id int64 1 7.14k | link stringlengths 75 84 | no int64 1 14 | problem stringlengths 14 5.33k | solution stringlengths 21 6.43k | answer int64 0 999 |
|---|---|---|---|---|---|
3,801 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12B_Problems/Problem_1 | 3 | What is the value of $1234 + 2341 + 3412 + 4123$
$\textbf{(A)}\: 10000\qquad\textbf{(B)} \: 10010\qquad\textbf{(C)} \: 10110\qquad\textbf{(D)} \: 11000\qquad\textbf{(E)} \: 11110$ | We see that the units digit must be $0$ , since $4+3+2+1$ is $0$ . But every digit from there, will be a $1$ since we have that each time afterwards, we must carry the $1$ from the previous sum. The answer choice that satisfies these conditions is $\boxed{11110}$ | 110 |
3,802 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12B_Problems/Problem_1 | 4 | What is the value of $1234 + 2341 + 3412 + 4123$
$\textbf{(A)}\: 10000\qquad\textbf{(B)} \: 10010\qquad\textbf{(C)} \: 10110\qquad\textbf{(D)} \: 11000\qquad\textbf{(E)} \: 11110$ | We can simply add the numbers. $1234 + 2341 + 3412 + 4123 = 11110 \implies \boxed{11110}$ | 110 |
3,803 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12B_Problems/Problem_2 | 1 | What is the area of the shaded figure shown below? [asy] size(200); defaultpen(linewidth(0.4)+fontsize(12)); pen s = linewidth(0.8)+fontsize(8); pair O,X,Y; O = origin; X = (6,0); Y = (0,5); fill((1,0)--(3,5)--(5,0)--(3,2)--cycle, palegray+opacity(0.2)); for (int i=1; i<7; ++i) { draw((i,0)--(i,5), gray+dashed); label... | The line of symmetry divides the shaded figure into two congruent triangles, each with base $3$ and height $2.$
Therefore, the area of the shaded figure is \[2\cdot\left(\frac12\cdot3\cdot2\right)=2\cdot3=\boxed{6}.\] ~MRENTHUSIASM ~Wilhelm Z | 6 |
3,804 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12B_Problems/Problem_2 | 2 | What is the area of the shaded figure shown below? [asy] size(200); defaultpen(linewidth(0.4)+fontsize(12)); pen s = linewidth(0.8)+fontsize(8); pair O,X,Y; O = origin; X = (6,0); Y = (0,5); fill((1,0)--(3,5)--(5,0)--(3,2)--cycle, palegray+opacity(0.2)); for (int i=1; i<7; ++i) { draw((i,0)--(i,5), gray+dashed); label... | 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 $5$ ): \[\frac12\cdot4\cdot5-\frac12\cdot4\cdot2=10-4=\boxed{6}.\] ~MRENTHUSIASM ~Steven Chen (www.professorchenedu.com) | 6 |
3,805 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12B_Problems/Problem_2 | 4 | What is the area of the shaded figure shown below? [asy] size(200); defaultpen(linewidth(0.4)+fontsize(12)); pen s = linewidth(0.8)+fontsize(8); pair O,X,Y; O = origin; X = (6,0); Y = (0,5); fill((1,0)--(3,5)--(5,0)--(3,2)--cycle, palegray+opacity(0.2)); for (int i=1; i<7; ++i) { draw((i,0)--(i,5), gray+dashed); label... | We have $4$ lattice points in the interior and $6$ lattice points on the boundary. By Pick's Theorem , the area of the shaded figure is \[4+\frac{6}{2}-1 = 4+3-1 = \boxed{6}.\] ~danprathab | 6 |
3,806 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12B_Problems/Problem_3 | 1 | 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... | At noon on a certain day, let $M$ and $L$ be the temperatures (in degrees) in Minneapolis and St. Louis, respectively. It follows that $M=L+N.$
At $4{:}00,$ we get \begin{align*} |(M-5)-(L+3)| &= 2 \\ |M-L-8| &= 2 \\ |N-8| &= 2. \end{align*} We have two cases:
Together, the product of all possible values of $N$ is $10\... | 60 |
3,807 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12B_Problems/Problem_3 | 2 | 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... | At noon on a certain day, the difference of temperatures in Minneapolis and St. Louis is $N$ degrees.
At $4{:}00,$ the difference of temperatures in Minneapolis and St. Louis is $N-8$ degrees.
It follows that \[|N-8|=2.\] We continue with the casework in Solution 1 to get the answer $\boxed{60}.$ | 60 |
3,808 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12B_Problems/Problem_5 | 2 | Call a fraction $\frac{a}{b}$ , 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... | Let $a=15-b,$ so the special fraction is \[\frac ab = \frac{15-b}{b} = \frac{15}{b}-1.\] We can ignore the $-1$ part and only focus on $\frac{15}{b}.$
The integers are $\frac{15}{1},\frac{15}{3},\frac{15}{5},$ which are $15,5,3,$ respectively. We get $30,20,18,10,8,6$ from this group of numbers.
The halves are $\frac{1... | 11 |
3,809 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12B_Problems/Problem_5 | 3 | Call a fraction $\frac{a}{b}$ , 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... | We split this up into two cases:
Case 1: integer + integer
The whole numbers we have are $\frac{10}{5}$ (or $2$ ), $\frac{12}{3}$ (or $4$ ), and $\frac{14}{1}$ (or $14$ ). There are $\dbinom{3}{2}=3$ ways to choose different-numbered pairs and $3$ ways to choose the same-numbered pairs. So, $3+3=6$
Case 2: fraction + f... | 11 |
3,810 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12B_Problems/Problem_6 | 1 | The greatest prime number that is a divisor 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 $16383$
$\textbf{(A)} \: 3\qquad\textbf{(B)} \: 7\qquad\textbf{(C)} \: 10\qquad\textbf{(D)} \: 16\qquad\textbf{(E)} \: 22$ | 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 which can divide $16383$ . The sum of $127$ 's digits is $1+2+7=\boxed{10}$ | 10 |
3,811 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12B_Problems/Problem_8 | 1 | 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\te... | Let the lengths of the two congruent sides of the triangle be $x$ , then the product desired is $x^2$
Notice that the product of the base and twice the height is $4$ times the area of the triangle.
Set the vertex angle to be $a$ , we derive the equation:
$x^2=4\left(\frac{1}{2}x^2\sin(a)\right)$
$\sin(a)=\frac{1}{2}$
A... | 150 |
3,812 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12B_Problems/Problem_8 | 2 | 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\te... | Denote by $a$ the length of each congruent side. Denote by $\theta$ the degree measure of each acute angle.
Denote by $\phi$ the degree measure of the obtuse angle.
Hence, this problem tells us the following relationship: \[ a^2 = 2 a \cos \theta \cdot 2 a \sin \theta . \]
Hence, \begin{align*} 1 & = 2 \cdot 2 \sin \th... | 150 |
3,813 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12B_Problems/Problem_9 | 1 | 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)} \: 2... | Construct the circle that passes through $A$ $O$ , and $C$ , centered at $X$
Also notice that $\overline{OA}$ and $\overline{OC}$ are the angle bisectors of angle $\angle BAC$ and $\angle BCA$ respectively. We then deduce $\angle AOC=120^\circ$
Consider another point $M$ on Circle $X$ opposite to point $O$
As $AOCM$ is... | 12 |
3,814 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12B_Problems/Problem_9 | 2 | 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)} \: 2... | We have $\angle AOC = 120^\circ$
Denote by $R$ the circumradius of $\triangle AOC$ .
In $\triangle AOC$ , the law of sines implies \[ 2 R = \frac{AC}{\sin \angle AOC} = 4 \sqrt{3} . \]
Hence, the area of the circumcircle of $\triangle AOC$ is \[ \pi R^2 = 12 \pi . \]
Therefore, the answer is $\boxed{12}$ | 12 |
3,815 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12B_Problems/Problem_9 | 3 | 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)} \: 2... | As in the previous solution, construct the circle that passes through $A$ $O$ , and $C$ , centered at $X$ . Let $Y$ be the intersection of $\overline{OX}$ and $\overline{AB}$
Note that since $\overline{OA}$ is the angle bisector of $\angle BAC$ that $\angle OAC=30^\circ$ . Also by symmetry, $\overline{OX}$ $\perp$ $\ov... | 12 |
3,816 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12B_Problems/Problem_9 | 4 | 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)} \: 2... | The semiperimeter is $\frac{6+6+6}{2}=9$ units.
