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/2019_AIME_II_Problems/Problem_10 | null | 547 | There is a unique angle $\theta$ between $0^{\circ}$ and $90^{\circ}$ such that for nonnegative integers $n$ , the value of $\tan{\left(2^{n}\theta\right)}$ is positive when $n$ is a multiple of $3$ , and negative otherwise. The degree measure of $\theta$ is $\tfrac{p}{q}$ , where $p$ and $q$ are relatively prime integ... | [
"Note that if $\\tan \\theta$ is positive, then $\\theta$ is in the first or third quadrant, so $0^{\\circ} < \\theta < 90^{\\circ} \\pmod{180^{\\circ}}$\nFurthermore, the only way $\\tan{\\left(2^{n}\\theta\\right)}$ can be positive for all $n$ that are multiples of $3$ is when: \\[2^0\\theta \\equiv 2^3\\theta \\... |
https://artofproblemsolving.com/wiki/index.php/2020_AMC_12A_Problems/Problem_10 | E | 13 | There is a unique positive integer $n$ such that \[\log_2{(\log_{16}{n})} = \log_4{(\log_4{n})}.\] What is the sum of the digits of $n?$
$\textbf{(A) } 4 \qquad \textbf{(B) } 7 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 11 \qquad \textbf{(E) } 13$ | [
"We can use the fact that $\\log_{a^b} c = \\frac{1}{b} \\log_a c.$ This can be proved by using change of base formula to base $a.$\nSo, the original equation $\\log_2{(\\log_{2^4}{n})} = \\log_{2^2}{(\\log_{2^2}{n})}$ becomes \\[\\log_2\\left({\\frac{1}{4}\\log_{2}{n}}\\right) = \\frac{1}{2}\\log_2\\left({\\frac{1... |
https://artofproblemsolving.com/wiki/index.php/2020_AIME_I_Problems/Problem_2 | null | 17 | There is a unique positive real number $x$ such that the three numbers $\log_8{2x}$ $\log_4{x}$ , and $\log_2{x}$ , in that order, form a geometric progression with positive common ratio. The number $x$ can be written as $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$ | [
"Since these form a geometric series, $\\frac{\\log_2{x}}{\\log_4{x}}$ is the common ratio. Rewriting this, we get $\\frac{\\log_x{4}}{\\log_x{2}} = \\log_2{4} = 2$ by base change formula. Therefore, the common ratio is 2. Now $\\frac{\\log_4{x}}{\\log_8{2x}} = 2 \\implies \\log_4{x} = 2\\log_8{2} + 2\\log_8{x} \\i... |
https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_25 | C | 1 | There is a unique sequence of integers $a_1, a_2, \cdots a_{2023}$ such that \[\tan2023x = \frac{a_1 \tan x + a_3 \tan^3 x + a_5 \tan^5 x + \cdots + a_{2023} \tan^{2023} x}{1 + a_2 \tan^2 x + a_4 \tan^4 x \cdots + a_{2022} \tan^{2022} x}\] whenever $\tan 2023x$ is defined. What is $a_{2023}?$
$\textbf{(A) } -2023 \qqua... | [
"\\begin{align*} \\cos 2023 x + i \\sin 2023 x &= (\\cos x + i \\sin x)^{2023}\\\\ &= \\cos^{2023} x + \\binom{2023}{1} \\cos^{2022} x (i\\sin x) + \\binom{2023}{2} \\cos^{2021} x (i \\sin x)^{2} +\\binom{2023}{3} \\cos^{2023} x (i \\sin x)^{3}\\\\ &+ \\dots + \\binom{2023}{2022} \\cos x (i \\sin x)^{2022} + (i \\s... |
https://artofproblemsolving.com/wiki/index.php/2006_AIME_II_Problems/Problem_8 | null | 336 | There is an unlimited supply of congruent equilateral triangles made of colored paper. Each triangle is a solid color with the same color on both sides of the paper. A large equilateral triangle is constructed from four of these paper triangles. Two large triangles are considered distinguishable if it is not possible t... | [
"If two of our big equilateral triangles have the same color for their center triangle and the same multiset of colors for their outer three triangles, we can carry one onto the other by a combination of rotation and reflection. Thus, to make two triangles distinct, they must differ either in their center triangle... |
https://artofproblemsolving.com/wiki/index.php/2007_AMC_8_Problems/Problem_1 | D | 12 | Theresa's parents have agreed to buy her tickets to see her favorite band if she spends an average of $10$ hours per week helping around the house for $6$ weeks. For the first $5$ weeks she helps around the house for $8$ $11$ $7$ $12$ and $10$ hours. How many hours must she work for the final week to earn the tickets?
