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/2018_AMC_12A_Problems/Problem_23 | E | 80 | In $\triangle PAT,$ $\angle P=36^{\circ},$ $\angle A=56^{\circ},$ and $PA=10.$ Points $U$ and $G$ lie on sides $\overline{TP}$ and $\overline{TA},$ respectively, so that $PU=AG=1.$ Let $M$ and $N$ be the midpoints of segments $\overline{PA}$ and $\overline{UG},$ respectively. What is the degree measure of the acute ang... | [
"Let $P$ be the origin, and $PA$ lie on the $x$ -axis.\nWe can find $U=\\left(\\cos(36), \\sin(36)\\right)$ and $G=\\left(10-\\cos(56), \\sin(56)\\right)$\nThen, we have $M=(5, 0)$ and $N$ is the midpoint of $U$ and $G$ , or $\\left(\\frac{10+\\cos(36)-\\cos(56)}{2}, \\frac{\\sin(36)+\\sin(56)}{2}\\right)$\nNotice ... |
https://artofproblemsolving.com/wiki/index.php/2019_AIME_I_Problems/Problem_3 | null | 120 | In $\triangle PQR$ $PR=15$ $QR=20$ , and $PQ=25$ . Points $A$ and $B$ lie on $\overline{PQ}$ , points $C$ and $D$ lie on $\overline{QR}$ , and points $E$ and $F$ lie on $\overline{PR}$ , with $PA=QB=QC=RD=RE=PF=5$ . Find the area of hexagon $ABCDEF$ | [
"We know the area of the hexagon $ABCDEF$ to be $\\triangle PQR- \\triangle PAF- \\triangle BCQ- \\triangle RED$ . Since $PR^2+RQ^2=PQ^2$ , we know that $\\triangle PRQ$ is a right triangle. Thus the area of $\\triangle PQR$ is $150$ . Another way to compute the area is \\[\\frac12 \\cdot PQ\\cdot RQ \\sin \\angle ... |
https://artofproblemsolving.com/wiki/index.php/2014_AIME_II_Problems/Problem_11 | null | 56 | In $\triangle RED$ $\measuredangle DRE=75^{\circ}$ and $\measuredangle RED=45^{\circ}$ $RD=1$ . Let $M$ be the midpoint of segment $\overline{RD}$ . Point $C$ lies on side $\overline{ED}$ such that $\overline{RC}\perp\overline{EM}$ . Extend segment $\overline{DE}$ through $E$ to point $A$ such that $CA=AR$ . Then $AE=\... | [
"Let $P$ be the foot of the perpendicular from $A$ to $\\overline{CR}$ , so $\\overline{AP}\\parallel\\overline{EM}$ . Since triangle $ARC$ is isosceles, $P$ is the midpoint of $\\overline{CR}$ , and $\\overline{PM}\\parallel\\overline{CD}$ . Thus, $APME$ is a parallelogram and $AE = PM = \\frac{CD}{2}$ . We can th... |
https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_19 | A | 44 | In $\triangle{ABC}$ medians $\overline{AD}$ and $\overline{BE}$ intersect at $G$ and $\triangle{AGE}$ is equilateral. Then $\cos(C)$ can be written as $\frac{m\sqrt p}n$ , where $m$ and $n$ are relatively prime positive integers and $p$ is a positive integer not divisible by the square of any prime. What is $m+n+p?$
$\... | [
"Let $AG = AE = GE = CE = 1$ . Since $G$ is the centroid, $DG = \\frac12$ $BG = 2$\n\\[\\angle BGD = \\angle AGE = 60^{\\circ}\\]\nBy the Law of Cosine in $\\triangle BGD$\n\\[BD^2 = BG^2 + DG^2 - 2 \\cdot BG \\cdot DG \\cdot \\cos \\angle BGD\\]\n\\[BD = \\sqrt {2^2 + \\left( \\frac{1}{2} \\right)^2 - 2 \\cdot 2 \... |
https://artofproblemsolving.com/wiki/index.php/2010_AIME_I_Problems/Problem_15 | null | 45 | In $\triangle{ABC}$ with $AB = 12$ $BC = 13$ , and $AC = 15$ , let $M$ be a point on $\overline{AC}$ such that the incircles of $\triangle{ABM}$ and $\triangle{BCM}$ have equal radii . Then $\frac{AM}{CM} = \frac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p + q$ | [
"Let $AM = x$ , then $CM = 15 - x$ . Also let $BM = d$ Clearly, $\\frac {[ABM]}{[CBM]} = \\frac {x}{15 - x}$ . We can also express each area by the rs formula. Then $\\frac {[ABM]}{[CBM]} = \\frac {p(ABM)}{p(CBM)} = \\frac {12 + d + x}{28 + d - x}$ . Equating and cross-multiplying yields $25x + 2dx = 15d + 180$ or ... |
https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_21 | E | 72 | In $\triangle{ABC}$ with side lengths $AB = 13$ $AC = 12$ , and $BC = 5$ , let $O$ and $I$ denote the circumcenter and incenter, respectively. A circle with center $M$ is tangent to the legs $AC$ and $BC$ and to the circumcircle of $\triangle{ABC}$ . What is the area of $\triangle{MOI}$
$\textbf{(A)}\ \frac52\qquad\tex... | [
"In this solution, let the brackets denote areas.\nWe place the diagram in the coordinate plane: Let $A=(12,0),B=(0,5),$ and $C=(0,0).$\nSince $\\triangle ABC$ is a right triangle with $\\angle ACB=90^\\circ,$ its circumcenter is the midpoint of $\\overline{AB},$ from which $O=\\left(6,\\frac52\\right).$ Note that ... |
https://artofproblemsolving.com/wiki/index.php/2014_AIME_II_Problems/Problem_14 | null | 77 | In $\triangle{ABC}, AB=10, \angle{A}=30^\circ$ , and $\angle{C=45^\circ}$ . Let $H, D,$ and $M$ be points on the line $BC$ such that $AH\perp{BC}$ $\angle{BAD}=\angle{CAD}$ , and $BM=CM$ . Point $N$ is the midpoint of the segment $HM$ , and point $P$ is on ray $AD$ such that $PN\perp{BC}$ . Then $AP^2=\dfrac{m}{n}$ , w... | [
"Let us just drop the perpendicular from $B$ to $AC$ and label the point of intersection $O$ . We will use this point later in the problem.\nAs we can see,\n$M$ is the midpoint of $BC$ and $N$ is the midpoint of $HM$\n$AHC$ is a $45-45-90$ triangle, so $\\angle{HAB}=15^\\circ$\n$AHD$ is $30-60-90$ triangle.\n$AH$ a... |
https://artofproblemsolving.com/wiki/index.php/2011_AMC_10A_Problems/Problem_19 | E | 62 | In 1991 the population of a town was a perfect square. Ten years later, after an increase of 150 people, the population was 9 more than a perfect square. Now, in 2011, with an increase of another 150 people, the population is once again a perfect square. Which of the following is closest to the percent growth of the to... | [
"Let the population of the town in $1991$ be $p^2$ . Let the population in $2001$ be $q^2+9$ . It follows that $p^2+150=q^2+9$ . Rearrange this equation to get $141=q^2-p^2=(q-p)(q+p)$ . Since $q$ and $p$ are both positive integers with $q>p$ $(q-p)$ and $(q+p)$ also must be, and thus, they are both factors of $141... |
https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_10 | B | 414 | In January $1980$ the Mauna Loa Observatory recorded carbon dioxide $(CO2)$ levels of $338$ ppm (parts per million). Over the years the average $CO2$ reading has increased by about $1.515$ ppm each year. What is the expected $CO2$ level in ppm in January $2030$ ? Round your answer to the nearest integer.
$\textbf{(A)}\... | [
"This is a time period of $50$ years, so we can expect the ppm to increase by $50*1.515=75.75~76$ $76+338=\\boxed{414}$",
"For each year that has passed, the ppm will increase by $1.515$ . In $2030$ , the CO2 would have increased by $50\\cdot 1.515 \\approx. 76,$ so the total ppm of CO2 will be $76 + 338 = \\boxe... |
https://artofproblemsolving.com/wiki/index.php/1992_AIME_Problems/Problem_4 | null | 62 | In Pascal's Triangle, each entry is the sum of the two entries above it. The first few rows of the triangle are shown below.
\[\begin{array}{c@{\hspace{8em}} c@{\hspace{6pt}}c@{\hspace{6pt}}c@{\hspace{6pt}}c@{\hspace{4pt}}c@{\hspace{2pt}} c@{\hspace{2pt}}c@{\hspace{2pt}}c@{\hspace{2pt}}c@{\hspace{3pt}}c@{\hspace{6pt}} ... | [
"Consider what the ratio means. Since we know that they are consecutive terms, we can say \\[\\frac{\\dbinom{n}{k-1}}{3} = \\frac{\\dbinom{n}{k}}{4} = \\frac{\\dbinom{n}{k+1}}{5}.\\]\nTaking the first part, and using our expression for $n$ choose $k$ \\[\\frac{n!}{3(k-1)!(n-k+1)!} = \\frac{n!}{4k!(n-k)!}\\] \\[\\fr... |
https://artofproblemsolving.com/wiki/index.php/2008_AMC_8_Problems/Problem_15 | B | 40 | In Theresa's first $8$ basketball games, she scored $7, 4, 3, 6, 8, 3, 1$ and $5$ points. In her ninth game, she scored fewer than $10$ points and her points-per-game average for the nine games was an integer. Similarly in her tenth game, she scored fewer than $10$ points and her points-per-game average for the $10$ ga... | [
"The total number of points from the first $8$ games is $7+4+3+6+8+3+1+5=37$ . We have to make this a multiple of $9$ by scoring less than $10$ points. The closest multiple of $9$ is $45$ $45-37=8$ . Now we have to add a number to get a multiple of 10. The next multiple is $50$ we added $5$ , multiplying these toge... |
https://artofproblemsolving.com/wiki/index.php/2007_AIME_I_Problems/Problem_10 | null | 860 | In a 6 x 4 grid (6 rows, 4 columns), 12 of the 24 squares are to be shaded so that there are two shaded squares in each row and three shaded squares in each column. Let $N$ be the number of shadings with this property. Find the remainder when $N$ is divided by 1000.
