link stringlengths 75 84 | letter stringclasses 5
values | answer float64 0 2,935,363,332B | problem stringlengths 14 5.33k | solution listlengths 1 13 |
|---|---|---|---|---|
https://artofproblemsolving.com/wiki/index.php/2014_AMC_10B_Problems/Problem_5 | A | 26 | Doug constructs a square window using $8$ equal-size panes of glass, as shown. The ratio of the height to width for each pane is $5 : 2$ , and the borders around and between the panes are $2$ inches wide. In inches, what is the side length of the square window?
[asy] fill((0,0)--(2,0)--(2,26)--(0,26)--cycle,gray); fill... | [
"We note that the total length must be the same as the total height, as it is given in the problem. Calling the width of each small rectangle $2x$ , and the height $5x$ , we can see that the length is composed of $4$ widths and $5$ bars of length $2$ . This is equal to two heights of the small rectangles as well as... |
https://artofproblemsolving.com/wiki/index.php/2014_AMC_12B_Problems/Problem_5 | A | 26 | Doug constructs a square window using $8$ equal-size panes of glass, as shown. The ratio of the height to width for each pane is $5 : 2$ , and the borders around and between the panes are $2$ inches wide. In inches, what is the side length of the square window?
[asy] fill((0,0)--(2,0)--(2,26)--(0,26)--cycle,gray); fill... | [
"Let the height of the panes equal $5x$ , and let the width of the panes equal $2x$ . Now notice that the total width of the borders equals $10$ , and the total height of the borders is $6$ . We have \\[10 + 4(2x) = 6 + 2(5x)\\] \\[x = 2\\] Now, the total side length of the window equals \\[10+ 4(2x) = 10 + 16 = ... |
https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_6 | B | 55 | Driving along a highway, Megan noticed that her odometer showed $15951$ (miles). This number is a palindrome-it reads the same forward and backward. Then $2$ hours later, the odometer displayed the next higher palindrome. What was her average speed, in miles per hour, during this $2$ -hour period?
$\textbf{(A)}\ 50 \qq... | [
"In order to get the smallest palindrome greater than $15951$ , we need to raise the middle digit. If we were to raise any of the digits after the middle, we would be forced to also raise a digit before the middle to keep it a palindrome, making it unnecessarily larger.\nSo we raise $9$ to the next largest value, $... |
https://artofproblemsolving.com/wiki/index.php/2017_AMC_12A_Problems/Problem_13 | B | 135 | Driving at a constant speed, Sharon usually takes $180$ minutes to drive from her house to her mother's house. One day Sharon begins the drive at her usual speed, but after driving $\frac{1}{3}$ of the way, she hits a bad snowstorm and reduces her speed by $20$ miles per hour. This time the trip takes her a total of $2... | [
"Let total distance be $x$ . Her speed in miles per minute is $\\tfrac{x}{180}$ . Then, the distance that she drove before hitting the snowstorm is $\\tfrac{x}{3}$ . Her speed in snowstorm is reduced $20$ miles per hour, or $\\tfrac{1}{3}$ miles per minute. Knowing it took her $276$ minutes in total, we create equa... |
https://artofproblemsolving.com/wiki/index.php/1993_AIME_Problems/Problem_2 | null | 580 | During a recent campaign for office, a candidate made a tour of a country which we assume lies in a plane. On the first day of the tour he went east, on the second day he went north, on the third day west, on the fourth day south, on the fifth day east, etc. If the candidate went $n^{2}_{}/2$ miles on the $n^{\mbox{th}... | [
"On the first day, the candidate moves $[4(0) + 1]^2/2\\ \\text{east},\\, [4(0) + 2]^2/2\\ \\text{north},\\, [4(0) + 3]^2/2\\ \\text{west},\\, [4(0) + 4]^2/2\\ \\text{south}$ , and so on. The E/W displacement is thus $1^2 - 3^2 + 5^2 \\ldots +37^2 - 39^2 = \\left|\\sum_{i=0}^9 \\frac{(4i+1)^2}{2} - \\sum_{i=0}^9 \\... |
https://artofproblemsolving.com/wiki/index.php/1992_AJHSME_Problems/Problem_4 | E | 80 | During the softball season, Judy had $35$ hits. Among her hits were $1$ home run, $1$ triple and $5$ doubles. The rest of her hits were single. What percent of her hits were single?
$\text{(A)}\ 28\% \qquad \text{(B)}\ 35\% \qquad \text{(C)}\ 70\% \qquad \text{(D)}\ 75\% \qquad \text{(E)}\ 80\%$ | [
"From the information given, $35-5-1-1=28$ hits were single. Thus, the percentage was \\[\\dfrac{28}{35}=80\\% \\rightarrow \\boxed{80}.\\]"
] |
https://artofproblemsolving.com/wiki/index.php/1997_AJHSME_Problems/Problem_21 | D | 54 | Each corner cube is removed from this $3\text{ cm}\times 3\text{ cm}\times 3\text{ cm}$ cube. The surface area of the remaining figure is
[asy] draw((2.7,3.99)--(0,3)--(0,0)); draw((3.7,3.99)--(1,3)--(1,0)); draw((4.7,3.99)--(2,3)--(2,0)); draw((5.7,3.99)--(3,3)--(3,0)); draw((0,0)--(3,0)--(5.7,0.99)); draw((0,1)--(3... | [
"The original cube has $6$ square surfaces that each have an area of $3^2 = 9$ , for a toal surface area of $6\\cdot 9 = 54$\nSince no two corner cubes touch, we can examine the effect of removing each corner cube individually.\nEach corner cube contribues $3$ faces each of surface area $1$ to the big cube, so the ... |
https://artofproblemsolving.com/wiki/index.php/1990_AJHSME_Problems/Problem_18 | C | 36 | Each corner of a rectangular prism is cut off. Two (of the eight) cuts are shown. How many edges does the new figure have?
[asy] draw((0,0)--(3,0)--(3,3)--(0,3)--cycle); draw((3,0)--(5,2)--(5,5)--(2,5)--(0,3)); draw((3,3)--(5,5)); draw((2,0)--(3,1.8)--(4,1)--cycle,linewidth(1)); draw((2,3)--(4,4)--(3,2)--cycle,linewi... | [
"In addition to the original $12$ edges, each original vertex contributes $3$ new edges.\nThere are $8$ original vertices, so there are $12+3\\times 8=36$ edges in the new figure $\\rightarrow \\boxed{36}$"
] |
https://artofproblemsolving.com/wiki/index.php/1996_AHSME_Problems/Problem_2 | A | 3 | Each day Walter gets $3$ dollars for doing his chores or $5$ dollars for doing them exceptionally well. After $10$ days of doing his chores daily, Walter has received a total of $36$ dollars. On how many days did Walter do them exceptionally well?
$\text{(A)}\ 3\qquad\text{(B)}\ 4\qquad\text{(C)}\ 5\qquad\text{(D)}\ 6\... | [
"If Walter had done his chores for $10$ days without doing any of them well, he would have earned $3 \\cdot 10 = 30$ dollars. He got $6$ dollars more than this.\nHe gets a $5 - 3 = 2$ dollar bonus every day he does his chores well. Thus, he did his chores exceptionally well $\\frac{6}{2} = 3$ days, and the answer... |
https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_23 | C | 25 | Each day for four days, Linda traveled for one hour at a speed that resulted in her traveling one mile in an integer number of minutes. Each day after the first, her speed decreased so that the number of minutes to travel one mile increased by $5$ minutes over the preceding day. Each of the four days, her distance trav... | [
"It is well known that $\\text{Distance}=\\text{Speed} \\cdot \\text{Time}$ . In the question, we want distance. From the question, we have that the time is $60$ minutes or $1$ hour. By the equation derived from $\\text{Distance}=\\text{Speed} \\cdot \\text{Time}$ , we have $\\text{Speed}=\\frac{\\text{Distance}}{... |
https://artofproblemsolving.com/wiki/index.php/2000_AMC_10_Problems/Problem_3 | B | 50 | Each day, Jenny ate $20\%$ of the jellybeans that were in her jar at the beginning of that day. At the end of the second day, $32$ remained. How many jellybeans were in the jar originally?
$\textbf{(A)} \ 40 \qquad \textbf{(B)} \ 50 \qquad \textbf{(C)} \ 55 \qquad \textbf{(D)} \ 60 \qquad \textbf{(E)} \ 75$ | [
"We can begin by labeling the number of initial jellybeans $x$ . If she ate $20\\%$ of the jellybeans, then $80\\%$ is remaining. Hence, after day 1, there are: $0.8 * x$\nAfter day 2, there are: $0.8 * 0.8 * x$ or $0.64x$ jellybeans. $0.64x = 32$ , so $x = \\boxed{50}$",
"Testing the answers choices out, we see ... |
https://artofproblemsolving.com/wiki/index.php/2000_AMC_12_Problems/Problem_3 | B | 50 | Each day, Jenny ate $20\%$ of the jellybeans that were in her jar at the beginning of that day. At the end of the second day, $32$ remained. How many jellybeans were in the jar originally?
