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/2013_AIME_I_Problems/Problem_11 | null | 148 | Ms. Math's kindergarten class has $16$ registered students. The classroom has a very large number, $N$ , of play blocks which satisfies the conditions:
(a) If $16$ $15$ , or $14$ students are present in the class, then in each case all the blocks can be distributed in equal numbers to each student, and
(b) There are th... | [
"$N$ must be some multiple of $\\text{lcm}(14, 15, 16)= 2^{4}\\cdot 3\\cdot 5\\cdot 7$ ; this $lcm$ is hereby denoted $k$ and $N = qk$\n$1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$ $10$ , and $12$ all divide $k$ , so $x, y, z = 9, 11, 13$\nWe have the following three modulo equations:\n$nk\\equiv 3\\pmod{9}$\n$nk\\equiv 3\\pmod... |
https://artofproblemsolving.com/wiki/index.php/2008_AMC_8_Problems/Problem_17 | D | 132 | Ms.Osborne asks each student in her class to draw a rectangle with integer side lengths and a perimeter of $50$ units. All of her students calculate the area of the rectangle they draw. What is the difference between the largest and smallest possible areas of the rectangles?
$\textbf{(A)}\ 76\qquad \textbf{(B)}\ 120\qq... | [
"A rectangle's area is maximized when its length and width are equivalent, or the two side lengths are closest together in this case with integer lengths. This occurs with the sides $12 \\times 13 = 156$ . Likewise, the area is smallest when the side lengths have the greatest difference, which is $1 \\times 24 = 24... |
https://artofproblemsolving.com/wiki/index.php/2023_AMC_8_Problems/Problem_11 | C | 60,000 | NASA’s Perseverance Rover was launched on July $30,$ $2020.$ After traveling $292526838$ miles, it landed on Mars in Jezero Crater about $6.5$ months later. Which of the following is closest to the Rover’s average interplanetary speed in miles per hour?
$\textbf{(A)}\ 6000 \qquad \textbf{(B)}\ 12000 \qquad \textbf{(C)}... | [
"Note that $6.5$ months is approximately $6.5\\cdot30\\cdot24$ hours. Therefore, the speed (in miles per hour) is \\[\\frac{292526838}{6.5\\cdot30\\cdot24} \\approx \\frac{300000000}{6.5\\cdot30\\cdot24} = \\frac{10000000}{6.5\\cdot24} \\approx \\frac{10000000}{6.4\\cdot25} = \\frac{10000000}{160} = 62500 \\approx ... |
https://artofproblemsolving.com/wiki/index.php/2023_AMC_8_Problems/Problem_14 | E | 55 | Nicolas is planning to send a package to his friend Anton, who is a stamp collector. To pay for the postage, Nicolas would like to cover the package with a large number of stamps. Suppose he has a collection of $5$ -cent, $10$ -cent, and $25$ -cent stamps, with exactly $20$ of each type. What is the greatest number of ... | [
"Let's use the most stamps to make $7.10.$ We have $20$ of each stamp, $5$ -cent (nickels), $10$ -cent (dimes), and $25$ -cent (quarters).\nIf we want the highest number of stamps, we must have the highest number of the smaller value stamps (like the coins above). We can use $20$ nickels and $20$ dimes to bring our... |
https://artofproblemsolving.com/wiki/index.php/2004_AMC_8_Problems/Problem_12 | B | 8 | Niki usually leaves her cell phone on. If her cell phone is on but
she is not actually using it, the battery will last for $24$ hours. If
she is using it constantly, the battery will last for only $3$ hours.
Since the last recharge, her phone has been on $9$ hours, and during
that time she has used it for $60$ minutes.... | [
"When not being used, the cell phone uses up $\\frac{1}{24}$ of its battery per hour. When being used, the cell phone uses up $\\frac{1}{3}$ of its battery per hour. Since Niki's phone has been on for $9$ hours, of those $8$ simply on and $1$ being used to talk, $8(\\frac{1}{24}) + 1(\\frac{1}{3}) = \\frac{2}{3}$ o... |
https://artofproblemsolving.com/wiki/index.php/2011_AIME_II_Problems/Problem_12 | null | 97 | Nine delegates, three each from three different countries, randomly select chairs at a round table that seats nine people. Let the probability that each delegate sits next to at least one delegate from another country be $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$ | [
"We use complementary counting.\nWe can order the $9$ people around a circle in $\\frac{9!}{9} = 8!$ ways. Now we count when there is at least one delegate surrounded by people from only his/her country.\nLet the countries be $A,B,C$ . Suppose that the group $XXX$ (for some country $X$ ) appears. To account for cir... |
https://artofproblemsolving.com/wiki/index.php/2012_AIME_I_Problems/Problem_3 | null | 216 | Nine people sit down for dinner where there are three choices of meals. Three people order the beef meal, three order the chicken meal, and three order the fish meal. The waiter serves the nine meals in random order. Find the number of ways in which the waiter could serve the meal types to the nine people so that exact... | [
"Call a beef meal $B,$ a chicken meal $C,$ and a fish meal $F.$ Now say the nine people order meals $\\text{BBBCCCFFF}$ respectively and say that the person who receives the correct meal is the first person. We will solve for this case and then multiply by $9$ to account for the $9$ different ways in which the pers... |
https://artofproblemsolving.com/wiki/index.php/1998_AIME_Problems/Problem_4 | null | 17 | Nine tiles are numbered $1, 2, 3, \cdots, 9,$ respectively. Each of three players randomly selects and keeps three of the tiles, and sums those three values. The probability that all three players obtain an odd sum is $m/n,$ where $m$ and $n$ are relatively prime positive integers . Find $m+n.$ | [
"In order for a player to have an odd sum, he must have an odd number of odd tiles: that is, he can either have three odd tiles, or two even tiles and an odd tile. Thus, since there are $5$ odd tiles and $4$ even tiles, the only possibility is that one player gets $3$ odd tiles and the other two players get $2$ ev... |
https://artofproblemsolving.com/wiki/index.php/1994_AIME_Problems/Problem_11 | null | 465 | Ninety-four bricks, each measuring $4''\times10''\times19'',$ are to be stacked one on top of another to form a tower 94 bricks tall. Each brick can be oriented so it contributes $4''\,$ or $10''\,$ or $19''\,$ to the total height of the tower. How many different tower heights can be achieved using all ninety-four of... | [
"We have the smallest stack, which has a height of $94 \\times 4$ inches. Now when we change the height of one of the bricks, we either add $0$ inches, $6$ inches, or $15$ inches to the height. Now all we need to do is to find the different change values we can get from $94$ $0$ 's, $6$ 's, and $15$ 's. Because $0$... |
https://artofproblemsolving.com/wiki/index.php/2010_AMC_10A_Problems/Problem_16 | B | 33 | Nondegenerate $\triangle ABC$ has integer side lengths, $\overline{BD}$ is an angle bisector, $AD = 3$ , and $DC=8$ . What is the smallest possible value of the perimeter?
$\textbf{(A)}\ 30 \qquad \textbf{(B)}\ 33 \qquad \textbf{(C)}\ 35 \qquad \textbf{(D)}\ 36 \qquad \textbf{(E)}\ 37$ | [
"By the Angle Bisector Theorem , we know that $\\frac{AB}{BC} = \\frac{3}{8}$ . If we use the lowest possible integer values for $AB$ and $BC$ (the lengths of $AD$ and $DC$ , respectively), then $AB + BC = AD + DC = AC$ , contradicting the Triangle Inequality . If we use the next lowest values ( $AB = 6$ and $BC = ... |
https://artofproblemsolving.com/wiki/index.php/2010_AMC_12A_Problems/Problem_14 | B | 33 | Nondegenerate $\triangle ABC$ has integer side lengths, $\overline{BD}$ is an angle bisector, $AD = 3$ , and $DC=8$ . What is the smallest possible value of the perimeter?
$\textbf{(A)}\ 30 \qquad \textbf{(B)}\ 33 \qquad \textbf{(C)}\ 35 \qquad \textbf{(D)}\ 36 \qquad \textbf{(E)}\ 37$ | [
"By the Angle Bisector Theorem , we know that $\\frac{AB}{BC} = \\frac{3}{8}$ . If we use the lowest possible integer values for $AB$ and $BC$ (the lengths of $AD$ and $DC$ , respectively), then $AB + BC = AD + DC = AC$ , contradicting the Triangle Inequality . If we use the next lowest values ( $AB = 6$ and $BC = ... |
https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_7 | B | 1 | Nonzero real numbers $x$ $y$ $a$ , and $b$ satisfy $x < a$ and $y < b$ . How many of the following inequalities must be true?