... | [
"Let $x$ be the number of hours she must work for the final week. We are looking for the average, so \\[\\frac{8 + 11 + 7 + 12 + 10 + x}{6} = 10\\] Solving gives: \\[\\frac{48 + x}{6} = 10\\]\n\\[48 + x = 60\\]\n\\[x = 12\\]\nSo, the answer is $\\boxed{12}$",
"Use average deviation:\nThe average is 10 hour per da... |
https://artofproblemsolving.com/wiki/index.php/2004_AMC_8_Problems/Problem_15 | C | 11 | Thirteen black and six white hexagonal tiles were used to create the figure below. If a new figure is created by attaching a border of white tiles with the same size and shape as the others, what will be the difference between the total number of white tiles and the total number of black tiles in the new figure?
[asy] ... | [
"The first ring around the middle tile has $6$ tiles, and the second has $12$ . From this pattern, the third ring has $18$ tiles. Of these, $6+18=24$ are white and $1+12=13$ are black, with a difference of $24-13 = \\boxed{11}$"
] |
https://artofproblemsolving.com/wiki/index.php/1990_AJHSME_Problems/Problem_24 | C | 3 | Three $\Delta$ 's and a $\diamondsuit$ will balance nine $\bullet$ 's. One $\Delta$ will balance a $\diamondsuit$ and a $\bullet$
[asy] unitsize(5.5); fill((0,0)--(-4,-2)--(4,-2)--cycle,black); draw((-12,2)--(-12,0)--(12,0)--(12,2)); draw(ellipse((-12,5),8,3)); draw(ellipse((12,5),8,3)); label("$\Delta \hspace{2 mm}\D... | [
"For simplicity, suppose $\\Delta = a$ $\\diamondsuit = b$ and $\\bullet = c$ . Then, \\[3a+b=9c\\] \\[a=b+c\\] and we want to know what $2b$ is in terms of $c$ . Substituting the second equation into the first, we have \\[4b=6c\\Rightarrow 2b=3c\\]\nThus, we need $3$ $\\bullet$ 's $\\rightarrow \\boxed{3}$"
] |
https://artofproblemsolving.com/wiki/index.php/2008_AMC_8_Problems/Problem_14 | C | 4 | Three $\text{A's}$ , three $\text{B's}$ , and three $\text{C's}$ are placed in the nine spaces so that each row and column contains one of each letter. If $\text{A}$ is placed in the upper left corner, how many arrangements are possible?
[asy] size((80)); draw((0,0)--(9,0)--(9,9)--(0,9)--(0,0)); draw((3,0)--(3,9)); dra... | [
"There are $2$ ways to place the remaining $\\text{As}$ $2$ ways to place the remaining $\\text{Bs}$ , and $1$ way to place the remaining $\\text{Cs}$ for a total of $(2)(2)(1) = \\boxed{4}$"
] |
https://artofproblemsolving.com/wiki/index.php/1985_AIME_Problems/Problem_15 | null | 864 | Three 12 cm $\times$ 12 cm squares are each cut into two pieces $A$ and $B$ , as shown in the first figure below, by joining the midpoints of two adjacent sides. These six pieces are then attached to a regular hexagon , as shown in the second figure, so as to fold into a polyhedron . What is the volume (in $\mathrm{cm}... | [
"Note that gluing two of the given polyhedra together along a hexagonal face (rotated $60^\\circ$ from each other) yields a cube , so the volume is $\\frac12 \\cdot 12^3 = 864$ , so our answer is $\\boxed{864}$"
] |
https://artofproblemsolving.com/wiki/index.php/1997_AJHSME_Problems/Problem_13 | A | 31 | Three bags of jelly beans contain 26, 28, and 30 beans. The ratios of yellow beans to all beans in each of these bags are $50\%$ $25\%$ , and $20\%$ , respectively. All three bags of candy are dumped into one bowl. Which of the following is closest to the ratio of yellow jelly beans to all beans in the bowl?
