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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("${"+string(i)+"}$", (i,0), 2*S); if (i<6) { draw((0,i)--(6,i), gray+dashed); label("${"+string(i)+"}$", (0,i), 2*W); } } label("$0$", O, 2*SW); draw(O--X+(0.35,0), black+1.5, EndArrow(10)); draw(O--Y+(0,0.35), black+1.5, EndArrow(10)); draw((1,0)--(3,5)--(5,0)--(3,2)--(1,0), black+1.5); [/asy]
$\textbf{(A)}\: 4\qquad\textbf{(B)} \: 6\qquad\textbf{(C)} \: 8\qquad\textbf{(D)} \: 10\qquad\textbf{(E)} \: 12$
|
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("${"+string(i)+"}$", (i,0), 2*S); if (i<6) { draw((0,i)--(6,i), gray+dashed); label("${"+string(i)+"}$", (0,i), 2*W); } } label("$0$", O, 2*SW); draw(O--X+(0.35,0), black+1.5, EndArrow(10)); draw(O--Y+(0,0.35), black+1.5, EndArrow(10)); draw((1,0)--(3,5)--(5,0)--(3,2)--(1,0), black+1.5); [/asy]
$\textbf{(A)}\: 4\qquad\textbf{(B)} \: 6\qquad\textbf{(C)} \: 8\qquad\textbf{(D)} \: 10\qquad\textbf{(E)} \: 12$
|
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("${"+string(i)+"}$", (i,0), 2*S); if (i<6) { draw((0,i)--(6,i), gray+dashed); label("${"+string(i)+"}$", (0,i), 2*W); } } label("$0$", O, 2*SW); draw(O--X+(0.35,0), black+1.5, EndArrow(10)); draw(O--Y+(0,0.35), black+1.5, EndArrow(10)); draw((1,0)--(3,5)--(5,0)--(3,2)--(1,0), black+1.5); [/asy]
$\textbf{(A)}\: 4\qquad\textbf{(B)} \: 6\qquad\textbf{(C)} \: 8\qquad\textbf{(D)} \: 10\qquad\textbf{(E)} \: 12$
|
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 of $N?$
$\textbf{(A)}\: 10\qquad\textbf{(B)} \: 30\qquad\textbf{(C)} \: 60\qquad\textbf{(D)} \: 100\qquad\textbf{(E)} \: 120$
|
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 $10\cdot6=\boxed{60}.$
| 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 of $N?$
$\textbf{(A)}\: 10\qquad\textbf{(B)} \: 30\qquad\textbf{(C)} \: 60\qquad\textbf{(D)} \: 100\qquad\textbf{(E)} \: 120$
|
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?
$\textbf{(A)}\ 9 \qquad\textbf{(B)}\ 10 \qquad\textbf{(C)}\ 11 \qquad\textbf{(D)}\ 12 \qquad\textbf{(E)}\ 13$
|
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 $\frac{15}{2},\frac{15}{6},\frac{15}{10},$ which are $7\frac12,2\frac12,1\frac12,$ respectively. We get $15,10,9,5,4,3$ from this group of numbers.
The quarters are $\frac{15}{4},\frac{15}{12},$ which are $3\frac34,1\frac14,$ respectively. We get $5$ from this group of numbers.
Note that $10$ and $5$ each appear twice. Therefore, the answer is $\boxed{11}.$
| 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?
$\textbf{(A)}\ 9 \qquad\textbf{(B)}\ 10 \qquad\textbf{(C)}\ 11 \qquad\textbf{(D)}\ 12 \qquad\textbf{(E)}\ 13$
|
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 + fraction
The fractions we have are $\frac{5}{10}$ (or $\frac{1}{2}$ ), $\frac{9}{6}$ (or $\frac{3}{2}$ ), and $\frac{13}{2}$ . Similarly, there are $\dbinom{3}{2}=3$ ways to choose different-numbered pairs and $3$ ways to choose the same-numbered pairs. So, $3+3=6$
Thus, $6+6=12$
So now you would just go ahead and innocently choose $\textbf{(D) }12$ , right? No! We overcounted $8$ , as $\frac{13}{2}+\frac96=\frac{12}{3}+\frac{12}{3}=8$ . Therefore, the correct answer is actually $12-1=\boxed{11}$
| 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?
$\textbf{(A)} \: 105 \qquad\textbf{(B)} \: 120 \qquad\textbf{(C)} \: 135 \qquad\textbf{(D)} \: 150 \qquad\textbf{(E)} \: 165$
|
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:
$x^2=4\left(\frac{1}{2}x^2\sin(a)\right)$
$\sin(a)=\frac{1}{2}$
As the triangle is obtuse, $a=150^\circ$ only. We get $\boxed{150}.$
| 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?
$\textbf{(A)} \: 105 \qquad\textbf{(B)} \: 120 \qquad\textbf{(C)} \: 135 \qquad\textbf{(D)} \: 150 \qquad\textbf{(E)} \: 165$
|
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 \theta \cos \theta \\ & = 2 \sin 2 \theta \\ & = 2 \sin \left( 180^\circ - 2 \theta \right) \\ & = 2 \sin \phi . \end{align*}
Hence, $\phi = 150^\circ$
Therefore, the answer is $\boxed{150}$
| 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 $C$
$\textbf{(A)} \: 9\pi \qquad\textbf{(B)} \: 12\pi \qquad\textbf{(C)} \: 18\pi \qquad\textbf{(D)} \: 24\pi \qquad\textbf{(E)} \: 27\pi$
|
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 an inscribed quadrilateral of Circle $X$ $\angle AMC=180^\circ-120^\circ=60^\circ$
Afterward, deduce that $\angle AXC=2·\angle AMC=120^\circ$
By the Cosine Rule, we have the equation: (where $r$ is the radius of circle $X$
$2r^2(1-\cos(120^\circ))=6^2$
$r^2=12$
The area is therefore $\pi r^2 = \boxed{12}$
| 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 $C$
$\textbf{(A)} \: 9\pi \qquad\textbf{(B)} \: 12\pi \qquad\textbf{(C)} \: 18\pi \qquad\textbf{(D)} \: 24\pi \qquad\textbf{(E)} \: 27\pi$
|
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 $C$
$\textbf{(A)} \: 9\pi \qquad\textbf{(B)} \: 12\pi \qquad\textbf{(C)} \: 18\pi \qquad\textbf{(D)} \: 24\pi \qquad\textbf{(E)} \: 27\pi$
|
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, $\overline{OX}$ $\perp$ $\overline{AB}$ and $AY = 3$ . Thus $\tan(30^\circ) = \frac{OY}{3}$ so $OY = \sqrt{3}$
Let $r$ be the radius of circle $X$ , and note that $AX = OX = r$ . So $\triangle AYX$ is a right triangle with legs of length $3$ and $r - \sqrt{3}$ and hypotenuse $r$ . By Pythagoras, $3^2 + (r - \sqrt{3})^2 = r^2$ . So $r = 2\sqrt{3}$
Thus the area is $\pi r^2 = \boxed{12}$
| 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 $C$
$\textbf{(A)} \: 9\pi \qquad\textbf{(B)} \: 12\pi \qquad\textbf{(C)} \: 18\pi \qquad\textbf{(D)} \: 24\pi \qquad\textbf{(E)} \: 27\pi$
|
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 $M$ , forming two 30-60-90 triangles. As $CM=MA=3$ , we can solve for $OC=2\sqrt{3}$ . Now, the area of the circle is just $\pi*(2*\sqrt{3})^2 = 12\pi$ . Select $\boxed{12}$
| 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 $C$
$\textbf{(A)} \: 9\pi \qquad\textbf{(B)} \: 12\pi \qquad\textbf{(C)} \: 18\pi \qquad\textbf{(D)} \: 24\pi \qquad\textbf{(E)} \: 27\pi$
|
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 $OC=OA=OB$ and $OC$ is the hypotenuse of a $30-60-90$ right triangle, with the longer leg being $\frac{6}{2}=3$ so $OC=2\sqrt{3}$ . Applying Ptolemy's Theorem on cyclic quadrilateral $OCEA$ , we get $2({\sqrt{d^2-12}})\cdot{2\sqrt{3}}=6d$ . Squaring and solving we get $d^2=48 \Longrightarrow (2r)^2=48$ so $r^2=12$ . Therefore, the area of the circle is $\boxed{12}$
| 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?
$\textbf{(A)} \: 100 \qquad\textbf{(B)} \: 150 \qquad\textbf{(C)} \: 330 \qquad\textbf{(D)} \: 360 \qquad\textbf{(E)} \: 380$
|
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 \[\frac{\sin 3c \cdot \sin 6c \cdot \sin 9c \cdot \sin 12c \cdot \sin 15c}{\sin c \cdot \sin 2c \cdot \sin 3c \cdot \sin 4c \cdot \sin 5c}?\]
$\textbf{(A)}\ {-}1 \qquad\textbf{(B)}\ {-}\frac{\sqrt{11}}{5} \qquad\textbf{(C)}\ \frac{\sqrt{11}}{5} \qquad\textbf{(D)}\ \frac{10}{11} \qquad\textbf{(E)}\ 1$
|
Plugging in $c$ , we get \[\frac{\sin 3c \cdot \sin 6c \cdot \sin 9c \cdot \sin 12c \cdot \sin 15c}{\sin c \cdot \sin 2c \cdot \sin 3c \cdot \sin 4c \cdot \sin 5c}=\frac{\sin \frac{6\pi}{11} \cdot \sin \frac{12\pi}{11} \cdot \sin \frac{18\pi}{11} \cdot \sin \frac{24\pi}{11} \cdot \sin \frac{30\pi}{11}}{\sin \frac{2\pi}{11} \cdot \sin \frac{4\pi}{11} \cdot \sin \frac{6\pi}{11} \cdot \sin \frac{8\pi}{11} \cdot \sin \frac{10\pi}{11}}.\] Since $\sin(x+2\pi)=\sin(x),$ $\sin(2\pi-x)=\sin(-x),$ and $\sin(-x)=-\sin(x),$ we get \[\frac{\sin \frac{6\pi}{11} \cdot \sin \frac{12\pi}{11} \cdot \sin \frac{18\pi}{11} \cdot \sin \frac{24\pi}{11} \cdot \sin \frac{30\pi}{11}}{\sin \frac{2\pi}{11} \cdot \sin \frac{4\pi}{11} \cdot \sin \frac{6\pi}{11} \cdot \sin \frac{8\pi}{11} \cdot \sin \frac{10\pi}{11}}=\frac{\sin \frac{6\pi}{11} \cdot \sin \frac{-10\pi}{11} \cdot \sin \frac{-4\pi}{11} \cdot \sin \frac{2\pi}{11} \cdot \sin \frac{8\pi}{11}}{\sin \frac{2\pi}{11} \cdot \sin \frac{4\pi}{11} \cdot \sin \frac{6\pi}{11} \cdot \sin \frac{8\pi}{11} \cdot \sin \frac{10\pi}{11}}=\boxed{1}.\]
| 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 \[\frac{\sin 3c \cdot \sin 6c \cdot \sin 9c \cdot \sin 12c \cdot \sin 15c}{\sin c \cdot \sin 2c \cdot \sin 3c \cdot \sin 4c \cdot \sin 5c}?\]
$\textbf{(A)}\ {-}1 \qquad\textbf{(B)}\ {-}\frac{\sqrt{11}}{5} \qquad\textbf{(C)}\ \frac{\sqrt{11}}{5} \qquad\textbf{(D)}\ \frac{10}{11} \qquad\textbf{(E)}\ 1$
|
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 symbol is calculated using quadratic reciprocity which is $\left(\frac{p}{q}\right)\left(\frac{q}{p}\right)=(-1)^{\frac{p-1}{2}\frac{q-1}{2}}$ . The Legendre symbol $\left(\frac{3}{11}\right)=(-1)\left(\frac{11}{3}\right)=(-1)\left(\frac{-1}{3}\right)=(-1)(-1)=\boxed{1}$
| 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 of $N$
$\textbf{(A)}\: 0\qquad\textbf{(B)} \: 1\qquad\textbf{(C)} \: 2\qquad\textbf{(D)} \: 3\qquad\textbf{(E)} \: 5$
|
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 $N=\boxed{1}.$
| 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 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 $a+b+c$
$(\textbf{A})\: 75\qquad(\textbf{B}) \: 93\qquad(\textbf{C}) \: 96\qquad(\textbf{D}) \: 129\qquad(\textbf{E}) \: 147$
|
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, $AB = 2 AO \sin \frac{\angle AOB}{2} = 3 \left( \sqrt{3} - 1 \right)$
We notice that $\angle MAB = \angle MBA = 30^\circ$ .
