add $9$ to $R(n)$ to get $R(n+1)$, but must subtract $k$ for all $k|n+1$. Hence, we see that there are 5 ways to do that because $9=7+2=6+3=5+4=4+3+2$. Note that only $7+2$ is a plausible option, since $4+3+2$ indicates $n+1$ is also divisible by $6$, $5+4$ indicates that $n+1$ is also divisible by $2$, $6+3$ indicates $n+1$ is also divisible by $2$, and $9$ itself indicates divisibility by $3$, too. So, $14|n+1$ and $n+1$ is not divisible by any positive integers from $2$ to $10$, inclusive, except $2$ and $7$. We check and get that only $n+1=14 \cdot 1$ and $n+1=14 \cdot 7$ give possible solutions so our answer is $2$.