$\text{... | [
"In bag $A$ , there are $26$ jellybeans, and $50\\%$ are yellow. That means there are $26\\times 50\\% = 26\\times 0.50 = 13$ yellow jelly beans in this bag.\nIn bag $B$ , there are $28$ jellybeans, and $25\\%$ are yellow. That means there are $28\\times 25\\% = 28\\times 0.25 = 7$ yellow jelly beans in this bag.... |
https://artofproblemsolving.com/wiki/index.php/2021_AMC_12B_Problems/Problem_23 | A | 55 | Three balls are randomly and independantly tossed into bins numbered with the positive integers so that for each ball, the probability that it is tossed into bin $i$ is $2^{-i}$ for $i=1,2,3,....$ More than one ball is allowed in each bin. The probability that the balls end up evenly spaced in distinct bins is $\frac p... | [
"\"Evenly spaced\" just means the bins form an arithmetic sequence.\nSuppose the middle bin in the sequence is $x$ . There are $x-1$ different possibilities for the first bin, and these two bins uniquely determine the final bin. Now, the probability that these $3$ bins are chosen is $6\\cdot 2^{-3x} = 6\\cdot \\fra... |
https://artofproblemsolving.com/wiki/index.php/1998_AHSME_Problems/Problem_20 | null | 4 | Three cards, each with a positive integer written on it, are lying face-down on a table. Casey, Stacy, and Tracy are told that
First, Casey looks at the number on the leftmost card and says, "I don't have enough information to determine the other two numbers." Then Tracy looks at the number on the rightmost card and s... | [
"Initially, there are the following possibilities for the numbers on the cards: $(1,2,10)$ $(1,3,9)$ $(1,4,8)$ $(1,5,7)$ $(2,3,8)$ $(2,4,7)$ $(2,5,6)$ , and $(3,4,6)$\nIf Casey saw the number $3$ , she would have known the other two numbers. As she does not, we eliminated the possibility $(3,4,6)$\nAt this moment, ... |
https://artofproblemsolving.com/wiki/index.php/1984_AIME_Problems/Problem_6 | null | 24 | Three circles, each of radius $3$ , are drawn with centers at $(14, 92)$ $(17, 76)$ , and $(19, 84)$ . A line passing through $(17,76)$ is such that the total area of the parts of the three circles to one side of the line is equal to the total area of the parts of the three circles to the other side of it. What is the ... | [
"1984 AIME-6.png\nThe line passes through the center of the bottom circle; hence it is the circle's diameter and splits the circle into two equal areas. For the rest of the problem, we do not have to worry about that circle.\nDraw the midpoint of $\\overline{AC}$ (the centers of the other two circles), and call it ... |
https://artofproblemsolving.com/wiki/index.php/2000_AMC_8_Problems/Problem_19 | null | 50 | Three circular arcs of radius $5$ units bound the region shown. Arcs $AB$ and $AD$ are quarter-circles, and arc $BCD$ is a semicircle. What is the area, in square units, of the region?
[asy] pair A,B,C,D; A = (0,0); B = (-5,5); C = (0,10); D = (5,5); draw(arc((-5,0),A,B,CCW)); draw(arc((0,5),B,D,CW)); draw(arc((5,0),D,... | [
"Draw two squares: one that has opposing corners at $A$ and $B$ , and one that has opposing corners at $A$ and $D$ . These squares share side $\\overline{AO}$ , where $O$ is the center of the large semicircle.\nThese two squares have a total area of $2 \\cdot 5^2$ , but have two quarter circle \"bites\" of radius... |
https://artofproblemsolving.com/wiki/index.php/2004_AIME_II_Problems/Problem_6 | null | 408 | Three clever monkeys divide a pile of bananas. The first monkey takes some bananas from the pile, keeps three-fourths of them, and divides the rest equally between the other two. The second monkey takes some bananas from the pile, keeps one-fourth of them, and divides the rest equally between the other two. The third m... | [
"Denote the number of bananas the first monkey took from the pile as $b_1$ , the second $b_2$ , and the third $b_3$ ; the total is $b_1 + b_2 + b_3$ . Thus, the first monkey got $\\frac{3}{4}b_1 + \\frac{3}{8}b_2 + \\frac{11}{24}b_3$ , the second monkey got $\\frac{1}{8}b_1 + \\frac{1}{4}b_2 + \\frac{11}{24}b_3$ , ... |
https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_18 | A | 108 | Three concentric circles centered at $O$ have radii of $1$ $2$ , and $3$ . Points $B$ and $C$ lie on the largest circle. The region between the two smaller circles is shaded, as is the portion of the region between the two larger circles bounded by central angle $BOC$ , as shown in the figure below. Suppose the shaded ... | [
"Let $x=\\angle{BOC}$\nWe see that the shaded region is the inner ring plus a sector $x^\\circ$ of the outer ring. The area of this in terms of $x$ is $\\left( 4 \\pi - \\pi \\right)+\\frac{x}{360} \\left( 9 \\pi - 4 \\pi \\right)$ . This simplifies to $3 \\pi + \\frac{x}{360}(5 \\pi)$\nAlso, the unshaded portion i... |
https://artofproblemsolving.com/wiki/index.php/2012_AIME_I_Problems/Problem_13 | null | 41 | Three concentric circles have radii $3,$ $4,$ and $5.$ An equilateral triangle with one vertex on each circle has side length $s.$ The largest possible area of the triangle can be written as $a + \tfrac{b}{c} \sqrt{d},$ where $a,$ $b,$ $c,$ and $d$ are positive integers, $b$ and $c$ are relatively prime, and $d$ is not... | [
"Reinterpret the problem in the following manner. Equilateral triangle $ABC$ has a point $X$ on the interior such that $AX = 5,$ $BX = 4,$ and $CX = 3.$ $60^\\circ$ counter-clockwise rotation about vertex $A$ maps $X$ to $X'$ and $B$ to $C.$\nNote that angle $XAX'$ is $60$ and $XA = X'A = 5$ which tells us that tri... |
https://artofproblemsolving.com/wiki/index.php/1995_AJHSME_Problems/Problem_9 | C | 32 | Three congruent circles with centers $P$ $Q$ , and $R$ are tangent to the sides of rectangle $ABCD$ as shown. The circle centered at $Q$ has diameter $4$ and passes through points $P$ and $R$ . The area of the rectangle is
[asy] pair A,B,C,D,P,Q,R; A = (0,4); B = (8,4); C = (8,0); D = (0,0); P = (2,2); Q = (4,2); R =... | [
"If circle $Q$ has diameter $4$ , then so do congruent circles $P$ and $R$ . Draw a diameter through $P$ parallel to $AD$ . The diameter will be congruent to $AD$ , and thus $AD = 4$ , which is the height of the rectangle.\nDraw a horizontal line $PQR$ that extends to the sides of the rectangle. This line is $2$... |
https://artofproblemsolving.com/wiki/index.php/2008_AMC_12A_Problems/Problem_11 | C | 164 | Three cubes are each formed from the pattern shown. They are then stacked on a table one on top of another so that the $13$ visible numbers have the greatest possible sum. What is that sum?