Hence, $AM = \frac{AB}{2\cos \angle MAB} = 3 - \sqrt{3}$
Therefore, the area of the region that three squares cover is \begin{align*} & {\rm Area} \ ABCDEFGHIJKL - 12 {\rm Area} \ \triangle MAB \\ & = 12 {\rm Area} \ \triangle OAB - 12 {\rm Area} \ \triangle MAB \\ & = 12 \cdot \frac{1}{2} OA \cdot OB \sin \angle AOB - 12 \cdot \frac{1}{2} MA \cdot MB \sin \angle AMB \\ & = 6 OA^2 \sin \angle AOB - 6 MA^2 \sin \angle AMB \\ & = 108 - 36 \sqrt{3} . \end{align*}
Therefore, the answer is $\boxed{147}$
| 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 ${\rm gcd} \left( b , c \right)$ 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 $179 + 259 = \boxed{438}$
| 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 that $a$ $b$ , and $c$ have to all be divisible by $2$ . Therefore, $x$ cannot be equal to $1$
$x=1,y=3,z=5$ : Note that $c$ has to be both a multiple of $3$ and $5$ . Therefore, $c$ has to be a multiple of $15$ . The only solution for this is $a=5, b=3, c=15$
$x=1,y=4,z=4$ : No solutions. By $y$ and $z$ , we know that $a$ $b$ , and $c$ have to all be divisible by $4$ . Therefore, $x$ cannot be equal to $1$
$x=2,y=2,z=5$ : No solutions. By $x$ and $y$ , we know that $a$ $b$ , and $c$ have to all be divisible by $2$ . Therefore, $z$ cannot be equal to $5$
$x=2,y=3,z=4$ : No solutions. By $x$ and $z$ , we know that $a$ $b$ , and $c$ have to all be divisible by $2$ . Therefore, $y$ cannot be equal to $3$
$x=3,y=3,z=3$ : No solutions. As $a$ $b$ , and $c$ have to all be divisible by $3$ $a+b+c$ has to be divisible by $3$ . This contradicts the sum $a+b+c=23$
Putting these solutions together, we have $(7^2+9^2+7^2)+(5^2+3^2+15^2)=179+259=\boxed{438}$
| 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 generality, let $a\le b\le c$ . Since $a+b+c=23$ , then $a=7, b=7, c=9$ or $a=3, b=5, c=15$
$(7^2+7^2+9^2)+(3^2+5^2+15^2)=179+259=\boxed{438}$
| 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 \[|u_k-L| \le \frac{1}{2^{1000}}?\]
$\textbf{(A)}\: 10\qquad\textbf{(B)}\: 87\qquad\textbf{(C)}\: 123\qquad\textbf{(D)}\: 329\qquad\textbf{(E)}\: 401$
|
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 \[\frac12-\frac{1}{2^{1000}} \leq u_k \leq \frac12+\frac{1}{2^{1000}},\] and we only need to consider $\frac12-\frac{1}{2^{1000}} \leq u_k.$
We have \begin{alignat*}{8} u_0 &= \phantom{1}\frac14 &&= \frac{2^1-1}{2^2}, \\ u_1 &= \phantom{1}\frac38 &&= \frac{2^2-1}{2^3}, \\ u_2 &= \ \frac{15}{32} &&= \frac{2^4-1}{2^5}, \\ u_3 &= \frac{255}{512} &&= \frac{2^8-1}{2^9}, \\ & \phantom{1111} \vdots \end{alignat*} By induction, it can be proven that \[u_k=\frac{2^{2^k}-1}{2^{2^k+1}}=\frac12-\frac{1}{2^{2^k+1}}.\] We substitute this into the inequality, then solve for $k:$ \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 such value of $k$ is $\boxed{10}.$
| 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 \[|u_k-L| \le \frac{1}{2^{1000}}?\]
$\textbf{(A)}\: 10\qquad\textbf{(B)}\: 87\qquad\textbf{(C)}\: 123\qquad\textbf{(D)}\: 329\qquad\textbf{(E)}\: 401$
|
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} &= 2\cdot\left(\frac{1}{2}-\frac{1}{2^{n_{k}}}\right)-2\cdot\left(\frac{1}{2}-\frac{1}{2^{n_{k}}}\right)^2 \\ &= 1-\frac{1}{2^{n_k - 1}}-2\cdot\left(\frac{1}{4}-\frac{1}{2^{n_k}}+\frac{1}{2^{2 \cdot n_k}}\right)\\ &= 1-\frac{1}{2^{n_k - 1}}-\frac{1}{2}+\frac{1}{2^{n_k-1}}-\frac{1}{2^{2 \cdot n_k-1}} \\ &= \frac{1}{2}-\frac{1}{2^{2 \cdot n_k-1}}. \end{align*} This means that this series approaches $\frac{1}{2}$ , as the second term is decreasing. In addition, we find that $n_{k+1}=2 \cdot n_k-1$
We claim that $n_k = 2^k+1$ , which can be proven by induction:
Base Case
We have $n_0=2=2^0+1$
Induction Step
Assuming that the claim is true, we have $n_{k+1}=2 \cdot (2^k+1)-1=2^{k+1}+1$
It follows that $n_{10}=2^{10}+1>1000$ and $n_9=2^9+1<1000$ . Therefore, the least value of $k$ would be $\boxed{10}$
| 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 \[|u_k-L| \le \frac{1}{2^{1000}}?\]
$\textbf{(A)}\: 10\qquad\textbf{(B)}\: 87\qquad\textbf{(C)}\: 123\qquad\textbf{(D)}\: 329\qquad\textbf{(E)}\: 401$
|
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 $1$ and because our recursive relation for $v_k$ squares the previous term (and negates it), we see that as $k \rightarrow \infty, 2u_k - 1 = v_k \rightarrow 0$ . This means $u_k \rightarrow \frac{1}{2}$ , so $L = \frac{1}{2}$
Isolating $u_k$ in our relation $2u_k - 1 = v_k$ gives us $u_k = \frac{v_k + 1}{2}$ .
Substituting into the inequality, we have $\left|\frac{v_k + 1}{2}-\frac{1}{2}\right| \le \frac{1}{2^{1000}}$ . Rewriting this, we get $|v_k| \le \frac{1}{2^{999}}$
The sequence $\{v_k\}$ is much easier to handle because of its simple recursive relation. Writing out a few terms shows that $|v_k| = \frac{1}{2^{2^k}}$ . Now it just comes down to having $2^k > 999$ , so $k = \boxed{10}$
| 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?
$(\textbf{A})\: 52\qquad(\textbf{B}) \: 56\qquad(\textbf{C}) \: 60\qquad(\textbf{D}) \: 64\qquad(\textbf{E}) \: 68$
|
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. ~KingRavi)
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 $2\cdot(3\cdot5+2\cdot6+1\cdot7)=\boxed{68}$
| 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.)
$\textbf{(A)}\ 7 \qquad\textbf{(B)}\ 8 \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ 10 \qquad\textbf{(E)}\ 11$
|
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.
Topology.jpg
Here is how the $4$ blue unit cubes are arranged:
In Figure $(1)$ $4$ blue unit cubes are on the same layer (horizontal or vertical).
In Figure $(2)$ $4$ blue unit cubes are in $T$ shape.
In Figure $(3)$ and $(4)$ $4$ blue unit cubes are in $S$ shape.
In Figure $(5)$ $3$ blue unit cubes are in $L$ shape, and the other is isolated without a shared face.
In Figure $(6)$ $2$ pairs of neighboring blue unit cubes are isolated from each other without a shared face.
In Figure $(7)$ $4$ blue unit cubes are isolated from each other without a shared face.
So the answer is $\boxed{7}$
| 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.)
$\textbf{(A)}\ 7 \qquad\textbf{(B)}\ 8 \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ 10 \qquad\textbf{(E)}\ 11$
|
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 one possibility for the top layer - all of the other color - $\binom{4}{4}$ . Therefore there is 1 construction from this case.
Case 2: 3, 1
In this case, the top layer has four possibilities, because there are four different ways to arrange it so that it also has a 3, 1 color distribution - $\binom{4}{3}$ . Therefore there are 4 constructions from this case.
Case 3: 2, 2
In this case, the top layer has six possibilities of arrangement - $\binom{4}{2}$ . However, having adjacent colors one way can be rotated to having adjacent colors any other way, so there is only one construction for the adjacent colors subcase and similarly, only one for the diagonal color subcase. Therefore the total number of constructions for this case is 2.
The total number of constructions for the cube is thus $1+4+2=7=\boxed{7}$
| 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.)
$\textbf{(A)}\ 7 \qquad\textbf{(B)}\ 8 \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ 10 \qquad\textbf{(E)}\ 11$
|
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.)
$\textbf{(A)}\ 7 \qquad\textbf{(B)}\ 8 \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ 10 \qquad\textbf{(E)}\ 11$
|
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 $fix(g)$
3. the sum of the $fix(g)/|G|$ equals the # of orbit.
Let's start with defining the operator for a cube,
1. $\textbf{e (identity)}$
For identity, there are $\frac{8!}{4!4!} = 70$
2. ${\bf r^{1}, r^{2}, r^{3}}$ to be the rotation axis along three pair of opposite face,
each contains $r^{i}_{90}, r^{i}_{180}, r^{i}_{270}$ where $i= 1, 2, 3$
$fix(r^{i}_{90}) = fix(r^{i}_{270}) = 2\cdot1 = 2$
$fix(r^{i}_{180}) = \frac{4!}{2!\cdot2!} = 6$
therefore $fix(\bf r^{i}) = 2+2+6 = 10$ , and $fix(\bf r^{1})+fix(\bf r^{2})+fix(\bf r^{3}) = 30$
3. ${\bf r^{4}, r^{5}, r^{6}, r^{7}}$ to the rotation axis along four cube diagnals.
each contains $r^{i}_{120}, r^{i}_{240}$ where $i= 4, 5, 6, 7$
$fix(r^{i}_{120}) = fix(r^{i}_{240}) = 2\cdot1\cdot2\cdot1 = 4$
therefore $fix(\bf r^{i}) = 4+4 = 8$ , and $fix(\bf r^{4})+fix(\bf r^{5})+fix(\bf r^{6})+fix(\bf r^{7}) = 32$
4. ${\bf r^{8}, r^{9}, r^{10}, r^{11}, r^{12}, r^{13}}$ to be the rotation axis along 6 pairs of diagnally opposite sides
each contains $r^{i}_{180}$ where $i= 8, 9, 10, 11, 12, 13$
$fix(r^{i}_{180}) = \frac{4!}{2!\cdot2!} = 6$
therefore $fix(\bf r^{8})+fix(\bf r^{9})+fix(\bf r^{10})+fix(\bf r^{11})+fix(\bf r^{12})+fix(\bf r^{13}) = 36$
5. The total number of operators are
$|G| = 1 + 3\cdot3 + 4\cdot2 + 6\cdot1 = 24$
Based on 1, 2, 3, 4 the total number of stablizer is $70 + 30 + 32 + 36 = 168$
therefore the number of orbit $= \frac{168}{G=24} = \boxed{7}$
| 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.)
$\textbf{(A)}\ 7 \qquad\textbf{(B)}\ 8 \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ 10 \qquad\textbf{(E)}\ 11$
|
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 side with it, 3 1-by-1-by-1 cubes share a corner with it, and 1 1-by-1-by-1 cube does not touch the assigned cube at all, from the perspective of someone who can only see the cube's faces. We'll call the first 3 "adjacent", the second 3 "cornering", and the last one "opposite."
We can use a little bit of intuition to confirm that due to the rotation condition, we should treat all adjacents as indistinguishable, all cornerings as indistinguishable, and of course the opposite one is unique from all the others. Thus, we can list out like so (keeping in mind that there are 3 adjacents, 3 cornerings, and 1 opposite, and that we're choosing the positions of the remaining 3 blue cubes):
OAA, OAC, OCC, CCC, CAA, CCA, AAA.
This gives the answer to be $\boxed{7}$
| 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 $x$ is real number then $a$ must have magnitude of $1$ , but none of the solutions for $a$ have magnitude of $1$ , so the answer is $\boxed{0}$ ~lopkiloinm
| 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} \ \left( 1 + y - y^2 + y^3 \right) \\ & = \frac{1}{2i} \left( y - \bar y - y^2 + \bar y^2 + y^3 - \bar y^3 \right) \\ & = \frac{y - \bar y}{2i} \left( 1 - y - \bar y + y^2 + y \bar y + \bar y^2 \right) \\ & = {\rm Im} \ y \left( 1 - \left( y + \bar y \right) + \left( y + \bar y \right)^2 - y \bar y \right) \\ & = {\rm Im} \ y \left( 1 - 2 {\rm Re} \ y + 4 \left( {\rm Re} \ y \right)^2 - |y|^2 \right) \\ & = {\rm Im} \ y \left( 1 - 2 {\rm Re} \ y + 4 \left( {\rm Re} \ y \right)^2 - 1 \right) \\ & = 2 {\rm Im} \ y \cdot {\rm Re} \ y \left( 2 {\rm Re} \ y - 1 \right) \\ \end{align*} The sixth equality follows from the property that $|y| = 1$
Therefore, we have either ${\rm Re} \ y = 0$ or ${\rm Im} \ y = 0$ or $2 {\rm Re} \ y - 1 = 0$
Case 1: ${\rm Re} \ y = 0$
Because $|y| = 1$ $y = \pm i$
However, these solutions fail to satisfy Equation (1).
