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/2015_AMC_10A_Problems/Problem_12 | C | 2 | Points $( \sqrt{\pi} , a)$ and $( \sqrt{\pi} , b)$ are distinct points on the graph of $y^2 + x^4 = 2x^2 y + 1$ . What is $|a-b|$
$\textbf{(A)}\ 1 \qquad\textbf{(B)} \ \frac{\pi}{2} \qquad\textbf{(C)} \ 2 \qquad\textbf{(D)} \ \sqrt{1+\pi} \qquad\textbf{(E)} \ 1 + \sqrt{\pi}$ | [
"Since points on the graph make the equation true, substitute $\\sqrt{\\pi}$ in to the equation and then solve to find $a$ and $b$\n$y^2 + \\sqrt{\\pi}^4 = 2\\sqrt{\\pi}^2 y + 1$\n$y^2 + \\pi^2 = 2\\pi y + 1$\n$y^2 - 2\\pi y + \\pi^2 = 1$\n$(y-\\pi)^2 = 1$\n$y-\\pi = \\pm 1$\n$y = \\pi + 1$\n$y = \\pi - 1$\nThere a... |
https://artofproblemsolving.com/wiki/index.php/2001_AMC_12_Problems/Problem_20 | null | 10 | Points $A = (3,9)$ $B = (1,1)$ $C = (5,3)$ , and $D=(a,b)$ lie in the first quadrant and are the vertices of quadrilateral $ABCD$ . The quadrilateral formed by joining the midpoints of $\overline{AB}$ $\overline{BC}$ $\overline{CD}$ , and $\overline{DA}$ is a square. What is the sum of the coordinates of point $D$
$\te... | [
"\nWe already know two vertices of the square: $(A+B)/2 = (2,5)$ and $(B+C)/2 = (3,2)$\nThere are only two possibilities for the other vertices of the square: either they are $(6,3)$ and $(5,6)$ , or they are $(0,1)$ and $(-1,4)$ . The second case would give us $D$ outside the first quadrant, hence the first case i... |
https://artofproblemsolving.com/wiki/index.php/2001_AMC_8_Problems/Problem_11 | C | 18 | Points $A$ $B$ $C$ and $D$ have these coordinates: $A(3,2)$ $B(3,-2)$ $C(-3,-2)$ and $D(-3, 0)$ . The area of quadrilateral $ABCD$ is
[asy] for (int i = -4; i <= 4; ++i) { for (int j = -4; j <= 4; ++j) { dot((i,j)); } } draw((0,-4)--(0,4),linewidth(1)); draw((-4,0)--(4,0),linewidth(1)); for (int i = -4; i <= 4; ++i) {... | [
"\nThis quadrilateral is a trapezoid, because $AB\\parallel CD$ but $BC$ is not parallel to $AD$ . The area of a trapezoid is the product of its height and its median, where the median is the average of the side lengths of the bases. The two bases are $AB$ and $CD$ , which have lengths $2$ and $4$ , respectively, s... |
https://artofproblemsolving.com/wiki/index.php/2018_AIME_II_Problems/Problem_1 | null | 800 | Points $A$ $B$ , and $C$ lie in that order along a straight path where the distance from $A$ to $C$ is $1800$ meters. Ina runs twice as fast as Eve, and Paul runs twice as fast as Ina. The three runners start running at the same time with Ina starting at $A$ and running toward $C$ , Paul starting at $B$ and running tow... | [
"We know that in the same amount of time given, Ina will run twice the distance of Eve, and Paul would run quadruple the distance of Eve. Let's consider the time it takes for Paul to meet Eve: Paul would've run 4 times the distance of Eve, which we can denote as $d$ . Thus, the distance between $B$ and $C$ is $4d+d... |
https://artofproblemsolving.com/wiki/index.php/1996_AJHSME_Problems/Problem_8 | B | 3 | Points $A$ and $B$ are 10 units apart. Points $B$ and $C$ are 4 units apart. Points $C$ and $D$ are 3 units apart. If $A$ and $D$ are as close as possible, then the number of units between them is
$\text{(A)}\ 0 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 9 \qquad \text{(D)}\ 11 \qquad \text{(E)}\ 17$ | [
"If $AB = 10$ and $BC=4$ , then $(10 - 4) \\le AC \\le (10 + 4)$ by the triangle inequality . In the triangle inequality, the equality is only reached when the \"triangle\" $ABC$ is really a degenerate triangle, and $ABC$ are collinear.\nSimplifying, this means the smallest value $AC$ can be is $6$\nApplying the t... |
https://artofproblemsolving.com/wiki/index.php/2009_AMC_10B_Problems/Problem_16 | B | 12 | Points $A$ and $C$ lie on a circle centered at $O$ , each of $\overline{BA}$ and $\overline{BC}$ are tangent to the circle, and $\triangle ABC$ is equilateral. The circle intersects $\overline{BO}$ at $D$ . What is $\frac{BD}{BO}$
$\text{(A) } \frac {\sqrt2}{3} \qquad \text{(B) } \frac {1}{2} \qquad \text{(C) } \frac {... | [
"\nAs $\\triangle ABC$ is equilateral, we have $\\angle BAC = \\angle BCA = 60^\\circ$ , hence $\\angle OAC = \\angle OCA = 30^\\circ$ . Then $\\angle AOC = 120^\\circ$ , and from symmetry we have $\\angle AOB = \\angle COB = 60^\\circ$ . Thus, this gives us $\\angle ABO = \\angle CBO = 30^\\circ$\nWe know that $DO... |
https://artofproblemsolving.com/wiki/index.php/2009_AMC_10B_Problems/Problem_16 | null | 12 | Points $A$ and $C$ lie on a circle centered at $O$ , each of $\overline{BA}$ and $\overline{BC}$ are tangent to the circle, and $\triangle ABC$ is equilateral. The circle intersects $\overline{BO}$ at $D$ . What is $\frac{BD}{BO}$
$\text{(A) } \frac {\sqrt2}{3} \qquad \text{(B) } \frac {1}{2} \qquad \text{(C) } \frac {... | [
"\nAs in the previous solution, we find out that $\\angle AOB = \\angle COB = 60^\\circ$ . Hence $\\triangle AOD$ and $\\triangle COD$ are both equilateral.\nWe then have $\\angle SCD = \\angle SAD = 30^\\circ$ , hence $D$ is the incenter of $\\triangle ABC$ , and as $\\triangle ABC$ is equilateral, $D$ is also its... |
https://artofproblemsolving.com/wiki/index.php/2017_AMC_10B_Problems/Problem_8 | C | 49 | Points $A(11, 9)$ and $B(2, -3)$ are vertices of $\triangle ABC$ with $AB=AC$ . The altitude from $A$ meets the opposite side at $D(-1, 3)$ . What are the coordinates of point $C$
$\textbf{(A)}\ (-8, 9)\qquad\textbf{(B)}\ (-4, 8)\qquad\textbf{(C)}\ (-4, 9)\qquad\textbf{(D)}\ (-2, 3)\qquad\textbf{(E)}\ (-1, 0)$ | [
"Since $AB = AC$ , then $\\triangle ABC$ is isosceles, so $BD = CD$ . Therefore, the coordinates of $C$ are $(-1 - 3, 3 + 6) = \\boxed{4,9}$",
"Calculating the equation of the line running between points $B$ and $D$ $y = -2x + 1$ . The only coordinate of $C$ that is also on this line is $\\boxed{4,9}$",
"Simila... |
https://artofproblemsolving.com/wiki/index.php/2006_AMC_8_Problems/Problem_5 | D | 30 | Points $A, B, C$ and $D$ are midpoints of the sides of the larger square. If the larger square has area 60, what is the area of the smaller square?
[asy]size(100); draw((0,0)--(2,0)--(2,2)--(0,2)--cycle,linewidth(1)); draw((0,1)--(1,2)--(2,1)--(1,0)--cycle); label("$A$", (1,2), N); label("$B$", (2,1), E); label("$C$", ... | [
"Drawing segments $AC$ and $BD$ , the number of triangles outside square $ABCD$ is the same as the number of triangles inside the square. Thus areas must be equal so the area of $ABCD$ is half the area of the larger square which is $\\frac{60}{2}=\\boxed{30}$",
"If the side length of the larger square is $x$ , th... |
https://artofproblemsolving.com/wiki/index.php/2002_AMC_10A_Problems/Problem_23 | D | 9 | Points $A,B,C$ and $D$ lie on a line, in that order, with $AB = CD$ and $BC = 12$ . Point $E$ is not on the line, and $BE = CE = 10$ . The perimeter of $\triangle AED$ is twice the perimeter of $\triangle BEC$ . Find $AB$
$\text{(A)}\ 15/2 \qquad \text{(B)}\ 8 \qquad \text{(C)}\ 17/2 \qquad \text{(D)}\ 9 \qquad \text{(... | [
"Let $M$ be the foot of the altitude from $E$ to $BC.$ Then $MB=MC=6$ because $\\triangle BEC$ is isosceles. By the Pythagorean triple $(6,8,10)$ the altitude is $8.$ Since $(8,15,17)$ is the only primitive Pythagorean triple with leg $8,$ we test $AE=DE=17,AM=DM=15.$ Since $2(10+10+12)=(17+17+2\\cdot 15)$ this wor... |
https://artofproblemsolving.com/wiki/index.php/1999_AMC_8_Problems/Problem_25 | A | 6 | Points $B$ $D$ , and $J$ are midpoints of the sides of right triangle $ACG$ . Points $K$ $E$ $I$ are midpoints of the sides of triangle $JDG$ , etc. If the dividing and shading process is done 100 times (the first three are shown) and $AC=CG=6$ , then the total area of the shaded triangles is nearest
[asy] draw((0,0)--... | [
"Since $\\triangle FGH$ is fairly small relative to the rest of the diagram, we can make an underestimate by using the current diagram. All triangles are right-isosceles triangles.\n$CD = \\frac {CG}{2} = 3, DE = \\frac{CD}{2} = \\frac{3}{2}, EF = \\frac{DE}{2} = \\frac{3}{4}$\n$CB = CD = 3, DK = DE = \\frac{3}{2}... |
https://artofproblemsolving.com/wiki/index.php/2004_AMC_10A_Problems/Problem_20 | D | 2 | Points $E$ and $F$ are located on square $ABCD$ so that $\triangle BEF$ is equilateral. What is the ratio of the area of $\triangle DEF$ to that of $\triangle ABE$
$\mathrm{(A) \ } \frac{4}{3} \qquad \mathrm{(B) \ } \frac{3}{2} \qquad \mathrm{(C) \ } \sqrt{3} \qquad \mathrm{(D) \ } 2 \qquad \mathrm{(E) \ } 1+\sqrt{3}$ | [
"Since triangle $BEF$ is equilateral, $EA=FC$ , and $EAB$ and $FCB$ are $SAS$ congruent. Thus, triangle $DEF$ is an isosceles right triangle. So we let $DE=x$ . Thus $EF=EB=FB=x\\sqrt{2}$ . If we go angle chasing, we find out that $\\angle AEB=75^{\\circ}$ , thus $\\angle ABE=15^{\\circ}$ $\\frac{AE}{EB}=\\sin{15^{... |
https://artofproblemsolving.com/wiki/index.php/2000_AMC_10_Problems/Problem_5 | B | 1 | Points $M$ and $N$ are the midpoints of sides $PA$ and $PB$ of $\triangle PAB$ . As $P$ moves along a line that is parallel to side $AB$ , how many of the four quantities listed below change?
