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/2012_AIME_II_Problems/Problem_15 | null | 919 | Triangle $ABC$ is inscribed in circle $\omega$ with $AB=5$ $BC=7$ , and $AC=3$ . The bisector of angle $A$ meets side $\overline{BC}$ at $D$ and circle $\omega$ at a second point $E$ . Let $\gamma$ be the circle with diameter $\overline{DE}$ . Circles $\omega$ and $\gamma$ meet at $E$ and a second point $F$ . Then $AF^... | [
"Take a force-overlaid inversion about $A$ and note $D$ and $E$ map to each other. As $DE$ was originally the diameter of $\\gamma$ $DE$ is still the diameter of $\\gamma$ . Thus $\\gamma$ is preserved. Note that the midpoint $M$ of $BC$ lies on $\\gamma$ , and $BC$ and $\\omega$ are swapped. Thus points $F$ and $M... |
https://artofproblemsolving.com/wiki/index.php/1995_AIME_Problems/Problem_9 | null | 616 | Triangle $ABC$ is isosceles , with $AB=AC$ and altitude $AM=11.$ Suppose that there is a point $D$ on $\overline{AM}$ with $AD=10$ and $\angle BDC=3\angle BAC.$ Then the perimeter of $\triangle ABC$ may be written in the form $a+\sqrt{b},$ where $a$ and $b$ are integers. Find $a+b.$
[asy] import graph; size(5cm); real... | [
"Let $x=\\angle CAM$ , so $3x=\\angle CDM$ . Then, $\\frac{\\tan 3x}{\\tan x}=\\frac{CM/1}{CM/11}=11$ . Expanding $\\tan 3x$ using the angle sum identity gives \\[\\tan 3x=\\tan(2x+x)=\\frac{3\\tan x-\\tan^3x}{1-3\\tan^2x}.\\] Thus, $\\frac{3-\\tan^2x}{1-3\\tan^2x}=11$ . Solving, we get $\\tan x= \\frac 12$ . Hence... |
https://artofproblemsolving.com/wiki/index.php/2003_AIME_I_Problems/Problem_10 | null | 83 | Triangle $ABC$ is isosceles with $AC = BC$ and $\angle ACB = 106^\circ.$ Point $M$ is in the interior of the triangle so that $\angle MAC = 7^\circ$ and $\angle MCA = 23^\circ.$ Find the number of degrees in $\angle CMB.$ | [
"Take point $N$ inside $\\triangle ABC$ such that $\\angle CBN = 7^\\circ$ and $\\angle BCN = 23^\\circ$\n$\\angle MCN = 106^\\circ - 2\\cdot 23^\\circ = 60^\\circ$ . Also, since $\\triangle AMC$ and $\\triangle BNC$ are congruent (by ASA), $CM = CN$ . Hence $\\triangle CMN$ is an equilateral triangle , so $\\angle... |
https://artofproblemsolving.com/wiki/index.php/2005_AIME_I_Problems/Problem_10 | null | 47 | Triangle $ABC$ lies in the cartesian plane and has an area of $70$ . The coordinates of $B$ and $C$ are $(12,19)$ and $(23,20),$ respectively, and the coordinates of $A$ are $(p,q).$ The line containing the median to side $BC$ has slope $-5.$ Find the largest possible value of $p+q.$ | [
"The midpoint $M$ of line segment $\\overline{BC}$ is $\\left(\\frac{35}{2}, \\frac{39}{2}\\right)$ . The equation of the median can be found by $-5 = \\frac{q - \\frac{39}{2}}{p - \\frac{35}{2}}$ . Cross multiply and simplify to yield that $-5p + \\frac{35 \\cdot 5}{2} = q - \\frac{39}{2}$ , so $q = -5p + 107$\nUs... |
https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_24 | D | 75 | Triangle $ABC$ with $AB=50$ and $AC=10$ has area $120$ . Let $D$ be the midpoint of $\overline{AB}$ , and let $E$ be the midpoint of $\overline{AC}$ . The angle bisector of $\angle BAC$ intersects $\overline{DE}$ and $\overline{BC}$ at $F$ and $G$ , respectively. What is the area of quadrilateral $FDBG$
$\textbf{(A) }6... | [
"Let $BC = a$ $BG = x$ $GC = y$ , and the length of the perpendicular from $BC$ through $A$ be $h$ . By angle bisector theorem, we have that \\[\\frac{50}{x} = \\frac{10}{y},\\] where $y = -x+a$ . Therefore substituting we have that $BG=\\frac{5a}{6}$ . By similar triangles, we have that $DF=\\frac{5a}{12}$ , and t... |
https://artofproblemsolving.com/wiki/index.php/2018_AMC_12A_Problems/Problem_18 | D | 75 | Triangle $ABC$ with $AB=50$ and $AC=10$ has area $120$ . Let $D$ be the midpoint of $\overline{AB}$ , and let $E$ be the midpoint of $\overline{AC}$ . The angle bisector of $\angle BAC$ intersects $\overline{DE}$ and $\overline{BC}$ at $F$ and $G$ , respectively. What is the area of quadrilateral $FDBG$
$\textbf{(A) }6... | [
"Let $BC = a$ $BG = x$ $GC = y$ , and the length of the perpendicular from $BC$ through $A$ be $h$ . By angle bisector theorem, we have that \\[\\frac{50}{x} = \\frac{10}{y},\\] where $y = -x+a$ . Therefore substituting we have that $BG=\\frac{5a}{6}$ . By similar triangles, we have that $DF=\\frac{5a}{12}$ , and t... |
https://artofproblemsolving.com/wiki/index.php/2010_AIME_II_Problems/Problem_14 | null | 7 | Triangle $ABC$ with right angle at $C$ $\angle BAC < 45^\circ$ and $AB = 4$ . Point $P$ on $\overline{AB}$ is chosen such that $\angle APC = 2\angle ACP$ and $CP = 1$ . The ratio $\frac{AP}{BP}$ can be represented in the form $p + q\sqrt{r}$ , where $p$ $q$ $r$ are positive integers and $r$ is not divisible by the squa... | [
"Let $O$ be the circumcenter of $ABC$ and let the intersection of $CP$ with the circumcircle be $D$ . It now follows that $\\angle{DOA} = 2\\angle ACP = \\angle{APC} = \\angle{DPB}$ . Hence $ODP$ is isosceles and $OD = DP = 2$\nDenote $E$ the projection of $O$ onto $CD$ . Now $CD = CP + DP = 3$ . By the Pythagorean... |
https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_5 | null | 182 | Triangle $ABC_0$ has a right angle at $C_0$ . Its side lengths are pairwise relatively prime positive integers, and its perimeter is $p$ . Let $C_1$ be the foot of the altitude to $\overline{AB}$ , and for $n \geq 2$ , let $C_n$ be the foot of the altitude to $\overline{C_{n-2}B}$ in $\triangle C_{n-2}C_{n-1}B$ . The s... | [
"Do note that by counting the area in 2 ways, the first altitude is $x = \\frac{ab}{c}$ . By similar triangles, the common ratio is $\\rho = \\frac{a}{c}$ for each height, so by the geometric series formula, we have \\begin{align} 6p=\\frac{x}{1-\\rho} = \\frac{ab}{c-a}. \\end{align} Writing $p=a+b+c$ and clear... |
https://artofproblemsolving.com/wiki/index.php/2013_AIME_I_Problems/Problem_13 | null | 961 | Triangle $AB_0C_0$ has side lengths $AB_0 = 12$ $B_0C_0 = 17$ , and $C_0A = 25$ . For each positive integer $n$ , points $B_n$ and $C_n$ are located on $\overline{AB_{n-1}}$ and $\overline{AC_{n-1}}$ , respectively, creating three similar triangles $\triangle AB_nC_n \sim \triangle B_{n-1}C_nC_{n-1} \sim \triangle AB_{... | [
"Note that every $B_nC_n$ is parallel to each other for any nonnegative $n$ . Also, the area we seek is simply the ratio $k=\\frac{[B_0B_1C_1]}{[B_0B_1C_1]+[C_1C_0B_0]}$ , because it repeats in smaller and smaller units. Note that the area of the triangle, by Heron's formula, is 90.\nFor ease, all ratios I will use... |
https://artofproblemsolving.com/wiki/index.php/2020_AMC_10A_Problems/Problem_12 | C | 96 | Triangle $AMC$ is isosceles with $AM = AC$ . Medians $\overline{MV}$ and $\overline{CU}$ are perpendicular to each other, and $MV=CU=12$ . What is the area of $\triangle AMC?$
[asy] draw((-4,0)--(4,0)--(0,12)--cycle); draw((-2,6)--(4,0)); draw((2,6)--(-4,0)); label("M", (-4,0), W); label("C", (4,0), E); label("A", (0, ... | [
"Since quadrilateral $UVCM$ has perpendicular diagonals, its area can be found as half of the product of the length of the diagonals. Also note that $\\triangle AUV$ has $\\frac 14$ the area of triangle $AMC$ by similarity, so $[UVCM]=\\frac 34\\cdot [AMC].$ Thus, \\[\\frac 12 \\cdot 12\\cdot 12=\\frac 34 \\cdot [A... |
https://artofproblemsolving.com/wiki/index.php/1956_AHSME_Problems/Problem_49 | null | 70 | Triangle $PAB$ is formed by three tangents to circle $O$ and $\angle APB = 40^{\circ}$ ; then $\angle AOB$ equals:
$\textbf{(A)}\ 45^{\circ}\qquad \textbf{(B)}\ 50^{\circ}\qquad \textbf{(C)}\ 55^{\circ}\qquad \textbf{(D)}\ 60^{\circ}\qquad \textbf{(E)}\ 70^{\circ}$ | [
"First, from triangle $ABO$ $\\angle AOB = 180^\\circ - \\angle BAO - \\angle ABO$ . Note that $AO$ bisects $\\angle BAT$ (to see this, draw radii from $O$ to $AB$ and $AT,$ creating two congruent right triangles), so $\\angle BAO = \\angle BAT/2$ . Similarly, $\\angle ABO = \\angle ABR/2$\nAlso, $\\angle BAT = 180... |
https://artofproblemsolving.com/wiki/index.php/2000_AMC_8_Problems/Problem_15 | C | 15 | Triangles $ABC$ $ADE$ , and $EFG$ are all equilateral. Points $D$ and $G$ are midpoints of $\overline{AC}$ and $\overline{AE}$ , respectively. If $AB = 4$ , what is the perimeter of figure $ABCDEFG$
[asy] pair A,B,C,D,EE,F,G; A = (4,0); B = (0,0); C = (2,2*sqrt(3)); D = (3,sqrt(3)); EE = (5,sqrt(3)); F = (5.5,sqrt(3)/2... | [
"The large triangle $ABC$ has sides of length $4$ . The medium triangle has sides of length $2$ . The small triangle has sides of length $1$ . There are $3$ segment sizes, and all segments depicted are one of these lengths.\nStarting at $A$ and going clockwise, the perimeter is:\n$AB + BC + CD + DE + EF + FG + G... |