The area of the triangle is $9\sqrt{3}$ units squared. By the formula that says that the area of the triangle is its semiperimeter times its inradius, the inradius $r=\sqrt{3}$ . As $\angle{AOC}=120^\circ$ , we can form an altitude from point $O$ to side $AC$ at point $M... | 12 |
3,817 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12B_Problems/Problem_9 | 5 | 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)} \: 2... | Call the diameter of the circle $d$ . If we extend points $A$ and $C$ to meet at a point on the circle and call it $E$ , then $\bigtriangleup OAE=\bigtriangleup OCE$ . Note that both triangles are right, since their hypotenuse is the diameter of the circle. Therefore, $CE=AE=\sqrt{d^2-12}$ . We know this since $OC=OA=O... | 12 |
3,818 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12B_Problems/Problem_10 | 1 | 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 \[(\cos 40^\circ,\sin 40^\circ), (\cos 60^\circ,\sin 60^\circ), \text{ and } (\cos t^\circ,\sin t^\circ)\] is isosceles?
$\textbf{(A)} \: 100 \qquad\textbf{(B)} \: 150 \qquad\textbf{(C)}... | Let $A = (\cos 40^{\circ}, \sin 40^{\circ}), B = (\cos 60^{\circ}, \sin 60^{\circ}),$ and $C = (\cos t^{\circ}, \sin t^{\circ}).$ We apply casework to the legs of isosceles $\triangle ABC:$
Together, the sum of all such possible values of $t$ is $20+80+50+230=\boxed{380}.$ | 380 |
3,819 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12B_Problems/Problem_13 | 1 | Let $c = \frac{2\pi}{11}.$ 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)}\ {-}\frac{\sqrt{11}}{5} \qquad\textbf{(C)}\ \frac{\sqrt{11}}{5} \qquad\textbf{(D)}\ \frac{1... | 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}... | 1 |
3,820 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12B_Problems/Problem_13 | 2 | Let $c = \frac{2\pi}{11}.$ 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)}\ {-}\frac{\sqrt{11}}{5} \qquad\textbf{(C)}\ \frac{\sqrt{11}}{5} \qquad\textbf{(D)}\ \frac{1... | Eisenstein used such a quotient in his proof of quadratic reciprocity . Let $c=\frac{2\pi}{p}$ where $p$ is an odd prime number and $q$ is any integer.
Then $\dfrac{\sin(qc)\sin(2qc)\cdots\sin(\frac{p-1}{2}qc)}{\sin(c)\sin(2c)\cdots\sin(\frac{p-1}{2}c)}$ is the Legendre symbol $\left(\frac{q}{p}\right)$ . Legendre symb... | 1 |
3,821 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12B_Problems/Problem_14 | 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... | 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 $N = 1$ :... | 1 |
3,822 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12B_Problems/Problem_15 | 2 | 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 polygo... | As shown in Image:2021_AMC_12B_(Nov)_Problem_15,_sol.png , all 12 vertices of three squares form a regular dodecagon (12-gon).
Denote by $O$ the center of this dodecagon.
Hence, $\angle AOB = \frac{360^\circ}{12} = 30^\circ$
Because the length of a side of a square is 6, $AO = 3 \sqrt{2}$
Hence, $AB = 2 AO \sin \frac{\... | 147 |
3,823 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12B_Problems/Problem_16 | 1 | 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 $a^2+b^2+c^2$
$\textbf{(A)}\: 259\qquad\textbf{(B)} \: 438\qquad\textbf{(C)} \: 516\qquad\textbf{(D)} \: 625\qquad\textbf{(E)} \: 687$ | 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... | 438 |
3,824 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12B_Problems/Problem_16 | 2 | 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 $a^2+b^2+c^2$
$\textbf{(A)}\: 259\qquad\textbf{(B)} \: 438\qquad\textbf{(C)} \: 516\qquad\textbf{(D)} \: 625\qquad\textbf{(E)} \: 687$ | Let $\gcd(a,b)=x$ $\gcd(b,c)=y$ $\gcd(c,a)=z$ . Without the loss of generality, let $x \le y \le z$ . We can split this off into cases:
$x=1,y=1,z=7$ : let $a=7A, c=7C,$ we can try all possibilities of $A$ and $C$ to find that $a=7, b=9, c=7$ is the only solution.
$x=1,y=2,z=6$ : No solutions. By $y$ and $z$ , we know ... | 438 |
3,825 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12B_Problems/Problem_16 | 3 | 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 $a^2+b^2+c^2$
$\textbf{(A)}\: 259\qquad\textbf{(B)} \: 438\qquad\textbf{(C)} \: 516\qquad\textbf{(D)} \: 625\qquad\textbf{(E)} \: 687$ | Since $a+b+c=23$ $\gcd(a,b,c)=23$ or $\gcd(a,b,c)=1$
As $\gcd(a,b)+\gcd(b,c)+\gcd(c,a)=9$ , it is impossible for $\gcd(a,b,c)=23$ , so $\gcd(a,b,c)=1$
This means that $\gcd(a,b)$ $\gcd(b,c)$ , and $\gcd(c,a)$ must all be coprime. The only possible ways for this to be true are $1+1+7=9$ and $1+3+5=9$
Without loss of gen... | 438 |
3,826 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12B_Problems/Problem_18 | 1 | Set $u_0 = \frac{1}{4}$ , 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... | Note that terms of the sequence $(u_k)$ lie in the interval $\left(0,\frac12\right),$ strictly increasing.
Since the sequence $(u_k)$ tends to the limit $L,$ we set $u_{k+1}=u_k=L>0.$
The given equation becomes \[L=2L-2L^2,\] from which $L=\frac12.$
The given inequality becomes \[\frac12-\frac{1}{2^{1000}} \leq u_k \le... | 10 |
3,827 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12B_Problems/Problem_18 | 2 | Set $u_0 = \frac{1}{4}$ , 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... | If we list out the first few values of $k$ , we get the series $\frac{1}{4}, \frac{3}{8}, \frac{15}{32}, \frac{255}{512}$ , which always seems to be a negative power of $2$ away from $\frac{1}{2}$ . We can test this out by setting $u_k=\frac{1}{2}-\frac{1}{2^{n_k}}$ , where $n_0=2$
Now, we get \begin{align*} u_{k+1} &=... | 10 |
3,828 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12B_Problems/Problem_18 | 3 | Set $u_0 = \frac{1}{4}$ , 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... | We are given $u_{k+1} = 2u_k - 2{u_k}^2$ . Multiply this equation by $2$ and subtract $1$ from both sides. The equations can then be written nicely as $2u_{k+1} - 1 = -(2u_k-1)^2$ . Let $v_k = 2u_k - 1$ so that $v_{k+1} = -(v_k)^2$
Clearly, $v_0 = 2u_0 - 1 = -\frac{1}{2}$ . Since the magnitude of $v_0$ is less than $1$... | 10 |
3,829 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12B_Problems/Problem_19 | 1 | 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(\t... | 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 t... | 68 |
3,830 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12B_Problems/Problem_20 | 1 | 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... | 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 u... | 7 |
3,831 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12B_Problems/Problem_20 | 2 | 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... | Let’s split the cube into two layers; a bottom and top. Note that there must be four of each color, so however many number of one color are in the bottom, there will be four minus that number of the color on the top. We do casework on the color distribution of the bottom layer.
Case 1: 4, 0
In this case, there is only ... | 7 |
3,832 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12B_Problems/Problem_20 | 3 | 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... | Divide the $2 \times 2 \times 2$ cube into two layers, say, front and back. Any possible construction can be rotated such that the front layer has the same or greater number of white cubes than blue cubes, so we only need to count the number of cases given that is true.
Therefore, our answer is $6+1+0=\boxed{7}$ | 7 |
3,833 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12B_Problems/Problem_20 | 4 | 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... | Burnside lemma is used to counting number of orbit where the element on the same orbit can be achieved by the defined operator, naming rotation, reflection and etc.
The fact for Burnside lemma are
1. the sum of stablizer on the same orbit equals to the # of operators;
2. the sum of stablizer can be counted as $fix(g)$
... | 7 |
3,834 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12B_Problems/Problem_20 | 5 | 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... | Since rotations of a single pattern are considered indistinguishable, we can assume that the forward upper right corner of the 2-by-2-by-2 cube is a blue cube (since we can always rotate the big cube to place a blue cube in that spot).
