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/2008_AMC_10A_Problems/Problem_15 | D | 150 | Yesterday Han drove 1 hour longer than Ian at an average speed 5 miles per hour faster than Ian. Jan drove 2 hours longer than Ian at an average speed 10 miles per hour faster than Ian. Han drove 70 miles more than Ian. How many more miles did Jan drive than Ian?
$\mathrm{(A)}\ 120\qquad\mathrm{(B)}\ 130\qquad\mathrm{(... | [
"Set the time Ian traveled as $I$ , and set Han's speed as $H$ . Therefore, Jan's speed is $H+5.$\nWe get the following equation for how much Han is ahead of Ian: $H+5I = 70.$\nThe expression for how much Jan is ahead of Ian is: $2(H+5)+10I.$\nThis simplifies to: $2H+10+10I.$\nHowever, this is just $2(H+5I)+10.$\nS... |
https://artofproblemsolving.com/wiki/index.php/2004_AMC_10A_Problems/Problem_1 | A | 250 | You and five friends need to raise $1500$ dollars in donations for a charity, dividing the fundraising equally. How many dollars will each of you need to raise?
$\mathrm{(A) \ } 250\qquad \mathrm{(B) \ } 300 \qquad \mathrm{(C) \ } 1500 \qquad \mathrm{(D) \ } 7500 \qquad \mathrm{(E) \ } 9000$ | [
"There are $6$ people to split the $1500$ dollars among, so each person must raise $\\frac{1500}6=250$ dollars. $\\Rightarrow\\boxed{250}$"
] |
https://artofproblemsolving.com/wiki/index.php/1960_AHSME_Problems/Problem_33 | A | 0 | You are given a sequence of $58$ terms; each term has the form $P+n$ where $P$ stands for the product $2 \times 3 \times 5 \times\ldots \times 61$ of all prime numbers less than or equal to $61$ , and $n$ takes, successively, the values $2, 3, 4,\ldots, 59$ .
Let $N$ be the number of primes appearing in this sequence.... | [
"First, note that $n$ does not have a prime number larger than $61$ as one of its factors. Also, note that $n$ does not equal $1$\nTherefore, since the prime factorization of $n$ only has primes from $2$ to $59$ $n$ and $P$ share at least one common factor other than $1$ . Therefore $P+n$ is not prime for any $n$... |
https://artofproblemsolving.com/wiki/index.php/2023_AMC_10B_Problems/Problem_10 | C | 4 | You are playing a game. A $2 \times 1$ rectangle covers two adjacent squares (oriented either horizontally or vertically) of a $3 \times 3$ grid of squares, but you are not told which two squares are covered. Your goal is to find at least one square that is covered by the rectangle. A "turn" consists of you guessing a ... | [
"First, note that since the rectangle covers 2 squares, we only need to guess squares that are not adjacent to any of our other guesses. To minimize the amount of guesses, each of our guessed squares should try to touch another guess on one vertex and one vertex only. There are only two ways to do this: one with $5... |
https://artofproblemsolving.com/wiki/index.php/2023_AMC_10B_Problems/Problem_10 | null | 4 | You are playing a game. A $2 \times 1$ rectangle covers two adjacent squares (oriented either horizontally or vertically) of a $3 \times 3$ grid of squares, but you are not told which two squares are covered. Your goal is to find at least one square that is covered by the rectangle. A "turn" consists of you guessing a ... | [
"The $3 \\times 3$ grid can be colored like a checkerboard with alternating black and white squares.\nLet the top left square be white, and the rest of the squares alternate colors.\nEach $2 \\times 1$ rectangle always covers $1$ white square and $1$ black square.\nYou can ensure that at least one of your guessed s... |
https://artofproblemsolving.com/wiki/index.php/2023_AMC_12B_Problems/Problem_5 | C | 4 | You are playing a game. A $2 \times 1$ rectangle covers two adjacent squares (oriented either horizontally or vertically) of a $3 \times 3$ grid of squares, but you are not told which two squares are covered. Your goal is to find at least one square that is covered by the rectangle. A "turn" consists of you guessing a ... | [
"First, note that since the rectangle covers 2 squares, we only need to guess squares that are not adjacent to any of our other guesses. To minimize the amount of guesses, each of our guessed squares should try to touch another guess on one vertex and one vertex only. There are only two ways to do this: one with $5... |
https://artofproblemsolving.com/wiki/index.php/2023_AMC_12B_Problems/Problem_5 | null | 4 | You are playing a game. A $2 \times 1$ rectangle covers two adjacent squares (oriented either horizontally or vertically) of a $3 \times 3$ grid of squares, but you are not told which two squares are covered. Your goal is to find at least one square that is covered by the rectangle. A "turn" consists of you guessing a ... | [
"The $3 \\times 3$ grid can be colored like a checkerboard with alternating black and white squares.\nLet the top left square be white, and the rest of the squares alternate colors.\nEach $2 \\times 1$ rectangle always covers $1$ white square and $1$ black square.\nYou can ensure that at least one of your guessed s... |
https://artofproblemsolving.com/wiki/index.php/2000_AMC_8_Problems/Problem_20 | A | 1 | You have nine coins: a collection of pennies, nickels, dimes, and quarters having a total value of $ $1.02$ , with at least one coin of each type. How many dimes must you have?
$\text{(A)}\ 1\qquad\text{(B)}\ 2\qquad\text{(C)}\ 3\qquad\text{(D)}\ 4\qquad\text{(E)}\ 5$ | [
"Since you have one coin of each type, $1 + 5 + 10 + 25 = 41$ cents are already determined, leaving you with a total of $102 - 41 = 61$ cents remaining for $5$ coins.\nYou must have $1$ more penny. If you had more than $1$ penny, you must have at least $6$ pennies to leave a multiple of $5$ for the nickels, dimes,... |
https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_10 | C | 12 | Zara has a collection of $4$ marbles: an Aggie, a Bumblebee, a Steelie, and a Tiger. She wants to display them in a row on a shelf, but does not want to put the Steelie and the Tiger next to one another. In how many ways can she do this?
