AMC12 2022 A

We have \begin{align*} 3+\frac{1}{3+\frac{1}{3+\frac13}} &= 3+\frac{1}{3+\frac{1}{\left(\frac{10}{3}\right)}} \\ &= 3+\frac{1}{3+\frac{3}{10}} \\ &= 3+\frac{1}{\left(\frac{33}{10}\right)} \\ &= 3+\frac{10}{33} \\ &= \boxed{\textbf{(D)}\ \frac{109}{33}}. \end{align*}

Let $x$ be the third number. It follows that the first number is $6x,$ and the second number is $x+40.$ We have \[6x+(x+40)+x=8x+40=96,\] from which $x=7.$ Therefore, the first number is $42,$ and the second number is $47.$ Their absolute value of the difference is $|42-47|=\boxed{\textbf{(E) } 5}.$

The area of this square is equal to $6 + 8 + 30 + 14 + 6 = 64$, and thus its side lengths are $8$. The sum of the dimensions of the rectangles are $2 + 7 + 5 + 6 + 2 + 3 + 1 + 6 + 2 + 4 = 38$. Thus, because the perimeter of the square is $32$, the rectangle on the inside must have a perimeter of $6 \cdot 2 = 12$. The only rectangle that works is $\boxed{\textbf{(B) }B}$.

Note that \begin{align*} 18 &= 2\cdot3^2, \\ 180 &= 2^2\cdot3^2\cdot5, \\ 45 &= 3^2\cdot5 \\ 15 &= 3\cdot5. \end{align*} Let $n = 2^a\cdot3^b\cdot5^c.$ It follows that: 1. From the least common multiple condition, we have \[\operatorname{lcm}(n,18) = \operatorname{lcm}(2^a\cdot3^b\cdot5^c,2\cdot3^2) = 2^{\max(a,1)}\cdot3^{\max(b,2)}\cdot5^{\max(c,0)} = 2^2\cdot3^2\cdot5,\] from which $a=2, b\in\{0,1,2\},$ and $c=1.$ 2. From the greatest common divisor condition, we have \[\gcd(n,45) = \gcd(2^2\cdot3^b\cdot5,3^2\cdot5) = 2^{\min(2,0)}\cdot3^{\min(b,2)}\cdot5^{\min(1,1)} = 3\cdot5,\] from which $b=1.$ Together, we conclude that $n=2^2\cdot3\cdot5=60.$ The sum of its digits is $6+0=\boxed{\textbf{(B) } 6}.$

Let us consider the number of points for a certain $x$-coordinate. For any $x$, the viable points are in the range $[-20 + |x|, 20 - |x|]$. This means that our total sum is equal to \begin{align*} 1 + 3 + 5 + \cdots + 41 + 39 + 37 + \cdots + 1 &= (1 + 3 + 5 + \cdots + 39) + (1 + 3 + 5 + \cdots + 41) \\ & = 20^2 + 21^2 \\ & = 29^2 \\ &= \boxed{\textbf{(C)} \, 841}. \end{align*}

First, note that $1+7+5+2+5=20$. There are $3$ possible cases: Case 1: the mean is $5$. $X = 5 \cdot 6 - 20 = 10$. Case 2: the mean is $7$. $X = 7 \cdot 6 - 20 = 22$. Case 3: the mean is $X$. $X= \frac{20+X}{6} \Rightarrow X=4$. Therefore, the answer is $10+22+4=\boxed{\textbf{(D) }36}$.

The top left rectangle can be $5$ possible colors. Then the bottom left region can only be $4$ possible colors, and the bottom middle can only be $3$ colors since it is next to the top left and bottom left. Similarly, we have $3$ choices for the top right and $3$ choices for the bottom right, which gives us a total of $5\cdot4\cdot3\cdot3\cdot3=\boxed{\textbf{(D) }540}$.