Once we've assigned this cube to be blue, we note that 3 1-by-1-by-1 cubes share a s... | 7 |
3,835 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12B_Problems/Problem_21 | 1 | 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 \[P(x)=0?\]
$\textbf{(A)}\ 0 \qquad\textbf{(B)}\ 1 \qquad\textbf{(C)}\ 2 \qquad\textbf{(D)}\ 3 \qquad\textbf{(E)}\ 4$ | Let $a=\cos(x)+i\sin(x)$ . Now $P(a)=1+a-a^2+a^3$ $P(-1)=-2$ and $P(0)=1$ so there is a real root $a_1$ between $-1$ and $0$ . The other $a$ 's must be complex conjugates since all coefficients of the polynomial are real. The magnitude of those complex $a$ 's squared is $\frac{1}{a_1}$ which is greater than $1$ . If $x... | 0 |
3,836 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12B_Problems/Problem_21 | 2 | 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 \[P(x)=0?\]
$\textbf{(A)}\ 0 \qquad\textbf{(B)}\ 1 \qquad\textbf{(C)}\ 2 \qquad\textbf{(D)}\ 3 \qquad\textbf{(E)}\ 4$ | 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 $\cos(x)=\tfrac 12$ , 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 $\boxed{0}$ | 0 |
3,837 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12B_Problems/Problem_21 | 3 | 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 \[P(x)=0?\]
$\textbf{(A)}\ 0 \qquad\textbf{(B)}\ 1 \qquad\textbf{(C)}\ 2 \qquad\textbf{(D)}\ 3 \qquad\textbf{(E)}\ 4$ | We have \begin{align*} P \left( x \right) & = 1 + e^{ix} - e^{i 2x} + e^{i 3x} . \end{align*}
Denote $y = e^{i x}$ . Hence, this problem asks us to find the number of $y$ with $| y| = 1$ that satisfy \[ 1 + y - y^2 + y^3 = 0 . \hspace{1cm} (1) \]
Taking imaginary part of both sides, we have \begin{align*} 0 & = {\rm Im... | 0 |
3,838 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12B_Problems/Problem_21 | 4 | 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 \[P(x)=0?\]
$\textbf{(A)}\ 0 \qquad\textbf{(B)}\ 1 \qquad\textbf{(C)}\ 2 \qquad\textbf{(D)}\ 3 \qquad\textbf{(E)}\ 4$ | Let $a=\cos(x)+i\sin(x)$ , so by De Moivre $P(x)=a^3-a^2+a+1$ . The problem essentially asks for the number of real roots of $P$ which lie on the complex unit circle.
Let $|r|=1$ be a root of $P$ , and note that we can't have $r^3-r^2+r=0$ , else $P(r)=0$ . Thus, suppose henceforth that $r^3-r^2+r \neq 0$ . We then hav... | 0 |
3,839 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12B_Problems/Problem_21 | 5 | 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 \[P(x)=0?\]
$\textbf{(A)}\ 0 \qquad\textbf{(B)}\ 1 \qquad\textbf{(C)}\ 2 \qquad\textbf{(D)}\ 3 \qquad\textbf{(E)}\ 4$ | $P(x)$ can be written equivalently as $P(x) = 1 + cis(x) - cis(2x) + cis(3x).$ Thus, we aim to find $x$ such that the sum of the vectors $cis(x)$ $-cis(2x)$ , and $cis(3x)$ is -1. Notice that $cis(x)$ $-cis(2x)$ $cis(3x)$ all lie on the unit circle in the complex plane, and the vector $cis(x) + cis(3x)$ is collinear wi... | 0 |
3,840 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12B_Problems/Problem_24 | 1 | 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 $CF$
$\t... | By the Law of Cosine $\cos A = \frac{AC^2 + AB^2 - BC^2}{ 2 \cdot AC \cdot AB} = \frac{20^2 + 11^2 - 24^2}{2\cdot20\cdot11} = -\frac18$
As $ABEC$ is a cyclic quadrilateral, $\angle CEA = \angle CBA$ . As $BDEF$ is a cyclic quadrilateral, $\angle CBA = \angle FEA$
$\because \quad \angle CEA = \angle FEA \quad \text{and}... | 30 |
3,841 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12B_Problems/Problem_24 | 2 | 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 $CF$
$\t... | Construct the $E$ -antipode, $E^{\prime}\in(ABC)$ . Notice $\triangle CE^{\prime}A\stackrel{+}{\sim}\triangle CBF$ by spiral similarity at $C$ , thus $CF=\dfrac{CB\cdot CA}{CE^{\prime}}=\frac{480}{CE^{\prime}}$ . Let $CE^{\prime}=x$ ; by symmetry $BE^{\prime}=x$ as well and $\cos\angle BE^{\prime}C=\cos\angle A=\tfrac{... | 30 |
3,842 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12B_Problems/Problem_24 | 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 $CF$
$\t... | Applying Stewart's theorem on $\triangle ABC$ with cevian $\overline{CF}$ using the directed lengths $AF = AC = 20$ and $FB = 11-20 = -9$ , we obtain \begin{align*} (20)(-9)(11) + (CF)(11)(CF) &= (24)(20)(24) + (20)(-9)(20) \\ 11CF^{2} - 1980 &= 11520 - 3600\end{align*} so $CF=\sqrt{\frac{11520 - 3600 + 1980}{11}}=\sqr... | 30 |
3,843 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12B_Problems/Problem_24 | 4 | 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 $CF$
$\t... | Note that $\angle CAF = \angle CAB$ so we may plug into Law of Cosines to find the angle's cosine: \[AB^2+AC^2-2\cdot AB \cdot AC \cdot \cos(\angle CAB) = BC^2 \to \cos(\angle CAB) = -\frac{1}{8}.\]
So, we observe that we can use Law of Cosines again to find $CF$ \[CF^2 = AF^2+AC^2-2 \cdot AF \cdot AC \cdot \cos(\angle... | 30 |
3,844 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12B_Problems/Problem_24 | 5 | 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 $CF$
$\t... | This solution is based on this figure: 2021 AMC 12B (Nov) Problem 24, sol.png
Denote by $O$ the circumcenter of $\triangle BED$ .
Denote by $R$ the circumradius of $\triangle BED$
In $\triangle BCF$ , following from the law of cosines, we have \begin{align*} CF^2 & = BC^2 + BF^2 - 2 BC \cdot BF \cos \angle CBF \\ & = B... | 30 |
3,845 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12B_Problems/Problem_25 | 1 | 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{(... | Note that 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 four 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 divisible by $6$ $5+4$ indicates that $n+1$ is divisible by $2$ $6+3... | 2 |
3,846 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12B_Problems/Problem_25 | 2 | 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{(... | Denote by ${\rm Rem} \ \left( n, k \right)$ the remainder of $n$ divided by $k$ .
Define $\Delta \left( n, k \right) = {\rm Rem} \ \left( n + 1, k \right) - {\rm Rem} \ \left( n, k \right)$
Hence, \[ \Delta \left( n, k \right) = \left\{ \begin{array}{ll} 1 & \mbox{ if } n \not\equiv -1 \pmod{k} \\ - \left( k -1 \right... | 2 |
3,847 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_12A_Problems/Problem_1 | 1 | What is the value of \[2^{1+2+3}-(2^1+2^2+2^3)?\] $\textbf{(A) }0 \qquad \textbf{(B) }50 \qquad \textbf{(C) }52 \qquad \textbf{(D) }54 \qquad \textbf{(E) }57$ | We evaluate the given expression to get that \[2^{1+2+3}-(2^1+2^2+2^3)=2^6-(2^1+2^2+2^3)=64-2-4-8=50 \implies \boxed{50}\] | 50 |
3,848 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_12A_Problems/Problem_3 | 1 | The sum of two natural numbers is $17402$ . One of the two numbers is divisible by $10$ . If the units digit of that number is erased, the other number is obtained. What is the difference of these two numbers?
$\textbf{(A)} ~10272\qquad\textbf{(B)} ~11700\qquad\textbf{(C)} ~13362\qquad\textbf{(D)} ~14238\qquad\textbf{(... | The units digit of a multiple of $10$ will always be $0$ . We add a $0$ whenever we multiply by $10$ . So, removing the units digit is equal to dividing by $10$
Let the smaller number (the one we get after removing the units digit) be $a$ . This means the bigger number would be $10a$
We know the sum is $10a+a = 11a$ so... | 238 |
3,849 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_12A_Problems/Problem_3 | 2 | The sum of two natural numbers is $17402$ . One of the two numbers is divisible by $10$ . If the units digit of that number is erased, the other number is obtained. What is the difference of these two numbers?
$\textbf{(A)} ~10272\qquad\textbf{(B)} ~11700\qquad\textbf{(C)} ~13362\qquad\textbf{(D)} ~14238\qquad\textbf{(... | Since the unit's place of a multiple of $10$ is $0$ , the other integer must end with a $2$ , for both integers sum up to a number ending in a $2$ . Thus, the unit's place of the difference must be $10-2=8$ , and the only answer choice that ends with an $8$ is $\boxed{14238}$ | 238 |
3,850 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_12A_Problems/Problem_3 | 3 | The sum of two natural numbers is $17402$ . One of the two numbers is divisible by $10$ . If the units digit of that number is erased, the other number is obtained. What is the difference of these two numbers?