$\textbf{(A)} \ 40 \qquad \textbf{(B)} \ 50 \qquad \textbf{(C)} \ 55 \qquad \textbf{(D)} \ 60 \qquad \textbf{(E)} \ 75$ | [
"We can begin by labeling the number of initial jellybeans $x$ . If she ate $20\\%$ of the jellybeans, then $80\\%$ is remaining. Hence, after day 1, there are: $0.8 * x$\nAfter day 2, there are: $0.8 * 0.8 * x$ or $0.64x$ jellybeans. $0.64x = 32$ , so $x = \\boxed{50}$",
"Testing the answers choices out, we see ... |
https://artofproblemsolving.com/wiki/index.php/1959_AHSME_Problems/Problem_1 | B | 125 | Each edge of a cube is increased by $50$ %. The percent of increase of the surface area of the cube is: $\textbf{(A)}\ 50 \qquad\textbf{(B)}\ 125\qquad\textbf{(C)}\ 150\qquad\textbf{(D)}\ 300\qquad\textbf{(E)}\ 750$ | [
"Note that increasing the length of each edge by $50$ % with result in a cube that is similar to the original cube with scale factor $1.5$ . Therefore, the surface area will increase by a factor of $1.5^2$ , or $2.25$ . Converting this back into a percent, the percent increase will be $125$ %. Therefore, the answer... |
https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_13 | null | 125 | Each face of two noncongruent parallelepipeds is a rhombus whose diagonals have lengths $\sqrt{21}$ and $\sqrt{31}$ .
The ratio of the volume of the larger of the two polyhedra to the volume of the smaller is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$ . A parallelepiped is a ... | [
"Denote $\\alpha = \\tan^{-1} \\frac{\\sqrt{21}}{\\sqrt{31}}$ .\nDenote by $d$ the length of each side of a rhombus.\nNow, we put the solid to the 3-d coordinate space.\nWe put the bottom face on the $x-O-y$ plane.\nFor this bottom face, we put a vertex with an acute angle $2 \\alpha$ at the origin, denoted as $O$ ... |
https://artofproblemsolving.com/wiki/index.php/2001_AMC_8_Problems/Problem_24 | B | 5 | Each half of this figure is composed of 3 red triangles, 5 blue triangles and 8 white triangles. When the upper half is folded down over the centerline, 2 pairs of red triangles coincide, as do 3 pairs of blue triangles. There are 2 red-white pairs. How many white pairs coincide?
[asy] draw((0,0)--(4,4*sqrt(3))); draw(... | [
"Each half has $3$ red triangles, $5$ blue triangles, and $8$ white triangles. There are also $2$ pairs of red triangles, so $2$ red triangles on each side are used, leaving $1$ red triangle, $5$ blue triangles, and $8$ white triangles remaining on each half. Also, there are $3$ pairs of blue triangles, using $3$ b... |
https://artofproblemsolving.com/wiki/index.php/2009_AMC_10B_Problems/Problem_1 | B | 2 | Each morning of her five-day workweek, Jane bought either a $50$ -cent muffin or a $75$ -cent bagel. Her total cost for the week was a whole number of dollars. How many bagels did she buy?
$\textbf{(A) } 1\qquad\textbf{(B) } 2\qquad\textbf{(C) } 3\qquad\textbf{(D) } 4\qquad\textbf{(E) } 5$ | [
"If Jane bought one more bagel but one fewer muffin, then her total cost for the week would increase by $25$ cents.\nIf Jane bought $1$ bagel, then she bought $4$ muffins. Her total cost for the week would be $75\\cdot1+50\\cdot4=275$ cents, or $2.75$ dollars. Clearly, she bought one more bagel but one fewer muffin... |
https://artofproblemsolving.com/wiki/index.php/2009_AMC_12B_Problems/Problem_1 | B | 2 | Each morning of her five-day workweek, Jane bought either a $50$ -cent muffin or a $75$ -cent bagel. Her total cost for the week was a whole number of dollars. How many bagels did she buy?
$\textbf{(A) } 1\qquad\textbf{(B) } 2\qquad\textbf{(C) } 3\qquad\textbf{(D) } 4\qquad\textbf{(E) } 5$ | [
"If Jane bought one more bagel but one fewer muffin, then her total cost for the week would increase by $25$ cents.\nIf Jane bought $1$ bagel, then she bought $4$ muffins. Her total cost for the week would be $75\\cdot1+50\\cdot4=275$ cents, or $2.75$ dollars. Clearly, she bought one more bagel but one fewer muffin... |
https://artofproblemsolving.com/wiki/index.php/2010_AMC_10A_Problems/Problem_23 | A | 45 | Each of $2010$ boxes in a line contains a single red marble, and for $1 \le k \le 2010$ , the box in the $k\text{th}$ position also contains $k$ white marbles. Isabella begins at the first box and successively draws a single marble at random from each box, in order. She stops when she first draws a red marble. Let $P(n... | [
"The probability of drawing a white marble from box $k$ is $\\frac{k}{k + 1}$ , and the probability of drawing a red marble from box $k$ is $\\frac{1}{k+1}$\nTo stop after drawing $n$ marbles, we must draw a white marble from boxes $1, 2, \\ldots, n-1,$ and draw a red marble from box $n.$ Thus, \\[P(n) = \\left(\\f... |
https://artofproblemsolving.com/wiki/index.php/2010_AMC_12A_Problems/Problem_19 | A | 45 | Each of $2010$ boxes in a line contains a single red marble, and for $1 \le k \le 2010$ , the box in the $k\text{th}$ position also contains $k$ white marbles. Isabella begins at the first box and successively draws a single marble at random from each box, in order. She stops when she first draws a red marble. Let $P(n... | [
"The probability of drawing a white marble from box $k$ is $\\frac{k}{k + 1}$ , and the probability of drawing a red marble from box $k$ is $\\frac{1}{k+1}$\nTo stop after drawing $n$ marbles, we must draw a white marble from boxes $1, 2, \\ldots, n-1,$ and draw a red marble from box $n.$ Thus, \\[P(n) = \\left(\\f... |
https://artofproblemsolving.com/wiki/index.php/2023_AMC_10B_Problems/Problem_21 | E | 14 | Each of $2023$ balls is randomly placed into one of $3$ bins. Which of the following is closest to the probability that each of the bins will contain an odd number of balls?
$\textbf{(A) } \frac{2}{3} \qquad\textbf{(B) } \frac{3}{10} \qquad\textbf{(C) } \frac{1}{2} \qquad\textbf{(D) } \frac{1}{3} \qquad\textbf{(E) } \f... | [
"We first examine the possible arrangements for parity of number of balls in each box for $2022$ balls.\nIf a $E$ denotes an even number and a $O$ denotes an odd number, then the distribution of balls for $2022$ balls could be $EEE,EOO,OEO,$ or $OOE$ . With the insanely overpowered magic of cheese, we assume that e... |
https://artofproblemsolving.com/wiki/index.php/2023_AMC_12B_Problems/Problem_19 | E | 14 | Each of $2023$ balls is randomly placed into one of $3$ bins. Which of the following is closest to the probability that each of the bins will contain an odd number of balls?
$\textbf{(A) } \frac{2}{3} \qquad\textbf{(B) } \frac{3}{10} \qquad\textbf{(C) } \frac{1}{2} \qquad\textbf{(D) } \frac{1}{3} \qquad\textbf{(E) } \f... | [
"We first examine the possible arrangements for parity of number of balls in each box for $2022$ balls.\nIf a $E$ denotes an even number and a $O$ denotes an odd number, then the distribution of balls for $2022$ balls could be $EEE,EOO,OEO,$ or $OOE$ . With the insanely overpowered magic of cheese, we assume that e... |
https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_11 | E | 64 | Each of the $100$ students in a certain summer camp can either sing, dance, or act. Some students have more than one talent, but no student has all three talents. There are $42$ students who cannot sing, $65$ students who cannot dance, and $29$ students who cannot act. How many students have two of these talents?
$\tex... | [
"Let $a$ be the number of students that can only sing, $b$ can only dance, and $c$ can only act.\nLet $ab$ be the number of students that can sing and dance, $ac$ can sing and act, and $bc$ can dance and act.\nFrom the information given in the problem, $a + ab + b = 29, b + bc + c = 42,$ and $a + ac + c = 65$\nAddi... |
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10A_Problems/Problem_24 | E | 20 | Each of the $12$ edges of a cube is labeled $0$ or $1$ . Two labelings are considered different even if one can be obtained from the other by a sequence of one or more rotations and/or reflections. For how many such labelings is the sum of the labels on the edges of each of the $6$ faces of the cube equal to $2$
$\text... | [
"For simplicity, we will name this cube $ABCDEFGH$ by vertices, as shown below. Note that for each face of this cube, two edges are labeled $0,$ and two edges are labeled $1.$ For all twelve edges of this cube, we conclude that six edges are labeled $0,$ and six edges are labeled $1.$\nWe apply casework to face $A... |
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10A_Problems/Problem_21 | E | 16 | Each of the $20$ balls is tossed independently and at random into one of the $5$ bins. Let $p$ be the probability that some bin ends up with $3$ balls, another with $5$ balls, and the other three with $4$ balls each. Let $q$ be the probability that every bin ends up with $4$ balls. What is $\frac{p}{q}$
$\textbf{(A)}\ ... | [
"For simplicity purposes, we assume that the balls and the bins are both distinguishable.\nRecall that there are $5^{20}$ ways to distribute $20$ balls into $5$ bins. We have \\[p=\\frac{5\\cdot4\\cdot\\binom{20}{3,5,4,4,4}}{5^{20}} \\text{ and } q=\\frac{\\binom{20}{4,4,4,4,4}}{5^{20}}.\\] Therefore, the answer is... |
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12A_Problems/Problem_18 | E | 16 | Each of the $20$ balls is tossed independently and at random into one of the $5$ bins. Let $p$ be the probability that some bin ends up with $3$ balls, another with $5$ balls, and the other three with $4$ balls each. Let $q$ be the probability that every bin ends up with $4$ balls. What is $\frac{p}{q}$
$\textbf{(A)}\ ... | [
"For simplicity purposes, we assume that the balls and the bins are both distinguishable.\nRecall that there are $5^{20}$ ways to distribute $20$ balls into $5$ bins. We have \\[p=\\frac{5\\cdot4\\cdot\\binom{20}{3,5,4,4,4}}{5^{20}} \\text{ and } q=\\frac{\\binom{20}{4,4,4,4,4}}{5^{20}}.\\] Therefore, the answer is... |
https://artofproblemsolving.com/wiki/index.php/2001_AIME_II_Problems/Problem_2 | null | 298 | Each of the $2001$ students at a high school studies either Spanish or French, and some study both. The number who study Spanish is between $80$ percent and $85$ percent of the school population, and the number who study French is between $30$ percent and $40$ percent. Let $m$ be the smallest number of students who cou... | [
"Let $S$ be the percent of people who study Spanish, $F$ be the number of people who study French, and let $S \\cap F$ be the number of students who study both. Then $\\left\\lceil 80\\% \\cdot 2001 \\right\\rceil = 1601 \\le S \\le \\left\\lfloor 85\\% \\cdot 2001 \\right\\rfloor = 1700$ , and $\\left\\lceil 30\\%... |
https://artofproblemsolving.com/wiki/index.php/2008_AMC_8_Problems/Problem_11 | A | 7 | Each of the $39$ students in the eighth grade at Lincoln Middle School has one dog or one cat or both a dog and a cat. Twenty students have a dog and $26$ students have a cat. How many students have both a dog and a cat?