$\textbf{(I)}\ x+y < a+b\qquad$
$\textbf{(II)}\ x-y < a-b\qquad$
$\textbf{(III)}\ xy < ab\qquad$
$\textbf{(IV)}\ \frac{x}{y} < \frac{a}{b}$
$\textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)... | [
"Let us denote $a = x + k$ where $k > 0$ and $b = y + l$ where $l > 0$ . We can write that $x + y < x + y + k + l \\implies x + y < a + b$\nIt is important to note that $1$ counterexample fully disproves a claim. Let's try substituting $x=-3,y=-4,a=1,b=4$\n$\\textbf{(II)}$ states that $x-y<a-b \\implies -3 - (-4) <... |
https://artofproblemsolving.com/wiki/index.php/2003_AMC_12A_Problems/Problem_22 | C | 0.2 | Objects $A$ and $B$ move simultaneously in the coordinate plane via a sequence of steps, each of length one. Object $A$ starts at $(0,0)$ and each of its steps is either right or up, both equally likely. Object $B$ starts at $(5,7)$ and each of its steps is either to the left or down, both equally likely. Which of the ... | [
"We know that the sum of the vertical steps must be equal to $7$ . We also know that they must take $6$ steps each. Since moving vertically or horizontally is equally likely, we can write all the possible paths as a generating function:\n\\[P(x)=(x+1)^{12}\\]\nWhere we need to extract the $x^5$ coefficient. By the ... |
https://artofproblemsolving.com/wiki/index.php/2018_AIME_II_Problems/Problem_9 | null | 184 | Octagon $ABCDEFGH$ with side lengths $AB = CD = EF = GH = 10$ and $BC = DE = FG = HA = 11$ is formed by removing 6-8-10 triangles from the corners of a $23$ $\times$ $27$ rectangle with side $\overline{AH}$ on a short side of the rectangle, as shown. Let $J$ be the midpoint of $\overline{AH}$ , and partition the octago... | [
"We represent octagon $ABCDEFGH$ in the coordinate plane with the upper left corner of the rectangle being the origin. Then it follows that $A=(0,-6), B=(8, 0), C=(19,0), D=(27, -6), E=(27, -17), F=(19, -23), G=(8, -23), H=(0, -17), J=(0, -\\frac{23}{2})$ . Recall that the centroid is $\\frac{1}{3}$ way up each med... |
https://artofproblemsolving.com/wiki/index.php/2006_AMC_10A_Problems/Problem_15 | D | 47 | Odell and Kershaw run for $30$ minutes on a circular track. Odell runs clockwise at $250 m/min$ and uses the inner lane with a radius of $50$ meters. Kershaw runs counterclockwise at $300 m/min$ and uses the outer lane with a radius of $60$ meters, starting on the same radial line as Odell. How many times after the sta... | [
"Since $d = rt$ , we note that Odell runs one lap in $\\frac{2 \\cdot 50\\pi}{250} = \\frac{2\\pi}{5}$ minutes, while Kershaw also runs one lap in $\\frac{2 \\cdot 60\\pi}{300} = \\frac{2\\pi}{5}$ minutes. They take the same amount of time to run a lap, and since they are running in opposite directions they will me... |
https://artofproblemsolving.com/wiki/index.php/2006_AMC_10A_Problems/Problem_15 | null | 47 | Odell and Kershaw run for $30$ minutes on a circular track. Odell runs clockwise at $250 m/min$ and uses the inner lane with a radius of $50$ meters. Kershaw runs counterclockwise at $300 m/min$ and uses the outer lane with a radius of $60$ meters, starting on the same radial line as Odell. How many times after the sta... | [
"Since Odell's rate is $5/6$ that of Kershaw, but Kershaw's lap distance is $6/5$ that of Odell, they each run a lap in the same time. Hence they pass twice each time they circle the track. Odell runs \\[(30 \\ \\text{min})\\left(250\\frac{\\text{m}}{\\text{min}}\\right)\\left(\\frac{1}{100\\pi}\\frac{\\text{laps}}... |
https://artofproblemsolving.com/wiki/index.php/2010_AMC_8_Problems/Problem_12 | D | 100 | Of the $500$ balls in a large bag, $80\%$ are red and the rest are blue. How many of the red balls must be removed so that $75\%$ of the remaining balls are red?
$\textbf{(A)}\ 25\qquad\textbf{(B)}\ 50\qquad\textbf{(C)}\ 75\qquad\textbf{(D)}\ 100\qquad\textbf{(E)}\ 150$ | [
"Since 80 percent of the 500 balls are red, there are 400 red balls. Therefore, there must be 100 blue balls. For the 100 blue balls to be 25% or $\\dfrac{1}{4}$ of the bag, there must be 400 balls in the bag so 100 red balls must be removed. The answer is $\\boxed{100}$",
"We could also set up a proportion. Sinc... |
https://artofproblemsolving.com/wiki/index.php/2001_AMC_8_Problems/Problem_13 | D | 50 | Of the 36 students in Richelle's class, 12 prefer chocolate pie, 8 prefer apple, and 6 prefer blueberry. Half of the remaining students prefer cherry pie and half prefer lemon. For Richelle's pie graph showing this data, how many degrees should she use for cherry pie?
$\text{(A)}\ 10 \qquad \text{(B)}\ 20 \qquad \text{... | [
"There are $36$ students in the class: $12$ prefer chocolate pie, $8$ prefer apple pie, and $6$ prefer blueberry pie. Therefore, $36-12-8-6=10$ students prefer cherry pie or lemon pie. Half of these prefer each, so $5$ students prefer cherry pie. This means that $\\frac{5}{36}$ of the students prefer cherry pie, so... |
https://artofproblemsolving.com/wiki/index.php/1950_AHSME_Problems/Problem_18 | E | 1 | Of the following
(1) $a(x-y)=ax-ay$
(2) $a^{x-y}=a^x-a^y$
(3) $\log (x-y)=\log x-\log y$
(4) $\frac{\log x}{\log y}=\log{x}-\log{y}$
(5) $a(xy)=ax \cdot ay$
$\textbf{(A)}\text{Only 1 and 4 are true}\qquad\\\textbf{(B)}\ \text{Only 1 and 5 are true}\qquad\\\textbf{(C)}\ \text{Only 1 and 3 are true}\qquad\\\textbf{(D)}\ ... | [
"The distributive property doesn't apply to logarithms or in the ways illustrated, and only applies to addition and subtraction. Also, $a^{x-y} = \\frac{a^x}{a^y}$ , so $\\boxed{1}$"
] |
https://artofproblemsolving.com/wiki/index.php/2021_AMC_12A_Problems/Problem_13 | B | 3 | Of the following complex numbers $z$ , which one has the property that $z^5$ has the greatest real part?
$\textbf{(A) }{-}2 \qquad \textbf{(B) }{-}\sqrt3+i \qquad \textbf{(C) }{-}\sqrt2+\sqrt2 i \qquad \textbf{(D) }{-}1+\sqrt3 i\qquad \textbf{(E) }2i$ | [
"First, $\\textbf{(B)}$ is $2\\text{cis}(150)$ $\\textbf{(C)}$ is $2\\text{cis}(135)$ $\\textbf{(D)}$ is $2\\text{cis}(120)$\nTaking the real part of the $5$ th power of each we have:\n$\\textbf{(A): }(-2)^5=-32$\n$\\textbf{(B): }32\\cos(750)=32\\cos(30)=16\\sqrt{3}$\n$\\textbf{(C): }32\\cos(675)=32\\cos(-45)=16\\s... |
https://artofproblemsolving.com/wiki/index.php/2008_AIME_I_Problems/Problem_1 | null | 252 | Of the students attending a school party, $60\%$ of the students are girls, and $40\%$ of the students like to dance. After these students are joined by $20$ more boy students, all of whom like to dance, the party is now $58\%$ girls. How many students now at the party like to dance? | [
"Say that there were $3k$ girls and $2k$ boys at the party originally. $2k$ like to dance. Then, there are $3k$ girls and $2k + 20$ boys, and $2k + 20$ like to dance.\nThus, $\\dfrac{3k}{5k + 20} = \\dfrac{29}{50}$ , solving gives $k = 116$ . Thus, the number of people that like to dance is $2k + 20 = \\boxed{252}$... |
https://artofproblemsolving.com/wiki/index.php/2022_AMC_10A_Problems/Problem_12 | A | 7 | On Halloween $31$ children walked into the principal's office asking for candy. They
can be classified into three types: Some always lie; some always tell the truth; and
some alternately lie and tell the truth. The alternaters arbitrarily choose their first
response, either a lie or the truth, but each subsequent state... | [
"Note that:\nSuppose that there are $T$ truth-tellers, $L$ liars, and $A$ alternaters who responded lie-truth-lie.\nThe conditions of the first two questions imply that \\begin{align*} T+L+A&=22, \\\\ L+A&=15. \\end{align*} Subtracting the second equation from the first, we have $T=22-15=\\boxed{7}.$",
"Consider ... |
https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_9 | A | 7 | On Halloween $31$ children walked into the principal's office asking for candy. They
can be classified into three types: Some always lie; some always tell the truth; and
some alternately lie and tell the truth. The alternaters arbitrarily choose their first
response, either a lie or the truth, but each subsequent state... | [
"Note that:\nSuppose that there are $T$ truth-tellers, $L$ liars, and $A$ alternaters who responded lie-truth-lie.\nThe conditions of the first two questions imply that \\begin{align*} T+L+A&=22, \\\\ L+A&=15. \\end{align*} Subtracting the second equation from the first, we have $T=22-15=\\boxed{7}.$",
"Consider ... |
https://artofproblemsolving.com/wiki/index.php/2011_AMC_10B_Problems/Problem_6 | A | 30 | On Halloween Casper ate $\frac{1}{3}$ of his candies and then gave $2$ candies to his brother. The next day he ate $\frac{1}{3}$ of his remaining candies and then gave $4$ candies to his sister. On the third day he ate his final $8$ candies. How many candies did Casper have at the beginning?
$\textbf{(A)}\ 30 \qquad\te... | [
"Let $x$ represent the amount of candies Casper had at the beginning.\n\\begin{align*} \\frac{2}{3} \\left(\\frac{2}{3} x - 2\\right) - 4 - 8 &= 0\\\\ \\frac{2}{3} x - 2 &= 18\\\\ \\frac{2}{3} x &= 20\\\\ x &= \\boxed{30}",
"We work backwards. If he had 8 candies at the end, then before he gave candies to his sis... |
https://artofproblemsolving.com/wiki/index.php/2015_AMC_8_Problems/Problem_22 | C | 60 | On June 1, a group of students is standing in rows, with 15 students in each row. On June 2, the same group is standing with all of the students in one long row. On June 3, the same group is standing with just one student in each row. On June 4, the same group is standing with 6 students in each row. This process conti... | [
"The text suggests that the number of students in the group has $12$ factors, since each arrangement is a factor. The smallest integer with $12$ factors is $2^2\\cdot3\\cdot5=\\boxed{60}$",
"Since we know the number must be a multiple of $15$ , we can eliminate $A$ . We also know that after $12$ days, the student... |
https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_8 | D | 6 | On Monday Taye has $$2$ . Every day, he either gains $$3$ or doubles the amount of money he had on the previous day. How many different dollar amounts could Taye have on Thursday, $3$ days later?