AIME I 2007-10.png | [
"Consider the first column. There are ${6\\choose3} = 20$ ways that the rows could be chosen, but without loss of generality let them be the first three rows. (Change the order of the rows to make this true.) We will multiply whatever answer we get by 20 to get our final answer.\nNow consider the 3x3 that is nex... |
https://artofproblemsolving.com/wiki/index.php/2019_AIME_II_Problems/Problem_6 | null | 216 | In a Martian civilization, all logarithms whose bases are not specified are assumed to be base $b$ , for some fixed $b\ge2$ . A Martian student writes down \[3\log(\sqrt{x}\log x)=56\] \[\log_{\log x}(x)=54\] and finds that this system of equations has a single real number solution $x>1$ . Find $b$ | [
"Using change of base on the second equation to base b, \\[\\frac{\\log x}{\\log \\log_{b}{x}}=54\\] \\[\\log x = 54 \\cdot \\log\\log_{b}{x}\\] \\[x = (\\log_{b}{x})^{54}\\] Note by dolphin7 - you could also just rewrite the second equation in exponent form.\nSubstituting this into the $\\sqrt x$ of the first equa... |
https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_19 | A | 2 | In a certain card game, a player is dealt a hand of $10$ cards from a deck of $52$ distinct cards. The number of distinct (unordered) hands that can be dealt to the player can be written as $158A00A4AA0$ . What is the digit $A$
$\textbf{(A) } 2 \qquad\textbf{(B) } 3 \qquad\textbf{(C) } 4 \qquad\textbf{(D) } 6 \qquad\te... | [
"$158A00A4AA0 \\equiv 1+5+8+A+0+0+A+4+A+A+0 \\equiv 4A \\pmod{9}$\nWe're looking for the amount of ways we can get $10$ cards from a deck of $52$ , which is represented by $\\binom{52}{10}$\n$\\binom{52}{10}=\\frac{52\\cdot51\\cdot50\\cdot49\\cdot48\\cdot47\\cdot46\\cdot45\\cdot44\\cdot43}{10\\cdot9\\cdot8\\cdot7\\... |
https://artofproblemsolving.com/wiki/index.php/2001_AIME_I_Problems/Problem_13 | null | 174 | In a certain circle , the chord of a $d$ -degree arc is $22$ centimeters long, and the chord of a $2d$ -degree arc is $20$ centimeters longer than the chord of a $3d$ -degree arc, where $d < 120.$ The length of the chord of a $3d$ -degree arc is $- m + \sqrt {n}$ centimeters, where $m$ and $n$ are positive integers. F... | [
"Note that a cyclic quadrilateral in the form of an isosceles trapezoid can be formed from three chords of three $d$ -degree arcs and one chord of one $3d$ -degree arc. The diagonals of this trapezoid turn out to be two chords of two $2d$ -degree arcs. Let $AB$ $AC$ , and $BD$ be the chords of the $d$ -degree arcs,... |
https://artofproblemsolving.com/wiki/index.php/1989_AHSME_Problems/Problem_25 | null | 13 | In a certain cross country meet between 2 teams of 5 runners each, a runner who finishes in the $n$ th position contributes $n$ to his teams score. The team with the lower score wins. If there are no ties among the runners, how many different winning scores are possible?
(A) 10 (B) 13 (C) 27 (D) 120 (E) 126 | [
"The scores of all ten runners must sum to $55$ . So a winning score is anything between $1+2+3+4+5=15$ and $\\lfloor\\tfrac{55}{2}\\rfloor=27$ inclusive. It is easy to check that this range is covered by considering $1+2+3+4+x$ $1+2+3+x+10$ and $1+2+x+9+10$ , so the answer is $\\boxed{13}$"
] |
https://artofproblemsolving.com/wiki/index.php/2009_AMC_10B_Problems/Problem_8 | B | 17 | In a certain year the price of gasoline rose by $20\%$ during January, fell by $20\%$ during February, rose by $25\%$ during March, and fell by $x\%$ during April. The price of gasoline at the end of April was the same as it had been at the beginning of January. To the nearest integer, what is $x$
$\mathrm{(A)}\ 12\qqu... | [
"Let $p$ be the price at the beginning of January. The price at the end of March was $(1.2)(0.8)(1.25)p = 1.2p.$ Because the price at the end of April was $p$ , the price decreased by $0.2p$ during April, and the percent decrease was \\[x = 100 \\cdot \\frac{0.2p}{1.2p} = \\frac {100}{6} \\approx 16.7.\\] So to th... |
https://artofproblemsolving.com/wiki/index.php/2009_AMC_12B_Problems/Problem_7 | B | 17 | In a certain year the price of gasoline rose by $20\%$ during January, fell by $20\%$ during February, rose by $25\%$ during March, and fell by $x\%$ during April. The price of gasoline at the end of April was the same as it had been at the beginning of January. To the nearest integer, what is $x$
$\mathrm{(A)}\ 12\qqu... | [
"Let $p$ be the price at the beginning of January. The price at the end of March was $(1.2)(0.8)(1.25)p = 1.2p.$ Because the price at the end of April was $p$ , the price decreased by $0.2p$ during April, and the percent decrease was \\[x = 100 \\cdot \\frac{0.2p}{1.2p} = \\frac {100}{6} \\approx 16.7.\\] So to th... |
https://artofproblemsolving.com/wiki/index.php/1995_AIME_Problems/Problem_14 | null | 378 | In a circle of radius $42$ , two chords of length $78$ intersect at a point whose distance from the center is $18$ . The two chords divide the interior of the circle into four regions. Two of these regions are bordered by segments of unequal lengths, and the area of either of them can be expressed uniquely in the for... | [
"Let the center of the circle be $O$ , and the two chords be $\\overline{AB}, \\overline{CD}$ and intersecting at $E$ , such that $AE = CE < BE = DE$ . Let $F$ be the midpoint of $\\overline{AB}$ . Then $\\overline{OF} \\perp \\overline{AB}$\nBy the Pythagorean Theorem $OF = \\sqrt{OB^2 - BF^2} = \\sqrt{42^2 - 39^2... |
https://artofproblemsolving.com/wiki/index.php/1985_AIME_Problems/Problem_9 | null | 49 | In a circle parallel chords of lengths 2, 3, and 4 determine central angles of $\alpha$ $\beta$ , and $\alpha + \beta$ radians , respectively, where $\alpha + \beta < \pi$ . If $\cos \alpha$ , which is a positive rational number , is expressed as a fraction in lowest terms, what is the sum of its numerator and denomina... | [
"All chords of a given length in a given circle subtend the same arc and therefore the same central angle. Thus, by the given, we can re-arrange our chords into a triangle with the circle as its circumcircle\nThis triangle has semiperimeter $\\frac{2 + 3 + 4}{2}$ so by Heron's formula it has area $K = \\sqrt{\\fra... |
https://artofproblemsolving.com/wiki/index.php/1985_AHSME_Problems/Problem_22 | D | 5 | In a circle with center $O$ $AD$ is a diameter, $ABC$ is a chord, $BO = 5$ and $\angle ABO = \ \stackrel{\frown}{CD} \ = 60^{\circ}$ . Then the length of $BC$ is
[asy] defaultpen(linewidth(0.7)+fontsize(10)); pair O=origin, A=dir(35), C=dir(155), D=dir(215), B=intersectionpoint(dir(125)--O, A--C); draw(C--A--D^^B--O^^C... | [
"Since $\\angle CAD$ is an angle inscribed in a $60{^\\circ}$ arc, we obtain $\\angle CAD =\\frac{60^{\\circ}}{2} = 30^{\\circ}$ , so $\\triangle ABO$ is a $30^{\\circ}$ $60^{\\circ}$ $90^{\\circ}$ right triangle. This gives $AO = BO\\sqrt{3} = 5\\sqrt{3}$ and $AB = 2BO = 10$ , and now since $AD$ is a diameter, $AD... |
https://artofproblemsolving.com/wiki/index.php/2008_AMC_10A_Problems/Problem_12 | null | 3.4 | In a collection of red, blue, and green marbles, there are $25\%$ more red marbles than blue marbles, and there are $60\%$ more green marbles than red marbles. Suppose that there are $r$ red marbles. What is the total number of marbles in the collection?