(a) the length of the segment $MN$
(b) the perimeter of $\triangle PAB$
(c) the area of $\triangle PAB$
(d) the area of trapezoi... | [
"(a) Triangles $ABP$ and $MNP$ are similar, and since $PM=\\frac{1}{2}AP$ $MN=\\frac{1}{2}AB$\n(b) We see the perimeter changes. For example, imagine if P was extremely far to the left.\n(c) The area clearly doesn't change, as both the base $AB$ and its corresponding height remain the same.\n(d) The bases $AB$ and ... |
https://artofproblemsolving.com/wiki/index.php/1961_AHSME_Problems/Problem_23 | B | 70 | Points $P$ and $Q$ are both in the line segment $AB$ and on the same side of its midpoint. $P$ divides $AB$ in the ratio $2:3$ ,
and $Q$ divides $AB$ in the ratio $3:4$ . If $PQ=2$ , then the length of $AB$ is:
$\textbf{(A)}\ 60\qquad \textbf{(B)}\ 70\qquad \textbf{(C)}\ 75\qquad \textbf{(D)}\ 80\qquad \textbf{(E)}\ 8... | [
"\nDraw diagram as shown, where $P$ and $Q$ are on the same side. Let $AP = x$ and $QB = y$\nSince $P$ divides $AB$ in the ratio $2:3$ $\\frac{x}{y+2} = \\frac{2}{3}$ . Since $Q$ divides $AB$ in the ratio $3:4$ $\\frac{x+2}{y} = \\frac{3}{4}$ . Cross multiply to get a system of equations \\[3x = 2y+4\\] \\[4x+8=... |
https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_8 | D | 8 | Points $P$ and $Q$ lie in a plane with $PQ=8$ . How many locations for point $R$ in this plane are there such that the triangle with vertices $P$ $Q$ , and $R$ is a right triangle with area $12$ square units?
$\textbf{(A)}\ 2 \qquad\textbf{(B)}\ 4 \qquad\textbf{(C)}\ 6 \qquad\textbf{(D)}\ 8 \qquad\textbf{(E)}\ 12$ | [
"Let the brackets denote areas. We are given that \\[[PQR]=\\frac12\\cdot PQ\\cdot h_R=12.\\] Since $PQ=8,$ it follows that $h_R=3.$\nWe construct a circle with diameter $\\overline{PQ}.$ All such locations for $R$ are shown below:\n\nWe apply casework to the right angle of $\\triangle PQR:$\nTogether, there are $\... |
https://artofproblemsolving.com/wiki/index.php/2001_AMC_8_Problems/Problem_23 | D | 4 | Points $R$ $S$ and $T$ are vertices of an equilateral triangle, and points $X$ $Y$ and $Z$ are midpoints of its sides. How many noncongruent triangles can be
drawn using any three of these six points as vertices?
[asy] pair SS,R,T,X,Y,Z; SS = (2,2*sqrt(3)); R = (0,0); T = (4,0); X = (2,0); Y = (1,sqrt(3)); Z = (3,sqrt(... | [
"There are $6$ points in the figure, and $3$ of them are needed to form a triangle, so there are ${6\\choose{3}} =20$ possible triplets of the $6$ points. However, some of these created congruent triangles, and some don't even make triangles at all.\nCase 1: Triangles congruent to $\\triangle RST$ There is obviousl... |
https://artofproblemsolving.com/wiki/index.php/2002_AIME_I_Problems/Problem_15 | null | 163 | Polyhedron $ABCDEFG$ has six faces. Face $ABCD$ is a square with $AB = 12;$ face $ABFG$ is a trapezoid with $\overline{AB}$ parallel to $\overline{GF},$ $BF = AG = 8,$ and $GF = 6;$ and face $CDE$ has $CE = DE = 14.$ The other three faces are $ADEG, BCEF,$ and $EFG.$ The distance from $E$ to face $ABCD$ is 12. Given ... | [
"We let $A$ be the origin, or $(0,0,0)$ $B = (0,0,12)$ , and $D = (12,0,0)$ . Draw the perpendiculars from F and G to AB, and let their intersections be X and Y, respectively. By symmetry, $FX = GY = \\frac{12-6}2 = 3$ , so $G = (a,b,3)$ , where a and b are variables.\nWe can now calculate the coordinates of E. Dra... |
https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_2 | C | 1,950 | Portia's high school has $3$ times as many students as Lara's high school. The two high schools have a total of $2600$ students. How many students does Portia's high school have?
$\textbf{(A)} ~600 \qquad\textbf{(B)} ~650 \qquad\textbf{(C)} ~1950 \qquad\textbf{(D)} ~2000\qquad\textbf{(E)} ~2050$ | [
"The following system of equations can be formed with $P$ representing the number of students in Portia's high school and $L$ representing the number of students in Lara's high school: \\begin{align*} P&=3L, \\\\ P+L&=2600. \\end{align*} Substituting $P=3L$ gives $4L=2600.$ Solving for $L$ gives $L=650.$ Since we n... |
https://artofproblemsolving.com/wiki/index.php/2010_AIME_I_Problems/Problem_5 | null | 501 | Positive integers $a$ $b$ $c$ , and $d$ satisfy $a > b > c > d$ $a + b + c + d = 2010$ , and $a^2 - b^2 + c^2 - d^2 = 2010$ . Find the number of possible values of $a$ | [
"Using the difference of squares $2010 = (a^2 - b^2) + (c^2 - d^2) = (a + b)(a - b) + (c + d)(c - d) \\ge a + b + c + d = 2010$ , where equality must hold so $b = a - 1$ and $d = c - 1$ . Then we see $a = 1004$ is maximal and $a = 504$ is minimal, so the answer is $\\boxed{501}$",
"Since $a+b$ must be greater tha... |
https://artofproblemsolving.com/wiki/index.php/2013_AMC_10B_Problems/Problem_5 | B | 15 | Positive integers $a$ and $b$ are each less than $6$ . What is the smallest possible value for $2 \cdot a - a \cdot b$
$\textbf{(A)}\ -20\qquad\textbf{{(B)}}\ -15\qquad\textbf{{(C)}}\ -10\qquad\textbf{{(D)}}\ 0\qquad\textbf{{(E)}}\ 2$ | [
"Factoring the equation gives $a(2 - b)$ . From this we can see that to obtain the least possible value, $2 - b$ should be negative, and should be as small as possible. To do so, $b$ should be maximized. Because $2 - b$ is negative, we should maximize the positive value of $a$ as well. The maximum values of both $a... |
https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_21 | E | 8 | Positive integers $a$ and $b$ are such that the graphs of $y=ax+5$ and $y=3x+b$ intersect the $x$ -axis at the same point. What is the sum of all possible $x$ -coordinates of these points of intersection?
$\textbf{(A)}\ {-20}\qquad\textbf{(B)}\ {-18}\qquad\textbf{(C)}\ {-15}\qquad\textbf{(D)}\ {-12}\qquad\textbf{(E)}\ ... | [
"Note that when $y=0$ , the $x$ values of the equations should be equal by the problem statement. We have that \\[0 = ax + 5 \\implies x = -\\dfrac{5}{a}\\] \\[0 = 3x+b \\implies x= -\\dfrac{b}{3}\\] Which means that \\[-\\dfrac{5}{a} = -\\dfrac{b}{3} \\implies ab = 15\\] The only possible pairs $(a,b)$ then are $(... |
https://artofproblemsolving.com/wiki/index.php/2013_AIME_II_Problems/Problem_2 | null | 881 | Positive integers $a$ and $b$ satisfy the condition \[\log_2(\log_{2^a}(\log_{2^b}(2^{1000}))) = 0.\] Find the sum of all possible values of $a+b$ | [
"To simplify, we write this logarithmic expression as an exponential one. Just looking at the first log, it has a base of 2 and an argument of the expression in parenthesis. Therefore, we can make 2 the base, 0 the exponent, and the argument the result. That means $\\log_{2^a}(\\log_{2^b}(2^{1000}))=1$ (because $2... |
https://artofproblemsolving.com/wiki/index.php/2003_AMC_12B_Problems/Problem_24 | C | 1,002 | Positive integers $a,b,$ and $c$ are chosen so that $a<b<c$ , and the system of equations
has exactly one solution. What is the minimum value of $c$
$\mathrm{(A)}\ 668 \qquad\mathrm{(B)}\ 669 \qquad\mathrm{(C)}\ 1002 \qquad\mathrm{(D)}\ 2003 \qquad\mathrm{(E)}\ 2004$ | [
"Consider the graph of $f(x)=|x-a|+|x-b|+|x-c|$\nWhen $x<a$ , the slope is $-3$\nWhen $a<x<b$ , the slope is $-1$\nWhen $b<x<c$ , the slope is $1$\nWhen $c<x$ , the slope is $3$\nSetting $x=b$ gives $y=|b-a|+|b-b|+|b-c|=c-a$ , so $(b,c-a)$ is a point on $f(x)$ . In fact, it is the minimum of $f(x)$ considering the ... |
https://artofproblemsolving.com/wiki/index.php/2010_AIME_II_Problems/Problem_5 | null | 75 | Positive numbers $x$ $y$ , and $z$ satisfy $xyz = 10^{81}$ and $(\log_{10}x)(\log_{10} yz) + (\log_{10}y) (\log_{10}z) = 468$ . Find $\sqrt {(\log_{10}x)^2 + (\log_{10}y)^2 + (\log_{10}z)^2}$ | [
"Using the properties of logarithms, $\\log_{10}xyz = 81$ by taking the log base 10 of both sides, and $(\\log_{10}x)(\\log_{10}y) + (\\log_{10}x)(\\log_{10}z) + (\\log_{10}y)(\\log_{10}z)= 468$ by using the fact that $\\log_{10}ab = \\log_{10}a + \\log_{10}b$\nThrough further simplification, we find that $\\log_{1... |
https://artofproblemsolving.com/wiki/index.php/2019_AMC_12A_Problems/Problem_12 | B | 20 | Positive real numbers $x \neq 1$ and $y \neq 1$ satisfy $\log_2{x} = \log_y{16}$ and $xy = 64$ . What is $(\log_2{\tfrac{x}{y}})^2$
$\textbf{(A) } \frac{25}{2} \qquad\textbf{(B) } 20 \qquad\textbf{(C) } \frac{45}{2} \qquad\textbf{(D) } 25 \qquad\textbf{(E) } 32$ | [
"Let $\\log_2{x} = \\log_y{16}=k$ , so that $2^k=x$ and $y^k=16 \\implies y=2^{\\frac{4}{k}}$ . Then we have $(2^k)(2^{\\frac{4}{k}})=2^{k+\\frac{4}{k}}=2^6$\nWe therefore have $k+\\frac{4}{k}=6$ , and deduce $k^2-6k+4=0$ . The solutions to this are $k = 3 \\pm \\sqrt{5}$\nTo solve the problem, we now find \\begin{... |