$\textbf{(A) }6 \qquad \textbf{(B) }8 \qquad \textbf{(C) }12 \qquad \textbf{(D) }1... | [
"Let the Aggie, Bumblebee, Steelie, and Tiger, be referred to by $A,B,S,$ and $T$ , respectively. If we ignore the constraint that $S$ and $T$ cannot be next to each other, we get a total of $4!=24$ ways to arrange the 4 marbles. We now simply have to subtract out the number of ways that $S$ and $T$ can be next to ... |
https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_1 | null | 97 | Zou and Chou are practicing their $100$ -meter sprints by running $6$ races against each other. Zou wins the first race, and after that, the probability that one of them wins a race is $\frac23$ if they won the previous race but only $\frac13$ if they lost the previous race. The probability that Zou will win exactly $5... | [
"For the next five races, Zou wins four and loses one. Let $W$ and $L$ denote a win and a loss, respectively. There are five possible outcome sequences for Zou:\nWe proceed with casework:\nCase (1): Sequences #1-4, in which Zou does not lose the last race.\nThe probability that Zou loses a race is $\\frac13,$ and t... |
https://artofproblemsolving.com/wiki/index.php/1988_AHSME_Problems/Problem_9 | C | 13 | [asy] defaultpen(linewidth(0.7)+fontsize(10)); pair A=(0,0), B=(16,0), C=(16,16), D=(0,16), E=(32,0), F=(48,0), G=(48,16), H=(32,16), I=(0,8), J=(10,8), K=(10,16), L=(32,6), M=(40,6), N=(40,16); draw(A--B--C--D--A^^E--F--G--H--E^^I--J--K^^L--M--N); label("S", (18,8)); label("S", (50,8)); label("Figure 1", (A+B)/2, S); ... | [
"Small correction: The writer below has maximized the area of the rectangle (with sides parallel to the walls) that fits around the table, but there is a larger single dimension we can find in the table. The height or width is maximized when the diagonal of the table is horizontal or vertical. By the Pythagorean Th... |
https://artofproblemsolving.com/wiki/index.php/1979_AHSME_Problems/Problem_1 | D | 18 | [asy] draw((-2,1)--(2,1)--(2,-1)--(-2,-1)--cycle); draw((0,0)--(0,-1)--(-2,-1)--(-2,0)--cycle); label("$F$",(0,0),E); label("$A$",(-2,1),W); label("$B$",(2,1),E); label("$C$", (2,-1),E); label("$D$",(-2,-1),WSW); label("$E$",(-2,0),W); label("$G$",(0,-1),S); //Credit to TheMaskedMagician for the diagram [/asy]
If recta... | [
"Solution by e_power_pi_times_i\nSince the dimensions of $DEFG$ are half of the dimensions of $ABCD$ , the area of $DEFG$ is $\\dfrac{1}{2}\\cdot\\dfrac{1}{2}$ of $ABCD$ , so the area of $ABCD$ is $\\dfrac{1}{4}\\cdot72 = \\boxed{18}$"
] |
https://artofproblemsolving.com/wiki/index.php/1991_AHSME_Problems/Problem_19 | B | 128 | [asy] draw((0,0)--(0,3)--(4,0)--cycle,dot); draw((4,0)--(7,0)--(7,10)--cycle,dot); draw((0,3)--(7,10),dot); MP("C",(0,0),SW);MP("A",(0,3),NW);MP("B",(4,0),S);MP("E",(7,0),SE);MP("D",(7,10),NE); [/asy]
Triangle $ABC$ has a right angle at $C, AC=3$ and $BC=4$ . Triangle $ABD$ has a right angle at $A$ and $AD=12$ . Points... | [
"Solution by e_power_pi_times_i\nLet $F$ be the point such that $DF$ and $CF$ are parallel to $CE$ and $DE$ , respectively, and let $DE = x$ and $BE^2 = 169-x^2$ . Then, $[FDEC] = x(4+\\sqrt{169-x^2}) = [ABC] + [BED] + [ABD] + [AFD] = 6 + \\dfrac{x\\sqrt{169-x^2}}{2} + 30 + \\dfrac{(x-3)(4+\\sqrt{169-x^2})}{2}$ . S... |
https://artofproblemsolving.com/wiki/index.php/1988_AHSME_Problems/Problem_3 | A | 36 | [asy] draw((0,0)--(1,0)--(1,4)--(0,4)--(0,0)--(0,1)--(-1,1)--(-1,2)); draw((-1,2)--(0,2)--(0,4)--(-1,4)--(-1,5)--(1,5)--(1,6)--(0,6)); draw((0,6)--(0,5)--(3,5)--(3,6)--(4,6)--(4,2)--(5,2)); draw((5,2)--(5,1)--(1,1)--(3,1)--(3,0)--(4,0)--(4,1)); draw((1,4)--(3,4)--(3,2)--(1,2)--(4,2)--(3,2)--(3,6)); draw((3,6)--(4,6)--(... | [
"We first notice that the paper strips cover up part of the others. Since the width of the overlap is $1$ and the length of the overlap is $1$ , the area of \neach of the strips with the overlap is $(10\\cdot 1)-1=9$ . Since there are 4 strips, $4\\cdot 9=36 \\implies \\boxed{36}$"
] |
https://artofproblemsolving.com/wiki/index.php/1969_AHSME_Problems/Problem_26 | B | 15 | [asy] draw(arc((0,-1),2,30,150),dashed+linewidth(.75)); draw((-1.7,0)--(0,0)--(1.7,0),dot); draw((0,0)--(0,.98),dot); MP("A",(-1.7,0),W);MP("B",(1.7,0),E);MP("M",(0,0),S);MP("C",(0,1),N); [/asy]
A parabolic arch has a height of $16$ inches and a span of $40$ inches. The height, in inches, of the arch at the point $5$ i... | [
"Because the arch has a height of $16$ inches, an equation that models the arch is $y = ax^2 + 16$ , where $x$ is the horizontal distance from the center and $y$ is the height. The arch has a span of $40$ inches, so the arch meets the ground $20$ inches from the center. That means $0 = 400a + 16$ , so $a = -\\fra... |
https://artofproblemsolving.com/wiki/index.php/1979_AHSME_Problems/Problem_3 | C | 15 | [asy] real s=sqrt(3)/2; draw(box((0,0),(1,1))); draw((1+s,0.5)--(1,1)); draw((1+s,0.5)--(1,0)); draw((0,1)--(1+s,0.5)); label("$A$",(1,1),N); label("$B$",(1,0),S); label("$C$",(0,0),W); label("$D$",(0,1),W); label("$E$",(1+s,0.5),E); //Credit to TheMaskedMagician for the diagram[/asy]
In the adjoining figure, $ABCD$ is... | [
"Solution by e_power_pi_times_i\nNotice that $\\measuredangle DAE = 90^\\circ+60^\\circ = 150^\\circ$ and that $AD = AE$ . Then triangle $ADE$ is isosceles, so $\\measuredangle AED = \\dfrac{180^\\circ-150^\\circ}{2} = \\boxed{15}$",
"WLOG, let the side length of the square and the equilateral triangle be $1$ $\\... |
https://artofproblemsolving.com/wiki/index.php/1978_AHSME_Problems/Problem_26 | B | 4.8 | [asy] size(100); real a=4, b=3; // import cse5; pathpen=black; pair A=(a,0), B=(0,b), C=(0,0); D(MP("A",A)--MP("B",B,N)--MP("C",C,SW)--cycle); pair X=IP(B--A,(0,0)--(b,a)); D(CP((X+C)/2,C)); D(MP("R",IP(CP((X+C)/2,C),B--C),NW)--MP("Q",IP(CP((X+C)/2,C),A--C+(0.1,0)))); //Credit to chezbgone2 for the diagram [/asy]
In $\... | [
"We know that triangle $RCQ$ is similar to triangle $ABC$ . We draw a line to point $D$ on hypotenuse $AB$ such that $\\angle QDR$ is $90 ^\\circ$ and that $RDQC$ is a rectangle. Since triangle $RCQ$ is similar to triangle $ABC$ , let $RC$ be $4x$ and $RD/CQ$ be $3x$ . Now we have line segment $AQ$ $8-3x$ , and lin... |
https://artofproblemsolving.com/wiki/index.php/1977_AHSME_Problems/Problem_9 | B | 15 | [asy] size(120); path c = Circle((0, 0), 1); pair A = dir(20), B = dir(130), C = dir(240), D = dir(330); draw(c); pair F = 3(A-B) + B; pair G = 3(D-C) + C; pair E = intersectionpoints(B--F, C--G)[0]; draw(B--E--C--A); label("$A$", A, NE); label("$B$", B, NW); label("$C$", C, SW); label("$D$", D, SE); label("$E$", E, E)... | [
"Solution by e_power_pi_times_i\nIf arcs $AB$ $BC$ , and $CD$ are congruent, then $\\measuredangle ACB = \\measuredangle BDC = \\measuredangle CBD = \\theta$ . Because $ABCD$ is cyclic, $\\measuredangle CAD = \\measuredangle CBD = \\theta$ , and $\\measuredangle ADB = \\measuredangle ACB = \\theta$ . Then, $\\measu... |
https://artofproblemsolving.com/wiki/index.php/1979_AHSME_Problems/Problem_12 | B | 15 | [asy] size(200); pair A=(-2,0),B,C=(-1,0),D=(1,0),EE,O=(0,0); draw(arc(O,1, 0, 180)); EE=midpoint(arc(O,1, 0, 90)); draw(A--EE); draw(A--D); B=intersectionpoint(arc(O,1, 180, 0),EE--A); draw(O--EE); label("$A$",A,W); label("$B$",B,NW); label("$C$",C,S);label("$D$",D,E);label("$E$",EE,NE);label("$O$",O,S);label("$45^\ci... | [
"Solution by e_power_pi_times_i\nBecause $AB = OD$ , triangles $ABO$ and $BOE$ are isosceles. Denote $\\measuredangle BAO = \\measuredangle AOB = \\theta$ . Then $\\measuredangle ABO = 180^\\circ-2\\theta$ , and $\\measuredangle EBO = \\measuredangle OEB = 2\\theta$ , so $\\measuredangle BOE = 180^\\circ-4\\theta$ ... |
https://artofproblemsolving.com/wiki/index.php/1985_AJHSME_Problems/Problem_1 | A | 1 | [katex]\dfrac{3\times 5}{9\times 11}\times \dfrac{7\times 9\times 11}{3\times 5\times 7}=[/katex]
[katex]\text{(A)}\ 1 \qquad \text{(B)}\ 0 \qquad \text{(C)}\ 49 \qquad \text{(D)}\ \frac{1}{49} \qquad \text{(E)}\ 50[/katex] | [
"By the associative property , we can rearrange the numbers in the numerator and the denominator. [katex display=true]\\frac{3}{3}\\cdot \\frac{5}{5}\\cdot\\frac{7}{7}\\cdot\\frac{9}{9}\\cdot\\frac{11}{11}=1\\cdot1\\cdot1\\cdot1\\cdot1=\\boxed{1}[/katex]",
"Notice that the $9 \\times 11$ in the denominator of the... |
https://artofproblemsolving.com/wiki/index.php/2004_AIME_II_Problems/Problem_1 | null | 592 | chord of a circle is perpendicular to a radius at the midpoint of the radius. The ratio of the area of the larger of the two regions into which the chord divides the circle to the smaller can be expressed in the form $\frac{a\pi+b\sqrt{c}}{d\pi-e\sqrt{f}},$ where $a, b, c, d, e,$ and $f$ are positive integers $a$ and $... | [
"Let $r$ be the length of the radius of the circle. A right triangle is formed by half of the chord, half of the radius (since the chord bisects it), and the radius. Thus, it is a $30^\\circ$ $60^\\circ$ $90^\\circ$ triangle , and the area of two such triangles is $2 \\cdot \\frac{1}{2} \\cdot \\frac{r}{2} \\cdot ... |
https://artofproblemsolving.com/wiki/index.php/2002_AMC_8_Problems/Problem_1 | D | 5 | circle and two distinct lines are drawn on a sheet of paper. What is the largest possible number of points of intersection of these figures?