$\textbf{(A)} ~10272\qquad\textbf{(B)} ~11700\qquad\textbf{(C)} ~13362\qquad\textbf{(D)} ~14238\qquad\textbf{(... | Let the larger number be $\underline{ABCD0}.$ It follows that the smaller number is $\underline{ABCD}.$ Adding vertically, we have \[\begin{array}{cccccc} & A & B & C & D & 0 \\ +\quad & & A & B & C & D \\ \hline & & & & & \\ [-2.5ex] & 1 & 7 & 4 & 0 & 2 \\ \end{array}\] Working from right to left, we... | 238 |
3,851 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_12A_Problems/Problem_3 | 4 | The sum of two natural numbers is $17402$ . One of the two numbers is divisible by $10$ . If the units digit of that number is erased, the other number is obtained. What is the difference of these two numbers?
$\textbf{(A)} ~10272\qquad\textbf{(B)} ~11700\qquad\textbf{(C)} ~13362\qquad\textbf{(D)} ~14238\qquad\textbf{(... | We know that the larger number has a units digit of $0$ since it is divisible by 10. If $D$ is the ten's digit of the larger number, then $D$ is the units digit of the smaller number. Since the sum of the natural numbers has a unit's digit of $2$ $D=2$
The units digit of the larger number is $0$ and the units digit of ... | 238 |
3,852 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_12A_Problems/Problem_5 | 1 | When a student multiplied the number $66$ by the repeating decimal, \[\underline{1}.\underline{a} \ \underline{b} \ \underline{a} \ \underline{b}\ldots=\underline{1}.\overline{\underline{a} \ \underline{b}},\] where $a$ and $b$ are digits, he did not notice the notation and just multiplied $66$ times $\underline{1}.\un... | We are given that $66\Bigl(\underline{1}.\overline{\underline{a} \ \underline{b}}\Bigr)-0.5=66\Bigl(\underline{1}.\underline{a} \ \underline{b}\Bigr),$ from which \begin{align*} 66\Bigl(\underline{1}.\overline{\underline{a} \ \underline{b}}\Bigr)-66\Bigl(\underline{1}.\underline{a} \ \underline{b}\Bigr)&=0.5 \\ 66\Bigl... | 75 |
3,853 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_12A_Problems/Problem_5 | 2 | When a student multiplied the number $66$ by the repeating decimal, \[\underline{1}.\underline{a} \ \underline{b} \ \underline{a} \ \underline{b}\ldots=\underline{1}.\overline{\underline{a} \ \underline{b}},\] where $a$ and $b$ are digits, he did not notice the notation and just multiplied $66$ times $\underline{1}.\un... | It is known that $\underline{0}.\overline{\underline{a} \ \underline{b}}=\frac{\underline{a} \ \underline{b}}{99}$ and $\underline{0}.\underline{a} \ \underline{b}=\frac{\underline{a} \ \underline{b}}{100}.$
Let $x=\underline{a} \ \underline{b}.$ We have \[66\biggl(1+\frac{x}{99}\biggr)-66\biggl(1+\frac{x}{100}\biggr)=... | 75 |
3,854 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_12A_Problems/Problem_5 | 3 | When a student multiplied the number $66$ by the repeating decimal, \[\underline{1}.\underline{a} \ \underline{b} \ \underline{a} \ \underline{b}\ldots=\underline{1}.\overline{\underline{a} \ \underline{b}},\] where $a$ and $b$ are digits, he did not notice the notation and just multiplied $66$ times $\underline{1}.\un... | We have \[66 \cdot \left(1 + \frac{10a+b}{100}\right) + \frac{1}{2} = 66 \cdot \left(1+ \frac{10a+b}{99}\right).\] Expanding both sides, we have \[66 + \frac{33(10a+b)}{50} + \frac{1}{2} = 66 + \frac{2(10a+b)}{3}.\] Subtracting $66$ from both sides, we have \[\frac{33(10a+b)}{50} + \frac{1}{2} = \frac{2(10a+b)}{3}.\] M... | 75 |
3,855 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_12A_Problems/Problem_6 | 1 | A deck of cards has only red cards and black cards. The probability of a randomly chosen card being red is $\frac13$ . When $4$ black cards are added to the deck, the probability of choosing red becomes $\frac14$ . How many cards were in the deck originally?
$\textbf{(A) }6 \qquad \textbf{(B) }9 \qquad \textbf{(C) }12 ... | If the probability of choosing a red card is $\frac{1}{3}$ , the red and black cards are in ratio $1:2$ . This means at the beginning there are $x$ red cards and $2x$ black cards.
After $4$ black cards are added, there are $2x+4$ black cards. This time, the probability of choosing a red card is $\frac{1}{4}$ so the rat... | 12 |
3,856 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_12A_Problems/Problem_6 | 2 | A deck of cards has only red cards and black cards. The probability of a randomly chosen card being red is $\frac13$ . When $4$ black cards are added to the deck, the probability of choosing red becomes $\frac14$ . How many cards were in the deck originally?
$\textbf{(A) }6 \qquad \textbf{(B) }9 \qquad \textbf{(C) }12 ... | In terms of the number of cards, the original deck is $3$ times the red cards, and the final deck is $4$ times the red cards. So, the final deck is $\frac43$ times the original deck. We are given that adding $4$ cards to the original deck is the same as increasing the original deck by $\frac13$ of itself. Since $4$ car... | 12 |
3,857 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_12A_Problems/Problem_6 | 3 | A deck of cards has only red cards and black cards. The probability of a randomly chosen card being red is $\frac13$ . When $4$ black cards are added to the deck, the probability of choosing red becomes $\frac14$ . How many cards were in the deck originally?
$\textbf{(A) }6 \qquad \textbf{(B) }9 \qquad \textbf{(C) }12 ... | Suppose there were $x$ cards in the deck originally. Now, the deck has $x+4$ cards, which must be a multiple of $4.$
Only $12+4=16$ is a multiple of $4,$ so the answer is $x=\boxed{12}.$ | 12 |
3,858 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_12A_Problems/Problem_7 | 3 | What is the least possible value of $(xy-1)^2+(x+y)^2$ for real numbers $x$ and $y$
$\textbf{(A)} ~0\qquad\textbf{(B)} ~\frac{1}{4}\qquad\textbf{(C)} ~\frac{1}{2} \qquad\textbf{(D)} ~1 \qquad\textbf{(E)} ~2$ | Like solution 1, expand and simplify the original equation to $x^2+y^2+x^2y^2+1$ and let $f(x, y) = x^2+y^2+x^2y^2+1$ . To find local extrema, find where $\nabla f(x, y) = \boldsymbol{0}$ . First, find the first partial derivative with respect to x and y and find where they are $0$ \[\frac{\partial f}{\partial x} = 2x ... | 1 |
3,859 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_12A_Problems/Problem_10 | 1 | Two right circular cones with vertices facing down as shown in the figure below contain the same amount of liquid. The radii of the tops of the liquid surfaces are $3$ cm and $6$ cm. Into each cone is dropped a spherical marble of radius $1$ cm, which sinks to the bottom and is completely submerged without spilling any... | The heights of the cones are not given, so suppose the heights are very large (i.e. tending towards infinity) in order to approximate the cones as cylinders with base radii $3$ and $6$ and infinitely large height. Then the base area of the wide cylinder is $4$ times that of the narrow cylinder. Since we are dropping a ... | 4 |
3,860 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_12A_Problems/Problem_12 | 1 | All the roots of the polynomial $z^6-10z^5+Az^4+Bz^3+Cz^2+Dz+16$ are positive integers, possibly repeated. What is the value of $B$
$\textbf{(A) }{-}88 \qquad \textbf{(B) }{-}80 \qquad \textbf{(C) }{-}64 \qquad \textbf{(D) }{-}41\qquad \textbf{(E) }{-}40$ | By Vieta's formulas, the sum of the six roots is $10$ and the product of the six roots is $16$ . By inspection, we see the roots are $1, 1, 2, 2, 2,$ and $2$ , so the function is $(z-1)^2(z-2)^4=(z^2-2z+1)(z^4-8z^3+24z^2-32z+16)$ . Therefore, calculating just the $z^3$ terms, we get $B = -32 - 48 - 8 = \boxed{88}$ | 88 |
3,861 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_12A_Problems/Problem_12 | 2 | All the roots of the polynomial $z^6-10z^5+Az^4+Bz^3+Cz^2+Dz+16$ are positive integers, possibly repeated. What is the value of $B$
$\textbf{(A) }{-}88 \qquad \textbf{(B) }{-}80 \qquad \textbf{(C) }{-}64 \qquad \textbf{(D) }{-}41\qquad \textbf{(E) }{-}40$ | Using the same method as Solution 1, we find that the roots are $2, 2, 2, 2, 1,$ and $1$ . Note that $B$ is the negation of the 3rd symmetric sum of the roots. Using casework on the number of 1's in each of the $\binom {6}{3} = 20$ products $r_a \cdot r_b \cdot r_c,$ we obtain \[B= - \left(\binom {4}{3} \binom {2}{0} \... | 88 |
3,862 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_12A_Problems/Problem_13 | 1 | Of the following complex numbers $z$ , which one has the property that $z^5$ has the greatest real part?