https://artofproblemsolving.com/wiki/index.php/2020_AIME_II_Problems/Problem_4 | null | 108 | Triangles $\triangle ABC$ and $\triangle A'B'C'$ lie in the coordinate plane with vertices $A(0,0)$ $B(0,12)$ $C(16,0)$ $A'(24,18)$ $B'(36,18)$ $C'(24,2)$ . A rotation of $m$ degrees clockwise around the point $(x,y)$ where $0<m<180$ , will transform $\triangle ABC$ to $\triangle A'B'C'$ . Find $m+x+y$ | [
"After sketching, it is clear a $90^{\\circ}$ rotation is done about $(x,y)$ . Looking between $A$ and $A'$ $x+y=18$ . Thus $90+18=\\boxed{108}$ .\n~mn28407",
"\nWe first draw a diagram with the correct Cartesian coordinates and a center of rotation $P$ . Note that $PC=PC'$ because $P$ lies on the perpendicular b... |
https://artofproblemsolving.com/wiki/index.php/2016_AMC_10A_Problems/Problem_8 | C | 35 | Trickster Rabbit agrees with Foolish Fox to double Fox's money every time Fox crosses the bridge by Rabbit's house, as long as Fox pays $40$ coins in toll to Rabbit after each crossing. The payment is made after the doubling, Fox is excited about his good fortune until he discovers that all his money is gone after cros... | [
"If you started backwards you would get: \\[0\\Rightarrow (+40)=40 , \\Rightarrow \\left(\\frac{1}{2}\\right)=20 , \\Rightarrow (+40)=60 , \\Rightarrow \\left(\\frac{1}{2}\\right)=30 , \\Rightarrow (+40)=70 , \\Rightarrow \\left(\\frac{1}{2}\\right)=\\boxed{35}\\]",
"If you have $x$ as the amount of money Foolish... |
https://artofproblemsolving.com/wiki/index.php/1991_AIME_Problems/Problem_11 | null | 135 | Twelve congruent disks are placed on a circle $C^{}_{}$ of radius 1 in such a way that the twelve disks cover $C^{}_{}$ , no two of the disks overlap, and so that each of the twelve disks is tangent to its two neighbors. The resulting arrangement of disks is shown in the figure below. The sum of the areas of the twelv... | [
"We wish to find the radius of one circle, so that we can find the total area.\nNotice that for them to contain the entire circle, each pair of circles must be tangent on the larger circle. Now consider two adjacent smaller circles. This means that the line connecting the radii is a segment of length $2r$ that is t... |
https://artofproblemsolving.com/wiki/index.php/2004_AMC_8_Problems/Problem_3 | A | 8 | Twelve friends met for dinner at Oscar's Overstuffed Oyster House, and each ordered one meal. The portions were so large, there was enough food for $18$ people. If they shared, how many meals should they have ordered to have just enough food for the $12$ of them?
$\textbf{(A)}\ 8\qquad\textbf{(B)}\ 9\qquad\textbf{(C)}\... | [
"Set up the proportion $\\frac{12\\ \\text{meals}}{18\\ \\text{people}}=\\frac{x\\ \\text{meals}}{12\\ \\text{people}}$ . Solving for $x$ gives us $x= \\boxed{8}$"
] |
https://artofproblemsolving.com/wiki/index.php/2022_AIME_II_Problems/Problem_5 | null | 72 | Twenty distinct points are marked on a circle and labeled $1$ through $20$ in clockwise order. A line segment is drawn between every pair of points whose labels differ by a prime number. Find the number of triangles formed whose vertices are among the original $20$ points. | [
"Let $a$ $b$ , and $c$ be the vertex of a triangle that satisfies this problem, where $a > b > c$ \\[a - b = p_1\\] \\[b - c = p_2\\] \\[a - c = p_3\\]\n$p_3 = a - c = a - b + b - c = p_1 + p_2$ . Because $p_3$ is the sum of two primes, $p_1$ and $p_2$ $p_1$ or $p_2$ must be $2$ . Let $p_1 = 2$ , then $p_3 = p_2 + ... |
https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_7 | null | 57 | Twenty five of King Arthur's knights are seated at their customary round table. Three of them are chosen - all choices being equally likely - and are sent off to slay a troublesome dragon. Let $P$ be the probability that at least two of the three had been sitting next to each other. If $P$ is written as a fraction in l... | [
"We can use complementary counting , by finding the probability that none of the three knights are sitting next to each other and subtracting it from $1$\nImagine that the $22$ other (indistinguishable) people are already seated, and fixed into place.\nWe will place $A$ $B$ , and $C$ with and without the restrictio... |
https://artofproblemsolving.com/wiki/index.php/2009_AMC_10B_Problems/Problem_5 | D | 36 | Twenty percent less than 60 is one-third more than what number?
$\mathrm{(A)}\ 16\qquad \mathrm{(B)}\ 30\qquad \mathrm{(C)}\ 32\qquad \mathrm{(D)}\ 36\qquad \mathrm{(E)}\ 48$ | [
"Twenty percent less than 60 is $\\frac 45 \\cdot 60 = 48$ . One-third more than a number is $\\frac 43n$ . Therefore $\\frac 43n = 48$ and the number is $\\boxed{36}$"
] |
https://artofproblemsolving.com/wiki/index.php/2009_AMC_12B_Problems/Problem_3 | D | 36 | Twenty percent less than 60 is one-third more than what number?
$\mathrm{(A)}\ 16\qquad \mathrm{(B)}\ 30\qquad \mathrm{(C)}\ 32\qquad \mathrm{(D)}\ 36\qquad \mathrm{(E)}\ 48$ | [
"Twenty percent less than 60 is $\\frac 45 \\cdot 60 = 48$ . One-third more than a number is $\\frac 43n$ . Therefore $\\frac 43n = 48$ and the number is $\\boxed{36}$"
] |
https://artofproblemsolving.com/wiki/index.php/2005_AIME_I_Problems/Problem_9 | null | 74 | Twenty seven unit cubes are painted orange on a set of four faces so that two non-painted faces share an edge . The 27 cubes are randomly arranged to form a $3\times 3 \times 3$ cube. Given the probability of the entire surface area of the larger cube is orange is $\frac{p^a}{q^br^c},$ where $p,q,$ and $r$ are distinct... | [
"2005 I AIME-9.png\nWe can consider the orientation of each of the individual cubes independently.\nThe unit cube at the center of our large cube has no exterior faces, so all of its orientations work.\nFor the six unit cubes and the centers of the faces of the large cube, we need that they show an orange face. Th... |
https://artofproblemsolving.com/wiki/index.php/2009_AMC_8_Problems/Problem_19 | D | 165 | Two angles of an isosceles triangle measure $70^\circ$ and $x^\circ$ . What is the sum of the three possible values of $x$
$\textbf{(A)}\ 95 \qquad \textbf{(B)}\ 125 \qquad \textbf{(C)}\ 140 \qquad \textbf{(D)}\ 165 \qquad \textbf{(E)}\ 180$ | [
"There are 3 cases: where $x^\\circ$ is a base angle with the $70^\\circ$ as the other angle, where $x^\\circ$ is a base angle with $70^\\circ$ as the vertex angle, and where $x^\\circ$ is the vertex angle with $70^\\circ$ as a base angle.\nCase 1: $x^\\circ$ is a base angle with the $70^\\circ$ as the other angle:... |
https://artofproblemsolving.com/wiki/index.php/1950_AHSME_Problems/Problem_28 | B | 8 | Two boys $A$ and $B$ start at the same time to ride from Port Jervis to Poughkeepsie, $60$ miles away. $A$ travels $4$ miles an hour slower than $B$ $B$ reaches Poughkeepsie and at once turns back meeting $A$ $12$ miles from Poughkeepsie. The rate of $A$ was:
$\textbf{(A)}\ 4\text{ mph}\qquad \textbf{(B)}\ 8\text{ mph}... | [
"Let the speed of boy $A$ be $a$ , and the speed of boy $B$ be $b$ . Notice that $A$ travels $4$ miles per hour slower than boy $B$ , so we can replace $b$ with $a+4$\nNow let us see the distances that the boys each travel. Boy $A$ travels $60-12=48$ miles, and boy $B$ travels $60+12=72$ miles. Now, we can use $d=r... |
https://artofproblemsolving.com/wiki/index.php/1973_AHSME_Problems/Problem_29 | A | 13 | Two boys start moving from the same point A on a circular track but in opposite directions. Their speeds are 5 ft. per second and 9 ft. per second. If they start at the same time and finish when they first meet at the point A again, then the number of times they meet, excluding the start and finish, is
$\textbf{(A)}\ 1... | [
"Let $d$ be the length of the track in feet and $x$ be the number of laps that one of the boys did, so time one of the boys traveled before the two finish is $\\tfrac{dx}{5}$ . Since the time elapsed for both boys is equal, one boy ran $5(\\tfrac{dx}{5})$ feet while the other boy ran $9(\\tfrac{dx}{5})$ feet. Bec... |
https://artofproblemsolving.com/wiki/index.php/1994_AJHSME_Problems/Problem_14 | E | 36 | Two children at a time can play pairball. For $90$ minutes, with only two children playing at time, five children take turns so that each one plays the same amount of time. The number of minutes each child plays is
$\text{(A)}\ 9 \qquad \text{(B)}\ 10 \qquad \text{(C)}\ 18 \qquad \text{(D)}\ 20 \qquad \text{(E)}\ 36$ | [
"There are $2 \\times 90 = 180$ minutes of total playing time. Divided equally among the five children, each child gets $180/5 = \\boxed{36}$ minutes."