We can write $\sqrt[3]{10}$ as $10 ^ \frac{1}{3}$. Similarly, $\sqrt[3]{\sqrt[3]{10}} = (10 ^ \frac{1}{3}) ^ \frac{1}{3} = 10 ^ \frac{1}{3^2}$. By continuing this, we get the form \[10 ^ \frac{1}{3} \cdot 10 ^ \frac{1}{3^2} \cdot 10 ^ \frac{1}{3^3} \cdots,\] which is \[10 ^ {\frac{1}{3} + \frac{1}{3^2} + \frac{1}{3^3} + \cdots}.\] Using the formula for an infinite geometric series $S = \frac{a}{1-r}$, we get \[\frac{1}{3} + \frac{1}{3^2} + \frac{1}{3^3} + \cdots = \frac{\frac{1}{3}}{1-\frac{1}{3}} = \frac{1}{2}.\] Thus, our answer is $10 ^ \frac{1}{2} = \boxed{\textbf{(A) }\sqrt{10}}$.

Note that: - Truth-tellers would answer yes-no-no to the three questions in this order. - Liars would answer yes-yes-no to the three questions in this order. - Alternaters who responded truth-lie-truth would answer no-no-no to the three questions in this order. - Alternaters who responded lie-truth-lie would answer yes-yes-yes to the three questions in this order. Suppose that there are $T$ truth-tellers, $L$ liars, and $A$ alternaters who responded lie-truth-lie. The conditions of the first two questions imply that \begin{align*} T+L+A&=22, \\ L+A&=15. \end{align*} Subtracting the second equation from the first, we have $T=22-15=\boxed{\textbf{(A) } 7}.$ Remark The condition of the third question is extraneous. However, we know that $A=9$ and $L=6,$ so there are $9$ alternaters who responded lie-truth-lie, $6$ liars, and $9$ alternaters who responded truth-lie-truth from this condition.

Clearly, the integers from $8$ through $14$ must be in different pairs, so are the integers from $1$ through $7.$ Note that $7$ must pair with $14.$ We pair the numbers $1,2,3,4,5,6$ with the numbers $8,9,10,11,12,13$ systematically: - $6$ can pair with either $12$ or $13.$ - $5$ can pair with any of the three remaining numbers from $10,11,12,13.$ - $1,2,3,4$ can pair with the other four remaining numbers from $8,9,10,11,12,13$ without restrictions. Together, the answer is $2\cdot3\cdot4!=\boxed{\textbf{(E) } 144}.$

Let $a = 2 \cdot |\log_6 10 - 1| = |\log_6 9 - \log_6 x| = \left|\log_6 \frac{9}{x}\right|$. $\pm a = \log_6 \frac{9}{x} \implies 6^{\pm a} = b^{\pm 1} = \frac{9}{x} \implies x = 9 \cdot b^{\pm 1}$ $9b^1 \cdot 9b^{-1} = \boxed{81}$.

Without loss of generality, let the edge-length of $ABCD$ be $2.$ It follows that $MC=MD=\sqrt3.$ Let $O$ be the center of $\triangle ABD,$ so $\overline{CO}\perp\overline{MOD}.$ Note that $MO=\frac13 MD=\frac{\sqrt{3}}{3}.$ In right $\triangle CMO,$ we have \[\cos(\angle CMD)=\frac{MO}{MC}=\boxed{\textbf{(B) } \frac13}.\]

If $z$ is a complex number and $z = a + bi$, then the magnitude (length) of $z$ is $\sqrt{a^2 + b^2}$. Therefore, $z_1$ has a magnitude of 5. If $z_2$ has a magnitude of at most one, that means for each point on the segment given by $z_1$, the bounds of the region $\mathcal{R}$ could be at most 1 away. Alone the line, excluding the endpoints, a rectangle with a width of 2 and a length of 5, the magnitude, would be formed. At the endpoints, two semicircles will be formed with a radius of 1 for a total area of $\pi \approx 3$. Therefore, the total area is $5(2) + \pi \approx 10 + 3 = \boxed{\textbf{(A) } 13}$.