Therefore, there is no solution in this case.
Case 2: ${\rm Im} \ y = 0$
Because $|y| = 1$ $y = \pm 1$
However, these solutions fail to satisfy Equation (1).
Therefore, there is no solution in this case.
Case 3: $2 {\rm Re} \ y - 1 = 0$
Because $|y| = 1$ $y = \frac{1}{2} \pm \frac{\sqrt{3}}{2} i = e^{i \pm \frac{\pi}{3}}$
However, these solutions fail to satisfy Equation (1).
Therefore, there is no solution in this case.
All cases above imply that there is no solution in this problem.
Therefore, the answer is $\boxed{0}$
| 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 have $r^3-r^2+r=r^2(r+\tfrac{1}{r}-1)=a^2(2\mathrm{Re}(r)-1)$ , hence the argument of $r^3-r^2+r$ is either the argument of $a^2$ , or the argument of $-a^2$ . Since $r^3-r^2+r=-1$ is real, it follows that $a^2=\pm 1 \implies a \in \{1,i,-1,-i\}$ . Now, we can check all of these values and find that none of them work, yielding an answer of $\boxed{0}$
| 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 with $-cis(2x).$ Since $|-cis(2x)| = 1$ and we want the three vectors to sum to -1, we either have $-cis(2x) = 1$ and $cis(x) + cis(3x) = -2$ , or $-cis(2x) = -1$ and $cis(x) + cis(3x) = 0.$ If the first condition is true, $cis(x) = cis(3x)=-1.$ This will imply that $x= \pi.$ But then $-cis(2x) = -1$ , which violates the condition. Similarly, we can show that the second condition cannot be met either. Thus $P(x)$ does not have any solutions on the interval $[0, 2\pi].$ Therefore, the answer is $\boxed{0}$
| 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 $CF$
$\textbf{(A) } 28 \qquad \textbf{(B) } 20\sqrt{2} \qquad \textbf{(C) } 30 \qquad \textbf{(D) } 32 \qquad \textbf{(E) } 20\sqrt{3}$
|
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, $\angle CBA = \angle FEA$
$\because \quad \angle CEA = \angle FEA \quad \text{and} \quad \angle CAE = \angle FAE$
$\therefore \quad \triangle AFE \cong \triangle ACE$ by $ASA$
Hence, $AF = AC = 20$
By the Law of Cosine $CF = \sqrt{20^2 + 20^2 - 2 \cdot 20 \cdot 20 (-\frac18)} = \sqrt{900} = \boxed{30}$
| 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 $CF$
$\textbf{(A) } 28 \qquad \textbf{(B) } 20\sqrt{2} \qquad \textbf{(C) } 30 \qquad \textbf{(D) } 32 \qquad \textbf{(E) } 20\sqrt{3}$
|
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 $\cos\angle BE^{\prime}C=\cos\angle A=\tfrac{11^{2}+20^{2}-24^{2}}{2\cdot 11\cdot 20}=-\tfrac{1}{8}$ from Law of Cosines in $\triangle ABC$ , so by Law of Cosines in $\triangle BE^{\prime}C$ we have \[x^{2}+x^{2}+\left(2x^{2}\right)\left(-\dfrac{1}{8}\right)=24^{2}\] from which $x=16$ . Now, $CF=\dfrac{480}{16}=\boxed{30}$
| 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 $CF$
$\textbf{(A) } 28 \qquad \textbf{(B) } 20\sqrt{2} \qquad \textbf{(C) } 30 \qquad \textbf{(D) } 32 \qquad \textbf{(E) } 20\sqrt{3}$
|
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 $CF=\sqrt{\frac{11520 - 3600 + 1980}{11}}=\sqrt{\frac{9900}{11}}=\sqrt{900}=\boxed{30}$
| 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 $CF$
$\textbf{(A) } 28 \qquad \textbf{(B) } 20\sqrt{2} \qquad \textbf{(C) } 30 \qquad \textbf{(D) } 32 \qquad \textbf{(E) } 20\sqrt{3}$
|
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 $CF$ \[CF^2 = AF^2+AC^2-2 \cdot AF \cdot AC \cdot \cos(\angle CAF) = 900 \to CF=\boxed{30}\] both ways.
| 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 $CF$
$\textbf{(A) } 28 \qquad \textbf{(B) } 20\sqrt{2} \qquad \textbf{(C) } 30 \qquad \textbf{(D) } 32 \qquad \textbf{(E) } 20\sqrt{3}$
|
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 \\ & = BC^2 + BF^2 + 2 BC \cdot BF \cos \angle ABC . \hspace{1cm} (1) \end{align*} For $BF$ , we have \begin{align*} BF & = 2 R \cos \angle FBO \\ & = 2 R \cos \left( 180^\circ - \angle ABC - \angle CBO \right) \\ & = 2 R \cos \left( 180^\circ - \angle ABC - \frac{180^\circ - \angle BOD}{2} \right) \\ & = 2 R \cos \left( 180^\circ - \angle ABC - \frac{180^\circ - 2 \angle BED}{2} \right) \\ & = 2 R \cos \left( 180^\circ - \angle ABC - \frac{180^\circ - 2 \angle BCA}{2} \right) \\ & = 2 R \cos \left( 90^\circ - \angle ABC + \angle BCA \right) \\ & = 2 R \sin \left( \angle ABC - \angle BCA \right) \\ & = \frac{BD}{\sin \angle BED} \sin \left( \angle ABC - \angle BCA \right) \\ & = \frac{BD}{\sin \angle BCA} \sin \left( \angle ABC - \angle BCA \right) \\ & = BD \left( \sin \angle ABC \cot \angle BCA - \cos \angle ABC \right) . \hspace{1cm} (2) \end{align*} The fourth equality follows from the property that $B$ $D$ $E$ are concyclic.
The fifth and the ninth equalities follow from the property that $A$ $B$ $C$ $E$ are concyclic.
Because $AD$ bisects $\angle BAC$ , following from the angle bisector theorem, we have \[ \frac{BD}{CD} = \frac{AB}{AC} . \] Hence, $BD = \frac{24 \cdot 11}{31}$
In $\triangle ABC$ , following from the law of cosines, we have \begin{align*} \cos \angle ABC & = \frac{AB^2 + BC^2 - AC^2}{2 AB \cdot BC} \\ & = \frac{9}{16} \end{align*} and \begin{align*} \cos \angle BCA & = \frac{AC^2 + BC^2 - AB^2}{2 AC \cdot BC} \\ & = \frac{57}{64} . \end{align*} Hence, $\sin \angle ABC = \frac{5 \sqrt{7}}{16}$ and $\sin \angle BCA = \frac{11 \sqrt{7}}{64}$ .
Hence, $\cot \angle BCA = \frac{57}{11 \sqrt{7}}$
Now, we are ready to compute $BF$ whose expression is given in Equation (2).
We get $BF = 9$
Now, we can compute $CF$ whose expression is given in Equation (1).
We have $CF = 30$
Therefore, the answer is $\boxed{30}$
| 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 $R(n) = R(n+1)\,?$
$\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }3\qquad\textbf{(E) }4$
|
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 $2$ $6+3$ indicates $n+1$ is 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 $\boxed{2}$
| 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 $R(n) = R(n+1)\,?$
$\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }3\qquad\textbf{(E) }4$
|
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, \[ \Delta \left( n, k \right) = \left\{ \begin{array}{ll} 1 & \mbox{ if } n \not\equiv -1 \pmod{k} \\ - \left( k -1 \right) & \mbox{ if } n \equiv -1 \pmod{k} \end{array} \right.. \]
Hence, this problem asks us to find all $n \in \left\{ 10 , 11, \cdots , 99 \right\}$ , such that $\sum_{k = 2}^{10} \Delta \left( n, k \right) = 0$
$\textbf{Case 1}$ $\Delta \left( n, 10 \right) = - 9$
We have $\sum_{k = 2}^9 \Delta \left( n, k \right) \leq \sum_{k = 2}^9 1 = 8$
Therefore, there is no $n$ in this case.
$\textbf{Case 2}$ $\Delta \left( n, 10 \right) = 1$ and $\Delta \left( n, 9 \right) = -8$
The condition $\Delta \left( n, 9 \right) = -8$ implies $n \equiv - 1 \pmod{9}$ .
This further implies $n \equiv - 1 \pmod{3}$ .
Hence, $\Delta \left( n, 3 \right) = -2$
To get $\sum_{k = 2}^{10} \Delta \left( n, k \right) = 0$ , we have $\sum_{k \in \left\{ 2 , 4 , 5 , 6, 7, 8\right\}} \Delta \left( n, k \right) = 9$
However, we have $\sum_{k \in \left\{ 2 , 4 , 5 , 6, 7, 8\right\}} \Delta \left( n, k \right) \leq \sum_{k \in \left\{ 2 , 4 , 5 , 6, 7, 8\right\}} 1 = 6$
Therefore, there is no $n$ in this case.
$\textbf{Case 3}$ $\Delta \left( n, k \right) = 1$ for $k \in \left\{ 9 , 10 \right\}$ and $\Delta \left( n, 8 \right) = -7$
The condition $\Delta \left( n, 8 \right) = -7$ implies $n \equiv - 1 \pmod{k}$ with $k \in \left\{ 2, 4 \right\}$ .
Hence, $\Delta \left( n, 2 \right) = -1$ and $\Delta \left( n, 4 \right) = -3$
To get $\sum_{k = 2}^{10} \Delta \left( n, k \right) = 0$ , we have $\sum_{k \in \left\{ 3, 5, 6, 7 \right\}} \Delta \left( n, k \right) = 9$
However, we have $\sum_{k \in \left\{ 3, 5, 6, 7 \right\}} \Delta \left( n, k \right) \leq \sum_{k \in \left\{ 3, 5, 6, 7 \right\}} 1 = 4$
Therefore, there is no $n$ in this case.
$\textbf{Case 4}$ $\Delta \left( n, k \right) = 1$ for $k \in \left\{ 8, \cdots , 10 \right\}$ and $\Delta \left( n, 7 \right) = -6$
To get $\sum_{k = 2}^{10} \Delta \left( n, k \right) = 0$ , we have $\sum_{k \in \left\{ 2, 3, 4, 5, 6 \right\}} \Delta \left( n, k \right) = 3$
Hence, we must have $\Delta \left( n, 2 \right) = -1$ and $\Delta \left( n, k \right) = 1$ for $k \in \left\{ 3 , 4 , 5 , 6 \right\}$
Therefore, $n = 13, 97$
$\textbf{Case 5}$ $\Delta \left( n, k \right) = 1$ for $k \in \left\{ 7 , \cdots , 10 \right\}$ and $\Delta \left( n, 6 \right) = -5$
The condition $\Delta \left( n, 6 \right) = -5$ implies $n \equiv - 1 \pmod{k}$ with $k \in \left\{ 2, 3 \right\}$ .
Hence, $\Delta \left( n, 2 \right) = -1$ and $\Delta \left( n, 3 \right) = -2$
To get $\sum_{k = 2}^{10} \Delta \left( n, k \right) = 0$ , we have $\sum_{k \in \left\{ 4, 5 \right\}} \Delta \left( n, k \right) = 4$
However, we have $\sum_{k \in \left\{ 4, 5 \right\}} \Delta \left( n, k \right) \leq \sum_{k \in \left\{ 4, 5 \right\}} 1 = 2$
Therefore, there is no $n$ in this case.
$\textbf{Case 6}$ $\Delta \left( n, k \right) = 1$ for $k \in \left\{ 6 , \cdots , 10 \right\}$ and $\Delta \left( n, 5 \right) = -4$
To get $\sum_{k = 2}^{10} \Delta \left( n, k \right) = 0$ , we have $\sum_{k \in \left\{ 2, 3, 4 \right\}} \Delta \left( n, k \right) = -1$
This can be achieved if $\Delta \left( n, 2 \right) = 1$ $\Delta \left( n, 3 \right) = 1$ $\Delta \left( n, 4 \right) = -3$
However, $\Delta \left( n, 4 \right) = -3$ implies $n \equiv - 1 \pmod{4}$ . This implies $n \equiv -1 \pmod{2}$ . Hence, $\Delta \left( n, 2 \right) = -1$ .
We get a contradiction.
Therefore, there is no $n$ in this case.