[asy] unitsize(.8cm); pen p = linewidth(1); draw(shift(-2,0)*unitsquare,p); label("1",(-1.5,0.5)); draw(shift(-1,0)*unitsquare,p)... | [
"Conversely, maximize the sum. Two cubes have 4 exposed faces. Since $32>16+8+4+2$ , 32 must be on the side. There are two distinct (asymmetrical) configurations with 32 on the side, but $(32,16,2,1)$ is the greatest at 51. There are 2 such cubes, so 51*2. The top cube has one unexposed face, so use 1 as the unexpo... |
https://artofproblemsolving.com/wiki/index.php/2006_AMC_8_Problems/Problem_22 | D | 26 | Three different one-digit positive integers are placed in the bottom row of cells. Numbers in adjacent cells are added and the sum is placed in the cell above them. In the second row, continue the same process to obtain a number in the top cell. What is the difference between the largest and smallest numbers possible i... | [
"If the lower cells contain $A, B$ and $C$ , then the second row will contain $A + B$ and $B + C$ , and the top cell will contain $A + 2B + C$ . To obtain the\nsmallest sum, place $1$ in the center cell and $2$ and $3$ in the outer ones. The top\nnumber will be $7$ . For the largest sum, place $9$ in the center cel... |
https://artofproblemsolving.com/wiki/index.php/2009_AMC_10A_Problems/Problem_24 | null | 47 | Three distinct vertices of a cube are chosen at random. What is the probability that the plane determined by these three vertices contains points inside the cube?
$\mathrm{(A)}\ \frac{1}{4} \qquad \mathrm{(B)}\ \frac{3}{8} \qquad \mathrm{(C)}\ \frac{4}{7} \qquad \mathrm{(D)}\ \frac{5}{7} \qquad \mathrm{(E)}\ \frac{3}{4... | [
"We will try to use symmetry as much as possible.\nPick the first vertex $A$ , its choice clearly does not influence anything.\nPick the second vertex $B$ . With probability $3/7$ vertices $A$ and $B$ have a common edge, with probability $3/7$ they are in opposite corners of the same face, and with probability $1/7... |
https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_14 | B | 6 | Three equally spaced parallel lines intersect a circle, creating three chords of lengths $38,38,$ and $34$ . What is the distance between two adjacent parallel lines?
$\textbf{(A) }5\frac12 \qquad \textbf{(B) }6 \qquad \textbf{(C) }6\frac12 \qquad \textbf{(D) }7 \qquad \textbf{(E) }7\frac12$ | [
"\nSince two parallel chords have the same length ( $38$ ), they must be equidistant from the center of the circle. Let the perpendicular distance of each chord from the center of the circle be $d$ . Thus, the distance from the center of the circle to the chord of length $34$ is\n\\[2d + d = 3d\\]\nand the distance... |
https://artofproblemsolving.com/wiki/index.php/2021_AMC_12B_Problems/Problem_8 | B | 6 | Three equally spaced parallel lines intersect a circle, creating three chords of lengths $38,38,$ and $34$ . What is the distance between two adjacent parallel lines?
$\textbf{(A) }5\frac12 \qquad \textbf{(B) }6 \qquad \textbf{(C) }6\frac12 \qquad \textbf{(D) }7 \qquad \textbf{(E) }7\frac12$ | [
"\nSince two parallel chords have the same length ( $38$ ), they must be equidistant from the center of the circle. Let the perpendicular distance of each chord from the center of the circle be $d$ . Thus, the distance from the center of the circle to the chord of length $34$ is\n\\[2d + d = 3d\\]\nand the distance... |
https://artofproblemsolving.com/wiki/index.php/1999_AMC_8_Problems/Problem_9 | C | 1,150 | Three flower beds overlap as shown. Bed A has 500 plants, bed B has 450 plants, and bed C has 350 plants. Beds A and B share 50 plants, while beds A and C share 100. The total number of plants is
[asy] draw((0,0)--(3,0)--(3,1)--(0,1)--cycle); draw(circle((.3,-.1),.7)); draw(circle((2.8,-.2),.8)); label("A",(1.3,.5),N);... | [
"Plants shared by two beds have been counted\ntwice, so the total is $500 + 450 + 350 - 50 - 100 = \\boxed{1150}$",
"Bed A has $350$ plants it doesn't\nshare with B or C. Bed B has $400$ plants it doesn't\nshare with A or C. And C has $250$ it doesn't share\nwith A or B. The total is $350 + 400 + 250 + 50 + 100 =... |
https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_5 | C | 15 | Three fourths of a pitcher is filled with pineapple juice. The pitcher is emptied by pouring an equal amount of juice into each of $5$ cups. What percent of the total capacity of the pitcher did each cup receive?