Let $\text{log } 2 = x$. The expression then becomes \[(1+x)^3+(1-x)^3+(3x)(-2x)=\boxed{2}.\]

Let $a$, $b$, $c$ be the three roots of the polynomial. The lengthened prism's volume is \[V = (a+2)(b+2)(c+2) = abc+2ac+2ab+2bc+4a+4b+4c+8 = abc + 2(ab+ac+bc) + 4(a+b+c) + 8.\] By Vieta's formulas, we know that a cubic polynomial $Ax^3+Bx^2+Cx+D$ with roots $a$, $b$, $c$ satisfies: \begin{alignat*}{8} a+b+c &= -\frac{B}{A} &&= \frac{39}{10}, \\ ab+ac+bc &= \hspace{2mm}\frac{C}{A} &&= \frac{29}{10}, \\ abc &= -\frac{D}{A} &&= \frac{6}{10}. \end{alignat*} We can substitute these into the expression, obtaining \[V = \frac{6}{10} + 2\left(\frac{29}{10}\right) + 4\left(\frac{39}{10}\right) + 8 = \boxed{\textbf{(D) } 30}.\]

We have $t_n = \frac{n (n+1)}{2}$. If $t_n$ is a perfect square, then it can be written as $\frac{n (n+1)}{2} = k^2$, where $k$ is a positive integer. Thus, $n (n+1) = 2 k^2$. Rearranging, we get $(2n+1)^2-2(2k)^2=1$, a Pell equation (see https://artofproblemsolving.com/wiki/index.php/Pell_equation ). So $\frac{2n+1}{2k}$ must be a truncation of the continued fraction for $\sqrt{2}$: \begin{eqnarray*} 1+\frac12&=&\frac{2\cdot1+1}{2\cdot1}\\ 1+\frac1{2+\frac1{2+\frac12}}&=&\frac{2\cdot8+1}{2\cdot6}\\ 1+\frac1{2+\frac1{2+\frac1{2+\frac1{2+\frac12}}}}&=&\frac{2\cdot49+1}{2\cdot35}\\ 1+\frac1{2+\frac1{2+\frac1{2+\frac1{2+\frac1{2+\frac1{2+\frac12}}}}}}&=&\frac{2\cdot288+1}{2\cdot204} \end{eqnarray*} Therefore, $t_{288} = \frac{288\cdot289}{2} = 204^2 = 41616$, so the answer is $4+1+6+1+6=\boxed{\textbf{(D) 18}}$.

We are given that $a\cdot(\sin{x}+\sin{(2x)})=\sin{(3x)}$ Using the sine double angle formula combine with the fact that $\sin{(3x)} = \sin{x}\cdot(4\cos^2{x}-1)$, which can be derived using sine angle addition with $\sin{(2x + x)}$, we have \[a\cdot(\sin{x}+2\sin{x}\cos{x})=\sin{x}\cdot(4\cos^2{x}-1)\] Since $\sin{x} \ne 0$ as it is on the open interval $(0, \pi)$, we can divide out $\sin{x}$ from both sides, leaving us with \[a\cdot(1+2\cos{x})=4\cos^2{x}-1\] Now, distributing $a$ and rearranging, we achieve the equation \[4\cos^2{x} - 2a\cos{x} - (1+a) = 0\] which is a quadratic in $\cos{x}$. Applying the quadratic formula to solve for $\cos{x}$, we get \[\cos{x} =\frac{2a\pm\sqrt{4a^2+4(4)(1+a)}}{8}\] and expanding the terms under the radical, we get \[\cos{x} =\frac{2a\pm\sqrt{4a^2+16a+16}}{8}\] Factoring, since $4a^2+16a+16 = (2a+4)^2$, we can simplify our expression even further to \[\cos{x} =\frac{a\pm(a+2)}{4}\] Now, solving for our two solutions, $\cos{x} = -\frac{1}{2}$ and $\cos{x} = \frac{a+1}{2}$. Since $\cos{x} = -\frac{1}{2}$ yields a solution that is valid for all $a$, that being $x = \frac{2\pi}{3}$, we must now solve for the case where $\frac{a+1}{2}$ yields a valid value. As $x\in (0, \pi)$, $\cos{x}\in (-1, 1)$, and therefore $\frac{a+1}{2}\in (-1, 1)$, and $a\in(-3,1)$. There is one more case we must consider inside this interval though, the case where $\frac{a+1}{2} = -\frac{1}{2}$, as this would lead to a double root for $\cos{x}$, yielding only one valid solution for $x$. Solving for this case, $a \ne -2$. Therefore, combining this fact with our solution interval, $a\in(-3, -2) \cup (-2, 1)$, so the answer is $-3-2+1 = \boxed{\textbf{(A) } {-}4}$.