$\mathrm{(A)}\ 2.85r\qquad\mathrm{(B)}\ 3r\qquad\mathrm{(C)}\ 3.4... | [
"Let the number of red marbles be 100. You have $b = 80,$ and $g = 160.$\nThe answer is $\\frac{100+80+160}{100} = \\boxed{3.4}$"
] |
https://artofproblemsolving.com/wiki/index.php/2015_AIME_I_Problems/Problem_5 | null | 341 | In a drawer Sandy has $5$ pairs of socks, each pair a different color. On Monday Sandy selects two individual socks at random from the $10$ socks in the drawer. On Tuesday Sandy selects $2$ of the remaining $8$ socks at random and on Wednesday two of the remaining $6$ socks at random. The probability that Wednesday ... | [
"Let the fifth sock be arbitrary; the probability that the sixth sock matches in color is $\\dfrac{1}{9}$\nAssuming this, then let the first sock be arbitrary; the probability that the second sock does not match is $\\dfrac{6}{7}.$\nThe only \"hard\" part is the third and fourth sock. But that is simple casework. I... |
https://artofproblemsolving.com/wiki/index.php/1996_AIME_Problems/Problem_6 | null | 49 | In a five-team tournament, each team plays one game with every other team. Each team has a $50\%$ chance of winning any game it plays. (There are no ties.) Let $\dfrac{m}{n}$ be the probability that the tournament will produce neither an undefeated team nor a winless team, where $m$ and $n$ are relatively prime integer... | [
"We can use complementary counting : finding the probability that at least one team wins all games or at least one team loses all games.\nNo more than 1 team can win or lose all games, so at most one team can win all games and at most one team can lose all games.\nNow we use PIE\nThe probability that one team wins ... |
https://artofproblemsolving.com/wiki/index.php/1992_AIME_Problems/Problem_12 | null | 792 | In a game of Chomp , two players alternately take bites from a 5-by-7 grid of unit squares . To take a bite, a player chooses one of the remaining squares , then removes ("eats") all squares in the quadrant defined by the left edge (extended upward) and the lower edge (extended rightward) of the chosen square. For exam... | [
"By drawing possible examples of the subset, one can easily see that making one subset is the same as dividing the game board into two parts.\nOne can also see that it is the same as finding the shortest route from the upper left hand corner to the lower right hand corner; Such a route would require 5 lengths that ... |
https://artofproblemsolving.com/wiki/index.php/1981_AHSME_Problems/Problem_14 | A | 28 | In a geometric sequence of real numbers, the sum of the first $2$ terms is $7$ , and the sum of the first $6$ terms is $91$ . The sum of the first $4$ terms is
$\textbf{(A)}\ 28\qquad\textbf{(B)}\ 32\qquad\textbf{(C)}\ 35\qquad\textbf{(D)}\ 49\qquad\textbf{(E)}\ 84$ | [
"Denote the sum of the first $2$ terms as $x$ . Since we know that the sum of the first $6$ terms is $91$ which is $7 \\cdot 13$ , we have $x$ $xy$ $xy^2$ $13x$ because it is a geometric series. We can quickly see that $y$ $3$ , and therefore, the sum of the first $4$ terms is $4x = 4 \\cdot 7 = \\boxed{28}$"
] |
https://artofproblemsolving.com/wiki/index.php/2019_AMC_10B_Problems/Problem_10 | A | 0 | In a given plane, points $A$ and $B$ are $10$ units apart. How many points $C$ are there in the plane such that the perimeter of $\triangle ABC$ is $50$ units and the area of $\triangle ABC$ is $100$ square units?
$\textbf{(A) }0\qquad\textbf{(B) }2\qquad\textbf{(C) }4\qquad\textbf{(D) }8\qquad\textbf{(E) }\text{infini... | [
"Notice that whatever point we pick for $C$ $AB$ will be the base of the triangle. Without loss of generality, let points $A$ and $B$ be $(0,0)$ and $(10,0)$ , since for any other combination of points, we can just rotate the plane to make them $(0,0)$ and $(10,0)$ under a new coordinate system. When we pick point ... |
https://artofproblemsolving.com/wiki/index.php/2019_AMC_12B_Problems/Problem_6 | A | 0 | In a given plane, points $A$ and $B$ are $10$ units apart. How many points $C$ are there in the plane such that the perimeter of $\triangle ABC$ is $50$ units and the area of $\triangle ABC$ is $100$ square units?
$\textbf{(A) }0\qquad\textbf{(B) }2\qquad\textbf{(C) }4\qquad\textbf{(D) }8\qquad\textbf{(E) }\text{infini... | [
"Notice that whatever point we pick for $C$ $AB$ will be the base of the triangle. Without loss of generality, let points $A$ and $B$ be $(0,0)$ and $(10,0)$ , since for any other combination of points, we can just rotate the plane to make them $(0,0)$ and $(10,0)$ under a new coordinate system. When we pick point ... |
https://artofproblemsolving.com/wiki/index.php/1956_AHSME_Problems/Problem_6 | B | 7 | In a group of cows and chickens, the number of legs was 14 more than twice the number of heads. The number of cows was:
$\textbf{(A)}\ 5 \qquad\textbf{(B)}\ 7 \qquad\textbf{(C)}\ 10 \qquad\textbf{(D)}\ 12 \qquad\textbf{(E)}\ 14$ | [
"Let there be $x$ cows and $y$ chickens. Then, there are $4x+2y$ legs and $x+y$ heads. Writing the equation: \\[4x+2y=14+2(x+y)\\] \\[4x+2y=14+2x+2y\\] \\[2x=14\\] \\[x=\\boxed{7}\\]"
] |
https://artofproblemsolving.com/wiki/index.php/2012_AIME_II_Problems/Problem_14 | null | 16 | In a group of nine people each person shakes hands with exactly two of the other people from the group. Let $N$ be the number of ways this handshaking can occur. Consider two handshaking arrangements different if and only if at least two people who shake hands under one arrangement do not shake hands under the other ar... | [
"Given that each person shakes hands with two people, we can view all of these through graph theory as 'rings'. This will split it into four cases: Three rings of three, one ring of three and one ring of six, one ring of four and one ring of five, and one ring of nine. (All other cases that sum to nine won't work... |
https://artofproblemsolving.com/wiki/index.php/2019_AMC_10B_Problems/Problem_3 | B | 154 | In a high school with $500$ students, $40\%$ of the seniors play a musical instrument, while $30\%$ of the non-seniors do not play a musical instrument. In all, $46.8\%$ of the students do not play a musical instrument. How many non-seniors play a musical instrument?
$\textbf{(A) } 66 \qquad\textbf{(B) } 154 \qquad\tex... | [
"$60\\%$ of seniors do not play a musical instrument. If we denote $x$ as the number of seniors, then \\[\\frac{3}{5}x + \\frac{3}{10}\\cdot(500-x) = \\frac{468}{1000}\\cdot500\\]\n\\[\\frac{3}{5}x + 150 - \\frac{3}{10}x = 234\\] \\[\\frac{3}{10}x = 84\\] \\[x = 84\\cdot\\frac{10}{3} = 280\\]\nThus there are $500-x... |
https://artofproblemsolving.com/wiki/index.php/2012_AMC_8_Problems/Problem_19 | C | 9 | In a jar of red, green, and blue marbles, all but 6 are red marbles, all but 8 are green, and all but 4 are blue. How many marbles are in the jar?
$\textbf{(A)}\hspace{.05in}6\qquad\textbf{(B)}\hspace{.05in}8\qquad\textbf{(C)}\hspace{.05in}9\qquad\textbf{(D)}\hspace{.05in}10\qquad\textbf{(E)}\hspace{.05in}12$ | [
"$6$ are blue and green - $b+g=6$\n$8$ are red and blue - $r+b=8$\n$4$ are red and green- $r+g=4$\nWe can do trial and error. Let's make blue $5$ . That makes green $1$ and red $3$ because $6-5=1$ and $8-5=3$ . To check this, let's plug $1$ and $3$ into $r+g=4$ , which works. Now count the number of marbles - $5+3+... |
https://artofproblemsolving.com/wiki/index.php/1996_AIME_Problems/Problem_1 | null | 200 | In a magic square , the sum of the three entries in any row, column, or diagonal is the same value. The figure shows four of the entries of a magic square. Find $x$ | [
"Let's make a table.\n\\[\\begin{array}{|c|c|c|} \\multicolumn{3}{c}{\\text{Table}}\\\\\\hline x&19&96\\\\\\hline 1&a&b\\\\\\hline c&d&e\\\\\\hline \\end{array}\\]\n\\begin{eqnarray*} x+19+96=x+1+c\\Rightarrow c=19+96-1=114,\\\\ 114+96+a=x+1+114\\Rightarrow a=x-95 \\end{eqnarray*}\n\\[\\begin{array}{|c|c|c|} \\mult... |
https://artofproblemsolving.com/wiki/index.php/1985_AJHSME_Problems/Problem_24 | D | 39 | In a magic triangle, each of the six whole numbers $10-15$ is placed in one of the circles so that the sum, $S$ , of the three numbers on each side of the triangle is the same. The largest possible value for $S$ is
[asy] draw(circle((0,0),1)); draw(dir(60)--6*dir(60)); draw(circle(7*dir(60),1)); draw(8*dir(60)--13*dir... | [
"Let the number in the top circle be $a$ and then $b$ $c$ $d$ $e$ , and $f$ , going in clockwise order. Then, we have \\[S=a+b+c\\] \\[S=c+d+e\\] \\[S=e+f+a\\]\nAdding these equations together, we get\n\\begin{align*} 3S &= (a+b+c+d+e+f)+(a+c+e) \\\\ &= 75+(a+c+e) \\\\ \\end{align*}\nwhere the last step comes from ... |
https://artofproblemsolving.com/wiki/index.php/2010_AMC_10A_Problems/Problem_15 | D | 3 | In a magical swamp there are two species of talking amphibians: toads, whose statements are always true, and frogs, whose statements are always false. Four amphibians, Brian, Chris, LeRoy, and Mike live together in this swamp, and they make the following statements.
Brian: "Mike and I are different species."
Chris: "Le... | [
"Start with Brian. If he is a toad, he tells the truth, hence Mike is a frog. If Brian is a frog, he lies, hence Mike is a frog, too. Thus Mike must be a frog.\nAs Mike is a frog, his statement is false, hence there is at most one toad.\nAs there is at most one toad, at least one of Chris and LeRoy is a frog. But t... |
https://artofproblemsolving.com/wiki/index.php/2010_AMC_12A_Problems/Problem_12 | D | 3 | In a magical swamp there are two species of talking amphibians: toads, whose statements are always true, and frogs, whose statements are always false. Four amphibians, Brian, Chris, LeRoy, and Mike live together in this swamp, and they make the following statements.
Brian: "Mike and I are different species."
Chris: "Le... | [
"Start with Brian. If he is a toad, he tells the truth, hence Mike is a frog. If Brian is a frog, he lies, hence Mike is a frog, too. Thus Mike must be a frog.\nAs Mike is a frog, his statement is false, hence there is at most one toad.\nAs there is at most one toad, at least one of Chris and LeRoy is a frog. But t... |
https://artofproblemsolving.com/wiki/index.php/2002_AMC_8_Problems/Problem_17 | C | 7 | In a mathematics contest with ten problems, a student gains 5 points for a correct answer and loses 2 points for an incorrect answer. If Olivia answered every problem and her score was 29, how many correct answers did she have?
