AMC10 2023 A

This is a $d=st$ problem, so let $x$ be the time it takes to meet. We can write the following equation: \[12x+18x=45\] Solving gives us $x=1.5$. The $18x$ is Alicia so $18\times1.5=\boxed{\textbf{(E) 27}}$

Use a system of equations. Let $x$ be the weight of a pizza and $y$ be the weight of a cup of orange slices. We have \[\frac{1}{3}x+\frac{7}{2}y=\frac{3}{4}x+\frac{1}{2}y.\] Rearranging, we get \begin{align*} \frac{5}{12}x&=3y, \\ x&=\frac{36}{5}y. \end{align*} Plugging in $\frac{1}{4}$ pounds for $y$ by the given gives $\frac{9}{5}=\boxed{\textbf{(A) }1\frac{4}{5}}.$

Since $\left \lfloor{\sqrt{2023}}\right \rfloor = 44$, there are $\left \lfloor{\frac{44}{5}}\right \rfloor = \boxed{\textbf{(A) 8}}$ perfect squares less than 2023 that are divisible by 5.

Let's use the triangle inequality. We know that for a triangle, the sum of the 2 shorter sides must always be longer than the longest side. This is because if the longest side were to be as long as the sum of the other sides, or longer, we would only have a line. Similarly, for a convex quadrilateral, the sum of the shortest 3 sides must always be longer than the longest side. Thus, the answer is $\frac{26}{2}-1=13-1=\boxed {\textbf{(D) 12}}$ Sidenote: If there weren't a restriction on integer side lengths, the answer would be the decimal just less than 13, so the sum of the other 3 sides could be just more than 13. That would make the longest side 12.99999..., stopping at who knows how many 9's.

Prime factorizing this gives us $2^{15}\cdot3^{5}\cdot5^{15}=10^{15}\cdot3^5=243\cdot10^{15}$. $10^{15}$ has $16$ digits and $243$ = $2.43*10^{2}$ gives us $3$ more digits. 16+2=18 $2.43*10^{17}$ has $18$ digits

Each of the vertices is counted $3$ times because each vertex is shared by three different edges. Each of the edges is counted $2$ times because each edge is shared by two different faces. Since the sum of the integers assigned to all vertices is $21$, the final answer is $21\times3\times2=\boxed{\textbf{(D) } 126}$

There are $3$ cases where the running total will equal $3$: one roll; two rolls; or three rolls: Case 1: The chance of rolling a running total of $3$, namely $(3)$ in exactly one roll is $\frac{1}{6}$. Case 2: The chance of rolling a running total of $3$ in exactly two rolls, namely $(1, 2)$ and $(2, 1)$ is $\frac{1}{6}\cdot\frac{1}{6}\cdot2=\frac{1}{18}$. Case 3: The chance of rolling a running total of 3 in exactly three rolls, namely $(1, 1, 1)$ is $\frac{1}{6}\cdot\frac{1}{6}\cdot\frac{1}{6}=\frac{1}{216}$. Using the rule of sum we have $\frac{1}{6}+\frac{1}{18}+\frac{1}{216}=\boxed{\textbf{(B) }\frac{49}{216}}$.

To solve this question, you can use $f(x) = mx + b$ where the $x$ is Fahrenheit and the $y$ is Breadus. We have $(110,0)$ and $(350,100)$. We want to find the value of $y$ in $(200,y)$ that falls on this line. The slope for these two points is $\frac{5}{12}$; $y = \frac{5}{12}x + b$. Solving for $b$ using $(110, 0)$, $\frac{550}{12} = -b$. We get $b = \frac{-275}{6}$. Plugging in $(200, y), \frac{1000}{12}-\frac{550}{12}=y$. Simplifying, $\frac{450}{12} = \boxed{\textbf{(D) }37.5}$

Do careful casework by each month. Make sure to start with 2023. In the month and the date, we need a $0$, a $3$, and two digits repeated (which has to be $1$ and $2$ after consideration). After the casework, we get $\boxed{\textbf{(E)}~9}$. For curious readers, the numbers (in chronological order) are:

Let $a$ represent the amount of tests taken previously and $x$ the mean of the scores taken previously. We can write the following equations: \[\frac{ax+11}{a+1}=x+1\qquad (1)\] \[\frac{ax+33}{a+3}=x+2\qquad (2)\] Multiplying equation $(1)$ by $(a+1)$ and solving, we get: \[ax+11=ax+a+x+1\] \[11=a+x+1\] \[a+x=10\qquad (3)\] Multiplying equation $(2)$ by $(a+3)$ and solving, we get: \[ax+33=ax+2a+3x+6\] \[33=2a+3x+6\] \[2a+3x=27\qquad (4)\] Solving the system of equations for $(3)$ and $(4)$, we find that $a=3$ and $x=\boxed{\textbf{(D) }7}$.