Let $P=(r,\theta)$ be a point in polar coordinates, where $\theta$ is in degrees. Rotating $P$ by $k^{\circ}$ counterclockwise around the origin gives the transformation $(r,\theta)\rightarrow(r,\theta+k^{\circ}).$ Reflecting $P$ across the $y$-axis gives the transformation $(r,\theta)\rightarrow(r,180^{\circ}-\theta).$ Note that \begin{align*} T_k(P)&=(r,180^{\circ}-\theta-k^{\circ}), \\ T_{k+1}(T_k(P)) &= (r,\theta -1^{\circ}). \end{align*} We start with $(1,0^{\circ})$ in polar coordinates. For the sequence of transformations $T_1, T_2, T_3, \cdots, T_k,$ it follows that - After $T_1,$ we have $(1,179^{\circ}).$ - After $T_2,$ we have $(1,-1^{\circ}).$ - After $T_3,$ we have $(1,178^{\circ}).$ - After $T_4,$ we have $(1,-2^{\circ}).$ - After $T_5,$ we have $(1,177^{\circ}).$ - After $T_6,$ we have $(1,-3^{\circ}).$ - ... - After $T_{2k-1},$ we have $(1,180^{\circ}-k^{\circ}).$ - After $T_{2k},$ we have $(1,-k^{\circ}).$ The least such positive integer $k$ is $180.$ Therefore, the least such positive integer $n$ is $2k-1=\boxed{\textbf{(A) } 359}.$

For $1\leq k\leq 12,$ suppose that cards $1, 2, \ldots, k$ are picked up on the first pass. It follows that cards $k+1,k+2,\ldots,13$ are picked up on the second pass. Once we pick the spots for the cards on the first pass, there is only one way to arrange all $\boldsymbol{13}$ cards. For each value of $k,$ there are $\binom{13}{k}-1$ ways to pick the $k$ spots for the cards on the first pass: We exclude the arrangement in which the cards are arranged such that the first pass consists of all $13$ cards. Therefore, the answer is \[\sum_{k=1}^{12}\left[\binom{13}{k}-1\right] = \left[\sum_{k=1}^{12}\binom{13}{k}\right]-12 = \left[\sum_{k=0}^{13}\binom{13}{k}\right]-14 = 2^{13} - 14 = \boxed{\textbf{(D) } 8178}.\]

Consider the reflection $P^{\prime}$ of $P$ over the perpendicular bisector of $\overline{BC}$, creating two new isosceles trapezoids $DAPP^{\prime}$ and $CBPP^{\prime}$. Under this reflection, $P^{\prime}A=PD=4$, $P^{\prime}D=PA=1$, $P^{\prime}C=PB=2$, and $P^{\prime}B=PC=3$. Since $DAPP'$ and $CBPP'$ are isosceles trapezoids, they are cyclic. Using Ptolemy's theorem on $DAPP'$, we get that $(PP')(AD) + (PA)(P'D) = (AP')(PD)$, so \[PP' \cdot AD + 1 \cdot 1 = 4 \cdot 4.\] Then, using Ptolemy's theorem again on $CBPP'$, we get that $(BC)(PP') + (BP)(CP') = (BP')(CP)$, so \[PP' \cdot BC + 2 \cdot 2 = 3 \cdot 3.\] Thus, $PP^{\prime}\cdot AD=15$ and $PP^{\prime}\cdot BC=5$; dividing these two equations and taking the reciprocal yields $\frac{BC}{AD}=\boxed{\textbf{(B) }\frac{1}{3}}$. (diagram by cinnamon_e)

