AMC10 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*}

Mike's speed is $\frac{15}{57}=\frac{5}{19}$ laps per minute. In the first $27$ minutes, he completed approximately $\frac{5}{19}\cdot27\approx\frac{1}{4}\cdot28=\boxed{\textbf{(B) } 7}$ laps.

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 formula for fuel efficiency is \[\frac{\text{Distance}}{\text{Gas Consumption}}.\] Note that $1$ mile equals $\frac 1m$ kilometers. We have \[\frac{x\text{ miles}}{1\text{ gallon}} = \frac{\frac{x}{m}\text{ kilometers}}{l\text{ liters}} = \frac{1\text{ kilometer}}{\frac{lm}{x}\text{ liters}} = \frac{100\text{ kilometers}}{\frac{100lm}{x}\text{ liters}}.\] Therefore, the answer is $\boxed{\textbf{(E) } \frac{100lm}{x}}.$

Note that $BP=BQ=DR=DS=1-s.$ It follows that $\triangle BPQ$ and $\triangle DRS$ are congruent isosceles right triangles. In $\triangle BPQ,$ we have $PQ=BP\sqrt2,$ or \begin{align*} s &= (1-s)\sqrt2 \\ s &= \sqrt2 - s\sqrt2 \\ \left(\sqrt2+1\right)s &= \sqrt2 \\ s &= \frac{\sqrt2}{\sqrt2 + 1}. \end{align*} Therefore, the answer is \[s = \frac{\sqrt2}{\sqrt2 + 1}\cdot\frac{\sqrt2 - 1}{\sqrt2 - 1} = \boxed{\textbf{(C) } 2 - \sqrt{2}}.\]

We have \begin{align*} \left|a-2-\sqrt{(a-1)^2}\right| &= \left|a-2-|a-1|\right| \\ &=\left|a-2-(1-a)\right| \\ &=\left|2a-3\right| \\ &=\boxed{\textbf{(A) } 3-2a}. \end{align*}

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 $6+0=\boxed{\textbf{(B) } 6}.$

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 $5\cdot4\cdot3\cdot3\cdot3=\boxed{\textbf{(D) }540}$.

Label the bottom left corner of the larger rectangle (without the square cut out) as $A$ and the top right as $D$. $w$ is the width of the rectangle and $\ell$ is the length. Now we have vertices $E, F, G, H$ as vertices of the irregular octagon created by cutting out the squares. Let $I, J$ be the two closest vertices formed by the squares. The distance between the two closest vertices of the squares is thus $IJ=\left(4\sqrt{2}\right).$ Substituting, we get: \[(IJ)^2 = (w-2)^2 + (\ell-2)^2 = \left(4\sqrt{2}\right)^2 = 32 \implies w^2+\ell^2-4w-4\ell = 24.\] Using the fact that the diagonal of the rectangle is $8,$ we get: \[w^2+\ell^2 = 64.\] Subtracting the first equation from the second equation, we get: \[4w+4\ell=40 \implies w+\ell = 10.\] Squaring yields: \[w^2 + 2w\ell + \ell^2 = 100.\] Subtracting the second equation from this, we get: $2w\ell = 36,$ and thus the area of the original rectangle is $w\ell = \boxed{\textbf{(E) } 18}.$

We are given that \[2^m\cdot\sqrt{\frac{1}{4096}} = 2\cdot\sqrt[m]{\frac{1}{4096}}.\] Converting everything into powers of $2$ and equating exponents, we have \begin{align*} 2^m\cdot(2^{-12})^{\frac12} &= 2\cdot (2^{-12})^{\frac1m} \\ 2^{m-6} &= 2^{1-\frac{12}{m}} \\ m-6 &= 1-\frac{12}{m}. \end{align*} We multiply both sides by $m,$ then rearrange as \[m^2-7m+12=0.\] By Vieta's Formulas, the sum of such values of $m$ is $\boxed{\textbf{(C) } 7}.$ Note that $m=3$ or $m=4$ from the quadratic equation above.

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.

Suppose that $\overline{BD}$ intersects $\overline{AP}$ and $\overline{AC}$ at $X$ and $Y,$ respectively. By Angle-Side-Angle, we conclude that $\triangle ABX\cong\triangle AYX.$ Let $AB=AY=2x.$ By the Angle Bisector Theorem, we have $AC=3x,$ or $YC=x.$ By alternate interior angles, we get $\angle YAD=\angle YCB$ and $\angle YDA=\angle YBC.$ Note that $\triangle ADY \sim \triangle CBY$ by the Angle-Angle Similarity, with the ratio of similitude $\frac{AY}{CY}=2.$ It follows that $AD=2CB=2(BP+PC)=\boxed{\textbf{(C) } 10}.$

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 $2\cdot3\cdot4!=\boxed{\textbf{(E) } 144}.$