$\text {(A)}\ 2 \qquad \text {(B)}\ 3 \qquad {(C)}\ 4 \qquad {(D)}\ 5 \qquad {(E)}\ 6$ | [
"The two lines can both intersect the circle twice, and can intersect each other once, so $2+2+1= \\boxed{5}.$"
] |
https://artofproblemsolving.com/wiki/index.php/2000_AIME_II_Problems/Problem_10 | null | 647 | circle is inscribed in quadrilateral $ABCD$ tangent to $\overline{AB}$ at $P$ and to $\overline{CD}$ at $Q$ . Given that $AP=19$ $PB=26$ $CQ=37$ , and $QD=23$ , find the square of the radius of the circle. | [
"Call the center of the circle $O$ . By drawing the lines from $O$ tangent to the sides and from $O$ to the vertices of the quadrilateral, four pairs of congruent right triangles are formed.\nThus, $\\angle{AOP}+\\angle{POB}+\\angle{COQ}+\\angle{QOD}=180$ , or $(\\arctan(\\tfrac{19}{r})+\\arctan(\\tfrac{26}{r}))+(\... |
https://artofproblemsolving.com/wiki/index.php/2004_AIME_I_Problems/Problem_10 | null | 817 | circle of radius 1 is randomly placed in a 15-by-36 rectangle $ABCD$ so that the circle lies completely within the rectangle. Given that the probability that the circle will not touch diagonal $AC$ is $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$ | [
"The location of the center of the circle must be in the $34 \\times 13$ rectangle that is one unit away from the sides of rectangle $ABCD$ . We want to find the area of the right triangle with hypotenuse one unit away from $\\overline{AC}$ . Let this triangle be $A'B'C'$\nNotice that $ABC$ and $A'B'C'$ share the s... |
https://artofproblemsolving.com/wiki/index.php/2011_AIME_II_Problems/Problem_10 | null | 57 | circle with center $O$ has radius 25. Chord $\overline{AB}$ of length 30 and chord $\overline{CD}$ of length 14 intersect at point $P$ . The distance between the midpoints of the two chords is 12. The quantity $OP^2$ can be represented as $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find th... | [
"We begin as in the first solution. Once we see that $\\triangle EOF$ has side lengths $12$ $20$ , and $24$ , we can compute its area with Heron's formula:\n\\[K = \\sqrt{s(s-a)(s-b)(s-c)} = \\sqrt{28\\cdot 16\\cdot 8\\cdot 4} = 32\\sqrt{14}.\\]\nThus, the circumradius of triangle $\\triangle EOF$ is $R = \\frac{a... |
https://artofproblemsolving.com/wiki/index.php/2004_AIME_I_Problems/Problem_3 | null | 241 | convex polyhedron $P$ has $26$ vertices, $60$ edges, and $36$ faces, $24$ of which are triangular and $12$ of which are quadrilaterals. A space diagonal is a line segment connecting two non-adjacent vertices that do not belong to the same face. How many space diagonals does $P$ have? | [
"Every pair of vertices of the polyhedron determines either an edge , a face diagonal or a space diagonal. We have ${26 \\choose 2} = \\frac{26\\cdot25}2 = 325$ total line segments determined by the vertices. Of these, $60$ are edges. Each triangular face has $0$ face diagonals and each quadrilateral face has $2... |
https://artofproblemsolving.com/wiki/index.php/1988_AIME_Problems/Problem_10 | null | 840 | convex polyhedron has for its faces 12 squares , 8 regular hexagons , and 6 regular octagons . At each vertex of the polyhedron one square, one hexagon, and one octagon meet. How many segments joining vertices of the polyhedron lie in the interior of the polyhedron rather than along an edge or a face | [
"The polyhedron described looks like this, a truncated cuboctahedron.\nThe number of segments joining the vertices of the polyhedron is ${48\\choose2} = 1128$ . We must now subtract out those segments that lie along an edge or a face.\nSince every vertex of the polyhedron lies on exactly one vertex of a square/hexa... |
https://artofproblemsolving.com/wiki/index.php/1990_AIME_Problems/Problem_9 | null | 73 | fair coin is to be tossed $10_{}^{}$ times. Let $\frac{i}{j}^{}_{}$ , in lowest terms, be the probability that heads never occur on consecutive tosses. Find $i+j_{}^{}$ | [
"Clearly, at least $5$ tails must be flipped; any less, then by the Pigeonhole Principle there will be heads that appear on consecutive tosses.\nConsider the case when $5$ tails occur. The heads must fall between the tails such that no two heads fall between the same tails, and must fall in the positions labeled $(... |
https://artofproblemsolving.com/wiki/index.php/1984_AIME_Problems/Problem_12 | null | 401 | function $f$ is defined for all real numbers and satisfies $f(2+x)=f(2-x)$ and $f(7+x)=f(7-x)$ for all $x$ . If $x=0$ is a root for $f(x)=0$ , what is the least number of roots $f(x)=0$ must have in the interval $-1000\leq x \leq 1000$ | [
"If $f(2+x)=f(2-x)$ , then substituting $t=2+x$ gives $f(t)=f(4-t)$ . Similarly, $f(t)=f(14-t)$ . In particular, \\[f(t)=f(14-t)=f(14-(4-t))=f(t+10)\\]\nSince $0$ is a root, all multiples of $10$ are roots, and anything congruent to $4\\pmod{10}$ are also roots. To see that these may be the only integer roots, obse... |
https://artofproblemsolving.com/wiki/index.php/1991_AIME_Problems/Problem_14 | null | 384 | hexagon is inscribed in a circle . Five of the sides have length $81$ and the sixth, denoted by $\overline{AB}$ , has length $31$ . Find the sum of the lengths of the three diagonals that can be drawn from $A_{}^{}$ | [
"Let $x=AC=BF$ $y=AD=BE$ , and $z=AE=BD$\nPtolemy's Theorem on $ABCD$ gives $81y+31\\cdot 81=xz$ , and Ptolemy on $ACDF$ gives $x\\cdot z+81^2=y^2$ .\nSubtracting these equations give $y^2-81y-112\\cdot 81=0$ , and from this $y=144$ . Ptolemy on $ADEF$ gives $81y+81^2=z^2$ , and from this $z=135$ . Finally, pluggin... |
https://artofproblemsolving.com/wiki/index.php/1963_AHSME_Problems/Problem_2 | A | 14 | let $n=x-y^{x-y}$ . Find $n$ when $x=2$ and $y=-2$
$\textbf{(A)}\ -14 \qquad \textbf{(B)}\ 0 \qquad \textbf{(C)}\ 1 \qquad \textbf{(D)}\ 18 \qquad \textbf{(E)}\ 256$ | [
"Substitute the variables to determine the value of $n$ \\[n = 2 - (-2)^{2-(-2)}\\] \\[n = 2 - (-2)^4\\] \\[n = 2 - 16\\] \\[n = -14\\] The answer is $\\boxed{14}$"
] |
https://artofproblemsolving.com/wiki/index.php/2005_AMC_12A_Problems/Problem_12 | null | 8 | line passes through $A\ (1,1)$ and $B\ (100,1000)$ . How many other points with integer coordinates are on the line and strictly between $A$ and $B$
$(\mathrm {A}) \ 0 \qquad (\mathrm {B}) \ 2 \qquad (\mathrm {C})\ 3 \qquad (\mathrm {D}) \ 8 \qquad (\mathrm {E})\ 9$ | [