https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_10 | D | 36 | Positive real numbers $x$ and $y$ satisfy $y^3=x^2$ and $(y-x)^2=4y^2$ . What is $x+y$ $\textbf{(A) }12\qquad\textbf{(B) }18\qquad\textbf{(C) }24\qquad\textbf{(D) }36\qquad\textbf{(E) }42$ | [
"Because $y^3=x^2$ , set $x=a^3$ $y=a^2$ $a\\neq 0$ ). Put them in $(y-x)^2=4y^2$ we get $(a^2(a-1))^2=4a^4$ which implies $a^2-2a+1=4$ . Solve the equation to get $a=3$ or $-1$ . Since $x$ and $y$ are positive, $a=3$ and $x+y=3^3+3^2=\\boxed{36}$",
"Since $a^2 = |a|^2$ , we can rewrite the second equation as $(x... |
https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_10 | null | 36 | Positive real numbers $x$ and $y$ satisfy $y^3=x^2$ and $(y-x)^2=4y^2$ . What is $x+y$ $\textbf{(A) }12\qquad\textbf{(B) }18\qquad\textbf{(C) }24\qquad\textbf{(D) }36\qquad\textbf{(E) }42$ | [
"Let's take the second equation and square root both sides. This will obtain $y-x = \\pm2y$ . Solving the case where $y-x=+2y$ , we'd find that $x=-y$ . This is known to be false because both $x$ and $y$ have to be positive, and $x=-y$ implies that at least one of the variables is not positive. So we instead solve ... |
https://artofproblemsolving.com/wiki/index.php/2008_AMC_12B_Problems/Problem_6 | A | 2,500 | Postman Pete has a pedometer to count his steps. The pedometer records up to $99999$ steps, then flips over to $00000$ on the next step. Pete plans to determine his mileage for a year. On January $1$ Pete sets the pedometer to $00000$ . During the year, the pedometer flips from $99999$ to $00000$ forty-four times. On D... | [
"Every time the pedometer flips, Pete has walked $100,000$ steps. Therefore, Pete has walked a total of $100,000 \\cdot 44 + 50,000 = 4,450,000$ steps, which is $4,450,000/1,800 = 2472.2$ miles, which is the closest to the answer choice $\\boxed{2500}$"
] |
https://artofproblemsolving.com/wiki/index.php/2008_AMC_10B_Problems/Problem_12 | A | 2,500 | Postman Pete has a pedometer to count his steps. The pedometer records up to 99999 steps, then flips over to 00000 on the next step. Pete plans to determine his mileage for a year. On January 1 Pete sets the pedometer to 00000. During the year, the pedometer flips from 99999 to 00000 forty-four
times. On December 31 t... | [
"Every time the pedometer flips from $99999$ to\n$00000$ Pete has walked $100000$ steps.\nSo, if the pedometer flipped $44$ times\nPete walked $44*100000+50000=4450000$ steps.\nDividing by $1800$ steps per mile gives $2472.\\overline{2}$\nThis is closest to answer $\\boxed{2500}$"
] |
https://artofproblemsolving.com/wiki/index.php/2006_AMC_8_Problems/Problem_14 | B | 11,400 | Problems 14, 15 and 16 involve Mrs. Reed's English assignment.
A Novel Assignment
The students in Mrs. Reed's English class are reading the same $760$ -page novel. Three friends, Alice, Bob and Chandra, are in the class. Alice reads a page in 20 seconds, Bob reads a page in $45$ seconds and Chandra reads a page in $30$... | [
"The information is the same for Problems 14,15, and 16. Therefore, we shall only use the information we need. All we need for this problem is that there's 760 pages, Bob reads a page in 45 seconds and Chandra reads a page in 30 seconds. A lot of people will find how long it takes Bob to read the book, how long it ... |
https://artofproblemsolving.com/wiki/index.php/2006_AMC_8_Problems/Problem_16 | E | 7,200 | Problems 14, 15 and 16 involve Mrs. Reed's English assignment.
A Novel Assignment
The students in Mrs. Reed's English class are reading the same 760-page novel. Three friends, Alice, Bob and Chandra, are in the class. Alice reads a page in 20 seconds, Bob reads a page in 45 seconds and Chandra reads a page in 30 second... | [
"The amount of pages Bob, Chandra, and Alice will read is in the ratio 4:6:9. Therefore, Bob, Chandra, and Alice read 160, 240, and 360 pages respectively. They would also be reading for the same amount of time because the ratio of the pages read was based on the time it takes each of them to read a page. Therefore... |
https://artofproblemsolving.com/wiki/index.php/2006_AMC_8_Problems/Problem_15 | C | 456 | Problems 14, 15 and 16 involve Mrs. Reed's English assignment.
A Novel Assignment
The students in Mrs. Reed's English class are reading the same 760-page novel. Three friends, Alice, Bob and Chandra, are in the class. Alice reads a page in 20 seconds, Bob reads a page in 45 seconds and Chandra reads a page in 30 second... | [
"Same as the previous problem, we only use the information we need. Note that it's not just Chandra reads half of it and Bob reads the rest since they have different reading rates. In this case, we set up an equation and solve.\nLet $x$ be the number of pages that Chandra reads.\n$30x = 45(760-x)$ Distribute the $4... |
https://artofproblemsolving.com/wiki/index.php/2003_AMC_8_Problems/Problem_9 | C | 40 | Problems 8, 9 and 10 use the data found in the accompanying paragraph and figures
Four friends, Art, Roger, Paul and Trisha, bake cookies, and all cookies have the same thickness. The shapes of the cookies differ, as shown.
$\circ$ Art's cookies are trapezoids: [asy]size(80);defaultpen(linewidth(0.8));defaultpen(fontsi... | [
"The area of one of Art's cookies is $3 \\cdot 3 + \\frac{2 \\cdot 3}{2}=9+3=12$ . As he has $12$ cookies in a batch, the amount of dough each person used is $12 \\cdot 12=144$ . Roger's cookies have an area of $\\frac{144}{2 \\cdot 4}=\\frac{144}{8}= 18$ cookies in a batch. In total, the amount of money Art will ... |
https://artofproblemsolving.com/wiki/index.php/2003_AMC_8_Problems/Problem_10 | E | 24 | Problems 8, 9 and 10 use the data found in the accompanying paragraph and figures
Four friends, Art, Roger, Paul and Trisha, bake cookies, and all cookies have the same thickness. The shapes of the cookies differ, as shown.
$\circ$ Art's cookies are trapezoids: [asy]size(80);defaultpen(linewidth(0.8));defaultpen(fontsi... | [
"Art's cookies have areas of $3 \\cdot 3 + \\frac{2 \\cdot 3}{2}=9+3=12$ . There are 12 cookies in one of Art's batches so everyone used $12 \\cdot 12=144 \\text{ in}^2$ of dough. Trisha's cookies have an area of $\\frac{3 \\cdot 4}{2}=6$ so she has $\\frac{144}{6}=\\boxed{24}$ cookies per batch."
] |
https://artofproblemsolving.com/wiki/index.php/2002_AMC_8_Problems/Problem_10 | E | 5.5 | Problems 8,9 and 10 use the data found in the accompanying paragraph and table:
The average price of his '70s stamps is closest to
$\text{(A)}\ 3.5 \text{ cents} \qquad \text{(B)}\ 4 \text{ cents} \qquad \text{(C)}\ 4.5 \text{ cents} \qquad \text{(D)}\ 5 \text{ cents} \qquad \text{(E)}\ 5.5 \text{ cents}$ | [
"The price of all the stamps in the '70s together over the total number of stamps is equal to the average price.\n\\[\\frac{(12)(0.06)+(12)(0.06)+(6)(0.04)+(13)(0.05)}{12+12+6+13}\\\\ = \\frac{0.72+0.72+0.24+0.65}{43}\\\\ = \\frac{2.33}{43} \\approx \\boxed{5.5}\\]"
] |
https://artofproblemsolving.com/wiki/index.php/2002_AMC_8_Problems/Problem_8 | D | 24 | Problems 8,9 and 10 use the data found in the accompanying paragraph and table:
[asy] /* AMC8 2002 #8, 9, 10 Problem */ size(3inch, 1.5inch); for ( int y = 0; y <= 5; ++y ) { draw((0,y)--(18,y)); } draw((0,0)--(0,5)); draw((6,0)--(6,5)); draw((9,0)--(9,5)); draw((12,0)--(12,5)); draw((15,0)--(15,5)); draw((18,0)--(1... | [
"France and Spain are European countries. The number of '80s stamps from France is $15$ and the number of '80s stamps from Spain is $9$ . The total number of stamps is $15+9=\\boxed{24}$"
] |
https://artofproblemsolving.com/wiki/index.php/2002_AMC_8_Problems/Problem_9 | B | 1.06 | Problems 8,9 and 10 use the data found in the accompanying paragraph and table:
[asy] /* AMC8 2002 #8, 9, 10 Problem */ size(3inch, 1.5inch); for ( int y = 0; y <= 5; ++y ) { draw((0,y)--(18,y)); } draw((0,0)--(0,5)); draw((6,0)--(6,5)); draw((9,0)--(9,5)); draw((12,0)--(12,5)); draw((15,0)--(15,5)); draw((18,0)--(1... | [
"Brazil 50s and 60s total 11 stamps with each 6 cents, Peru 50s and 60s total 10 stamps with each 4 cents. So total $11*0.06+10*0.04 = \\boxed{1.06}$"
] |
https://artofproblemsolving.com/wiki/index.php/2018_AMC_8_Problems/Problem_16 | C | 5,760 | Professor Chang has nine different language books lined up on a bookshelf: two Arabic, three German, and four Spanish. How many ways are there to arrange the nine books on the shelf keeping the Arabic books together and keeping the Spanish books together?