$\textbf{(A) } 3\qquad\textbf{(B) } 4\qquad\textbf{(C) } 5\qquad\textbf{(D) } 6\qquad\textbf{(E) } 7$ | [
"How many values could be on the first day? Only $2$ dollars. The second day, you can either add $3$ dollars, or double, so you can have $5$ dollars, or $4$ . For each of these values, you have $2$ values for each. For $5$ dollars, you have $10$ dollars or $8$ , and for $4$ dollars, you have $8$ dollars or $ $7$ . ... |
https://artofproblemsolving.com/wiki/index.php/1996_AHSME_Problems/Problem_28 | C | 2.1 | On a $4\times 4\times 3$ rectangular parallelepiped, vertices $A$ $B$ , and $C$ are adjacent to vertex $D$ . The perpendicular distance from $D$ to the plane containing $A$ $B$ , and $C$ is closest to
[asy] size(120); import three; currentprojection=orthographic(1, 4/5, 1/3); draw(box(O, (4,4,3))); triple A=(0,4,3), B=... | [
"By placing the cube in a coordinate system such that $D$ is at the origin, $A(0,0,3)$ $B(4,0,0)$ , and $C(0,4,0)$ , we find that the equation of plane $ABC$ is:\n\\[\\frac{x}{4} + \\frac{y}{4} + \\frac{z}{3} = 1,\\] so $3x + 3y + 4z - 12 = 0.$ The equation for the distance of a point $(a,b,c)$ to a plane $Ax + By ... |
https://artofproblemsolving.com/wiki/index.php/2010_AMC_10B_Problems/Problem_15 | C | 29 | On a $50$ -question multiple choice math contest, students receive $4$ points for a correct answer, $0$ points for an answer left blank, and $-1$ point for an incorrect answer. Jesse’s total score on the contest was $99$ . What is the maximum number of questions that Jesse could have answered correctly?
$\textbf{(A)}\ ... | [
"Let $a$ be the amount of questions Jesse answered correctly, $b$ be the amount of questions Jesse left blank, and $c$ be the amount of questions Jesse answered incorrectly. Since there were $50$ questions on the contest, $a+b+c=50$ . Since his total score was $99$ $4a-c=99$ . Also, $a+c\\leq50 \\Rightarrow c\\leq5... |
https://artofproblemsolving.com/wiki/index.php/2005_AMC_10B_Problems/Problem_19 | B | 1 | On a certain math exam, $10\%$ of the students got $70$ points, $25\%$ got $80$ points, $20\%$ got $85$ points, $15\%$ got $90$ points, and the rest got $95$ points. What is the difference between the mean and the median score on this exam?
$\textbf{(A) }\ 0 \qquad \textbf{(B) }\ 1 \qquad \textbf{(C) }\ 2 \qquad \text... | [
"To begin, we see that the remaining $30\\%$ of the students got $95$ points. Assume that there are $20$ students; we see that $2$ students got $70$ points, $5$ students got $80$ points, $4$ students got $85$ points, $3$ students got $90$ points, and $6$ students got $95$ points. The median is $85$ , since the $10^... |
https://artofproblemsolving.com/wiki/index.php/2005_AMC_12B_Problems/Problem_9 | B | 1 | On a certain math exam, $10\%$ of the students got $70$ points, $25\%$ got $80$ points, $20\%$ got $85$ points, $15\%$ got $90$ points, and the rest got $95$ points. What is the difference between the mean and the median score on this exam?
$\textbf{(A) }\ 0 \qquad \textbf{(B) }\ 1 \qquad \textbf{(C) }\ 2 \qquad \text... | [
"To begin, we see that the remaining $30\\%$ of the students got $95$ points. Assume that there are $20$ students; we see that $2$ students got $70$ points, $5$ students got $80$ points, $4$ students got $85$ points, $3$ students got $90$ points, and $6$ students got $95$ points. The median is $85$ , since the $10^... |
https://artofproblemsolving.com/wiki/index.php/2008_AIME_I_Problems/Problem_12 | null | 375 | On a long straight stretch of one-way single-lane highway, cars all travel at the same speed and all obey the safety rule: the distance from the back of the car ahead to the front of the car behind is exactly one car length for each 15 kilometers per hour of speed or fraction thereof (Thus the front of a car traveling ... | [
"Let $n$ be the number of car lengths that separates each car (it is easy to see that this should be the same between each pair of consecutive cars.) Then their speed is at most $15n$ . Let a unit be the distance between the cars (front to front). Then the length of each unit is $4(n + 1)$ . To maximize, in a unit,... |
https://artofproblemsolving.com/wiki/index.php/2004_AMC_8_Problems/Problem_1 | B | 102 | On a map, a $12$ -centimeter length represents $72$ kilometers. How many kilometers does a $17$ -centimeter length represent?
$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 102\qquad\textbf{(C)}\ 204\qquad\textbf{(D)}\ 864\qquad\textbf{(E)}\ 1224$ | [
"We set up the proportion $\\frac{12 \\text{cm}}{72 \\text{km}}=\\frac{17 \\text{cm}}{x \\text{km}}$ . Thus $x=102 \\Rightarrow \\boxed{102}$"
] |
https://artofproblemsolving.com/wiki/index.php/2013_AMC_10B_Problems/Problem_3 | C | 5 | On a particular January day, the high temperature in Lincoln, Nebraska, was $16$ degrees higher than the low temperature, and the average of the high and low temperatures was $3$ . In degrees, what was the low temperature in Lincoln that day?
$\textbf{(A)}\ -13 \qquad \textbf{(B)}\ -8 \qquad \textbf{(C)}\ -5 \qquad \te... | [
"Let $L$ be the low temperature. The high temperature is $L+16$ . The average is $\\frac{L+(L+16)}{2}=3$ . Solving for $L$ , we get $L=\\boxed{5}$"
] |
https://artofproblemsolving.com/wiki/index.php/2013_AMC_12B_Problems/Problem_1 | C | 5 | On a particular January day, the high temperature in Lincoln, Nebraska, was $16$ degrees higher than the low temperature, and the average of the high and low temperatures was $3$ . In degrees, what was the low temperature in Lincoln that day?
$\textbf{(A)}\ -13 \qquad \textbf{(B)}\ -8 \qquad \textbf{(C)}\ -5 \qquad \te... | [
"Let $L$ be the low temperature. The high temperature is $L+16$ . The average is $\\frac{L+(L+16)}{2}=3$ . Solving for $L$ , we get $L=\\boxed{5}$"
] |
https://artofproblemsolving.com/wiki/index.php/2015_AMC_12A_Problems/Problem_11 | D | 5 | On a sheet of paper, Isabella draws a circle of radius $2$ , a circle of radius $3$ , and all possible lines simultaneously tangent to both circles. Isabella notices that she has drawn exactly $k \ge 0$ lines. How many different values of $k$ are possible?
$\textbf{(A)}\ 2 \qquad\textbf{(B)}\ 3 \qquad\textbf{(C)}\ 4 \... | [
"Isabella can get $0$ lines if the circles are concentric, $1$ if internally tangent, $2$ if overlapping, $3$ if externally tangent, and $4$ if non-overlapping and not externally tangent. There are $\\boxed{5}$ values of $k$"
] |
https://artofproblemsolving.com/wiki/index.php/2004_AMC_10B_Problems/Problem_7 | A | 5 | On a trip from the United States to Canada, Isabella took $d$ U.S. dollars. At the border she exchanged them all, receiving $10$ Canadian dollars for every $7$ U.S. dollars. After spending $60$ Canadian dollars, she had $d$ Canadian dollars left. What is the sum of the digits of $d$
$\mathrm{(A)\ }5\qquad\mathrm{(B)\ }... | [
"Isabella had $60+d$ Canadian dollars. Setting up an equation we get $d=\\frac{7}{10}\\cdot(60+d)$ , which solves to $d=140$ , and the sum of digits of $d$ is $\\boxed{5}$",
"Each time Isabella exchanges $7$ U.S. dollars, she gets $7$ Canadian dollars and $3$ Canadian dollars extra. Isabella received a total of $... |
https://artofproblemsolving.com/wiki/index.php/2004_AMC_12B_Problems/Problem_5 | A | 5 | On a trip from the United States to Canada, Isabella took $d$ U.S. dollars. At the border she exchanged them all, receiving $10$ Canadian dollars for every $7$ U.S. dollars. After spending $60$ Canadian dollars, she had $d$ Canadian dollars left. What is the sum of the digits of $d$
$\mathrm{(A)\ }5\qquad\mathrm{(B)\ }... | [
"Isabella had $60+d$ Canadian dollars. Setting up an equation we get $d=\\frac{7}{10}\\cdot(60+d)$ , which solves to $d=140$ , and the sum of digits of $d$ is $\\boxed{5}$",
"Each time Isabella exchanges $7$ U.S. dollars, she gets $7$ Canadian dollars and $3$ Canadian dollars extra. Isabella received a total of $... |
https://artofproblemsolving.com/wiki/index.php/2018_AMC_8_Problems/Problem_6 | C | 80 | On a trip to the beach, Anh traveled 50 miles on the highway and 10 miles on a coastal access road. He drove three times as fast on the highway as on the coastal road. If Anh spent 30 minutes driving on the coastal road, how many minutes did his entire trip take?
[mathjax]\textbf{(A) }50\qquad\textbf{(B) }70\qquad\text... | [
"Since Anh spends half an hour to drive 10 miles on the coastal road, his speed is [mathjax]r=\\dfrac dt=\\dfrac{10}{0.5}=20[/mathjax] mph. His speed on the highway then is [mathjax]60[/mathjax] mph. He drives [mathjax]50[/mathjax] miles, so he drives for [mathjax]\\dfrac{5}{6}[/mathjax] hours, which is equal to [m... |
https://artofproblemsolving.com/wiki/index.php/1992_AJHSME_Problems/Problem_18 | A | 45 | On a trip, a car traveled $80$ miles in an hour and a half, then was stopped in traffic for $30$ minutes, then traveled $100$ miles during the next $2$ hours. What was the car's average speed in miles per hour for the $4$ -hour trip?