"The slope of the line is $\\frac{1000-1}{100-1}=\\frac{111}{11},$ so all points on the line have the form $(1+11t, 1+111t)$ for some value of $t$ (the rise is 111 and the run is 11). Such a point has integer coordinates if and only if $t$ is an integer, and the point is strictly between $A$ and $B$ if and only if ... |
https://artofproblemsolving.com/wiki/index.php/1995_AJHSME_Problems/Problem_12 | E | 1,994 | lucky year is one in which at least one date, when written in the form month/day/year, has the following property: The product of the month times the day equals the last two digits of the year . For example, 1956 is a lucky year because it has the date 7/8/56 and $7\times 8 = 56$ . Which of the following is NOT a luc... | [
"We examine only the factors of $90, 91, 92, 93,$ and $94$ that are less than $13$ , because for a year to be lucky, it must have at least one factor between $1$ and $12$ to represent the month.\n$90$ has factors of $1, 2, 3, 5, 6, 9,$ and $10$ . Dividing $90$ by the last number $10$ , gives $9$ . Thus, $10/9/90$... |
https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_13 | A | 2 | palindrome is a number that has the same value when read from left to right or from right to left. (For example, 12321 is a palindrome.) Let $N$ be the least three-digit integer which is not a palindrome but which is the sum of three distinct two-digit palindromes. What is the sum of the digits of $N$
$\textbf{(A) }2\q... | [
"Note that the only positive 2-digit palindromes are multiples of 11, namely $11, 22, \\ldots, 99$ . Since $N$ is the sum of 2-digit palindromes, $N$ is necessarily a multiple of 11. The smallest 3-digit multiple of 11 which is not a palindrome is 110, so $N=110$ is a candidate solution. We must check that 110 can ... |
https://artofproblemsolving.com/wiki/index.php/2006_AMC_10A_Problems/Problem_8 | E | 11 | parabola with equation $y=x^2+bx+c$ passes through the points $(2,3)$ and $(4,3)$ . What is $c$
$\textbf{(A) } 2\qquad \textbf{(B) } 5\qquad \textbf{(C) } 7\qquad \textbf{(D) } 10\qquad \textbf{(E) } 11$ | [
"Substitute the points $(2,3)$ and $(4,3)$ into the given equation for $(x,y)$\nThen we get a system of two equations:\n$3=4+2b+c$\n$3=16+4b+c$\nSubtracting the first equation from the second we have:\n$0=12+2b$\n$b=-6$\nThen using $b=-6$ in the first equation:\n$0=1+-12+c$\n$c=\\boxed{11}$",
"Alternatively, noti... |
https://artofproblemsolving.com/wiki/index.php/1995_AHSME_Problems/Problem_22 | E | 745 | pentagon is formed by cutting a triangular corner from a rectangular piece of paper. The five sides of the pentagon have lengths $13, 19, 20, 25$ and $31$ , although this is not necessarily their order around the pentagon. The area of the pentagon is
$\mathrm{(A) \ 459 } \qquad \mathrm{(B) \ 600 } \qquad \mathrm{(C) \ ... | [
"Since the pentagon is cut from a rectangle, the cut-off triangle must be right. Since all of the lengths given are integers, it follows that this triangle is a Pythagorean Triple . We know that $31$ and either $25,\\, 20$ must be the dimensions of the rectangle, since they are the largest lengths. With some trial-... |
https://artofproblemsolving.com/wiki/index.php/1984_AIME_Problems/Problem_3 | null | 144 | point $P$ is chosen in the interior of $\triangle ABC$ such that when lines are drawn through $P$ parallel to the sides of $\triangle ABC$ , the resulting smaller triangles $t_{1}$ $t_{2}$ , and $t_{3}$ in the figure, have areas $4$ $9$ , and $49$ , respectively. Find the area of $\triangle ABC$ | [
"By the transversals that go through $P$ , all four triangles are similar to each other by the $AA$ postulate. Also, note that the length of any one side of the larger triangle is equal to the sum of the sides of each of the corresponding sides on the smaller triangles. We use the identity $K = \\dfrac{ab\\sin C}{2... |
https://artofproblemsolving.com/wiki/index.php/1992_AIME_Problems/Problem_2 | null | 502 | positive integer is called ascending if, in its decimal representation , there are at least two digits and each digit is less than any digit to its right. How many ascending positive integers are there? | [
"Note that an ascending number is exactly determined by its digits : for any set of digits (not including 0, since the only position for 0 is at the leftmost end of the number, i.e. a leading 0), there is exactly one ascending number with those digits.\nSo, there are nine digits that may be used: $1,2,3,4,5,6,7,8,9... |
https://artofproblemsolving.com/wiki/index.php/2016_AIME_I_Problems/Problem_9 | null | 744 | problem_id
10b89ea3f13aa2c628bafff65bf48904 Triangle $ABC$ has $AB=40,AC=31,$ and $\sin{A}...
10b89ea3f13aa2c628bafff65bf48904 It has been noted that this answer won't actua...
Name: Text, dtype: object | [
"We start by drawing a diagram;\n\nWe know that $\\sin \\alpha = \\frac{1}{5}$ . Since $\\sin \\alpha = \\cos (90- \\alpha)$ \\[\\cos (90- \\alpha) = \\frac{1}{5} \\implies \\cos (\\beta + \\gamma) = \\frac{1}{5}\\]\nUsing our angle sum identities, we expand this to $\\cos \\beta \\cdot \\cos \\gamma - \\sin \\bet... |
https://artofproblemsolving.com/wiki/index.php/2015_AMC_10B_Problems/Problem_20 | A | 6 | problem_id
18bf638a0cb797d222d008b988c16b6d Erin the ant starts at a given corner of a cub...
18bf638a0cb797d222d008b988c16b6d From the 2006 AMC 10A Problem 25, they look fo...
Name: Text, dtype: object | [
"\nWe label the vertices of the cube as different letters and numbers shown above. We label these so that Erin can only crawl from a number to a letter or a letter to a number (this can be seen as a coloring argument). The starting point is labeled $A$\nIf we define a \"move\" as each time Erin crawls along a singl... |
https://artofproblemsolving.com/wiki/index.php/2001_AMC_12_Problems/Problem_12 | B | 801 | problem_id
21b33a597f802b0a4ad56201b9aeba1e How many positive integers not exceeding $2001...
21b33a597f802b0a4ad56201b9aeba1e How many positive integers not exceeding $2001...
Name: Text, dtype: object | [
"We can solve this problem by finding the cases where the number is divisible by $3$ or $4$ , then subtract from the cases where none of those cases divide $5$ . To solve the ways the numbers divide $3$ or $4$ we find the cases where a number is divisible by $3$ and $4$ as separate cases. We apply the floor functio... |
https://artofproblemsolving.com/wiki/index.php/2001_AMC_12_Problems/Problem_3 | B | 32,000 | problem_id
227cbd9a094a48b5f95a026123843b8c The state income tax where Kristin lives is le...
227cbd9a094a48b5f95a026123843b8c The state income tax where Kristin lives is le...
