AMC12 2025 A

At $2{:}30$, Andy is $8$ miles ahead. For every hour that they both travel, Betsy gains $4$ miles on Andy. Therefore, it will take her $2$ more hours to be the same distance from the starting point as Andy, which occurs at $\boxed{\textbf{(E) } 4{:}30}$.

We are given $0.2(10) = 2$ pounds of cashews in the first box. Denote the pounds of nuts in the second nut mix as $x.$ \[5 + 0.2x = 0.4(10 + x)\] \[0.2x = 1\] \[x = 5\] Thus, we have $5$ pounds of the second mix. \[0.4(5) + 2 = 2 + 2 = \boxed{\text{(B) }4}\]

When Ash joins a team, the team's average is pulled towards his age. Let $A$ be Ash's age and $N$ be the number of people on the student team. This means that there are $15-N$ people in the teacher team. Let us write an expression for the change in the average for each team. The students originally had an average of $12$, which became $14$ when Ash joined, so there was an increase of $2$. The term $A-12$ represents how much older Ash is compared to the average of the students'. If we divide this by $N+1$, which is the number of people on the student team when Ash joins, we get the average change per team member once Ash is added. Therefore, \[\frac{A-12}{N+1} = 2.\] Similarly, for teachers, the average was originally $55$, which decreased by $3$ to become $52$ when Ash joined. Intuitively, $55-A$ represents how much younger Ash is than the average age of the teachers. Dividing this by the expression $(15-N)+1$, which is the new total number of people on the teacher team, represents the average change per team member once Ash joins. We can write the equation \[\frac{55-A}{16-N} = 3.\] To solve the system, multiply equation (1) by $N+1$, and similarly multiply equation (2) by $16-N$. Then add the equations together, canceling $A$, leaving equation $43=50-N$. From this we get $N=7$ and $A= \boxed{28}.$

We first number all the statements: 1) At least one of these statements is true. 2) At least two of these statements are true. 3) At least two of these statements are false. 4) At least one of these statements is false. We can immediately see that statement 4 must be true, as it would contradict itself if it were false. Similarly, statement 1 must be true, as all the other statements must be false if it were false, which is contradictory because statement 4 is true. Since both 1 and 4 are true, statement 2 has to be true. Therefore, statement 3 is the only false statement, making the answer $\boxed{\text{(B) }1}$.

Let the side length of the largest square be $a,$ so it has area $a^2.$ Hence, the second-largest square has area $a^2k^2,$ the third-largest has $a^2k^4,$ and so on. It follows that the total shaded area is \[a^2-a^2k^2+a^2k^4-a^2k^6+...=a^2(1-k^2+k^4-k^6+...)=a^2\dfrac{1}{1+k^2}.\] The ratio of the area of the shaded region to that of the original square is then \[\dfrac{a^2\frac{1}{1+k^2}}{a^2}=\dfrac{1}{1+k^2}=\dfrac{64}{100}\] \[\implies 64+64k^2=100\implies k^2=\dfrac{36}{64}\implies k=\boxed{\text{(D) }\dfrac{3}{4}}.\] We can just let $a=1$ because the question deals with ratios, meaning that there wouldn't be a loss of generality if we let the side length equal some value, getting the same answer $\boxed{D}$. This was done in solution 3.

Pair two students together and put them adjacent on any two seats. There are 6 ways to do this. Considering one of these cases (they are all the same), there are 4 seats left, in which we wish to arrange the teachers together. So pair the teachers together and put them adjacent on any two seats not already occupied by two of the students. There are 3 ways to do this. For all 6 cases, there are 6×3=18 favorable outcomes. The number of ways to arrange the 2 students and 2 teachers is $\binom{6}{2} \times \binom{4}{2} = 90$. Our probability is 1890= $\boxed{\textbf{(B) } \frac 15}$

