AMC12 2021 B

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 $\boxed{\textbf{(D)} ~19}$.

There are $46$ blue students paired with a blue partner. The other $11$ students wearing blue shirts must each be paired with a partner wearing a shirt of the opposite color. There are $64$ yellow students remaining. Therefore the requested number of pairs is $\tfrac{64}{2}=\boxed{\textbf{(B)} ~32}$

Subtracting $2$ from both sides and taking reciprocals gives $1+\frac{1}{2+\frac{2}{3+x}}=\frac{53}{38}$. Subtracting $1$ from both sides and taking reciprocals again gives $2+\frac{2}{3+x}=\frac{38}{15}$. Subtracting $2$ from both sides and taking reciprocals for the final time gives $\frac{x+3}{2}=\frac{15}{8}$ or $x=\frac{3}{4} \implies \boxed{\text{(A) } 3/4}$.

Let there be $3x$ students in the morning class and $4x$ students in the afternoon class. The total number of students is $3x + 4x = 7x$. The average is $\frac{3x\cdot84 + 4x\cdot70}{7x}=76$. Therefore, the answer is $\boxed{\textbf{(C)} ~76}$.

The final image of $P$ is $(-6,3)$. We know the reflection rule for reflecting over $y=-x$ is $(x,y) \rightarrow (-y, -x)$. So before the reflection and after rotation the point is $(-3,6)$. By definition of rotation, the slope between $(-3,6)$ and $(1,5)$ must be perpendicular to the slope between $(a,b)$ and $(1,5)$. The first slope is $\frac{5-6}{1-(-3)} = \frac{-1}{4}$. This means the slope of $P$ and $(1,5)$ is $4$. Rotations also preserve distance to the center of rotation, and since we only "travelled" up and down by the slope once to get from $(-3,6)$ to $(1,5)$ it follows we shall only use the slope once to travel from $(1,5)$ to $P$. Therefore point $P$ is located at $(1+1, 5+4) = (2,9)$. The answer is $9-2 = 7 = \boxed{\textbf{(D)} ~7}$.

The volume of a cone is $\frac{1}{3} \cdot\pi \cdot r^2 \cdot h$ where $r$ is the base radius and $h$ is the height. The water completely fills up the cone so the volume of the water is $\frac{1}{3}\cdot18\cdot144\pi = 6\cdot144\pi$. The volume of a cylinder is $\pi \cdot r^2 \cdot h$ so the volume of the water in the cylinder would be $24\cdot24\cdot\pi\cdot h$. We can equate these two expressions because the water volume stays the same like this $24\cdot24\cdot\pi\cdot h = 6\cdot144\pi$. We get $4h = 6$ and $h=\frac{6}{4}$. So the answer is $\boxed{\textbf{(A)} ~1.5}.$

Prime factorize $N$ to get $N=2^{3} \cdot 3^{5} \cdot 5\cdot 7\cdot 17^{2}$. For each odd divisor $n$ of $N$, there exist even divisors $2n, 4n, 8n$ of $N$, therefore the ratio is $1:(2+4+8)=\boxed{\textbf{(C)} ~1 : 14}$

Since two parallel chords have the same length ($38$), they must be equidistant from the center of the circle. Let the perpendicular distance of each chord from the center of the circle be $d$. Thus, the distance from the center of the circle to the chord of length $34$ is \[2d + d = 3d\] The distance between each of the chords is $2d$. Let the radius of the circle be $r$. Drawing radii to the points where the lines intersect the circle, we obtain two different right triangles: - One with base $\dfrac{38}{2} = 19$, height $d$, and hypotenuse $r$ ($\triangle RAO$ in the diagram) - Another with base $\dfrac{34}{2} = 17$, height $3d$, and hypotenuse $r$ ($\triangle LBO$ in the diagram) By the Pythagorean theorem, we obtain the following system of equations: \begin{align*} 19^2 + d^2 &= r^2, \\ 17^2 + (3d)^2 &= r^2. \end{align*} That is, \begin{align*} 361 + d^2 &= r^2, \\ 289 + 9d^2 &= r^2. \end{align*} Set the right-hand sides equal: \[ 361 + d^2 = 289 + 9d^2 \] \[ 361 - 289 = 9d^2 - d^2 \] \[ 72 = 8d^2 \] \[ d^2 = 9 \implies d = 3. \] Thus, \[ 2d = 6. \]