$\textbf{(A) }1440\qquad\textbf{(B) }2880\qquad\textbf{(C) }5760... | [
"Since the Arabic books and Spanish books have to be kept together, we can treat them both as just one book. That means we're trying to find the number of ways you can arrange one Arabic book, one Spanish book, and three German books, which is just $5$ factorial. Now, we multiply this product by $2!$ because there ... |
https://artofproblemsolving.com/wiki/index.php/1982_USAMO_Problems/Problem_4 | null | 2,935,363,332,000,000,000 | Prove that there exists a positive integer $k$ such that $k\cdot2^n+1$ is composite for every integer $n$ | [
"Indeed, $\\boxed{2935363331541925531}$ has the requisite property."
] |
https://artofproblemsolving.com/wiki/index.php/1995_AIME_Problems/Problem_12 | null | 5 | Pyramid $OABCD$ has square base $ABCD,$ congruent edges $\overline{OA}, \overline{OB}, \overline{OC},$ and $\overline{OD},$ and $\angle AOB=45^\circ.$ Let $\theta$ be the measure of the dihedral angle formed by faces $OAB$ and $OBC.$ Given that $\cos \theta=m+\sqrt{n},$ where $m_{}$ and $n_{}$ are integers, find $m+n.$ | [
"The angle $\\theta$ is the angle formed by two perpendiculars drawn to $BO$ , one on the plane determined by $OAB$ and the other by $OBC$ . Let the perpendiculars from $A$ and $C$ to $\\overline{OB}$ meet $\\overline{OB}$ at $P.$ Without loss of generality , let $AP = 1.$ It follows that $\\triangle OPA$ is a $45-... |
https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_16 | D | 110 | Qiang drives $15$ miles at an average speed of $30$ miles per hour. How many additional miles will he have to drive at $55$ miles per hour to average $50$ miles per hour for the entire trip?
$\textbf{(A) }45\qquad\textbf{(B) }62\qquad\textbf{(C) }90\qquad\textbf{(D) }110\qquad\textbf{(E) }135$ | [
"The only option that is easily divisible by $55$ is $110$ , which gives 2 hours of travel. And, the formula is $\\frac{15}{30} + \\frac{110}{55} = \\frac{5}{2}$\nAnd, $\\text{Average Speed}$ $\\frac{\\text{Total Distance}}{\\text{Total Time}}$\nThus, $\\frac{125}{50} = \\frac{5}{2}$\nBoth are equal and thus our an... |
https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_1 | null | 116 | Quadratic polynomials $P(x)$ and $Q(x)$ have leading coefficients $2$ and $-2,$ respectively. The graphs of both polynomials pass through the two points $(16,54)$ and $(20,53).$ Find $P(0) + Q(0).$ | [
"Let $R(x)=P(x)+Q(x).$ Since the $x^2$ -terms of $P(x)$ and $Q(x)$ cancel, we conclude that $R(x)$ is a linear polynomial.\nNote that \\begin{alignat*}{8} R(16) &= P(16)+Q(16) &&= 54+54 &&= 108, \\\\ R(20) &= P(20)+Q(20) &&= 53+53 &&= 106, \\end{alignat*} so the slope of $R(x)$ is $\\frac{106-108}{20-16}=-\\frac12.... |
https://artofproblemsolving.com/wiki/index.php/2008_AMC_10B_Problems/Problem_24 | C | 85 | Quadrilateral $ABCD$ has $AB = BC = CD$ $m\angle ABC = 70^\circ$ and $m\angle BCD = 170^\circ$ . What is the degree measure of $\angle BAD$
$\mathrm{(A)}\ 75\qquad\mathrm{(B)}\ 80\qquad\mathrm{(C)}\ 85\qquad\mathrm{(D)}\ 90\qquad\mathrm{(E)}\ 95$ | [
"\nLet the unknown $\\angle BAD$ be $x$\nFirst, we draw diagonal $BD$ and $AC$ $I$ is the intersection of the two diagonals. The diagonals each form two isosceles triangles, $\\triangle BCD$ and $\\triangle ABC$\nUsing this, we find: $\\angle DBC = \\angle CDB = 5^\\circ$ and $\\angle BAC = \\angle BCA = 55^\\circ$... |
https://artofproblemsolving.com/wiki/index.php/2008_AMC_10B_Problems/Problem_24 | null | 85 | Quadrilateral $ABCD$ has $AB = BC = CD$ $m\angle ABC = 70^\circ$ and $m\angle BCD = 170^\circ$ . What is the degree measure of $\angle BAD$
$\mathrm{(A)}\ 75\qquad\mathrm{(B)}\ 80\qquad\mathrm{(C)}\ 85\qquad\mathrm{(D)}\ 90\qquad\mathrm{(E)}\ 95$ | [
"To start off, draw a diagram like in solution two and label the points. Create lines $\\overline{AC}$ and $\\overline{BD}$ . We can call their intersection point $Y$ . Note that triangle $BCD$ is an isosceles triangle so angles $CDB$ and $CBD$ are each $5$ degrees. Since $AB$ equals $BC$ , angle $BAC$ equals $55$ ... |
https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_4 | D | 120 | Quadrilateral $ABCD$ is a rhombus with perimeter $52$ meters. The length of diagonal $\overline{AC}$ is $24$ meters. What is the area in square meters of rhombus $ABCD$
[asy] draw((-13,0)--(0,5)); draw((0,5)--(13,0)); draw((13,0)--(0,-5)); draw((0,-5)--(-13,0)); dot((-13,0)); dot((0,5)); dot((13,0)); dot((0,-5)); label... | [
"\nA rhombus has sides of equal length. Because the perimeter of the rhombus is $52$ , each side is $\\frac{52}{4}=13$ . In a rhombus, diagonals are perpendicular and bisect each other, which means $\\overline{AE}$ $12$ $\\overline{EC}$\nConsider one of the right triangles:\n\n$\\overline{AB}$ $13$ , and $\\overlin... |
https://artofproblemsolving.com/wiki/index.php/2011_AMC_8_Problems/Problem_20 | D | 750 | Quadrilateral $ABCD$ is a trapezoid, $AD = 15$ $AB = 50$ $BC = 20$ , and the altitude is $12$ . What is the area of the trapezoid?
[asy] pair A,B,C,D; A=(3,20); B=(35,20); C=(47,0); D=(0,0); draw(A--B--C--D--cycle); dot((0,0)); dot((3,20)); dot((35,20)); dot((47,0)); label("A",A,N); label("B",B,N); label("C",C,S); labe... | [
"\nIf you draw altitudes from $A$ and $B$ to $CD,$ the trapezoid will be divided into two right triangles and a rectangle. You can find the values of $a$ and $b$ with the Pythagorean theorem\n\\[a=\\sqrt{15^2-12^2}=\\sqrt{81}=9\\]\n\\[b=\\sqrt{20^2-12^2}=\\sqrt{256}=16\\]\n$ABYX$ is a rectangle so $XY=AB=50.$\n\\[C... |
https://artofproblemsolving.com/wiki/index.php/2015_AMC_12B_Problems/Problem_13 | B | 6 | Quadrilateral $ABCD$ is inscribed in a circle with $\angle BAC=70^{\circ}, \angle ADB=40^{\circ}, AD=4,$ and $BC=6$ . What is $AC$
$\textbf{(A)}\; 3+\sqrt{5} \qquad\textbf{(B)}\; 6 \qquad\textbf{(C)}\; \dfrac{9}{2}\sqrt{2} \qquad\textbf{(D)}\; 8-\sqrt{2} \qquad\textbf{(E)}\; 7$ | [
"$\\angle ADB$ and $\\angle ACB$ are both subtended by segment $AB$ , hence $\\angle ACB = \\angle ADB = 40^\\circ$ . By considering $\\triangle ABC$ , it follows that $\\angle ABC = 180^\\circ - (70^\\circ + 40^\\circ) = 70^\\circ$ . Hence $\\triangle ABC$ is isosceles, and $AC = BC = \\boxed{6}.$"
] |
https://artofproblemsolving.com/wiki/index.php/2017_AMC_12A_Problems/Problem_24 | A | 17 | Quadrilateral $ABCD$ is inscribed in circle $O$ and has side lengths $AB=3, BC=2, CD=6$ , and $DA=8$ . Let $X$ and $Y$ be points on $\overline{BD}$ such that $\frac{DX}{BD} = \frac{1}{4}$ and $\frac{BY}{BD} = \frac{11}{36}$ .
Let $E$ be the intersection of line $AX$ and the line through $Y$ parallel to $\overline{AD}$ ... | [
"Using the given ratios, note that $\\frac{XY}{BD} = 1 - \\frac{1}{4} - \\frac{11}{36} = \\frac{4}{9}.$\nBy AA Similarity, $\\triangle AXD \\sim \\triangle EXY$ with a ratio of $\\frac{DX}{XY} = \\frac{9}{16}$ and $\\triangle ACX \\sim \\triangle EFX$ with a ratio of $\\frac{AX}{XE} = \\frac{DX}{XY} = \\frac{9}{16}... |
https://artofproblemsolving.com/wiki/index.php/2020_AMC_10A_Problems/Problem_20 | D | 360 | Quadrilateral $ABCD$ satisfies $\angle ABC = \angle ACD = 90^{\circ}, AC=20,$ and $CD=30.$ Diagonals $\overline{AC}$ and $\overline{BD}$ intersect at point $E,$ and $AE=5.$ What is the area of quadrilateral $ABCD?$
$\textbf{(A) } 330 \qquad \textbf{(B) } 340 \qquad \textbf{(C) } 350 \qquad \textbf{(D) } 360 \qquad \tex... | [
"\nIt's crucial to draw a good diagram for this one. Since $AC=20$ and $CD=30$ , we get $[ACD]=300$ . Now we need to find $[ABC]$ to get the area of the whole quadrilateral. Drop an altitude from $B$ to $AC$ and call the point of intersection $F$ . Let $FE=x$ . Since $AE=5$ , then $AF=5-x$\nBy dropping this altitud... |
https://artofproblemsolving.com/wiki/index.php/2020_AMC_12A_Problems/Problem_18 | D | 360 | Quadrilateral $ABCD$ satisfies $\angle ABC = \angle ACD = 90^{\circ}, AC=20,$ and $CD=30.$ Diagonals $\overline{AC}$ and $\overline{BD}$ intersect at point $E,$ and $AE=5.$ What is the area of quadrilateral $ABCD?$
$\textbf{(A) } 330 \qquad \textbf{(B) } 340 \qquad \textbf{(C) } 350 \qquad \textbf{(D) } 360 \qquad \tex... | [
"\nIt's crucial to draw a good diagram for this one. Since $AC=20$ and $CD=30$ , we get $[ACD]=300$ . Now we need to find $[ABC]$ to get the area of the whole quadrilateral. Drop an altitude from $B$ to $AC$ and call the point of intersection $F$ . Let $FE=x$ . Since $AE=5$ , then $AF=5-x$\nBy dropping this altitud... |
https://artofproblemsolving.com/wiki/index.php/2022_AMC_10A_Problems/Problem_15 | D | 1,565 | Quadrilateral $ABCD$ with side lengths $AB=7, BC=24, CD=20, DA=15$ is inscribed in a circle. The area interior to the circle but exterior to the quadrilateral can be written in the form $\frac{a\pi-b}{c},$ where $a,b,$ and $c$ are positive integers such that $a$ and $c$ have no common prime factor. What is $a+b+c?$
$\t... | [
"Opposite angles of every cyclic quadrilateral are supplementary, so \\[\\angle B + \\angle D = 180^{\\circ}.\\] We claim that $AC=25.$ We can prove it by contradiction:\nBy the Inscribed Angle Theorem, we conclude that $\\overline{AC}$ is the diameter of the circle. So, the radius of the circle is $r=\\frac{AC}{2}... |
https://artofproblemsolving.com/wiki/index.php/2013_AMC_12A_Problems/Problem_15 | D | 204 | Rabbits Peter and Pauline have three offspring—Flopsie, Mopsie, and Cotton-tail. These five rabbits are to be distributed to four different pet stores so that no store gets both a parent and a child. It is not required that every store gets a rabbit. In how many different ways can this be done?