$\text{(A)}\ 5\qquad\text{(B)}\ 6\qquad\text{(C)}\ 7\qquad\text{(D)}\ 8\qquad\text{(E)}\ 9$ | [
"We can try to guess and check to find the answer. If she got five right, her score would be $(5*5)-(5*2)=15$ . If she got six right her score would be $(6*5)-(2*4)=22$ . That's close, but it's still not right! If she got 7 right, her score would be $(7*5)-(2*3)=29$ . Thus, our answer is $\\boxed{7}$ . ~avamarora",... |
https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_2 | null | 25 | In a new school $40$ percent of the students are freshmen, $30$ percent are sophomores, $20$ percent are juniors, and $10$ percent are seniors. All freshmen are required to take Latin, and $80$ percent of the sophomores, $50$ percent of the juniors, and $20$ percent of the seniors elect to take Latin. The probability t... | [
"We see that $40\\% \\cdot 100\\% + 30\\% \\cdot 80\\% + 20\\% \\cdot 50\\% + 10\\% \\cdot 20\\% = 76\\%$ of students are learning Latin. In addition, $30\\% \\cdot 80\\% = 24\\%$ of students are sophomores learning Latin. Thus, our desired probability is $\\dfrac{24}{76}=\\dfrac{6}{19}$ and our answer is $6+19=\\b... |
https://artofproblemsolving.com/wiki/index.php/1986_AIME_Problems/Problem_10 | null | 358 | In a parlor game, the magician asks one of the participants to think of a three digit number $(abc)$ where $a$ $b$ , and $c$ represent digits in base $10$ in the order indicated. The magician then asks this person to form the numbers $(acb)$ $(bca)$ $(bac)$ $(cab)$ , and $(cba)$ , to add these five numbers, and to reve... | [
"Let $m$ be the number $100a+10b+c$ . Observe that $3194+m=222(a+b+c)$ so\n\\[m\\equiv -3194\\equiv -86\\equiv 136\\pmod{222}\\]\nThis reduces $m$ to one of $136, 358, 580, 802$ . But also $a+b+c=\\frac{3194+m}{222}>\\frac{3194}{222}>14$ so $a+b+c\\geq 15$ . \nRecall that $a, b, c$ refer to the digits the three dig... |
https://artofproblemsolving.com/wiki/index.php/1982_USAMO_Problems/Problem_1 | null | 1,979 | In a party with $1982$ people, among any group of four there is at least one person who knows each of the other three. What is the minimum number of people in the party who know everyone else? | [
"We induct on $n$ to prove that in a party with $n$ people, there must be at least $(n-3)$ people who know everyone else. (Clearly this is achievable by having everyone know everyone else except three people $A, B, C$ , who do not know each other.)\nBase case: $n = 4$ is obvious.\nInductive step: Suppose in a party... |
https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_7 | D | 65 | In a plane, four circles with radii $1,3,5,$ and $7$ are tangent to line $\ell$ at the same point $A,$ but they may be on either side of $\ell$ . Region $S$ consists of all the points that lie inside exactly one of the four circles. What is the maximum possible area of region $S$
$\textbf{(A) }24\pi \qquad \textbf{(B) ... | [
"Suppose that line $\\ell$ is horizontal, and each circle lies either north or south to $\\ell.$ We construct the circles one by one:\nThe diagram below shows one possible configuration of the four circles: Together, the answer is $\\pi\\cdot7^2+\\pi\\cdot5^2-\\pi\\cdot3^2=\\boxed{65}.$"
] |
https://artofproblemsolving.com/wiki/index.php/2013_AMC_10A_Problems/Problem_9 | B | 18 | In a recent basketball game, Shenille attempted only three-point shots and two-point shots. She was successful on $20\%$ of her three-point shots and $30\%$ of her two-point shots. Shenille attempted $30$ shots. How many points did she score?
$\textbf{(A)}\ 12 \qquad\textbf{(B)}\ 18 \qquad\textbf{(C)}\ 24 \qquad\te... | [
"Let the number of attempted three-point shots be $x$ and the number of attempted two-point shots be $y$ . We know that $x+y=30$ , and we need to evaluate $3(0.2x) + 2(0.3y)$ , as we know that the three-point shots are worth $3$ points and that she made $20$ % of them and that the two-point shots are worth $2$ and... |
https://artofproblemsolving.com/wiki/index.php/2013_AMC_12A_Problems/Problem_6 | B | 18 | In a recent basketball game, Shenille attempted only three-point shots and two-point shots. She was successful on $20\%$ of her three-point shots and $30\%$ of her two-point shots. Shenille attempted $30$ shots. How many points did she score?
$\textbf{(A)}\ 12\qquad\textbf{(B)}\ 18\qquad\textbf{(C)}\ 24\qquad\textbf{(D... | [
"(similar to Solution 1, however a slightly more obvious way)\nSay that\nx = # of 2-pt shots\ny = # of 3-pt shots\nBecause the total number of shots is $30$ $x + y = 30$\nHowever, Shenille was only successful on $20\\%$ of the 3-pt shots, and $30\\%$ of the 2-pt shots, so\n$0.2x + 0.3y$ = #number of successful shot... |
https://artofproblemsolving.com/wiki/index.php/2001_AIME_I_Problems/Problem_11 | null | 149 | In a rectangular array of points, with 5 rows and $N$ columns, the points are numbered consecutively from left to right beginning with the top row. Thus the top row is numbered 1 through $N,$ the second row is numbered $N + 1$ through $2N,$ and so forth. Five points, $P_1, P_2, P_3, P_4,$ and $P_5,$ are selected so t... | [
"Let each point $P_i$ be in column $c_i$ . The numberings for $P_i$ can now be defined as follows. \\begin{align*}x_i &= (i - 1)N + c_i\\\\ y_i &= (c_i - 1)5 + i \\end{align*}\nWe can now convert the five given equalities. \\begin{align}x_1&=y_2 & \\Longrightarrow & & c_1 &= 5 c_2-3\\\\ x_2&=y_1 & \\Longrightarrow ... |
https://artofproblemsolving.com/wiki/index.php/2003_AIME_II_Problems/Problem_4 | null | 28 | In a regular tetrahedron the centers of the four faces are the vertices of a smaller tetrahedron. The ratio of the volume of the smaller tetrahedron to that of the larger is $m/n$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | [
"Embed the tetrahedron in 4-space to make calculations easier.\nIts vertices are $(1,0,0,0)$ $(0,1,0,0)$ $(0,0,1,0)$ $(0,0,0,1)$\nTo get the center of any face, we take the average of the three coordinates of that face. The vertices of the center of the faces are: $(\\frac{1}{3}, \\frac{1}{3}, \\frac{1}{3}, 0)$ $(\... |
https://artofproblemsolving.com/wiki/index.php/1959_AHSME_Problems/Problem_15 | C | 45 | In a right triangle the square of the hypotenuse is equal to twice the product of the legs. One of the acute angles of the triangle is: $\textbf{(A)}\ 15^{\circ} \qquad\textbf{(B)}\ 30^{\circ} \qquad\textbf{(C)}\ 45^{\circ} \qquad\textbf{(D)}\ 60^{\circ}\qquad\textbf{(E)}\ 75^{\circ}$ | [
"WLOG, by scaling, that the hypotenuse has length 1. Let $\\theta$ be an angle opposite from some leg. Then the two legs have length $\\sin\\theta$ and $\\cos\\theta$ respectively, so we have $2\\sin\\theta\\cos\\theta = 1^2$ . From trigonometry, we know that this equation is true when $\\theta = 45^{\\circ}$ , so ... |
https://artofproblemsolving.com/wiki/index.php/2010_AMC_8_Problems/Problem_20 | A | 3 | In a room, $2/5$ of the people are wearing gloves, and $3/4$ of the people are wearing hats. What is the minimum number of people in the room wearing both a hat and a glove?
$\textbf{(A)}\ 3 \qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 15\qquad\textbf{(E)}\ 20$ | [
"Let $x$ be the number of people wearing both a hat and a glove. Since the number of people wearing a hat or a glove must be whole numbers, the number of people in the room must be a multiple of (4,5), but we do lcm of 4,5 = 20. Since we are trying to find the minimum $x$ , we must use the smallest possible value f... |
https://artofproblemsolving.com/wiki/index.php/2012_AMC_10B_Problems/Problem_15 | D | 5 | In a round-robin tournament with 6 teams, each team plays one game against each other team, and each game results in one team winning and one team losing. At the end of the tournament, the teams are ranked by the number of games won. What is the maximum number of teams that could be tied for the most wins at the end of... | [
"The total number of games (and wins) in the tournament is $\\frac{6 \\times 5}{2}= 15$ . A six-way tie is impossible as this would imply each team has 2.5 wins, so the maximum number of tied teams is five. Here's a chart of 15 games where five teams each have 3 wins:\nThe \"X's\" are for when it is where a team is... |
https://artofproblemsolving.com/wiki/index.php/1986_AIME_Problems/Problem_13 | null | 560 | In a sequence of coin tosses, one can keep a record of instances in which a tail is immediately followed by a head, a head is immediately followed by a head, and etc. We denote these by TH HH , and etc. For example, in the sequence TTTHHTHTTTHHTTH of 15 coin tosses we observe that there are two HH , three HT , four TH ... | [
"Let's consider each of the sequences of two coin tosses as an operation instead; this operation takes a string and adds the next coin toss on (eg, THHTH HT THHTHT ). We examine what happens to the last coin toss. Adding HH or TT is simply an identity for the last coin toss, so we will ignore them for now. However,... |
https://artofproblemsolving.com/wiki/index.php/2023_AMC_8_Problems/Problem_22 | D | 5 | In a sequence of positive integers, each term after the second is the product of the previous two terms. The sixth term is $4000$ . What is the first term?