Opposite angles of every cyclic quadrilateral are supplementary, so \[\angle B + \angle D = 180^{\circ}.\] We claim that $AC=25.$ We can prove it by contradiction: - If $AC<25,$ then $\angle B$ and $\angle D$ are both acute angles. This arrives at a contradiction. - If $AC>25,$ then $\angle B$ and $\angle D$ are both obtuse angles. This arrives at a contradiction. By the Inscribed Angle Theorem, we conclude that $\overline{AC}$ is the diameter of the circle. So, the radius of the circle is $r=\frac{AC}{2}=\frac{25}{2}.$ The area of the requested region is \[\pi r^2 - \frac12\cdot AB\cdot BC - \frac12\cdot AD\cdot DC = \frac{625\pi}{4}-\frac{168}{2}-\frac{300}{2}=\frac{625\pi-936}{4}.\] Therefore, the answer is $a+b+c=\boxed{\textbf{(D) } 1565}.$

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 rewrite the given equation, then rearrange: \begin{align*} \frac{100a+10b+c}{999} &= \frac13\left(\frac a9 + \frac b9 + \frac c9\right) \\ 100a+10b+c &= 37a + 37b + 37c \\ 63a &= 27b+36c \\ 7a &= 3b+4c. \end{align*} Now, this problem is equivalent to counting the ordered triples $(a,b,c)$ that satisfies the equation. Clearly, the $9$ ordered triples $(a,b,c)=(1,1,1),(2,2,2),\ldots,(9,9,9)$ are solutions to this equation. The expression $3b+4c$ has the same value when: - $b$ increases by $4$ as $c$ decreases by $3.$ - $b$ decreases by $4$ as $c$ increases by $3.$ We find $4$ more solutions from the $9$ solutions above: $(a,b,c)=(4,8,1),(5,1,8),(5,9,2),(6,2,9).$ Note that all solutions are symmetric about $(a,b,c)=(5,5,5).$ Together, we have $9+4=\boxed{\textbf{(D) } 13}$ ordered triples $(a,b,c).$

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 $2k-1=\boxed{\textbf{(A) } 359}.$

Notice that $L_{17}$ contains the highest power of every prime below $17$ since higher primes cannot divide $L_{17}$. Thus, $L_{17}=16\cdot 9 \cdot 5 \cdot 7 \cdot 11 \cdot 13 \cdot 17$. When writing the sum under a common fraction, we multiply the denominators by $L_{17}$ divided by each denominator. However, since $L_{17}$ is a multiple of $17$, all terms will be a multiple of $17$ until we divide out $17$, and the only term that will do this is $\frac{1}{17}$. Thus, the remainder of all other terms when divided by $17$ will be $0$, so the problem is essentially asking us what the remainder of $\frac{L_{17}}{17} = L_{16}$ divided by $17$ is. This is equivalent to finding the remainder of $16 \cdot 9 \cdot 5 \cdot 7 \cdot 11 \cdot 13$ divided by $17$. We use modular arithmetic to simplify our answer: This is congruent to $-1 \cdot 9 \cdot 5 \cdot 7 \cdot 11 \cdot 13 \pmod{17}$. Evaluating, we get: \begin{align*} (-1) \cdot 9 \cdot 35 \cdot 11 \cdot 13 &\equiv (-1) \cdot 9 \cdot 1 \cdot 11 \cdot 13 \pmod{17} \\ &\equiv 9 \cdot 11 \cdot (-13) \pmod{17} \\ &\equiv 9 \cdot 11 \cdot 4\pmod{17} \\ &\equiv 2 \cdot 11 \pmod{17} \\ &\equiv 5\pmod{17} \end{align*} Therefore the remainder is $\boxed{\textbf{(C) } 5}$.

Let the arithmetic sequence be $a,a+d,a+2d,a+3d$ and the geometric sequence be $b,br,br^2,br^3.$ We are given that \begin{align*} a+b&=57, \\ a+d+br&=60, \\ a+2d+br^2&=91, \end{align*} and we wish to find $a+3d+br^3.$ Subtracting the first equation from the second and the second equation from the third, we get \begin{align*} d+b(r-1)&=3, \\ d+br(r-1)&=31. \end{align*} Subtract these results, we get \[b(r-1)^2=28.\] Note that either $(r-1)^2=1$ or $(r-1)^2=4.$ We proceed with casework: - If $(r-1)^2=1,$ then $r=2,b=28,a=29,$ and $d=-25.$ The arithmetic sequence is $29,4,-21,-46,$ arriving at a contradiction. - If $(r-1)^2=4,$ then $r=3,b=7,a=50,$ and $d=-11.$ The arithmetic sequence is $50,39,28,17,$ and the geometric sequence is $7,21,63,189.$ This case is valid. Therefore, The answer is $a+3d+br^3=17+189=\boxed{\textbf{(E) } 206}.$