$\textbf{(A) }{-}2 \qquad \textbf{(B) }{-}\sqrt3+i \qquad \textbf{(C) }{-}\sqrt2+\sqrt2 i \qquad \textbf{(D) }{-}1+\sqrt3 i\qquad \textbf{(E) }2i$ | First, $\textbf{(B)}$ is $2\text{cis}(150)$ $\textbf{(C)}$ is $2\text{cis}(135)$ $\textbf{(D)}$ is $2\text{cis}(120)$
Taking the real part of the $5$ th power of each we have:
$\textbf{(A): }(-2)^5=-32$
$\textbf{(B): }32\cos(750)=32\cos(30)=16\sqrt{3}$
$\textbf{(C): }32\cos(675)=32\cos(-45)=16\sqrt{2}$
$\textbf{(D): }3... | 3 |
3,863 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_12A_Problems/Problem_13 | 2 | Of the following complex numbers $z$ , which one has the property that $z^5$ has the greatest real part?
$\textbf{(A) }{-}2 \qquad \textbf{(B) }{-}\sqrt3+i \qquad \textbf{(C) }{-}\sqrt2+\sqrt2 i \qquad \textbf{(D) }{-}1+\sqrt3 i\qquad \textbf{(E) }2i$ | We rewrite each answer choice to the polar form \[z=r\operatorname{cis}\theta=r(\cos\theta+i\sin\theta),\] where $r$ is the magnitude of $z$ such that $r\geq0,$ and $\theta$ is the argument of $z$ such that $0\leq\theta<2\pi.$
By De Moivre's Theorem , the real part of $z^5$ is \[\operatorname{Re}\left(z^5\right)=r^5\co... | 3 |
3,864 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_12A_Problems/Problem_13 | 3 | Of the following complex numbers $z$ , which one has the property that $z^5$ has the greatest real part?
$\textbf{(A) }{-}2 \qquad \textbf{(B) }{-}\sqrt3+i \qquad \textbf{(C) }{-}\sqrt2+\sqrt2 i \qquad \textbf{(D) }{-}1+\sqrt3 i\qquad \textbf{(E) }2i$ | To find the real part of $z^5,$ we only need the terms with even powers of $i:$ \begin{align*} \operatorname{Re}\left(z^5\right)&=\operatorname{Re}\left((a+bi)^5\right) \\ &=\sum_{k=0}^{2}\binom{5}{2k}a^{5-2k}(bi)^{2k} \\ &=\binom50 a^{5}(bi)^{0} + \binom52 a^{3}(bi)^{2} + \binom54 a^{1}(bi)^{4} \\ &=a^5 - 10a^3b^2 + 5... | 3 |
3,865 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_12A_Problems/Problem_13 | 4 | Of the following complex numbers $z$ , which one has the property that $z^5$ has the greatest real part?
$\textbf{(A) }{-}2 \qquad \textbf{(B) }{-}\sqrt3+i \qquad \textbf{(C) }{-}\sqrt2+\sqrt2 i \qquad \textbf{(D) }{-}1+\sqrt3 i\qquad \textbf{(E) }2i$ | The full expansion of $z^5$ is \begin{align*} z^5&=(a+bi)^5 \\ &=\sum_{k=0}^{5}\binom5k a^{5-k}(bi)^k \\ &=\binom50 a^{5}(bi)^0+\binom51 a^{4}(bi)^1+\binom52 a^{3}(bi)^2+\binom53 a^{2}(bi)^3+\binom54 a^{1}(bi)^4+\binom55 a^{0}(bi)^5 \\ &=a^5+5a^4bi-10a^3b^2-10a^2b^3i+5ab^4+b^5i \\ &=\left(a^5-10a^3b^2+5ab^4\right) + \l... | 3 |
3,866 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_12A_Problems/Problem_14 | 1 | What is the value of \[\left(\sum_{k=1}^{20} \log_{5^k} 3^{k^2}\right)\cdot\left(\sum_{k=1}^{100} \log_{9^k} 25^k\right)?\]
$\textbf{(A) }21 \qquad \textbf{(B) }100\log_5 3 \qquad \textbf{(C) }200\log_3 5 \qquad \textbf{(D) }2200\qquad \textbf{(E) }21000$ | We will apply the following logarithmic identity: \[\log_{p^n}{q^n}=\log_{p}{q},\] which can be proven by the Change of Base Formula: \[\log_{p^n}{q^n}=\frac{\log_{p}{q^n}}{\log_{p}{p^n}}=\frac{n\log_{p}{q}}{n}=\log_{p}{q}.\] Now, we simplify the expressions inside the summations: \begin{align*} \log_{5^k}{{3^k}^2}&=\l... | 0 |
3,867 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_12A_Problems/Problem_14 | 2 | What is the value of \[\left(\sum_{k=1}^{20} \log_{5^k} 3^{k^2}\right)\cdot\left(\sum_{k=1}^{100} \log_{9^k} 25^k\right)?\]
$\textbf{(A) }21 \qquad \textbf{(B) }100\log_5 3 \qquad \textbf{(C) }200\log_3 5 \qquad \textbf{(D) }2200\qquad \textbf{(E) }21000$ | First, we can get rid of the $k$ exponents using properties of logarithms: \[\log_{5^k} 3^{k^2} = k^2 \cdot \frac{1}{k} \cdot \log_{5} 3 = k\log_{5} 3 = \log_{5} 3^k.\] (Leaving the single $k$ in the exponent will come in handy later). Similarly, \[\log_{9^k} 25^{k} = k \cdot \frac{1}{k} \cdot \log_{9} 25 = \log_{9} 5^... | 0 |
3,868 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_12A_Problems/Problem_14 | 3 | What is the value of \[\left(\sum_{k=1}^{20} \log_{5^k} 3^{k^2}\right)\cdot\left(\sum_{k=1}^{100} \log_{9^k} 25^k\right)?\]
$\textbf{(A) }21 \qquad \textbf{(B) }100\log_5 3 \qquad \textbf{(C) }200\log_3 5 \qquad \textbf{(D) }2200\qquad \textbf{(E) }21000$ | In $\sum_{k=1}^{20} \log_{5^k} 3^{k^2},$ note that the addends are greater than $1$ for all $k\geq2.$
In $\sum_{k=1}^{100} \log_{9^k} 25^k,$ note that the addends are greater than $1$ for all $k\geq1.$
We have the inequality \[\left(\sum_{k=1}^{20} \log_{5^k} 3^{k^2}\right)\cdot\left(\sum_{k=1}^{100} \log_{9^k} 25^k\ri... | 0 |
3,869 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_12A_Problems/Problem_14 | 4 | What is the value of \[\left(\sum_{k=1}^{20} \log_{5^k} 3^{k^2}\right)\cdot\left(\sum_{k=1}^{100} \log_{9^k} 25^k\right)?\]
$\textbf{(A) }21 \qquad \textbf{(B) }100\log_5 3 \qquad \textbf{(C) }200\log_3 5 \qquad \textbf{(D) }2200\qquad \textbf{(E) }21000$ | Using the identity \[\log_{p^n}{q^n}=\log_{p}{q},\] simplify \begin{align*} \log_{5^k}{{3^k}^2}&=\log_{5^k}{(3^k)^k} \\ &=\log_{5}{3^k} \\ \end{align*} and \begin{align*} \log_{9^k}{25^k}&=\log_{3^{2k}}{5^{2k}} \\ &=\log_{3}{5}. \end{align*} . Now we have the product: \[\left(\sum_{k=1}^{20} \log_{5} 3^{k}\right)\cdot\... | 0 |
3,870 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_12A_Problems/Problem_16 | 1 | In the following list of numbers, the integer $n$ appears $n$ times in the list for $1 \leq n \leq 200$ \[1, 2, 2, 3, 3, 3, 4, 4, 4, 4, \ldots, 200, 200, \ldots , 200\] What is the median of the numbers in this list?
$\textbf{(A)} ~100.5 \qquad\textbf{(B)} ~134 \qquad\textbf{(C)} ~142 \qquad\textbf{(D)} ~150.5 \qquad\t... | The $x$ th number of this sequence is $\left\lceil\frac{-1\pm\sqrt{1+8x}}{2}\right\rceil$ via the quadratic formula. We can see that if we halve $x$ we end up getting $\left\lceil\frac{-1\pm\sqrt{1+4x}}{2}\right\rceil$ . This is approximately the number divided by $\sqrt{2}$ $\frac{200}{\sqrt{2}} = 141.4$ and since $14... | 142 |
3,871 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_12A_Problems/Problem_16 | 2 | In the following list of numbers, the integer $n$ appears $n$ times in the list for $1 \leq n \leq 200$ \[1, 2, 2, 3, 3, 3, 4, 4, 4, 4, \ldots, 200, 200, \ldots , 200\] What is the median of the numbers in this list?