$\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 4 \qquad \textbf{(D)}\ 5 \qquad \textbf{(E)}\ 10$ | [
"In this solution, we will use trial and error to solve. $4000$ can be expressed as $200 \\times 20$ . We divide $200$ by $20$ and get $10$ , divide $20$ by $10$ and get $2$ , and divide $10$ by $2$ to get $\\boxed{5}$ . No one said that they have to be in ascending order!",
"Consider the first term is $a$ and th... |
https://artofproblemsolving.com/wiki/index.php/1990_AIME_Problems/Problem_8 | null | 560 | In a shooting match, eight clay targets are arranged in two hanging columns of three targets each and one column of two targets. A marksman is to break all the targets according to the following rules:
1) The marksman first chooses a column from which a target is to be broken.
2) The marksman must then break the lowest... | [
"Clearly, the marksman must shoot the left column three times, the middle column two times, and the right column three times.\nFrom left to right, suppose that the columns are labeled $L,M,$ and $R,$ respectively. We consider the string $LLLMMRRR:$\nSince the letter arrangements of $LLLMMRRR$ and the shooting order... |
https://artofproblemsolving.com/wiki/index.php/2018_AMC_8_Problems/Problem_19 | C | 8 | In a sign pyramid a cell gets a "+" if the two cells below it have the same sign, and it gets a "-" if the two cells below it have different signs. The diagram below illustrates a sign pyramid with four levels. How many possible ways are there to fill the four cells in the bottom row to produce a "+" at the top of the ... | [
"You could just make out all of the patterns that make the top positive. In this case, you would have the following patterns:\n+−−+, −++−, −−−−, ++++, −+−+, +−+−, ++−−, −−++. There are 8 patterns and so the answer is $\\boxed{8}$",
"The top box is fixed by the problem.\nChoose the left 3 bottom-row boxes freely. ... |
https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_16 | B | 36 | In a table tennis tournament every participant played every other participant exactly once. Although there were twice as many right-handed players as left-handed players, the number of games won by left-handed players was $40\%$ more than the number of games won by right-handed players. (There were no ties and no ambid... | [
"We know that the total amount of games must be the sum of games won by left and right handed players. Then, we can write $g = l + r$ , and since $l = 1.4r$ $g = 2.4r$ . Given that $r$ and $g$ are both integers, $g/2.4$ also must be an integer. From here we can see that $g$ must be divisible by 12, leaving only ans... |
https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_13 | B | 36 | In a table tennis tournament every participant played every other participant exactly once. Although there were twice as many right-handed players as left-handed players, the number of games won by left-handed players was $40\%$ more than the number of games won by right-handed players. (There were no ties and no ambid... | [
"We know that the total amount of games must be the sum of games won by left and right handed players. Then, we can write $g = l + r$ , and since $l = 1.4r$ $g = 2.4r$ . Given that $r$ and $g$ are both integers, $g/2.4$ also must be an integer. From here we can see that $g$ must be divisible by 12, leaving only ans... |
https://artofproblemsolving.com/wiki/index.php/1985_AIME_Problems/Problem_14 | null | 25 | In a tournament each player played exactly one game against each of the other players. In each game the winner was awarded $1$ point, the loser got $0$ points, and each of the two players earned $\frac{1}{2}$ point if the game was a tie. After the completion of the tournament, it was found that exactly half of the poin... | [
"Let us suppose for convenience that there were $n + 10$ players overall. Among the $n$ players not in the weakest 10 there were $n \\choose 2$ games played and thus $n \\choose 2$ points earned. By the givens, this means that these $n$ players also earned $n \\choose 2$ points against our weakest 10. Now, the 1... |
https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_19 | C | 24 | In a tournament there are six teams that play each other twice. A team earns $3$ points for a win, $1$ point for a draw, and $0$ points for a loss. After all the games have been played it turns out that the top three teams earned the same number of total points. What is the greatest possible number of total points for ... | [
"This isn't finished\nto another. This gives equality, as each team wins once and loses once as well. For a win, we have $3$ points, so a team gets $3\\times2=6$ points if they each win a game and lose a game. This case brings a total of $18+6=24$ points.\nTherefore, we use Case 2 since it brings the greater amoun... |
https://artofproblemsolving.com/wiki/index.php/2011_AMC_8_Problems/Problem_6 | D | 306 | In a town of $351$ adults, every adult owns a car, motorcycle, or both. If $331$ adults own cars and $45$ adults own motorcycles, how many of the car owners do not own a motorcycle?
$\textbf{(A)}\ 20 \qquad \textbf{(B)}\ 25 \qquad \textbf{(C)}\ 45 \qquad \textbf{(D)}\ 306 \qquad \textbf{(E)}\ 351$ | [
"By PIE , the number of adults who own both cars and motorcycles is $331+45-351=25.$ Out of the $331$ car owners, $25$ of them own motorcycles and $331-25=\\boxed{306}$ of them don't.",
"There are $351$ total adults, and $45$ own a motorcycle. The number of adults that don't own a motorcycle is $351 - 45 = 306$ .... |
https://artofproblemsolving.com/wiki/index.php/2006_AMC_10B_Problems/Problem_10 | A | 43 | In a triangle with integer side lengths, one side is three times as long as a second side, and the length of the third side is $15$ . What is the greatest possible perimeter of the triangle?
$\textbf{(A) } 43\qquad \textbf{(B) } 44\qquad \textbf{(C) } 45\qquad \textbf{(D) } 46\qquad \textbf{(E) } 47$ | [
"Let $x$ be the length of the first side.\nThe lengths of the sides are: $x$ $3x$ , and $15$\nBy the Triangle Inequality\n$3x < x + 15$\n$2x < 15$\n$x < \\frac{15}{2}$\nThe greatest integer satisfying this inequality is $7$\nSo the greatest possible perimeter is $7 + 3\\cdot7 + 15 =\\boxed{43}$"
] |
https://artofproblemsolving.com/wiki/index.php/2006_AMC_12B_Problems/Problem_10 | A | 43 | In a triangle with integer side lengths, one side is three times as long as a second side, and the length of the third side is 15. What is the greatest possible perimeter of the triangle?
$\text {(A) } 43 \qquad \text {(B) } 44 \qquad \text {(C) } 45 \qquad \text {(D) } 46 \qquad \text {(E) } 47$ | [
"If the second size has length x, then the first side has length 3x, and we have the third side which has length 15. By the triangle inequality, we have: \\[\\\\ x+15>3x \\Rightarrow 2x<15 \\Rightarrow x<7.5\\] Now, since we want the greatest perimeter, we want the greatest integer x, and if $x<7.5$ then $x=7$ . Th... |
https://artofproblemsolving.com/wiki/index.php/1981_AHSME_Problems/Problem_21 | null | 60 | In a triangle with sides of lengths $a$ $b$ , and $c$ $(a+b+c)(a+b-c) = 3ab$ . The measure of the angle opposite the side length $c$ is
$\textbf{(A)}\ 15^\circ\qquad\textbf{(B)}\ 30^\circ\qquad\textbf{(C)}\ 45^\circ\qquad\textbf{(D)}\ 60^\circ\qquad\textbf{(E)}\ 150^\circ$ | [
"\\[(a+b+c)(a+b-c)=3ab\\] \\[a^2+2ab+b^2-c^2=3ab\\] \\[a^2+b^2-c^2=ab\\] \\[c^2=a^2+b^2-ab\\] This looks a lot like Law of Cosines, which is $c^2=a^2+b^2-2ab\\cos{c}$ \\[c^2=a^2+b^2-ab=a^2+b^2-2ab\\cos{c}\\] \\[ab=2ab\\cos{c}\\] \\[\\frac{1}{2}=\\cos{c}\\] $\\cos{c}$ is $\\frac{1}{2}$ , so the angle opposite side $... |
https://artofproblemsolving.com/wiki/index.php/2019_AIME_II_Problems/Problem_15 | null | 574 | In acute triangle $ABC$ points $P$ and $Q$ are the feet of the perpendiculars from $C$ to $\overline{AB}$ and from $B$ to $\overline{AC}$ , respectively. Line $PQ$ intersects the circumcircle of $\triangle ABC$ in two distinct points, $X$ and $Y$ . Suppose $XP=10$ $PQ=25$ , and $QY=15$ . The value of $AB\cdot AC$ can b... | [
"First we have $a\\cos A=PQ=25$ , and $(a\\cos A)(c\\cos C)=(a\\cos C)(c\\cos A)=AP\\cdot PB=10(25+15)=400$ by PoP. Similarly, $(a\\cos A)(b\\cos B)=15(10+25)=525,$ and dividing these each by $a\\cos A$ gives $b\\cos B=21,c\\cos C=16$\nIt is known that the sides of the orthic triangle are $a\\cos A,b\\cos B,c\\cos ... |
https://artofproblemsolving.com/wiki/index.php/1998_AHSME_Problems/Problem_21 | null | 2 | In an $h$ -meter race, Sunny is exactly $d$ meters ahead of Windy when Sunny finishes the race. The next time they race, Sunny sportingly starts $d$ meters behind Windy, who is at the starting line. Both runners run at the same constant speed as they did in the first race. How many meters ahead is Sunny when Sunny fini... | [
"Let $s$ and $w$ be the speeds of Sunny and Windy. From the first race we know that $\\frac sw = \\frac h{h-d}$ . In the second race, Sunny's track length is $h+d$ . She will finish this track in $\\frac{h+d}s$ . In this time, Windy will run the distance $w\\cdot \\frac{h+d}s = \\frac{(h+d)(h-d)}h$ . This is less t... |
https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_18 | C | 43 | In an All-Area track meet, $216$ sprinters enter a $100-$ meter dash competition. The track has $6$ lanes, so only $6$ sprinters can compete at a time. At the end of each race, the five non-winners are eliminated, and the winner will compete again in a later race. How many races are needed to determine the champion spr... | [
"From any $n-$ th race, only $\\frac{1}{6}$ will continue on. Since we wish to find the total number of races, a column representing the races over time is ideal.\nStarting with the first race: \\[\\frac{216}{6}=36\\] \\[\\frac{36}{6}=6\\] \\[\\frac{6}{6}=1\\] Adding all of the numbers in the second column yields $... |
https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_3 | C | 8 | In an after-school program for juniors and seniors, there is a debate team with an equal number of students from each class on the team. Among the $28$ students in the program, $25\%$ of the juniors and $10\%$ of the seniors are on the debate team. How many juniors are in the program?