Name: Text, dtype: object | [
"Let $A$ $T$ be Kristin's annual income and the income tax total, respectively. Notice that \\begin{align*} T &= p\\%\\cdot28000 + (p + 2)\\%\\cdot(A - 28000) \\\\ &= [p\\%\\cdot28000 + p\\%\\cdot(A - 28000)] + 2\\%\\cdot(A - 28000) \\\\ &= p\\%\\cdot A + 2\\%\\cdot(A - 28000) \\end{align*} We are also given that \... |
https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_10 | B | 4 | problem_id
319997b31b053c975a9c9d5ff9b8fbcc Five positive consecutive integers starting wi...
319997b31b053c975a9c9d5ff9b8fbcc Five positive consecutive integers starting wi...
Name: Text, dtype: object | [
"Let $a=1$ . Our list is $\\{1,2,3,4,5\\}$ with an average of $15\\div 5=3$ . Our next set starting with $3$ is $\\{3,4,5,6,7\\}$ . Our average is $25\\div 5=5$\nTherefore, we notice that $5=1+4$ which means that the answer is $\\boxed{4}$",
"We are given that \\[b=\\frac{a+a+1+a+2+a+3+a+4}{5}\\] \\[\\implies b =... |
https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_10 | null | 4 | problem_id
319997b31b053c975a9c9d5ff9b8fbcc Five positive consecutive integers starting wi...
319997b31b053c975a9c9d5ff9b8fbcc Five positive consecutive integers starting wi...
Name: Text, dtype: object | [
"We know from experience that the average of $5$ consecutive numbers is the $3^\\text{rd}$ one or the $1^\\text{st} + 2$ . With the logic, we find that $b=a+2$ $b+2=(a+2)+2=\\boxed{4}$",
"We know from experience that the average of $5$ consecutive numbers is the $3^\\text{rd}$ one or the $1^\\text{st} + 2$ . With... |
https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_5 | C | 3 | problem_id
396f2f976638a27e600105ba37c3a839 On an algebra quiz, $10\%$ of the students sco...
396f2f976638a27e600105ba37c3a839 On an algebra quiz, $10\%$ of the students sco...
Name: Text, dtype: object | [
"Without loss of generality, let there be $20$ students (the least whole number possible) who took the test. We have $2$ students score $70$ points, $7$ students score $80$ points, $6$ students score $90$ points and $5$ students score $100$ points. Therefore, the mean\nis $87$ and the median is $90$\nThus, the solu... |
https://artofproblemsolving.com/wiki/index.php/2001_AMC_12_Problems/Problem_11 | D | 35 | problem_id
44dac98b900fb2d03612e3e20d26762f A box contains exactly five chips, three red a...
44dac98b900fb2d03612e3e20d26762f A box contains exactly five chips, three red a...
Name: Text, dtype: object | [
"Let's assume we don't stop picking until all of the chips are picked. To satisfy this condition, we have to arrange the letters: $W, W, R, R, R$ such that both $W$ 's appear in the first $4$ . We find the number of ways to arrange the white chips in the first $4$ and divide that by the total ways to choose all the... |
https://artofproblemsolving.com/wiki/index.php/1972_AHSME_Problems/Problem_24 | null | 15 | problem_id
4ca2bf599f1693b695997dd52d9e4774 A man walked a certain distance at a constant ...
4ca2bf599f1693b695997dd52d9e4774 A man walked a certain distance at a constant ...
Name: Text, dtype: object | [
"We can make three equations out of the information, and since the distances are the same, we can equate these equations.\n\\[\\frac{4t}{5}(x+\\frac{1}{2})=xt=(t+\\frac{5}{2})(x-\\frac{1}{2})\\] where $x$ is the man's rate and $t$ is the time it takes him.\nLooking at the first two parts of the equations,\n\\[\\fra... |
https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_20 | D | 991 | problem_id
6a2cc5541e1749d3d0bb739aacabadf4 The product $(8)(888\dots8)$ , where the secon...
6a2cc5541e1749d3d0bb739aacabadf4 The product $(8)(888\dots8)$ , where the secon...
Name: Text, dtype: object | [
"We can list the first few numbers in the form $8 \\cdot (8....8)$\n(Hard problem to do without the multiplication, but you can see the pattern early on)\n$8 \\cdot 8 = 64$\n$8 \\cdot 88 = 704$\n$8 \\cdot 888 = 7104$\n$8 \\cdot 8888 = 71104$\n$8 \\cdot 88888 = 711104$\nBy now it's clear that the numbers will be in ... |
https://artofproblemsolving.com/wiki/index.php/2001_AMC_12_Problems/Problem_6 | E | 8 | problem_id
74b973e4f94621e9337c1a9c0077ccfc A telephone number has the form $\text{ABC-DEF...
74b973e4f94621e9337c1a9c0077ccfc A telephone number has the form $\text{ABC-DEF...
Name: Text, dtype: object | [
"We start by noting that there are $10$ letters, meaning there are $10$ digits in total. Listing them all out, we have $0, 1, 2, 3, 4, 5, 6, 7, 8, 9$ . Clearly, the most restrictive condition is the consecutive odd digits, so we create casework based on that.\nCase 1: $G$ $H$ $I$ , and $J$ are $7$ $5$ $3$ , and $1$... |
https://artofproblemsolving.com/wiki/index.php/2014_AMC_10B_Problems/Problem_24 | B | 2 | problem_id
77e0fabb4f8f99b4274e980292ddff5d The numbers $1, 2, 3, 4, 5$ are to be arranged...
77e0fabb4f8f99b4274e980292ddff5d The numbers $1, 2, 3, 4, 5$ are to be arranged...
Name: Text, dtype: object | [
"We see that there are $5!$ total ways to arrange the numbers. However, we can always rotate these numbers so that, for example, the number $1$ is always at the top of the circle. Thus, there are only $4!$ ways under rotation. Every case has exactly $1$ reflection, so that gives us only $4!/2$ , or $12$ cases, whic... |
https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_14 | null | 125 | problem_id
891fbd11f453d2b468075929a7f4cfd8 For any positive integer $a, \sigma(a)$ denote...
891fbd11f453d2b468075929a7f4cfd8 Warning: This solution doesn't explain why $43...
Name: Text, dtype: object | [
"We first claim that $n$ must be divisible by $42$ . Since $\\sigma(a^n)-1$ is divisible by $2021$ for all positive integers $a$ , we can first consider the special case where $a$ is prime and $a \\neq 0,1 \\pmod{43}$ . By Dirichlet's Theorem (Refer to the Remark section.), such $a$ always exists.\nThen $\\sigma(a^... |
https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_25 | B | 279 | problem_id
8dd58d71d622cdca7859918e42e278d8 The number $5^{867}$ is between $2^{2013}$ and...
8dd58d71d622cdca7859918e42e278d8 The number $5^{867}$ is between $2^{2013}$ and...
Name: Text, dtype: object | [
"The problem is asking for between how many consecutive powers of $5$ are there $3$ power of $2$\nThere can be either $2$ or $3$ powers of $2$ between any two consecutive powers of $5$ $5^n$ and $5^{n+1}$\nThe first power of $2$ is between $5^n$ and $2 \\cdot 5^n$\nThe second power of $2$ is between $2 \\cdot 5^n$ ... |
https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_4 | B | 3 | problem_id
9f71a48715f97d1d12960b17799f7db2 Walking down Jane Street, Ralph passed four ho...
9f71a48715f97d1d12960b17799f7db2 Walking down Jane Street, Ralph passed four ho...
Name: Text, dtype: object | [
"Let's use casework on the yellow house. The yellow house $(\\text{Y})$ is either the $3^\\text{rd}$ house or the last house.\nCase 1: $\\text{Y}$ is the $3^\\text{rd}$ house.\nThe only possible arrangement is $\\text{B}-\\text{O}-\\text{Y}-\\text{R}$\nCase 2: $\\text{Y}$ is the last house.\nThere are two possible ... |
https://artofproblemsolving.com/wiki/index.php/2012_USAMO_Problems/Problem_1 | null | 13 | problem_id
a45915bc778c3b967b94f7cee4faa46d Find all integers $n \ge 3$ such that among an...
a45915bc778c3b967b94f7cee4faa46d Find all integers $n \ge 3$ such that among an...