$\textbf{Case 7}$ $\Delta \left( n, k \right) = 1$ for $k \in \left\{ 5 , \cdots , 10 \right\}$ and $\Delta \left( n, 4 \right) = -3$
The condition $\Delta \left( n, 4 \right) = -3$ implies $n \equiv - 1 \pmod{k}$ with $k = 2$ .
Hence, $\Delta \left( n, 2 \right) = -1$
To get $\sum_{k = 2}^{10} \Delta \left( n, k \right) = 0$ , we have $\Delta \left( n, 3 \right) = - 2$ . This implies $n \equiv - 1 \pmod{3}$
Because $n \equiv - 1 \pmod{2}$ and $n \equiv - 1 \pmod{3}$ , we have $n \equiv - 1 \pmod{6}$ .
Hence, $\Delta \left( n, 6 \right) = - 5$ .
However, in this case, we assume $\Delta \left( n, 6 \right) = 1$ .
We get a contradiction.
Therefore, there is no $n$ in this case.
$\textbf{Case 8}$ $\Delta \left( n, k \right) = 1$ for $k \in \left\{ 4 , \cdots , 10 \right\}$ and $\Delta \left( n, 3 \right) = -2$
To get $\sum_{k = 2}^{10} \Delta \left( n, k \right) = 0$ , we have $\Delta \left( n, 2 \right) = - 5$ . This is infeasible.
Therefore, there is no $n$ in this case.
$\textbf{Case 9}$ $\Delta \left( n, k \right) = 1$ for $k \in \left\{3 , \cdots , 10 \right\}$
To get $\sum_{k = 2}^{10} \Delta \left( n, k \right) = 0$ , we have $\Delta \left( n, 2 \right) = - 8$ . This is infeasible.
Therefore, there is no $n$ in this case.
Putting all cases together, the answer is $\boxed{2}$
| 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?
$\textbf{(A)} ~10272\qquad\textbf{(B)} ~11700\qquad\textbf{(C)} ~13362\qquad\textbf{(D)} ~14238\qquad\textbf{(E)} ~15426$
|
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 $11a=17402$ . So $a=1582$ . The difference is $10a-a = 9a$ . So, the answer is $9(1582) = \boxed{14238}$
| 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?
$\textbf{(A)} ~10272\qquad\textbf{(B)} ~11700\qquad\textbf{(C)} ~13362\qquad\textbf{(D)} ~14238\qquad\textbf{(E)} ~15426$
|
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?
$\textbf{(A)} ~10272\qquad\textbf{(B)} ~11700\qquad\textbf{(C)} ~13362\qquad\textbf{(D)} ~14238\qquad\textbf{(E)} ~15426$
|
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 get \[D=2\implies C=8 \implies B=5 \implies A=1.\] The larger number is $15820$ and the smaller number is $1582.$ Their difference is $15820-1582=\boxed{14238}.$
| 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?
$\textbf{(A)} ~10272\qquad\textbf{(B)} ~11700\qquad\textbf{(C)} ~13362\qquad\textbf{(D)} ~14238\qquad\textbf{(E)} ~15426$
|
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 the smaller number is $2$ , so the positive difference between the numbers is 8. There is only one answer choice that has this units digit, and that is $\boxed{14238}.$
| 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 $\underline{1}.\underline{a} \ \underline{b}.$ Later he found that his answer is $0.5$ less than the correct answer. What is the $2$ -digit number $\underline{a} \ \underline{b}?$
$\textbf{(A) }15 \qquad \textbf{(B) }30 \qquad \textbf{(C) }45 \qquad \textbf{(D) }60 \qquad \textbf{(E) }75$
|
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(\underline{1}.\overline{\underline{a} \ \underline{b}} - \underline{1}.\underline{a} \ \underline{b}\Bigr)&=0.5 \\ 66\Bigl(\underline{0}.\underline{0} \ \underline{0} \ \overline{\underline{a} \ \underline{b}}\Bigr)&=0.5 \\ 66\left(\frac{1}{100}\cdot\underline{0}.\overline{\underline{a} \ \underline{b}}\right)&=\frac12 \\ \underline{0}.\overline{\underline{a} \ \underline{b}}&=\frac{25}{33} \\ \underline{0}.\overline{\underline{a} \ \underline{b}}&=0.\overline{75} \\ \underline{a} \ \underline{b}&=\boxed{75} ~MRENTHUSIASM
| 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 $\underline{1}.\underline{a} \ \underline{b}.$ Later he found that his answer is $0.5$ less than the correct answer. What is the $2$ -digit number $\underline{a} \ \underline{b}?$
$\textbf{(A) }15 \qquad \textbf{(B) }30 \qquad \textbf{(C) }45 \qquad \textbf{(D) }60 \qquad \textbf{(E) }75$
|
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 \[66\biggl(1+\frac{x}{99}\biggr)-66\biggl(1+\frac{x}{100}\biggr)=0.5.\] Expanding and simplifying give $\frac{x}{150}=0.5,$ so $x=\boxed{75}.$
| 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 $\underline{1}.\underline{a} \ \underline{b}.$ Later he found that his answer is $0.5$ less than the correct answer. What is the $2$ -digit number $\underline{a} \ \underline{b}?$
$\textbf{(A) }15 \qquad \textbf{(B) }30 \qquad \textbf{(C) }45 \qquad \textbf{(D) }60 \qquad \textbf{(E) }75$
|
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 \[\frac{33(10a+b)}{50} + \frac{1}{2} = \frac{2(10a+b)}{3}.\] Multiplying both sides by $50 \cdot 3 = 150,$ we have \[99(10a+b) + 75 = 100(10a+b).\] Thus, the answer is $10a+b = \boxed{75}.$
| 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?
$\textbf{(A) }6 \qquad \textbf{(B) }9 \qquad \textbf{(C) }12 \qquad \textbf{(D) }15 \qquad \textbf{(E) }18$
|
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 ratio of red to black cards is $1:3$ . This means in the new deck the number of black cards is also $3x$ for the same $x$ red cards.
So, $3x = 2x + 4$ and $x=4$ meaning there are $4$ red cards in the deck at the start and $2(4) = 8$ black cards.
So, the answer is $8+4 = 12 = \boxed{12}$
| 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?
$\textbf{(A) }6 \qquad \textbf{(B) }9 \qquad \textbf{(C) }12 \qquad \textbf{(D) }15 \qquad \textbf{(E) }18$
|
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 $4$ cards are equal to $\frac13$ of the original deck, the original deck has $4\cdot3=\boxed{12}$ cards.
| 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?
$\textbf{(A) }6 \qquad \textbf{(B) }9 \qquad \textbf{(C) }12 \qquad \textbf{(D) }15 \qquad \textbf{(E) }18$
|
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 $0$ \[\frac{\partial f}{\partial x} = 2x + 2xy^{2} = 2x(1 + y^{2}) = 0 \implies x = 0\] \[\frac{\partial f}{\partial y} = 2y + 2yx^{2} = 2y(1 + x^{2}) = 0 \implies y = 0\] Thus, there is a local extremum at $(0, 0)$ . Because this is the only extremum, we can assume that this is a minimum because the problem asks for the minimum (though this can also be proven using the partial second derivative test) and the global minimum since it's the only minimum, meaning $f(0, 0)$ is the minimum of $f(x, y)$ . Plugging $(0, 0)$ into $f(x, y)$ , we find 1 $\implies \boxed{1}$
| 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 liquid. What is the ratio of the rise of the liquid level in the narrow cone to the rise of the liquid level in the wide cone?
[asy] size(350); defaultpen(linewidth(0.8)); real h1 = 10, r = 3.1, s=0.75; pair P = (r,h1), Q = (-r,h1), Pp = s * P, Qp = s * Q; path e = ellipse((0,h1),r,0.9), ep = ellipse((0,h1*s),r*s,0.9); draw(ellipse(origin,r*(s-0.1),0.8)); fill(ep,gray(0.8)); fill(origin--Pp--Qp--cycle,gray(0.8)); draw((-r,h1)--(0,0)--(r,h1)^^e); draw(subpath(ep,0,reltime(ep,0.5)),linetype("4 4")); draw(subpath(ep,reltime(ep,0.5),reltime(ep,1))); draw(Qp--(0,Qp.y),Arrows(size=8)); draw(origin--(0,12),linetype("4 4")); draw(origin--(r*(s-0.1),0)); label("$3$",(-0.9,h1*s),N,fontsize(10)); real h2 = 7.5, r = 6, s=0.6, d = 14; pair P = (d+r-0.05,h2-0.15), Q = (d-r+0.05,h2-0.15), Pp = s * P + (1-s)*(d,0), Qp = s * Q + (1-s)*(d,0); path e = ellipse((d,h2),r,1), ep = ellipse((d,h2*s+0.09),r*s,1); draw(ellipse((d,0),r*(s-0.1),0.8)); fill(ep,gray(0.8)); fill((d,0)--Pp--Qp--cycle,gray(0.8)); draw(P--(d,0)--Q^^e); draw(subpath(ep,0,reltime(ep,0.5)),linetype("4 4")); draw(subpath(ep,reltime(ep,0.5),reltime(ep,1))); draw(Qp--(d,Qp.y),Arrows(size=8)); draw((d,0)--(d,10),linetype("4 4")); draw((d,0)--(d+r*(s-0.1),0)); label("$6$",(d-r/4,h2*s-0.06),N,fontsize(10)); [/asy]
$\textbf{(A) }1:1 \qquad \textbf{(B) }47:43 \qquad \textbf{(C) }2:1 \qquad \textbf{(D) }40:13 \qquad \textbf{(E) }4:1$
|
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 ball of the same volume into each cylinder, the water level in the narrow cone/cylinder should rise $\boxed{4}$ times as much.