The side lengths of the inner square and outer square are $\sqrt{2}$ and $\sqrt{3}$ respectively. Let the shorter side of our triangle be $x$, thus the longer leg is $\sqrt{3}-x$. Hence, by the Pythagorean Theorem, we have \begin{align*} (\sqrt{3}-x)^2+x^2&=(\sqrt{2})^2 \\ 3-2\sqrt{3}x+x^2+x^2&=2 \\ 2x^2-2\sqrt{3}x+1&=0. \end{align*} By the quadratic formula, we find that $x=\frac{\sqrt{3}\pm1}{2}$, so the answer is $\frac{\sqrt{3}-1}{\sqrt{3}+1}=\boxed{\textbf{(C) }2-\sqrt3}.$

Multiples of $5$ will always end in $0$ or $5$, and since the numbers have to be a three-digit numbers, it cannot start with 0 (otherwise it would be a two-digit number), narrowing our choices to 3-digit numbers starting with $5$. Since the numbers must be divisible by 7, all possibilities have to be in the range from $7 \cdot 72$ to $7 \cdot 85$ inclusive(504 to 595). (Add 1 to include 72) $85 - 72 + 1 = 14$. $\boxed{\textbf{(B) } 14}$. You can also take 497 away from each of the numbers(removing the hundreds digit and adding three to each of the numbers), resulting in the numbers {7, 14, 21..., 84, 91, 98}. Dividing each of them by 7, you get the numbers {1, 2, 3..., 12, 13, 14}. Therefore, the answer is $\boxed{\textbf{(B) 14}}$

Error creating thumbnail: Unable to save thumbnail to destination Let $\theta=\angle ACB$ and $x=\overline{AB}$. According to the Law of Sines, we know that $\dfrac{\sin\theta}x=\dfrac{\sin60^\circ}{48}=\dfrac{\sqrt3}{96}$. Rearranging, we get that $x=\dfrac{\sin\theta}{\frac{\sqrt3}{96}}=32\sqrt3\sin\theta$ where $x$ is a function of $\theta$. We want to maximize $x$. We know that the maximum value of $\sin\theta=1$, so this yields $x=32\sqrt3\implies x^2=\boxed{\textbf{(C) }3072.}$ A quick check verifies that $\theta=90^\circ$ indeed works.

In order for the divisor chosen to be a multiple of $11$, the original number chosen must also be a multiple of $11$. Among the first $100$ positive integers, there are 9 multiples of 11; 11, 22, 33, 44, 55, 66, 77, 88, 99. We can now perform a little casework on the probability of choosing a divisor which is a multiple of 11 for each of these 9, and see that the probability is 1/2 for each. The probability of choosing these 9 multiples in the first place is $\frac{9}{100}$, so the final probability is $\frac{9}{100} \cdot \frac{1}{2} = \frac{9}{200}$, so the answer is $\boxed{\textbf{(B)}~\frac{9}{200}}.$ $11 = 1, 11 \Rightarrow \frac{1}{2}\\ 22 = 2 \times 11: 1, 2, 11, 22 \Rightarrow \frac{1}{2}\\ 33 = 3 \times 11: 1, 3, 11, 33 \Rightarrow \frac{1}{2}\\ 44 = 2^2 \times 11: 1, 2, 4, 11, 22, 44 \Rightarrow \frac{1}{2}\\ 55 = 5 \times 11: 1, 5, 11, 55 \Rightarrow \frac{1}{2}\\ 66 = 2 \times 3 \times 11: 1, 2, 3, 6, 11, 22, 33, 66 \Rightarrow \frac{1}{2}\\ 77 = 7 \times 11: 1, 7, 11, 77 \Rightarrow \frac{1}{2}\\ 88 = 2^3 \times 11: 1, 2, 4, 8, 11, 22, 44, 88 \Rightarrow \frac{1}{2}\\ 99 = 3^2 \times 11: 1, 3, 9, 11, 33, 99 \Rightarrow \frac{1}{2}$

Notice that the area of the shaded region is $(2^2\pi-1^2\pi)+(4^2\pi-3^2\pi)+(6^2\pi-5^2\pi)+ \cdots + (n^2\pi-(n-1)^2 \pi)$ for any even number $n$. Using the difference of squares, this simplifies to $(1+2+3+4+\cdots+n) \pi$. So, we are basically finding the smallest $n$ such that $\frac{n(n+1)}{2}>2023 \Rightarrow n(n+1) > 4046$. Since $60(61) > 60^2=3600$, the only option higher than $60$ is $\boxed{\textbf{(E) } 64}$.