$\text{(A)}\ 45 \qquad \text{(B)}\ 50 \qquad \text{(C)}\ 60 \qquad \text{(D)}\ 75 \qq... | [
"The average speed is given by the total distance traveled divided by the total time traveled. \\[\\frac{80+0+100}{4} = \\boxed{45}\\]"
] |
https://artofproblemsolving.com/wiki/index.php/2001_AMC_8_Problems/Problem_22 | E | 97 | On a twenty-question test, each correct answer is worth 5 points, each unanswered question is worth 1 point and each incorrect answer is worth 0 points. Which of the following scores is NOT possible?
$\text{(A)}\ 90 \qquad \text{(B)}\ 91 \qquad \text{(C)}\ 92 \qquad \text{(D)}\ 95 \qquad \text{(E)}\ 97$ | [
"The highest possible score is if you get every answer right, to get $5(20)=100$ . The second highest possible score is if you get $19$ questions right and leave the remaining one blank, to get a $5(19)+1(1)=96$ . Therefore, no score between $96$ and $100$ , exclusive, is possible, so $97$ is not possible, $\\boxed... |
https://artofproblemsolving.com/wiki/index.php/2009_AMC_8_Problems/Problem_2 | D | 49 | On average, for every 4 sports cars sold at the local dealership, 7 sedans are sold. The dealership predicts that it will sell 28 sports cars next month. How many sedans does it expect to sell?
$\textbf{(A)}\ 7\qquad\textbf{(B)}\ 32\qquad\textbf{(C)}\ 35\qquad\textbf{(D)}\ 49\qquad\textbf{(E)}\ 112$ | [
"This means the ratio is $4:7$ . If the ratio now is $28:x$ , then that means $x= \\boxed{49}$"
] |
https://artofproblemsolving.com/wiki/index.php/2015_AMC_8_Problems/Problem_9 | D | 400 | On her first day of work, Janabel sold one widget. On day two, she sold three widgets. On day three, she sold five widgets, and on each succeeding day, she sold two more widgets than she had sold on the previous day. How many widgets in total had Janabel sold after working $20$ days?
$\textbf{(A) }39\qquad\textbf{(B) }... | [
"First, we have to find how many widgets she makes on Day $20$ . We can write the linear equation $y=-1+2x$ to represent this situation. Then, we can plug in $20$ for $x$ $y=-1+2(20)$ -- $y=-1+40$ -- $y=39$ . The sum of $1,3,5, ... 39$ is $\\dfrac{(1 + 39)(20)}{2}= \\boxed{400}$",
"The sum is just the first $20$ ... |
https://artofproblemsolving.com/wiki/index.php/2014_AIME_I_Problems/Problem_13 | null | 850 | On square $ABCD$ , points $E,F,G$ , and $H$ lie on sides $\overline{AB},\overline{BC},\overline{CD},$ and $\overline{DA},$ respectively, so that $\overline{EG} \perp \overline{FH}$ and $EG=FH = 34$ . Segments $\overline{EG}$ and $\overline{FH}$ intersect at a point $P$ , and the areas of the quadrilaterals $AEPH, BFPE,... | [
"$269+275+405+411=1360$ , a multiple of $17$ . In addition, $EG=FH=34$ , which is $17\\cdot 2$ .\nTherefore, we suspect the square of the \"hypotenuse\" of a right triangle, corresponding to $EG$ and $FH$ must be a multiple of $17$ . All of these triples are primitive:\n\\[17=1^2+4^2\\] \\[34=3^2+5^2\\] \\[51=\\emp... |
https://artofproblemsolving.com/wiki/index.php/2006_AMC_8_Problems/Problem_2 | C | 13 | On the AMC 8 contest Billy answers 13 questions correctly, answers 7 questions incorrectly and doesn't answer the last 5. What is his score?
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 13\qquad\textbf{(D)}\ 19\qquad\textbf{(E)}\ 26$ | [
"As the AMC 8 only rewards 1 point for each correct answer, everything is irrelevant except the number Billy answered correctly, $\\boxed{13}$"
] |
https://artofproblemsolving.com/wiki/index.php/2009_AMC_8_Problems/Problem_23 | B | 28 | On the last day of school, Mrs. Wonderful gave jelly beans to her class. She gave each boy as many jelly beans as there were boys in the class. She gave each girl as many jelly beans as there were girls in the class. She brought $400$ jelly beans, and when she finished, she had six jelly beans left. There were two more... | [
"If there are $x$ girls, then there are $x+2$ boys. She gave each girl $x$ jellybeans and each boy $x+2$ jellybeans, for a total of $x^2 + (x+2)^2$ jellybeans. She gave away $400-6=394$ jellybeans.\n\\begin{align*} x^2+(x+2)^2 &= 394\\\\ x^2+x^2+4x+4 &= 394\\\\ 2x^2 + 4x &= 390\\\\ x^2 + 2x &= 195\\\\ \\end{align*}... |
https://artofproblemsolving.com/wiki/index.php/2000_AIME_II_Problems/Problem_6 | null | 181 | One base of a trapezoid is $100$ units longer than the other base. The segment that joins the midpoints of the legs divides the trapezoid into two regions whose areas are in the ratio $2: 3$ . Let $x$ be the length of the segment joining the legs of the trapezoid that is parallel to the bases and that divides the trape... | [
"Let the shorter base have length $b$ (so the longer has length $b+100$ ), and let the height be $h$ . The length of the midline of the trapezoid is the average of its bases, which is $\\frac{b+b+100}{2} = b+50$ . The two regions which the midline divides the trapezoid into are two smaller trapezoids, both with hei... |
https://artofproblemsolving.com/wiki/index.php/2009_AMC_10A_Problems/Problem_1 | E | 11 | One can holds $12$ ounces of soda, what is the minimum number of cans needed to provide a gallon ( $128$ ounces) of soda?
$\textbf{(A)}\ 7\qquad \textbf{(B)}\ 8\qquad \textbf{(C)}\ 9\qquad \textbf{(D)}\ 10\qquad \textbf{(E)}\ 11$ | [
"$10$ cans would hold $120$ ounces, but $128>120$ , so $11$ cans are required. Thus, the answer is $\\mathrm{\\boxed{11}$"
] |
https://artofproblemsolving.com/wiki/index.php/1988_AIME_Problems/Problem_1 | null | 770 | One commercially available ten-button lock may be opened by pressing -- in any order -- the correct five buttons. The sample shown below has $\{1,2,3,6,9\}$ as its combination . Suppose that these locks are redesigned so that sets of as many as nine buttons or as few as one button could serve as combinations. How many ... | [
"Currently there are ${10 \\choose 5}$ possible combinations.\nWith any integer $x$ from $1$ to $9$ , the number of ways to choose a set of $x$ buttons is $\\sum^{9}_{k=1}{10 \\choose k}$ .\nNow we can use the identity $\\sum^{n}_{k=0}{n \\choose k}=2^{n}$ .\nSo the number of additional combinations is just $2^{10}... |
https://artofproblemsolving.com/wiki/index.php/2009_AMC_10A_Problems/Problem_11 | D | 125 | One dimension of a cube is increased by $1$ , another is decreased by $1$ , and the third is left unchanged. The volume of the new rectangular solid is $5$ less than that of the cube. What was the volume of the cube?
$\textbf{(A)}\ 8 \qquad \textbf{(B)}\ 27 \qquad \textbf{(C)}\ 64 \qquad \textbf{(D)}\ 125 \qquad \textb... | [
"Let the original cube have edge length $a$ . Then its volume is $a^3$ .\nThe new box has dimensions $a-1$ $a$ , and $a+1$ , hence its volume is $(a-1)a(a+1) = a^3-a$ . \nThe difference between the two volumes is $a$ . As we are given that the difference is $5$ , we have $a=5$ , and the volume of the original cube ... |
https://artofproblemsolving.com/wiki/index.php/2009_AMC_12A_Problems/Problem_5 | D | 125 | One dimension of a cube is increased by $1$ , another is decreased by $1$ , and the third is left unchanged. The volume of the new rectangular solid is $5$ less than that of the cube. What was the volume of the cube?
$\textbf{(A)}\ 8 \qquad \textbf{(B)}\ 27 \qquad \textbf{(C)}\ 64 \qquad \textbf{(D)}\ 125 \qquad \textb... | [
"Let the original cube have edge length $a$ . Then its volume is $a^3$ .\nThe new box has dimensions $a-1$ $a$ , and $a+1$ , hence its volume is $(a-1)a(a+1) = a^3-a$ . \nThe difference between the two volumes is $a$ . As we are given that the difference is $5$ , we have $a=5$ , and the volume of the original cube ... |
https://artofproblemsolving.com/wiki/index.php/1992_AJHSME_Problems/Problem_25 | D | 9 | One half of the water is poured out of a full container. Then one third of the remainder is poured out. Continue the process: one fourth of the remainder for the third pouring, one fifth of the remainder for the fourth pouring, etc. After how many pourings does exactly one tenth of the original water remain?
$\text{... | [
"1)Model the amount left in the container as follows:\nAfter the first pour $\\frac12$ remains, after the second $\\frac12 \\times \\frac23$ remains, etc.\nThis becomes the product $\\frac12 \\times \\frac23 \\times \\frac34 \\times \\cdots \\times \\frac{9}{10}$\nNote that the terms cancel out leaving $\\frac{1}{1... |
https://artofproblemsolving.com/wiki/index.php/2003_AIME_I_Problems/Problem_2 | null | 301 | One hundred concentric circles with radii $1, 2, 3, \dots, 100$ are drawn in a plane. The interior of the circle of radius $1$ is colored red, and each region bounded by consecutive circles is colored either red or green, with no two adjacent regions the same color. The ratio of the total area of the green regions to t... | [
"To get the green area, we can color all the circles of radius $100$ or below green, then color all those with radius $99$ or below red, then color all those with radius $98$ or below green, and so forth. This amounts to adding the area of the circle of radius $100$ , but subtracting the circle of radius $99$ , th... |
https://artofproblemsolving.com/wiki/index.php/1991_AHSME_Problems/Problem_16 | D | 125 | One hundred students at Century High School participated in the AHSME last year, and their mean score was 100. The number of non-seniors taking the AHSME was $50\%$ more than the number of seniors, and the mean score of the seniors was $50\%$ higher than that of the non-seniors. What was the mean score of the seniors?