$\textbf{(A)} \ 96 \qqua... | [
"We tackle the problem by sorting it by how many stores are involved in the transaction.\n1) 2 stores are involved. There are $\\binom{4}{2} = 6$ ways to choose which of the stores are involved and 2 ways to choose which store recieves the parents. $6 \\cdot 2 = 12$ total arrangements.\n2) 3 stores are involved. Th... |
https://artofproblemsolving.com/wiki/index.php/2012_AMC_8_Problems/Problem_1 | E | 9 | Rachelle uses 3 pounds of meat to make 8 hamburgers for her family. How many pounds of meat does she need to make 24 hamburgers for a neighbourhood picnic?
$\textbf{(A)}\hspace{.05in}6 \qquad \textbf{(B)}\hspace{.05in}6\dfrac23 \qquad \textbf{(C)}\hspace{.05in}7\dfrac12 \qquad \textbf{(D)}\hspace{.05in}8 \qquad \textbf... | [
"Since Rachelle uses $3$ pounds of meat to make $8$ hamburgers, she uses $\\frac{3}{8}$ pounds of meat to make one hamburger. She'll need 24 times that amount of meat for 24 hamburgers, or $\\frac{3}{8} \\cdot 24 = \\boxed{9}$"
] |
https://artofproblemsolving.com/wiki/index.php/2015_AMC_8_Problems/Problem_20 | D | 7 | Ralph went to the store and bought 12 pairs of socks for a total of $$24$ . Some of the socks he bought cost $$1$ a pair, some of the socks he bought cost $$3$ a pair, and some of the socks he bought cost $$4$ a pair. If he bought at least one pair of each type, how many pairs of $$1$ socks did Ralph buy?
$\textbf{(A) ... | [
"So, let there be $x$ pairs of $$1$ socks, $y$ pairs of $$3$ socks, and $z$ pairs of $$4$ socks.\nWe have $x+y+z=12$ $x+3y+4z=24$ , and $x,y,z \\ge 1$\nNow, we subtract to find $2y+3z=12$ , and $y,z \\ge 1$ .\nIt follows that $2y$ is a multiple of $3$ and $3z$ is a multiple of $3$ . Since sum of 2 multiples of 3 = ... |
https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_17 | C | 4 | Ravon, Oscar, Aditi, Tyrone, and Kim play a card game. Each person is given $2$ cards out of a set of $10$ cards numbered $1,2,3, \dots,10.$ The score of a player is the sum of the numbers of their cards. The scores of the players are as follows: Ravon-- $11,$ Oscar-- $4,$ Aditi-- $7,$ Tyrone-- $16,$ Kim-- $17.$ Which ... | [
"By logical deduction, we consider the scores from lowest to highest: \\begin{align*} \\text{Oscar's score is 4.} &\\implies \\text{Oscar is given cards 1 and 3.} \\\\ &\\implies \\text{Aditi is given cards 2 and 5.} \\\\ &\\implies \\text{Ravon is given cards 4 and 7.} && (\\bigstar) \\\\ &\\implies \\text{Tyrone ... |
https://artofproblemsolving.com/wiki/index.php/2013_AMC_10B_Problems/Problem_8 | B | 16 | Ray's car averages $40$ miles per gallon of gasoline, and Tom's car averages $10$ miles per gallon of gasoline. Ray and Tom each drive the same number of miles. What is the cars' combined rate of miles per gallon of gasoline? $\textbf{(A)}\ 10 \qquad \textbf{(B)}\ 16 \qquad \textbf{(C)}\ 25 \qquad \textbf{(D)}\ 30 \qqu... | [
"Let Ray and Tom drive 40 miles. Ray's car would require $\\frac{40}{40}=1$ gallon of gas and Tom's car would require $\\frac{40}{10}=4$ gallons of gas. They would have driven a total of $40+40=80$ miles, on $1+4=5$ gallons of gas, for a combined rate of $\\frac{80}{5}=$ $\\boxed{16}$",
"Taking the harmonic mean ... |
https://artofproblemsolving.com/wiki/index.php/2013_AMC_12B_Problems/Problem_4 | B | 16 | Ray's car averages $40$ miles per gallon of gasoline, and Tom's car averages $10$ miles per gallon of gasoline. Ray and Tom each drive the same number of miles. What is the cars' combined rate of miles per gallon of gasoline? $\textbf{(A)}\ 10 \qquad \textbf{(B)}\ 16 \qquad \textbf{(C)}\ 25 \qquad \textbf{(D)}\ 30 \qqu... | [
"Let Ray and Tom drive 40 miles. Ray's car would require $\\frac{40}{40}=1$ gallon of gas and Tom's car would require $\\frac{40}{10}=4$ gallons of gas. They would have driven a total of $40+40=80$ miles, on $1+4=5$ gallons of gas, for a combined rate of $\\frac{80}{5}=$ $\\boxed{16}$",
"Taking the harmonic mean ... |
https://artofproblemsolving.com/wiki/index.php/2014_AIME_II_Problems/Problem_5 | null | 420 | Real numbers $r$ and $s$ are roots of $p(x)=x^3+ax+b$ , and $r+4$ and $s-3$ are roots of $q(x)=x^3+ax+b+240$ . Find the sum of all possible values of $|b|$ | [
"Because the coefficient of $x^2$ in both $p(x)$ and $q(x)$ is 0, the remaining root of $p(x)$ is $-(r+s)$ , and the remaining root of $q(x)$ is $-(r+s+1)$ . The coefficients of $x$ in $p(x)$ and $q(x)$ are both equal to $a$ , and equating the two coefficients gives \\[rs-(r+s)^2 = (r+4)(s-3)-(r+s+1)^2\\] from whic... |
https://artofproblemsolving.com/wiki/index.php/2012_AMC_10A_Problems/Problem_25 | D | 10 | Real numbers $x$ $y$ , and $z$ are chosen independently and at random from the interval $[0,n]$ for some positive integer $n$ . The probability that no two of $x$ $y$ , and $z$ are within 1 unit of each other is greater than $\frac {1}{2}$ . What is the smallest possible value of $n$
$\textbf{(A)}\ 7\qquad\textbf{(B)}\... | [
"Since $x,y,z$ are all reals located in $[0, n]$ , the number of choices for each one is continuous so we use geometric probability.\nWLOG( Without loss of generality ), assume that $n\\geq x \\geq y \\geq z \\geq 0$ . Then the set of points $(x,y,z)$ is a tetrahedron, or a triangular pyramid. The point $(x,y,z)$ d... |
https://artofproblemsolving.com/wiki/index.php/2012_AMC_10A_Problems/Problem_25 | null | 10 | Real numbers $x$ $y$ , and $z$ are chosen independently and at random from the interval $[0,n]$ for some positive integer $n$ . The probability that no two of $x$ $y$ , and $z$ are within 1 unit of each other is greater than $\frac {1}{2}$ . What is the smallest possible value of $n$
$\textbf{(A)}\ 7\qquad\textbf{(B)}\... | [
"Imagine Points $x$ $y$ $z$ as the \"starting points\" of three \"blocks\" of real numbers that have length $1$ . We are just trying to find the probability that those three \"blocks\" do not overlap. To do this we can set each unit of $1$ into $\\mu$ equal little increments, and take the limit of the probability a... |
https://artofproblemsolving.com/wiki/index.php/2020_AMC_10A_Problems/Problem_14 | D | 440 | Real numbers $x$ and $y$ satisfy $x + y = 4$ and $x \cdot y = -2$ . What is the value of
\[x + \frac{x^3}{y^2} + \frac{y^3}{x^2} + y?\]
$\textbf{(A)}\ 360\qquad\textbf{(B)}\ 400\qquad\textbf{(C)}\ 420\qquad\textbf{(D)}\ 440\qquad\textbf{(E)}\ 480$ | [
"\\[x + \\frac{x^3}{y^2} + \\frac{y^3}{x^2} + y=x+\\frac{x^3}{y^2}+y+\\frac{y^3}{x^2}=\\frac{x^3}{x^2}+\\frac{y^3}{x^2}+\\frac{y^3}{y^2}+\\frac{x^3}{y^2}\\]\nContinuing to combine \\[\\frac{x^3+y^3}{x^2}+\\frac{x^3+y^3}{y^2}=\\frac{(x^2+y^2)(x^3+y^3)}{x^2y^2}=\\frac{(x^2+y^2)(x+y)(x^2-xy+y^2)}{x^2y^2}\\] From the g... |
https://artofproblemsolving.com/wiki/index.php/2013_AMC_10B_Problems/Problem_11 | B | 2 | Real numbers $x$ and $y$ satisfy the equation $x^2+y^2=10x-6y-34$ . What is $x+y$
$\text{(A)}\ 1 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 3 \qquad \text{(D)}\ 6 \qquad \text{(E)}\ 8$ | [
"If we move every term dependent on $x$ or $y$ to the LHS, we get $x^2 - 10x + y^2 + 6y = -34$ . Adding $34$ to both sides, we have $x^2 - 10x + y^2 + 6y + 34 = 0$ . We can split the $34$ into $25$ and $9$ to get $(x - 5)^2 + (y + 3)^2 = 0$ . Notice this is a circle with radius $0$ , which only contains one point. ... |