$\textbf{(A) }5 \qquad \textbf{(B) }10 \qquad \textbf{(C) }15 \qquad \textbf{(D) }20 \qquad \textbf{(E) }25... | [
"Each cup is filled with $\\frac{3}{4} \\cdot \\frac{1}{5} = /frac{3}{20}$ of the amount of juice in the pitcher, so the percentage is $\\frac{3}{20} \\cdot 100 = \\boxed{15}$",
"The pitcher is $\\frac{3}{4}$ full, i.e. $75\\%$ full. Therefore each cup receives $\\frac{75}{5}=\\boxed{15}$ percent of the total cap... |
https://artofproblemsolving.com/wiki/index.php/2004_AMC_8_Problems/Problem_17 | null | 4 | Three friends have a total of $6$ identical pencils, and each one has at least one pencil. In how many ways can this happen?
$\textbf{(A)}\ 1\qquad \textbf{(B)}\ 3\qquad \textbf{(C)}\ 6\qquad \textbf{(D)}\ 10 \qquad \textbf{(E)}\ 12$ | [
"Like in solution 1 and solution 2, assign one pencil to each of the three friends so that you have $3$ left. In partitioning the remaining $3$ pencils into $3$ distinct groups use casework. \nLet the three friends be $a$ $b$ $c$ repectively.\n$a + b + c = 3$\nCase $1:a=0$\n$b + c = 3$\n$b = 0,1,2,3$\n$c = 3,2,1,0$... |
https://artofproblemsolving.com/wiki/index.php/2004_AMC_8_Problems/Problem_17 | D | 10 | Three friends have a total of $6$ identical pencils, and each one has at least one pencil. In how many ways can this happen?
$\textbf{(A)}\ 1\qquad \textbf{(B)}\ 3\qquad \textbf{(C)}\ 6\qquad \textbf{(D)}\ 10 \qquad \textbf{(E)}\ 12$ | [
"For each person to have at least one pencil, assign one pencil to each of the three friends so that you have $3$ left. In partitioning the remaining $3$ pencils into $3$ distinct groups, use Ball-and-urn to find the number of possibilities is $\\binom{3+3-1}{3-1} = \\binom{5}{2} = \\boxed{10}$",
"Like in solutio... |
https://artofproblemsolving.com/wiki/index.php/1998_AJHSME_Problems/Problem_25 | D | 252 | Three generous friends, each with some money, redistribute the money as followed:
Amy gives enough money to Jan and Toy to double each amount has.
Jan then gives enough to Amy and Toy to double their amounts.
Finally, Toy gives enough to Amy and Jan to double their amounts.
If Toy had 36 dollars at the beginning and 3... | [
"If Toy had $36$ dollars at the beginning, then after Amy doubles his money, he has $36 \\times 2 = 72$ dollars after the first step.\nThen Jan doubles his money, and Toy has $72 \\times 2 = 144$ dollars after the second step.\nThen Toy doubles whatever Amy and Jan have. Since Toy ended up with $36$ , he spent $14... |
https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_4 | B | 37 | Three hexagons of increasing size are shown below. Suppose the dot pattern continues so that each successive hexagon contains one more band of dots. How many dots are in the next hexagon?