Substituting $v$ in the equation where $n=5$, we have: $\log(k \cdot 5^a m^b)$ $=$ $4+2\log m$. Using logarithmic properties, we can write this as: $\log(k \cdot 5^a m^b)$ $=$ $\log(10^4 \cdot m^2)$ We can do this with the other equation where m=25: $\log(k \cdot n^a 25^b)$ $=$ $\log(10^4 \cdot n^4)$ Now we can get rid of the logs on both sides and are left with the following system of equations: $k \cdot 5^am^b = 10^4 m^2$ $k \cdot n^a25^b = 10^4 n^4$ Notice that in the first equation, we can change $m$ arbitrarily, so we know that the exponent of $m$ must be the same - hence $b=2$. Similarly, from the second equation, we get $a=4$. $10^4$ can be written as $2^4 \cdot 5^4$, which means that $k=2^4 = 16$. Thus the answer is $2+4+16 = \boxed{(C) 22}$.

We will scale down the diagram by a factor of $3$ so that $AB = 3$ and $AD = 8.$ Since $\angle BEC = 30^{\circ},$ it follows that $\angle BAC = \angle BDC = 30^{\circ}$ as they all subtend the same arc. Similarly, since $\angle CED = 30^{\circ},$ it follows that $\angle CAD = \angle CBD = 30^{\circ}$ as well. We obtain the following diagram: Note that $\triangle ABD$ has $\angle BAD = 60^{\circ}.$ Applying Law of Cosines, we get \begin{align*} BD^2 &= AB^2+AD^2-2AB\cdot AD \cdot\cos{60^{\circ}} \\ &= 9 + 64 - 2 \cdot 3 \cdot 8 \cdot \frac{1}{2} \\ &= 49, \end{align*} from which $BD = 7.$ From here, we wish to find $BF.$ As $AF$ is the angle bisector of $\angle BAD,$ we apply the Angle Bisector Theorem: \begin{align*} \frac{AB}{BF} &= \frac{AD}{DF} \\ \frac{3}{BF} &= \frac{8}{7-BF}. \end{align*} Solving for $BF,$ we get $BF = \frac{21}{11}.$ Remember to scale the figure back up by a factor of $3,$ so our answer is $\frac{21}{11}\cdot 3 = \boxed{\textbf{(E) } \frac{63}{11}}.$

We begin by calculating $w^2$: \[w^2 = (2+i)^2 = 4+4i-1 = 3+4i.\] Values on the complex plane can easily be represented as points on the Cartesian plane, so we go ahead and do that so we are in a more familiar place. Translating onto the Cartesian plane, we have the points $(2,1)$ and $(3,4)$. The slope of the line passing through these points is $\frac{4-1}{3-2} = 3$, so the equation of this line is \begin{align*} y &= 3(x-2)+1 \\ y &= 3x-5. \end{align*} We want the real number that passes through this line, which is equivalent to the $x-$intercept. This occurs when $y=0$, so the $x$-intercept of this line is $x=\boxed{\textbf{(E)}~\frac53}.$

The ratio between the radius and the arc length is constant. For the inner circle, the radius, which we will denote as $r$, has a corresponding arc length of $2\pi r - 2\pi$ (the circumference minus the major arc length). For the outer circle, the radius, which is $r + 2\pi$, has a corresponding arc length of $2\pi$. We therefore write the equation \[\frac{r}{2\pi r - 2\pi} = \frac{r+2\pi}{2\pi},\] which simplifies to \[r = (r+2\pi)(r-1)\] \[0 = r^2 + (2\pi-2)r - 2\pi.\] Applying the Quadratic Formula, we get that $r =\boxed{\textbf{(B)} 1 - \pi + \sqrt{1 +\pi^2}}$

Altitudes are perpendicular to sides and pass through a vertex. We have the coordinates of the vertices and the slope of the sides can be found, meaning we have all the information needed to find the equations of the lines of the altitudes. Then, we only need to find the intersections of the lines Since $B$ and $C$ form a horizontal line, the altitude to $BC$ from $A$ is a vertical line, so its equation must be $x=2$. Then, we need to find the equation of one more altitude to solve a system of equations, giving us the coordinates. Suppose we choose the second altitude to be the one from $C$ to segment $AB$. The slope of segment $AB$ is $-\frac{2}{3}$, so the slope of the altitude, which is perpendicular to $AB$, is $\frac{3}{2}$. Since the altitude passes through $C$, by point-slope form, we have the equation of the altitude to be $y-27=\frac{3}{2}(x-18)$. Solving the two equations gives the coordinates of the orthocenter to be $(2, 3)$. The answer is therefore $2+3=\boxed{\text{(A) }5}$.