Name: Text, dtype: object | [
"Without loss of generality, assume that the set $\\{a\\}$ is ordered from least to greatest so that the bounding condition becomes $a_n \\le n \\cdot a_1.$ Now set $b_i \\equiv \\frac{a_i}{a_1},$ and since a triangle with sidelengths from $\\{a\\}$ will be similar to the corresponding triangle from $\\{b\\},$ we s... |
https://artofproblemsolving.com/wiki/index.php/2001_AMC_12_Problems/Problem_4 | D | 30 | problem_id
ae79010feec50f73241383732e6c476e The mean of three numbers is $10$ more than th...
ae79010feec50f73241383732e6c476e The mean of three numbers is $10$ more than th...
Name: Text, dtype: object | [
"Let $m$ be the mean of the three numbers. Then the least of the numbers is $m-10$ and the greatest is $m + 15$ . The middle of the three numbers is the median, $5$ . So $\\dfrac{1}{3}[(m-10) + 5 + (m + 15)] = m$ , which can be solved to get $m=10$ .\nHence, the sum of the three numbers is $3\\cdot 10 = \\boxed{30}... |
https://artofproblemsolving.com/wiki/index.php/2001_AMC_12_Problems/Problem_7 | A | 782 | problem_id
afa106734f55c02711ecd5e8bbf4e8d3 A charity sells $140$ benefit tickets for a to...
afa106734f55c02711ecd5e8bbf4e8d3 A charity sells $140$ benefit tickets for a to...
Name: Text, dtype: object | [
"Let's multiply ticket costs by $2$ , then the half price becomes an integer, and the charity sold $140$ tickets worth a total of $4002$ dollars.\nLet $h$ be the number of half price tickets, we then have $140-h$ full price tickets. The cost of $140-h$ full price tickets is equal to the cost of $280-2h$ half price ... |
https://artofproblemsolving.com/wiki/index.php/2006_AMC_10A_Problems/Problem_22 | C | 30 | problem_id
b36e53894fd42003fe068f1aec8a38d0 Two farmers agree that pigs are worth $300$ do...
b36e53894fd42003fe068f1aec8a38d0 Let us simplify this problem. Dividing by $30...
Name: Text, dtype: object | [
"The problem can be restated as an equation of the form $300p + 210g = x$ , where $p$ is the number of pigs, $g$ is the number of goats, and $x$ is the positive debt. The problem asks us to find the lowest possible. $p$ and $g$ must be integers, which makes the equation a Diophantine equation Bezout's Lemma tells u... |
https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_11 | C | 219.95 | problem_id
bdf1157d287c23856113e8dc3c9bdb32 A customer who intends to purchase an applianc...
bdf1157d287c23856113e8dc3c9bdb32 A customer who intends to purchase an applianc...
Name: Text, dtype: object | [
"Let the listed price be $x$ . Since all the answer choices are above $\\textdollar100$ , we can assume $x > 100$ . Thus the discounts after the coupons are used will be as follows:\nCoupon 1: $x\\times10\\%=.1x$\nCoupon 2: $20$\nCoupon 3: $18\\%\\times(x-100)=.18x-18$\nFor coupon $1$ to give a greater price reduct... |
https://artofproblemsolving.com/wiki/index.php/2001_AMC_12_Problems/Problem_10 | D | 56 | problem_id
c1c2900151c908ac390988a490c7e35c The plane is tiled by congruent squares and co...
c1c2900151c908ac390988a490c7e35c The plane is tiled by congruent squares and co...
Name: Text, dtype: object | [
"Consider any single tile:\n\nIf the side of the small square is $a$ , then the area of the tile is $9a^2$ , with $4a^2$ covered by squares and $5a^2$ by pentagons.\nHence exactly $5/9$ of any tile are covered by pentagons, and therefore pentagons cover $5/9$ of the plane. When expressed as a percentage, this is $5... |
https://artofproblemsolving.com/wiki/index.php/2007_AMC_12B_Problems/Problem_24 | null | 4 | problem_id
c84575ab947ea92b2fa92f55386966ac Also refer to the 2007 AMC 10B #25 (same problem)
c84575ab947ea92b2fa92f55386966ac How many pairs of positive integers $(a,b)$ ar...
Name: Text, dtype: object | [
"Rewrite the equation\\[ $\\frac{a}{b}+\\frac{14b}{9a}=k$ \\]in two different forms. First, multiply both sides by $b$ and subtract $a$ to obtain\\[ $\\frac{14b^2}{9a}=bk-a.$ \\]Because $a$ $b$ , and $k$ are integers, $14b^2$ must be a multiple of $a$ , and because $a$ and $b$ have no common factors greater than 1,... |
https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_15 | C | 210 | problem_id
d52fac580a189f0980006e4b2e9b254b David drives from his home to the airport to c...
d52fac580a189f0980006e4b2e9b254b David drives from his home to the airport to c...
Name: Text, dtype: object | [
"Note that he drives at $50$ miles per hour after the first hour and continues doing so until he arrives.\nLet $d$ be the distance still needed to travel after $1$ hour. We have that $\\dfrac{d}{50}+1.5=\\dfrac{d}{35}$ , where the $1.5$ comes from $1$ hour late decreased to $0.5$ hours early.\nSimplifying gives $7d... |
https://artofproblemsolving.com/wiki/index.php/2014_USAMO_Problems/Problem_4 | null | 6 | problem_id
f25c446de036ffaadaf5676a0b0756b1 Let $k$ be a positive integer. Two players $A$...
f25c446de036ffaadaf5676a0b0756b1 Let $k$ be a positive integer. Two players $A$...
Name: Text, dtype: object | [
"We claim that the minimum $k$ such that A cannot create a $k$ in a row is $\\boxed{6}$",
"We claim that the minimum $k$ such that A cannot create a $k$ in a row is $\\boxed{6}$"
] |
https://artofproblemsolving.com/wiki/index.php/2000_AMC_8_Problems/Problem_8 | D | 41 | problem_id
f530163b6184f697cd4b5054402e0ccf Three dice with faces numbered $1$ through $6$...
f530163b6184f697cd4b5054402e0ccf Three dice with faces numbered 1 through 6 are...
Name: Text, dtype: object | [
"The numbers on one die total $1+2+3+4+5+6 = 21$ , so the numbers\non the three dice total $63$ . Numbers $1, 1, 2, 3, 4, 5, 6$ are visible, and these total $22$ .\nThis leaves $63 - 22 = \\boxed{41}$ not seen."
] |
https://artofproblemsolving.com/wiki/index.php/2006_AMC_12A_Problems/Problem_14 | C | 30 | problem_id
ff8b260c6e48fc087b54f3971592eb5d Two farmers agree that pigs are worth $300$ do...
ff8b260c6e48fc087b54f3971592eb5d Let us simplify this problem. Dividing by $30...