$\textbf{(A)} ~100.5 \qquad\textbf{(B)} ~134 \qquad\textbf{(C)} ~142 \qquad\textbf{(D)} ~150.5 \qquad\t... | We can look at answer choice $\textbf{(C)}$ , which is $142$ first. That means that the number of numbers from $1$ to $142$ is roughly the number of numbers from $143$ to $200$
The number of numbers from $1$ to $142$ is $\frac{142(142+1)}{2}$ which is approximately $10000.$ The number of numbers from $143$ to $200$ is ... | 142 |
3,872 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_12A_Problems/Problem_16 | 3 | In the following list of numbers, the integer $n$ appears $n$ times in the list for $1 \leq n \leq 200$ \[1, 2, 2, 3, 3, 3, 4, 4, 4, 4, \ldots, 200, 200, \ldots , 200\] What is the median of the numbers in this list?
$\textbf{(A)} ~100.5 \qquad\textbf{(B)} ~134 \qquad\textbf{(C)} ~142 \qquad\textbf{(D)} ~150.5 \qquad\t... | We can arrange the numbers in the following pattern: \[ \begin{array}{cccccc} \ &\ &\ &\ &\ 200 & \\ \ &\ &\ &\ 199 & \ 200 & \\ \ &\ &\ \iddots& \ \vdots& \ \vdots& \\ \ &\ 2& \ \cdots& \ 199& \ 200& \\ 1 & \ 2 & \ \cdots& \ 199& \ 200& \end{array} \]
We can see this as a isosceles right triangle, with legs of length ... | 142 |
3,873 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_12A_Problems/Problem_17 | 1 | Trapezoid $ABCD$ has $\overline{AB}\parallel\overline{CD},BC=CD=43$ , and $\overline{AD}\perp\overline{BD}$ . Let $O$ be the intersection of the diagonals $\overline{AC}$ and $\overline{BD}$ , and let $P$ be the midpoint of $\overline{BD}$ . Given that $OP=11$ , the length of $AD$ can be written in the form $m\sqrt{n}$... | Angle chasing* reveals that $\triangle BPC\sim\triangle BDA$ , therefore \[2=\frac{BD}{BP}=\frac{AB}{BC}=\frac{AB}{43},\] or $AB=86$
Additional angle chasing shows that $\triangle ABO\sim\triangle CDO$ , therefore \[2=\frac{AB}{CD}=\frac{BO}{OD}=\frac{BP+11}{BP-11},\] or $BP=33$ and $BD=66$
Since $\triangle ADB$ is rig... | 194 |
3,874 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_12A_Problems/Problem_17 | 2 | Trapezoid $ABCD$ has $\overline{AB}\parallel\overline{CD},BC=CD=43$ , and $\overline{AD}\perp\overline{BD}$ . Let $O$ be the intersection of the diagonals $\overline{AC}$ and $\overline{BD}$ , and let $P$ be the midpoint of $\overline{BD}$ . Given that $OP=11$ , the length of $AD$ can be written in the form $m\sqrt{n}$... | Since $\triangle BCD$ is isosceles with base $\overline{BD},$ it follows that median $\overline{CP}$ is also an altitude. Let $OD=x$ and $CP=h,$ so $PB=x+11.$
Since $\angle AOD=\angle COP$ by vertical angles, we conclude that $\triangle AOD\sim\triangle COP$ by AA, from which $\frac{AD}{CP}=\frac{OD}{OP},$ or \[AD=CP\c... | 194 |
3,875 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_12A_Problems/Problem_17 | 3 | Trapezoid $ABCD$ has $\overline{AB}\parallel\overline{CD},BC=CD=43$ , and $\overline{AD}\perp\overline{BD}$ . Let $O$ be the intersection of the diagonals $\overline{AC}$ and $\overline{BD}$ , and let $P$ be the midpoint of $\overline{BD}$ . Given that $OP=11$ , the length of $AD$ can be written in the form $m\sqrt{n}$... | Let $CP = y$ $CP$ a is perpendicular bisector of $DB.$ Then, let $DO = x,$ thus $DP = PB = 11+x.$
(1) $\triangle CPO \sim \triangle ADO,$ so we get $\frac{AD}{x} = \frac{y}{11},$ or $AD = \frac{xy}{11}.$
(2) Applying Pythagorean Theorem on $\triangle CDP$ gives $(11+x)^2 + y^2 = 43^2.$
(3) $\triangle BPC \sim \triangle... | 194 |
3,876 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_12A_Problems/Problem_17 | 4 | Trapezoid $ABCD$ has $\overline{AB}\parallel\overline{CD},BC=CD=43$ , and $\overline{AD}\perp\overline{BD}$ . Let $O$ be the intersection of the diagonals $\overline{AC}$ and $\overline{BD}$ , and let $P$ be the midpoint of $\overline{BD}$ . Given that $OP=11$ , the length of $AD$ can be written in the form $m\sqrt{n}$... | Observe that $\triangle BPC$ is congruent to $\triangle DPC$ ; both are similar to $\triangle BDA$ . Let's extend $\overline{AD}$ and $\overline{BC}$ past points $D$ and $C$ respectively, such that they intersect at a point $E$ . Observe that $\angle BDE$ is $90$ degrees, and that $\angle DBE \cong \angle PBC \cong \an... | 194 |
3,877 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_12A_Problems/Problem_17 | 5 | Trapezoid $ABCD$ has $\overline{AB}\parallel\overline{CD},BC=CD=43$ , and $\overline{AD}\perp\overline{BD}$ . Let $O$ be the intersection of the diagonals $\overline{AC}$ and $\overline{BD}$ , and let $P$ be the midpoint of $\overline{BD}$ . Given that $OP=11$ , the length of $AD$ can be written in the form $m\sqrt{n}$... | Since $P$ is the midpoint of isosceles triangle $BCD$ , it would be pretty easy to see that $CP\perp BD$ . Since $AD\perp BD$ as well, $AD\parallel CP$ . Connecting $AP$ , it’s obvious that $[ADC]=[ADP]$ . Since $DP=BP$ $[APB]=[ADC]$
Since $P$ is the midpoint of $BD$ , the height of $\triangle APB$ on side $AB$ is half... | 194 |
3,878 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_12A_Problems/Problem_17 | 6 | Trapezoid $ABCD$ has $\overline{AB}\parallel\overline{CD},BC=CD=43$ , and $\overline{AD}\perp\overline{BD}$ . Let $O$ be the intersection of the diagonals $\overline{AC}$ and $\overline{BD}$ , and let $P$ be the midpoint of $\overline{BD}$ . Given that $OP=11$ , the length of $AD$ can be written in the form $m\sqrt{n}$... | Let $D$ be the origin of the cartesian coordinate plane, $B$ lie on the positive $x$ -axis, and $A$ lie on the negative $y$ -axis. Then let the coordinates of $B = (2a,0), A = (0, -2b).$ Then the slope of $AB$ is $\frac{b}{a}.$ Since $AB \parallel CD$ the slope of $CD$ is the same. Note that as $\triangle DCB$ is isosc... | 194 |
3,879 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_12A_Problems/Problem_19 | 1 | How many solutions does the equation $\sin \left( \frac{\pi}2 \cos x\right)=\cos \left( \frac{\pi}2 \sin x\right)$ have in the closed interval $[0,\pi]$
$\textbf{(A) }0 \qquad \textbf{(B) }1 \qquad \textbf{(C) }2 \qquad \textbf{(D) }3\qquad \textbf{(E) }4$ | The ranges of $\frac{\pi}2 \sin x$ and $\frac{\pi}2 \cos x$ are both $\left[-\frac{\pi}2, \frac{\pi}2 \right],$ which is included in the range of $\arcsin,$ so we can use it with no issues. \begin{align*} \frac{\pi}2 \cos x &= \arcsin \left( \cos \left( \frac{\pi}2 \sin x\right)\right) \\ \frac{\pi}2 \cos x &= \arcsin ... | 2 |
3,880 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_12A_Problems/Problem_19 | 2 | How many solutions does the equation $\sin \left( \frac{\pi}2 \cos x\right)=\cos \left( \frac{\pi}2 \sin x\right)$ have in the closed interval $[0,\pi]$
$\textbf{(A) }0 \qquad \textbf{(B) }1 \qquad \textbf{(C) }2 \qquad \textbf{(D) }3\qquad \textbf{(E) }4$ | By the Cofunction Identity $\cos\theta=\sin\left(\frac{\pi}{2}-\theta\right),$ we rewrite the given equation: \[\sin \left(\frac{\pi}2 \cos x\right) = \sin \left(\frac{\pi}2 - \frac{\pi}2 \sin x\right).\] Recall that if $\sin\theta=\sin\phi,$ then $\theta=\phi+2n\pi$ or $\theta=\pi-\phi+2n\pi$ for some integer $n.$ The... | 2 |
3,881 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_12A_Problems/Problem_19 | 3 | How many solutions does the equation $\sin \left( \frac{\pi}2 \cos x\right)=\cos \left( \frac{\pi}2 \sin x\right)$ have in the closed interval $[0,\pi]$
$\textbf{(A) }0 \qquad \textbf{(B) }1 \qquad \textbf{(C) }2 \qquad \textbf{(D) }3\qquad \textbf{(E) }4$ | This problem is equivalent to counting the intersections of the graphs of $y=\sin\left(\frac{\pi}{2}\cos x\right)$ and $y=\cos\left(\frac{\pi}{2}\sin x\right)$ in the closed interval $[0,\pi].$ We construct a table of values, as shown below: \[\begin{array}{c|ccc} & & & \\ [-2ex] & \boldsymbol{x=0} & \boldsymbol{x=\fra... | 2 |