We will need to determine the sum of the reciprocals of the roots. To find the sum of the reciprocals of the roots $p,q$ of the quadratic $ax^2+bx+c$, we use Vieta's formulas. Recall that $p+q = -b/a$ and $pq = c/a$. Therefore, \[\frac{1}{p} + \frac{1}{q} = \frac{p+q}{pq} = \frac{\frac{-b}{a}}{\frac{c}{a}} = \frac{-b}{a} \cdot \frac{a}{c} = \frac{-b}{c},\] which doesn't depend on $a$. The sum of the reciprocals of the roots of the quadratic $x^2-4x-3$ is $\frac{-(-4)}{-3} = -4/3.$ The same is true for every quadratic in the form $ax^2-4x-3$. The sum of all the reciprocals of the roots of $ax^2+bx+c$ is $2025 \cdot \left(-\frac{4}{3}\right).$ Because we have $2025$ quadratics, there are $2 \cdot 2025 = 4050$ total roots. Our answer is $\frac{1}{\frac{1}{4050}\cdot \frac{-4\cdot 2025}{3}} =\boxed{(B)\ -\dfrac{3}{2}}$.

We first find what $N$ is by figuring out how many numbers we need to take out of the set so that the set does not contain $5$ consecutive integers. Since $N$ must be maximized, we must minimize what numbers are removed, and we quickly find that taking two numbers out works. Consider taking out $5$ and $10$. You are left with $\{1,2,3,4,6,7,8,9,11,12,13\}$, which does not have a string of $5$ consecutive integers. There are only $3$ ways to take out two integers such that the resulting set meets our condition ($5$ and $10$, $5$ and $9$, or $4$ and $9$), and ${\binom{13}{2}}=78$ total ways to choose such integers. Therefore, the probability is $\boxed{\dfrac{1}{26}}$. Minor edits ~aashrithm29

Let the outer ellipse be ellipse 1, and the inner ellipse be ellipse 2. Let $a_1$, $b_1$, and $c_1$, correspond to the semimajor axis, semiminor axis, and focal distance of ellipse $1$, respectively. Similarly, let $a_2$, $b_2$, and $c_2$ correspond to the semimajor axis, semiminor axis, and focal distance of ellipse $2$, respectively. Ellipses with the same eccentricity are similar, so $\frac{a_1}{a_2} = \sqrt{2025} = 45$ because the ratio of semimajor axes between similar ellipses is equal to the square root of the ratio between their areas. Notice how \[a_1 = c_1 + c_2 + a_2.\] Substituting from the eccentricity equation, \[a_1 = ea_1 + ea_2 + a_2.\] Rearranging gives us \[a_1 - ea_1 = ea_2 + a_2\] \[a_1(1-e) = a_2(1+e)\] \[\frac{a_1}{a_2} = \frac{1+e}{1-e} = 45.\] Solving for $e$ then yields $\boxed{\text{(D) }\dfrac{22}{23}}.$

Let our subset be $\{11,12,13,...,20\}.$ If we add any element from the set $\{1,2,3,...,10\}$ to our current subset, we will have to remove at least one element from our subset. Hence, the maximum size of our subset is $\boxed{\text{(C) }10}$.