] |
https://artofproblemsolving.com/wiki/index.php/2010_AMC_10B_Problems/Problem_20 | D | 81 | Two circles lie outside regular hexagon $ABCDEF$ . The first is tangent to $\overline{AB}$ , and the second is tangent to $\overline{DE}$ . Both are tangent to lines $BC$ and $FA$ . What is the ratio of the area of the second circle to that of the first circle?
$\textbf{(A)}\ 18 \qquad \textbf{(B)}\ 27 \qquad \textbf{... | [
"A good diagram is very helpful.\nThe first circle is in red, the second in blue.\nWith this diagram, we can see that the first circle is inscribed in equilateral triangle $GBA$ while the second circle is inscribed in $GKJ$ .\nFrom this, it's evident that the ratio of the blue area to the red area is equal to the r... |
https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_15 | D | 69 | Two circles of radius $5$ are externally tangent to each other and are internally tangent to a circle of radius $13$ at points $A$ and $B$ , as shown in the diagram. The distance $AB$ can be written in the form $\tfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. What is $m+n$
[asy] draw(circle((0... | [
"\nLet the center of the surrounding circle be $X$ . The circle that is tangent at point $A$ will have point $Y$ as the center. Similarly, the circle that is tangent at point $B$ will have point $Z$ as the center. Connect $AB$ $YZ$ $XA$ , and $XB$ . Now observe that $\\triangle XYZ$ is similar to $\\triangle XAB$ b... |
https://artofproblemsolving.com/wiki/index.php/2014_AMC_10B_Problems/Problem_19 | D | 13 | Two concentric circles have radii $1$ and $2$ . Two points on the outer circle are chosen independently and uniformly at random. What is the probability that the chord joining the two points intersects the inner circle?
$\textbf{(A)}\ \frac{1}{6}\qquad \textbf{(B)}\ \frac{1}{4}\qquad \textbf{(C)}\ \frac{2-\sqrt{2}}{2}\... | [
"Let the center of the two circles be $O$ . Now pick an arbitrary point $A$ on the boundary of the circle with radius $2$ . We want to find the range of possible places for the second point, $A'$ , such that $AA'$ passes through the circle of radius $1$ . To do this, first draw the tangents from $A$ to the circle o... |
https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_23 | C | 120 | Two congruent circles centered at points $A$ and $B$ each pass through the other circle's center. The line containing both $A$ and $B$ is extended to intersect the circles at points $C$ and $D$ . The circles intersect at two points, one of which is $E$ . What is the degree measure of $\angle CED$
$\textbf{(A) }90\qquad... | [
"Observe that $\\triangle{EAB}$ is equilateral. Therefore, $m\\angle{AEB}=m\\angle{EAB}=m\\angle{EBA} = 60^{\\circ}$ . Since $CD$ is a straight line, we conclude that $m\\angle{EBD} = 180^{\\circ}-60^{\\circ}=120^{\\circ}$ . Since $BE=BD$ (both are radii of the same circle), $\\triangle{BED}$ is isosceles, meaning ... |
https://artofproblemsolving.com/wiki/index.php/2020_AIME_II_Problems/Problem_7 | null | 298 | Two congruent right circular cones each with base radius $3$ and height $8$ have the axes of symmetry that intersect at right angles at a point in the interior of the cones a distance $3$ from the base of each cone. A sphere with radius $r$ lies within both cones. The maximum possible value of $r^2$ is $\frac{m}{n}$ , ... | [
"Using the diagram above, we notice that the desired length is simply the distance between the point $C$ and $\\overline{AB}$ . We can mark $C$ as $(3,3)$ since it is $3$ units away from each of the bases. Point $B$ is $(8,3)$ . Thus, line $\\overline{AB}$ is $y = \\frac{3}{8}x \\Rightarrow 3x + 8y = 0$ . We can us... |
https://artofproblemsolving.com/wiki/index.php/2011_AMC_8_Problems/Problem_13 | C | 20 | Two congruent squares, $ABCD$ and $PQRS$ , have side length $15$ . They overlap to form the $15$ by $25$ rectangle $AQRD$ shown. What percent of the area of rectangle $AQRD$ is shaded?
[asy] filldraw((0,0)--(25,0)--(25,15)--(0,15)--cycle,white,black); label("D",(0,0),S); label("R",(25,0),S); label("Q",(25,15),N); label... | [
"The overlap length is $2(15)-25=5$ , so the shaded area is $5 \\cdot 15 =75$ . The area of the whole shape is $25 \\cdot 15 = 375$ . The fraction $\\dfrac{75}{375}$ reduces to $\\dfrac{1}{5}$ or 20%. Therefore, the answer is $\\boxed{20}$"
] |
https://artofproblemsolving.com/wiki/index.php/2016_AIME_I_Problems/Problem_2 | null | 71 | Two dice appear to be normal dice with their faces numbered from $1$ to $6$ , but each die is weighted so that the probability of rolling the number $k$ is directly proportional to $k$ . The probability of rolling a $7$ with this pair of dice is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. ... | [
"It is easier to think of the dice as $21$ sided dice with $6$ sixes, $5$ fives, etc. Then there are $21^2=441$ possible rolls. There are $2\\cdot(1\\cdot 6+2\\cdot 5+3\\cdot 4)=56$ rolls that will result in a seven. The odds are therefore $\\frac{56}{441}=\\frac{8}{63}$ . The answer is $8+63=\\boxed{071}$",
"... |
https://artofproblemsolving.com/wiki/index.php/2016_AMC_10B_Problems/Problem_12 | D | 0.7 | Two different numbers are selected at random from $\{1, 2, 3, 4, 5\}$ and multiplied together. What is the probability that the product is even?
$\textbf{(A)}\ 0.2\qquad\textbf{(B)}\ 0.4\qquad\textbf{(C)}\ 0.5\qquad\textbf{(D)}\ 0.7\qquad\textbf{(E)}\ 0.8$ | [
"The product will be even if at least one selected number is even, and odd if none are. Using complementary counting, the chance that both numbers are odd is $\\frac{\\tbinom32}{\\tbinom52}=\\frac3{10}$ , so the answer is $1-0.3$ which is $\\boxed{0.7}$",
"There are $2$ cases to get an even number. Case 1: $\\tex... |
https://artofproblemsolving.com/wiki/index.php/2019_AIME_II_Problems/Problem_1 | null | 59 | Two different points, $C$ and $D$ , lie on the same side of line $AB$ so that $\triangle ABC$ and $\triangle BAD$ are congruent with $AB = 9$ $BC=AD=10$ , and $CA=DB=17$ . The intersection of these two triangular regions has area $\tfrac mn$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | [
" - Diagram by Brendanb4321\nExtend $AB$ to form a right triangle with legs $6$ and $8$ such that $AD$ is the hypotenuse and connect the points $CD$ so \nthat you have a rectangle. (We know that $\\triangle ADE$ is a $6-8-10$ , since $\\triangle DEB$ is an $8-15-17$ .) The base $CD$ of the rectangle will be $9+6+6=... |
https://artofproblemsolving.com/wiki/index.php/2002_AMC_12A_Problems/Problem_13 | C | 5 | Two different positive numbers $a$ and $b$ each differ from their reciprocals by $1$ . What is $a+b$
$\text{(A) }1 \qquad \text{(B) }2 \qquad \text{(C) }\sqrt 5 \qquad \text{(D) }\sqrt 6 \qquad \text{(E) }3$ | [
"Each of the numbers $a$ and $b$ is a solution to $\\left| x - \\frac 1x \\right| = 1$\nHence it is either a solution to $x - \\frac 1x = 1$ , or to $\\frac 1x - x = 1$ . Then it must be a solution either to $x^2 - x - 1 = 0$ , or to $x^2 + x - 1 = 0$\nThere are in total four such values of $x$ , namely $\\frac{\\p... |
https://artofproblemsolving.com/wiki/index.php/2000_AMC_10_Problems/Problem_11 | C | 119 | Two different prime numbers between $4$ and $18$ are chosen. When their sum is subtracted from their product, which of the following numbers could be obtained?
$\textbf{(A)}\ 22 \qquad\textbf{(B)}\ 60 \qquad\textbf{(C)}\ 119 \qquad\textbf{(D)}\ 180 \qquad\textbf{(E)}\ 231$ | [
"Any two prime numbers between 4 and 18 have an odd product and an even sum. Any odd number minus an even number is an odd number, so we can eliminate A, B, and D. Since the highest two prime numbers we can pick are 13 and 17, the highest number we can make is $(13)(17)-(13+17) = 221 - 30 = 191$ . Thus, we can elim... |
https://artofproblemsolving.com/wiki/index.php/2000_AMC_12_Problems/Problem_6 | C | 119 | Two different prime numbers between $4$ and $18$ are chosen. When their sum is subtracted from their product, which of the following numbers could be obtained?