[asy] // diagram by SirCalcsALot, edited by MRENTHUSIASM size(250); path p = scale(0.8)*unitcircle; pair[] A; pen grey1 = rgb(100/25... | [
"Looking at the rows of each hexagon, we see that the first hexagon has $1$ dot, the second has $2+3+2$ dots, and the third has $3+4+5+4+3$ dots. Given the way the hexagons are constructed, it is clear that this pattern continues. Hence, the fourth hexagon has $4+5+6+7+6+5+4=\\boxed{37}$ dots.",
"The dots in the ... |
https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_2 | E | 150 | Three identical rectangles are put together to form rectangle $ABCD$ , as shown in the figure below. Given that the length of the shorter side of each of the smaller rectangles is 5 feet, what is the area in square feet of rectangle $ABCD$
[asy] draw((0,0)--(3,0)); draw((0,0)--(0,2)); draw((0,2)--(3,2)); draw((3,2)--(... | [
"We can see that there are $2$ rectangles lying on top of the other and that is the same as the length of one rectangle. Now we know that the shorter side is $5$ , so the bigger side is $10$ , if we do $5 \\cdot 2 = 10$ . Now we get the sides of the big rectangle being $15$ and $10$ , so the area is $\\boxed{150}$ ... |
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12B_Problems/Problem_15 | E | 147 | Three identical square sheets of paper each with side length $6$ are stacked on top of each other. The middle sheet is rotated clockwise $30^\circ$ about its center and the top sheet is rotated clockwise $60^\circ$ about its center, resulting in the $24$ -sided polygon shown in the figure below. The area of this polygo... | [
" The $24$ -sided polygon is made out of $24$ shapes like $\\triangle ABC$ . Then $\\angle BAC=360^\\circ/24=15^\\circ$ , and $\\angle EAC = 45^\\circ$ , so $\\angle{EAB} = 30^{\\circ}$ . Then $EB=AE\\tan 30^\\circ = \\sqrt{3}$ ; therefore $BC=EC-EB=3-\\sqrt{3}$ . Thus \\[[ABC] = \\frac{BC}{EC}\\cdot [ACE] = \\frac... |
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10B_Problems/Problem_18 | E | 147 | Three identical square sheets of paper each with side length $6{ }$ are stacked on top of each other. The middle sheet is rotated clockwise $30^\circ$ about its center and the top sheet is rotated clockwise $60^\circ$ about its center, resulting in the $24$ -sided polygon shown in the figure below. The area of this pol... | [
"First note the useful fact that if $R$ is the circumradius of a dodecagon ( $12$ -gon) the area of the figure is $3R^2.$ If we connect the vertices of the $3$ squares we get a dodecagon. The radius of circumcircle of the dodecagon is simply half the diagonal of the square, which is $3\\sqrt{2}.$ Thus the area of t... |
https://artofproblemsolving.com/wiki/index.php/1962_AHSME_Problems/Problem_24 | A | 23 | Three machines $\text{P, Q, and R,}$ working together, can do a job in $x$ hours. When working alone, $\text{P}$ needs an additional $6$ hours to do the job; $\text{Q}$ , one additional hour; and $R$ $x$ additional hours. The value of $x$ is:
$\textbf{(A)}\ \frac{2}3\qquad\textbf{(B)}\ \frac{11}{12}\qquad\textbf{(C)}\ ... | [
"Machine P takes $x+6$ hours, machine Q takes $x+1$ hours, and machine R takes $2x$ hours.\nWe also know that all three working together take $x$ hours. \nNow the time it takes for all three machines to complete the job is the harmonic mean of the times the three machines take individually; that is, \\[x=\\frac1{\\... |
https://artofproblemsolving.com/wiki/index.php/1999_AHSME_Problems/Problem_26 | null | 21 | Three non-overlapping regular plane polygons, at least two of which are congruent, all have sides of length $1$ . The polygons meet at a point $A$ in such a way that the sum of the three interior angles at $A$ is $360^{\circ}$ . Thus the three polygons form a new polygon with $A$ as an interior point. What is the large... | [
"We are looking for three regular polygons such that the sum of their internal angle sizes is exactly $360^{\\circ}$\nLet the number of sides in our polygons be $3\\leq a,b,c$ . From each of the polygons, two sides touch the other two, and the remaining sides are on the perimeter. Therefore the answer to our proble... |
https://artofproblemsolving.com/wiki/index.php/1963_AHSME_Problems/Problem_16 | C | 12 | Three numbers $a,b,c$ , none zero, form an arithmetic progression. Increasing $a$ by $1$ or increasing $c$ by $2$ results
in a geometric progression. Then $b$ equals:
$\textbf{(A)}\ 16 \qquad \textbf{(B)}\ 14 \qquad \textbf{(C)}\ 12 \qquad \textbf{(D)}\ 10 \qquad \textbf{(E)}\ 8$ | [
"Let $d$ be the common difference of the arithmetic sequence , so $a = b-d$ and $c = b+d$\nSince increasing $a$ by $1$ or $c$ by $2$ results in a geometric sequence \\[\\frac{b}{b-d+1} = \\frac{b+d}{b}\\] \\[\\frac{b}{b-d} = \\frac{b+d+2}{b}\\] Cross-multiply in both equations to get a system of equations \\[b^2 = ... |
https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_23 | C | 12 | Three numbers in the interval $\left[0,1\right]$ are chosen independently and at random. What is the probability that the chosen numbers are the side lengths of a triangle with positive area?