\[\frac{\log_{2}{80}}{\log_{40}{2}}-\frac{\log_{2}{160}}{\log_{20}{2}}\] Note that $\log_{40}{2}=\frac{1}{\log_{2}{40}}$, and similarly $\log_{20}{2}=\frac{1}{\log_{2}{20}}$ \[= \log_{2}{80}\cdot \log_{2}{40}-\log_{2}{160}\cdot \log_{2}{20}\] \[=(\log_{2}{4}+\log_{2}{20})(\log_{2}{2}+\log_{2}{20})-(\log_{2}{8}+\log_{2}{20})\log_{2}{20}\] \[=(2+\log_{2}{20})(1+\log_{2}{20})-(3+\log_{2}{20})\log_{2}{20}\] Expanding, \[2+2\log_{2}{20}+\log_{2}{20}+(\log_{2}{20})^2-3\log_{2}{20}-(\log_{2}{20})^2\] All the log terms cancel, so the answer is $2\implies\boxed{\text{(D)}}$.

The sum of the first $n$ integers is given by $\frac{n(n+1)}{2}$, so $\frac{37(37+1)}{2}=703$. Therefore, $703-x-y=xy$ Rearranging, $xy+x+y=703$. We can factor this equation by SFFT to get $(x+1)(y+1)=704$ Looking at the possible divisors of $704 = 2^6\cdot11$, $22$ and $32$ are within the constraints of $0 < x \leq y \leq 37$ so we try those: $(x+1)(y+1) = 22\cdot32$ $x+1=22, y+1 = 32$ $x = 21, y = 31$ Therefore, the difference $y-x=31-21=\boxed{\textbf{(E) }10}$.

Toss on the Cartesian plane with $A=(5, 12), B=(0, 0),$ and $C=(14, 0)$. Then $\overline{AD}\parallel\overline{BC}, \overline{AB}\parallel\overline{CE}$ by the trapezoid condition, where $D, E\in\overline{BP}$. Since $PC=10$, point $P$ is $\tfrac{10}{15}=\tfrac{2}{3}$ of the way from $C$ to $A$ and is located at $(8, 8)$. Thus line $BP$ has equation $y=x$. Since $\overline{AD}\parallel\overline{BC}$ and $\overline{BC}$ is parallel to the ground, we know $D$ has the same $y$-coordinate as $A$, except it'll also lie on the line $y=x$. Therefore, $D=(12, 12). \, \blacksquare$ To find the location of point $E$, we need to find the intersection of $y=x$ with a line parallel to $\overline{AB}$ passing through $C$. The slope of this line is the same as the slope of $\overline{AB}$, or $\tfrac{12}{5}$, and has equation $y=\tfrac{12}{5}x-\tfrac{168}{5}$. The intersection of this line with $y=x$ is $(24, 24)$. Therefore point $E$ is located at $(24, 24). \, \blacksquare$ The distance $DE$ is equal to the distance between $(12, 12)$ and $(24, 24)$, which is $\boxed{\textbf{(D) }12\sqrt2}$.

We can then say that A+S(n)n+1=32, S(n)n=35, and B+S(n)n+1=40. Expanding gives us A+S(n)=32n+32, S(n)=35n, and B+S(n)=40n+40. Substituting S(n)=35n to all gives us A+35n=32n+32 and B+35n=40n+40. Solving for A and B gives A=−3n+32 and B=5n+40. We now need to find S(n)+A+Bn+2. We substitute everything to get 35n+(−3n+32)+(5n+40)n+2, or 37n+72n+2. Say that the answer to this is Z. Then, Z needs to be a number that makes n a positive integer. The only options that work is $\boxed{\textbf{(C) }36.6}$ and $\boxed{\textbf{(D) }36.8}$. However, if 36.6 is an option, we get n=3. So that means that A is 23 and B is 55, and S(n)=105. But if there is 3 terms, then the middle number is 105, but we said that B is the largest number in the set, so therefore our answer cannot be $\boxed{\textbf{(C) }36.6}$ and is instead $\boxed{\textbf{(D) }36.8}$ and now, we're finished! If A is smaller than B by 72 therefore from the equation on the top you can find out that N=8 using substitution the plug it in to the equation 37n+72n+2 then you will get that Z = $\boxed{\textbf{(D) }36.8}$.