| 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 \[B= - \left(\binom {4}{3} \binom {2}{0} \cdot 2^{3} + \binom {4}{2} \binom{2}{1} \cdot 2^{2} \cdot 1 + \binom {4}{1} \binom {2}{2} \cdot 2 \right) = -\left(32+48+8 \right) = \boxed{88}.\] ~ike.chen
| 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:
$\textbf{(A): }(-2)^5=-32$
$\textbf{(B): }32\cos(750)=32\cos(30)=16\sqrt{3}$
$\textbf{(C): }32\cos(675)=32\cos(-45)=16\sqrt{2}$
$\textbf{(D): }32\cos(600)=32\cos(240)<0$
$\textbf{(E): }(2i)^5=32i$ , whose real part is $0$
Thus, the answer is $\boxed{3}$
| 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 \[\operatorname{Re}\left(z^5\right)=r^5\cos{(5\theta)}.\] We construct a table as follows: \[\begin{array}{c|ccc|ccc|cclclclcc} & & & & & & & & & & & & & & & \\ [-2ex] \textbf{Choice} & & \boldsymbol{r} & & & \boldsymbol{\theta} & & & & & & \multicolumn{1}{c}{\boldsymbol{\operatorname{Re}\left(z^5\right)}} & & & & \\ [0.5ex] \hline & & & & & & & & & & & & & & & \\ [-1ex] \textbf{(A)} & & 2 & & & \pi & & & &32\cos{(5\pi)}&=&32\cos\pi&=&32(-1)& & \\ [2ex] \textbf{(B)} & & 2 & & & \tfrac{5\pi}{6} & & & &32\cos{\tfrac{25\pi}{6}}&=&32\cos{\tfrac{\pi}{6}}&=&32\left(\tfrac{\sqrt3}{2}\right)& & \\ [2ex] \textbf{(C)} & & 2 & & & \tfrac{3\pi}{4} & & & &32\cos{\tfrac{15\pi}{4}}&=&32\cos{\tfrac{7\pi}{4}}&=&32\left(\tfrac{\sqrt2}{2}\right)& & \\ [2ex] \textbf{(D)} & & 2 & & & \tfrac{2\pi}{3} & & & &32\cos{\tfrac{10\pi}{3}}&=&32\cos{\tfrac{4\pi}{3}}&=&32\left(-\tfrac{1}{2}\right)& & \\ [2ex] \textbf{(E)} & & 2 & & & \tfrac{\pi}{2} & & & &32\cos{\tfrac{5\pi}{2}}&=&32\cos{\tfrac{\pi}{2}}&=&32\left(0\right)& & \\ [1ex] \end{array}\] Clearly, the answer is $\boxed{3}.$
| 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 + 5ab^4. \end{align*} We find the real parts of $\textbf{(B)},\textbf{(C)},$ and $\textbf{(D)}$ directly:
Therefore, the answer is $\boxed{3}.$
| 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) + \left(5a^4b-10a^2b^3+b^5\right)i. \end{align*} We find the full expansions of $\textbf{(B)},\textbf{(C)},$ and $\textbf{(D)},$ then extract their real parts:
Therefore, the answer is $\boxed{3}.$
| 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}&=\log_{5^k}{(3^k)^k} \\ &=k\log_{5^k}{3^k} \\ &=k\log_{5}{3}, \end{align*} and \begin{align*} \log_{9^k}{25^k}&=\log_{3^{2k}}{5^{2k}} \\ &=\log_{3}{5}. \end{align*} Using these results, we evaluate the original expression: \begin{align*} \left(\sum_{k=1}^{20} \log_{5^k} 3^{k^2}\right)\cdot\left(\sum_{k=1}^{100} \log_{9^k} 25^k\right)&=\left(\sum_{k=1}^{20} k\log_{5}{3}\right)\cdot\left(\sum_{k=1}^{100} \log_{3}{5}\right) \\ &= \left(\log_{5}{3}\cdot\sum_{k=1}^{20} k\right)\cdot\left(\log_{3}{5}\cdot\sum_{k=1}^{100} 1\right) \\ &= \left(\sum_{k=1}^{20} k\right)\cdot\left(\sum_{k=1}^{100} 1\right) \\ &= \frac{21\cdot20}{2}\cdot100 \\ &= \boxed{21000} ~MRENTHUSIASM (Solution)
| 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, \[\log_{9^k} 25^{k} = k \cdot \frac{1}{k} \cdot \log_{9} 25 = \log_{9} 5^2.\] Then, evaluating the first few terms in each parentheses, we can find the simplified expanded forms of each sum using the additive property of logarithms: \begin{align*} \sum_{k=1}^{20} \log_{5} 3^k &= \log_{5} 3^1 + \log_{5} 3^2 + \dots + \log_{5} 3^{20} \\ &= \log_{5} 3^{(1 + 2 + \dots + 20)} \\ &= \log_{5} 3^{\frac{20(20+1)}{2}} &&\hspace{15mm}(*) \\ &= \log_{5} 3^{210}, \\ \sum_{k=1}^{100} \log_{9} 5^2 &= \log_{9} 5^2 + \log_{9} 5^2 + \dots + \log_{9} 5^2 \\ &= \log_{9} 5^{2(100)} \\ &= \log_{9} 5^{200}. \end{align*} In $(*),$ we use the triangular numbers equation: \[1 + 2 + \dots + n = \frac{n(n+1)}{2} = \frac{20(20+1)}{2} = 210.\] Finally, multiplying the two logarithms together, we can use the chain rule property of logarithms to simplify: \[\log_{a} b\log_{x} y = \log_{a} y\log_{x} b.\] Thus, \begin{align*} \left(\log_{5} 3^{210}\right)\left(\log_{3^2} 5^{200}\right) &= \left(\log_{5} 5^{200}\right)\left(\log_{3^2} 3^{210}\right) \\ &= \left(\log_{5} 5^{200}\right)\left(\log_{3} 3^{105}\right) \\ &= (200)(105) \\ &= \boxed{21000} ~Joeya (Solution)
| 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 \[\left(\sum_{k=1}^{20} \log_{5^k} 3^{k^2}\right)\cdot\left(\sum_{k=1}^{100} \log_{9^k} 25^k\right)>\left(\sum_{k=2}^{20} 1\right)\cdot\left(\sum_{k=1}^{100} 1\right)=19\cdot100=1900,\] which eliminates choices $\textbf{(A)}, \textbf{(B)},$ and $\textbf{(C)}.$ We get the answer $\boxed{21000}$ by either an educated guess or a continued approximation:
| 0
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3,869
|
https://artofproblemsolving.com/wiki/index.php/2021_AMC_12A_Problems/Problem_14
| 4
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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: \[\left(\sum_{k=1}^{20} \log_{5} 3^{k}\right)\cdot\left(\sum_{k=1}^{100} \log_{3} 5\right)\] \begin{align*} \sum_{k=1}^{20} \log_{5} 3^k &= \log_{5} 3^1 + \log_{5} 3^2 + \dots + \log_{5} 3^{20} \\ &= \log_{5} 3^{(1 + 2 + \dots + 20)} \\ &= \log_{5} 3^{\frac{20(20+1)}{2}} \\ &= \log_{5} 3^{210} \\ &= {210}\cdot\log_{5} {3}, \\ \sum_{k=1}^{100}\log_{3} {5} &= {100}\cdot\log_{3} {5}. \end{align*} With the reciprocal rule the logarithms cancel out leaving: $\boxed{21000}.$
| 0
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3,870
|
https://artofproblemsolving.com/wiki/index.php/2021_AMC_12A_Problems/Problem_16
| 1
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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?
$\textbf{(A)} ~100.5 \qquad\textbf{(B)} ~134 \qquad\textbf{(C)} ~142 \qquad\textbf{(D)} ~150.5 \qquad\textbf{(E)} ~167$
|
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 $142$ looks like the only number close to it, it is answer $\boxed{142}$ ~Lopkiloinm
| 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?
$\textbf{(A)} ~100.5 \qquad\textbf{(B)} ~134 \qquad\textbf{(C)} ~142 \qquad\textbf{(D)} ~150.5 \qquad\textbf{(E)} ~167$
|
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 $\frac{200(200+1)}{2}-\frac{142(142+1)}{2}$ which is approximately $10000$ as well. Therefore, we can be relatively sure the answer choice is $\boxed{142}.$
| 142
|
3,872
|
https://artofproblemsolving.com/wiki/index.php/2021_AMC_12A_Problems/Problem_16
| 3
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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?
$\textbf{(A)} ~100.5 \qquad\textbf{(B)} ~134 \qquad\textbf{(C)} ~142 \qquad\textbf{(D)} ~150.5 \qquad\textbf{(E)} ~167$
|
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 $200.$ [asy]draw((0,0)--(200,200)--(200,0)--cycle); draw((142,0)--(142,142)); label("$x$",(142,0)--(142,142),E); label("$x$",(0,0)--(142,0),S); label("$200$",(200,0)--(200,200),E); [/asy]
Let $x$ be the side length such that both sides of the triangle have the same area. The desired answer is then around $x$ because about half of the numbers in the list fall on each side.
Solving for $x$ yields: \begin{align*} \frac{x^2}{2} =& \:\frac{1}{2} \cdot \frac{200^2}{2} \\ x^2 =& \:\frac{1}{2}\cdot 200^2 \\ x =& \:\frac{200}{\sqrt{2}} = \: 100\sqrt{2} \approx 141. \end{align*} We see that $\boxed{142}$ is the closest to $x$ by far, and thus, can be relatively certain this is the answer.
| 142
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3,873
|
https://artofproblemsolving.com/wiki/index.php/2021_AMC_12A_Problems/Problem_17
| 1
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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 $m\sqrt{n}$ , where $m$ and $n$ are positive integers and $n$ is not divisible by the square of any prime. What is $m+n$
$\textbf{(A) }65 \qquad \textbf{(B) }132 \qquad \textbf{(C) }157 \qquad \textbf{(D) }194\qquad \textbf{(E) }215$
|
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 right, the Pythagorean theorem implies that \[AD=\sqrt{86^2-66^2}=4\sqrt{190}.\] The answer is $4+190=\boxed{194}$
| 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 $m\sqrt{n}$ , where $m$ and $n$ are positive integers and $n$ is not divisible by the square of any prime. What is $m+n$
$\textbf{(A) }65 \qquad \textbf{(B) }132 \qquad \textbf{(C) }157 \qquad \textbf{(D) }194\qquad \textbf{(E) }215$
|
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 \[AD=CP\cdot\frac{OD}{OP}=h\cdot\frac{x}{11}.\] Let the brackets denote areas. Notice that $[AOD]=[BOC]$ (By the same base and height, we deduce that $[ACD]=[BDC].$ Subtracting $[OCD]$ from both sides gives $[AOD]=[BOC].$ ). Doubling both sides produces \begin{align*} 2[AOD]&=2[BOC] \\ OD\cdot AD&=OB\cdot CP \\ x\left(\frac{hx}{11}\right)&=(x+22)h \\ x^2&=11(x+22). \end{align*} Rearranging and factoring result in $(x-22)(x+11)=0,$ from which $x=22.$
Applying the Pythagorean Theorem to right $\triangle CPB,$ we have \[h=\sqrt{43^2-33^2}=\sqrt{(43+33)(43-33)}=\sqrt{760}=2\sqrt{190}.\] Finally, we get \[AD=h\cdot\frac{x}{11}=4\sqrt{190},\] so the answer is $4+190=\boxed{194}.$
| 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 $m\sqrt{n}$ , where $m$ and $n$ are positive integers and $n$ is not divisible by the square of any prime. What is $m+n$
$\textbf{(A) }65 \qquad \textbf{(B) }132 \qquad \textbf{(C) }157 \qquad \textbf{(D) }194\qquad \textbf{(E) }215$
|
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) $\triangle BPC \sim \triangle BDA$ with ratio $1:2,$ so $AD = 2y$ using the fact that $P$ is the midpoint of $BD$
Thus, $\frac{xy}{11} = 2y,$ or $x = 22.$ And $y = \sqrt{43^2 - 33^2} = 2 \sqrt{190},$ so $AD = 4 \sqrt{190}$ and the answer is $4+190=\boxed{194}.$
| 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 $m\sqrt{n}$ , where $m$ and $n$ are positive integers and $n$ is not divisible by the square of any prime. What is $m+n$
$\textbf{(A) }65 \qquad \textbf{(B) }132 \qquad \textbf{(C) }157 \qquad \textbf{(D) }194\qquad \textbf{(E) }215$
|
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 $\angle DBE \cong \angle PBC \cong \angle DBA \implies \angle DBE \cong \angle DBA$ . Thus, by ASA, we know that $\triangle ABD \cong \triangle EBD$ , thus, $AD = ED$ , meaning $D$ is the midpoint of $AE$ .