$\textbf{(A)}\ 22 \qquad\textbf{(B)}\ 60 \qquad\textbf{(C)}\ 119 \qquad\textbf{(D)}\ 180 \qquad\textbf{(E)}\ 231$ | [
"Any two prime numbers between 4 and 18 have an odd product and an even sum. Any odd number minus an even number is an odd number, so we can eliminate A, B, and D. Since the highest two prime numbers we can pick are 13 and 17, the highest number we can make is $(13)(17)-(13+17) = 221 - 30 = 191$ . Thus, we can elim... |
https://artofproblemsolving.com/wiki/index.php/2004_AMC_10A_Problems/Problem_21 | B | 7 | Two distinct lines pass through the center of three concentric circles of radii 3, 2, and 1. The area of the shaded region in the diagram is $\frac{8}{13}$ of the area of the unshaded region. What is the radian measure of the acute angle formed by the two lines? (Note: $\pi$ radians is $180$ degrees.)
$\mathrm{(A) \ } ... | [
"Let the area of the shaded region be $S$ , the area of the unshaded region be $U$ , and the acute angle that is formed by the two lines be $\\theta$ . We can set up two equations between $S$ and $U$\n$S+U=9\\pi$\n$S=\\dfrac{8}{13}U$\nThus $\\dfrac{21}{13}U=9\\pi$ , and $U=\\dfrac{39\\pi}{7}$ , and thus $S=\\dfrac{... |
https://artofproblemsolving.com/wiki/index.php/2021_AMC_12B_Problems/Problem_10 | E | 10 | Two distinct numbers are selected from the set $\{1,2,3,4,\dots,36,37\}$ so that the sum of the remaining $35$ numbers is the product of these two numbers. What is the difference of these two numbers?
$\textbf{(A) }5 \qquad \textbf{(B) }7 \qquad \textbf{(C) }8\qquad \textbf{(D) }9 \qquad \textbf{(E) }10$ | [
"The sum of the first $n$ integers is given by $\\frac{n(n+1)}{2}$ , so $\\frac{37(37+1)}{2}=703$\nTherefore, $703-x-y=xy$\nRearranging, $xy+x+y=703$ . We can factor this equation by SFFT to get\n$(x+1)(y+1)=704$\nLooking at the possible divisors of $704 = 2^6\\cdot11$ $22$ and $32$ are within the constraints of $0... |
https://artofproblemsolving.com/wiki/index.php/2002_AIME_II_Problems/Problem_11 | null | 518 | Two distinct, real, infinite geometric series each have a sum of $1$ and have the same second term. The third term of one of the series is $1/8$ , and the second term of both series can be written in the form $\frac{\sqrt{m}-n}p$ , where $m$ $n$ , and $p$ are positive integers and $m$ is not divisible by the square of ... | [
"Let the second term of each series be $x$ . Then, the common ratio is $\\frac{1}{8x}$ , and the first term is $8x^2$\nSo, the sum is $\\frac{8x^2}{1-\\frac{1}{8x}}=1$ . Thus, $64x^3-8x+1 = (4x-1)(16x^2+4x-1) = 0 \\Rightarrow x = \\frac{1}{4}, \\frac{-1 \\pm \\sqrt{5}}{8}$\nThe only solution in the appropriate form... |
https://artofproblemsolving.com/wiki/index.php/1952_AHSME_Problems/Problem_8 | A | 1 | Two equal circles in the same plane cannot have the following number of common tangents.
$\textbf{(A) \ }1 \qquad \textbf{(B) \ }2 \qquad \textbf{(C) \ }3 \qquad \textbf{(D) \ }4 \qquad \textbf{(E) \ }\text{none of these}$ | [
"Two congruent coplanar circles will either be tangent to one another (resulting in $3$ common tangents), intersect one another (resulting in $2$ common tangents), or be separate from one another (resulting in $4$ common tangents).\nHaving only $\\boxed{1}$ common tangent is impossible, unless the circles are non-c... |
https://artofproblemsolving.com/wiki/index.php/2022_AIME_II_Problems/Problem_15 | null | 140 | Two externally tangent circles $\omega_1$ and $\omega_2$ have centers $O_1$ and $O_2$ , respectively. A third circle $\Omega$ passing through $O_1$ and $O_2$ intersects $\omega_1$ at $B$ and $C$ and $\omega_2$ at $A$ and $D$ , as shown. Suppose that $AB = 2$ $O_1O_2 = 15$ $CD = 16$ , and $ABO_1CDO_2$ is a convex hexago... | [
"First observe that $AO_2 = O_2D$ and $BO_1 = O_1C$ . Let points $A'$ and $B'$ be the reflections of $A$ and $B$ , respectively, about the perpendicular bisector of $\\overline{O_1O_2}$ . Then quadrilaterals $ABO_1O_2$ and $B'A'O_2O_1$ are congruent, so hexagons $ABO_1CDO_2$ and $A'B'O_1CDO_2$ have the same area. F... |
https://artofproblemsolving.com/wiki/index.php/2021_AMC_12B_Problems/Problem_19 | B | 17 | Two fair dice, each with at least $6$ faces are rolled. On each face of each die is printed a distinct integer from $1$ to the number of faces on that die, inclusive. The probability of rolling a sum of $7$ is $\frac34$ of the probability of rolling a sum of $10,$ and the probability of rolling a sum of $12$ is $\frac{... | [
"Suppose the dice have $a$ and $b$ faces, and WLOG $a\\geq{b}$ . Since each die has at least $6$ faces, there will always be $6$ ways to sum to $7$ . As a result, there must be $\\tfrac{4}{3}\\cdot6=8$ ways to sum to $10$ . There are at most nine distinct ways to get a sum of $10$ , which are possible whenever $a,b... |
https://artofproblemsolving.com/wiki/index.php/2012_AIME_II_Problems/Problem_2 | null | 363 | Two geometric sequences $a_1, a_2, a_3, \ldots$ and $b_1, b_2, b_3, \ldots$ have the same common ratio, with $a_1 = 27$ $b_1=99$ , and $a_{15}=b_{11}$ . Find $a_9$ | [
"Call the common ratio $r.$ Now since the $n$ th term of a geometric sequence with first term $x$ and common ratio $y$ is $xy^{n-1},$ we see that $a_1 \\cdot r^{14} = b_1 \\cdot r^{10} \\implies r^4 = \\frac{99}{27} = \\frac{11}{3}.$ But $a_9$ equals $a_1 \\cdot r^8 = a_1 \\cdot (r^4)^2=27\\cdot {\\left(\\frac{11}{... |
https://artofproblemsolving.com/wiki/index.php/1952_AHSME_Problems/Problem_2 | B | 74 | Two high school classes took the same test. One class of $20$ students made an average grade of $80\%$ ; the other class of $30$ students made an average grade of $70\%$ . The average grade for all students in both classes is:
$\textbf{(A)}\ 75\%\qquad \textbf{(B)}\ 74\%\qquad \textbf{(C)}\ 72\%\qquad \textbf{(D)}\ 77\... | [
"The desired average can be found by dividing the total number of points earned by the total number of students. There are $20\\cdot 80+30\\cdot 70=3700$ points earned and $20+30=50$ students. Thus, our answer is $\\frac{3700}{50}$ , or $\\boxed{74}$"
] |
https://artofproblemsolving.com/wiki/index.php/2023_AMC_8_Problems/Problem_20 | D | 60 | Two integers are inserted into the list $3, 3, 8, 11, 28$ to double its range. The mode and median remain unchanged. What is the maximum possible sum of the two additional numbers?
$\textbf{(A) } 56 \qquad \textbf{(B) } 57 \qquad \textbf{(C) } 58 \qquad \textbf{(D) } 60 \qquad \textbf{(E) } 61$ | [
"To double the range, we must find the current range, which is $28 - 3 = 25$ , to then double to: $2(25) = 50$ . Since we do not want to change the median, we need to get a value less than $8$ (as $8$ would change the mode) for the smaller, making $53$ fixed for the larger. Remember, anything less than $3$ is not b... |
https://artofproblemsolving.com/wiki/index.php/2012_AMC_12B_Problems/Problem_5 | A | 1 | Two integers have a sum of $26$ . when two more integers are added to the first two, the sum is $41$ . Finally, when two more integers are added to the sum of the previous $4$ integers, the sum is $57$ . What is the minimum number of even integers among the $6$ integers?
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\text... | [
"Since, $x + y = 26$ $x$ can equal $15$ , and $y$ can equal $11$ , so no even integers are required to make 26. To get to $41$ , we have to add $41 - 26 = 15$ . If $a+b=15$ , at least one of $a$ and $b$ must be even because two odd numbers sum to an even number. Therefore, one even integer is required when transiti... |
https://artofproblemsolving.com/wiki/index.php/2012_AMC_10B_Problems/Problem_9 | A | 1 | Two integers have a sum of 26. When two more integers are added to the first two integers the sum is 41. Finally when two more integers are added to the sum of the previous four integers the sum is 57. What is the minimum number of odd integers among the 6 integers?