We rearrange to get \[5\cos3\theta = 3\sin\theta-1.\] We can graph two functions in this case: $y=5\cos{3x}$ and $y=3\sin{x} -1$. Using transformation of functions, we know that $5\cos{3x}$ is just a cosine function with amplitude $5$ and period $\frac{2\pi}{3}$. Similarly, $3\sin{x} -1$ is just a sine function with amplitude $3$ and shifted $1$ unit downward: So, we have $\boxed{\textbf{(D) }6}$ solutions.

Let $MA=a$ and $MD=d.$ It follows that $MC=a+2$ and $MB=a+4.$ As shown below, note that $\triangle MAD$ and $\triangle MBC$ are both right triangles. By the Pythagorean Theorem, we have \begin{alignat*}{6} AD^2 &= MA^2 - MD^2 &&= a^2 - d^2, \\ BC^2 &= MB^2 - MC^2 &&= (a+4)^2 - (a+2)^2. \end{alignat*} Since $AD=BC$ in rectangle $ABCD,$ we equate the expressions for $AD^2$ and $BC^2,$ then rearrange and factor: \begin{align*} a^2 - d^2 &= (a+4)^2 - (a+2)^2 \\ a^2 - d^2 &= 4a + 12 \\ a^2 - 4a - d^2 &= 12 \\ (a-2)^2 - d^2 &= 16 \\ (a+d-2)(a-d-2) &= 16. \end{align*} As $a+d-2$ and $a-d-2$ have the same parity, we get $a+d-2=8$ and $a-d-2=2,$ from which $(a,d)=(7,3).$ Applying the Pythagorean Theorem to right $\triangle MAD$ and right $\triangle MCD,$ we obtain $AD=2\sqrt{10}$ and $CD=6\sqrt2,$ respectively. Let the brackets denote areas. Together, the volume of pyramid $MABCD$ is \[\frac13\cdot [ABCD]\cdot MD = \frac13\cdot (AD\cdot CD)\cdot MD = \boxed{\textbf{(A) }24\sqrt5}.\]

Draw diagonals $AC$ and $AD$ to split the pentagon into three parts. We can compute the area for each triangle and sum them up at the end. For triangles $ABC$ and $ADE$, they each have area $2\cdot\frac{1}{2}\cdot\frac{4\sqrt{3}}{4}=\sqrt{3}$. For triangle $ACD$, we can see that $AC=AD=2\sqrt{3}$ and $CD=2$. Using Pythagorean Theorem, the altitude for this triangle is $\sqrt{11}$, so the area is $\sqrt{11}$. Adding each part up, we get $2\sqrt{3}+\sqrt{11}=\sqrt{12}+\sqrt{11} \implies \boxed{\textbf{(D)} ~23}$.

Note that $f(1/x)$ has the same roots as $g(x)$, if it is multiplied by some monomial so that the leading term is $x^3$ they will be equal. We have \[f(1/x) = \frac{1}{x^3} + \frac{a}{x^2}+\frac{b}{x} + c\] so we can see that \[g(x) = \frac{x^3}{c}f(1/x)\] Therefore \[g(1) = \frac{1}{c}f(1) = \boxed{\textbf{(A) }\frac{1+a+b+c}c}\]

Without the loss of generality, let $\mathcal T$ have vertices $A$, $B$, $C$, and $D$, with $AB = r$ and $CD = s$. Also denote by $P$ the point in the interior of $\mathcal T$. Let $X$ and $Y$ be the feet of the perpendiculars from $P$ to $AB$ and $CD$, respectively. Observe that $PX = \tfrac 8r$ and $PY = \tfrac 4s$. Now using the formula for the area of a trapezoid yields \[14 = \frac12\cdot XY\cdot (AB+CD) = \frac12\left(\frac 8r + \frac 4s\right)(r+s) = 6 + 2\cdot\frac rs + 4\cdot\frac sr.\] Thus, the ratio $\rho := \tfrac rs$ satisfies $\rho + 2\rho^{-1} = 4$; solving yields $\rho = \boxed{\textbf{(B)}\: 2+\sqrt{2}}$. (Observe that the given areas of $3$ and $5$ are irrelevant to the ratio $\frac{AB}{CD}$.)