... | [
"Solution by e_power_pi_times_i\nLet $s$ and $\\dfrac{3}{2}s$ denote the numbers of seniors and non-seniors, respectively. Then $\\dfrac{5}{2}s = 100$ , so $s = 40$ $\\dfrac{3}{2}s = 60$ . Let $m$ and $\\dfrac{2}{3}m$ denote the mean score of seniors and non-seniors, respectively. Then $40m + 60(\\dfrac{2}{3}m) = 4... |
https://artofproblemsolving.com/wiki/index.php/2000_AMC_10_Problems/Problem_22 | C | 5 | One morning each member of Angela's family drank an 8-ounce mixture of coffee with milk. The amounts of coffee and milk varied from cup to cup, but were never zero. Angela drank a quarter of the total amount of milk and a sixth of the total amount of coffee. How many people are in the family?
$\text {(A)}\ 3 \qquad \te... | [
"The number of ounces that Angela's family drank has to be a multiple of 8, so we can find the right answer by guessing random values for the number of ounces of coffee and milk Angela drank. With Angela drinking 4 ounces of milk and 4 ounces of coffee, we get 40 total ounces Angela's family drank. Dividing that by... |
https://artofproblemsolving.com/wiki/index.php/2000_AMC_10_Problems/Problem_22 | null | 5 | One morning each member of Angela's family drank an 8-ounce mixture of coffee with milk. The amounts of coffee and milk varied from cup to cup, but were never zero. Angela drank a quarter of the total amount of milk and a sixth of the total amount of coffee. How many people are in the family?
$\text {(A)}\ 3 \qquad \te... | [
"If there were 4 people in the family, and each of them drank exactly the same amount of coffee and milk as Angela, there would be too much coffee. If there were 6 people in the family, and each of them drank exactly the same amount of coffee and milk as Angela, there would be not enough milk. Thus, it has to be $\... |
https://artofproblemsolving.com/wiki/index.php/2000_AMC_12_Problems/Problem_13 | C | 5 | One morning each member of Angela's family drank an 8-ounce mixture of coffee with milk. The amounts of coffee and milk varied from cup to cup, but were never zero. Angela drank a quarter of the total amount of milk and a sixth of the total amount of coffee. How many people are in the family?
$\text {(A)}\ 3 \qquad \te... | [
"The number of ounces that Angela's family drank has to be a multiple of 8, so we can find the right answer by guessing random values for the number of ounces of coffee and milk Angela drank. With Angela drinking 4 ounces of milk and 4 ounces of coffee, we get 40 total ounces Angela's family drank. Dividing that by... |
https://artofproblemsolving.com/wiki/index.php/2000_AMC_12_Problems/Problem_13 | null | 5 | One morning each member of Angela's family drank an 8-ounce mixture of coffee with milk. The amounts of coffee and milk varied from cup to cup, but were never zero. Angela drank a quarter of the total amount of milk and a sixth of the total amount of coffee. How many people are in the family?
$\text {(A)}\ 3 \qquad \te... | [
"If there were 4 people in the family, and each of them drank exactly the same amount of coffee and milk as Angela, there would be too much coffee. If there were 6 people in the family, and each of them drank exactly the same amount of coffee and milk as Angela, there would be not enough milk. Thus, it has to be $\... |
https://artofproblemsolving.com/wiki/index.php/1989_AIME_Problems/Problem_9 | null | 144 | One of Euler's conjectures was disproved in the 1960s by three American mathematicians when they showed there was a positive integer such that \[133^5+110^5+84^5+27^5=n^{5}.\] Find the value of $n$ | [
"Taking the given equation modulo $2,3,$ and $5,$ respectively, we have \\begin{align*} n^5&\\equiv0\\pmod{2}, \\\\ n^5&\\equiv0\\pmod{3}, \\\\ n^5&\\equiv4\\pmod{5}. \\end{align*} By either Fermat's Little Theorem (FLT) or inspection, we get \\begin{align*} n&\\equiv0\\pmod{2}, \\\\ n&\\equiv0\\pmod{3}, \\\\ n&\\e... |
https://artofproblemsolving.com/wiki/index.php/1953_AHSME_Problems/Problem_50 | B | 13 | One of the sides of a triangle is divided into segments of $6$ and $8$ units by the point of tangency of the inscribed circle. If the radius of the circle is $4$ , then the length of the shortest side is
$\textbf{(A) \ } 12 \mathrm{\ units} \qquad \textbf{(B) \ } 13 \mathrm{\ units} \qquad \textbf{(C) \ } 14 \mathrm{\ ... | [
"Let the triangle have side lengths $14, 6+x,$ and $8+x$ . The area of this triangle can be computed two ways. We have $A = rs$ , and $A = \\sqrt{s(s-a)(s-b)(s-c)}$ , where $s = 14+x$ is the semiperimeter. Therefore, $4(14+x)=\\sqrt{(14+x)(x)(8)(6)}$ . Solving gives $x = 7$ as the only valid solution. This triangle... |
https://artofproblemsolving.com/wiki/index.php/1990_AJHSME_Problems/Problem_13 | C | 1.18 | One proposal for new postage rates for a letter was $30$ cents for the first ounce and $22$ cents for each additional ounce (or fraction of an ounce). The postage for a letter weighing $4.5$ ounces was
$\text{(A)}\ \text{96 cents} \qquad \text{(B)}\ \text{1.07 dollars} \qquad \text{(C)}\ \text{1.18 dollars} \qquad \te... | [
"After the first ounce, there are $3.5$ ounces left. Since each additional ounce or fraction of an ounce adds $22$ cents to the total cost, we need to add $4\\times 22$ to the cost for the first ounce.\nSo, the total price is $30+4\\times 22=118$ cents. The answer is choice $\\boxed{1.18}$"
] |
https://artofproblemsolving.com/wiki/index.php/1973_AHSME_Problems/Problem_2 | C | 488 | One thousand unit cubes are fastened together to form a large cube with edge length 10 units; this is painted and then separated into the original cubes. The number of these unit cubes which have at least one face painted is
$\textbf{(A)}\ 600\qquad\textbf{(B)}\ 520\qquad\textbf{(C)}\ 488\qquad\textbf{(D)}\ 480\qquad\t... | [
"The total number of cubes is $10^3$ or $1000$ . Because each surface of the large cube is one cube deep, the number of the unpainted cubes is $8^3 = 512$ , since we subtract two from the side lengths of the cube itself, and cube it to find the volume of that cube. So there are $1000-512=\\boxed{488}$ cubes that ha... |
https://artofproblemsolving.com/wiki/index.php/2015_AMC_8_Problems/Problem_25 | C | 15 | One-inch squares are cut from the corners of this 5 inch square. What is the area in square inches of the largest square that can be fitted into the remaining space?
[asy] size(75); draw((0,0)--(0,5)--(5,5)--(5,0)--cycle); filldraw((0,4)--(1,4)--(1,5)--(0,5)--cycle, gray); filldraw((0,0)--(1,0)--(1,1)--(0,1)--cycle, g... | [
"Proceed as in solution 1 to get $x^2 - 5x + 5 = 0$\nNotice that the side length of the square is $\\sqrt{x^2+(5-x)^2}$ by the Pythagorean Theorem. Thus, the area of the square is \\[\\left(\\sqrt{x^2+(5-x)^2}\\right)^2\\] \\[= x^2 + x^2 - 10x + 25\\] \\[= (2x^2 - 10x + 10) + 15\\] \\[= 2(x^2 - 5x + 5) + 15\\] \\[=... |
https://artofproblemsolving.com/wiki/index.php/2015_AMC_8_Problems/Problem_1 | A | 12 | Onkon wants to cover his room's floor with his favourite red carpet. How many square yards of red carpet are required to cover a rectangular floor that is $12$ feet long and $9$ feet wide? (There are 3 feet in a yard.)
$\textbf{(A) }12\qquad\textbf{(B) }36\qquad\textbf{(C) }108\qquad\textbf{(D) }324\qquad \textbf{(E) }... | [
"First, we multiply $12\\cdot9$ . To get that, we need $108$ square feet of carpet to cover the room's floor. Since there are $9$ square feet in a square yard, you divide $108$ by $9$ to get $12$ square yards, so our answer is $\\bold{\\boxed{12}$",
"Since there are $3$ feet in a yard, we divide $9$ by $3$ to ge... |
https://artofproblemsolving.com/wiki/index.php/2014_AMC_12B_Problems/Problem_2 | C | 36 | Orvin went to the store with just enough money to buy $30$ balloons. When he arrived he discovered that the store had a special sale on balloons: buy $1$ balloon at the regular price and get a second at $\frac{1}{3}$ off the regular price. What is the greatest number of balloons Orvin could buy?