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C... | [
"Out of the first two integers, it's possible for both to be even: for example, $10 + 16 = 26.$ But the next two integers, when added, increase the sum by $15,$ which is odd, so one of them must be odd and the other must be even: for example, $3 + 12 = 15.$ Finally, the next two integers increase the sum by $16,$ w... |
https://artofproblemsolving.com/wiki/index.php/2019_AMC_10B_Problems/Problem_11 | A | 5 | Two jars each contain the same number of marbles, and every marble is either blue or green. In Jar $1$ the ratio of blue to green marbles is $9:1$ , and the ratio of blue to green marbles in Jar $2$ is $8:1$ . There are $95$ green marbles in all. How many more blue marbles are in Jar $1$ than in Jar $2$
$\textbf{(A) } ... | [
"Call the number of marbles in each jar $x$ (because the problem specifies that they each contain the same number). Thus, $\\frac{x}{10}$ is the number of green marbles in Jar $1$ , and $\\frac{x}{9}$ is the number of green marbles in Jar $2$ . Since $\\frac{x}{9}+\\frac{x}{10}=\\frac{19x}{90}$ , we have $\\frac{19... |
https://artofproblemsolving.com/wiki/index.php/2019_AMC_10B_Problems/Problem_11 | B | 5 | Two jars each contain the same number of marbles, and every marble is either blue or green. In Jar $1$ the ratio of blue to green marbles is $9:1$ , and the ratio of blue to green marbles in Jar $2$ is $8:1$ . There are $95$ green marbles in all. How many more blue marbles are in Jar $1$ than in Jar $2$
$\textbf{(A) } ... | [
"Writing out to ratios, we have $9:1$ in jar $1$ and $8:1$ in jar $2$ . Since the jar must have to same amount of marbles, let's make a variable $a$ and $b$ for each of the ratios to be multiplied by. Now we would have $9a + a = 8b + b \\rightarrow 10a = 9b$ . We can take the most obvious values of $a$ and $b$ and ... |
https://artofproblemsolving.com/wiki/index.php/2019_AMC_10A_Problems/Problem_7 | C | 6 | Two lines with slopes $\dfrac{1}{2}$ and $2$ intersect at $(2,2)$ . What is the area of the triangle enclosed by these two lines and the line $x+y=10 ?$
$\textbf{(A) } 4 \qquad\textbf{(B) } 4\sqrt{2} \qquad\textbf{(C) } 6 \qquad\textbf{(D) } 8 \qquad\textbf{(E) } 6\sqrt{2}$ | [
"Let's first work out the slope-intercept form of all three lines: $(x,y)=(2,2)$ and $y=\\frac{x}{2} + b$ implies $2=\\frac{2}{2} +b=1+b$ so $b=1$ , while $y=2x + c$ implies $2= 2 \\cdot 2+c=4+c$ so $c=-2$ . Also, $x+y=10$ implies $y=-x+10$ . Thus the lines are $y=\\frac{x}{2} +1, y=2x-2,$ and $y=-x+10$ . \nNow we ... |
https://artofproblemsolving.com/wiki/index.php/2019_AMC_12A_Problems/Problem_5 | C | 6 | Two lines with slopes $\dfrac{1}{2}$ and $2$ intersect at $(2,2)$ . What is the area of the triangle enclosed by these two lines and the line $x+y=10 ?$
$\textbf{(A) } 4 \qquad\textbf{(B) } 4\sqrt{2} \qquad\textbf{(C) } 6 \qquad\textbf{(D) } 8 \qquad\textbf{(E) } 6\sqrt{2}$ | [
"Let's first work out the slope-intercept form of all three lines: $(x,y)=(2,2)$ and $y=\\frac{x}{2} + b$ implies $2=\\frac{2}{2} +b=1+b$ so $b=1$ , while $y=2x + c$ implies $2= 2 \\cdot 2+c=4+c$ so $c=-2$ . Also, $x+y=10$ implies $y=-x+10$ . Thus the lines are $y=\\frac{x}{2} +1, y=2x-2,$ and $y=-x+10$ . \nNow we ... |
https://artofproblemsolving.com/wiki/index.php/2007_AIME_II_Problems/Problem_11 | null | 179 | Two long cylindrical tubes of the same length but different diameters lie parallel to each other on a flat surface . The larger tube has radius $72$ and rolls along the surface toward the smaller tube, which has radius $24$ . It rolls over the smaller tube and continues rolling along the flat surface until it comes to ... | [
"2007 AIME II-11.png\nIf it weren’t for the small tube, the larger tube would travel $144\\pi$ . Consider the distance from which the larger tube first contacts the smaller tube, until when it completely loses contact with the smaller tube.\nDrawing the radii as shown in the diagram, notice that the hypotenuse of t... |
https://artofproblemsolving.com/wiki/index.php/1963_AHSME_Problems/Problem_20 | D | 4 | Two men at points $R$ and $S$ $76$ miles apart, set out at the same time to walk towards each other.
The man at $R$ walks uniformly at the rate of $4\tfrac{1}{2}$ miles per hour; the man at $S$ walks at the constant
rate of $3\tfrac{1}{4}$ miles per hour for the first hour, at $3\tfrac{3}{4}$ miles per hour for the s... | [
"First, find the number of hours it takes for the two to meet together. After $h$ hours, the person at $R$ walks $4.5h$ miles. In the same amount of time, the person at $S$ has been walking at $3.25+0.5(h-1)$ mph for the past hour, so the person walks $\\frac{h(6.5+0.5(h-1))}{2}$ miles.\nIn order for both to meet... |
https://artofproblemsolving.com/wiki/index.php/2013_AMC_10B_Problems/Problem_21 | C | 104 | Two non-decreasing sequences of nonnegative integers have different first terms. Each sequence has the property that each term beginning with the third is the sum of the previous two terms, and the seventh term of each sequence is $N$ . What is the smallest possible value of $N$
$\textbf{(A)}\ 55 \qquad \textbf{(B)}\ 8... | [
"Let the first two terms of the first sequence be $x_{1}$ and $x_{2}$ and the first two of the second sequence be $y_{1}$ and $y_{2}$ . Computing the seventh term, we see that $5x_{1} + 8x_{2} = 5y_{1} + 8y_{2}$ . Note that this means that $x_{1}$ and $y_{1}$ must have the same value modulo $8$ . To minimize, le... |
https://artofproblemsolving.com/wiki/index.php/2013_AMC_12B_Problems/Problem_14 | C | 104 | Two non-decreasing sequences of nonnegative integers have different first terms. Each sequence has the property that each term beginning with the third is the sum of the previous two terms, and the seventh term of each sequence is $N$ . What is the smallest possible value of $N$
$\textbf{(A)}\ 55 \qquad \textbf{(B)}\ 8... | [
"Let the first two terms of the first sequence be $x_{1}$ and $x_{2}$ and the first two of the second sequence be $y_{1}$ and $y_{2}$ . Computing the seventh term, we see that $5x_{1} + 8x_{2} = 5y_{1} + 8y_{2}$ . Note that this means that $x_{1}$ and $y_{1}$ must have the same value modulo $8$ . To minimize, le... |
https://artofproblemsolving.com/wiki/index.php/2010_AIME_II_Problems/Problem_12 | null | 676 | Two noncongruent integer-sided isosceles triangles have the same perimeter and the same area. The ratio of the lengths of the bases of the two triangles is $8: 7$ . Find the minimum possible value of their common perimeter. | [
"Let $s$ be the semiperimeter of the two triangles. Also, let the base of the longer triangle be $16x$ and the base of the shorter triangle be $14x$ for some arbitrary factor $x$ . Then, the dimensions of the two triangles must be $s-8x,s-8x,16x$ and $s-7x,s-7x,14x$ . By Heron's Formula, we have\nSince $15$ and $33... |
https://artofproblemsolving.com/wiki/index.php/2020_AMC_12B_Problems/Problem_7 | C | 32 | Two nonhorizontal, non vertical lines in the $xy$ -coordinate plane intersect to form a $45^{\circ}$ angle. One line has slope equal to $6$ times the slope of the other line. What is the greatest possible value of the product of the slopes of the two lines?
$\textbf{(A)}\ \frac16 \qquad\textbf{(B)}\ \frac23 \qquad\text... | [
"Intersect at the origin and select a point on each line to define vectors $\\mathbf{v}_{i}=(x_{i},y_{i})$ .\nNote that $\\theta=45^{\\circ}$ gives equal magnitudes of the vector products \\[\\mathbf{v}_1\\cdot\\mathbf{v}_2 = v_{1}v_{2}\\cos\\theta \\quad\\mathrm{and}\\quad |\\mathbf{v}_1\\times\\mathbf{v}_2| = v_{... |
https://artofproblemsolving.com/wiki/index.php/1964_AHSME_Problems/Problem_23 | D | 48 | Two numbers are such that their difference, their sum, and their product are to one another as $1:7:24$ . The product of the two numbers is:
$\textbf{(A)}\ 6\qquad \textbf{(B)}\ 12\qquad \textbf{(C)}\ 24\qquad \textbf{(D)}\ 48\qquad \textbf{(E)}\ 96$ | [
"Set the two numbers as $x$ and $y$ . Therefore, $x+y=7(x-y), xy=24(x-y)$ , and $24(x+y)=7xy$ . Simplifying the first equation gives $y=\\frac{3}{4}x$ . Substituting for $y$ in the second equation gives $\\frac{3}{4}x^2=6x.$ Solving yields $x=8$ or $x=0$ . Substituting $x=0$ back into the first equation yields $1=... |
https://artofproblemsolving.com/wiki/index.php/2015_AMC_12A_Problems/Problem_2 | E | 72 | Two of the three sides of a triangle are 20 and 15. Which of the following numbers is not a possible perimeter of the triangle?
$\textbf{(A)}\ 52\qquad\textbf{(B)}\ 57\qquad\textbf{(C)}\ 62\qquad\textbf{(D)}\ 67\qquad\textbf{(E)}\ 72$ | [
"Letting $x$ be the third side, then by the triangle inequality, $20-15 < x < 20+15$ , or $5 < x < 35$ . Therefore the perimeter must be greater than 40 but less than 70. 72 is not in this range, so $\\boxed{72}$ is our answer."
] |
https://artofproblemsolving.com/wiki/index.php/2003_AIME_II_Problems/Problem_10 | null | 156 | Two positive integers differ by $60$ . The sum of their square roots is the square root of an integer that is not a perfect square. What is the maximum possible sum of the two integers? | [
"Call the two integers $b$ and $b+60$ , so we have $\\sqrt{b}+\\sqrt{b+60}=\\sqrt{c}$ . Square both sides to get $2b+60+2\\sqrt{b^2+60b}=c$ . Thus, $b^2+60b$ must be a square, so we have $b^2+60b=n^2$ , and $(b+n+30)(b-n+30)=900$ . The sum of these two factors is $2b+60$ , so they must both be even. To maximize $b$... |
https://artofproblemsolving.com/wiki/index.php/2010_AMC_12A_Problems/Problem_25 | C | 568 | Two quadrilaterals are considered the same if one can be obtained from the other by a rotation and a translation. How many different convex cyclic quadrilaterals are there with integer sides and perimeter equal to 32?