$\textbf{(A)} ~5 \qquad\textbf{(B)... | [
"Say there are $j$ juniors and $s$ seniors in the program. Converting percentages to fractions, $\\frac{j}{4}$ and $\\frac{s}{10}$ are on the debate team, and since an equal number of juniors and seniors are on the debate team, $\\frac{j}{4} = \\frac{s}{10}.$\nCross-multiplying and simplifying we get $5j=2s.$ Addit... |
https://artofproblemsolving.com/wiki/index.php/2003_AIME_I_Problems/Problem_8 | null | 129 | In an increasing sequence of four positive integers, the first three terms form an arithmetic progression , the last three terms form a geometric progression , and the first and fourth terms differ by $30$ . Find the sum of the four terms. | [
"Denote the first term as $a$ , and the common difference between the first three terms as $d$ . The four numbers thus are in the form $a,\\ a+d,\\ a+2d,\\ \\frac{(a + 2d)^2}{a + d}$\nSince the first and fourth terms differ by $30$ , we have that $\\frac{(a + 2d)^2}{a + d} - a = 30$ . Multiplying out by the denomin... |
https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_4 | null | 18 | In an isosceles trapezoid, the parallel bases have lengths $\log 3$ and $\log 192$ , and the altitude to these bases has length $\log 16$ . The perimeter of the trapezoid can be written in the form $\log 2^p 3^q$ , where $p$ and $q$ are positive integers. Find $p + q$ | [
"Call the trapezoid $ABCD$ with $AB$ as the smaller base and $CD$ as the longer. Let the point where an altitude intersects the larger base be $E$ , where $E$ is closer to $D$\nSubtract the two bases and divide to find that $ED$ is $\\log 8$ . The altitude can be expressed as $\\frac{4}{3} \\log 8$ . Therefore, the... |
https://artofproblemsolving.com/wiki/index.php/1988_AIME_Problems/Problem_15 | null | 704 | In an office at various times during the day, the boss gives the secretary a letter to type, each time putting the letter on top of the pile in the secretary's inbox. When there is time, the secretary takes the top letter off the pile and types it. There are nine letters to be typed during the day, and the boss deliver... | [
"Re-stating the problem for clarity, let $S$ be a set arranged in increasing order. At any time an element can be appended to the end of $S$ , or the last element of $S$ can be removed. The question asks for the number of different orders in which all of the remaining elements of $S$ can be removed, given that $8$ ... |
https://artofproblemsolving.com/wiki/index.php/2013_AMC_10A_Problems/Problem_19 | C | 13 | In base $10$ , the number $2013$ ends in the digit $3$ . In base $9$ , on the other hand, the same number is written as $(2676)_{9}$ and ends in the digit $6$ . For how many positive integers $b$ does the base- $b$ -representation of $2013$ end in the digit $3$
$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 9\qquad\textbf{(C)}\... | [
"We want the integers $b$ such that $2013\\equiv 3\\pmod{b} \\Rightarrow b$ is a factor of $2010$ . Since $2010=2 \\cdot 3 \\cdot 5 \\cdot 67$ , it has $(1+1)(1+1)(1+1)(1+1)=16$ factors. Since $b$ cannot equal $1, 2,$ or $3$ , as these cannot have the digit $3$ in their base representations, our answer is $16-3=\\b... |
https://artofproblemsolving.com/wiki/index.php/1955_AHSME_Problems/Problem_23 | null | 11 | In checking the petty cash a clerk counts $q$ quarters, $d$ dimes, $n$ nickels, and $c$ cents. Later he discovers that $x$ of the nickels were counted as quarters and $x$ of the dimes were counted as cents. To correct the total obtained the clerk must:
$\textbf{(A)}\ \text{make no correction}\qquad\textbf{(B)}\ \text{s... | [
"If the clerk mistook $x$ nickels as quarters, then every mistake inflates the total by $20$ cents. In order to correct this, we have to subtract $20$ cents $x$ times, for a total of $20x$ cents. We can do the same for the $x$ dimes that were turned into pennies (or cents). This exchange would increase the total va... |
https://artofproblemsolving.com/wiki/index.php/2006_AIME_II_Problems/Problem_1 | null | 46 | In convex hexagon $ABCDEF$ , all six sides are congruent, $\angle A$ and $\angle D$ are right angles , and $\angle B, \angle C, \angle E,$ and $\angle F$ are congruent . The area of the hexagonal region is $2116(\sqrt{2}+1).$ Find $AB$ | [
"Let the side length be called $x$ , so $x=AB=BC=CD=DE=EF=AF$\n2006 II AIME-1.png\nThe diagonal $BF=\\sqrt{AB^2+AF^2}=\\sqrt{x^2+x^2}=x\\sqrt{2}$ . Then the areas of the triangles AFB and CDE in total are $\\frac{x^2}{2}\\cdot 2$ ,\nand the area of the rectangle BCEF equals $x\\cdot x\\sqrt{2}=x^2\\sqrt{2}$\nThen w... |
https://artofproblemsolving.com/wiki/index.php/2003_AIME_I_Problems/Problem_12 | null | 777 | In convex quadrilateral $ABCD, \angle A \cong \angle C, AB = CD = 180,$ and $AD \neq BC.$ The perimeter of $ABCD$ is $640$ . Find $\lfloor 1000 \cos A \rfloor.$ (The notation $\lfloor x \rfloor$ means the greatest integer that is less than or equal to $x.$ | [
"By the Law of Cosines on $\\triangle ABD$ at angle $A$ and on $\\triangle BCD$ at angle $C$ (note $\\angle C = \\angle A$ ),\n\\[180^2 + AD^2 - 360 \\cdot AD \\cos A = 180^2 + BC^2 - 360 \\cdot BC \\cos A\\] \\[(AD^2 - BC^2) = 360(AD - BC) \\cos A\\] \\[(AD - BC)(AD + BC) = 360(AD - BC) \\cos A\\] \\[(AD + BC) = 3... |
https://artofproblemsolving.com/wiki/index.php/2019_AIME_I_Problems/Problem_6 | null | 90 | In convex quadrilateral $KLMN$ side $\overline{MN}$ is perpendicular to diagonal $\overline{KM}$ , side $\overline{KL}$ is perpendicular to diagonal $\overline{LN}$ $MN = 65$ , and $KL = 28$ . The line through $L$ perpendicular to side $\overline{KN}$ intersects diagonal $\overline{KM}$ at $O$ with $KO = 8$ . Find $MO$ | [
"Let $\\angle MKN=\\alpha$ and $\\angle LNK=\\beta$ . Let $P$ be the project of $L$ onto line $NK$ . Note $\\angle KLP=\\beta$\nThen, $KP=28\\sin\\beta=8\\cos\\alpha$ .\nFurthermore, $KN=\\frac{65}{\\sin\\alpha}=\\frac{28}{\\sin\\beta} \\Rightarrow 65\\sin\\beta=28\\sin\\alpha$\nDividing the equations gives \\[\\fr... |
https://artofproblemsolving.com/wiki/index.php/1963_AHSME_Problems/Problem_19 | B | 210 | In counting $n$ colored balls, some red and some black, it was found that $49$ of the first $50$ counted were red.
Thereafter, $7$ out of every $8$ counted were red. If, in all, $90$ % or more of the balls counted were red, the maximum value of $n$ is:
$\textbf{(A)}\ 225 \qquad \textbf{(B)}\ 210 \qquad \textbf{(C)}\ 2... | [
"The desired percentage of red balls is more than $90$ percent, so write an inequality\n\\[\\frac{49+7x}{50+8x} \\ge 0.9\\]\nSince $x >0$ , the sign does not need to be swapped after multiplying both sides by $50+8x$\n\\[49+7x \\ge 45+7.2x\\] \\[4 \\ge 0.2x\\] \\[20 \\ge x\\]\nThus, up to $20$ batches of balls can ... |
https://artofproblemsolving.com/wiki/index.php/2018_AIME_II_Problems/Problem_4 | null | 23 | In equiangular octagon $CAROLINE$ $CA = RO = LI = NE =$ $\sqrt{2}$ and $AR = OL = IN = EC = 1$ . The self-intersecting octagon $CORNELIA$ encloses six non-overlapping triangular regions. Let $K$ be the area enclosed by $CORNELIA$ , that is, the total area of the six triangular regions. Then $K =$ $\dfrac{a}{b}$ , where... | [
"We can draw $CORNELIA$ and introduce some points.\n2018 AIME II Problem 4.png\nThe diagram is essentially a 3x3 grid where each of the 9 squares making up the grid have a side length of 1.\nIn order to find the area of $CORNELIA$ , we need to find 4 times the area of $\\bigtriangleup$ $ACY$ and 2 times the area of... |
https://artofproblemsolving.com/wiki/index.php/2013_AIME_II_Problems/Problem_5 | null | 20 | In equilateral $\triangle ABC$ let points $D$ and $E$ trisect $\overline{BC}$ . Then $\sin(\angle DAE)$ can be expressed in the form $\frac{a\sqrt{b}}{c}$ , where $a$ and $c$ are relatively prime positive integers, and $b$ is an integer that is not divisible by the square of any prime. Find $a+b+c$ | [
"\nWithout loss of generality, assume the triangle sides have length 3. Then the trisected side is partitioned into segments of length 1, making your computation easier.\nLet $M$ be the midpoint of $\\overline{DE}$ . Then $\\Delta MCA$ is a 30-60-90 triangle with $MC = \\dfrac{3}{2}$ $AC = 3$ and $AM = \\dfrac{3\\s... |
https://artofproblemsolving.com/wiki/index.php/2011_AMC_8_Problems/Problem_24 | A | 0 | In how many ways can $10001$ be written as the sum of two primes?