Let $CD \perp AB$ with foot $D$. Right triangles $ACD$ and $BCD$ give $AC^2 = AD^2+CD^2$, $BC^2 = BD^2+CD^2$, $AC^2-BC^2 = AD^2-BD^2 =(AD-BD)(AD+BD)$. Since $AD+BD = AB = 80$ and $AC^2-BC^2 = 75^2-45^2 = 3600$, we get the equation $3600 = 80(AD-BD)$. This equation simplifies to $45 = AD - BD$. We can solve the system of equations $AD + BD = 80$ and $AD - BD = 45$ easily via elimination, and we get $AD = \frac{125}{2}$, $BD = \frac{35}{2}$. $CD^2 = AC^2-AD^2 = 75^2-\left(\frac{125}{2}\right)^2 = \frac{6875}{4}$, $CD = \frac{25\sqrt{11}}{2}$. By Angle Bisector Theorem, $\frac{DP}{PC} = \frac{DB}{BC} = \frac{\frac{35}{2}}{45} = \frac{7}{18}$, $PC = CD-DP$ thus, $18DP = 7(CD-DP)$, $25DP = 7CD$, $DP = \left(\frac{7}{25}\right)CD = \left(\frac{7}{25}\right)\left(\frac{25\sqrt{11}}{2}\right) = \frac{7\sqrt{11}}{2}$. $BP^2 = BD^2+DP^2 = \left(\frac{35}{2}\right)^2+\left(\frac{7\sqrt{11}}{2}\right)^2 = \frac{1225}{4}+\frac{49(11)}{4} = \frac{1764}{4} = 441$, thus $BP = \boxed{\text{(D) }21}.$

Noticing the symmetry, we begin with a substitution: $w = z+2i$. We now have the polynomial $(w-i)(w)(w+i)+10$. Expanding, we get \[w^3+w+10.\] Using the Rational Root Theorem, we notice that $-2$ is a root of this polynomial. Upon dividing the polynomial by $w+2$, we get that \[w^3+w+10 = (w+2)(w^2-2w+5).\] Using the Quadratic Formula upon $w^2-2w+5$, we get that the other two roots are $1+2i$ and $1-2i$. From here, notice that the area of the triangle formed by the roots of this polynomial will be equal to that of the original polynomial because the substitution only shifted the graph $2i$ up, not affecting the distances between each root. Graphing the roots onto the complex plane, the vertical side of the triangle has length $4$, with the altitude to that side having length $3$. Therefore, the triangle has area $\frac{4 \cdot 3}{2} = \boxed{6}.$

Let $0<x<y<z \le 8$; $x$ cannot be $1$ because it makes $xy>z$ $\rightarrow$ $y>z$; $x=2$, $y=3$, $z$ can be $4$, $5$ but not others; $x=2$, $y=4$, $z$ can be $5$, $6$, $7$; $x=2$, $y=5$, $z$ can be $6$, $7$, $8$; $x=2$, $y=6$, $z$ can be $7$, $8$; $x=2$, $y=7$, $z$ can be $8$; for $x=2$, total $11$ cases; Similarly, for $x=3$, $y=4$, $5$, $6$, $7$, total $10$ cases; for $x=4$, $y=5$, $6$, $7$, total $6$ cases; $x=5$, $y=6$, $7$, $3$ cases; $x=6$, $y=7$, $z=8$, $1$ cases; Total $= 11 + 10 + 6 + 3 + 1 = 31$. Permutate $x$, $y$, $z$ for ordered triple, it is $31 \cdot 6=186$, $\boxed{C}$.

We begin by factoring: a3b2+a2b3+b3c2+b2c3+c3a2+c2a3=a2b2(a+b)+b2c2(b+c)+c2a2(c+a)=(a2b2+b2c2+c2a2)(a+b+c)−a2b2c−b2c2a−c2a2b=(a2b2+b2c2+c2a2)(a+b+c)−(abc)(ab+bc+ca). From Vieta's Formulas, we know that $a+b+c = 0$, $ab+bc+ca = k$, and $abc = -1$. Therefore, the answer equals $(a^2b^2+b^2c^2+c^2a^2)(0) - (-1)(k) = \boxed{k}.$