$\textbf{(A)}\ 7\qquad \textbf{(B)}\ 13\qquad \textbf{(C)}\ 19\qquad \textbf{(D)}\ 39\qquad \textbf... | [
"The union of two sets is equal to the sum of each set minus their intersection. The number of students that have both a dog and a cat is $20+26-39 = \\boxed{7}$",
"We create a diagram: \nLet $x$ be the number of students with both a dog and a cat.\nTherefore, we have\n\\[26+20-x = 39\\] \\[46-x = 39\\] \\[x = \\... |
https://artofproblemsolving.com/wiki/index.php/2012_AMC_8_Problems/Problem_16 | C | 87,431 | Each of the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9 is used only once to make two five-digit numbers so that they have the largest possible sum. Which of the following could be one of the numbers?
$\textbf{(A)}\hspace{.05in}76531\qquad\textbf{(B)}\hspace{.05in}86724\qquad\textbf{(C)}\hspace{.05in}87431\qquad\textbf{(D)... | [
"In order to maximize the sum of the numbers, the numbers must have their digits ordered in decreasing value. There are only two numbers from the answer choices with this property: $76531$ and $87431$ . To determine the answer we will have to use estimation and the first two digits of the numbers.\nFor $76531$ the ... |
https://artofproblemsolving.com/wiki/index.php/1955_AHSME_Problems/Problem_30 | B | 3 | Each of the equations $3x^2-2=25, (2x-1)^2=(x-1)^2, \sqrt{x^2-7}=\sqrt{x-1}$ has:
$\textbf{(A)}\ \text{two integral roots}\qquad\textbf{(B)}\ \text{no root greater than 3}\qquad\textbf{(C)}\ \text{no root zero}\\ \textbf{(D)}\ \text{only one root}\qquad\textbf{(E)}\ \text{one negative root and one positive root}$
Solut... | [
"Since the question asks us about the unifying characteristic of all three equations' roots, we have to first determine them.\n$3x^2-2 = 25$ can be rewritten as $3x^2 - 27 = 0$ , which gives the following roots $+3$ and $-3$\n$(2x-1)^2 = (x-1)^2$ can be expanded to $4x^2-4x+1=x^2-2x+1$ , which in turn leads to $3x^... |
https://artofproblemsolving.com/wiki/index.php/1999_AMC_8_Problems/Problem_11 | D | 24 | Each of the five numbers 1, 4, 7, 10, and 13 is placed in one of the five squares so that the sum of the three numbers in the horizontal row equals the sum of the three numbers in the vertical column. The largest possible value for the horizontal or vertical sum is
[asy] draw((0,0)--(3,0)--(3,1)--(0,1)--cycle); draw((1... | [
"The largest sum occurs when $13$ is placed in the center. This sum is $13 + 10 + 1 = 13 + 7 + 4 = \\boxed{24}$ . Note: Two other common sums, $18$ and $21$ , are also possible.",
"Since the horizontal sum equals the vertical sum, twice this sum will be the sum\nof the five numbers plus the number in the center. ... |
https://artofproblemsolving.com/wiki/index.php/2011_AMC_8_Problems/Problem_7 | C | 25 | Each of the following four large congruent squares is subdivided into combinations of congruent triangles or rectangles and is partially bolded . What percent of the total area is partially bolded? [asy] import graph; size(7.01cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real x... | [
"Assume that the area of each square is $1$ . Then, the area of the bolded region in the top left square is $\\dfrac{1}{4}$ . The area of the top right bolded region is $\\dfrac{1}{8}$ . The area of the bottom left bolded region is $\\dfrac{3}{8}$ . And the area of the bottom right bolded region is $\\dfrac{1}{... |
https://artofproblemsolving.com/wiki/index.php/1998_AJHSME_Problems/Problem_10 | E | 7 | Each of the letters $\text{W}$ $\text{X}$ $\text{Y}$ , and $\text{Z}$ represents a different integer in the set $\{ 1,2,3,4\}$ , but not necessarily in that order. If $\dfrac{\text{W}}{\text{X}} - \dfrac{\text{Y}}{\text{Z}}=1$ , then the sum of $\text{W}$ and $\text{Y}$ is
$\text{(A)}\ 3 \qquad \text{(B)}\ 4 \qquad \t... | [
"There are different ways to approach this problem, and I'll start with the different factor of the numbers of the set $\\{ 1,2,3,4\\}$\n$1$ has factor $1$\n$2$ has factors $1$ and $2$\n$3$ has factors $1$ and $3$\n$4$ has factors $1$ $2$ , and $4$\nFrom here, we note that even though all numbers have the factor $1... |
https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_16 | E | 5 | Each of the points $A,B,C,D,E,$ and $F$ in the figure below represents a different digit from $1$ to $6.$ Each of the five lines shown passes through some of these points. The digits along each line are added to produce five sums, one for each line. The total of the five sums is $47.$ What is the digit represented by $... | [
"We can form the following expressions for the sum along each line: \\[\\begin{dcases}A+B+C\\\\A+E+F\\\\C+D+E\\\\B+D\\\\B+F\\end{dcases}\\] Adding these together, we must have $2A+3B+2C+2D+2E+2F=47$ , i.e. $2(A+B+C+D+E+F)+B=47$ . Since $A,B,C,D,E,F$ are unique integers between $1$ and $6$ , we obtain $A+B+C+D+E+F=1... |
https://artofproblemsolving.com/wiki/index.php/2008_AMC_10A_Problems/Problem_10 | E | 4 | Each of the sides of a square $S_1$ with area $16$ is bisected, and a smaller square $S_2$ is constructed using the bisection points as vertices. The same process is carried out on $S_2$ to construct an even smaller square $S_3$ . What is the area of $S_3$
$\mathrm{(A)}\ \frac{1}{2}\qquad\mathrm{(B)}\ 1\qquad\mathrm{(C... | [
"Since the length ratio is $\\frac{1}{\\sqrt{2}}$ , then the area ratio is $\\frac{1}{2}$ (since the area ratio between two similar 2-dimensional objects is equal to the side ratio of those objects squared). This means that $S_2 = 8$ and $S_3 = \\boxed{4}$"
] |
https://artofproblemsolving.com/wiki/index.php/2002_AMC_10A_Problems/Problem_5 | C | 2 | Each of the small circles in the figure has radius one. The innermost circle is tangent to the six circles that surround it, and each of those circles is tangent to the large circle and to its small-circle neighbors. Find the area of the shaded region.
[asy] unitsize(.3cm); path c=Circle((0,2),1); filldraw(Circle((0,0)... | [
"The outer circle has radius $1+1+1=3$ , and thus area $9\\pi$ . The little circles have area $\\pi$ each; since there are 7, their total area is $7\\pi$ . Thus, our answer is $9\\pi-7\\pi=\\boxed{2}$"
] |
https://artofproblemsolving.com/wiki/index.php/2002_AMC_12A_Problems/Problem_5 | C | 2 | Each of the small circles in the figure has radius one. The innermost circle is tangent to the six circles that surround it, and each of those circles is tangent to the large circle and to its small-circle neighbors. Find the area of the shaded region.