$\textbf{(A)}\ 33\qquad... | [
"If every balloon costs $n$ dollars, then Orvin has $30n$ dollars. For every balloon he buys for $n$ dollars, he can buy another for $\\frac{2n}{3}$ dollars. This means it costs him $\\frac{5n}{3}$ dollars to buy a bundle of $2$ balloons. With $30n$ dollars, he can buy $\\frac{30n}{\\frac{5n}{3}} = 18$ sets of t... |
https://artofproblemsolving.com/wiki/index.php/2017_AMC_10A_Problems/Problem_2 | D | 13 | Pablo buys popsicles for his friends. The store sells single popsicles for $$1$ each, $3$ -popsicle boxes for $$2$ each, and $5$ -popsicle boxes for $$3$ . What is the greatest number of popsicles that Pablo can buy with $$8$
$\textbf{(A)}\ 8\qquad\textbf{(B)}\ 11\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 13\qquad\textb... | [
"$$3$ boxes give us the most popsicles/dollar, so we want to buy as many of those as possible. After buying $2$ , we have $$2$ left. We cannot buy a third $$3$ box, so we opt for the $$2$ box instead (since it has a higher popsicles/dollar ratio than the $$1$ pack). We're now out of money. We bought $5+5+3=13$ pops... |
https://artofproblemsolving.com/wiki/index.php/2017_AMC_12A_Problems/Problem_1 | D | 13 | Pablo buys popsicles for his friends. The store sells single popsicles for $$1$ each, 3-popsicle boxes for $$2$ , and 5-popsicle boxes for $$3$ . What is the greatest number of popsicles that Pablo can buy with $$8$
$\textbf{(A)}\ 8\qquad\textbf{(B)}\ 11\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 13\qquad\textbf{(E)}\ 15... | [
"We can take two 5-popsicle boxes and one 3-popsicle box with $$8$ . Note that it is optimal since one popsicle is at the rate of $$1$ per popsicle, three popsicles at $$\\frac{2}{3}$ per popsicle and finally, five popsicles at $$\\frac{3}{5}$ per popsicle, hence we want as many $$3$ sets as possible. It is clear t... |
https://artofproblemsolving.com/wiki/index.php/2009_AMC_12B_Problems/Problem_22 | C | 784 | Parallelogram $ABCD$ has area $1,\!000,\!000$ . Vertex $A$ is at $(0,0)$ and all other vertices are in the first quadrant. Vertices $B$ and $D$ are lattice points on the lines $y = x$ and $y = kx$ for some integer $k > 1$ , respectively. How many such parallelograms are there? (A lattice point is any point whose coordi... | [
"The area of any parallelogram $ABCD$ can be computed as the size of the vector product of $\\overrightarrow{AB}$ and $\\overrightarrow{AD}$\nIn our setting where $A=(0,0)$ $B=(s,s)$ , and $D=(t,kt)$ this is simply $s\\cdot kt - s\\cdot t = (k-1)st$\nIn other words, we need to count the triples of integers $(k,s,t)... |
https://artofproblemsolving.com/wiki/index.php/2003_AMC_12B_Problems/Problem_20 | B | 2 | Part of the graph of $f(x) = ax^3 + bx^2 + cx + d$ is shown. What is $b$
2003 12B AMC-20.png
$\mathrm{(A)}\ -4 \qquad\mathrm{(B)}\ -2 \qquad\mathrm{(C)}\ 0 \qquad\mathrm{(D)}\ 2 \qquad\mathrm{(E)}\ 4$ | [
"The roots of this equation are $-1, 1, \\text{ and } x$ , letting $x$ be the root not shown in the graph. By Vieta, we know that $-1+1+x=x=-\\frac{b}{a}$ and $-1\\cdot 1\\cdot x=-x=-\\frac{d}{a}$ . Therefore, $x=\\frac{d}{a}$ . Setting the two equations for $x$ equal to each other, $\\frac{d}{a}=-\\frac{b}{a}$ . W... |
https://artofproblemsolving.com/wiki/index.php/1993_AJHSME_Problems/Problem_22 | D | 119 | Pat Peano has plenty of 0's, 1's, 3's, 4's, 5's, 6's, 7's, 8's and 9's, but he has only twenty-two 2's. How far can he number the pages of his scrapbook with these digits?
$\text{(A)}\ 22 \qquad \text{(B)}\ 99 \qquad \text{(C)}\ 112 \qquad \text{(D)}\ 119 \qquad \text{(E)}\ 199$ | [
"There is $1$ two in the one-digit numbers.\nThe number of two-digit numbers with a two in the tens place is $10$ and the number with a two in the ones place is $9$ . Thus the digit two is used $10+9=19$ times for the two digit numbers.\nNow, Pat Peano only has $22-1-19=2$ remaining twos. You must subtract 1 becaus... |
https://artofproblemsolving.com/wiki/index.php/1994_AHSME_Problems/Problem_5 | E | 1,000 | Pat intended to multiply a number by $6$ but instead divided by $6$ . Pat then meant to add $14$ but instead subtracted $14$ . After these mistakes, the result was $16$ . If the correct operations had been used, the value produced would have been
$\textbf{(A)}\ \text{less than 400} \qquad\textbf{(B)}\ \text{between 400... | [
"We reverse the operations that he did and then use the correct operations. His end result is $16$ . Before that, he subtracted $14$ which means that his number after the first operation was $30$ . He divided by $6$ so his number was $180$\nNow, we multiply $180$ by $6$ to get $1080$ . Finally, $1080+14=1094$ . Sin... |
https://artofproblemsolving.com/wiki/index.php/2001_AMC_10_Problems/Problem_19 | D | 15 | Pat wants to buy four donuts from an ample supply of three types of donuts: glazed, chocolate, and powdered. How many different selections are possible?
$\textbf{(A)}\ 6 \qquad \textbf{(B)}\ 9 \qquad \textbf{(C)}\ 12 \qquad \textbf{(D)}\ 15 \qquad \textbf{(E)}\ 18$ | [
"Let's use stars and bars .\nLet the donuts be represented by $O$ s. We wish to find all possible combinations of glazed, chocolate, and powdered donuts that give us $4$ in all. The four donuts we want can be represented as $OOOO$ . Notice that we can add two \"dividers\" to divide the group of donuts into three di... |
https://artofproblemsolving.com/wiki/index.php/2002_AIME_II_Problems/Problem_4 | null | 803 | Patio blocks that are hexagons $1$ unit on a side are used to outline a garden by placing the blocks edge to edge with $n$ on each side. The diagram indicates the path of blocks around the garden when $n=5$
AIME 2002 II Problem 4.gif
If $n=202$ , then the area of the garden enclosed by the path, not including the path ... | [
"When $n>1$ , the path of blocks has $6(n-1)$ blocks total in it. When $n=1$ , there is just one lonely block. Thus, the area of the garden enclosed by the path when $n=202$ is\n\\[(1+6+12+18+\\cdots +1200)A=(1+6(1+2+3...+200))A\\]\nwhere $A$ is the area of one block. Then, because $n(n+1)/2$ is equal to the sum of... |
https://artofproblemsolving.com/wiki/index.php/2004_AMC_10B_Problems/Problem_15 | A | 1.15 | Patty has $20$ coins consisting of nickels and dimes. If her nickels were dimes and her dimes were nickels, she would have $70$ cents more. How much are her coins worth?
$\textbf{(A)}\ \textdollar 1.15\qquad\textbf{(B)}\ \textdollar 1.20\qquad\textbf{(C)}\ \textdollar 1.25\qquad\textbf{(D)}\ \textdollar 1.30\qquad\text... | [
"She has $n$ nickels and $d=20-n$ dimes. Their total cost is $5n+10d=5n+10(20-n)=200-5n$ cents. If the dimes were nickels and vice versa, she would have $10n+5d=10n+5(20-n)=100+5n$ cents. This value should be $70$ cents more than the previous one. We get $200-5n+70=100+5n$ , which solves to $n=17$ . Her coins are w... |
https://artofproblemsolving.com/wiki/index.php/2014_AMC_8_Problems/Problem_2 | E | 5 | Paul owes Paula $35$ cents and has a pocket full of $5$ -cent coins, $10$ -cent coins, and $25$ -cent coins that he can use to pay her. What is the difference between the largest and the smallest number of coins he can use to pay her?
$\textbf{(A) }1\qquad\textbf{(B) }2\qquad\textbf{(C) }3\qquad\textbf{(D) }4\qquad \te... | [
"The fewest amount of coins that can be used is $2$ (a quarter and a dime). The greatest amount is $7$ , if he only uses nickels. Therefore we have $7-2=\\boxed{5}$"
] |
https://artofproblemsolving.com/wiki/index.php/2012_AMC_10A_Problems/Problem_19 | D | 48 | Paula the painter and her two helpers each paint at constant, but different, rates. They always start at 8:00 AM, and all three always take the same amount of time to eat lunch. On Monday the three of them painted 50% of a house, quitting at 4:00 PM. On Tuesday, when Paula wasn't there, the two helpers painted only 24%... | [
"Let Paula work at a rate of $p$ , the two helpers work at a combined rate of $h$ , and the time it takes to eat lunch be $L$ , where $p$ and $h$ are in house/hours and L is in hours. Then the labor on Monday, Tuesday, and Wednesday can be represented by the three following equations:\n\\[(8-L)(p+h)=50\\]\n\\[(6.2-... |
https://artofproblemsolving.com/wiki/index.php/2012_AMC_12A_Problems/Problem_13 | D | 48 | Paula the painter and her two helpers each paint at constant, but different, rates. They always start at 8:00 AM, and all three always take the same amount of time to eat lunch. On Monday the three of them painted 50% of a house, quitting at 4:00 PM. On Tuesday, when Paula wasn't there, the two helpers painted only 24%... | [
"Let Paula work at a rate of $p$ , the two helpers work at a combined rate of $h$ , and the time it takes to eat lunch be $L$ , where $p$ and $h$ are in house/hours and L is in hours. Then the labor on Monday, Tuesday, and Wednesday can be represented by the three following equations:\n\\[(8-L)(p+h)=50\\]\n\\[(6.2-... |
https://artofproblemsolving.com/wiki/index.php/2009_AMC_10B_Problems/Problem_3 | C | 15 | Paula the painter had just enough paint for 30 identically sized rooms. Unfortunately, on the way to work, three cans of paint fell off her truck, so she had only enough paint for 25 rooms. How many cans of paint did she use for the 25 rooms?