3,882 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_12A_Problems/Problem_21 | 1 | The five solutions to the equation \[(z-1)(z^2+2z+4)(z^2+4z+6)=0\] may be written in the form $x_k+y_ki$ for $1\le k\le 5,$ where $x_k$ and $y_k$ are real. Let $\mathcal E$ be the unique ellipse that passes through the points $(x_1,y_1),(x_2,y_2),(x_3,y_3),(x_4,y_4),$ and $(x_5,y_5)$ . The eccentricity of $\mathcal E$ ... | The solutions to this equation are $z = 1$ $z = -1 \pm i\sqrt 3$ , and $z = -2\pm i\sqrt 2$ . Consider the five points $(1,0)$ $\left(-1,\pm\sqrt 3\right)$ , and $\left(-2,\pm\sqrt 2\right)$ ; these are the five points which lie on $\mathcal E$ . Note that since these five points are symmetric about the $x$ -axis, so m... | 7 |
3,883 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_12A_Problems/Problem_21 | 2 | The five solutions to the equation \[(z-1)(z^2+2z+4)(z^2+4z+6)=0\] may be written in the form $x_k+y_ki$ for $1\le k\le 5,$ where $x_k$ and $y_k$ are real. Let $\mathcal E$ be the unique ellipse that passes through the points $(x_1,y_1),(x_2,y_2),(x_3,y_3),(x_4,y_4),$ and $(x_5,y_5)$ . The eccentricity of $\mathcal E$ ... | Completing the square in the original equation, we have \[(z-1)\left((z+1)^2+3\right)\left((z+2)^2+2\right)=0,\] from which $z=1,-1\pm\sqrt{3}i,-2\pm\sqrt{2}i.$
Now, we will find the equation of an ellipse $\mathcal E$ that passes through $(1,0),\left(-1,\pm\sqrt3\right),$ and $\left(-2,\pm\sqrt2\right)$ in the $xy$ -p... | 7 |
3,884 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_12A_Problems/Problem_21 | 3 | The five solutions to the equation \[(z-1)(z^2+2z+4)(z^2+4z+6)=0\] may be written in the form $x_k+y_ki$ for $1\le k\le 5,$ where $x_k$ and $y_k$ are real. Let $\mathcal E$ be the unique ellipse that passes through the points $(x_1,y_1),(x_2,y_2),(x_3,y_3),(x_4,y_4),$ and $(x_5,y_5)$ . The eccentricity of $\mathcal E$ ... | Starting from this system of equations from Solution 2: \begin{align*} \frac{(1-h)^2}{a^2}&=1, \\ \frac{(-1-h)^2}{a^2}+\frac{{\sqrt3}^2}{b^2}&=1, \\ \frac{(-2-h)^2}{a^2}+\frac{{\sqrt2}^2}{b^2}&=1. \end{align*} Let $A=a^{-2}$ and $B=b^{-2}$ . Therefore, the system can be rewritten as: \begin{align*} (h^2-2h+1)A&=1, &(1)... | 7 |
3,885 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_12A_Problems/Problem_21 | 4 | The five solutions to the equation \[(z-1)(z^2+2z+4)(z^2+4z+6)=0\] may be written in the form $x_k+y_ki$ for $1\le k\le 5,$ where $x_k$ and $y_k$ are real. Let $\mathcal E$ be the unique ellipse that passes through the points $(x_1,y_1),(x_2,y_2),(x_3,y_3),(x_4,y_4),$ and $(x_5,y_5)$ . The eccentricity of $\mathcal E$ ... | The five roots are $1,-1+i\sqrt{3},-1-i\sqrt{3},-2+i\sqrt{2},-2-i\sqrt{2}.$
So, we express this conic in the form $ax^2+by^2+cx+z=0.$ Note that this conic cannot have the $ky$ term since the roots are symmetric about the $x$ -axis.
Now we have equations \begin{align*} a+c+z&=0, \\ a+3b-c+z&=0, \\ 4a+2b-2c+z&=0, \end{al... | 7 |
3,886 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_12A_Problems/Problem_21 | 5 | The five solutions to the equation \[(z-1)(z^2+2z+4)(z^2+4z+6)=0\] may be written in the form $x_k+y_ki$ for $1\le k\le 5,$ where $x_k$ and $y_k$ are real. Let $\mathcal E$ be the unique ellipse that passes through the points $(x_1,y_1),(x_2,y_2),(x_3,y_3),(x_4,y_4),$ and $(x_5,y_5)$ . The eccentricity of $\mathcal E$ ... | After calculating the $5$ points that lie on $\mathcal E$ , we try to find a transformation that sends $\mathcal E$ to the unit circle. Scaling about $(1, 0)$ works, since $(1, 0)$ is already on the unit circle and such a transformation will preserve the ellipse's symmetry about the $x$ -axis. If $2a$ and $2b$ are the ... | 7 |
3,887 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_12A_Problems/Problem_24 | 1 | Semicircle $\Gamma$ has diameter $\overline{AB}$ of length $14$ . Circle $\Omega$ lies tangent to $\overline{AB}$ at a point $P$ and intersects $\Gamma$ at points $Q$ and $R$ . If $QR=3\sqrt3$ and $\angle QPR=60^\circ$ , then the area of $\triangle PQR$ equals $\tfrac{a\sqrt{b}}{c}$ , where $a$ and $c$ are relatively p... | Let $O=\Gamma$ be the center of the semicircle and $X=\Omega$ be the center of the circle.
Applying the Extended Law of Sines to $\triangle PQR,$ we find the radius of $\odot X:$ \[XP=\frac{QR}{2\cdot\sin \angle QPR}=\frac{3\sqrt3}{2\cdot\frac{\sqrt3}{2}}=3.\] Alternatively, by the Inscribed Angle Theorem, $\triangle Q... | 122 |
3,888 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_12A_Problems/Problem_24 | 2 | Semicircle $\Gamma$ has diameter $\overline{AB}$ of length $14$ . Circle $\Omega$ lies tangent to $\overline{AB}$ at a point $P$ and intersects $\Gamma$ at points $Q$ and $R$ . If $QR=3\sqrt3$ and $\angle QPR=60^\circ$ , then the area of $\triangle PQR$ equals $\tfrac{a\sqrt{b}}{c}$ , where $a$ and $c$ are relatively p... | [asy] size(150); draw(circle((7,0),7)); pair A = (0, 0); pair B = (14, 0); draw(A--B); draw(circle((11,3),3)); label("$C$", (7, 0), S); label("$O$", (11, 3), E); label("$P$", (11, 0), S); pair C = (7, 0); pair O = (11, 3); pair P = (11, 0); pair Q = intersectionpoints(circle(C, 7), circle(O, 3))[1]; pair R = intersecti... | 122 |
3,889 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_12A_Problems/Problem_24 | 3 | Semicircle $\Gamma$ has diameter $\overline{AB}$ of length $14$ . Circle $\Omega$ lies tangent to $\overline{AB}$ at a point $P$ and intersects $\Gamma$ at points $Q$ and $R$ . If $QR=3\sqrt3$ and $\angle QPR=60^\circ$ , then the area of $\triangle PQR$ equals $\tfrac{a\sqrt{b}}{c}$ , where $a$ and $c$ are relatively p... | [asy] size(300); pair C = (7, 0); draw(arc(C, 7, 0, 180)); pair A = (0, 0), B = (14, 0); draw(A--B); draw(circle((11,3),3)); label("$A$", A, SSE); label("$B$", B, SSW); label("$C$", (A+B)/2, S); label("$O$", (11, 3), E); label("$P$", (11, 0), S); pair O = (11, 3), P = (11, 0), Q = intersectionpoints(circle(C, 7), circ... | 122 |
3,890 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_12A_Problems/Problem_24 | 4 | Semicircle $\Gamma$ has diameter $\overline{AB}$ of length $14$ . Circle $\Omega$ lies tangent to $\overline{AB}$ at a point $P$ and intersects $\Gamma$ at points $Q$ and $R$ . If $QR=3\sqrt3$ and $\angle QPR=60^\circ$ , then the area of $\triangle PQR$ equals $\tfrac{a\sqrt{b}}{c}$ , where $a$ and $c$ are relatively p... | Let $O_{1}$ be the center of $\odot\Gamma, O_2$ be the center of $\odot\Omega,$ and $M$ be the midpoint of $\overline{QR}.$ We have $O_{1}M=\sqrt{7^2-\left(\frac{3\sqrt3}{2}\right)^2}=\frac{13}{2}$ and by Extended Law of Sines, the radius of $\odot\Omega$ is $\frac{3\sqrt3}{2\sin 60^\circ}=3$ so $O_{2}M=3\cos 60^\circ=... | 122 |
3,891 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_12A_Problems/Problem_24 | 5 | Semicircle $\Gamma$ has diameter $\overline{AB}$ of length $14$ . Circle $\Omega$ lies tangent to $\overline{AB}$ at a point $P$ and intersects $\Gamma$ at points $Q$ and $R$ . If $QR=3\sqrt3$ and $\angle QPR=60^\circ$ , then the area of $\triangle PQR$ equals $\tfrac{a\sqrt{b}}{c}$ , where $a$ and $c$ are relatively p... | By the Law of Sine in $\triangle PQR$ and its circumcircle $\odot \Omega$ $2r_{\Omega} = \frac{QR}{ \sin 60^{\circ} } = \frac{ 3\sqrt{3} }{ \frac{ \sqrt{3} }{2} } = 6$ $r_{\Omega} = 3$
\[\Gamma \Omega = \sqrt{r_{\Gamma}^2 - \left( \frac{ PQ }{2}\right)^2} - \sqrt{r_{\Omega} - \left( \frac{PQ}{2}\right)^2} = \sqrt{7^2 -... | 122 |
3,892 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_12A_Problems/Problem_24 | 6 | Semicircle $\Gamma$ has diameter $\overline{AB}$ of length $14$ . Circle $\Omega$ lies tangent to $\overline{AB}$ at a point $P$ and intersects $\Gamma$ at points $Q$ and $R$ . If $QR=3\sqrt3$ and $\angle QPR=60^\circ$ , then the area of $\triangle PQR$ equals $\tfrac{a\sqrt{b}}{c}$ , where $a$ and $c$ are relatively p... | Following Solution 4, We have $O_{1}$ (0,0) , $O_{2}$ (4,3).