[asy] unitsize(.3cm); path c=Circle((0,2),1); filldraw(Circle((0,0)... | [
"The outer circle has radius $1+1+1=3$ , and thus area $9\\pi$ . The little circles have area $\\pi$ each; since there are 7, their total area is $7\\pi$ . Thus, our answer is $9\\pi-7\\pi=\\boxed{2}$"
] |
https://artofproblemsolving.com/wiki/index.php/2003_AMC_8_Problems/Problem_18 | D | 6 | Each of the twenty dots on the graph below represents one of Sarah's classmates. Classmates who are friends are connected with a line segment. For her birthday party, Sarah is inviting only the following: all of her friends and all of those classmates who are friends with at least one of her friends. How many class... | [
"There are $3$ people who are friends with only each other who won't be invited, plus $1$ person who has no friends, and $2$ people who are friends of friends of friends who won’t be invited. So the answer is $\\boxed{6}$"
] |
https://artofproblemsolving.com/wiki/index.php/2000_AIME_I_Problems/Problem_5 | null | 26 | Each of two boxes contains both black and white marbles, and the total number of marbles in the two boxes is $25.$ One marble is taken out of each box randomly. The probability that both marbles are black is $27/50,$ and the probability that both marbles are white is $m/n,$ where $m$ and $n$ are relatively prime positi... | [
"If we work with the problem for a little bit, we quickly see that there is no direct combinatorics way to calculate $m/n$ . The Principle of Inclusion-Exclusion still requires us to find the individual probability of each box.\nLet $a, b$ represent the number of marbles in each box, and without loss of generality ... |
https://artofproblemsolving.com/wiki/index.php/2019_AMC_10B_Problems/Problem_7 | B | 21 | Each piece of candy in a store costs a whole number of cents. Casper has exactly enough money to buy either $12$ pieces of red candy, $14$ pieces of green candy, $15$ pieces of blue candy, or $n$ pieces of purple candy. A piece of purple candy costs $20$ cents. What is the smallest possible value of $n$
$\textbf{(A) } ... | [
"If he has enough money to buy $12$ pieces of red candy, $14$ pieces of green candy, and $15$ pieces of blue candy, then the smallest amount of money he could have is $\\text{lcm}{(12,14,15)} = 420$ cents. Since a piece of purple candy costs $20$ cents, the smallest possible value of $n$ is $\\frac{420}{20} = \\box... |
https://artofproblemsolving.com/wiki/index.php/2019_AMC_12B_Problems/Problem_5 | B | 21 | Each piece of candy in a store costs a whole number of cents. Casper has exactly enough money to buy either $12$ pieces of red candy, $14$ pieces of green candy, $15$ pieces of blue candy, or $n$ pieces of purple candy. A piece of purple candy costs $20$ cents. What is the smallest possible value of $n$
$\textbf{(A) } ... | [
"If he has enough money to buy $12$ pieces of red candy, $14$ pieces of green candy, and $15$ pieces of blue candy, then the smallest amount of money he could have is $\\text{lcm}{(12,14,15)} = 420$ cents. Since a piece of purple candy costs $20$ cents, the smallest possible value of $n$ is $\\frac{420}{20} = \\box... |
https://artofproblemsolving.com/wiki/index.php/2000_AMC_8_Problems/Problem_5 | C | 4 | Each principal of Lincoln High School serves exactly one $3$ -year term. What is the maximum number of principals this school could have during an $8$ -year period?
$\text{(A)}\ 2 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 4 \qquad \text{(D)}\ 5 \qquad \text{(E)}\ 8$ | [
"If the first year of the $8$ -year period was the final year of a principal's\nterm, then in the next six years two more principals would serve, and the last year of the\nperiod would be the first year of the fourth principal's term. Therefore, the maximum\nnumber of principals who can serve during an $8$ -year pe... |
https://artofproblemsolving.com/wiki/index.php/2004_AMC_10B_Problems/Problem_1 | C | 363 | Each row of the Misty Moon Amphitheater has $33$ seats. Rows $12$ through $22$ are reserved for a youth club. How many seats are reserved for this club?
$\mathrm{(A) \ } 297 \qquad \mathrm{(B) \ } 330\qquad \mathrm{(C) \ } 363\qquad \mathrm{(D) \ } 396\qquad \mathrm{(E) \ } 726$ | [
"There are $22-12+1=11$ rows of $33$ seats, giving $11\\times 33=\\boxed{363}$ seats."
] |
https://artofproblemsolving.com/wiki/index.php/1961_AHSME_Problems/Problem_10 | D | 63 | Each side of $\triangle ABC$ is $12$ units. $D$ is the foot of the perpendicular dropped from $A$ on $BC$ ,
and $E$ is the midpoint of $AD$ . The length of $BE$ , in the same unit, is:
$\textbf{(A)}\ \sqrt{18} \qquad \textbf{(B)}\ \sqrt{28} \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ \sqrt{63} \qquad \textbf{(E)}\ \sq... | [
" Note that $\\triangle ABC$ is an equilateral triangle . From the Pythagorean Theorem (or by using 30-60-90 triangles), $AD = 6\\sqrt{3}$ . That means $DE = 3\\sqrt{3}$ . Using the Pythagorean Theorem again, $BE = \\sqrt{63}$ , which is answer choice $\\boxed{63}$"
] |
https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_20 | D | 72 | Each square in a $3\times3$ grid of squares is colored red, white, blue, or green so that every $2\times2$ square contains one square of each color. One such coloring is shown on the right below. How many different colorings are possible?
[asy] unitsize(0.5cm, 0.5cm); draw((0,0)--(9,0)--(9,9)--(0,9)--cycle); draw((0,3)... | [
"Let a \"tile\" denote a $1\\times1$ square and \"square\" refer to $2\\times2$\nWe first have $4!=24$ possible ways to fill out the top left square. We then fill out the bottom right tile. In the bottom right square, we already have one corner filled out (from our initial coloring), and we now have $3$ options lef... |
https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_19 | C | 22 | Each square in a $5 \times 5$ grid is either filled or empty, and has up to eight adjacent neighboring squares, where neighboring squares share either a side or a corner. The grid is transformed by the following rules:
A sample transformation is shown in the figure below. [asy] import geometry; unitsize... | [
"There are two cases for the initial configuration:\nTogether, the answer is $2+20=\\boxed{22}.$"
] |
https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_18 | C | 22 | Each square in a $5 \times 5$ grid is either filled or empty, and has up to eight adjacent neighboring squares, where neighboring squares share either a side or a corner. The grid is transformed by the following rules:
A sample transformation is shown in the figure below. [asy] import geometry; unitsize... | [
"There are two cases for the initial configuration:\nTogether, the answer is $2+20=\\boxed{22}.$"
] |
https://artofproblemsolving.com/wiki/index.php/2012_AMC_10B_Problems/Problem_1 | C | 64 | Each third-grade classroom at Pearl Creek Elementary has $18$ students and $2$ pet rabbits. How many more students than rabbits are there in all $4$ of the third-grade classrooms?
$\textbf{(A)}\ 48\qquad\textbf{(B)}\ 56\qquad\textbf{(C)}\ 64\qquad\textbf{(D)}\ 72\qquad\textbf{(E)}\ 80$ | [
"In each class, there are $18-2=16$ more students than rabbits. So for all classrooms, the difference between students and rabbits is $16 \\times 4 = \\boxed{64}$"
] |
https://artofproblemsolving.com/wiki/index.php/2012_AMC_12B_Problems/Problem_1 | C | 64 | Each third-grade classroom at Pearl Creek Elementary has $18$ students and $2$ pet rabbits. How many more students than rabbits are there in all $4$ of the third-grade classrooms?
$\textbf{(A)}\ 48\qquad\textbf{(B)}\ 56\qquad\textbf{(C)}\ 64\qquad\textbf{(D)}\ 72\qquad\textbf{(E)}\ 80$ | [
"Multiplying $18$ and $2$ by $4$ we get $72$ students and $8$ rabbits. We then subtract: $72 - 8 = \\boxed{64}.$",
"In each class, there are $18-2=16$ more students than rabbits. So for all classrooms, the difference between students and rabbits is $16 \\times 4 = \\boxed{64}$"
] |
https://artofproblemsolving.com/wiki/index.php/2001_AIME_II_Problems/Problem_9 | null | 929 | Each unit square of a 3-by-3 unit-square grid is to be colored either blue or red. For each square, either color is equally likely to be used. The probability of obtaining a grid that does not have a 2-by-2 red square is $\frac {m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$ | [
"We can use complementary counting , counting all of the colorings that have at least one red $2\\times 2$ square.\nBy the Principle of Inclusion-Exclusion , there are (alternatively subtracting and adding) $128-40+8-1=95$ ways to have at least one red $2 \\times 2$ square.\nThere are $2^9=512$ ways to paint the $3... |
https://artofproblemsolving.com/wiki/index.php/1973_AHSME_Problems/Problem_14 | C | 1.2 | Each valve $A$ $B$ , and $C$ , when open, releases water into a tank at its own constant rate. With all three valves open, the tank fills in 1 hour, with only valves $A$ and $C$ open it takes 1.5 hours, and with only valves $B$ and $C$ open it takes 2 hours. The number of hours required with only valves $A$ and $B$ ope... | [
"Let the rate of water flowing through valve $A$ be $a$ , the rate of water flowing through valve $B$ be $b$ , and the rate of water flowing through valve $C$ be $c$ WLOG , let the volume of the tank be 1 liter, and let the units for the rates be liters per hour. With this information, we can write three equations... |
https://artofproblemsolving.com/wiki/index.php/2016_AMC_10A_Problems/Problem_18 | C | 6 | Each vertex of a cube is to be labeled with an integer $1$ through $8$ , with each integer being used once, in such a way that the sum of the four numbers on the vertices of a face is the same for each face. Arrangements that can be obtained from each other through rotations of the cube are considered to be the same. H... | [
"Note that the sum of the numbers on each face must be 18, because $\\frac{1+2+\\cdots+8}{2}=18$\nSo now consider the opposite edges (two edges which are parallel but not on same face of the cube);\nthey must have the same sum value too.\nNow think about the points $1$ and $8$ . If they are not on the same edge, th... |
https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_14 | C | 6 | Each vertex of a cube is to be labeled with an integer $1$ through $8$ , with each integer being used once, in such a way that the sum of the four numbers on the vertices of a face is the same for each face. Arrangements that can be obtained from each other through rotations of the cube are considered to be the same. H... | [
"Note that the sum of the numbers on each face must be 18, because $\\frac{1+2+\\cdots+8}{2}=18$\nSo now consider the opposite edges (two edges which are parallel but not on same face of the cube);\nthey must have the same sum value too.\nNow think about the points $1$ and $8$ . If they are not on the same edge, th... |
https://artofproblemsolving.com/wiki/index.php/2023_AIME_II_Problems/Problem_7 | null | 64 | Each vertex of a regular dodecagon ( $12$ -gon) is to be colored either red or blue, and thus there are $2^{12}$ possible colorings. Find the number of these colorings with the property that no four vertices colored the same color are the four vertices of a rectangle. | [
"\\[\\text{First, we notice that a rectangle is made from two pairs of vertices 1/2 turn away from each other.}\\]\n\\[\\textit{Note: The image is }\\frac{\\textit{280}}{\\textit{841}}\\approx\\frac{\\textit{1}}{\\textit{3}}\\textit{ size.}\\]\n\\[\\text{For there to be no rectangles, there can be at most one same-... |
https://artofproblemsolving.com/wiki/index.php/2023_AIME_II_Problems/Problem_7 | null | 928 | Each vertex of a regular dodecagon ( $12$ -gon) is to be colored either red or blue, and thus there are $2^{12}$ possible colorings. Find the number of these colorings with the property that no four vertices colored the same color are the four vertices of a rectangle. | [
"Note that the condition is equivalent to stating that there are no 2 pairs of oppositely spaced vertices with the same color.\nCase 1: There are no pairs. This yields $2$ options for each vertices 1-6, and the remaining vertices 7-12 are set, yielding $2^6=64$ cases.\nCase 2: There is one pair. Again start with 2 ... |
https://artofproblemsolving.com/wiki/index.php/2011_AMC_12A_Problems/Problem_16 | C | 3,120 | Each vertex of convex pentagon $ABCDE$ is to be assigned a color. There are $6$ colors to choose from, and the ends of each diagonal must have different colors. How many different colorings are possible?