Notice that the triangular faces have a slant height of $\sqrt{13^2-4^2}=\sqrt{153}$ and that the height is therefore $\sqrt{153-(\frac{13-7}{2})^2} = 12$. Then we can split the pentahedron into a triangular prism and two pyramids. The pyramids each have a volume of $\frac{1}{3}(3)(8)(12) = 96$ and the prism has a volume of $\frac{1}{2}(8)(12)(7) = 336$. Thus the answer is $336+96 \cdot 2 = \boxed{\textbf{(C) } 528}$

The numerator can be written as $2^{2a} +2^{2a+2k}+...+2^{2a+2mk}$, which is actually a sum of geometric series. This can be expressed as $2^{2a} \cdot \frac{1-2^{2k(m+1)}}{1-2^{2k}}$. The denominator in the same way can be expressed as $2^a \cdot \frac{1-2^{k(m+1)}}{1-2^k}$. Doing some algebra on the top and bottom we get: 2a⋅1+2k(m+1)1+2k=964 The prime factorization of $964$ is $241 \cdot 2^2$. Equating $2^a = 2^2$, we get $a = 2$. Next since $241$ is a prime number we equate the latter half of the product to $241$. 1+2k(m+1)1+2k=241 Doing some algebra we get, that $2^k(2^{km}-241) = 240$ The closest power of $2$ to $241$ is $256$ which is $2^8$. So setting $km = 8$, we get $2^8-241 = 15$ $2^k(15) = 240$, $2^k = 16$, $k = 4$. This we know that if $k = 4$, then $4m = 8$, so $m = 2$. Finally, we conclude that $a = 2$, $m = 2$, and $k = 4$. $4+2+2 = 8$, so the answer is $\boxed{A}$

We can solve the problem by approaching it geometrically by mapping each possible triple to a coordinate $(a, b, c)$ in a unit cube. Now, we just have to find the volume of the solution set over $1$. WLOG, assume that $a$ is the greatest number. The intersection between $a>2b, a>2c$ and the unit cube is an oblique square pyramid with apex $(0,0,0)$ and base vertices $(1,0,0), (1,0,\frac12), (1,\frac12,0), \text{ and } (1,\frac12, \frac12)$. It helps to visualize the two planes cutting into the cube and leaving triangular traces in the $xy$ and $xz$ planes. The square base $b=s^2=(\frac12) ^2=\frac14$, and the height $h=1$, so the volume $v=\frac13 bh=\frac13(\frac14)(1)=\frac{1}{12}$. From here, we can multiply this volume by $3$ to account for $b,c$ being the greatest (all mutually exclusive). Alternatively, we can find $P(a>2b, 2c|a>b,c)$ by taking $\frac{1}{12}$ over the volume of the interesction between $a>b, a>c$ and the unit cube (a pyramid wth base $1$ and height $1$). Either way, we get the final probability of $\boxed{\text{(E) } \frac{1}{4}}$.

To satisfy the conditions, a $\textit{fair}$ integer must have no digit be a local minimum. Let's say we have $n$ distinct digits, with each digit being a number from $1$ to $9$. To create a $\textit{fair}$ integer, we begin by placing the largest digit. For the second-largest digit, we can either place this digit to the right or to the left of the string already created. We have these $2$ options for the third-largest digit, and so on. Therefore, there are $2^{n-1}$ valid permutations to create a $\textit{fair}$ integer. We must also choose which digits will be in the permutation. If you are creating an $n$-digit long $\textit{fair}$ integer, there are $9\choose{n}$ ways to pick which digits will be in the number. Therefore, for each $n \in \{1,2,\dots, 9\}$, the number of fair integers of length $n$ is: \[\binom{9}{n} \cdot 2^{n-1}.\] Summing over all $n$: \[\sum_{n=1}^9{\binom{9}{n} \cdot 2^{n-1}}=\frac{1}{2}\left(\sum_{n=0}^9{\binom{9}{n}}2^n -1\right)=\frac{1}{2}\left((1+2)^9 -1 \right) = \frac{1}{2}(19682) = \boxed{9841}.\]