$\mathrm{(A)}\ 10\qquad \mathrm{(B)}\ 12\qquad \mathrm{(C)}\ 15\qquad \math... | [
"Losing three cans of paint corresponds to being able to paint five fewer rooms. So $\\frac 35 \\cdot 25 = \\boxed{15}$"
] |
https://artofproblemsolving.com/wiki/index.php/2009_AMC_12B_Problems/Problem_2 | C | 15 | Paula the painter had just enough paint for 30 identically sized rooms. Unfortunately, on the way to work, three cans of paint fell off her truck, so she had only enough paint for 25 rooms. How many cans of paint did she use for the 25 rooms?
$\mathrm{(A)}\ 10\qquad \mathrm{(B)}\ 12\qquad \mathrm{(C)}\ 15\qquad \math... | [
"Losing three cans of paint corresponds to being able to paint five fewer rooms. So $\\frac 35 \\cdot 25 = \\boxed{15}$"
] |
https://artofproblemsolving.com/wiki/index.php/1994_AJHSME_Problems/Problem_17 | D | 7 | Pauline Bunyan can shovel snow at the rate of $20$ cubic yards for the first hour, $19$ cubic yards for the second, $18$ for the third, etc., always shoveling one cubic yard less per hour than the previous hour. If her driveway is $4$ yards wide, $10$ yards long, and covered with snow $3$ yards deep, then the number o... | [
"Her driveway has $(4)(10)(3)=120$ cubic yards of snow. After the first hour she would have $120-20=100$ cubic yards, then $100-19=81$ $81-18=63$ $63-17=46$ $46-16=30$ $30-15=15$ , and $15-14=1$ cubic yard after the seventh hour. It will take her a little more than seven hours to shovel it clean, which is closest t... |
https://artofproblemsolving.com/wiki/index.php/1985_AHSME_Problems/Problem_13 | E | 6 | Pegs are put in a board $1$ unit apart both horizontally and vertically. A rubber band is stretched over $4$ pegs as shown in the figure, forming a quadrilateral. Its area in square units is
[asy] int i,j; for(i=0; i<5; i=i+1) { for(j=0; j<4; j=j+1) { dot((i,j)); }} draw((0,1)--(1,3)--(4,1)--(3,0)--cycle, linewidth(0.7... | [
" We draw in the rectangle bounding the given quadrilateral and label the points as shown. The area of rectangle $ABCD$ is $(3)(4) = 12$ , while the areas of the triangles $AEH$ $EBF$ $FCG$ , and $GDH$ are, respectively, \\begin{align*}&\\text{area of } \\triangle AEH = \\frac{1}{2}(1)(2) = 1, \\\\ &\\text{area of ... |
https://artofproblemsolving.com/wiki/index.php/2003_AMC_12B_Problems/Problem_7 | D | 64 | Penniless Pete's piggy bank has no pennies in it, but it has 100 coins, all nickels,dimes, and quarters, whose total value is $8.35. It does not necessarily contain coins of all three types. What is the difference between the largest and smallest number of dimes that could be in the bank?
$\text {(A) } 0 \qquad \text {... | [
"Where $a,b,c$ is the number of nickels, dimes, and quarters, respectively, we can set up two equations:\n\\[(1)\\ 5a+10b+25c=835\\ \\ \\ \\ (2)\\ a+b+c=100\\]\nEliminate $a$ by subtracting $(2)$ from $(1)/5$ to get $b+4c=67$ . Of the integer solutions $(b,c)$ to this equation, the number of dimes $b$ is least in $... |
https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_13 | B | 1 | Peter, Emma, and Kyler played chess with each other. Peter won 4 games and lost 2 games. Emma won 3 games and lost 3 games. If Kyler lost 3 games, how many games did he win?
$\textbf{(A) }0\quad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }3\qquad\textbf{(E) }4$ | [
"Given $n$ games, there must be a total of $n$ wins and $n$ losses. Hence, $4 + 3 + K = 2 + 3 + 3$ where $K$ is Kyler's wins. $K = 1$ , so our final answer is $\\boxed{1}.$ ~CHECKMATE2021"
] |
https://artofproblemsolving.com/wiki/index.php/2007_AMC_8_Problems/Problem_19 | C | 79 | Pick two consecutive positive integers whose sum is less than $100$ . Square both
of those integers and then find the difference of the squares. Which of the
following could be the difference?
$\mathrm{(A)}\ 2 \qquad \mathrm{(B)}\ 64 \qquad \mathrm{(C)}\ 79 \qquad \mathrm{(D)}\ 96 \qquad \mathrm{(E)}\ 131$ | [
"Let the smaller of the two numbers be $x$ . Then, the problem states that $(x+1)+x<100$ $(x+1)^2-x^2=x^2+2x+1-x^2=2x+1$ $2x+1$ is obviously odd, so only answer choices C and E need to be considered.\n$2x+1=131$ contradicts the fact that $2x+1<100$ , so the answer is $\\boxed{79}$",
"Since for two consecutive num... |
https://artofproblemsolving.com/wiki/index.php/1988_AJHSME_Problems/Problem_16 | E | 6 | Placing no more than one $\text{X}$ in each small square , what is the greatest number of $\text{X}$ 's that can be put on the grid shown without getting three $\text{X}$ 's in a row vertically, horizontally, or diagonally?
$\text{(A)}\ 2 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 4 \qquad \text{(D)}\ 5 \qquad \text{(E)}\... | [
"By the Pigeonhole Principle , if there are at least $7$ $\\text{X}$ 's, then there will be some row with $3$ $\\text{X}$ 's. We can put in $6$ by leaving out the three boxes in one of the main diagonals.\n$\\rightarrow \\boxed{6}$"
] |
https://artofproblemsolving.com/wiki/index.php/2015_AIME_I_Problems/Problem_6 | null | 58 | Point $A,B,C,D,$ and $E$ are equally spaced on a minor arc of a circle. Points $E,F,G,H,I$ and $A$ are equally spaced on a minor arc of a second circle with center $C$ as shown in the figure below. The angle $\angle ABD$ exceeds $\angle AHG$ by $12^\circ$ . Find the degree measure of $\angle BAG$
[asy] pair A,B,C,D,E,F... | [
"Let $O$ be the center of the circle with $ABCDE$ on it.\nLet $x$ be the degree measurement of $\\overarc{ED}=\\overarc{DC}=\\overarc{CB}=\\overarc{BA}$ in circle $O$\nand $y$ be the degree measurement of $\\overarc{EF}=\\overarc{FG}=\\overarc{GH}=\\overarc{HI}=\\overarc{IA}$ in circle $C$\n$\\angle ECA$ is, theref... |
https://artofproblemsolving.com/wiki/index.php/1997_AIME_Problems/Problem_6 | null | 42 | Point $B$ is in the exterior of the regular $n$ -sided polygon $A_1A_2\cdots A_n$ , and $A_1A_2B$ is an equilateral triangle . What is the largest value of $n$ for which $A_1$ $A_n$ , and $B$ are consecutive vertices of a regular polygon? | [
"1997 AIME-6.png\nLet the other regular polygon have $m$ sides. Using the interior angle of a regular polygon formula, we have $\\angle A_2A_1A_n = \\frac{(n-2)180}{n}$ $\\angle A_nA_1B = \\frac{(m-2)180}{m}$ , and $\\angle A_2A_1B = 60^{\\circ}$ . Since those three angles add up to $360^{\\circ}$\n\\begin{eqnarray... |
https://artofproblemsolving.com/wiki/index.php/2003_AIME_I_Problems/Problem_7 | null | 380 | Point $B$ is on $\overline{AC}$ with $AB = 9$ and $BC = 21.$ Point $D$ is not on $\overline{AC}$ so that $AD = CD,$ and $AD$ and $BD$ are integers . Let $s$ be the sum of all possible perimeters of $\triangle ACD$ . Find $s.$ | [
"Denote the height of $\\triangle ACD$ as $h$ $x = AD = CD$ , and $y = BD$ . Using the Pythagorean theorem , we find that $h^2 = y^2 - 6^2$ and $h^2 = x^2 - 15^2$ . Thus, $y^2 - 36 = x^2 - 225 \\Longrightarrow x^2 - y^2 = 189$ . The LHS is difference of squares , so $(x + y)(x - y) = 189$ . As both $x,\\ y$ are int... |
https://artofproblemsolving.com/wiki/index.php/2015_AIME_I_Problems/Problem_4 | null | 507 | Point $B$ lies on line segment $\overline{AC}$ with $AB=16$ and $BC=4$ . Points $D$ and $E$ lie on the same side of line $AC$ forming equilateral triangles $\triangle ABD$ and $\triangle BCE$ . Let $M$ be the midpoint of $\overline{AE}$ , and $N$ be the midpoint of $\overline{CD}$ . The area of $\triangle BMN$ is $x$ .... | [
"Let $A$ be the origin, so $B=(16,0)$ and $C=(20,0).$ Using equilateral triangle properties tells us that $D=(8,8\\sqrt3)$ and $E=(18,2\\sqrt3)$ as well. Therefore, $M=(9,\\sqrt3)$ and $N=(14,4\\sqrt3).$ Applying the Shoelace Theorem to triangle $BMN$ gives\n\\[x=\\dfrac 1 2 |16\\sqrt3+36\\sqrt3+0-(0+14\\sqrt3+64\\... |
https://artofproblemsolving.com/wiki/index.php/1983_AHSME_Problems/Problem_19 | A | 2 | Point $D$ is on side $CB$ of triangle $ABC$ . If $\angle{CAD} = \angle{DAB} = 60^\circ\mbox{, }AC = 3\mbox{ and }AB = 6$ ,
then the length of $AD$ is