We know that the total amount of games must be the sum of games won by left and right handed players. Then, we can write $g = l + r$, and since $l = 1.4r$, $g = 2.4r$. Given that $r$ and $g$ are both integers, $g/2.4$ also must be an integer. From here we can see that $g$ must be divisible by 12, leaving only answers B and D. Now we know the formula for how many games are played in this tournament is $n(n-1)/2$, the sum of the first $n-1$ numbers. Now, setting 36 and 48 equal to the equation will show that two consecutive numbers must have a product of 72 or 96. Clearly, $72=8*9$, so the answer is $\boxed{\textbf{(B) }36}$.

We know that all side lengths are integers, so we can test Pythagorean triples for all triangles. First, we focus on $\triangle{ABP}$. The length of $AB$ is $30$, and the possible (small enough) Pythagorean triples $\triangle{ABP}$ can be are $(3, 4, 5), (5, 12, 13), (8, 15, 17),$ where the length of the longer leg is a factor of $30$. Testing these, we get that only $(8, 15, 17)$ is a valid solution. Thus, we know that $BP = 16$ and $AP = 34$. Next, we move on to $\triangle{QDA}$. The length of $AD$ is $28$, and the small enough triples are $(3, 4, 5)$ and $(7, 24, 25)$. Testing again, we get that $(3, 4, 5)$ is our triple. We get the value of $DQ = 21$, and $AQ = 35$. We know that $CQ = CD - DQ$ which is $9$, and $CP = BC - BP$ which is $12$. $\triangle{CPQ}$ is therefore a right triangle with side length ratios ${3, 4, 5}$, and the hypotenuse is equal to $15$. $\triangle{APQ}$ has side lengths $34, 35,$ and $15,$ so the perimeter is equal to $34 + 35 + 15 = \boxed{\textbf{(A) } 84}.$

Note Euler's formula where $\text{Vertices}+\text{Faces}-\text{Edges}=2$. There are $12$ faces. There are $24$ edges, because there are 12 faces each with four edges and each edge is shared by two faces. Now we know that there are $2-12+24=14$ vertices. Now note that the sum of the degrees of all the points is $48$(the number of edges times 2). Let $x=$ the number of vertices with $3$ edges. Now we know $\frac{3x+4(14-x)}{2}=24$. Solving this equation gives $x = \boxed{\textbf{(D) }8}$. ~aiden22gao ~zgahzlkw (LaTeX) ~ESAOPS (Simplified)

Due to rotations preserving an equal distance, we can bash the answer with the distance formula. $D(A, P) = D(A', P)$, and $D(B, P) = D(B',P)$. Thus we will square our equations to yield: $(1-r)^2+(2-s)^2=(3-r)^2+(1-s)^2$, and $(3-r)^2+(3-s)^2=(4-r)^2+(3-s)^2$. Canceling $(3-s)^2$ from the second equation makes it clear that $r$ equals $3.5$. Substituting will yield \begin{align*}(2.5)^2+(2-s)^2 &= (-0.5)^2+(1-s)^2 \\ 6.25+4-4s+s^2 &= 0.25+1-2s+s^2 \\ 2s &= 9 \\ s &=4.5 \\ \end{align*}. Now $|r-s| = |3.5-4.5| = \boxed{\textbf{(E) } 1}$.

Let a "tile" denote a 1×1 square, and a "square" refer to a 2×2 square. We have 4!=24 possible ways to fill out the top-left square. Next, we fill out the bottom-right corner tile. In the bottom-right square, one corner is already filled (the central tile) from our initial coloring), so we have 3 color options remaining for this. Now considering the remaining tiles, all of these only have one way to be filled (Try it yourself if you don’t believe). For example, the right tile of the middle row is part of two squares: the top-right and the bottom-right. Among these squares, 3 colors have already been used, leaving us with only 1 remaining option. Similarly, every other remaining square has only one available option for coloring. Thus, the total number of ways is 3×4!=(D) 72. A quick version of this method is used when you use for example B as the top left corner block. You can see that $B$ in the top left corner has $6$ possible ways. Now, see that there are $3$ possible corners for a center cell and there are $4$ possible center cells. We get $6$ times $4$ times $3 = D$, the answer is (D) 72.