https://artofproblemsolving.com/wiki/index.php/2013_AMC_12B_Problems/Problem_6 | B | 2 | Real numbers $x$ and $y$ satisfy the equation $x^2+y^2=10x-6y-34$ . What is $x+y$
$\text{(A)}\ 1 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 3 \qquad \text{(D)}\ 6 \qquad \text{(E)}\ 8$ | [
"If we move every term dependent on $x$ or $y$ to the LHS, we get $x^2 - 10x + y^2 + 6y = -34$ . Adding $34$ to both sides, we have $x^2 - 10x + y^2 + 6y + 34 = 0$ . We can split the $34$ into $25$ and $9$ to get $(x - 5)^2 + (y + 3)^2 = 0$ . Notice this is a circle with radius $0$ , which only contains one point. ... |
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12A_Problems/Problem_15 | D | 208 | Recall that the conjugate of the complex number $w = a + bi$ , where $a$ and $b$ are real numbers and $i = \sqrt{-1}$ , is the complex number $\overline{w} = a - bi$ . For any complex number $z$ , let $f(z) = 4i\hspace{1pt}\overline{z}$ . The polynomial \[P(z) = z^4 + 4z^3 + 3z^2 + 2z + 1\] has four complex roots: $z_1... | [
"By Vieta's formulas, $z_1z_2+z_1z_3+\\dots+z_3z_4=3$ , and $B=(4i)^2\\left(\\overline{z}_1\\,\\overline{z}_2+\\overline{z}_1\\,\\overline{z}_3+\\dots+\\overline{z}_3\\,\\overline{z}_4\\right).$\nSince $\\overline{a}\\cdot\\overline{b}=\\overline{ab},$ \\[B=(4i)^2\\left(\\overline{z_1z_2}+\\overline{z_1z_3}+\\overl... |
https://artofproblemsolving.com/wiki/index.php/2010_AMC_12A_Problems/Problem_3 | null | 10 | Rectangle $ABCD$ , pictured below, shares $50\%$ of its area with square $EFGH$ . Square $EFGH$ shares $20\%$ of its area with rectangle $ABCD$ . What is $\frac{AB}{AD}$
$\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 8 \qquad \textbf{(E)}\ 10$ | [
"From the problem statement, we know that \\[\\frac{[ABCD]}{2} = \\frac{[EFGH]}{5} \\Rightarrow [ABCD]=\\frac{2[EFGH]}{5}\\]\nIf we let $a = EF$ and $b = AD$ , we see \\[[ABCD] = 2ab = \\frac{2a^2}{5} \\Rightarrow b = \\frac{a}{5}\\] . Hence, $\\frac{AB}{AD} = \\frac{2a}{b} = 2a(\\frac{5}{a}) = \\boxed{10}$"
] |
https://artofproblemsolving.com/wiki/index.php/2010_AIME_I_Problems/Problem_13 | null | 69 | Rectangle $ABCD$ and a semicircle with diameter $AB$ are coplanar and have nonoverlapping interiors. Let $\mathcal{R}$ denote the region enclosed by the semicircle and the rectangle. Line $\ell$ meets the semicircle, segment $AB$ , and segment $CD$ at distinct points $N$ $U$ , and $T$ , respectively. Line $\ell$ divide... | [
"The center of the semicircle is also the midpoint of $AB$ . Let this point be O. Let $h$ be the length of $AD$\nRescale everything by 42, so $AU = 2, AN = 3, UB = 4$ . Then $AB = 6$ so $OA = OB = 3$\nSince $ON$ is a radius of the semicircle, $ON = 3$ . Thus $OAN$ is an equilateral triangle.\nLet $X$ $Y$ , and $Z$ ... |
https://artofproblemsolving.com/wiki/index.php/2014_AMC_8_Problems/Problem_14 | B | 13 | Rectangle $ABCD$ and right triangle $DCE$ have the same area. They are joined to form a trapezoid, as shown. What is $DE$
[asy] size(250); defaultpen(linewidth(0.8)); pair A=(0,5),B=origin,C=(6,0),D=(6,5),E=(18,0); draw(A--B--E--D--cycle^^C--D); draw(rightanglemark(D,C,E,30)); label("$A$",A,NW); label("$B$",B,SW); labe... | [
"The area of $\\bigtriangleup CDE$ is $\\frac{DC\\cdot CE}{2}$ . The area of $ABCD$ is $AB\\cdot AD=5\\cdot 6=30$ , which also must be equal to the area of $\\bigtriangleup CDE$ , which, since $DC=5$ , must in turn equal $\\frac{5\\cdot CE}{2}$ . Through transitivity, then, $\\frac{5\\cdot CE}{2}=30$ , and $CE=12$ ... |
https://artofproblemsolving.com/wiki/index.php/2011_AMC_10B_Problems/Problem_18 | E | 75 | Rectangle $ABCD$ has $AB = 6$ and $BC = 3$ . Point $M$ is chosen on side $AB$ so that $\angle AMD = \angle CMD$ . What is the degree measure of $\angle AMD$
$\textbf{(A)}\ 15 \qquad\textbf{(B)}\ 30 \qquad\textbf{(C)}\ 45 \qquad\textbf{(D)}\ 60 \qquad\textbf{(E)}\ 75$ | [
"It is given that $\\angle AMD \\sim \\angle CMD$ . Since $\\angle AMD$ and $\\angle CDM$ are alternate interior angles and $\\overline{AB} \\parallel \\overline{DC}$ $\\angle AMD \\cong \\angle CDM \\longrightarrow \\angle CMD \\cong \\angle CDM$ . Use the Base Angle Theorem to show $\\overline{DC} \\cong \\overli... |
https://artofproblemsolving.com/wiki/index.php/2011_AMC_10B_Problems/Problem_18 | null | 75 | Rectangle $ABCD$ has $AB = 6$ and $BC = 3$ . Point $M$ is chosen on side $AB$ so that $\angle AMD = \angle CMD$ . What is the degree measure of $\angle AMD$
$\textbf{(A)}\ 15 \qquad\textbf{(B)}\ 30 \qquad\textbf{(C)}\ 45 \qquad\textbf{(D)}\ 60 \qquad\textbf{(E)}\ 75$ | [
"After finding $MC = 6,$ we can continue using trigonometry as follows.\nWe know that $\\angle{BMC} = 180-2x$ and so $\\sin (180-2x) = \\frac{3}{6} = \\frac{1}{2}$\nIt is obvious that $\\sin (30) = \\frac{1}{2}$ and so $180-2x=30.$\nSolving, we have $x = \\boxed{75}$",
"Let $\\angle{DMC} = \\angle{AMD} = \\theta$... |
https://artofproblemsolving.com/wiki/index.php/2011_AMC_12B_Problems/Problem_10 | E | 75 | Rectangle $ABCD$ has $AB=6$ and $BC=3$ . Point $M$ is chosen on side $AB$ so that $\angle AMD=\angle CMD$ . What is the degree measure of $\angle AMD$
$\textrm{(A)}\ 15 \qquad \textrm{(B)}\ 30 \qquad \textrm{(C)}\ 45 \qquad \textrm{(D)}\ 60 \qquad \textrm{(E)}\ 75$ | [
"Since $AB \\parallel CD$ $\\angle AMD = \\angle CDM$ , so $\\angle AMD = \\angle CMD = \\angle CDM$ , so $\\bigtriangleup CMD$ is isosceles, and hence $CM=CD=6$ . Therefore, $\\angle BMC = 30^\\circ$ . Therefore $\\angle AMD=\\boxed{75}$"
] |
https://artofproblemsolving.com/wiki/index.php/2017_AIME_II_Problems/Problem_10 | null | 546 | Rectangle $ABCD$ has side lengths $AB=84$ and $AD=42$ . Point $M$ is the midpoint of $\overline{AD}$ , point $N$ is the trisection point of $\overline{AB}$ closer to $A$ , and point $O$ is the intersection of $\overline{CM}$ and $\overline{DN}$ . Point $P$ lies on the quadrilateral $BCON$ , and $\overline{BP}$ bisects ... | [
" Impose a coordinate system on the diagram where point $D$ is the origin. Therefore $A=(0,42)$ $B=(84,42)$ $C=(84,0)$ , and $D=(0,0)$ . Because $M$ is a midpoint and $N$ is a trisection point, $M=(0,21)$ and $N=(28,42)$ . The equation for line $DN$ is $y=\\frac{3}{2}x$ and the equation for line $CM$ is $\\frac{1}{... |
https://artofproblemsolving.com/wiki/index.php/2014_AMC_8_Problems/Problem_20 | B | 4 | Rectangle $ABCD$ has sides $CD=3$ and $DA=5$ . A circle of radius $1$ is centered at $A$ , a circle of radius $2$ is centered at $B$ , and a circle of radius $3$ is centered at $C$ . Which of the following is closest to the area of the region inside the rectangle but outside all three circles?
[asy] draw((0,0)--(5,0)--... | [
"The area in the rectangle but outside the circles is the area of the rectangle minus the area of all three of the quarter circles in the rectangle.\nThe area of the rectangle is $3\\cdot5 =15$ . The area of all 3 quarter circles is $\\frac{\\pi}{4}+\\frac{\\pi(2)^2}{4}+\\frac{\\pi(3)^2}{4} = \\frac{14\\pi}{4} = \\... |
https://artofproblemsolving.com/wiki/index.php/1987_AIME_Problems/Problem_6 | null | 193 | Rectangle $ABCD$ is divided into four parts of equal area by five segments as shown in the figure, where $XY = YB + BC + CZ = ZW = WD + DA + AX$ , and $PQ$ is parallel to $AB$ . Find the length of $AB$ (in cm) if $BC = 19$ cm and $PQ = 87$ cm.