$\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }3\qquad\textbf{(E) }4$ | [
"For the sum of two numbers to be odd, one must be odd and the other must be even, because all odd numbers are of the form $2n+1$ where n is an integer, and all even numbers are of the form $2m$ where m is an integer. \\[2n + 1 + 2m = 2m + 2n + 1 = 2(m+n) + 1\\] and $m+n$ is an integer because $m$ and $n$ are both ... |
https://artofproblemsolving.com/wiki/index.php/2016_AMC_10B_Problems/Problem_18 | E | 7 | In how many ways can $345$ be written as the sum of an increasing sequence of two or more consecutive positive integers?
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 7$ | [
"Factor $345=3\\cdot 5\\cdot 23$\nSuppose we take an odd number $k$ of consecutive integers, with the median as $m$ . Then $mk=345$ with $\\tfrac12k<m$ .\nLooking at the factors of $345$ , the possible values of $k$ are $3,5,15,23$ with medians as $115,69,23,15$ respectively.\nSuppose instead we take an even number... |
https://artofproblemsolving.com/wiki/index.php/2016_AMC_12B_Problems/Problem_16 | E | 7 | In how many ways can $345$ be written as the sum of an increasing sequence of two or more consecutive positive integers?
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 7$ | [
"We proceed with this problem by considering two cases, when: 1) There are an odd number of consecutive numbers, 2) There are an even number of consecutive numbers.\nFor the first case, we can cleverly choose the convenient form of our sequence to be \\[a-n,\\cdots, a-1, a, a+1, \\cdots, a+n\\]\nbecause then our su... |
https://artofproblemsolving.com/wiki/index.php/2016_AMC_12B_Problems/Problem_16 | null | 7 | In how many ways can $345$ be written as the sum of an increasing sequence of two or more consecutive positive integers?
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 7$ | [
"By the sum of an arithmetic sequence... this ultimately comes to $n+n+1+n+2....+n+p=345=(2n+p)(p+1)=690=23\\cdot3\\cdot5\\cdot2$\nQuick testing (would take you roughly a minute)\nWe see that the first 7 values of $p$ that work are\n$p=1,2,4,5,9,14,22$\nWe see that each one of them works.\nHence, the answer is $\\b... |
https://artofproblemsolving.com/wiki/index.php/1989_AJHSME_Problems/Problem_16 | A | 0 | In how many ways can $47$ be written as the sum of two primes
$\text{(A)}\ 0 \qquad \text{(B)}\ 1 \qquad \text{(C)}\ 2 \qquad \text{(D)}\ 3 \qquad \text{(E)}\ \text{more than 3}$ | [
"For $47$ to be written as the sum of two integers , one must be odd and the other must be even. There is only one even prime, namely $2$ , so one of the numbers must be $2$ , making the other $45$\nHowever, $45$ is not prime, so there are no ways to write $47$ as the sum of two primes $\\rightarrow \\boxed{0}$"
] |
https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_14 | D | 24 | In how many ways can the letters in $\textbf{BEEKEEPER}$ be rearranged so that two or more $\textbf{E}$ s do not appear together?
$\textbf{(A) } 1 \qquad \textbf{(B) } 4 \qquad \textbf{(C) } 12 \qquad \textbf{(D) } 24 \qquad \textbf{(E) } 120$ | [
"All valid arrangements of the letters must be of the form \\[\\textbf{E\\underline{\\hspace{3mm}}E\\underline{\\hspace{3mm}}E\\underline{\\hspace{3mm}}E\\underline{\\hspace{3mm}}E}.\\] The problem is equivalent to counting the arrangements of $\\textbf{B},\\textbf{K},\\textbf{P},$ and $\\textbf{R}$ into the four b... |
https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_20 | D | 32 | In how many ways can the sequence $1,2,3,4,5$ be rearranged so that no three consecutive terms are increasing and no three consecutive terms are decreasing?
$\textbf{(A)} ~10\qquad\textbf{(B)} ~18\qquad\textbf{(C)} ~24 \qquad\textbf{(D)} ~32 \qquad\textbf{(E)} ~44$ | [
"We write out the $5!=120$ cases, then filter out the valid ones:\n$13254,14253,14352,15243,15342,21435,21534,23154,24153,24351,25143,25341,\\linebreak 31425,31524,32415,32514,34152,34251,35142,35241,41325,41523,42315,42513,\\linebreak 43512,45132,45231,51324,51423,52314,52413,53412.$\nWe count these out and get $... |
https://artofproblemsolving.com/wiki/index.php/1998_AJHSME_Problems/Problem_17 | C | 100 | In how many years, approximately, from 1998 will the population of Nisos be as much as Queen Irene has proclaimed that the islands can support?
$\text{(A)}\ 50\text{ yrs.} \qquad \text{(B)}\ 75\text{ yrs.} \qquad \text{(C)}\ 100\text{ yrs.} \qquad \text{(D)}\ 125\text{ yrs.} \qquad \text{(E)}\ 150\text{ yrs.}$ | [
"We can divide the total area by how much will be occupied per person:\n$\\frac{24900 \\text{ acres}}{1.5 \\text{ acres per person}}=16600 \\text{ people}$ can stay on the island at its maximum capacity.\nWe can divide 16600 by the current population in $1998$ which is 200 to see by what factor the population incre... |
https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_3 | null | 242 | In isosceles trapezoid $ABCD$ , parallel bases $\overline{AB}$ and $\overline{CD}$ have lengths $500$ and $650$ , respectively, and $AD=BC=333$ . The angle bisectors of $\angle{A}$ and $\angle{D}$ meet at $P$ , and the angle bisectors of $\angle{B}$ and $\angle{C}$ meet at $Q$ . Find $PQ$ | [
"We have the following diagram: \nLet $X$ and $W$ be the points where $AP$ and $BQ$ extend to meet $CD$ , and $YZ$ be the height of $\\triangle AZB$ . As proven in Solution 2, triangles $APD$ and $DPW$ are congruent right triangles. Therefore, $AD = DW = 333$ . We can apply this logic to triangles $BCQ$ and $XCQ$ a... |
https://artofproblemsolving.com/wiki/index.php/2007_AIME_I_Problems/Problem_12 | null | 875 | In isosceles triangle $\triangle ABC$ $A$ is located at the origin and $B$ is located at $(20,0)$ . Point $C$ is in the first quadrant with $AC = BC$ and angle $BAC = 75^{\circ}$ . If triangle $ABC$ is rotated counterclockwise about point $A$ until the image of $C$ lies on the positive $y$ -axis, the area of the regi... | [
"Redefine the points in the same manner as the last time ( $\\triangle AB'C'$ , intersect at $D$ $E$ , and $F$ ). This time, notice that $[ADEF] = [\\triangle AB'C'] - ([\\triangle ADC'] + [\\triangle EFB'])$\nThe area of $[\\triangle AB'C'] = [\\triangle ABC]$ . The altitude of $\\triangle ABC$ is clearly $10 \\ta... |
https://artofproblemsolving.com/wiki/index.php/2011_AMC_10B_Problems/Problem_5 | E | 224 | In multiplying two positive integers $a$ and $b$ , Ron reversed the digits of the two-digit number $a$ . His erroneous product was $161$ . What is the correct value of the product of $a$ and $b$
$\textbf{(A)}\ 116 \qquad\textbf{(B)}\ 161 \qquad\textbf{(C)}\ 204 \qquad\textbf{(D)}\ 214 \qquad\textbf{(E)}\ 224$ | [
"We have $161 = 7 \\cdot 23.$ Since $a$ has two digits, the factors must be $23$ and $7,$ so $a = 32$ and $b = 7.$ Then, $ab = 7 \\times 32 = \\boxed{224}.$"
] |
https://artofproblemsolving.com/wiki/index.php/2011_AMC_12B_Problems/Problem_4 | E | 224 | In multiplying two positive integers $a$ and $b$ , Ron reversed the digits of the two-digit number $a$ . His erroneous product was $161.$ What is the correct value of the product of $a$ and $b$
$\textbf{(A)}\ 116 \qquad \textbf{(B)}\ 161 \qquad \textbf{(C)}\ 204 \qquad \textbf{(D)}\ 214 \qquad \textbf{(E)}\ 224$ | [
"Taking the prime factorization of $161$ reveals that it is equal to $23*7.$ Therefore, the only ways to represent $161$ as a product of two positive integers is $161*1$ and $23*7.$ Because neither $161$ nor $1$ is a two-digit number, we know that $a$ and $b$ are $23$ and $7.$ Because $23$ is a two-digit number, we... |
https://artofproblemsolving.com/wiki/index.php/2000_AMC_8_Problems/Problem_16 | C | 400 | In order for Mateen to walk a kilometer (1000m) in his rectangular backyard, he must walk the length 25 times or walk its perimeter 10 times. What is the area of Mateen's backyard in square meters?