Let $M$ be the midpoint of $\overline{DE}$ . Note that $\triangle CME$ is congruent to $\triangle BPC$ , thus $BC = CE$ , meaning $C$ is the midpoint of $\overline{BE}.$
Therefore, $\overline{AC}$ and $\overline{BD}$ are both medians of $\triangle ABE$ . This means that $O$ is the centroid of $\triangle ABE$ ; therefore, because the centroid divides the median in a 2:1 ratio, $\frac{BO}{2} = DO = \frac{BD}{3}$ . Recall that $P$ is the midpoint of $BD$ $DP = \frac{BD}{2}$ . The question tells us that $OP = 11$ $DP-DO=11$ ; we can write this in terms of $DB$ $\frac{DB}{2}-\frac{DB}{3} = \frac{DB}{6} = 11 \implies DB = 66$
We are almost finished. Each side length of $\triangle ABD$ is twice as long as the corresponding side length $\triangle CBP$ or $\triangle CPD$ , since those triangles are similar; this means that $AB = 2 \cdot 43 = 86$ . Now, by Pythagorean theorem on $\triangle ABD$ $AB^{2} - BD^{2} = AD^{2} \implies 86^{2}-66^{2} = AD^{2} \implies AD = \sqrt{3040} \implies AD = 4 \sqrt{190}$
The answer is $4+190 = \boxed{194}$
| 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 $m\sqrt{n}$ , where $m$ and $n$ are positive integers and $n$ is not divisible by the square of any prime. What is $m+n$
$\textbf{(A) }65 \qquad \textbf{(B) }132 \qquad \textbf{(C) }157 \qquad \textbf{(D) }194\qquad \textbf{(E) }215$
|
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 that of $\triangle ADC$ on $CD$ . Since $[APB]=[ADC]$ $AB=2CD$
As a basic property of a trapezoid, $\triangle AOB \sim \triangle COD$ , so $\frac{OB}{OD}=\frac{AB}{CD}=2$ , or $OB=2OD$ . Letting $OD=x$ , then $PB=DP=11+x$ , and $OB=22+x$ . Hence $22+x=2x$ and $x=22$
Since $\triangle AOD \sim \triangle COP$ $\frac{AD}{PC}=\frac{OD}{OP}=2$ . Since $PD=11+22=33$ $PC=\sqrt{43^2-33^2}=\sqrt{760}$
So, $AD=2\sqrt{760}=4\sqrt{190}$ . The correct answer is $\boxed{194}$
| 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 $m\sqrt{n}$ , where $m$ and $n$ are positive integers and $n$ is not divisible by the square of any prime. What is $m+n$
$\textbf{(A) }65 \qquad \textbf{(B) }132 \qquad \textbf{(C) }157 \qquad \textbf{(D) }194\qquad \textbf{(E) }215$
|
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 isosceles $C$ lies on $x = a.$ Thus since $CD$ has equation $y = \frac{b}{a}x$ $D$ is the origin), $C = (a,b).$ Therefore $AC$ has equation $y = \frac{3b}{a}x - 2b$ and intersects $BD$ $x$ -axis) at $O =\left(\frac{2}{3}a, 0\right).$ The midpoint of $BD$ is $P = (a,0),$ so $OP = \frac{a}{3} = 11,$ from which $a = 33.$ Then by Pythagorean theorem on $\triangle DPC$ $\triangle DBC$ is isosceles), we have $b = \sqrt{43^2 - 33^2} = 2\sqrt{190},$ so $2b=4\sqrt{190}.$
Finally, the answer is $4+190=\boxed{194}.$
| 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 \left( \sin \left( \frac{\pi}2 - \frac{\pi}2 \sin x\right)\right) \\ \frac{\pi}2 \cos x &= \frac{\pi}2 - \frac{\pi}2 \sin x \\ \cos x &= 1 - \sin x \\ \cos x + \sin x &= 1. \end{align*} This only happens at $x = 0, \frac{\pi}2$ on the interval $[0,\pi],$ because one of $\sin$ and $\cos$ must be $1$ and the other $0.$ Therefore, the answer is $\boxed{2}.$
| 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 $n.$ Therefore, we have two cases:
Together, we obtain $\boxed{2}.$
| 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: \[\begin{array}{c|ccc} & & & \\ [-2ex] & \boldsymbol{x=0} & \boldsymbol{x=\frac{\pi}{2}} & \boldsymbol{x=\pi} \\ [1.5ex] \hline & & & \\ [-1ex] \boldsymbol{\cos x} & 1 & 0 & -1 \\ [1.5ex] \boldsymbol{\frac{\pi}{2}\cos x} & \frac{\pi}{2} & 0 & -\frac{\pi}{2} \\ [1.5ex] \boldsymbol{\sin\left(\frac{\pi}{2}\cos x\right)} & 1 & 0 & -1 \\ [1.5ex] \hline & & & \\ [-1ex] \boldsymbol{\sin x} & 0 & 1 & 0 \\ [1.5ex] \boldsymbol{\frac{\pi}{2}\sin x} & 0 & \frac{\pi}{2} & 0 \\ [1.5ex] \boldsymbol{\cos\left(\frac{\pi}{2}\sin x\right)} & 1 & 0 & 1 \\ [1ex] \end{array}\] For $x\in[0,\pi],$ note that:
For the graphs to intersect, we need $\sin\left(\frac{\pi}{2}\cos x\right)\in[0,1].$ This occurs when $\frac{\pi}{2}\cos x\in\left[0,\frac{\pi}{2}\right].$
By the Cofunction Identity $\cos\theta=\sin\left(\frac{\pi}{2}-\theta\right),$ we rewrite the given equation: \[\sin\left(\frac{\pi}{2}\cos x\right) = \sin\left(\frac{\pi}{2}-\frac{\pi}{2}\sin x\right).\] Since $\frac{\pi}{2}\cos x\in\left[0,\frac{\pi}{2}\right]$ and $\frac{\pi}{2}\sin x\in\left[0,\frac{\pi}{2}\right],$ it follows that $x\in\left[0,\frac{\pi}{2}\right]$ and $\frac{\pi}{2}-\frac{\pi}{2}\sin x\in\left[0,\frac{\pi}{2}\right].$
We can apply the arcsine function to both sides, then rearrange and simplify: \begin{align*} \frac{\pi}{2}\cos x &= \frac{\pi}{2}-\frac{\pi}{2}\sin x \\ \sin x + \cos x &= 1. \end{align*} From Case 1 in Solution 2, we conclude that $(0,1)$ and $\left(\frac{\pi}{2},0\right)$ are the only points of intersection, as shown below: [asy] /* Made by MRENTHUSIASM */ size(600,200); real f(real x) { return sin(pi/2*cos(x)); } real g(real x) { return cos(pi/2*sin(x)); } draw(graph(f,0,pi),red,"$y=\sin\left(\frac{\pi}{2}\cos x\right)$"); draw(graph(g,0,pi),blue,"$y=\cos\left(\frac{\pi}{2}\sin x\right)$"); real xMin = 0; real xMax = 5/4*pi; real yMin = -2; real yMax = 2; //Draws the horizontal gridlines void horizontalLines() { for (real i = yMin+1; i < yMax; ++i) { draw((xMin,i)--(xMax,i), mediumgray+linewidth(0.4)); } } //Draws the vertical gridlines void verticalLines() { for (real i = xMin+pi/2; i < xMax; i+=pi/2) { draw((i,yMin)--(i,yMax), mediumgray+linewidth(0.4)); } } //Draws the horizontal ticks void horizontalTicks() { for (real i = yMin+1; i < yMax; ++i) { draw((-1/8,i)--(1/8,i), black+linewidth(1)); } } //Draws the vertical ticks void verticalTicks() { for (real i = xMin+pi/2; i < xMax; i+=pi/2) { draw((i,-1/8)--(i,1/8), black+linewidth(1)); } } horizontalLines(); verticalLines(); horizontalTicks(); verticalTicks(); draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(5)); draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(5)); label("$x$",(xMax,0),(2,0)); label("$y$",(0,yMax),(0,2)); pair A[]; A[0] = (pi/2,0); A[1] = (pi,0); A[2] = (0,1); A[3] = (0,0); A[4] = (0,-1); label("$\frac{\pi}{2}$",A[0],(0,-2.5)); label("$\pi$",A[1],(0,-2.5)); label("$1$",A[2],(-2.5,0)); label("$0$",A[3],(-2.5,0)); label("$-1$",A[4],(-2.5,0)); dot((0,1),linewidth(5)); dot((pi/2,0),linewidth(5)); add(legend(),point(E),40E,UnFill); [/asy] Therefore, the answer is $\boxed{2}.$
| 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 $\mathcal E$ can be written in the form $\sqrt{\frac mn}$ , where $m$ and $n$ are relatively prime positive integers. What is $m+n$ ? (Recall that the eccentricity of an ellipse $\mathcal E$ is the ratio $\frac ca$ , where $2a$ is the length of the major axis of $\mathcal E$ and $2c$ is the is the distance between its two foci.)
$\textbf{(A) }7 \qquad \textbf{(B) }9 \qquad \textbf{(C) }11 \qquad \textbf{(D) }13\qquad \textbf{(E) }15$
|
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 must $\mathcal E$
Now let $r=b/a$ denote the ratio of the length of the minor axis of $\mathcal E$ to the length of its major axis. Remark that if we perform a transformation of the plane which scales every $x$ -coordinate by a factor of $r$ $\mathcal E$ is sent to a circle $\mathcal E'$ . Thus, the problem is equivalent to finding the value of $r$ such that $(r,0)$ $\left(-r,\pm\sqrt 3\right)$ , and $\left(-2r,\pm\sqrt 2\right)$ all lie on a common circle; equivalently, it suffices to determine the value of $r$ such that the circumcenter of the triangle formed by the points $P_1 = (r,0)$ $P_2 = \left(-r,\sqrt 3\right)$ , and $P_3 = \left(-2r,\sqrt 2\right)$ lies on the $x$ -axis.
Recall that the circumcenter of a triangle $ABC$ is the intersection point of the perpendicular bisectors of its three sides. The equations of the perpendicular bisectors of the segments $\overline{P_1P_2}$ and $\overline{P_1P_3}$ are \[y = \tfrac{\sqrt 3}2 + \tfrac{2r}{\sqrt 3}x\qquad\text{and}\qquad y = \tfrac{\sqrt 2}2 + \tfrac{3r}{\sqrt 2}\left(x + \tfrac r2\right)\] respectively. These two lines have different slopes for $r\neq 0$ , so indeed they will intersect at some point $(x_0,y_0)$ ; we want $y_0 = 0$ . Plugging $y = 0$ into the first equation yields $x = -\tfrac{3}{4r}$ , and so plugging $y=0$ into the second equation and simplifying yields \[-\tfrac{1}{3r} = x + \tfrac r2 = -\tfrac{3}{4r} + \tfrac{r}{2}.\] Solving yields $r=\sqrt{\tfrac 56}$
Finally, recall that the lengths $a$ $b$ , and $c$ (where $c$ is the distance between the foci of $\mathcal E$ ) satisfy $c = \sqrt{a^2 - b^2}$ . Thus the eccentricity of $\mathcal E$ is $\tfrac ca = \sqrt{1 - \left(\tfrac ba\right)^2} = \sqrt{\tfrac 16}$ and the requested answer is $\boxed{7}$
| 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 $\mathcal E$ can be written in the form $\sqrt{\frac mn}$ , where $m$ and $n$ are relatively prime positive integers. What is $m+n$ ? (Recall that the eccentricity of an ellipse $\mathcal E$ is the ratio $\frac ca$ , where $2a$ is the length of the major axis of $\mathcal E$ and $2c$ is the is the distance between its two foci.)
$\textbf{(A) }7 \qquad \textbf{(B) }9 \qquad \textbf{(C) }11 \qquad \textbf{(D) }13\qquad \textbf{(E) }15$
|
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 $xy$ -plane. By symmetry, the center of $\mathcal E$ must be on the $x$ -axis.
The formula of $\mathcal E$ is \[\frac{(x-h)^2}{a^2}+\frac{y^2}{b^2}=1, \hspace{44.5mm} (\bigstar)\] with the center $(h,0)$ and the axes' lengths $2a$ and $2b.$
Plugging the points $(1,0),\left(-1,\sqrt3\right),$ and $\left(-2,\sqrt2\right)$ into $(\bigstar),$ respectively, we have the following system of equations: \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*} Since $t^2=(-t)^2$ holds for all real numbers $t,$ we clear fractions and simplify: \begin{align*} (1-h)^2&=a^2, \hspace{30.25mm} &(1)\\ b^2(1+h)^2 + 3a^2 &= a^2b^2, &(2)\\ b^2(2+h)^2 + 2a^2 &= a^2b^2. &(3) \end{align*} Applying the Transitive Property to $(2)$ and $(3),$ we isolate $a^2:$ \begin{align*} b^2(1+h)^2 + 3a^2 &= b^2(2+h)^2 + 2a^2 \\ a^2 &= b^2\left((2+h)^2-(1+h)^2\right) \\ a^2 &= b^2(2h+3). \hspace{26.75mm} (*) \end{align*} Substituting $(1)$ and $(*)$ into $(2),$ we solve for $h:$ \begin{align*} b^2(1+h)^2 + 3\underbrace{b^2(2h+3)}_{\text{by }(*)} &= \underbrace{(1-h)^2}_{\text{by }(1)}b^2 \\ (1+h)^2+3(2h+3)&=(1-h)^2 \\ 1+2h+h^2+6h+9&=1-2h+h^2 \\ 10h&=-9 \\ h&=-\frac{9}{10}. \end{align*} Substituting this into $(1),$ we get $a^2=\frac{361}{100}.$
Substituting the current results into $(*),$ we get $b^2=\frac{361}{120}.$
Finally, we obtain \[c^2 = a^2-b^2 = 361\left(\frac{1}{100}-\frac{1}{120}\right) = \frac{361}{600},\] from which \[\frac{c}{a}=\sqrt{\frac{c^2}{a^2}}=\sqrt{\frac{361/600}{361/100}}=\sqrt{\frac 16}.\] The answer is $1+6=\boxed{7}.$
| 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 $\mathcal E$ can be written in the form $\sqrt{\frac mn}$ , where $m$ and $n$ are relatively prime positive integers. What is $m+n$ ? (Recall that the eccentricity of an ellipse $\mathcal E$ is the ratio $\frac ca$ , where $2a$ is the length of the major axis of $\mathcal E$ and $2c$ is the is the distance between its two foci.)
$\textbf{(A) }7 \qquad \textbf{(B) }9 \qquad \textbf{(C) }11 \qquad \textbf{(D) }13\qquad \textbf{(E) }15$
|
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)\\ (h^2+2h+1)A+3B&=1, &(2)\\ (h^2+4h+4)A+2B&=1. &(3) \end{align*} Subtracting $(1)$ from $(2)$ and $(3)$ , we get \[4hA+3B=0\quad\text{and}\quad 3A-6hA+3B=0.\] Plugging the former into the latter and simplifying yields $6A=5B$ . Hence $a^2:b^2=6:5$ . Since $c^2=a^2-b^2$ , we get $a^2=6c^2$ , so the eccentricity is $\frac ca=\sqrt{\frac16}$
Therefore, the answer is $1+6=\boxed{7}$
| 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 $\mathcal E$ can be written in the form $\sqrt{\frac mn}$ , where $m$ and $n$ are relatively prime positive integers. What is $m+n$ ? (Recall that the eccentricity of an ellipse $\mathcal E$ is the ratio $\frac ca$ , where $2a$ is the length of the major axis of $\mathcal E$ and $2c$ is the is the distance between its two foci.)