$\textbf{(A)} \ 2 \qquad \textbf{(B)} \ 2.5 \qquad \textbf{(C)} \ 3 \qquad \textbf{(D)} \ 3.5 \qquad \textbf{(E)} \ 4$ | [
"Let $AD = y$ . Since $AD$ bisects $\\angle{BAC}$ , the Angle Bisector Theorem gives $\\frac{DB}{CD} = \\frac{AB}{AC} = 2$ , so let $CD = x$ and $DB = 2x$ . Applying the Law of Cosines to $\\triangle CAD$ gives $x^2 = 3^2 + y^2 - 3y$ , and to $\\triangle DAB$ gives $(2x)^2 = 6^2 + y^2 - 6y$ . Subtracting $4$ times ... |
https://artofproblemsolving.com/wiki/index.php/2020_AIME_I_Problems/Problem_13 | null | 36 | Point $D$ lies on side $\overline{BC}$ of $\triangle ABC$ so that $\overline{AD}$ bisects $\angle BAC.$ The perpendicular bisector of $\overline{AD}$ intersects the bisectors of $\angle ABC$ and $\angle ACB$ in points $E$ and $F,$ respectively. Given that $AB=4,BC=5,$ and $CA=6,$ the area of $\triangle AEF$ can be writ... | [
"Points are defined as shown. It is pretty easy to show that $\\triangle AFE \\sim \\triangle AGH$ by spiral similarity at $A$ by some short angle chasing. Now, note that $AD$ is the altitude of $\\triangle AFE$ , as the altitude of $AGH$ . We need to compare these altitudes in order to compare their areas. Note th... |
https://artofproblemsolving.com/wiki/index.php/1981_AHSME_Problems/Problem_2 | C | 3 | Point $E$ is on side $AB$ of square $ABCD$ . If $EB$ has length one and $EC$ has length two, then the area of the square is
$\textbf{(A)}\ \sqrt{3}\qquad\textbf{(B)}\ \sqrt{5}\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 2\sqrt{3}\qquad\textbf{(E)}\ 5$ | [
"Note that $\\triangle BCE$ is a right triangle. Thus, we do Pythagorean theorem to find that side $BC=\\sqrt{3}$ . Since this is the side length of the square, the area of $ABCD$ is $\\boxed{3}$"
] |
https://artofproblemsolving.com/wiki/index.php/2018_AMC_8_Problems/Problem_22 | B | 108 | Point $E$ is the midpoint of side $\overline{CD}$ in square $ABCD,$ and $\overline{BE}$ meets diagonal $\overline{AC}$ at $F.$ The area of quadrilateral $AFED$ is $45.$ What is the area of $ABCD?$
[asy] size(5cm); draw((0,0)--(6,0)--(6,6)--(0,6)--cycle); draw((0,6)--(6,0)); draw((3,0)--(6,6)); label("$A$",(0,6),NW); la... | [
"Let the area of $\\triangle CEF$ be $x$ . Thus, the area of triangle $\\triangle ACD$ is $45+x$ and the area of the square is $2(45+x) = 90+2x$\nBy AA similarity, $\\triangle CEF \\sim \\triangle ABF$ with a 1:2 ratio, so the area of triangle $\\triangle ABF$ is $4x$ . Now, consider trapezoid $ABED$ . Its area is ... |
https://artofproblemsolving.com/wiki/index.php/2018_AMC_8_Problems/Problem_22 | null | 108 | Point $E$ is the midpoint of side $\overline{CD}$ in square $ABCD,$ and $\overline{BE}$ meets diagonal $\overline{AC}$ at $F.$ The area of quadrilateral $AFED$ is $45.$ What is the area of $ABCD?$
[asy] size(5cm); draw((0,0)--(6,0)--(6,6)--(0,6)--cycle); draw((0,6)--(6,0)); draw((3,0)--(6,6)); label("$A$",(0,6),NW); la... | [
"Solution with Cartesian and Barycentric Coordinates:\nWe start with the following:\nClaim: Given a square $ABCD$ , let $E$ be the midpoint of $\\overline{DC}$ and let $BE\\cap AC = F$ . Then $\\frac {AF}{FC}=2$\nProof: We use Cartesian coordinates. Let $D$ be the origin, $A=(0,1),C=(0,1),B=(1,1)$ . We have that $\... |
https://artofproblemsolving.com/wiki/index.php/1963_AHSME_Problems/Problem_25 | A | 12 | Point $F$ is taken in side $AD$ of square $ABCD$ . At $C$ a perpendicular is drawn to $CF$ , meeting $AB$ extended at $E$ .
The area of $ABCD$ is $256$ square inches and the area of $\triangle CEF$ is $200$ square inches. Then the number of inches in $BE$ is:
[asy] size(6cm); pair A = (0, 0), B = (1, 0), C = (1, 1), D... | [
"Because $ABCD$ is a square $DC = CB = 16$ $DC \\perp DA$ , and $CB \\perp BA$ . Also, because $\\angle DCF + \\angle FCB = \\angle FCB + \\angle BCE = 90^\\circ$ $\\angle DCF = \\angle BCE$ . Thus, by ASA Congruency, $\\triangle DCF \\cong \\triangle BCF$\nFrom the congruency, $CF = CE$ . Using the area formula... |
https://artofproblemsolving.com/wiki/index.php/1963_AHSME_Problems/Problem_38 | E | 16 | Point $F$ is taken on the extension of side $AD$ of parallelogram $ABCD$ $BF$ intersects diagonal $AC$ at $E$ and side $DC$ at $G$ .
If $EF = 32$ and $GF = 24$ , then $BE$ equals:
[asy] size(7cm); pair A = (0, 0), B = (7, 0), C = (10, 5), D = (3, 5), F = (5.7, 9.5); pair G = intersectionpoints(B--F, D--C)[0]; pair E =... | [
"Let $BE = x$ and $BC = y$ . Since $AF \\parallel BC$ , by AA Similarity, $\\triangle AFE \\sim \\triangle CBE$ . That means $\\frac{AF}{CB} = \\frac{FE}{BE}$ . Substituting in values results in \\[\\frac{AF}{y} = \\frac{32}{x}\\] Thus, $AF = \\frac{32y}{x}$ , so $FD = \\frac{32y - xy}{x}$\nIn addition, $DC \\pa... |
https://artofproblemsolving.com/wiki/index.php/1991_AHSME_Problems/Problem_10 | null | 12 | Point $P$ is $9$ units from the center of a circle of radius $15$ . How many different chords of the circle contain $P$ and have integer lengths?
(A) 11 (B) 12 (C) 13 (D) 14 (E) 29 | [
"Let $O$ be the center of the circle, and let the chord passing through $P$ that is perpendicular to $OP$ intersect the circle at $Q$ and $R$ . Then $OP = 9$ and $OQ = 15$ , so by the Pythagorean Theorem, $PQ = 12$ . By symmetry, $PR = 12$\n Let $AB$ be the diameter passing through $P$. Then the longest chord o... |
https://artofproblemsolving.com/wiki/index.php/1989_AIME_Problems/Problem_15 | null | 108 | Point $P$ is inside $\triangle ABC$ . Line segments $APD$ $BPE$ , and $CPF$ are drawn with $D$ on $BC$ $E$ on $AC$ , and $F$ on $AB$ (see the figure below). Given that $AP=6$ $BP=9$ $PD=6$ $PE=3$ , and $CF=20$ , find the area of $\triangle ABC$ | [
"Let $[RST]$ be the area of polygon $RST$ . We'll make use of the following fact: if $P$ is a point in the interior of triangle $XYZ$ , and line $XP$ intersects line $YZ$ at point $L$ , then $\\dfrac{XP}{PL} = \\frac{[XPY] + [ZPX]}{[YPZ]}.$\nThis is true because triangles $XPY$ and $YPL$ have their areas in ratio ... |
https://artofproblemsolving.com/wiki/index.php/1983_AHSME_Problems/Problem_2 | B | 2 | Point $P$ is outside circle $C$ on the plane. At most how many points on $C$ are $3$ cm from $P$
$\textbf{(A)} \ 1 \qquad \textbf{(B)} \ 2 \qquad \textbf{(C)} \ 3 \qquad \textbf{(D)} \ 4 \qquad \textbf{(E)} \ 8$ | [
"The points $3$ cm away from $P$ can be represented as a circle centered at $P$ with radius $3$ cm. The maximum number of intersection points of two circles is $\\boxed{2}$"
] |
https://artofproblemsolving.com/wiki/index.php/2011_AIME_II_Problems/Problem_13 | null | 96 | Point $P$ lies on the diagonal $AC$ of square $ABCD$ with $AP > CP$ . Let $O_{1}$ and $O_{2}$ be the circumcenters of triangles $ABP$ and $CDP$ respectively. Given that $AB = 12$ and $\angle O_{1}PO_{2} = 120^{\circ}$ , then $AP = \sqrt{a} + \sqrt{b}$ , where $a$ and $b$ are positive integers. Find $a + b$ | [
"This is a combination of Solutions 1 and 2.\nFirst, draw $O_1P,O_2P,BP,DP$ . Then, observe that $\\angle BAP=45$ implies that $\\angle BO_1P=90$ . So, $\\triangle BO_1P$ is a $45-90-45$ triangle. Similarly, observe that $DO_2P$ is too. So, a rotation of $\\angle O_1PO_2$ to $\\angle BPO_2$ adds $45$ degrees. Then,... |
https://artofproblemsolving.com/wiki/index.php/1999_AIME_Problems/Problem_14 | null | 463 | Point $P_{}$ is located inside triangle $ABC$ so that angles $PAB, PBC,$ and $PCA$ are all congruent. The sides of the triangle have lengths $AB=13, BC=14,$ and $CA=15,$ and the tangent of angle $PAB$ is $m/n,$ where $m_{}$ and $n_{}$ are relatively prime positive integers. Find $m+n.$ | [
"Drop perpendiculars from $P$ to the three sides of $\\triangle ABC$ and let them meet $\\overline{AB}, \\overline{BC},$ and $\\overline{CA}$ at $D, E,$ and $F$ respectively.\nLet $BE = x, CF = y,$ and $AD = z$ . We have that \\begin{align*}DP&=z\\tan\\theta\\\\ EP&=x\\tan\\theta\\\\ FP&=y\\tan\\theta\\end{align*}... |
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