$P(x) = x^{2022} + x^{1011} + 1$ is equal to $\frac{x^{3033}-1}{x^{1011}-1}$ by difference of powers. Therefore, the answer is a polynomial that divides $x^{3033}-1$ but not $x^{1011}-1$. Note that any polynomial $x^m-1$ divides $x^n-1$ if and only if $m$ is a factor of $n$. The prime factorizations of $1011$ and $3033$ are $3*337$ and $3^2*337$, respectively. Hence, $x^9-1$ is a divisor of $x^{3033}-1$ but not $x^{1011}-1$. By difference of powers, $x^9-1=(x^3-1)(x^6+x^3+1)$. Therefore, the answer is $\boxed{E}$.

Because $c$ is real, $z_2 = \bar z_1$. We have \begin{align*} 10 & = z_1 z_2 \\ & = z_1 \bar z_1 \\ & = |z_1|^2 , \end{align*} where the first equality follows from Vieta's formula. Thus, $|z_1| = \sqrt{10}$. We have \begin{align*} c & = z_1 + z_2 \\ & = z_1 + \bar z_1 \\ & = 2 {\rm Re}(z_1), \end{align*} where the first equality follows from Vieta's formula. Thus, ${\rm Re}(z_1) = \frac{c}{2}$. We have \begin{align*} \frac{1}{z_1} & = \frac{1}{10}\cdot\frac{10}{z_1} \\ & = \frac{1}{10}\cdot\frac{z_1 z_2}{z_1} \\ & = \frac{z_2}{10} \\ & = \frac{\bar z_1}{10}. \end{align*} where the second equality follows from Vieta's formula. We have \begin{align*} \frac{1}{z_2} & = \frac{1}{10}\cdot\frac{10}{z_2} \\ & = \frac{1}{10}\cdot\frac{z_1 z_2}{z_2} \\ & = \frac{z_1}{10}. \end{align*} where the second equality follows from Vieta's formula. Therefore, \begin{align*} {\rm Area} \ Q & = \frac{1}{2} \left| {\rm Re}(z_1) \right| \cdot 2 \left| {\rm Im}(z_1) \right| \cdot \left( 1 - \frac{1}{10^2} \right) \\ & = \frac{1}{2} |c| \sqrt{10 - \frac{c^2}{4}} \left( 1 - \frac{1}{10^2} \right) \\ & = \frac{1 - \frac{1}{10^2}}{4} \sqrt{c^2 \left( 40 - c^2 \right)} \\ & \leq \frac{1 - \frac{1}{10^2}}{4} \cdot \frac{c^2 + \left( 40 - c^2 \right)}{2} \\ & = \frac{1 - \frac{1}{10^2}}{4} \cdot 20 , \end{align*} where the inequality follows from the AM-GM inequality, and it is augmented to an equality if and only if $c^2 = 40 - c^2$. Thus, $|c| = 2 \sqrt{5} \approx \boxed{\textbf{(A) 4.5}}$.