Name: Text, dtype: object | [
"The problem can be restated as an equation of the form $300p + 210g = x$ , where $p$ is the number of pigs, $g$ is the number of goats, and $x$ is the positive debt. The problem asks us to find the lowest possible. $p$ and $g$ must be integers, which makes the equation a Diophantine equation Bezout's Lemma tells u... |
https://artofproblemsolving.com/wiki/index.php/2007_AIME_II_Problems/Problem_8 | null | 896 | rectangular piece of paper measures 4 units by 5 units. Several lines are drawn parallel to the edges of the paper. A rectangle determined by the intersections of some of these lines is called basic if
Given that the total length of all lines drawn is exactly 2007 units, let $N$ be the maximum possible number of basic ... | [
"Denote the number of horizontal lines drawn as $x$ , and the number of vertical lines drawn as $y$ . The number of basic rectangles is $(x - 1)(y - 1)$ $5x + 4y = 2007 \\Longrightarrow y = \\frac{2007 - 5x}{4}$ . Substituting, we find that $(x - 1)\\left(-\\frac 54x + \\frac{2003}4\\right)$\nFOIL this to get a qua... |
https://artofproblemsolving.com/wiki/index.php/1990_AIME_Problems/Problem_12 | null | 720 | regular 12-gon is inscribed in a circle of radius 12. The sum of the lengths of all sides and diagonals of the 12-gon can be written in the form $a + b \sqrt{2} + c \sqrt{3} + d \sqrt{6},$ where $a^{}_{}$ $b^{}_{}$ $c^{}_{}$ , and $d^{}_{}$ are positive integers. Find $a + b + c + d^{}_{}$ | [
"1990 AIME-12.png\nThe easiest way to do this seems to be to find the length of each of the sides and diagonals. To do such, draw the radii that meet the endpoints of the sides/diagonals; this will form isosceles triangles . Drawing the altitude of those triangles and then solving will yield the respective lengths.... |
https://artofproblemsolving.com/wiki/index.php/2008_AIME_II_Problems/Problem_13 | null | 29 | regular hexagon with center at the origin in the complex plane has opposite pairs of sides one unit apart. One pair of sides is parallel to the imaginary axis. Let $R$ be the region outside the hexagon, and let $S = \left\lbrace\frac{1}{z}|z \in R\right\rbrace$ . Then the area of $S$ has the form $a\pi + \sqrt{b}$ , wh... | [
"If a point $z = r\\text{cis}\\,\\theta$ is in $R$ , then the point $\\frac{1}{z} = \\frac{1}{r} \\text{cis}\\, \\left(-\\theta\\right)$ is in $S$ (where cis denotes $\\text{cis}\\, \\theta = \\cos \\theta + i \\sin \\theta$ ). Since $R$ is symmetric every $60^{\\circ}$ about the origin, it suffices to consider the... |
https://artofproblemsolving.com/wiki/index.php/2016_AIME_I_Problems/Problem_3 | null | 810 | regular icosahedron is a $20$ -faced solid where each face is an equilateral triangle and five triangles meet at every vertex. The regular icosahedron shown below has one vertex at the top, one vertex at the bottom, an upper pentagon of five vertices all adjacent to the top vertex and all in the same horizontal plane, ... | [
"Think about each plane independently. From the top vertex, we can go down to any of 5 different points in the second plane. From that point, there are 9 ways to go to another point within the second plane: rotate as many as 4 points clockwise or as many 4 counterclockwise, or stay in place. Then, there are 2 path... |
https://artofproblemsolving.com/wiki/index.php/2023_AMC_8_Problems/Problem_17 | A | 1 | regular octahedron has eight equilateral triangle faces with four faces meeting at each vertex. Jun will make the regular octahedron shown on the right by folding the piece of paper shown on the left. Which numbered face will end up to the right of $Q$
[asy] // Diagram by TheMathGuyd import graph; // The Solid // To sa... | [
"We color face $6$ red and face $5$ yellow. Note that from the octahedron, face $5$ and face $?$ do not share anything in common. From the net, face $5$ shares at least one vertex with all other faces except face $1,$ which is shown in green: Therefore, the answer is $\\boxed{1}.$",
"We label the octohedron goin... |
https://artofproblemsolving.com/wiki/index.php/2004_AIME_II_Problems/Problem_11 | null | 625 | right circular cone has a base with radius $600$ and height $200\sqrt{7}.$ A fly starts at a point on the surface of the cone whose distance from the vertex of the cone is $125$ , and crawls along the surface of the cone to a point on the exact opposite side of the cone whose distance from the vertex is $375\sqrt{2}.$ ... | [
"The easiest way is to unwrap the cone into a circular sector. Center the sector at the origin with one radius on the positive $x$ -axis and the angle $\\theta$ going counterclockwise. The circumference of the base is $C=1200\\pi$ . The sector's radius (cone's sweep) is $R=\\sqrt{r^2+h^2}=\\sqrt{600^2+(200\\sqrt{7}... |
https://artofproblemsolving.com/wiki/index.php/2005_AIME_I_Problems/Problem_11 | null | 544 | semicircle with diameter $d$ is contained in a square whose sides have length 8. Given the maximum value of $d$ is $m - \sqrt{n},$ find $m+n.$ | [
"We can just look at a quarter circle inscribed in a $45-45-90$ right triangle. We can then extend a radius, $r$ to one of the sides creating an $r,r, r\\sqrt{2}$ right triangle. This means that we have $r + r\\sqrt{2} = 8\\sqrt{2}$ so $r = \\frac{8\\sqrt{2}}{1+\\sqrt{2}} = 16 - 8\\sqrt{2}$ . Then the diameter is $... |
https://artofproblemsolving.com/wiki/index.php/2006_AIME_II_Problems/Problem_11 | null | 834 | sequence is defined as follows $a_1=a_2=a_3=1,$ and, for all positive integers $n, a_{n+3}=a_{n+2}+a_{n+1}+a_n.$ Given that $a_{28}=6090307, a_{29}=11201821,$ and $a_{30}=20603361,$ find the remainder when $\sum^{28}_{k=1} a_k$ is divided by 1000. | [
"Define the sum as $s$ . Since $a_n\\ = a_{n + 3} - a_{n + 2} - a_{n + 1}$ , the sum will be:\nThus $s = \\frac{a_{28} + a_{30}}{2}$ , and $a_{28},\\,a_{30}$ are both given; the last four digits of their sum is $3668$ , and half of that is $1834$ . Therefore, the answer is $\\boxed{834}$ .−",
"Since the problem o... |
https://artofproblemsolving.com/wiki/index.php/2007_AIME_I_Problems/Problem_14 | null | 224 | sequence is defined over non-negative integral indexes in the following way: $a_{0}=a_{1}=3$ $a_{n+1}a_{n-1}=a_{n}^{2}+2007$
Find the greatest integer that does not exceed $\frac{a_{2006}^{2}+a_{2007}^{2}}{a_{2006}a_{2007}}.$ | [
"Define a function $f(n)$ on the non-negative integers, as \\[f(n) = \\frac{a_n^2 + a_{n+1}^2}{a_na_{n+1}} = \\frac{a_n}{a_{n+1}}+\\frac{a_{n+1}}{a_{n}}\\] We want $\\left\\lfloor f(2006) \\right\\rfloor$\nConsider the relation $a_{n+1}a_{n-1}=a_{n}^{2}+2007$ . Dividing through by $a_{n}a_{n-1}$ , we get \\[.\\phan... |
https://artofproblemsolving.com/wiki/index.php/1985_AIME_Problems/Problem_5 | null | 986 | sequence of integers $a_1, a_2, a_3, \ldots$ is chosen so that $a_n = a_{n - 1} - a_{n - 2}$ for each $n \ge 3$ . What is the sum of the first 2001 terms of this sequence if the sum of the first 1492 terms is 1985, and the sum of the first 1985 terms is 1492? | [
"The problem gives us a sequence defined by a recursion , so let's calculate a few values to get a feel for how it acts. We aren't given initial values, so let $a_1 = a$ and $a_2 = b$ . Then $a_3 = b - a$ $a_4 = (b - a) - b = -a$ $a_5 = -a - (b - a) = -b$ $a_6 = -b - (-a) = a - b$ $a_7 = (a - b) - (-b) = a$ and $... |
https://artofproblemsolving.com/wiki/index.php/2000_AIME_I_Problems/Problem_10 | null | 173 | sequence of numbers $x_{1},x_{2},x_{3},\ldots,x_{100}$ has the property that, for every integer $k$ between $1$ and $100,$ inclusive, the number $x_{k}$ is $k$ less than the sum of the other $99$ numbers. Given that $x_{50} = m/n,$ where $m$ and $n$ are relatively prime positive integers, find $m + n$ | [
"Let the sum of all of the terms in the sequence be $\\mathbb{S}$ . Then for each integer $k$ $x_k = \\mathbb{S}-x_k-k \\Longrightarrow \\mathbb{S} - 2x_k = k$ . Summing this up for all $k$ from $1, 2, \\ldots, 100$\n\\begin{align*}100\\mathbb{S}-2(x_1 + x_2 + \\cdots + x_{100}) &= 1 + 2 + \\cdots + 100\\\\ 100\\ma... |
https://artofproblemsolving.com/wiki/index.php/2004_AIME_II_Problems/Problem_9 | null | 973 | sequence of positive integers with $a_1=1$ and $a_9+a_{10}=646$ is formed so that the first three terms are in geometric progression , the second, third, and fourth terms are in arithmetic progression , and, in general, for all $n\ge1,$ the terms $a_{2n-1}, a_{2n}, a_{2n+1}$ are in geometric progression, and the terms ... | [
"Let $x = a_2$ ; then solving for the next several terms, we find that $a_3 = x^2,\\ a_4 = x(2x-1),\\ a_5$ $= (2x-1)^2,\\ a_6$ $= (2x-1)(3x-2)$ , and in general, $a_{2n} = f(n-1)f(n)$ $a_{2n+1} = f(n)^2$ , where $f(n) = nx - (n-1)$\nFrom \\[a_9 + a_{10} = f(4)^2 + f(4)f(5) = (4x-3)(9x-7) = 646 = 2\\cdot 17 \\cdot 1... |
https://artofproblemsolving.com/wiki/index.php/2004_AMC_10A_Problems/Problem_18 | A | 1 | sequence of three real numbers forms an arithmetic progression with a first term of $9$ . If $2$ is added to the second term and $20$ is added to the third term, the three resulting numbers form a geometric progression . What is the smallest possible value for the third term in the geometric progression?