$\textbf{(A)}\ 560 \qquad \textbf{(B)}\ 564 \qquad \textbf{(C)}\ 568 \qquad \textbf{(D)}\ 1498 \qquad ... | [
"It should first be noted that given any quadrilateral of fixed side lengths, there is exactly one way to manipulate the angles so that the quadrilateral becomes cyclic.\nProof. Given a quadrilateral $ABCD$ where all sides are fixed (in a certain order), we can construct the diagonal $\\overline{BD}$ . When $BD$ is... |
https://artofproblemsolving.com/wiki/index.php/2017_AIME_I_Problems/Problem_8 | null | 41 | Two real numbers $a$ and $b$ are chosen independently and uniformly at random from the interval $(0, 75)$ . Let $O$ and $P$ be two points on the plane with $OP = 200$ . Let $Q$ and $R$ be on the same side of line $OP$ such that the degree measures of $\angle POQ$ and $\angle POR$ are $a$ and $b$ respectively, and $\ang... | [
"Noting that $\\angle OQP$ and $\\angle ORP$ are right angles, we realize that we can draw a semicircle with diameter $\\overline{OP}$ and points $Q$ and $R$ on the semicircle. Since the radius of the semicircle is $100$ , if $\\overline{QR} \\leq 100$ , then $\\overarc{QR}$ must be less than or equal to $60^{\\cir... |
https://artofproblemsolving.com/wiki/index.php/2011_AMC_10B_Problems/Problem_13 | D | 59 | Two real numbers are selected independently at random from the interval $[-20, 10]$ . What is the probability that the product of those numbers is greater than zero?
$\textbf{(A)}\ \frac{1}{9} \qquad\textbf{(B)}\ \frac{1}{3} \qquad\textbf{(C)}\ \frac{4}{9} \qquad\textbf{(D)}\ \frac{5}{9} \qquad\textbf{(E)}\ \frac{2}{3}... | [
"We will use complementary counting. The probability that the product is negative can be found by finding the probability that one number is positive and the other number is negative. The probability of a positive number being selected is $\\frac13$ , and the probability of a negative number being selected is $\\fr... |
https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_12 | E | 4 | Two right circular cones with vertices facing down as shown in the figure below contain the same amount of liquid. The radii of the tops of the liquid surfaces are $3$ cm and $6$ cm. Into each cone is dropped a spherical marble of radius $1$ cm, which sinks to the bottom and is completely submerged without spilling any... | [
"The heights of the cones are not given, so suppose the heights are very large (i.e. tending towards infinity) in order to approximate the cones as cylinders with base radii $3$ and $6$ and infinitely large height. Then the base area of the wide cylinder is $4$ times that of the narrow cylinder. Since we are droppi... |
https://artofproblemsolving.com/wiki/index.php/2021_AMC_12A_Problems/Problem_10 | E | 4 | Two right circular cones with vertices facing down as shown in the figure below contain the same amount of liquid. The radii of the tops of the liquid surfaces are $3$ cm and $6$ cm. Into each cone is dropped a spherical marble of radius $1$ cm, which sinks to the bottom and is completely submerged without spilling any... | [
"The heights of the cones are not given, so suppose the heights are very large (i.e. tending towards infinity) in order to approximate the cones as cylinders with base radii $3$ and $6$ and infinitely large height. Then the base area of the wide cylinder is $4$ times that of the narrow cylinder. Since we are droppi... |
https://artofproblemsolving.com/wiki/index.php/2015_AMC_10A_Problems/Problem_9 | D | 21 | Two right circular cylinders have the same volume. The radius of the second cylinder is $10\%$ more than the radius of the first. What is the relationship between the heights of the two cylinders?
$\textbf{(A)}\ \text{The second height is } 10\% \text{ less than the first.} \\ \textbf{(B)}\ \text{The first height is ... | [
"Let the radius of the first cylinder be $r_1$ and the radius of the second cylinder be $r_2$ . Also, let the height of the first cylinder be $h_1$ and the height of the second cylinder be $h_2$ . We are told \\[r_2=\\frac{11r_1}{10}\\] \\[\\pi r_1^2h_1=\\pi r_2^2h_2\\] Substituting the first equation into the seco... |
https://artofproblemsolving.com/wiki/index.php/2015_AMC_12A_Problems/Problem_7 | D | 21 | Two right circular cylinders have the same volume. The radius of the second cylinder is $10\%$ more than the radius of the first. What is the relationship between the heights of the two cylinders?
$\textbf{(A)}\ \text{The second height is } 10\% \text{ less than the first.} \\ \textbf{(B)}\ \text{The first height is ... | [
"Let the radius of the first cylinder be $r_1$ and the radius of the second cylinder be $r_2$ . Also, let the height of the first cylinder be $h_1$ and the height of the second cylinder be $h_2$ . We are told \\[r_2=\\frac{11r_1}{10}\\] \\[\\pi r_1^2h_1=\\pi r_2^2h_2\\] Substituting the first equation into the seco... |
https://artofproblemsolving.com/wiki/index.php/2013_AMC_10A_Problems/Problem_15 | D | 12 | Two sides of a triangle have lengths $10$ and $15$ . The length of the altitude to the third side is the average of the lengths of the altitudes to the two given sides. How long is the third side?
$\textbf{(A)}\ 6 \qquad\textbf{(B)}\ 8 \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ 12 \qquad\textbf{(E)}\ 18$ | [
"The shortest side length has the longest altitude perpendicular to it. The average of the two altitudes given will be between the lengths of the two altitudes,\ntherefore the length of the side perpendicular to that altitude will be between $10$ and $15$ . The only answer choice that meets this requirement is $\\b... |
https://artofproblemsolving.com/wiki/index.php/2013_AMC_10A_Problems/Problem_15 | null | 12 | Two sides of a triangle have lengths $10$ and $15$ . The length of the altitude to the third side is the average of the lengths of the altitudes to the two given sides. How long is the third side?
$\textbf{(A)}\ 6 \qquad\textbf{(B)}\ 8 \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ 12 \qquad\textbf{(E)}\ 18$ | [
"Let $x$ be the height of triangle when the base is $10$ and $y$ is the height of the triangles when the base is $15$ . This means the height for when the triangles has the $3$ rd side length, the height would be $\\dfrac{(x+y)}{2}$ giving us the following equation:\n$\\dfrac{10x}{2}=\\dfrac{15y}{2}=\\dfrac{n(x+y)}... |
https://artofproblemsolving.com/wiki/index.php/1989_AIME_Problems/Problem_6 | null | 160 | Two skaters, Allie and Billie, are at points $A$ and $B$ , respectively, on a flat, frozen lake. The distance between $A$ and $B$ is $100$ meters. Allie leaves $A$ and skates at a speed of $8$ meters per second on a straight line that makes a $60^\circ$ angle with $AB$ . At the same time Allie leaves $A$ , Billie leave... | [
"Label the point of intersection as $C$ . Since $d = rt$ $AC = 8t$ and $BC = 7t$ . According to the law of cosines\n\\begin{align*}(7t)^2 &= (8t)^2 + 100^2 - 2 \\cdot 8t \\cdot 100 \\cdot \\cos 60^\\circ\\\\ 0 &= 15t^2 - 800t + 10000 = 3t^2 - 160t + 2000\\\\ t &= \\frac{160 \\pm \\sqrt{160^2 - 4\\cdot 3 \\cdot 2000... |
https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_10 | null | 335 | Two spheres with radii $36$ and one sphere with radius $13$ are each externally tangent to the other two spheres and to two different planes $\mathcal{P}$ and $\mathcal{Q}$ . The intersection of planes $\mathcal{P}$ and $\mathcal{Q}$ is the line $\ell$ . The distance from line $\ell$ to the point where the sphere with ... | [
"This solution refers to the Diagram section.\nAs shown below, let $O_1,O_2,O_3$ be the centers of the spheres (where sphere $O_3$ has radius $13$ ) and $T_1,T_2,T_3$ be their respective points of tangency to plane $\\mathcal{P}.$ Let $\\mathcal{R}$ be the plane that is determined by $O_1,O_2,$ and $O_3.$ Suppose $... |
https://artofproblemsolving.com/wiki/index.php/1996_AIME_Problems/Problem_7 | null | 300 | Two squares of a $7\times 7$ checkerboard are painted yellow, and the rest are painted green. Two color schemes are equivalent if one can be obtained from the other by applying a rotation in the plane board. How many inequivalent color schemes are possible? | [
"There are ${49 \\choose 2}$ possible ways to select two squares to be painted yellow. There are four possible ways to rotate each board. Given an arbitrary pair of yellow squares, these four rotations will either yield two or four equivalent but distinct boards.\nNote that a pair of yellow squares will only yield ... |
https://artofproblemsolving.com/wiki/index.php/2008_AMC_10A_Problems/Problem_23 | B | 40 | Two subsets of the set $S=\lbrace a,b,c,d,e\rbrace$ are to be chosen so that their union is $S$ and their intersection contains exactly two elements. In how many ways can this be done, assuming that the order in which the subsets are chosen does not matter?
$\mathrm{(A)}\ 20\qquad\mathrm{(B)}\ 40\qquad\mathrm{(C)}\ 60\... | [
"First, choose the two letters to be repeated in each set. $\\dbinom{5}{2}=10$ . Now we have three remaining elements that we wish to place into two separate subsets. There are $2^3 = 8$ ways to do so because each of the three remaining letters can be placed either into the first or second subset. Both of those su... |
https://artofproblemsolving.com/wiki/index.php/1960_AHSME_Problems/Problem_34 | C | 20 | Two swimmers, at opposite ends of a $90$ -foot pool, start to swim the length of the pool,
one at the rate of $3$ feet per second, the other at $2$ feet per second.
They swim back and forth for $12$ minutes. Allowing no loss of times at the turns, find the number of times they pass each other.
$\textbf{(A)}\ 24\qquad... | [
"First, note that it will take $30$ seconds for the first swimmer to reach the other side and $45$ seconds for the second swimmer to reach the other side. Also, note that after $180$ seconds (or $3$ minutes), both swimmers will complete an even number of laps, essentially returning to their starting point.\n\nAt t... |
https://artofproblemsolving.com/wiki/index.php/1961_AHSME_Problems/Problem_11 | C | 40 | Two tangents are drawn to a circle from an exterior point $A$ ; they touch the circle at points $B$ and $C$ respectively.