Using the fact $z\bar{z}=|z|^2$, the equation rewrites itself as \begin{align*} 12z\bar{z}&=2(z+2)(\bar{z}+2)+(z^2+1)(\bar{z}^2+1)+31 \\ -12z\bar{z}+2z\bar{z}+4(z+\bar{z})+8+z^2\bar{z}^2+(z^2+\bar{z}^2)+32&=0 \\ \left((z^2+2z\bar{z}+\bar{z}^2)+4(z+\bar{z})+4\right)+\left(z^2\bar{z}^2-12z\bar{z}+36\right)&=0 \\ (z+\bar{z}+2)^2+(z\bar{z}-6)^2&=0. \end{align*} As the two quantities in the parentheses are real, both quantities must equal $0$ so \[z+\frac6z=z+\bar{z}=\boxed{\textbf{(A) }-2}.\]

Suppose the dice have $a$ and $b$ faces, and WLOG $a\geq{b}$. Since each die has at least $6$ faces, there will always be $6$ ways to sum to $7$. As a result, there must be $\tfrac{4}{3}\cdot6=8$ ways to sum to $10$. There are at most nine distinct ways to get a sum of $10$, which are possible whenever $a,b\geq{9}$. To achieve exactly eight ways, $b$ must have $8$ faces, and $a\geq9$. Let $n$ be the number of ways to obtain a sum of $12$, then $\tfrac{n}{8a}=\tfrac{1}{12}\implies n=\tfrac{2}{3}a$. Since $b=8$, $n\leq8\implies a\leq{12}$. In addition to $3\mid{a}$, we only have to test $a=9,12$, of which both work. Taking the smaller one, our answer becomes $a+b=9+8=\boxed{\textbf{(B)}\ 17}$.

Let $z=s$ be a root of $z^2+z+1$ so that $s^2+s+1=0.$ It follows that \[(s-1)\left(s^2+s+1\right)=s^3-1=0,\] from which $s^3=1,$ but $s\neq1.$ Note that \begin{align*} s^{2021}+1 &= s^{3\cdot673+2}+1 \\ &= (s^3)^{673}\cdot s^2+1 \\ &= s^2+1 \\ &= \left(s^2+s+1\right)-s \\ &= -s. \end{align*} Since $z^{2021}+1=-z$ for each root $z=s$ of $z^2+z+1,$ the remainder when $z^{2021}+1$ is divided by $z^2+z+1$ is $R(z)=\boxed{\textbf{(A) }{-}z}.$

Note that \begin{align*} x^{2^{\sqrt{2}}} &= {\sqrt{2}}^{2^x} \\ 2^{\sqrt{2}} \log_2 x &= 2^{x} \log_2 \sqrt{2}. \end{align*} (At this point we see by inspection that $x=\sqrt{2}$ is a solution.) We simplify the RHS, then take the base-$2$ logarithm for both sides: \begin{align*} 2^{\sqrt{2}} \log_2 x &= 2^{x-1} \\ \log_2{\left(2^{\sqrt{2}} \log_2 x\right)} &= x-1 \\ \sqrt{2} + \log_2 \log_2 x &= x-1 \\ \log_2 \log_2 x &= x - 1 - \sqrt{2}. \end{align*} The RHS is a line; the LHS is a concave curve that looks like a logarithm and has $x$ intercept at $(2,0).$ There are at most two solutions, one of which is $\sqrt{2}.$ But note that at $x=2,$ we have $\log_2 \log_2 {2} = 0 > 2 - 1 - \sqrt{2},$ meaning that the log log curve is above the line, so it must intersect the line again at a point $x > 2.$ Now we check $x=4$ and see that $\log_2 \log_2 {4} = 1 < 4 - 1 - \sqrt{2},$ which means at $x=4$ the line is already above the log log curve. Thus, the second solution lies in the interval $(2,4).$ The answer is $\boxed{\textbf{(D) }2\le S<6}.$