$\text {(A)}\ ... | [
"Let $d$ be the common difference. Then $9$ $9+d+2=11+d$ $9+2d+20=29+2d$ are the terms of the geometric progression. Since the middle term is the geometric mean of the other two terms, $(11+d)^2 = 9(2d+29)$ $\\Longrightarrow d^2 + 4d - 140$ $= (d+14)(d-10) = 0$ . The smallest possible value occurs when $d = -14$ , ... |
https://artofproblemsolving.com/wiki/index.php/2004_AMC_12A_Problems/Problem_14 | A | 1 | sequence of three real numbers forms an arithmetic progression with a first term of $9$ . If $2$ is added to the second term and $20$ is added to the third term, the three resulting numbers form a geometric progression . What is the smallest possible value for the third term in the geometric progression?
$\text {(A)}\ ... | [
"Let $d$ be the common difference. Then $9$ $9+d+2=11+d$ $9+2d+20=29+2d$ are the terms of the geometric progression. Since the middle term is the geometric mean of the other two terms, $(11+d)^2 = 9(2d+29)$ $\\Longrightarrow d^2 + 4d - 140$ $= (d+14)(d-10) = 0$ . The smallest possible value occurs when $d = -14$ , ... |
https://artofproblemsolving.com/wiki/index.php/2004_AIME_I_Problems/Problem_11 | null | 512 | solid in the shape of a right circular cone is 4 inches tall and its base has a 3-inch radius. The entire surface of the cone, including its base, is painted. A plane parallel to the base of the cone divides the cone into two solids, a smaller cone-shaped solid $C$ and a frustum -shaped solid $F,$ in such a way that th... | [
"Our original solid has volume equal to $V = \\frac13 \\pi r^2 h = \\frac13 \\pi 3^2\\cdot 4 = 12 \\pi$ and has surface area $A = \\pi r^2 + \\pi r \\ell$ , where $\\ell$ is the slant height of the cone. Using the Pythagorean Theorem , we get $\\ell = 5$ and $A = 24\\pi$\nLet $x$ denote the radius of the small con... |
https://artofproblemsolving.com/wiki/index.php/2007_AMC_10A_Problems/Problem_21 | C | 8 | sphere is inscribed in a cube that has a surface area of $24$ square meters. A second cube is then inscribed within the sphere. What is the surface area in square meters of the inner cube?
$\text{(A)}\ 3 \qquad \text{(B)}\ 6 \qquad \text{(C)}\ 8 \qquad \text{(D)}\ 9 \qquad \text{(E)}\ 12$ | [
"We rotate the smaller cube around the sphere such that two opposite vertices of the cube are on opposite faces of the larger cube. Thus the main diagonal of the smaller cube is the side length of the outer square.\nLet $S$ be the surface area of the inner square. The ratio of the areas of two similar figures is eq... |
https://artofproblemsolving.com/wiki/index.php/2007_AMC_10A_Problems/Problem_21 | null | 8 | sphere is inscribed in a cube that has a surface area of $24$ square meters. A second cube is then inscribed within the sphere. What is the surface area in square meters of the inner cube?
$\text{(A)}\ 3 \qquad \text{(B)}\ 6 \qquad \text{(C)}\ 8 \qquad \text{(D)}\ 9 \qquad \text{(E)}\ 12$ | [
"Since the surface area of the original cube is 24 square meters, each face of the cube has a surface area of $24/6 = 4$ square meters, and the side length of this cube is 2 meters. The sphere inscribed within the cube has diameter 2 meters, which is also the length of the diagonal of the cube inscribed in the sphe... |
https://artofproblemsolving.com/wiki/index.php/2001_AIME_I_Problems/Problem_12 | null | 5 | sphere is inscribed in the tetrahedron whose vertices are $A = (6,0,0), B = (0,4,0), C = (0,0,2),$ and $D = (0,0,0).$ The radius of the sphere is $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$ | [
"\nThe center $I$ of the insphere must be located at $(r,r,r)$ where $r$ is the sphere's radius. $I$ must also be a distance $r$ from the plane $ABC$\nThe signed distance between a plane and a point $I$ can be calculated as $\\frac{(I-G) \\cdot P}{|P|}$ , where G is any point on the plane, and P is a vector perpend... |
https://artofproblemsolving.com/wiki/index.php/2005_AMC_10B_Problems/Problem_25 | C | 62 | subset $B$ of the set of integers from $1$ to $100$ , inclusive, has the property that no two elements of $B$ sum to $125$ . What is the maximum possible number of elements in $B$
$\textbf{(A) } 50 \qquad \textbf{(B) } 51 \qquad \textbf{(C) } 62 \qquad \textbf{(D) } 65 \qquad \textbf{(E) } 68$ | [
"The question asks for the maximum possible number of elements. The integers from $1$ to $24$ can be included because you cannot make $125$ with integers from $1$ to $24$ without the other number being greater than $100$ . The integers from $25$ to $100$ are left. They can be paired so the sum is $125$ $25+100$ $26... |
https://artofproblemsolving.com/wiki/index.php/1990_AIME_Problems/Problem_7 | null | 89 | triangle has vertices $P_{}^{}=(-8,5)$ $Q_{}^{}=(-15,-19)$ , and $R_{}^{}=(1,-7)$ . The equation of the bisector of $\angle P$ can be written in the form $ax+2y+c=0_{}^{}$ . Find $a+c_{}^{}$ | [
"Use the angle bisector theorem to find that the angle bisector of $\\angle P$ divides $QR$ into segments of length $\\frac{25}{x} = \\frac{15}{20 -x} \\Longrightarrow x = \\frac{25}{2},\\ \\frac{15}{2}$ . It follows that $\\frac{QP'}{RP'} = \\frac{5}{3}$ , and so $P' = \\left(\\frac{5x_R + 3x_Q}{8},\\frac{5y_R + 3... |
https://artofproblemsolving.com/wiki/index.php/2008_AIME_I_Problems/Problem_6 | null | 17 | triangular array of numbers has a first row consisting of the odd integers $1,3,5,\ldots,99$ in increasing order. Each row below the first has one fewer entry than the row above it, and the bottom row has a single entry. Each entry in any row after the top row equals the sum of the two entries diagonally above it in ... | [
"Let the $k$ th number in the $n$ th row be $a(n,k)$ . Writing out some numbers, we find that $a(n,k) = 2^{n-1}(n+2k-2)$\nWe wish to find all $(n,k)$ such that $67| a(n,k) = 2^{n-1} (n+2k-2)$ . Since $2^{n-1}$ and $67$ are relatively prime , it follows that $67|n+2k-2$ . Since every row has one less element than th... |
https://artofproblemsolving.com/wiki/index.php/2007_AIME_II_Problems/Problem_13 | null | 640 | triangular array of squares has one square in the first row, two in the second, and in general, $k$ squares in the $k$ th row for $1 \leq k \leq 11.$ With the exception of the bottom row, each square rests on two squares in the row immediately below (illustrated in the given diagram). In each square of the eleventh row... | [
"Label each of the bottom squares as $x_0, x_1 \\ldots x_9, x_{10}$\nThrough induction , we can find that the top square is equal to ${10\\choose0}x_0 + {10\\choose1}x_1 + {10\\choose2}x_2 + \\ldots {10\\choose10}x_{10}$ . (This also makes sense based on a combinatorial argument: the number of ways a number can \"t... |
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