Let the center of the large circle be $O$ and the centers of the $12$ circles be $A_1, A_2, A_3, \dots, A_{12}$. Triangle $OA_1A_2$ has side lengths $r+1, r+1, 2$, with the angle opposite $2$ being $360/12 = 30$. Note that the line connecting $A_1$ and $A_2$ go through their common point of tangency, by definition, which causes $A_1A_2$ to have a length of $2$. Drawing the angle bisector of the $30$ degree angle, we split $OA_1A_2$ into two congruent right triangles, each with hypotenuse $r+1$ and side opposite the $15$ degree angle $1$. From here, note that $\sin{15} = \frac{\sqrt{6}-\sqrt{2}}{4}$, which be derived using the trigonometric identity $\sin{(A-B)} = \sin{A} \cos{B} - \sin{B} \cos{A}$, with $A=45$ and $B=30$. In our right triangle, $\sin{15} = \frac{1}{r+1} = \frac{\sqrt{6}-\sqrt{2}}{4}$. Let $x=r+1$. Solving for $x$, we get $x = \frac{4}{\sqrt{6}-\sqrt{2}}$. Rationalizing, we get that $x = \sqrt{6}+\sqrt{2}$. Remember $x = r+1 = \sqrt{6}+\sqrt{2}$, so $r = \sqrt{6}+\sqrt{2} - 1$. Therefore, our answer is $6+2-1 = \boxed{7}.$

None of the answer choices on the official test (which asked for the number of possible functions $f(x)$) were correct, but choice E would be correct if this problem asked for the number of pairs of functions $(P(x), Q(x))$. Let $R(x) = \frac{P(x)}{Q(x)}$. Since $R(x) \leq 0$ on $[a, b]$ but not for values slightly less than $a$ or slightly more than $b$, $P(x) = 0$ at $x = a$ and $x = b$. Therefore, $P(x) = (x-a)(x-b)(x-r)$ for some $r \in \{1, 2, 3, 4, 5\}$. Since $R(x) \leq 0$ on $(c, d)$ but not at $x = c$ or $x = d$, $R(x)$ is not continuous at $x = c$ or $x = d$. Therefore, $R(x)$ must be undefined at $x = c$ and $x = d$, that is, $Q(x) = 0$ at $x = c$ and $x = d$. So $Q(x) = (x-c)(x-d)(x-s)$ for some $s \in \{1, 2, 3, 4, 5\}$. Therefore, $R(x) = \frac{(x-a)(x-b)(x-r)}{(x-c)(x-d)(x-s)}$. Notice that $\frac{(x-a)(x-b)}{(x-c)(x-d)} \leq 0$ only on $[a, b]$ and $(c, d)$. Therefore, $\frac{x-r}{x-s}$ must be nonnegative for all $x \notin \{a, b, c, d, r, s\}$. This only happens if $r = s$. Thus $R(x) = \frac{(x-a)(x-b)(x-r)}{(x-c)(x-d)(x-r)}$, which is the same as $\frac{(x-a)(x-b)}{(x-c)(x-d)}$ except that it is undefined at $x = r$. Thus $R(x)$ satisfies the desired property as long as $r \notin [a, b] \cup (c, d)$. Note that each quintuple $(a, b, c, d, r)$ defines a unique pair of functions $(P(x), Q(x))$. If $(a, b, c, d) = (1, 2, 3, 4)$, $r$ can be $3$, $4$, or $5$. If $(a, b, c, d) = (1, 2, 3, 5)$, $r$ can be $3$ or $5$. If $(a, b, c, d) = (1, 2, 4, 5)$, $r$ can be $3$, $4$, or $5$. If $(a, b, c, d) = (1, 3, 4, 5)$, $r$ can be $4$ or $5$. If $(a, b, c, d) = (2, 3, 4, 5)$, $r$ can be $1$, $4$, or $5$. Therefore, there are $\boxed{(\textbf{E})\ 13}$ possible pairs of functions $(P(x), Q(x))$.