$\textbf{(A) }7 \qquad \textbf{(B) }9 \qquad \textbf{(C) }11 \qquad \textbf{(D) }13\qquad \textbf{(E) }15$
|
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{align*} from which $a:b:c=5:6:9.$
So, the conic can be written in the form $5x^2+6y^2+9x=14.$ If it is written in the form of $\frac{(x-m)^2}{r^2}+\frac{y^2}{s^2}=1,$ then $r^2:s^2=6:5.$
Therefore, the desired eccentricity is $\sqrt{\frac{\sqrt{6-5}}{6}}=\sqrt{\frac{1}{6}},$ and the answer is $1+6=\boxed{7}.$
| 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 $\mathcal E$ can be written in the form $\sqrt{\frac mn}$ , where $m$ and $n$ are relatively prime positive integers. What is $m+n$ ? (Recall that the eccentricity of an ellipse $\mathcal E$ is the ratio $\frac ca$ , where $2a$ is the length of the major axis of $\mathcal E$ and $2c$ is the is the distance between its two foci.)
$\textbf{(A) }7 \qquad \textbf{(B) }9 \qquad \textbf{(C) }11 \qquad \textbf{(D) }13\qquad \textbf{(E) }15$
|
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 lengths of the major and minor axes, respectively, then the ellipse will be scaled by a factor of $r := \frac1a$ in the $x$ -dimension and $s := \frac1b$ in the $y$ -dimension.
The transformation then sends the points $\left(-1,\pm\sqrt 3\right)$ and $\left(-2,\pm\sqrt 2\right)$ to the points $\left(1-2r, \pm s\sqrt 3\right)$ and $\left(1-3r, \pm s\sqrt 2\right)$ , respectively. These points are on the unit circle, so \[(1-2r)^2 + 3s^2 = 1 \quad \text{and} \quad (1-3r)^2 + 2s^2 = 1.\] This yields \[4r^2 + 3s^2 = 4r \quad \text{and} \quad 9r^2 + 2s^2 = 6r,\] from which \begin{align*} 12r^2 + 9s^2 &= 18r^2 + 4s^2 \\ \frac{r^2}{s^2} &= \frac56. \end{align*} Recalling that $r = \frac1a$ and $s = \frac1b$ , this implies $\frac{b^2}{a^2} = \frac56$ . From this, we get \[\frac{c^2}{a^2} = \frac{a^2-b^2}{a^2} = 1 - \frac{b^2}{a^2} = \frac{1}{6},\] so $\frac ca = \sqrt{\frac16}$ , giving an answer of $1 + 6 = \boxed{7}$
| 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 prime positive integers, and $b$ is a positive integer not divisible by the square of any prime. What is $a+b+c$
$\textbf{(A) }110 \qquad \textbf{(B) }114 \qquad \textbf{(C) }118 \qquad \textbf{(D) }122\qquad \textbf{(E) }126$
|
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, $\triangle QRX$ is a $30^\circ\text{-}30^\circ\text{-}120^\circ$ triangle with base $QR=3\sqrt3.$ Dividing $\triangle QRX$ into two congruent $30^\circ\text{-}60^\circ\text{-}90^\circ$ triangles, we get that the radius of $\odot X$ is $XQ=XR=3$ by the side-length ratios.
Let $M$ be the midpoint of $\overline{QR}.$ By the Perpendicular Chord Bisector Converse, we have $\overline{OM}\perp\overline{QR}$ and $\overline{XM}\perp\overline{QR}.$ Together, points $O, X,$ and $M$ must be collinear.
By the SAS Congruence, we have $\triangle QXM\cong\triangle RXM,$ both of which are $30^\circ\text{-}60^\circ\text{-}90^\circ$ triangles. By the side-length ratios, we obtain $RM=\frac{3\sqrt3}{2}, RX=3,$ and $XM=\frac{3}{2}.$ By the Pythagorean Theorem on right $\triangle ORM,$ we get $OM=\frac{13}{2}$ and $OX=OM-XM=5.$ By the Pythagorean Theorem on right $\triangle OXP,$ we get $OP=4.$
Let $C$ be the foot of the perpendicular from $P$ to $\overline{QR},$ and $D$ be the foot of the perpendicular from $X$ to $\overline{PC},$ as shown below: [asy] /* Made by MRENTHUSIASM */ size(300); pair O, X, A, B, P, Q, R, M, C, D; O = (0,0); X = (4,3); A = (-7,0); B = (7,0); P = (4,0); Q = intersectionpoints(Circle(O,7),Circle(X,3))[0]; R = intersectionpoints(Circle(O,7),Circle(X,3))[1]; M = midpoint(Q--R); C = foot(P,Q,R); D = foot(X,P,C); fill(P--Q--R--cycle,yellow); dot("$O$",O,S); dot("$X$",X,N); dot("$A$",A,SW); dot("$B$",B,SE); dot("$P$",P,S); dot("$Q$",Q,E); dot("$R$",R,N); dot("$M$",M,dir(M)); dot("$C$",C,NE); dot("$D$",D,SE); markscalefactor=0.0375; draw(rightanglemark(O,M,R),red); draw(rightanglemark(P,C,M),red); draw(rightanglemark(P,D,X),red); draw(rightanglemark(O,P,X),red); draw(P--Q--R--cycle); draw(arc(O, 7, 0, 180)^^A--B^^Circle(X,3)); draw(O--M^^X--P); draw(P--C^^X--D,dashed); [/asy] Clearly, quadrilateral $XDCM$ is a rectangle. Since $\angle XPD=\angle OXP$ by alternate interior angles, we have $\triangle XPD\sim\triangle OXP$ by the AA Similarity, with the ratio of similitude $\frac{XP}{OX}=\frac 35.$ Therefore, we get $PD=\frac 95$ and $PC=PD+DC=PD+XM=\frac 95 + \frac 32 = \frac{33}{10}.$
The area of $\triangle PQR$ is \[\frac12\cdot QR\cdot PC=\frac12\cdot3\sqrt3\cdot\frac{33}{10}=\frac{99\sqrt3}{20},\] from which the answer is $99+3+20=\boxed{122}.$
| 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 prime positive integers, and $b$ is a positive integer not divisible by the square of any prime. What is $a+b+c$
$\textbf{(A) }110 \qquad \textbf{(B) }114 \qquad \textbf{(C) }118 \qquad \textbf{(D) }122\qquad \textbf{(E) }126$
|
[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 = intersectionpoints(circle(C, 7), circle(O, 3))[0]; draw(C--O); draw(C--Q); draw(C--R); draw(Q--R); draw(O--P); draw(O--Q); draw(O--R); draw(P--Q); draw(P--R); label("$Q$", Q, N); label("$R$", R, E); [/asy]
Suppose we label the points as shown in the diagram above, where $C$ is the center of the semicircle and $O$ is the center of the circle tangent to $\overline{AB}$ . Since $\angle QPR = 60^{\circ}$ , we have $\angle QOR = 2\cdot 60^{\circ}=120^{\circ}$ and $\triangle QOR$ is a $30-30-120$ triangle, which can be split into two $30-60-90$ triangles by the altitude from $O$ . Since $QR=3\sqrt{3},$ we know $OQ=OR=\tfrac{3\sqrt{3}}{\sqrt{3}}=3$ by $30-60-90$ triangles. The area of this part of $\triangle PQR$ is $\frac{1}{2}bh=\tfrac{3\sqrt{3}}{2}\cdot\tfrac{3}{2}=\tfrac{9\sqrt{3}}{4}$ . We would like to add this value to the sum of the areas of the other two parts of $\triangle PQR$
To find the areas of the other two parts of $\triangle PQR$ using the $\sin$ area formula, we need the sides and included angles. Here we know the sides but what we don't know are the angles. So it seems like we will have to use an angle from another triangle and combine them with the angles we already know to find these angles easily. We know that $\angle QOR = 120^{\circ}$ and triangles $\triangle COQ$ and $\triangle COR$ are congruent as they share a side, $CQ=CR,$ and $OQ=OR$ . Therefore $\angle COQ = \angle COR = 120^{\circ}$ . Suppose $CO=x$ . Then $3^{2}+x^{2}-6x\cos{120^{\circ}}=7^{2}$ , and since $\cos{120^{\circ}}=-\tfrac{1}{2}$ , this simplifies to $x^{2}+3x=7^{2}-3^{2}\rightarrow x^{2}+3x-40=0$ . This factors nicely as $(x-5)(x+8)=0$ , so $x=5$ as $x$ can't be $-8$ . Since $CO=5, OP=3$ and $\angle OPC=90^{\circ}$ , we now know that $\triangle OPC$ is a $3-4-5$ right triangle. This may be useful info for later as we might use an angle in this triangle to find the areas of the other two parts of $\triangle PQR$
Let $\angle POC = \alpha$ . Then $\sin\alpha = \tfrac{4}{5}, \cos\alpha = \tfrac{3}{5}, \angle QOP = 120+\alpha,$ and $\angle POR = 120-\alpha$ . The sum of the areas of $\triangle QOP$ and $\triangle POR$ is $3\cdot 3\cdot\tfrac{1}{2}\cdot\left[\sin(120-\alpha)+\sin(120+\alpha)\right]=\tfrac{9}{2}\left[\sin(120-\alpha)+\sin(120+\alpha)\right],$ which we will add to $\tfrac{9\sqrt{3}}{4}$ to get the area of $\triangle PQR$ . Observe that \[\sin(120-\alpha) = \sin 120\cos\alpha-\sin\alpha\cos 120 = \tfrac{\sqrt{3}}{2}\cdot\tfrac{3}{5}-\tfrac{4}{5}\cdot\tfrac{-1}{2}=\tfrac{3\sqrt{3}}{10}+\tfrac{4}{10}=\tfrac{3\sqrt{3}+4}{10}\] and similarly $\sin(120+\alpha)=\tfrac{3\sqrt{3}-4}{10}$ . Adding these two gives $\tfrac{3\sqrt{3}}{5}$ and multiplying that by $\tfrac{9}{2}$ gets us $\tfrac{27\sqrt{3}}{10},$ which we add to $\tfrac{9\sqrt{3}}{4}$ to get $\tfrac{54\sqrt{3}+45\sqrt{3}}{20}=\tfrac{99\sqrt{3}}{20}$ . The answer is $99+3+20=102+20=\boxed{122}.$
| 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 prime positive integers, and $b$ is a positive integer not divisible by the square of any prime. What is $a+b+c$
$\textbf{(A) }110 \qquad \textbf{(B) }114 \qquad \textbf{(C) }118 \qquad \textbf{(D) }122\qquad \textbf{(E) }126$
|
[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), circle(O, 3))[1], R = intersectionpoints(circle(C, 7), circle(O, 3))[0], S = (Q+R)/2, N = (121/8, 0), T = (8/11)*N + (3/11)*R, X = (4/7)*T + (3/7)*S; draw(C--O, blue); draw(O--S, red); draw(C--Q); draw(C--R); draw(Q--N--B); draw(O--P); draw(O--Q); draw(O--R); draw(P--Q--R--cycle); draw(B--T); draw(P--X); label("$Q$", Q, NNE); label("$R$", R, E); label("$S$", S, ENE); label("$N$", N, SSE); label("$T$", T, ENE); label("$X$", X, NE); draw(rightanglemark(P, X, Q)); draw(rightanglemark(B, T, R)); draw(rightanglemark(C, S, Q)); [/asy] Define points as shown above, where $N=\overleftrightarrow{PA}\cap\overleftrightarrow{QR}$ . The area of $\triangle PQR$ is simply \[\dfrac{1}{2}PX\cdot QR=\dfrac{3\sqrt{3}}{2}PX;\] it remains to compute the value of $PX$ . Note that $PX$ is simply a weighted average of $BT$ and $CS;$ it is $\dfrac{CP}{BP}$ times closer to $BT$ than it is to $CS$ . Observe that \[CS=\sqrt{CQ^{2}-\left(\dfrac{1}{2}QR\right)^{2}}=\sqrt{7^{2}-\left(\dfrac{3\sqrt{3}}{2}\right)^{2}}=6.5\] since the radius of $\Gamma$ is $7$ as its diameter is $14$ . Note also by the Extended Law of Sines the radius of $\Omega$ is $\dfrac{3\sqrt{3}}{2\sin 60^{\circ}}=3,$ so $OS=3\cos 60^{\circ}=1.5$ . Since $C, O,$ and $S$ are collinear by symmetry we have $CO=CS-OS=5,$ so $CP=\sqrt{5^{2}-3^{2}}=4$ and $BP=7-4=3$ . Therefore, $\triangle OPC$ is a $3\text{-}4\text{-}5$ right triangle; $\triangle OPC\sim\triangle NSC$ since $\angle OPC=\angle CSN=90^{\circ}$ and $\angle OCP=\angle NCS=\sin^{-1}\left(\dfrac{3}{5}\right)$ . Therefore $\dfrac{CN}{CS}=\cfrac{CO}{CP}=\dfrac{5}{4}$ so $CN=\dfrac{5}{4}CS=\dfrac{65}{8}$ . Since $\triangle BTN\sim\triangle CSN,$ we have $\dfrac{BT}{BN}=\dfrac{CS}{CN}=\dfrac{4}{5}$ . Therefore \[BT=\dfrac{4}{5}BN=\dfrac{4}{5}\left(CN-7\right)=\dfrac{4}{5}\cdot\dfrac{9}{8}=\dfrac{36}{40}=0.9;\] so $PX$ is $\dfrac{4}{3}$ times as close to $0.9$ as to $6.5;$ we can compute $PX=\dfrac{4}{7}BT+\dfrac{3}{7}CS=\dfrac{4}{7}\cdot0.9+\dfrac{3}{7}\cdot6.5=3.3$ . The area of $\triangle PQR$ is \[\dfrac{3\sqrt{3}}{2}\cdot 3.3=\dfrac{99\sqrt{3}}{20}\] and $99+3+20=\boxed{122}$