We extend line segments $\ell,m,$ and $n$ to their point of concurrency, as shown below: We claim that lines $\ell,m,$ and $n$ are concurrent: In the lateral faces of the bowl, we know that lines $\ell$ and $m$ must intersect, and lines $\ell$ and $n$ must intersect. In addition, line $\ell$ intersects the top plane of the bowl at exactly one point. Since lines $m$ and $n$ are both in the top plane of the bowl, we conclude that lines $\ell,m,$ and $n$ are concurrent. In the lateral faces of the bowl, the dashed red line segments create equilateral triangles. So, the dashed red line segments all have length $1.$ In the top plane of the bowl, we know that $\overleftrightarrow{m}\perp\overleftrightarrow{n}.$ So, the dashed red line segments create an isosceles triangle with leg-length $1.$ Note that octagon has four pairs of parallel sides, and the successive side-lengths are $1,\sqrt2,1,\sqrt2,1,\sqrt2,1,\sqrt2,$ as shown below: The area of the octagon is \[3^2-4\cdot\left(\frac12\cdot1^2\right)=\boxed{\textbf{(B) }7}.\]

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)

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.

Let $r$ be the number of lattice points on the side length of square $R$, $s$ be the number of lattice points on the side length of square $S$, and $t$ be the number of lattice points on the side length of square $T$. Note that the actual lengths of the side lengths are the number of lattice points minus $1$, so we can work in terms of $r, s, t$ and subtract $3$ to get the actual answer at the end. Furthermore, note that the number of lattice points inside a rectangular region is equal to the number of lattice points in its width times the number of lattice points along its length. Using this fact, the number of lattice points in $R$ is $r^2$, the number of lattice points in $S$ is $s^2$, and the number of lattice points in $T$ is $t^2$. Now, by the first condition, we have \[r^2=\frac{9}{4}\cdot s^2 \implies r = \frac{3}{2}s \quad \quad \quad \quad \quad (1)\] The second condition, the number of lattice points contained in $T$ is a fourth of the number of lattice points contained in $R \cup S$. The number of lattice points in $R \cup S$ is equal to the sum of the lattice points in their individually bounded regions, but the lattice points along the y-axis for the full length of square $S$ is shared by both of them, so we need to subtract that out. In all, this condition yields us $t^2 = \frac{1}{4}\cdot(r^2 + s^2 - s )\implies t^2 = \frac{1}{4}\cdot\left(\frac{9}{4}\cdot s^2 + s^2 - s \right)$ $\implies t^2=\frac{1}{4}\cdot\frac{13s^2-4s}{4} \implies 16t^2= s(13s-4)$ Note from $(1)$ that $s$ is a multiple of $2$. We can write $s=2j$ and substitute: $16t^2=2j(26j-4) \implies 4t^2=j(13j-2)$. Note that $j$ must be divisible by two for the product to be divisible by 4. Thus we make another substitution, $j=2k$: \[4t^2=2k(26k-2) \implies t^2 = k(13k-1) \quad \quad \quad \quad \quad (2)\] Finally we look at the last condition; that the fraction of the lattice points inside $S$ that are inside $S \cap T$ is $27$ times the fraction of lattice points inside $r$ that are inside $R \cap T$. Let $x$ be the number of lattice points along the bottom of the rectangle formed by $S \cap T$, and $y$ be the number of lattice points along the bottom of the the rectangle formed by $R \cap T$. Therefore, the number of lattice points in $S\cap T$ is $xt$ and the number of lattice points in $R \cap T$ is $yt$. Thus by this condition, $\frac{xt}{s^2} = 27 \cdot \frac{yt}{r^2} \implies \frac{x}{s^2} = 27 \cdot \frac{y}{\frac{9}{4}\cdot s^2} \implies x= 12y$ Finally, notice that $t=x+y-1=12y+y-1$ (subtracting overlap), and so we have \[t=13y-1 \quad \quad \quad \quad \quad (3)\] Now notice that by $(3)$ , $t\equiv -1 \pmod{13}\implies t^2 \equiv 1 \pmod{13}$. However, by $(2)$ , $t^2 \equiv k \cdot -1 \pmod{13}$. Therefore, $-k \equiv 1 \pmod{13} \implies k \equiv -1 \pmod{13}$ Also, by $(2)$ , we know $k$ must be a perfect square since $k$ is relatively prime to $13k-1$ (Euclids algorithm) and the two must multiply to a perfect square. Hence we know two conditions on $k$, and we can now guess and check to find the smallest that satisfies both. We check $k=12$ first since its one less than a multiple of $13$, but this does not work. Next, we have $k=25$ which works because $25$ is a perfect square. Thus we have found the smallest $k$, and therefore the smallest $r, s, t$. Now we just work backwards: $j= 2k = 50$ and $s=2j=100$. Then $r=\frac{3}{2}\cdot 100 = 150$. Finally, from $(2)$ , $t^2=25(13\cdot25-1) \implies t^2 = 25 \cdot 324 \implies t=5\cdot 18=90$. Finally, the sum of each square’s side lengths is $r+s+t-3=340-3=337=\boxed{\textbf{(B) }337}$.