AIME 1987 Problem 6.png | [
"Since $XY = WZ$ $PQ = PQ$ and the areas of the trapezoids $PQZW$ and $PQYX$ are the same, then the heights of the trapezoids are the same. Thus both trapezoids have area $\\frac{1}{2} \\cdot \\frac{19}{2}(XY + PQ) = \\frac{19}{4}(XY + 87)$ . This number is also equal to one quarter the area of the entire rectang... |
https://artofproblemsolving.com/wiki/index.php/2007_AIME_II_Problems/Problem_9 | null | 259 | Rectangle $ABCD$ is given with $AB=63$ and $BC=448.$ Points $E$ and $F$ lie on $AD$ and $BC$ respectively, such that $AE=CF=84.$ The inscribed circle of triangle $BEF$ is tangent to $EF$ at point $P,$ and the inscribed circle of triangle $DEF$ is tangent to $EF$ at point $Q.$ Find $PQ.$ | [
"Several Pythagorean triples exist amongst the numbers given. $BE = DF = \\sqrt{63^2 + 84^2} = 21\\sqrt{3^2 + 4^2} = 105$ . Also, the length of $EF = \\sqrt{63^2 + (448 - 2\\cdot84)^2} = 7\\sqrt{9^2 + 40^2} = 287$\nUse the Two Tangent Theorem on $\\triangle BEF$ . Since both circles are inscribed in congruent trian... |
https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_18 | A | 240 | Rectangle $ABCD$ is inscribed in a semicircle with diameter $\overline{FE},$ as shown in the figure. Let $DA=16,$ and let $FD=AE=9.$ What is the area of $ABCD?$
[asy] draw(arc((0,0),17,180,0)); draw((-17,0)--(17,0)); fill((-8,0)--(-8,15)--(8,15)--(8,0)--cycle, 1.5*grey); draw((-8,0)--(-8,15)--(8,15)--(8,0)--cycle); do... | [
"\nLet $O$ be the center of the semicircle. The diameter of the semicircle is $9+16+9=34$ , so $OC = 17$ . By symmetry, $O$ is the midpoint of $DA$ , so $OD=OA=\\frac{16}{2}= 8$ . By the Pythagorean theorem in right-angled triangle $ODC$ (or $OBA$ ), we have that $CD$ (or $AB$ ) is $\\sqrt{17^2-8^2}=15$ . According... |
https://artofproblemsolving.com/wiki/index.php/1991_AIME_Problems/Problem_2 | null | 840 | Rectangle $ABCD_{}^{}$ has sides $\overline {AB}$ of length 4 and $\overline {CB}$ of length 3. Divide $\overline {AB}$ into 168 congruent segments with points $A_{}^{}=P_0, P_1, \ldots, P_{168}=B$ , and divide $\overline {CB}$ into 168 congruent segments with points $C_{}^{}=Q_0, Q_1, \ldots, Q_{168}=B$ . For $1_{}^{}... | [
"The length of the diagonal is $\\sqrt{3^2 + 4^2} = 5$ (a 3-4-5 right triangle ). For each $k$ $\\overline{P_kQ_k}$ is the hypotenuse of a $3-4-5$ right triangle with sides of $3 \\cdot \\frac{168-k}{168}, 4 \\cdot \\frac{168-k}{168}$ . Thus, its length is $5 \\cdot \\frac{168-k}{168}$ . Let $a_k=\\frac{5(168-k)}{1... |
https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_22 | C | 3 | Rectangle $DEFA$ below is a $3 \times 4$ rectangle with $DC=CB=BA=1$ . The area of the "bat wings" (shaded area) is
[asy] draw((0,0)--(3,0)--(3,4)--(0,4)--(0,0)--(2,4)--(3,0)); draw((3,0)--(1,4)--(0,0)); fill((0,0)--(1,4)--(1.5,3)--cycle, black); fill((3,0)--(2,4)--(1.5,3)--cycle, black); label("$A$",(3.05,4.2)); label... | [
"Let G be the midpoint B and C\nDraw H, J, K beneath C, G, B, respectively.\n\nLet us take a look at rectangle CDEH. I have labeled E' for convenience. First of all, we can see that EE'H and CE'B are similar triangles because all their three angles are the same. Furthermore, since EH=CB, we can confirm that EE'H an... |
https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_25 | A | 651 | Rectangles $R_1$ and $R_2,$ and squares $S_1,\,S_2,\,$ and $S_3,$ shown below, combine to form a rectangle that is 3322 units wide and 2020 units high. What is the side length of $S_2$ in units?
[asy] draw((0,0)--(5,0)--(5,3)--(0,3)--(0,0)); draw((3,0)--(3,1)--(0,1)); draw((3,1)--(3,2)--(5,2)); draw((3,2)--(2,2)--(2,1)... | [
"Let the side length of each square $S_k$ be $s_k$ . Then, from the diagram, we can line up the top horizontal lengths of $S_1$ $S_2$ , and $S_3$ to cover the top side of the large rectangle, so $s_{1}+s_{2}+s_{3}=3322$ . Similarly, the short side of $R_2$ will be $s_1-s_2$ , and lining this up with the left side o... |
https://artofproblemsolving.com/wiki/index.php/2019_AIME_II_Problems/Problem_13 | null | 504 | Regular octagon $A_1A_2A_3A_4A_5A_6A_7A_8$ is inscribed in a circle of area $1.$ Point $P$ lies inside the circle so that the region bounded by $\overline{PA_1},\overline{PA_2},$ and the minor arc $\widehat{A_1A_2}$ of the circle has area $\tfrac{1}{7},$ while the region bounded by $\overline{PA_3},\overline{PA_4},$ an... | [
"The actual size of the diagram doesn't matter. To make calculation easier, we discard the original area of the circle, \\(1\\), and assume the side length of the octagon is \\(2\\).\nLet \\(r\\) denote the radius of the circle, \\(O\\) be the center of the circle. Then: \\[r^2= 1^2 + \\left(\\sqrt{2}+1\\right)^2= ... |
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10B_Problems/Problem_21 | E | 68 | Regular polygons with $5,6,7,$ and $8$ sides are inscribed in the same circle. No two of the polygons share a vertex, and no three of their sides intersect at a common point. At how many points inside the circle do two of their sides intersect?
$(\textbf{A})\: 52\qquad(\textbf{B}) \: 56\qquad(\textbf{C}) \: 60\qquad(\t... | [
"Imagine we have $2$ regular polygons with $m$ and $n$ sides and $m>n$ inscribed in a circle without sharing a vertex. We see that each side of the polygon with $n$ sides (the polygon with fewer sides) will be intersected twice.\n(We can see this because to have a vertex of the $m$ -gon on an arc subtended by a sid... |
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12B_Problems/Problem_19 | E | 68 | Regular polygons with $5,6,7,$ and $8$ sides are inscribed in the same circle. No two of the polygons share a vertex, and no three of their sides intersect at a common point. At how many points inside the circle do two of their sides intersect?
$(\textbf{A})\: 52\qquad(\textbf{B}) \: 56\qquad(\textbf{C}) \: 60\qquad(\t... | [
"Imagine we have $2$ regular polygons with $m$ and $n$ sides and $m>n$ inscribed in a circle without sharing a vertex. We see that each side of the polygon with $n$ sides (the polygon with fewer sides) will be intersected twice.\n(We can see this because to have a vertex of the $m$ -gon on an arc subtended by a sid... |
https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_8 | null | 125 | Rhombus $ABCD$ has $\angle BAD < 90^\circ.$ There is a point $P$ on the incircle of the rhombus such that the distances from $P$ to the lines $DA,AB,$ and $BC$ are $9,$ $5,$ and $16,$ respectively. Find the perimeter of $ABCD.$ | [
"This solution refers to the Diagram section.\nLet $O$ be the incenter of $ABCD$ for which $\\odot O$ is tangent to $\\overline{DA},\\overline{AB},$ and $\\overline{BC}$ at $X,Y,$ and $Z,$ respectively. Moreover, suppose that $R,S,$ and $T$ are the feet of the perpendiculars from $P$ to $\\overleftrightarrow{DA},\\... |
https://artofproblemsolving.com/wiki/index.php/2006_AMC_10B_Problems/Problem_15 | C | 8 | Rhombus $ABCD$ is similar to rhombus $BFDE$ . The area of rhombus $ABCD$ is $24$ and $\angle BAD = 60^\circ$ . What is the area of rhombus $BFDE$
[asy] defaultpen(linewidth(0.7)+fontsize(10)); size(120); pair A=origin, B=(2,0), C=(3, sqrt(3)), D=(1, sqrt(3)), E=(1, 1/sqrt(3)), F=(2, 2/sqrt(3)); pair point=(3/2, sqrt(3)... | [
"Using the property that opposite angles are equal in a rhombus $\\angle DAB = \\angle DCB = 60 ^\\circ$ and $\\angle ADC = \\angle ABC = 120 ^\\circ$ . It is easy to see that rhombus $ABCD$ is made up of equilateral triangles $DAB$ and $DCB$ . Let the lengths of the sides of rhombus $ABCD$ be $s$\nThe longer diago... |
https://artofproblemsolving.com/wiki/index.php/1991_AIME_Problems/Problem_12 | null | 677 | Rhombus $PQRS^{}_{}$ is inscribed in rectangle $ABCD^{}_{}$ so that vertices $P^{}_{}$ $Q^{}_{}$ $R^{}_{}$ , and $S^{}_{}$ are interior points on sides $\overline{AB}$ $\overline{BC}$ $\overline{CD}$ , and $\overline{DA}$ , respectively. It is given that $PB^{}_{}=15$ $BQ^{}_{}=20$ $PR^{}_{}=30$ , and $QS^{}_{}=40$ . L... | [
"Let $O$ be the center of the rhombus. Via parallel sides and alternate interior angles, we see that the opposite triangles are congruent $\\triangle BPQ \\cong \\triangle DRS$ $\\triangle APS \\cong \\triangle CRQ$ ). Quickly we realize that $O$ is also the center of the rectangle.\nBy the Pythagorean Theorem , we... |
https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_8 | C | 8,072 | Ricardo has $2020$ coins, some of which are pennies ( $1$ -cent coins) and the rest of which are nickels ( $5$ -cent coins). He has at least one penny and at least one nickel. What is the difference in cents between the greatest possible and least amounts of money that Ricardo can have?
$\textbf{(A) }\text{806} \qquad ... | [
"Clearly, the amount of money Ricardo has will be maximized when he has the maximum number of nickels. Since he must have at least one penny, the greatest number of nickels he can have is $2019$ , giving a total of $(2019\\cdot 5 + 1)$ cents. Analogously, the amount of money he has will be least when he has the gre... |
https://artofproblemsolving.com/wiki/index.php/2002_AMC_10B_Problems/Problem_24 | D | 10 | Riders on a Ferris wheel travel in a circle in a vertical plane. A particular wheel has radius $20$ feet and revolves at the constant rate of one revolution per minute. How many seconds does it take a rider to travel from the bottom of the wheel to a point $10$ vertical feet above the bottom?