$\textbf{(A)}\ 2520 \qquad \textbf{(B)}\ 2880 \qquad \textbf{(C)}\ 3120 \qquad \textbf{(D)}\ 3250 \qquad \textbf{(E)... | [
"This solution essentially explains other ways of thinking about the cases stated in Solution 2.\nCase 1:\n${6\\choose5} \\cdot 5!$\n5 colors need to be chosen from the group of 6. Those 5 colors have 5! distinct arrangements on the pentagon's vertices.\nCase 2:\n${6\\choose4} \\cdot4\\cdot5\\cdot3!$\n4 colors need... |
https://artofproblemsolving.com/wiki/index.php/2011_AMC_10A_Problems/Problem_22 | C | 3,120 | Each vertex of convex pentagon $ABCDE$ is to be assigned a color. There are $6$ colors to choose from, and the ends of each diagonal must have different colors. How many different colorings are possible?
$\textbf{(A)}\ 2520\qquad\textbf{(B)}\ 2880\qquad\textbf{(C)}\ 3120\qquad\textbf{(D)}\ 3250\qquad\textbf{(E)}\ 3750$ | [
"Let vertex $A$ be any vertex, then vertex $B$ be one of the diagonal vertices to $A$ $C$ be one of the diagonal vertices to $B$ , and so on. We consider cases for this problem.\nIn the case that $C$ has the same color as $A$ $D$ has a different color from $A$ and so $E$ has a different color from $A$ and $D$ . In ... |
https://artofproblemsolving.com/wiki/index.php/2014_AMC_12B_Problems/Problem_6 | D | 40 | Ed and Ann both have lemonade with their lunch. Ed orders the regular size. Ann gets the large lemonade, which is 50% more than the regular. After both consume $\frac{3}{4}$ of their drinks, Ann gives Ed a third of what she has left, and 2 additional ounces. When they finish their lemonades they realize that they both ... | [
"Let the size of Ed's drink equal $x$ ounces, and let the size of Ann's drink equal $\\frac{3}{2}x$ ounces. After both consume $\\frac{3}{4}$ of their drinks, Ed and Ann have $\\frac{x}{4}$ and $\\frac{3x}{8}$ ounces of their drinks remaining. Ann gives away $\\frac{x}{8} + 2$ ounces to Ed.\nIn the end, Ed drank ... |
https://artofproblemsolving.com/wiki/index.php/2008_AIME_I_Problems/Problem_3 | null | 314 | Ed and Sue bike at equal and constant rates. Similarly, they jog at equal and constant rates, and they swim at equal and constant rates. Ed covers $74$ kilometers after biking for $2$ hours, jogging for $3$ hours, and swimming for $4$ hours, while Sue covers $91$ kilometers after jogging for $2$ hours, swimming for $... | [
"Let the biking rate be $b$ , swimming rate be $s$ , jogging rate be $j$ , all in km/h.\nWe have $2b + 3j + 4s = 74,2j + 3s + 4b = 91$ . Subtracting the second from twice the first gives $4j + 5s = 57$ . Mod 4, we need $s\\equiv1\\pmod{4}$ . Thus, $(j,s) = (13,1),(8,5),(3,9)$\n$(13,1)$ and $(3,9)$ give non-integral... |
https://artofproblemsolving.com/wiki/index.php/2011_AIME_II_Problems/Problem_7 | null | 3 | Ed has five identical green marbles, and a large supply of identical red marbles. He arranges the green marbles and some of the red ones in a row and finds that the number of marbles whose right hand neighbor is the same color as themselves is equal to the number of marbles whose right hand neighbor is the other color.... | [
"We are limited by the number of marbles whose right hand neighbor is not the same color as the marble. By surrounding every green marble with red marbles - RGRGRGRGRGR. That's 10 \"not the same colors\" and 0 \"same colors.\" Now, for every red marble we add, we will add one \"same color\" pair and keep all 10 ... |
https://artofproblemsolving.com/wiki/index.php/1992_AJHSME_Problems/Problem_22 | C | 18 | Eight $1\times 1$ square tiles are arranged as shown so their outside edges form a polygon with a perimeter of $14$ units. Two additional tiles of the same size are added to the figure so that at least one side of each tile is shared with a side of one of the squares in the original figure. Which of the following cou... | [
"One such figure would be\n\nThe perimeter of this figure is $\\boxed{18}$"
] |
https://artofproblemsolving.com/wiki/index.php/2006_AIME_I_Problems/Problem_10 | null | 65 | Eight circles of diameter 1 are packed in the first quadrant of the coordinate plane as shown. Let region $\mathcal{R}$ be the union of the eight circular regions. Line $l,$ with slope 3, divides $\mathcal{R}$ into two regions of equal area. Line $l$ 's equation can be expressed in the form $ax=by+c,$ where $a, b,$ and... | [
"The line passing through the tangency point of the bottom left circle and the one to its right and through the tangency of the top circle in the middle column and the one beneath it is the line we are looking for: a line passing through the tangency of two circles cuts congruent areas, so our line cuts through the... |
https://artofproblemsolving.com/wiki/index.php/2024_AIME_I_Problems/Problem_8 | null | 197 | Eight circles of radius $34$ are sequentially tangent, and two of the circles are tangent to $AB$ and $BC$ of triangle $ABC$ , respectively. $2024$ circles of radius $1$ can be arranged in the same manner. The inradius of triangle $ABC$ can be expressed as $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive... | [
"Draw an altitude from both end circles of the diagram with the circles of radius one, and call the lengths you get drawing the altitudes of the circles down to $BC$ $a$ and $b$ . Now we have the length of side $BC$ of being $(2)(2022)+1+1+a+b$ . However, the side $BC$ can also be written as $(6)(68)+34+34+34a+34b$... |
https://artofproblemsolving.com/wiki/index.php/2000_AMC_12_Problems/Problem_25 | E | 1,680 | Eight congruent equilateral triangles , each of a different color, are used to construct a regular octahedron . How many distinguishable ways are there to construct the octahedron? (Two colored octahedrons are distinguishable if neither can be rotated to look just like the other.)
$\textbf {(A)}\ 210 \qquad \textbf {(B... | [
"This problem can be approached by Graph Theory . Note that each face of the octahedron is connected to 3 other faces. We use the above graph to represent the problem. Each vertex represents a face of the octahedron, each edge represent the octahedron's edge.\nNow the problem becomes how many distinguishable ways t... |
https://artofproblemsolving.com/wiki/index.php/2013_AMC_8_Problems/Problem_4 | C | 140 | Eight friends ate at a restaurant and agreed to share the bill equally. Because Judi forgot her money, each of her seven friends paid an extra $2.50 to cover her portion of the total bill. What was the total bill?
$\textbf{(A)}\ \text{\textdollar}120\qquad\textbf{(B)}\ \text{\textdollar}128\qquad\textbf{(C)}\ \text{\te... | [
"Since Judi's 7 friends had to pay $2.50 extra each to cover the total amount that Judi should have paid, we multiply $2.50\\cdot7=17.50$ is the bill Judi would have paid if she had money. Hence, to calculate the total amount, we multiply $17.50\\cdot8=\\boxed{140}$ to find the total the 8 friends paid.",
"Let $m... |
https://artofproblemsolving.com/wiki/index.php/2010_AMC_10A_Problems/Problem_22 | A | 28 | Eight points are chosen on a circle, and chords are drawn connecting every pair of points. No three chords intersect in a single point inside the circle. How many triangles with all three vertices in the interior of the circle are created?