A third tangent intersects segment $AB$ in $P$ and $AC$ in $R$ , and touches the circle at $Q$ . If $AB=20$ , then the perimeter of $\triangle APR$ is
$\textbf{(A)}\ 42\qquad \textbf{(B)}\ 40.5 \qq... | [
"Since $Q$ can be anywhere on the circle between $A$ and $B$ , it can basically be \"on top\" of $A$ . Then $R$ will be at the same point as $A$ , so $APR$ form a degenerate triable with side length $AB=20$ . So its perimeter will be $40$ . Since $BP=PQ$ and $QR=CR$ by power of a point, as $AP$ and $AR$ decrease in... |
https://artofproblemsolving.com/wiki/index.php/2011_AMC_12B_Problems/Problem_6 | C | 36 | Two tangents to a circle are drawn from a point $A$ . The points of contact $B$ and $C$ divide the circle into arcs with lengths in the ratio $2 : 3$ . What is the degree measure of $\angle{BAC}$
$\textbf{(A)}\ 24 \qquad \textbf{(B)}\ 30 \qquad \textbf{(C)}\ 36 \qquad \textbf{(D)}\ 48 \qquad \textbf{(E)}\ 60$ | [
"In order to solve this problem, use of the tangent-tangent intersection theorem (Angle of intersection between two tangents dividing a circle into arc length A and arc length B = 1/2 (Arc A° - Arc B°).\nIn order to utilize this theorem, the degree measures of the arcs must be found. First, set A (Arc length A) eq... |
https://artofproblemsolving.com/wiki/index.php/1993_AIME_Problems/Problem_9 | null | 118 | Two thousand points are given on a circle . Label one of the points $1$ . From this point, count $2$ points in the clockwise direction and label this point $2$ . From the point labeled $2$ , count $3$ points in the clockwise direction and label this point $3$ . (See figure.) Continue this process until the labels $1,2,... | [
"The label $1993$ will occur on the $\\frac12(1993)(1994) \\pmod{2000}$ th point around the circle. (Starting from 1) A number $n$ will only occupy the same point on the circle if $\\frac12(n)(n + 1)\\equiv \\frac12(1993)(1994) \\pmod{2000}$\nSimplifying this expression, we see that $(1993)(1994) - (n)(n + 1) = (19... |
https://artofproblemsolving.com/wiki/index.php/1991_AIME_Problems/Problem_10 | null | 532 | Two three-letter strings, $aaa^{}_{}$ and $bbb^{}_{}$ , are transmitted electronically. Each string is sent letter by letter. Due to faulty equipment, each of the six letters has a 1/3 chance of being received incorrectly, as an $a^{}_{}$ when it should have been a $b^{}_{}$ , or as a $b^{}_{}$ when it should be an $a^... | [
"Let us make a chart of values in alphabetical order, where $P_a,\\ P_b$ are the probabilities that each string comes from $aaa$ and $bbb$ multiplied by $27$ , and $S_b$ denotes the partial sums of $P_b$ (in other words, $S_b = \\sum_{n=1}^{b} P_b$ ): \\[\\begin{array}{|r||r|r|r|} \\hline \\text{String}&P_a&P_b&S_b... |
https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_5 | null | 90 | Two unit squares are selected at random without replacement from an $n \times n$ grid of unit squares. Find the least positive integer $n$ such that the probability that the two selected unit squares are horizontally or vertically adjacent is less than $\frac{1}{2015}$ | [
"Call the given grid \"Grid A\". Consider Grid B, where the vertices of Grid B fall in the centers of the squares of Grid A; thus, Grid B has dimensions $(n-1) \\times (n-1)$ . There is a one-to-one correspondence between the edges of Grid B and the number of adjacent pairs of unit squares in Grid A. The number of ... |
https://artofproblemsolving.com/wiki/index.php/2002_AMC_12P_Problems/Problem_9 | D | 4 | Two walls and the ceiling of a room meet at right angles at point $P.$ A fly is in the air one meter from one wall, eight meters from the other wall, and nine meters from point $P$ . How many meters is the fly from the ceiling?
$\text{(A) }\sqrt{13} \qquad \text{(B) }\sqrt{14} \qquad \text{(C) }\sqrt{15} \qquad \text{(... | [
"We can use the formula for the diagonal of the rectangle, or $d=\\sqrt{a^2+b^2+c^2}$ The problem gives us $a=1, b=8,$ and $c=9.$ Solving gives us $9=\\sqrt{1^2 + 8^2 + c^2} \\implies c^2=9^2-8^2-1^2 \\implies c^2=16 \\implies c=\\boxed{4}.$"
] |
https://artofproblemsolving.com/wiki/index.php/2015_AMC_10A_Problems/Problem_8 | B | 4 | Two years ago Pete was three times as old as his cousin Claire. Two years before that, Pete was four times as old as Claire. In how many years will the ratio of their ages be $2$ $1$
$\textbf{(A)}\ 2 \qquad\textbf{(B)} \ 4 \qquad\textbf{(C)} \ 5 \qquad\textbf{(D)} \ 6 \qquad\textbf{(E)} \ 8$ | [
"This problem can be converted to a system of equations. Let $p$ be Pete's current age and $c$ be Claire's current age.\nThe first statement can be written as $p-2=3(c-2)$ . The second statement can be written as $p-4=4(c-4)$\nTo solve the system of equations:\n$p=3c-4$\n$p=4c-12$\n$3c-4=4c-12$\n$c=8$\n$p=20.$\nLet... |
https://artofproblemsolving.com/wiki/index.php/2015_AMC_12A_Problems/Problem_6 | B | 4 | Two years ago Pete was three times as old as his cousin Claire. Two years before that, Pete was four times as old as Claire. In how many years will the ratio of their ages be $2$ $1$
$\textbf{(A)}\ 2 \qquad\textbf{(B)} \ 4 \qquad\textbf{(C)} \ 5 \qquad\textbf{(D)} \ 6 \qquad\textbf{(E)} \ 8$ | [
"This problem can be converted to a system of equations. Let $p$ be Pete's current age and $c$ be Claire's current age.\nThe first statement can be written as $p-2=3(c-2)$ . The second statement can be written as $p-4=4(c-4)$\nTo solve the system of equations:\n$p=3c-4$\n$p=4c-12$\n$3c-4=4c-12$\n$c=8$\n$p=20.$\nLet... |
https://artofproblemsolving.com/wiki/index.php/2004_AMC_8_Problems/Problem_20 | D | 27 | Two-thirds of the people in a room are seated in three-fourths of the chairs. The rest of the people are standing. If there are $6$ empty chairs, how many people are in the room?
$\textbf{(A)}\ 12\qquad \textbf{(B)}\ 18\qquad \textbf{(C)}\ 24\qquad \textbf{(D)}\ 27\qquad \textbf{(E)}\ 36$ | [
"Working backwards, if $3/4$ of the chairs are taken and $6$ are empty, then there are three times as many taken chairs as empty chairs, or $3 \\cdot 6 = 18$ . If $x$ is the number of people in the room and $2/3$ are seated, then $\\frac23 x = 18$ and $x = \\boxed{27}$"
] |
https://artofproblemsolving.com/wiki/index.php/2001_AMC_8_Problems/Problem_14 | C | 72 | Tyler has entered a buffet line in which he chooses one kind of meat, two different vegetables and one dessert. If the order of food items is not important, how many different meals might he choose?
$\text{(A)}\ 4 \qquad \text{(B)}\ 24 \qquad \text{(C)}\ 72 \qquad \text{(D)}\ 80 \qquad \text{(E)}\ 144$ | [
"There are $3$ possibilities for the meat and $4$ possibilites for the dessert, for a total of $4\\times3=12$ possibilities for the meat and the dessert. There are $4$ possibilities for the first vegetable and $3$ possibilities for the second, but order doesn't matter, so we overcounted by a factor of $2$ . For exa... |
https://artofproblemsolving.com/wiki/index.php/2010_AMC_10A_Problems/Problem_3 | D | 25 | Tyrone had $97$ marbles and Eric had $11$ marbles. Tyrone then gave some of his marbles to Eric so that Tyrone ended with twice as many marbles as Eric. How many marbles did Tyrone give to Eric?
$\mathrm{(A)}\ 3 \qquad \mathrm{(B)}\ 13 \qquad \mathrm{(C)}\ 18 \qquad \mathrm{(D)}\ 25 \qquad \mathrm{(E)}\ 29$ | [
"Let $x$ be the number of marbles Tyrone gave to Eric. Then, $97-x = 2\\cdot(11+x)$ . Solving for $x$ yields $75=3x$ and $x = 25$ . The answer is $\\boxed{25}$",
"Since the number of balls Tyrone and Eric have a specific ratio when Tyrone gives some of his balls to Eric, we can divide the total amount of balls by... |
https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_15 | A | 56 | Usain is walking for exercise by zigzagging across a $100$ -meter by $30$ -meter rectangular field, beginning at point $A$ and ending on the segment $\overline{BC}$ . He wants to increase the distance walked by zigzagging as shown in the figure below $(APQRS)$ . What angle $\theta = \angle PAB=\angle QPC=\angle RQB=\cd... | [
"Drop an altitude from $P$ to $AB$ and let its base be $x$ . Note that if we repeat this for $Q$ and $R$ , all four right triangles (including $\\triangle{RSC}$ ) will have the same trig ratios. By proportion, the hypotenuse $AP$ is $\\frac{x}{100}(120) = \\frac65 x$ , so $\\cos\\theta = \\frac{x}{(\\frac65x)} = \\... |
https://artofproblemsolving.com/wiki/index.php/1986_AHSME_Problems/Problem_6 | C | 30 | Using a table of a certain height, two identical blocks of wood are placed as shown in Figure 1. Length $r$ is found to be $32$ inches. After rearranging the blocks as in Figure 2, length $s$ is found to be $28$ inches. How high is the table?