We say that a game state is an N-position if it is winning for the next player (the player to move), and a P-position if it is winning for the other player. We are trying to find which of the given states is a P-position. First we note that symmetrical positions are P-positions, as the second player can win by mirroring the first player's moves. It follows that $(6, 1, 1)$ is an N-position, since we can win by moving to $(2, 2, 1, 1)$; this rules out $\textbf{(A)}$. We next look at $(6, 2, 1)$. The possible next states are \[(6, 2), (6, 1, 1), (6, 1), (5, 2, 1), (4, 2, 1, 1),(4, 2, 1), (3, 2, 2, 1), (3, 2, 1, 1), (2, 2, 2, 1).\] None of these are symmetrical, so we might reasonably suspect that they are all N-positions. Indeed, it just so happens that for all of these states except $(6, 2)$ and $(6, 1)$, we can win by moving to $(2, 2, 1, 1)$; it remains to check that $(6, 2)$ and $(6, 1)$ are N-positions. To save ourselves work, it would be nice if we could find a single P-position directly reachable from both $(6, 2)$ and $(6, 1)$. We notice that $(3, 2, 1)$ is directly reachable from both states, so it would suffice to show that $(3, 2, 1)$ is a P-position. Indeed, the possible next states are \[(3, 2), (3, 1, 1), (3, 1), (2, 2, 1), (2, 1, 1, 1), (2, 1, 1),\] which allow for the following refutations: \begin{align*} &(3, 2) \to (2, 2), && &&(3, 1, 1) \to (1, 1, 1, 1), && &&(3, 1) \to (1, 1), \\ &(2, 2, 1) \to (2, 2), && &&(2, 1, 1, 1) \to (1, 1, 1, 1), && &&(2, 1, 1) \to (1, 1). \end{align*} Hence, $(3, 2, 1)$ is a P-position, so $(6, 2)$ and $(6, 1)$ are both N-positions, along with all other possible next states from $(6, 2, 1)$ as noted before. Thus, $(6, 2, 1)$ is a P-position, so our answer is $\boxed{\textbf{(B)}}$. (For completeness, we could also rule out $\textbf{(C)}$, $\textbf{(D)}$ and $\textbf{(E)}$ as in Solution 2.)

"Evenly spaced" just means the bins form an arithmetic sequence. Suppose the middle bin in the sequence is $x$. There are $x-1$ different possibilities for the first bin, and these two bins uniquely determine the final bin. Now, the probability that these $3$ bins are chosen is $6\cdot 2^{-3x} = 6\cdot \frac{1}{8^x}$, so the probability $x$ is the middle bin is $6\cdot\frac{x-1}{8^x}$. Then, we want the sum \begin{align*} 6\sum_{x=2}^{\infty}\frac{x-1}{8^x} &= \frac{6}{8}\left[\frac{1}{8} + \frac{2}{8^2} + \frac{3}{8^3}\cdots\right]\\ &= \frac34\left[\left(\frac{1}{8} + \frac{1}{8^2} + \frac{1}{8^3}+\cdots \right) + \left(\frac{1}{8^2} + \frac{1}{8^3} + \frac{1}{8^4} + \cdots \right) + \cdots\right]\\ &= \frac34\left[\frac17\cdot \left(1 + \frac{1}{8} + \frac{1}{8^2} + \frac{1}{8^3} + \cdots \right)\right]\\ &= \frac34\cdot \frac{8}{49}\\ &= \frac{6}{49} \end{align*} The answer is $6+49=\boxed{\textbf{(A) }55}.$