$\textbf{(A) }\frac16\qquad\textbf{(B) }\frac13\qquad\textbf{(C) }\frac12\qquad\textbf{(D) }\frac23\qquad\textbf{(E) }\frac56$ | [
"Because we can let the the sides of the triangle be any variable we want, to make it easier for us when solving, let’s let the side lengths be $x,y,$ and $a$ . WLOG assume $a$ is the largest. Then, $x+y>a$ , meaning the solution is $\\boxed{12}$ , as shown in the graph below. ",
"WLOG, let the largest of the thr... |
https://artofproblemsolving.com/wiki/index.php/1993_AIME_Problems/Problem_7 | null | 5 | Three numbers, $a_1, a_2, a_3$ , are drawn randomly and without replacement from the set $\{1, 2, 3,\ldots, 1000\}$ . Three other numbers, $b_1, b_2, b_3$ , are then drawn randomly and without replacement from the remaining set of $997$ numbers. Let $p$ be the probability that, after suitable rotation, a brick of dimen... | [
"There is a total of $P(1000,6)$ possible ordered $6$ -tuples $(a_1,a_2,a_3,b_1,b_2,b_3).$\nThere are $C(1000,6)$ possible sets $\\{a_1,a_2,a_3,b_1,b_2,b_3\\}.$ We have five valid cases for the increasing order of these six elements:\nNote that the $a$ 's are different from each other, as there are $3!=6$ ways to p... |
https://artofproblemsolving.com/wiki/index.php/2007_AIME_I_Problems/Problem_4 | null | 105 | Three planets orbit a star circularly in the same plane. Each moves in the same direction and moves at constant speed. Their periods are 60, 84, and 140 years. The three planets and the star are currently collinear . What is the fewest number of years from now that they will all be collinear again? | [
"Denote the planets $A, B, C$ respectively. Let $a(t), b(t), c(t)$ denote the angle which each of the respective planets makes with its initial position after $t$ years. These are given by $a(t) = \\frac{t \\pi}{30}$ $b(t) = \\frac{t \\pi}{42}$ $c(t) = \\frac{t \\pi}{70}$\nIn order for the planets and the central... |
https://artofproblemsolving.com/wiki/index.php/2013_AMC_10B_Problems/Problem_9 | D | 160 | Three positive integers are each greater than $1$ , have a product of $27000$ , and are pairwise relatively prime. What is their sum?
$\textbf{(A)}\ 100\qquad\textbf{(B)}\ 137\qquad\textbf{(C)}\ 156\qquad\textbf{(D)}\ 160\qquad\textbf{(E)}\ 165$ | [
"The prime factorization of $27000$ is $2^3*3^3*5^3$ . These three factors are pairwise relatively prime, so the sum is $2^3+3^3+5^3=8+27+125=$ $\\boxed{160}$"
] |
https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_6 | C | 6 | Three positive integers are equally spaced on a number line. The middle number is $15,$ and the largest number is $4$ times the smallest number. What is the smallest of these three numbers?
$\textbf{(A) } 4 \qquad \textbf{(B) } 5 \qquad \textbf{(C) } 6 \qquad \textbf{(D) } 7 \qquad \textbf{(E) } 8$ | [
"Let the smallest number be $x.$ It follows that the largest number is $4x.$\nSince $x,15,$ and $4x$ are equally spaced on a number line, we have \\begin{align*} 4x-15 &= 15-x \\\\ 5x &= 30 \\\\ x &= \\boxed{6} ~MRENTHUSIASM",
"Let the common difference of the arithmetic sequence be $d$ . Consequently, the smalle... |
https://artofproblemsolving.com/wiki/index.php/1983_AHSME_Problems/Problem_3 | A | 2 | Three primes $p,q$ , and $r$ satisfy $p+q = r$ and $1 < p < q$ . Then $p$ equals
$\textbf{(A)}\ 2\qquad \textbf{(B)}\ 3\qquad \textbf{(C)}\ 7\qquad \textbf{(D)}\ 13\qquad \textbf{(E)}\ 17$ | [
"We are given that $p,q$ and $r$ are primes. In order for $p$ and $q$ to sum to another prime, either $p$ or $q$ has to be even, because the sum of two odd numbers would be even, and the only even prime is $2$ (but $p + q = 2$ would have, as the only solution in positive integers, $p = q = 1$ , and $1$ is not prime... |
https://artofproblemsolving.com/wiki/index.php/2008_AMC_10B_Problems/Problem_22 | C | 16 | Three red beads, two white beads, and one blue bead are placed in line in random order. What is the probability that no two neighboring beads are the same color?