$\textbf{(A)}\ 28 \qquad \textbf{(B)}\ 56 \qquad \textbf{(C)}\ 70 \qquad \textbf... | [
"To choose a chord, we know that two points must be chosen. This implies that for three chords to create a triangle and not intersect at a single point, six points need to be chosen. We also know that for any six points we pick, there is only $1$ way to connect the points such that a triangle is formed in the circl... |
https://artofproblemsolving.com/wiki/index.php/2008_AMC_8_Problems/Problem_19 | B | 27 | Eight points are spaced around at intervals of one unit around a $2 \times 2$ square, as shown. Two of the $8$ points are chosen at random. What is the probability that the two points are one unit apart? [asy] size((50)); dot((5,0)); dot((5,5)); dot((0,5)); dot((-5,5)); dot((-5,0)); dot((-5,-5)); dot((0,-5)); dot((5,-5... | [
"The two points are one unit apart at $8$ places around the edge of the square. There are $8 \\choose 2$ $= 28$ ways to choose two points. The probability is\n\\[\\frac{8}{28} = \\boxed{27}\\]",
"Arbitrarily pick a point in the grid. Clearly, we see two options for the other point to be placed, so the answer is $... |
https://artofproblemsolving.com/wiki/index.php/1998_AIME_Problems/Problem_10 | null | 152 | Eight spheres of radius 100 are placed on a flat surface so that each sphere is tangent to two others and their centers are the vertices of a regular octagon . A ninth sphere is placed on the flat surface so that it is tangent to each of the other eight spheres. The radius of this last sphere is $a +b\sqrt {c},$ wher... | [
"The key is to realize the significance that the figures are spheres, not circles . The 2D analogue of the diagram onto the flat surface will not contain 8 circles tangent to a ninth one; instead the circles will overlap since the middle sphere has a larger radius and will sort of “bulge” out.\n1998 AIME-10a.png\nL... |
https://artofproblemsolving.com/wiki/index.php/2014_AMC_8_Problems/Problem_8 | D | 3 | Eleven members of the Middle School Math Club each paid the same amount for a guest speaker to talk about problem solving at their math club meeting. They paid their guest speaker $\textdollar\underline{1} \underline{A} \underline{2}$ . What is the missing digit $A$ of this $3$ -digit number?
$\textbf{(A) }0\qquad\text... | [
"Since all the eleven members paid the same amount, that means that the total must be divisible by $11$ . We can do some trial-and-error to get $A=3$ , so our answer is $\\boxed{3}$ ~SparklyFlowers",
"We know that a number is divisible by $11$ if the odd digits added together minus the even digits added together ... |
https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_9 | null | 247 | Ellina has twelve blocks, two each of red ( $\textbf{R}$ ), blue ( $\textbf{B}$ ), yellow ( $\textbf{Y}$ ), green ( $\textbf{G}$ ), orange ( $\textbf{O}$ ), and purple ( $\textbf{P}$ ). Call an arrangement of blocks $\textit{even}$ if there is an even number of blocks between each pair of blocks of the same color. For ... | [
"Consider this position chart: \\[\\textbf{1 2 3 4 5 6 7 8 9 10 11 12}\\] Since there has to be an even number of spaces between each pair of the same color, spots $1$ $3$ $5$ $7$ $9$ , and $11$ contain some permutation of all $6$ colored balls. Likewise, so do the even spots, so the number of even configurations i... |
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10A_Problems/Problem_6 | B | 8 | Elmer the emu takes $44$ equal strides to walk between consecutive telephone poles on a rural road. Oscar the ostrich can cover the same distance in $12$ equal leaps. The telephone poles are evenly spaced, and the $41$ st pole along this road is exactly one mile ( $5280$ feet) from the first pole. How much longer, in f... | [
"There are $41-1=40$ gaps between the $41$ telephone poles, so the distance of each gap is $5280\\div40=132$ feet.\nEach of Oscar's leaps covers $132\\div12=11$ feet, and each of Elmer's strides covers $132\\div44=3$ feet.\nTherefore, Oscar's leap is $11-3=\\boxed{8}$ feet longer than Elmer's stride.",
"There are... |
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12A_Problems/Problem_5 | B | 8 | Elmer the emu takes $44$ equal strides to walk between consecutive telephone poles on a rural road. Oscar the ostrich can cover the same distance in $12$ equal leaps. The telephone poles are evenly spaced, and the $41$ st pole along this road is exactly one mile ( $5280$ feet) from the first pole. How much longer, in f... | [
"There are $41-1=40$ gaps between the $41$ telephone poles, so the distance of each gap is $5280\\div40=132$ feet.\nEach of Oscar's leaps covers $132\\div12=11$ feet, and each of Elmer's strides covers $132\\div44=3$ feet.\nTherefore, Oscar's leap is $11-3=\\boxed{8}$ feet longer than Elmer's stride.",
"There are... |
https://artofproblemsolving.com/wiki/index.php/2017_AMC_10B_Problems/Problem_12 | A | 20 | Elmer's new car gives $50\%$ percent better fuel efficiency, measured in kilometers per liter, than his old car. However, his new car uses diesel fuel, which is $20\%$ more expensive per liter than the gasoline his old car used. By what percent will Elmer save money if he uses his new car instead of his old car for a l... | [
"Suppose that his old car runs at $x$ km per liter. Then his new car runs at $\\frac{3}{2}x$ km per liter, or $x$ km per $\\frac{2}{3}$ of a liter. Let the cost of the old car's fuel be $c$ , so the trip in the old car takes $xc$ dollars, while the trip in the new car takes $\\frac{2}{3}\\cdot\\frac{6}{5}xc = \\fra... |
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10A_Problems/Problem_11 | A | 70 | Emily sees a ship traveling at a constant speed along a straight section of a river. She walks parallel to the riverbank at a uniform rate faster than the ship. She counts $210$ equal steps walking from the back of the ship to the front. Walking in the opposite direction, she counts $42$ steps of the same size from the... | [
"Let $x$ be the length of the ship.\nThen, in the time that Emily walks $210$ steps, the ship moves $210-x$ steps.\nAlso, in the time that Emily walks $42$ steps, the ship moves $x-42$ steps.\nSince the ship and Emily have the same ratio of absolute speeds in either direction, $\\frac{210}{210-x} = \\frac{42}{x-42}... |
https://artofproblemsolving.com/wiki/index.php/2010_AMC_10A_Problems/Problem_19 | E | 6 | Equiangular hexagon $ABCDEF$ has side lengths $AB=CD=EF=1$ and $BC=DE=FA=r$ . The area of $\triangle ACE$ is $70\%$ of the area of the hexagon. What is the sum of all possible values of $r$
$\textbf{(A)}\ \frac{4\sqrt{3}}{3} \qquad \textbf{(B)} \frac{10}{3} \qquad \textbf{(C)}\ 4 \qquad \textbf{(D)}\ \frac{17}{4} \qqua... | [
"It is clear that $\\triangle ACE$ is an equilateral triangle. From the Law of Cosines on $\\triangle ABC$ , we get that $AC^2 = r^2+1^2-2r\\cos{\\frac{2\\pi}{3}} = r^2+r+1$ . Therefore, the area of $\\triangle ACE$ is $\\frac{\\sqrt{3}}{4}(r^2+r+1)$ by area of an equilateral triangle.\nIf we extend $BC$ $DE$ and $... |
https://artofproblemsolving.com/wiki/index.php/2010_AMC_12A_Problems/Problem_17 | E | 6 | Equiangular hexagon $ABCDEF$ has side lengths $AB=CD=EF=1$ and $BC=DE=FA=r$ . The area of $\triangle ACE$ is $70\%$ of the area of the hexagon. What is the sum of all possible values of $r$
$\textbf{(A)}\ \frac{4\sqrt{3}}{3} \qquad \textbf{(B)} \frac{10}{3} \qquad \textbf{(C)}\ 4 \qquad \textbf{(D)}\ \frac{17}{4} \qqua... | [
"It is clear that $\\triangle ACE$ is an equilateral triangle. From the Law of Cosines on $\\triangle ABC$ , we get that $AC^2 = r^2+1^2-2r\\cos{\\frac{2\\pi}{3}} = r^2+r+1$ . Therefore, the area of $\\triangle ACE$ is $\\frac{\\sqrt{3}}{4}(r^2+r+1)$ by area of an equilateral triangle.\nIf we extend $BC$ $DE$ and $... |
https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_14 | null | 450 | Equilateral $\triangle ABC$ has side length $600$ . Points $P$ and $Q$ lie outside the plane of $\triangle ABC$ and are on opposite sides of the plane. Furthermore, $PA=PB=PC$ , and $QA=QB=QC$ , and the planes of $\triangle PAB$ and $\triangle QAB$ form a $120^{\circ}$ dihedral angle (the angle between the two planes).... | [