$\text{(A)}\ 40 \qquad \text{(B)}\ 200 \qquad \text{(C)}\ 400 \qquad \text{(D)}\ 500 \qquad \text{(E)}\ 1000$ | [
"The length $L$ of the rectangle is $\\frac{1000}{25}=40$ meters. The perimeter $P$ is $\\frac{1000}{10}=100$ meters. Since $P_{rect} = 2L + 2W$ , we plug values in to get:\n$100 = 2\\cdot 40 + 2W$\n$100 = 80 + 2W$\n$2W = 20$\n$W = 10$ meters\nSince $A_{rect} = LW$ , the area is $40\\cdot 10=400$ square meters or $... |
https://artofproblemsolving.com/wiki/index.php/2004_AIME_II_Problems/Problem_5 | null | 766 | In order to complete a large job, $1000$ workers were hired, just enough to complete the job on schedule. All the workers stayed on the job while the first quarter of the work was done, so the first quarter of the work was completed on schedule. Then $100$ workers were laid off, so the second quarter of the work was co... | [
"A train is traveling at $1000$ miles per hour and has one hour to reach its destination $1000$ miles away. After $15$ minutes and $250$ miles it slows to $900$ mph, and thus takes $\\frac{250}{900}(60)=\\frac{50}{3}$ minutes to travel the next $250$ miles. Then it slows to $800$ mph, so the next quarter takes $\\f... |
https://artofproblemsolving.com/wiki/index.php/1955_AHSME_Problems/Problem_50 | C | 5 | In order to pass $B$ going $40$ mph on a two-lane highway, $A$ , going $50$ mph, must gain $30$ feet.
Meantime, $C, 210$ feet from $A$ , is headed toward him at $50$ mph. If $B$ and $C$ maintain their speeds,
then, in order to pass safely, $A$ must increase his speed by:
$\textbf{(A)}\ \text{30 mph}\qquad\textbf{(B)}... | [
"Let $V_A, V_B, V_C$ be $A, B, C$ 's velocity, respectively. We want to pass $B$ before we collide with $C$ . Since $A$ and $B$ are going in the same direction and $V_A>V_B$ $A$ will pass $B$ in $\\frac{30\\mathrm{ft}}{V_A-V_B}$ time. Since $A$ and $C$ are going in opposite directions, their relative velocity is $V... |
https://artofproblemsolving.com/wiki/index.php/1956_AHSME_Problems/Problem_31 | E | 110 | In our number system the base is ten. If the base were changed to four you would count as follows: $1,2,3,10,11,12,13,20,21,22,23,30,\ldots$ The twentieth number would be:
$\textbf{(A)}\ 20 \qquad\textbf{(B)}\ 38 \qquad\textbf{(C)}\ 44 \qquad\textbf{(D)}\ 104 \qquad\textbf{(E)}\ 110$ | [
"The $20^{\\text{th}}$ number will be the value of $20_{10}$ in base $4$ . Thus, we see \\[20_{10} = (1) \\cdot 4^2 + (1) \\cdot 4^1 + 0 \\cdot 4^0\\] \\[= 110_{4}\\]\n$\\boxed{110}$"
] |
https://artofproblemsolving.com/wiki/index.php/1996_AIME_Problems/Problem_15 | null | 777 | In parallelogram $ABCD$ , let $O$ be the intersection of diagonals $\overline{AC}$ and $\overline{BD}$ . Angles $CAB$ and $DBC$ are each twice as large as angle $DBA$ , and angle $ACB$ is $r$ times as large as angle $AOB$ . Find $\lfloor 1000r \rfloor$ | [
"Let $\\theta = \\angle DBA$ . Then $\\angle CAB = \\angle DBC = 2\\theta$ $\\angle AOB = 180 - 3\\theta$ , and $\\angle ACB = 180 - 5\\theta$ . Since $ABCD$ is a parallelogram, it follows that $OA = OC$ . By the Law of Sines on $\\triangle ABO,\\, \\triangle BCO$\nDividing the two equalities yields\n\\[\\frac{\\si... |
https://artofproblemsolving.com/wiki/index.php/2009_AIME_I_Problems/Problem_4 | null | 177 | In parallelogram $ABCD$ , point $M$ is on $\overline{AB}$ so that $\frac {AM}{AB} = \frac {17}{1000}$ and point $N$ is on $\overline{AD}$ so that $\frac {AN}{AD} = \frac {17}{2009}$ . Let $P$ be the point of intersection of $\overline{AC}$ and $\overline{MN}$ . Find $\frac {AC}{AP}$ | [
"One of the ways to solve this problem is to make this parallelogram a straight line.\nSo the whole length of the line is $APC$ $AMC$ or $ANC$ ), and $ABC$ is $1000x+2009x=3009x.$\n$AP$ $AM$ or $AN$ ) is $17x.$\nSo the answer is $3009x/17x = \\boxed{177}$",
"Draw a diagram with all the given points and lines invo... |
https://artofproblemsolving.com/wiki/index.php/2009_AMC_10A_Problems/Problem_12 | null | 13 | In quadrilateral $ABCD$ $AB = 5$ $BC = 17$ $CD = 5$ $DA = 9$ , and $BD$ is an integer. What is $BD$
$\textbf{(A)}\ 11 \qquad \textbf{(B)}\ 12 \qquad \textbf{(C)}\ 13 \qquad \textbf{(D)}\ 14 \qquad \textbf{(E)}\ 15$ | [
"By the triangle inequality we have $BD < DA + AB = 9 + 5 = 14$ , and also $BD + CD > BC$ , hence $BD > BC - CD = 17 - 5 = 12$\nWe get that $12 < BD < 14$ , and as we know that $BD$ is an integer, we must have $BD=\\boxed{13}$"
] |
https://artofproblemsolving.com/wiki/index.php/2009_AMC_12A_Problems/Problem_10 | null | 13 | In quadrilateral $ABCD$ $AB = 5$ $BC = 17$ $CD = 5$ $DA = 9$ , and $BD$ is an integer. What is $BD$
$\textbf{(A)}\ 11 \qquad \textbf{(B)}\ 12 \qquad \textbf{(C)}\ 13 \qquad \textbf{(D)}\ 14 \qquad \textbf{(E)}\ 15$ | [
"By the triangle inequality we have $BD < DA + AB = 9 + 5 = 14$ , and also $BD + CD > BC$ , hence $BD > BC - CD = 17 - 5 = 12$\nWe get that $12 < BD < 14$ , and as we know that $BD$ is an integer, we must have $BD=\\boxed{13}$"
] |
https://artofproblemsolving.com/wiki/index.php/2006_AIME_I_Problems/Problem_1 | null | 84 | In quadrilateral $ABCD$ $\angle B$ is a right angle diagonal $\overline{AC}$ is perpendicular to $\overline{CD}$ $AB=18$ $BC=21$ , and $CD=14$ . Find the perimeter of $ABCD$ | [
"From the problem statement, we construct the following diagram:\nUsing the Pythagorean Theorem\nSubstituting $(AB)^2 + (BC)^2$ for $(AC)^2$\nPlugging in the given information:\nSo the perimeter is $18+21+14+31=84$ , and the answer is $\\boxed{084}$"
] |
https://artofproblemsolving.com/wiki/index.php/2001_AIME_II_Problems/Problem_13 | null | 69 | In quadrilateral $ABCD$ $\angle{BAD}\cong\angle{ADC}$ and $\angle{ABD}\cong\angle{BCD}$ $AB = 8$ $BD = 10$ , and $BC = 6$ . The length $CD$ may be written in the form $\frac {m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$ | [
"Extend $\\overline{AD}$ and $\\overline{BC}$ to meet at $E$ . Then, since $\\angle BAD = \\angle ADC$ and $\\angle ABD = \\angle DCE$ , we know that $\\triangle ABD \\sim \\triangle DCE$ . Hence $\\angle ADB = \\angle DEC$ , and $\\triangle BDE$ is isosceles . Then $BD = BE = 10$\nUsing the similarity, we have:\n\... |
https://artofproblemsolving.com/wiki/index.php/2005_AMC_8_Problems/Problem_9 | D | 17 | In quadrilateral $ABCD$ , sides $\overline{AB}$ and $\overline{BC}$ both have length 10, sides $\overline{CD}$ and $\overline{DA}$ both have length 17, and the measure of angle $ADC$ is $60^\circ$ . What is the length of diagonal $\overline{AC}$
[asy]draw((0,0)--(17,0)); draw(rotate(301, (17,0))*(0,0)--(17,0)); picture... | [
"Because $\\overline{AD} = \\overline{CD}$ $\\triangle ADC$ is an isosceles triangle with $\\angle DAC = \\angle DCA$ . Angles in a triangle add up to $180^\\circ$ , and since $\\angle ADC=60^\\circ$ , the other two angles are also $60^\\circ$ , and $\\triangle ADC$ is an equilateral triangle. Therefore $\\overline... |
https://artofproblemsolving.com/wiki/index.php/1967_AHSME_Problems/Problem_32 | E | 166 | In quadrilateral $ABCD$ with diagonals $AC$ and $BD$ , intersecting at $O$ $BO=4$ $OD = 6$ $AO=8$ $OC=3$ , and $AB=6$ . The length of $AD$ is: $\textbf{(A)}\ 9\qquad \textbf{(B)}\ 10\qquad \textbf{(C)}\ 6\sqrt{3}\qquad \textbf{(D)}\ 8\sqrt{2}\qquad \textbf{(E)}\ \sqrt{166}$ | [
"After drawing the diagram, we see that we actually have a lot of lengths to work with. Considering triangle ABD, we know values of $AB, BD(BD = BO + OD)$ , but we want to find the value of $AD$ . We can apply stewart's theorem now, letting $m = 4, n = 6, AD = X, AB = 6$ , and we have $10 \\cdot 6 \\cdot 4 + 8 \\cd... |
https://artofproblemsolving.com/wiki/index.php/2005_AIME_I_Problems/Problem_7 | null | 150 | In quadrilateral $ABCD,\ BC=8,\ CD=12,\ AD=10,$ and $m\angle A= m\angle B = 60^\circ.$ Given that $AB = p + \sqrt{q},$ where $p$ and $q$ are positive integers , find $p+q.$ | [
"\nDraw line segment $DE$ such that line $DE$ is concurrent with line $BC$ . Then, $ABED$ is an isosceles trapezoid so $AD=BE=10$ , and $BC=8$ and $EC=2$ . We are given that $DC=12$ . Since $\\angle CED = 120^{\\circ}$ , using Law of Cosines on $\\bigtriangleup CED$ gives \\[12^2=DE^2+4-2(2)(DE)(\\cos 120^{\\circ})... |
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