Let $X$ denote the intersection point of the diagonals $AC$ and $BD$. Remark that by symmetry $X$ is the midpoint of both $\overline{PQ}$ and $\overline{RS}$, so $XP = XQ = 3$ and $XR = XS = 4$. Now note that since $\angle APB = \angle ARB = 90^\circ$, quadrilateral $ARPB$ is cyclic, and so \[XR\cdot XA = XP\cdot XB,\]which implies $\tfrac{XA}{XB} = \tfrac{XP}{XR} = \tfrac34$. Thus let $x> 0$ be such that $XA = 3x$ and $XB = 4x$. Then Pythagorean Theorem on $\triangle APX$ yields $AP = \sqrt{AX^2 - XP^2} = 3\sqrt{x^2-1}$, and so\[[ABCD] = 2[ABD] = AP\cdot BD = 3\sqrt{x^2-1}\cdot 8x = 24x\sqrt{x^2-1}=15\]Solving this for $x^2$ yields $x^2 = \tfrac12 + \tfrac{\sqrt{41}}8$, and so\[(8x)^2 = 64x^2 = 64\left(\tfrac12 + \tfrac{\sqrt{41}}8\right) = 32 + 8\sqrt{41}.\]The requested answer is $32 + 8 + 41 = \boxed{\textbf{(A)} ~81}$.

First, we find a numerical representation for the number of lattice points in $S$ that are under the line $y=mx.$ For any value of $x,$ the highest lattice point under $y=mx$ is $\lfloor mx \rfloor.$ Because every lattice point from $(x, 1)$ to $(x, \lfloor mx \rfloor)$ is under the line, the total number of lattice points under the line is $\sum_{x=1}^{30}(\lfloor mx \rfloor).$ Now, we proceed by finding lower and upper bounds for $m.$ To find the lower bound, we start with an approximation. If $300$ lattice points are below the line, then around $\frac{1}{3}$ of the area formed by $S$ is under the line. By using the formula for a triangle's area, we find that when $x=30, y \approx 20.$ Solving for $m$ assuming that $(30, 20)$ is a point on the line, we get $m = \frac{2}{3}.$ Plugging in $m$ to $\sum_{x=1}^{30}(\lfloor mx \rfloor),$ we get: \[\sum_{x=1}^{30}(\lfloor \frac{2}{3}x \rfloor) = 0 + 1 + 2 + 2 + 3 + \cdots + 18 + 18 + 19 + 20\] We have a repeat every $3$ values (every time $y=\frac{2}{3}x$ goes through a lattice point). Thus, we can use arithmetic sequences to calculate the value above: \[\sum_{x=1}^{30}(\lfloor \frac{2}{3}x \rfloor) = 0 + 1 + 2 + 2 + 3 + \cdots + 18 + 18 + 19 + 20\]\[=\frac{20(21)}{2} + 2 + 4 + 6 + \cdots + 18\]\[=210 + \frac{20}{2}\cdot 9\]\[=300\] This means that $\frac{2}{3}$ is a possible value of $m.$ Furthermore, it is the lower bound for $m.$ This is because $y=\frac{2}{3}x$ goes through many points (such as $(21, 14)$). If $m$ was lower, $y=mx$ would no longer go through some of these points, and there would be less than $300$ lattice points under it. Now, we find an upper bound for $m.$ Imagine increasing $m$ slowly and rotating the line $y=mx,$ starting from the lower bound of $m=\frac{2}{3}.$The upper bound for $m$ occurs when $y=mx$ intersects a lattice point again (look at this problem to get a better idea of what's happening: https://artofproblemsolving.com/wiki/index.php/2011_AMC_10B_Problems/Problem_24). In other words, we are looking for the first $m > \frac{2}{3}$ that is expressible as a ratio of positive integers $\frac{p}{q}$ with $q \le 30.$ For each $q=1,\dots,30$, the smallest multiple of $\frac{1}{q}$ which exceeds $\frac{2}{3}$ is $1, \frac{2}{2}, \frac{3}{3}, \frac{3}{4}, \frac{4}{5}, \cdots , \frac{19}{27}, \frac{19}{28}, \frac{20}{29}, \frac{21}{30}$ respectively, and the smallest of these is $\frac{19}{28}.$ Alternatively, see the method of finding upper bounds in solution 2. The lower bound is $\frac{2}{3}$ and the upper bound is $\frac{19}{28}.$ Their difference is $\frac{1}{84},$ so the answer is $1 + 84 = \boxed{85}.$