| 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 prime positive integers, and $b$ is a positive integer not divisible by the square of any prime. What is $a+b+c$
$\textbf{(A) }110 \qquad \textbf{(B) }114 \qquad \textbf{(C) }118 \qquad \textbf{(D) }122\qquad \textbf{(E) }126$
|
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 $O_{2}M=3\cos 60^\circ=\frac{3}{2}.$ Therefore $O_{1}O_{2}=O_{1}M-O_{2}M=5$ and $O_{1}P=\sqrt{5^2 - 3^2}=4.$
Let $X=\overline{AB}\cap\overline{QR}.$ Obviously \[\angle O_{1}PO_{2}=\angle O_{1}MX=90^\circ~\text{and}~\angle PO_{1}O_{2}=\angle MO_{1}X=\arcsin\left(\frac{3}{5}\right)\] so $\triangle PO_{1}O_{2}\sim\triangle MO_{1}X$ with ratio $\frac{PO_{1}}{MO_{1}}=\frac{4}{\tfrac{13}{2}}=\frac{8}{13}.$ Therefore $O_1X=\frac{13}{8}\cdot O_{1}O_{2}=\frac{13}{8}\cdot5=\frac{65}{8}$ and $MX=\frac{13}{8}\cdot PO_{2}=\frac{13}{8}\cdot3=\frac{39}{8}.$
Let $H$ denote the foot of the altitude from $P$ to $\overline{QR}.$ Because $\overline{PH}\parallel\overline{O_{1}M},$ it follows that $\triangle PHX\sim\triangle O_{1}MX.$ This similarity has ratio \[\frac{PX}{O_{1}X}=1-\frac{O_{1}P}{O_{1}X}=1-\frac{4}{\tfrac{65}{8}}=1-\frac{32}{65}=\frac{33}{65}.\] We therefore have $PH=\frac{33}{65}\cdot O_{1}M=\frac{33}{65}\cdot\frac{13}{2}=\frac{33}{10}.$
Finally, the area of $\triangle PQR$ is \[\frac{1}{2}\cdot QR\cdot PH=\frac{1}{2}\cdot3\sqrt3\cdot\frac{33}{10}=\frac{99\sqrt3}{20},\] so the answer is $99+3+20=\boxed{122}.$
| 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 prime positive integers, and $b$ is a positive integer not divisible by the square of any prime. What is $a+b+c$
$\textbf{(A) }110 \qquad \textbf{(B) }114 \qquad \textbf{(C) }118 \qquad \textbf{(D) }122\qquad \textbf{(E) }126$
|
By the Law of Sine in $\triangle PQR$ and its circumcircle $\odot \Omega$ $2r_{\Omega} = \frac{QR}{ \sin 60^{\circ} } = \frac{ 3\sqrt{3} }{ \frac{ \sqrt{3} }{2} } = 6$ $r_{\Omega} = 3$
\[\Gamma \Omega = \sqrt{r_{\Gamma}^2 - \left( \frac{ PQ }{2}\right)^2} - \sqrt{r_{\Omega} - \left( \frac{PQ}{2}\right)^2} = \sqrt{7^2 - \left( \frac{ 3 \sqrt{3} }{2}\right)^2} - \sqrt{3^2 - \left( \frac{ 3 \sqrt{3} }{2}\right)^2} = \frac{13}{2} - \frac32 = 5, \quad \Gamma P = \sqrt{5^2 - 3^2} = 4\]
By Power of a Point in $\odot \Gamma$ $PQ \cdot PS = PA \cdot PB = (7+4)(7-4) = 33$
By the Law of Sine in $\triangle PRS$ $\frac{PR}{PS} = \frac{ \sin \angle PSR }{ \sin \angle PRS }$
By the Law of Sine in $\triangle QRS$ and its circumcircle $\odot \Gamma$ $\frac{QR}{ \sin \angle PSR } = 14$ $\frac{ 3\sqrt{3} }{ \sin \angle PSR } = 14$ $\sin \angle PSR = \frac{ 3\sqrt{3} }{14}$ $\cos \angle PSR = \frac{ 13 }{14}$
\[\sin \angle PRS = \sin ( 60^{\circ} - \angle PSR ) = \sin 60^{\circ} \cos \angle PSR - \sin \angle PSR \cos 60^{\circ} = \frac{ \sqrt{3} }{2} \cdot \frac{ 13 }{14} - \frac{ 3\sqrt{3} }{14} \cdot \frac12 = \frac{ 5\sqrt{3} }{14}\]
\[\frac{PR}{PS} = \frac{ \frac{ 3\sqrt{3} }{14} }{ \frac{ 5\sqrt{3} }{14} } = \frac35, \quad PQ \cdot PR = PQ \cdot PS \cdot \frac{PR}{PS} = 33 \cdot \frac35 = \frac{99}{5}\]
\[[PQR] = \frac12 \cdot \sin 60^{\circ} \cdot PQ \cdot PR = \frac12 \cdot \frac12 \cdot \frac{99}{5} = \frac{ 99\sqrt{3} }{20}, \quad 99 + 3 + 20 = \boxed{122}\]
| 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 prime positive integers, and $b$ is a positive integer not divisible by the square of any prime. What is $a+b+c$
$\textbf{(A) }110 \qquad \textbf{(B) }114 \qquad \textbf{(C) }118 \qquad \textbf{(D) }122\qquad \textbf{(E) }126$
|
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 $y$ value, in other words, (3) is the linear equation that go through $R$ and $Q$ .
Let the height drops from $P$ to $QR$ at $H$ . Therefore, we have \[Area \triangle PQR =\frac{1}{2}\cdot{QR}\cdot{PH}\] So \[QR=3\sqrt{3}\] And by distance formula, $PH$ is the distance from $P$ (4,0) to $\overline{QR}$ \[PH={\frac{|8\cdot4+6\cdot0-65|}{\sqrt{8^{2}+6^{2}}}=\frac{33}{10}}\] Thus, We get \[Area \triangle PQR =\frac{1}{2}\cdot3\sqrt{3}\cdot\frac{33}{10}=\frac{99\sqrt{3}}{20}\] So the answer is $99+3+20=\boxed{122}.$
| 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 $n\ne N$ . What is the sum of the digits of $N?$
$\textbf{(A) }5 \qquad \textbf{(B) }6 \qquad \textbf{(C) }7 \qquad \textbf{(D) }8\qquad \textbf{(E) }9$
|
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 $f(n)>0$ for all positive integers $n,$ note that $f(a)>f(b)$ if and only if $f(a)^3>f(b)^3$ for all positive integers $a$ and $b.$ So, $f(n)$ is maximized if and only if \[f(n)^3=\prod_{i=1}^{k}\frac{(e_i+1)^3}{p_i^{{e_i}}}\] is maximized.
For each independent factor $\frac{(e_i+1)^3}{p_i^{e_i}}$ with a fixed prime $p_i,$ where $1\leq i\leq k,$ the denominator grows faster than the numerator, as exponential functions always grow faster than polynomial functions. Therefore, for each prime $p_i$ with $\left(p_1,p_2,p_3,p_4,\ldots\right)=\left(2,3,5,7,\ldots\right),$ we look for the nonnegative integer $e_i$ such that $\frac{(e_i+1)^3}{p_i^{e_i}}$ is a relative maximum: \[\begin{array}{c|c|c|c|c} & & & & \\ [-2.25ex] \boldsymbol{i} & \boldsymbol{p_i} & \boldsymbol{e_i} & \boldsymbol{\dfrac{(e_i+1)^3}{p_i^{e_i}}} & \textbf{Max?} \\ [2.5ex] \hline\hline & & & & \\ [-2ex] 1 & 2 & 0 & 1 & \\ & & 1 & 4 & \\ & & 2 & 27/4 &\\ & & 3 & 8 & \checkmark\\ & & 4 & 125/16 & \\ [0.5ex] \hline & & & & \\ [-2ex] 2 & 3 & 0 & 1 &\\ & & 1 & 8/3 & \\ & & 2 & 3 & \checkmark\\ & & 3 & 64/27 & \\ [0.5ex] \hline & & & & \\ [-2ex] 3 & 5 & 0 & 1 & \\ & & 1 & 8/5 & \checkmark\\ & & 2 & 27/25 & \\ [0.5ex] \hline & & & & \\ [-2ex] 4 & 7 & 0 & 1 & \\ & & 1 & 8/7 & \checkmark\\ & & 2 & 27/49 & \\ [0.5ex] \hline & & & & \\ [-2ex] \geq5 & \geq11 & 0 & 1 & \checkmark \\ & & \geq1 & \leq8/11 & \\ [0.5ex] \end{array}\] Finally, the positive integer we seek is $N=2^3\cdot3^2\cdot5^1\cdot7^1=2520.$ The sum of its digits is $2+5+2+0=\boxed{9}.$
| 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 $n\ne N$ . What is the sum of the digits of $N?$
$\textbf{(A) }5 \qquad \textbf{(B) }6 \qquad \textbf{(C) }7 \qquad \textbf{(D) }8\qquad \textbf{(E) }9$
|
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 multiply by a factor of $a+1$ ; this is because the number $2^{a}$ has $a+1$ divisors. The denominator, $\sqrt[3]{N}$ , will simply multiply by $\sqrt[3]{2^{a}}$ . Therefore, the entire function multiplies by a factor of $\frac{a+1}{\sqrt[3]{2^{a}}}$ . We want to find the integer value of $a$ that maximizes this value. By inspection, this is $3$ . Therefore, we multiply $N$ by $8$ ; right now, $N$ is $8$
Next, we will find what power of $3$ to multiply $N$ by. Similar to the previous step, we wish to find the integer value of $a$ that maximizes $\frac{a+1}{\sqrt[3]{3^{a}}}$ . This value, also by inspection, is $2$
We can repeat this step on the rest of the primes to get \[N = 2^{3} \cdot 3^{2} \cdot 5 \cdot 7\] but from $11$ on, $a=0$ will maximize the value of the function, so the prime is not a factor in $N$ .
We evaluate $N$ to be $2520$ , so the answer is $\boxed{9}$
| 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 $n\ne N$ . What is the sum of the digits of $N?$
$\textbf{(A) }5 \qquad \textbf{(B) }6 \qquad \textbf{(C) }7 \qquad \textbf{(D) }8\qquad \textbf{(E) }9$
|
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 $d(9n) = 3d(n)$ . Then \begin{align*} f(3n) &= \frac{2d(n)}{\sqrt[3]{3n}} \approx 1.38 f(n)\\ f(9n) &= \frac{3d(n)}{\sqrt[3]{9n}} \approx 1.44 f(n) \end{align*} Note that the values $\frac{2}{\sqrt[3]{3}}$ and $\frac{3}{\sqrt[3]{9}}$ do not have to be explicitly computed; we only need the fact that $\frac{3}{\sqrt[3]{9}} > \frac{2}{\sqrt[3]{3}} > 1$ which is easy to show by hand.
The above claim automatically implies $N$ is a multiple of 9: if $N$ was not divisible by 9, then $f(9N) > f(N)$ which is a contradiction, and if $N$ was divisible by 3 and not 9, then $f(3N) > f(N) > f\left(\frac{N}{3}\right)$ , also a contradiction. Then the sum of digits of $N$ must be a multiple of 9, so only choice $\boxed{9}$ works.
| 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 $n\ne N$ . What is the sum of the digits of $N?$
$\textbf{(A) }5 \qquad \textbf{(B) }6 \qquad \textbf{(C) }7 \qquad \textbf{(D) }8\qquad \textbf{(E) }9$
|
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 answer is $\boxed{19}$
| 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 $9$ . There are a total of \[9-(-9) + 1 = \boxed{19}.\] PureSwag
| 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 zero is neither negative nor positive, but we still need to add it into the solution. Hence, the answer is $9+9+1=18+1=\boxed{19}.$
| 19
|
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