[asy] size(300); defaultpen(linewidth(0.8)+fontsize(13pt)); path table = o... | [
"Let $h$ $l$ , and $w$ represent the height of the table and the length and width of the wood blocks, respectively, in inches. From Figure 1, we have $l+h-w=32$ , and from Figure 2, $w+h-l=28$ . Adding the equations gives $2h=60 \\implies h=30$ , which is $\\boxed{30}$"
] |
https://artofproblemsolving.com/wiki/index.php/2010_AMC_8_Problems/Problem_7 | B | 10 | Using only pennies, nickels, dimes, and quarters, what is the smallest number of coins Freddie would need so he could pay any amount of money less than a dollar?
$\textbf{(A)}\ 6 \qquad\textbf{(B)}\ 10\qquad\textbf{(C)}\ 15\qquad\textbf{(D)}\ 25\qquad\textbf{(E)}\ 99$ | [
"You need $2$ dimes, $1$ nickel, and $4$ pennies for the first $25$ cents. From $26$ cents to $50$ cents, you only need to add $1$ quarter. From $51$ cents to $75$ cents, you also only need to add $1$ quarter. The same for $76$ cents to $99$ cents. Notice that instead of $100$ , it is $99$ . We are left with $3$ qu... |
https://artofproblemsolving.com/wiki/index.php/1986_AJHSME_Problems/Problem_9 | E | 6 | Using only the paths and the directions shown, how many different routes are there from $\text{M}$ to $\text{N}$
[asy] draw((0,0)--(3,0),MidArrow); draw((3,0)--(6,0),MidArrow); draw(6*dir(60)--3*dir(60),MidArrow); draw(3*dir(60)--(0,0),MidArrow); draw(3*dir(60)--(3,0),MidArrow); draw(5.1961524227066318805823390245176*d... | [
"There is 1 way to get from C to N. There is only one way to get from D to N, which is DCN.\nSince A can only go to C or D, which each only have 1 way to get to N each, there are $1+1=2$ ways to get from A to N.\nSince B can only go to A, C or N, and A only has 2 ways to get to N, C only has 1 way and to get from B... |
https://artofproblemsolving.com/wiki/index.php/2002_AMC_10A_Problems/Problem_15 | E | 190 | Using the digits 1, 2, 3, 4, 5, 6, 7, and 9, form 4 two-digit prime numbers, using each digit only once. What is the sum of the 4 prime numbers?
$\text{(A)}\ 150 \qquad \text{(B)}\ 160 \qquad \text{(C)}\ 170 \qquad \text{(D)}\ 180 \qquad \text{(E)}\ 190$ | [
"Since a multiple-digit prime number is not divisible by either 2 or 5, it must end with 1, 3, 7, or 9 in the units place. The remaining digits given must therefore appear in the tens place. Hence our answer is $20 + 40 + 50 + 60 + 1 + 3 + 7 + 9 = 190\\Rightarrow\\boxed{190}$"
] |
https://artofproblemsolving.com/wiki/index.php/2002_AMC_10B_Problems/Problem_9 | D | 115 | Using the letters $A$ $M$ $O$ $S$ , and $U$ , we can form five-letter "words". If these "words" are arranged in alphabetical order, then the "word" $USAMO$ occupies position
$\mathrm{(A) \ } 112\qquad \mathrm{(B) \ } 113\qquad \mathrm{(C) \ } 114\qquad \mathrm{(D) \ } 115\qquad \mathrm{(E) \ } 116$ | [
"There are $4!\\cdot 4$ \"words\" beginning with each of the first four letters alphabetically. From there, there are $3!\\cdot 3$ with $U$ as the first letter and each of the first three letters alphabetically. After that, the next \"word\" is $USAMO$ , hence our answer is $4\\cdot 4!+3\\cdot 3!+1=\\boxed{115}$",
... |
https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_15 | C | 90 | Values for $A,B,C,$ and $D$ are to be selected from $\{1, 2, 3, 4, 5, 6\}$ without replacement (i.e. no two letters have the same value). How many ways are there to make such choices so that the two curves $y=Ax^2+B$ and $y=Cx^2+D$ intersect? (The order in which the curves are listed does not matter; for example, the c... | [
"Visualizing the two curves, we realize they are both parabolas with the same axis of symmetry. Now assume that the first equation is above the second, since order doesn't matter. Then $C>A$ and $B>D$ . Therefore the number of ways to choose the four integers is $\\tbinom{6}{2}\\tbinom{4}{2}=90$ , and the answer is... |
https://artofproblemsolving.com/wiki/index.php/2023_AMC_8_Problems/Problem_15 | B | 4.2 | Viswam walks half a mile to get to school each day. His route consists of $10$ city blocks of equal length and he takes $1$ minute to walk each block. Today, after walking $5$ blocks, Viswam discovers he has to make a detour, walking $3$ blocks of equal length instead of $1$ block to reach the next corner. From the tim... | [
"Note that Viswam walks at a constant speed of $60$ blocks per hour as he takes $1$ minute to walk each block. After walking $5$ blocks, he has taken $5$ minutes, and he has $5$ minutes remaining, to walk $7$ blocks. Therefore, he must walk at a speed of $7 \\cdot 60 \\div 5 = 84$ blocks per hour to get to school o... |
https://artofproblemsolving.com/wiki/index.php/1997_AJHSME_Problems/Problem_8 | B | 60 | Walter gets up at 6:30 a.m., catches the school bus at 7:30 a.m., has 6 classes that last 50 minutes each, has 30 minutes for lunch, and has 2 hours additional time at school. He takes the bus home and arrives at 4:00 p.m. How many minutes has he spent on the bus?
$\text{(A)}\ 30 \qquad \text{(B)}\ 60 \qquad \text{(C... | [
"There are $4\\frac{1}{2}$ hours from 7:30 a.m. to noon, and $4$ hours from noon to 4 p.m., meaning Walter was away from home for a total of $8\\frac{1}{2}$ hours, or $8.5\\times 60 = 510$ minutes.\nTallying up the times he was not on the bus, he has $6\\times 50 = 300$ minutes in classes, $30$ minutes at lunch, an... |
https://artofproblemsolving.com/wiki/index.php/1995_AJHSME_Problems/Problem_1 | D | 41 | Walter has exactly one penny, one nickel, one dime and one quarter in his pocket. What percent of one dollar is in his pocket?
$\text{(A)}\ 4\% \qquad \text{(B)}\ 25\% \qquad \text{(C)}\ 40\% \qquad \text{(D)}\ 41\% \qquad \text{(E)}\ 59\%$ | [
"Walter has $1 + 5 + 10 + 25 = 41$ cents in his pocket. There are $100$ cents in a dollar. Therefore, Walter has $\\frac{41}{100}\\cdot 100\\% = 41\\%$ of a dollar in his pocket, and the right answer is $\\boxed{41}$"
] |
https://artofproblemsolving.com/wiki/index.php/2001_AMC_10_Problems/Problem_8 | B | 84 | Wanda, Darren, Beatrice, and Chi are tutors in the school math lab. Their schedule is as follows: Darren works every third school day, Wanda works every fourth school day, Beatrice works every sixth school day, and Chi works every seventh school day. Today they are all working in the math lab. In how many school days f... | [
"We need to find the least common multiple of the four numbers given.\n$\\textrm{LCM}(3, 4, 6, 7) = \\textrm{LCM}(3, 2^2, 2 \\cdot 3, 7) = 2^2 \\cdot 3 \\cdot 7 = 84$\nSo the answer is $\\boxed{84}$"
] |
https://artofproblemsolving.com/wiki/index.php/2015_AMC_10B_Problems/Problem_10 | C | 5 | What are the sign and units digit of the product of all the odd negative integers strictly greater than $-2015$ $\textbf{(A) }$ It is a negative number ending with a 1. $\textbf{(B) }$ It is a positive number ending with a 1. $\textbf{(C) }$ It is a negative number ending with a 5. $\textbf{(D) }$ It is a positive numb... | [
"Since $-5>-2015$ , the product must end with a $5$\nThe multiplicands are the odd negative integers from $-1$ to $-2013$ . There are $\\frac{|-2013+1|}2+1=1006+1$ of these numbers. Since $(-1)^{1007}=-1$ , the product is negative.\nTherefore, the answer must be $\\boxed{5.}$"
] |
https://artofproblemsolving.com/wiki/index.php/2006_AMC_10B_Problems/Problem_1 | C | 0 | What is $(-1)^{1} + (-1)^{2} + ... + (-1)^{2006}$
$\textbf{(A)} -2006\qquad \textbf{(B)} -1\qquad \textbf{(C) } 0\qquad \textbf{(D) } 1\qquad \textbf{(E) } 2006$ | [
"Since $-1$ raised to an odd integer is $-1$ and $-1$ raised to an even integer exponent is $1$\n$(-1)^{1} + (-1)^{2} + ... + (-1)^{2006} = (-1) + (1) + ... + (-1)+(1) = \\boxed{0}.$"
] |
https://artofproblemsolving.com/wiki/index.php/2014_AMC_10B_Problems/Problem_2 | E | 64 | What is $\frac{2^3 + 2^3}{2^{-3} + 2^{-3}}$
$\textbf {(A) } 16 \qquad \textbf {(B) } 24 \qquad \textbf {(C) } 32 \qquad \textbf {(D) } 48 \qquad \textbf {(E) } 64$ | [
"We can synchronously multiply ${2^3}$ to the expresions both above and below the fraction bar. Thus, \\[\\frac{2^3+2^3}{2^{-3}+2^{-3}}\\\\=\\frac{2^6+2^6}{1+1}\\\\={2^6}.\\] Hence, the fraction equals to $\\boxed{64}$",
"We have \\[\\frac{2^3+2^3}{2^{-3}+2^{-3}} = \\frac{8 + 8}{\\frac{1}{8} + \\frac{1}{8}} = \\f... |
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