"The inradius of $\\triangle ABC$ is $100\\sqrt 3$ and the circumradius is $200 \\sqrt 3$ . Now, consider the line perpendicular to plane $ABC$ through the circumcenter of $\\triangle ABC$ . Note that $P,Q,O$ must lie on that line to be equidistant from each of the triangle's vertices. Also, note that since $P, Q, ... |
https://artofproblemsolving.com/wiki/index.php/2012_AIME_II_Problems/Problem_13 | null | 677 | Equilateral $\triangle ABC$ has side length $\sqrt{111}$ . There are four distinct triangles $AD_1E_1$ $AD_1E_2$ $AD_2E_3$ , and $AD_2E_4$ , each congruent to $\triangle ABC$ ,
with $BD_1 = BD_2 = \sqrt{11}$ . Find $\sum_{k=1}^4(CE_k)^2$ | [
"Screen Shot 2020-02-17 at 2.24.50 PM.png\nWe create a diagram. Each of the red quadrilaterals are actually 60 degree rotations about point A. We easily get that $\\overline{C E_1} = \\sqrt{11}$ , and $\\overline{C E_3} = \\sqrt{11}$ . If we set $\\angle B A D_2 = \\theta$ , we can start angle chasing. In particula... |
https://artofproblemsolving.com/wiki/index.php/2006_AIME_II_Problems/Problem_12 | null | 865 | Equilateral $\triangle ABC$ is inscribed in a circle of radius $2$ . Extend $\overline{AB}$ through $B$ to point $D$ so that $AD=13,$ and extend $\overline{AC}$ through $C$ to point $E$ so that $AE = 11.$ Through $D,$ draw a line $l_1$ parallel to $\overline{AE},$ and through $E,$ draw a line $l_2$ parallel to $\overli... | [
"Notice that $\\angle{E} = \\angle{BGC} = 120^\\circ$ because $\\angle{A} = 60^\\circ$ . Also, $\\angle{GBC} = \\angle{GAC} = \\angle{FAE}$ because they both correspond to arc ${GC}$ . So $\\Delta{GBC} \\sim \\Delta{EAF}$\n\\[[EAF] = \\frac12 (AE)(EF)\\sin \\angle AEF = \\frac12\\cdot11\\cdot13\\cdot\\sin{120^\\ci... |
https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_2 | null | 336 | Equilateral triangle $ABC$ has side length $840$ . Point $D$ lies on the same side of line $BC$ as $A$ such that $\overline{BD} \perp \overline{BC}$ . The line $\ell$ through $D$ parallel to line $BC$ intersects sides $\overline{AB}$ and $\overline{AC}$ at points $E$ and $F$ , respectively. Point $G$ lies on $\ell$ suc... | [
"By angle chasing, we conclude that $\\triangle AGF$ is a $30^\\circ\\text{-}30^\\circ\\text{-}120^\\circ$ triangle, and $\\triangle BED$ is a $30^\\circ\\text{-}60^\\circ\\text{-}90^\\circ$ triangle.\nLet $AF=x.$ It follows that $FG=x$ and $EB=FC=840-x.$ By the side-length ratios in $\\triangle BED,$ we have $DE=\... |
https://artofproblemsolving.com/wiki/index.php/2009_AIME_II_Problems/Problem_5 | null | 32 | Equilateral triangle $T$ is inscribed in circle $A$ , which has radius $10$ . Circle $B$ with radius $3$ is internally tangent to circle $A$ at one vertex of $T$ . Circles $C$ and $D$ , both with radius $2$ , are internally tangent to circle $A$ at the other two vertices of $T$ . Circles $B$ $C$ , and $D$ are all exter... | [
"Let $X$ be the intersection of the circles with centers $B$ and $E$ , and $Y$ be the intersection of the circles with centers $C$ and $E$ . Since the radius of $B$ is $3$ $AX =4$ . Assume $AE$ $p$ . Then $EX$ and $EY$ are radii of circle $E$ and have length $4+p$ $AC = 8$ , and angle $CAE = 60$ degrees because we ... |
https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_8 | null | 378 | Equilateral triangle $\triangle ABC$ is inscribed in circle $\omega$ with radius $18.$ Circle $\omega_A$ is tangent to sides $\overline{AB}$ and $\overline{AC}$ and is internally tangent to $\omega.$ Circles $\omega_B$ and $\omega_C$ are defined analogously. Circles $\omega_A,$ $\omega_B,$ and $\omega_C$ meet in six po... | [
"We can extend $AB$ and $AC$ to $B'$ and $C'$ respectively such that circle $\\omega_A$ is the incircle of $\\triangle AB'C'$ Since the diameter of the circle is the height of this triangle, the height of this triangle is $36$ . We can use inradius or equilateral triangle properties to get the inradius of this tri... |
https://artofproblemsolving.com/wiki/index.php/1988_AHSME_Problems/Problem_7 | D | 4 | Estimate the time it takes to send $60$ blocks of data over a communications channel if each block consists of $512$ "chunks" and the channel can transmit $120$ chunks per second.
$\textbf{(A)}\ 0.04 \text{ seconds}\qquad \textbf{(B)}\ 0.4 \text{ seconds}\qquad \textbf{(C)}\ 4 \text{ seconds}\qquad \textbf{(D)}\ 4\text... | [
"We want to figure out the number of chunks in $60$ blocks, so we have $60\\cdot 512 \\approx 30000$ . We divide this by $120$ to determine the number of seconds necessary to transmit. $30000/120 \\approx 250$ , which means that it takes approximately $4$ minutes to transmit. Thus, the answer is $\\boxed{4}$",
"T... |
https://artofproblemsolving.com/wiki/index.php/1998_AJHSME_Problems/Problem_16 | B | 2,075 | Estimate the year in which the population of Nisos will be approximately 6,000.
$\text{(A)}\ 2050 \qquad \text{(B)}\ 2075 \qquad \text{(C)}\ 2100 \qquad \text{(D)}\ 2125 \qquad \text{(E)}\ 2150$ | [
"We could triple the population every $25$ years and make a chart:\nYear: 2000\nPopulation: 200\nYear: 2025\nPopulation: 600\nYear: 2050\nPopulation: 1800\nYear: 2075\nPopulation: 5400\nYear: 2100\nPopulation: 16200\nThe closest year is 2075, or $\\boxed{2075}$",
"We could find out how many periods of 25 years we... |
https://artofproblemsolving.com/wiki/index.php/1989_AJHSME_Problems/Problem_4 | E | 2,000 | Estimate to determine which of the following numbers is closest to $\frac{401}{.205}$
$\text{(A)}\ .2 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 20 \qquad \text{(D)}\ 200 \qquad \text{(E)}\ 2000$ | [
"$401$ is around $400$ and $.205$ is around $.2$ so the fraction is approximately \\[\\frac{400}{.2}=2000\\rightarrow \\boxed{2000}\\]"
] |
https://artofproblemsolving.com/wiki/index.php/1993_AIME_Problems/Problem_10 | null | 250 | Euler's formula states that for a convex polyhedron with $V$ vertices $E$ edges , and $F$ faces $V-E+F=2$ . A particular convex polyhedron has 32 faces, each of which is either a triangle or a pentagon . At each of its $V$ vertices, $T$ triangular faces and $P$ pentagonal faces meet. What is the value of $100P+10T+V$ | [
"The convex polyhedron of the problem can be easily visualized; it corresponds to a dodecahedron (a regular solid with $12$ equilateral pentagons) in which the $20$ vertices have all been truncated to form $20$ equilateral triangles with common vertices. The resulting solid has then $p=12$ smaller equilateral penta... |
https://artofproblemsolving.com/wiki/index.php/1982_AHSME_Problems/Problem_3 | C | 256 | Evaluate $(x^x)^{(x^x)}$ at $x = 2$
$\text{(A)} \ 16 \qquad \text{(B)} \ 64 \qquad \text{(C)} \ 256 \qquad \text{(D)} \ 1024 \qquad \text{(E)} \ 65,536$ | [
"Plugging in $2$ as $x$ gives $4^4$ , which is merely $\\boxed{256}$"
] |
https://artofproblemsolving.com/wiki/index.php/1986_AIME_Problems/Problem_2 | null | 104 | Evaluate the product \[\left(\sqrt{5}+\sqrt{6}+\sqrt{7}\right)\left(\sqrt{5}+\sqrt{6}-\sqrt{7}\right)\left(\sqrt{5}-\sqrt{6}+\sqrt{7}\right)\left(-\sqrt{5}+\sqrt{6}+\sqrt{7}\right).\] | [
"More generally, let $(x,y,z)=\\left(\\sqrt5,\\sqrt6,\\sqrt7\\right)$ so that $\\left(x^2,y^2,z^2\\right)=(5,6,7).$\nWe rewrite the original expression in terms of $x,y,$ and $z,$ then apply the difference of squares repeatedly: \\begin{align*} (x+y+z)(x+y-z)(x-y+z)(-x+y+z) &= \\left[((x+y)+z)((x+y)-z)\\right]\\lef... |
https://artofproblemsolving.com/wiki/index.php/1997_AIME_Problems/Problem_10 | null | 117 | Every card in a deck has a picture of one shape - circle, square, or triangle, which is painted in one of the three colors - red, blue, or green. Furthermore, each color is applied in one of three shades - light, medium, or dark. The deck has 27 cards, with every shape-color-shade combination represented. A set of thre... | [
"Adding the cases up, we get $27 + 54 + 36 = \\boxed{117}$",
"Let's say we have picked two cards. We now compare their attributes to decide how we can pick the third card to make a complement set. For each of the three attributes, should the two values be the same we have one option - choose a card with the same ... |
https://artofproblemsolving.com/wiki/index.php/2010_AMC_10B_Problems/Problem_17 | B | 23 | Every high school in the city of Euclid sent a team of $3$ students to a math contest. Each participant in the contest received a different score. Andrea's score was the median among all students, and hers was the highest score on her team. Andrea's teammates Beth and Carla placed $37$ th and $64$ th , respectively. Ho... | [