$\mathrm{(A) \ } 5\qquad \... | [
"We can let this circle represent the ferris wheel with center $O,$ and $C$ represent the desired point $10$ feet above the bottom. Draw a diagram like the one above. We find out $\\triangle OBC$ is a $30-60-90$ triangle. That means $\\angle BOC = 60^\\circ$ and the ferris wheel has made $\\frac{60}{360} = \\frac{1... |
https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_16 | D | 13 | Right triangle $ABC$ has leg lengths $AB=20$ and $BC=21$ . Including $\overline{AB}$ and $\overline{BC}$ , how many line segments with integer length can be drawn from vertex $B$ to a point on hypotenuse $\overline{AC}$
$\textbf{(A) }5 \qquad \textbf{(B) }8 \qquad \textbf{(C) }12 \qquad \textbf{(D) }13 \qquad \textbf{(... | [
"\nAs the problem has no diagram, we draw a diagram. The hypotenuse has length $29$ . Let $P$ be the foot of the altitude from $B$ to $AC$ . Note that $BP$ is the shortest possible length of any segment. Writing the area of the triangle in two ways, we can solve for $BP=\\dfrac{20\\cdot 21}{29}$ , which is between... |
https://artofproblemsolving.com/wiki/index.php/2005_AIME_I_Problems/Problem_5 | null | 630 | Robert has 4 indistinguishable gold coins and 4 indistinguishable silver coins. Each coin has an engraving of one face on one side, but not on the other. He wants to stack the eight coins on a table into a single stack so that no two adjacent coins are face to face. Find the number of possible distinguishable arrangeme... | [
"There are two separate parts to this problem: one is the color (gold vs silver), and the other is the orientation.\nThere are ${8\\choose4} = 70$ ways to position the gold coins in the stack of 8 coins, which determines the positions of the silver coins.\nCreate a string of letters H and T to denote the orientatio... |
https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_23 | D | 7,000 | Rodrigo has a very large sheet of graph paper. First he draws a line segment connecting point $(0,4)$ to point $(2,0)$ and colors the $4$ cells whose interiors intersect the segment, as shown below. Next Rodrigo draws a line segment connecting point $(2000,3000)$ to point $(5000,8000)$ . How many cells will he color th... | [
"Let $f(x, y)$ be the number of cells the line segment from $(0, 0)$ to $(x, y)$ passes through. The problem is then equivalent to finding \\[f(5000-2000, 8000-3000)=f(3000, 5000).\\] Sometimes the segment passes through lattice points in between the endpoints, which happens $\\text{gcd}(3000, 5000)-1=999$ times. ... |
https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_12 | E | 26 | Rohan keeps a total of 90 guppies in 4 fish tanks.
How many guppies are in the 4th tank?
$\textbf{(A)}\ 20 \qquad \textbf{(B)}\ 21 \qquad \textbf{(C)}\ 23 \qquad \textbf{(D)}\ 24 \qquad \textbf{(E)}\ 26$ | [
"Let $x$ denote the number of guppies in the first tank.\nThen, we have the following for the number of guppies in the rest of the tanks:\nThe number of guppies in all of the tanks combined is 90, so we can write the equation\n$x + x + 1 + x + 1 + 2 + x + 1 + 2 + 3 = 90$\nSimplifying the equation gives\n$4x + 10 = ... |
https://artofproblemsolving.com/wiki/index.php/2006_AMC_10A_Problems/Problem_12 | C | 4 | Rolly wishes to secure his dog with an 8-foot rope to a square shed that is 16 feet on each side. His preliminary drawings are shown.
[asy] size(150); pathpen = linewidth(0.6); defaultpen(fontsize(10)); D((0,0)--(16,0)--(16,-16)--(0,-16)--cycle); D((16,-8)--(24,-8)); label('Dog', (24, -8), SE); MP('I', (8,-8), (0,0));... | [
" Let us first examine the area of both possible arrangements. The rope outlines a circular boundary that the dog may dwell in. Arrangement $I$ allows the dog $\\frac12\\cdot(\\pi\\cdot8^2) = 32\\pi$ square feet of area. Arrangement $II$ allows $32\\pi$ square feet plus a little more on the top part of the fence. S... |
https://artofproblemsolving.com/wiki/index.php/2003_AMC_10B_Problems/Problem_4 | A | 108 | Rose fills each of the rectangular regions of her rectangular flower bed with a different type of flower. The lengths, in feet, of the rectangular regions in her flower bed are as shown in the figure. She plants one flower per square foot in each region. Asters cost $$ $1$ each, begonias $$ $1.50$ each, cannas $$ $2$ e... | [
"The areas of the five regions from greatest to least are $21,20,15,6$ and $4$\nIf we want to minimize the cost, we want to maximize the area of the cheapest flower and minimize the area of the most expensive flower. Doing this, the cost is $1\\cdot21+1.50\\cdot20+2\\cdot15+2.50\\cdot6+3\\cdot4$ , which simplifies ... |
https://artofproblemsolving.com/wiki/index.php/2003_AMC_12B_Problems/Problem_3 | A | 108 | Rose fills each of the rectangular regions of her rectangular flower bed with a different type of flower. The lengths, in feet, of the rectangular regions in her flower bed are as shown in the figure. She plants one flower per square foot in each region. Asters cost $$ $1$ each, begonias $$ $1.50$ each, cannas $$ $2$ e... | [
"The areas of the five regions from greatest to least are $21,20,15,6$ and $4$\nIf we want to minimize the cost, we want to maximize the area of the cheapest flower and minimize the area of the most expensive flower. Doing this, the cost is $1\\cdot21+1.50\\cdot20+2\\cdot15+2.50\\cdot6+3\\cdot4$ , which simplifies ... |
https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_20 | C | 5 | Rows 1, 2, 3, 4, and 5 of a triangular array of integers are shown below.
[asy] size(4.5cm); label("$1$", (0,0)); label("$1$", (-0.5,-2/3)); label("$1$", (0.5,-2/3)); label("$1$", (-1,-4/3)); label("$3$", (0,-4/3)); label("$1$", (1,-4/3)); label("$1$", (-1.5,-2)); label("$5$", (-0.5,-2)); label("$5$", (0.5,-2)); label(... | [
"First, let $R(n)$ be the sum of the $n$ th row. Now, with some observation and math instinct, we can guess that $R(n) = 2^n - n$\nNow we try to prove it by induction,\n$R(1) = 2^n - n = 2^1 - 1 = 1$ (works for base case)\n$R(k) = 2^k - k$\n$R(k+1) = 2^{k+1} - (k + 1) = 2(2^k) - k - 1$\nBy definition from the quest... |
https://artofproblemsolving.com/wiki/index.php/2011_AMC_10A_Problems/Problem_15 | C | 440 | Roy bought a new battery-gasoline hybrid car. On a trip the car ran exclusively on its battery for the first $40$ miles, then ran exclusively on gasoline for the rest of the trip, using gasoline at a rate of $0.02$ gallons per mile. On the whole trip he averaged $55$ miles per gallon. How long was the trip in miles?
$\... | [
"We know that $\\frac{\\text{total miles}}{\\text{total gas}}=55$ . Let $x$ be the distance in miles the car traveled during the time it ran on gasoline, then the amount of gas used is $0.02x$ . The total distance traveled is $40+x$ , so we get $\\frac{40+x}{0.02x}=55$ . Solving this equation, we get $x=400$ , so t... |
https://artofproblemsolving.com/wiki/index.php/2011_AMC_10A_Problems/Problem_15 | null | 440 | Roy bought a new battery-gasoline hybrid car. On a trip the car ran exclusively on its battery for the first $40$ miles, then ran exclusively on gasoline for the rest of the trip, using gasoline at a rate of $0.02$ gallons per mile. On the whole trip he averaged $55$ miles per gallon. How long was the trip in miles?
$\... | [
"Let $d$ be the length of the trip in miles. Roy used no gasoline for the 40 first miles, then used 0.02 gallons of gasoline per mile on the remaining $d - 40$ miles, for a total of $0.02 (d - 40)$ gallons. Hence, his average mileage was \\[\\frac{d}{0.02 (d - 40)} = 55.\\] Multiplying both sides by $0.02 (d - 40)$... |
https://artofproblemsolving.com/wiki/index.php/2008_AIME_II_Problems/Problem_2 | null | 620 | Rudolph bikes at a constant rate and stops for a five-minute break at the end of every mile. Jennifer bikes at a constant rate which is three-quarters the rate that Rudolph bikes, but Jennifer takes a five-minute break at the end of every two miles. Jennifer and Rudolph begin biking at the same time and arrive at the $... | [
"Let $r$ be the time Rudolph takes disregarding breaks and $\\frac{4}{3}r$ be the time Jennifer takes disregarding breaks. We have the equation \\[r+5\\left(49\\right)=\\frac{4}{3}r+5\\left(24\\right)\\] \\[125=\\frac13r\\] \\[r=375.\\] Thus, the total time they take is $375 + 5(49) = \\boxed{620}$ minutes.",
"Le... |
https://artofproblemsolving.com/wiki/index.php/2010_AMC_8_Problems/Problem_9 | D | 84 | Ryan got $80\%$ of the problems correct on a $25$ -problem test, $90\%$ on a $40$ -problem test, and $70\%$ on a $10$ -problem test. What percent of all the problems did Ryan answer correctly?
$\textbf{(A)}\ 64 \qquad\textbf{(B)}\ 75\qquad\textbf{(C)}\ 80\qquad\textbf{(D)}\ 84\qquad\textbf{(E)}\ 86$ | [
"Ryan answered $(0.8)(25)=20$ problems correct on the first test, $(0.9)(40)=36$ on the second, and $(0.7)(10)=7$ on the third. This amounts to a total of $20+36+7=63$ problems correct. The total number of problems is $25+40+10=75.$ Therefore, the percentage is $\\dfrac{63}{75} = 84\\% \\rightarrow \\boxed{84}$"
] |
https://artofproblemsolving.com/wiki/index.php/2003_AMC_10A_Problems/Problem_24 | E | 12 | Sally has five red cards numbered $1$ through $5$ and four blue cards numbered $3$ through $6$ . She stacks the cards so that the colors alternate and so that the number on each red card divides evenly into the number on each neighboring blue card. What is the sum of the numbers on the middle three cards?
$\mathrm{(A) ... | [
"Let $R_i$ and $B_j$ designate the red card numbered $i$ and the blue card numbered $j$ , respectively.\n$B_5$ is the only blue card that $R_5$ evenly divides, so $R_5$ must be at one end of the stack and $B_5$ must be the card next to it.\n$R_1$ is the only other red card that evenly divides $B_5$ , so $R_1$ must ... |
https://artofproblemsolving.com/wiki/index.php/2003_AMC_12A_Problems/Problem_12 | E | 12 | Sally has five red cards numbered $1$ through $5$ and four blue cards numbered $3$ through $6$ . She stacks the cards so that the colors alternate and so that the number on each red card divides evenly into the number on each neighboring blue card. What is the sum of the numbers on the middle three cards?
$\mathrm{(A) ... | [
"Let $R_i$ and $B_j$ designate the red card numbered $i$ and the blue card numbered $j$ , respectively.\n$B_5$ is the only blue card that $R_5$ evenly divides, so $R_5$ must be at one end of the stack and $B_5$ must be the card next to it.\n$R_1$ is the only other red card that evenly divides $B_5$ , so $R_1$ must ... |
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