We can write the equation of the two circles as: \[\odot\Gamma : x^{2}+y^{2}=7^{2}...(1)\] \[\odot\Omega : (x-4)^{2}+(y-3)^{2}=3^{2}...(2)\] By substituting (1) into (2), we get \[8x+6y-65=0...(3)\] Notice (3) is the relationship between $x$ value and $y$... | 122 |
3,893 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_12A_Problems/Problem_25 | 1 | Let $d(n)$ denote the number of positive integers that divide $n$ , including $1$ and $n$ . For example, $d(1)=1,d(2)=2,$ and $d(12)=6$ . (This function is known as the divisor function.) Let \[f(n)=\frac{d(n)}{\sqrt [3]n}.\] There is a unique positive integer $N$ such that $f(N)>f(n)$ for all positive integers $n\ne N... | We consider the prime factorization of $n:$ \[n=\prod_{i=1}^{k}p_i^{e_i}.\] By the Multiplication Principle, we have \[d(n)=\prod_{i=1}^{k}(e_i+1).\] Now, we rewrite $f(n)$ as \[f(n)=\frac{d(n)}{\sqrt [3]n}=\frac{\prod_{i=1}^{k}(e_i+1)}{\prod_{i=1}^{k}p_i^{e_i/3}}=\prod_{i=1}^{k}\frac{e_i+1}{p_i^{{e_i}/3}}.\] As $f(n)>... | 9 |
3,894 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_12A_Problems/Problem_25 | 2 | Let $d(n)$ denote the number of positive integers that divide $n$ , including $1$ and $n$ . For example, $d(1)=1,d(2)=2,$ and $d(12)=6$ . (This function is known as the divisor function.) Let \[f(n)=\frac{d(n)}{\sqrt [3]n}.\] There is a unique positive integer $N$ such that $f(N)>f(n)$ for all positive integers $n\ne N... | The question statement asks for the value of $N$ that maximizes $f(N)$ . Let $N$ start out at $1$ ; we will find what factors to multiply $N$ by, in order for $N$ to maximize the function.
First, we will find what power of $2$ to multiply $N$ by. If we multiply $N$ by $2^{a}$ , the numerator of $f$ $d(N)$ , will multip... | 9 |
3,895 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_12A_Problems/Problem_25 | 3 | Let $d(n)$ denote the number of positive integers that divide $n$ , including $1$ and $n$ . For example, $d(1)=1,d(2)=2,$ and $d(12)=6$ . (This function is known as the divisor function.) Let \[f(n)=\frac{d(n)}{\sqrt [3]n}.\] There is a unique positive integer $N$ such that $f(N)>f(n)$ for all positive integers $n\ne N... | Using the answer choices to our advantage, we can show that $N$ must be divisible by 9 without explicitly computing $N$ , by exploiting the following fact:
Claim : If $n$ is not divisible by 3, then $f(9n) > f(3n) > f(n)$
Proof : Since $d(\cdot)$ is a multiplicative function , we have $d(3n) = d(3)d(n) = 2d(n)$ and $d(... | 9 |
3,896 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_12A_Problems/Problem_25 | 4 | Let $d(n)$ denote the number of positive integers that divide $n$ , including $1$ and $n$ . For example, $d(1)=1,d(2)=2,$ and $d(12)=6$ . (This function is known as the divisor function.) Let \[f(n)=\frac{d(n)}{\sqrt [3]n}.\] There is a unique positive integer $N$ such that $f(N)>f(n)$ for all positive integers $n\ne N... | The problem mentions the sum of digits - recall that if a number is divisible by 9, then so is the sum of its digits. Guess that the answer must therefore be $\boxed{9}$ | 9 |
3,897 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_12B_Problems/Problem_1 | 1 | How many integer values of $x$ satisfy $|x|<3\pi$
$\textbf{(A)} ~9 \qquad\textbf{(B)} ~10 \qquad\textbf{(C)} ~18 \qquad\textbf{(D)} ~19 \qquad\textbf{(E)} ~20$ | Since $3\pi\approx9.42$ , we multiply $9$ by $2$ for the integers from $1$ to $9$ and the integers from $-1$ to $-9$ and add $1$ to account for $0$ to get $\boxed{19}$ | 19 |
3,898 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_12B_Problems/Problem_1 | 2 | How many integer values of $x$ satisfy $|x|<3\pi$
$\textbf{(A)} ~9 \qquad\textbf{(B)} ~10 \qquad\textbf{(C)} ~18 \qquad\textbf{(D)} ~19 \qquad\textbf{(E)} ~20$ | $|x|<3\pi$ $\iff$ $-3\pi<x<3\pi$ . Since $\pi$ is approximately $3.14$ $3\pi$ is approximately $9.42$ . We are trying to solve for $-9.42<x<9.42$ , where $x\in\mathbb{Z}$ . Hence, $-9.42<x<9.42$ $\implies$ $-9\leq x\leq9$ , for $x\in\mathbb{Z}$ . The number of integer values of $x$ is $9-(-9)+1=19$ . Therefore, the ans... | 19 |
3,899 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_12B_Problems/Problem_1 | 3 | How many integer values of $x$ satisfy $|x|<3\pi$
$\textbf{(A)} ~9 \qquad\textbf{(B)} ~10 \qquad\textbf{(C)} ~18 \qquad\textbf{(D)} ~19 \qquad\textbf{(E)} ~20$ | $3\pi \approx 9.4.$ There are two cases here.
When $x>0, |x|>0,$ and $x = |x|.$ So then $x<9.4$
When $x<0, |x|>0,$ and $x = -|x|.$ So then $-x<9.4$ . Dividing by $-1$ and flipping the sign, we get $x>-9.4.$
From case 1 and 2, we know that $-9.4 < x < 9.4$ . Since $x$ is an integer, we must have $x$ between $-9$ and $9$... | 19 |
3,900 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_12B_Problems/Problem_1 | 4 | How many integer values of $x$ satisfy $|x|<3\pi$
$\textbf{(A)} ~9 \qquad\textbf{(B)} ~10 \qquad\textbf{(C)} ~18 \qquad\textbf{(D)} ~19 \qquad\textbf{(E)} ~20$ | Looking at the problem, we see that instead of directly saying $x$ , we see that it is $|x|.$ That means all the possible values of $x$ in this case are positive and negative. Rounding $\pi$ to $3$ we get $3(3)=9.$ There are $9$ positive solutions and $9$ negative solutions: $9+9=18.$ But what about zero? Even though z... | 19 |
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