"There are $x$ schools. This means that there are $3x$ people. Because no one's score was the same as another person's score, that means that there could only have been $1$ median score. This implies that $x$ is an odd number. $x$ cannot be less than $23$ , because there wouldn't be a $64$ th place if x was. $x$ ca... |
https://artofproblemsolving.com/wiki/index.php/2010_AMC_10B_Problems/Problem_17 | null | 23 | Every high school in the city of Euclid sent a team of $3$ students to a math contest. Each participant in the contest received a different score. Andrea's score was the median among all students, and hers was the highest score on her team. Andrea's teammates Beth and Carla placed $37$ th and $64$ th , respectively. Ho... | [
"Let $a$ be Andrea's place. We know that she was the highest on her team, so $a < 37$\nSince $a$ is the median, there are $a-1$ to the left and right of the median, so the total number of people is $2a-1$ and the number of schools is $(2a-1)/3$ . This implies that $2a-1 \\equiv 0 \\pmod{3} \\implies a \\equiv 2 \\p... |
https://artofproblemsolving.com/wiki/index.php/2010_AMC_12B_Problems/Problem_8 | B | 23 | Every high school in the city of Euclid sent a team of $3$ students to a math contest. Each participant in the contest received a different score. Andrea's score was the median among all students, and hers was the highest score on her team. Andrea's teammates Beth and Carla placed $37$ th and $64$ th , respectively. Ho... | [
"There are $x$ schools. This means that there are $3x$ people. Because no one's score was the same as another person's score, that means that there could only have been $1$ median score. This implies that $x$ is an odd number. $x$ cannot be less than $23$ , because there wouldn't be a $64$ th place if x was. $x$ ca... |
https://artofproblemsolving.com/wiki/index.php/2010_AMC_12B_Problems/Problem_8 | null | 23 | Every high school in the city of Euclid sent a team of $3$ students to a math contest. Each participant in the contest received a different score. Andrea's score was the median among all students, and hers was the highest score on her team. Andrea's teammates Beth and Carla placed $37$ th and $64$ th , respectively. Ho... | [
"Let $a$ be Andrea's place. We know that she was the highest on her team, so $a < 37$\nSince $a$ is the median, there are $a-1$ to the left and right of the median, so the total number of people is $2a-1$ and the number of schools is $(2a-1)/3$ . This implies that $2a-1 \\equiv 0 \\pmod{3} \\implies a \\equiv 2 \\p... |
https://artofproblemsolving.com/wiki/index.php/2000_AIME_II_Problems/Problem_14 | null | 495 | Every positive integer $k$ has a unique factorial base expansion $(f_1,f_2,f_3,\ldots,f_m)$ , meaning that $k=1!\cdot f_1+2!\cdot f_2+3!\cdot f_3+\cdots+m!\cdot f_m$ , where each $f_i$ is an integer, $0\le f_i\le i$ , and $0<f_m$ . Given that $(f_1,f_2,f_3,\ldots,f_j)$ is the factorial base expansion of $16!-32!+48!-64... | [
"Note that $1+\\sum_{k=1}^{n-1} {k\\cdot k!} = 1+\\sum_{k=1}^{n-1} {((k+1)\\cdot k!- k!)} = 1+\\sum_{k=1}^{n-1} {((k+1)!- k!)} = 1 + ((2! - 1!) + (3! - 2!) + \\cdots + (n! - (n-1)!)) = n!$\nThus for all $m\\in\\mathbb{N}$\n$(32m+16)!-(32m)! = \\left(1+\\sum_{k=1}^{32m+15} {k\\cdot k!}\\right)-\\left(1+\\sum_{k=1}^{... |
https://artofproblemsolving.com/wiki/index.php/2017_AMC_10A_Problems/Problem_14 | D | 23 | Every week Roger pays for a movie ticket and a soda out of his allowance. Last week, Roger's allowance was $A$ dollars. The cost of his movie ticket was $20\%$ of the difference between $A$ and the cost of his soda, while the cost of his soda was $5\%$ of the difference between $A$ and the cost of his movie ticket. To ... | [
"Let $m$ = cost of movie ticket Let $s$ = cost of soda\nWe can create two equations:\n\\[m = \\frac{1}{5}(A - s)\\] \\[s = \\frac{1}{20}(A - m)\\]\nSubstituting we get:\n\\[m = \\frac{1}{5}(A - \\frac{1}{20}(A - m))\\] which yields: \\[m = \\frac{19}{99}A\\]\nNow we can find s and we get:\n\\[s = \\frac{4}{99}A\\]... |
https://artofproblemsolving.com/wiki/index.php/2010_AMC_8_Problems/Problem_25 | E | 24 | Everyday at school, Jo climbs a flight of $6$ stairs. Jo can take the stairs $1$ $2$ , or $3$ at a time. For example, Jo could climb $3$ , then $1$ , then $2$ . In how many ways can Jo climb the stairs?
$\textbf{(A)}\ 13 \qquad\textbf{(B)}\ 18\qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 22\qquad\textbf{(E)}\ 24$ | [
"A dynamics programming approach is quick and easy. The number of ways to climb one stair is $1$ . There are $2$ ways to climb two stairs: $1$ $1$ or $2$ . For 3 stairs, there are $4$ ways: \n( $1$ $1$ $1$ )\n( $1$ $2$ )\n( $2$ $1$ )\n( $3$\nFor four stairs, consider what step they came from to land on the fourth s... |
https://artofproblemsolving.com/wiki/index.php/1985_AHSME_Problems/Problem_14 | C | 6 | Exactly three of the interior angles of a convex polygon are obtuse. What is the maximum number of sides of such a polygon?
$\mathrm{(A)\ } 4 \qquad \mathrm{(B) \ }5 \qquad \mathrm{(C) \ } 6 \qquad \mathrm{(D) \ } 7 \qquad \mathrm{(E) \ }8$ | [
"Suppose that such a polygon has $n$ sides. Let the three obtuse angle measures, in degrees, be $o_1$ $o_2$ , and $o_3$ and the $(n-3)$ acute angle measures, again in degrees, be $a_1,a_2,a_3, \\dotsc a_{n-3}$\nSince $90 < o_i < 180$ for each $i$ , we have \\[3(90) = 270 < o_1+o_2+o_3 < 3(180) = 540,\\] and similar... |
https://artofproblemsolving.com/wiki/index.php/1998_AIME_Problems/Problem_8 | null | 618 | Except for the first two terms, each term of the sequence $1000, x, 1000 - x,\ldots$ is obtained by subtracting the preceding term from the one before that. The last term of the sequence is the first negative term encounted. What positive integer $x$ produces a sequence of maximum length? | [
"We can start to write out some of the inequalities now:\nAnd in general,\nIt is apparent that the bounds are slowly closing in on $x$ , so we can just calculate $x$ for some large value of $n$ (randomly, 10, 11):\n$x < \\frac{F_{7}}{F_{8}} \\cdot 1000 = \\frac{13}{21} \\cdot 1000 = 618.\\overline{18}$\n$x > \\frac... |
https://artofproblemsolving.com/wiki/index.php/1991_AIME_Problems/Problem_3 | null | 166 | Expanding $(1+0.2)^{1000}_{}$ by the binomial theorem and doing no further manipulation gives
${1000 \choose 0}(0.2)^0+{1000 \choose 1}(0.2)^1+{1000 \choose 2}(0.2)^2+\cdots+{1000 \choose 1000}(0.2)^{1000}$ $= A_0 + A_1 + A_2 + \cdots + A_{1000},$ where $A_k = {1000 \choose k}(0.2)^k$ for $k = 0,1,2,\ldots,1000$ . For ... | [
"Let $0<x_{}^{}<1$ . Then we may write $A_{k}^{}={N\\choose k}x^{k}=\\frac{N!}{k!(N-k)!}x^{k}=\\frac{(N-k+1)!}{k!}x^{k}$ . Taking logarithms in both sides of this last equation and using the well-known fact $\\log(a_{}^{}b)=\\log a + \\log b$ (valid if $a_{}^{},b_{}^{}>0$ ), we have\n$\\log(A_{k})=\\log\\left[\\fra... |
https://artofproblemsolving.com/wiki/index.php/2012_AMC_10A_Problems/Problem_11 | D | 12 | Externally tangent circles with centers at points $A$ and $B$ have radii of lengths $5$ and $3$ , respectively. A line externally tangent to both circles intersects ray $AB$ at point $C$ . What is $BC$
$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 4.8\qquad\textbf{(C)}\ 10.2\qquad\textbf{(D)}\ 12\qquad\textbf{(E)}\ 14.4$ | [
"Let $D$ and $E$ be the points of tangency on circles $A$ and $B$ with line $CD$ $AB=8$ . Also, let $BC=x$ . As $\\angle ADC$ and $\\angle BEC$ are right angles (a radius is perpendicular to a tangent line at the point of tangency) and both triangles share $\\angle ACD$ $\\triangle ADC \\sim \\triangle BEC$ . From ... |
https://artofproblemsolving.com/wiki/index.php/1992_AIME_Problems/Problem_7 | null | 320 | Faces $ABC^{}_{}$ and $BCD^{}_{}$ of tetrahedron $ABCD^{}_{}$ meet at an angle of $30^\circ$ . The area of face $ABC^{}_{}$ is $120^{}_{}$ , the area of face $BCD^{}_{}$ is $80^{}_{}$ , and $BC=10^{}_{}$ . Find the volume of the tetrahedron. | [
"Since the area $BCD=80=\\frac{1}{2}\\cdot10\\cdot16$ , the perpendicular from $D$ to $BC$ has length $16$\nThe perpendicular from $D$ to $ABC$ is $16 \\cdot \\sin 30^\\circ=8$ . Therefore, the volume is $\\frac{8\\cdot120}{3}=\\boxed{320}$"
] |
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