From the problem statement, we find $P(2-2)=0$, $P(9)=0$ and $4P(4)=0$. Therefore, we know that $0$, $9$, and $4$ are roots. So, we can factor $P(x)$ as $x(x - 9)(x - 4)(x - a)$, where $a$ is the unknown root. Since $P(x) - 1 = 0$, we plug in $x = 1$ which gives $1(-8)(-3)(1 - a) = 1$, so $24(1 - a) = 1 \implies 1 - a = 1/24 \implies a = 23/24$. Therefore, our answer is $23 + 24 =\boxed{\textbf{(D) }47}$

Let $O$ be the center of the midpoint of the line segment connecting both the centers, say $A$ and $B$. Let the point of tangency with the inscribed circle and the right larger circles be $T$. Then $OT = BO + BT = BO + AT - \frac{1}{2} = \frac{1}{4} + 1 - \frac{1}{2} = \frac{3}{4}.$ Since $C_4$ is internally tangent to $C_1$, center of $C_4$, $C_1$ and their tangent point must be on the same line. Now, if we connect centers of $C_4$, $C_3$ and $C_1$/$C_2$, we get a right angled triangle. Let the radius of $C_4$ equal $r$. With the pythagorean theorem on our triangle, we have \[\left(r+\frac{3}{4}\right)^2+\left(\frac{1}{4}\right)^2=(1-r)^2\] Solving this equation gives us \[r = \boxed{\textbf{(D) } \frac{3}{28}}\]

Consider positive integers $a, b$ with a difference of $20$. Suppose $b = a-20$. Then, we have $(a)(a-20) = n$. If there is another pair of two integers that multiply to $n$ but have a difference of 23, one integer must be greater than $a$, and the other must be smaller than $a-20$. We can create two cases and set both equal. We have $(a)(a-20) = (a+1)(a-22) \text{ or } (a+2)(a-21).$ Note that if we go further to $(a+3)(a-20)$ and beyond, that would violate the condition that one of the two integers must be smaller than $a-20.$ Starting with the first case, we have $a^2-20a = a^2-21a-22$, or $0=-a-22$, which gives $a=-22$, which is not possible. The other case is $a^2-20a = a^2-19a-42$, so $a=42$. Thus, our product is $(42)(22) = (44)(21)$, so $n = 924$. Adding the digits, we have $9+2+4 = \boxed{\textbf{(C) } 15}$.

Examining the red isosceles trapezoid with $1$ and $\dfrac{3}{7}$ as two bases, we know that the side lengths are $\dfrac{4}{7}$ from $30-60-90$ triangle. We can conclude that the big hexagon has side length 3. Thus the target area is: area of the big hexagon - 6 * area of the small hexagon. $\dfrac{3\sqrt{3}}{2}(3^2-6\cdot1^2) = \dfrac{3\sqrt{3}}{2}(3) = \boxed{\textbf{(C)}~\frac{9 \sqrt{3}}{2}}$ We can extend the line of the parallelogram to the end until it touches the next hexagon and it will make a small equilateral triangle and a longer parallelogram. We can prove that one side of the tiny equilateral triangle is 4/7 by playing around with angles and the parallelogram because it is parallel, we can then use the whole side of the hexagon which is one and subtract 3/7 which is one side of the equilateral triangle which is 4/7. That means the whole side of the big hexagon length is 3 and we can continue with solution 1.

To find the total amount of vertices we first find the amount of edges, and that is $\frac{20 \times 3}{2}$. Next, to find the amount of vertices we can use Euler's characteristic, $V - E + F = 2$, and therefore the amount of vertices is $12$ So there are $\binom{12}{3} = 220$ ways to choose 3 distinct points. Now, the furthest distance we can get from one point to another point in an icosahedron is 3. Which gives us a range of $1 \leq d(Q, R), d(R, S) \leq 3$ With some case work, we get two cases: Case 1: $d(Q, R) = 3; d(R, S) = 1, 2$ Since we have only one way to choose Q, that is, the opposite point from R, we have one option for Q and any of the other points could work for S. Then, we get $12 \times 1 \times 10 = 120$ (ways to choose R × ways to choose Q × ways to choose S) Case 2: $d(Q, R) = 2; d(R, S) = 1$ We can visualize the icosahedron as 4 rows, first row with 1 vertex, second row with 5 vertices, third row with 5 vertices and fourth row with 1 vertex. We set R as the one vertex on the first row, and we have 12 options for R. Then, Q can be any of the 5 points on the third row and finally S can be one of the 5 points on the second row. Therefore, we have $12 \times 5 \times 5 = 300$ (ways to choose R × ways to choose Q × ways to choose S) In total, we have $420$ ways, but we must divide by $3!$ to account for permutations, giving us $70/220 = \boxed{\textbf{(A) } 7/22}.$