We are given that \[\sum_{i=1}^{n}\frac1i = \frac{1}{L_n}\sum_{i=1}^{n}\frac{L_n}{i} = \frac{h_n}{k_n}.\] Since $k_n < L_n,$ we need $\gcd\left(\sum_{i=1}^{n}\frac{L_n}{i}, L_n\right)>1.$ For all primes $p$ such that $p\leq n,$ let $\nu_p(L_n)=e\geq1$ be the exponent of the largest power of $p$ that divides $L_n.$ It is clear that $L_n\equiv0\pmod{p},$ so we test whether $\sum_{i=1}^{n}\frac{L_n}{i}\equiv0\pmod{p}.$ Note that \[\sum_{i=1}^{n}\frac{L_n}{i} \equiv \sum_{i=1}^{\left\lfloor\tfrac{n}{p^e}\right\rfloor}\frac{L_n}{ip^e} \ (\operatorname{mod} \ p^e) \equiv \sum_{i=1}^{\left\lfloor\tfrac{n}{p^e}\right\rfloor}\frac{L_n}{ip^e} \ (\operatorname{mod} \ p).\] We construct the following table for $\nu_p(L_n)=e:$ Note that: 1. If the Sum column has only one term, then it is never congruent to $0$ modulo $p.$ 2. If $p$ and $q$ are positive integers such that $p\geq q,$ then $L_p$ is a multiple of $L_q.$ Therefore, for a specific case, if the sum is congruent to $0$ modulo $p$ for the smallest element in the interval of $n,$ then it is also congruent to $0$ modulo $p$ for all other elements in the interval of $n.$ Together, there are $\boxed{\textbf{(D) }8}$ such integers $n,$ namely \[6,7,8,18,19,20,21,22.\]

For some $n$, let there be $n+1$ parking spaces counterclockwise in a circle. Consider a string of $n$ integers $c_1c_2 \ldots c_n$ each between $0$ and $n$, and let $n$ cars come into this circle so that the $i$th car tries to park at spot $c_i$, but if it is already taken then it instead keeps going counterclockwise and takes the next available spot. After this process, exactly one spot will remain empty. Then the strings of $n$ numbers between $0$ and $n-1$ that contain at least $k$ integers $<k$ for $1 \leq k \leq n$ are exactly the set of strings that leave spot $n$ empty. Also note for any string $c_1c_2 \ldots c_n$, we can add $1$ to each $c_i$ (mod $n+1$) to shift the empty spot counterclockwise, meaning for each string there exists exactly one $j$ with $0 \leq j \leq n$ so that $(c_1+j)(c_2+j) \ldots (c_n+j)$ leaves spot $n$ empty. This gives there are $\frac{(n+1)^{n}}{n+1} = (n+1)^{n-1}$ such strings. Plugging in $n = 5$ gives $\boxed{\textbf{(E) }1296}$ such strings.

Suppose that with a pair $(a_i,b_i)$ the circle is an excircle. Then notice that the hypotenuse must be $(r-x)+(r-y)$, so it must be the case that \[a_i^2+b_i^2=(2r-a_i-b_i)^2.\] Similarly, if with a pair $(a_i,b_i)$ the circle is an incircle, the hypotenuse must be $(x-r)+(y-r)$, leading to the same equation. Notice that this equation can be simplified through SFFT to \[(a_i-2r)(b_i-2r)=2r^2.\] Thus, we want the smallest $r$ such that this equation has at least $14$ distinct pairs $(a_i,b_i)$ for which this holds. The obvious choice to check is $r=6$. In this case, since $2r^2=2^3\cdot 3^2$ has $12$ positive factors, we get $12$ pairs, and we get another two if the factors are $-8,-9$ or vice versa. One can check that for smaller values of $r$, we do not even get close to $14$ possible pairs. When $r=6$, the smallest possible $c$-value is clearly when the factors are negative. When this occurs, $a_i=4, b_i=3$ (or vice versa), so the mimimal $c$ is $5$. The largest possible $c$-value occurs when the largest of $a_i$ and $b_i$ are maximized. This occurs when the factors are $72$ and $1$, leading to $a_i=84, b_i=13$ (or vice-versa), leading to a maximal $c$ of $85$. Hence the answer is $\frac{85}5=\boxed{17}$.