$\mathrm{(A)}\ 1/12\qquad\mathrm{(B)}\ 1/10\qquad\mathrm{(C)}\ 1/6\qquad\mathrm{(D)}\ 1/3\qquad\mathrm{(E)}\ 1/2$ | [
"There are $\\frac{6!}{3!\\cdot2!\\cdot1!}=60$ total orderings.\nSuppose we order the red and white beads first. If these two colors are ordered first, we must make sure that no neighboring beads are the same color, or only one pair of neighboring beads are the same color. There are five possible orderings:\n$R\\ W... |
https://artofproblemsolving.com/wiki/index.php/2012_AMC_10A_Problems/Problem_16 | C | 2,500 | Three runners start running simultaneously from the same point on a 500-meter circular track. They each run clockwise around the course maintaining constant speeds of 4.4, 4.8, and 5.0 meters per second. The runners stop once they are all together again somewhere on the circular course. How many seconds do the runners ... | [
"First consider the first two runners. The faster runner will lap the slower runner exactly once, or run 500 meters farther. Let $x$ be the time these runners run in seconds.\n\\[4.8x-4.4x=500 \\Rightarrow x=1250\\]\nBecause $4.4(1250)=5500$ is a multiple of 500, it turns out they just meet back at the start line.\... |
https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_10 | null | 756 | Three spheres with radii $11$ $13$ , and $19$ are mutually externally tangent. A plane intersects the spheres in three congruent circles centered at $A$ $B$ , and $C$ , respectively, and the centers of the spheres all lie on the same side of this plane. Suppose that $AB^2 = 560$ . Find $AC^2$ | [
"This solution refers to the Diagram section.\nWe let $\\ell$ be the plane that passes through the spheres and $O_A$ and $O_B$ be the centers of the spheres with radii $11$ and $13$ . We take a cross-section that contains $A$ and $B$ , which contains these two spheres but not the third, as shown below: Because the... |
https://artofproblemsolving.com/wiki/index.php/1972_AHSME_Problems/Problem_34 | A | 42 | Three times Dick's age plus Tom's age equals twice Harry's age.
Double the cube of Harry's age is equal to three times the cube of Dick's age added to the cube of Tom's age.
Their respective ages are relatively prime to each other. The sum of the squares of their ages is
$\textbf{(A) }42\qquad \textbf{(B) }46\qquad \... | [
"\\[t=2h-3d\\] \\[3d^3+t^3=2h^3\\]\nFirst, substitute in t into the second equation and get $3d^3+8h^3-36h^2d+54hd^2-27d^3=2h^3$ . That turns into $h^3-6h^2d+9hd^2-4d^3=0$ which is factored into $(h-4d)(h-d)^2 =0.$ WLOG, $d=1$ and consequently $h=4$ . Then $t=8-3=5$ . Everything appears to be relatively prime alrea... |
https://artofproblemsolving.com/wiki/index.php/2012_AMC_10A_Problems/Problem_15 | B | 15 | Three unit squares and two line segments connecting two pairs of vertices are shown. What is the area of $\triangle ABC$
$\textbf{(A)}\ \frac16 \qquad\textbf{(B)}\ \frac15 \qquad\textbf{(C)}\ \frac29 \qquad\textbf{(D)}\ \frac13 \qquad\textbf{(E)}\ \frac{\sqrt{2}}{4}$ | [
"$AC$ intersects $BC$ at a right angle, (this can be proved by noticing that the slopes of the two lines are negative reciprocals of each other) so $\\triangle ABC \\sim \\triangle BED$ . The hypotenuse of right triangle BED is $\\sqrt{1^2+2^2}=\\sqrt{5}$\n\\[\\frac{AC}{BC}=\\frac{BD}{ED} \\Rightarrow \\frac{AC}{BC... |
https://artofproblemsolving.com/wiki/index.php/1963_AHSME_Problems/Problem_12 | E | 9 | Three vertices of parallelogram $PQRS$ are $P(-3,-2), Q(1,-5), R(9,1)$ with $P$ and $R$ diagonally opposite.
The sum of the coordinates of vertex $S$ is:
$\textbf{(A)}\ 13 \qquad \textbf{(B)}\ 12 \qquad \textbf{(C)}\ 11 \qquad \textbf{(D)}\ 10 \qquad \textbf{(E)}\ 9$ | [
" Graph the three points on the coordinate grid. Noting that the opposite sides of a parallelogram are congruent and parallel, count boxes to find out that point $S$ is on $(5,4)$ . The sum of the x-coordinates and y-coordinates is $9$ , so the answer is $\\boxed{9}$"
] |
https://artofproblemsolving.com/wiki/index.php/2018_AMC_10B_Problems/Problem_18 | D | 96 | Three young brother-sister pairs from different families need to take a trip in a van. These six children will occupy the second and third rows in the van, each of which has three seats. To avoid disruptions, siblings may not sit right next to each other in the same row, and no child may sit directly in front of his or... | [
"We can begin to put this into cases. Let's call the pairs $a$ $b$ and $c$ , and assume that a member of pair $a$ is sitting in the leftmost seat of the second row. We can have the following cases then.\nCase $1$ : \nSecond Row: a b c\nThird Row: b c a\nCase $2$ :\nSecond Row: a c b\nThird Row: c b a\nCase $3$ : \n... |
https://artofproblemsolving.com/wiki/index.php/2000_AMC_8_Problems/Problem_9 | D | 6 | Three-digit powers of $2$ and $5$ are used in this "cross-number" puzzle. What is the only possible digit for the outlined square? \[\begin{array}{lcl} \textbf{ACROSS} & & \textbf{DOWN} \\ \textbf{2}.~ 2^m & & \textbf{1}.~ 5^n \end{array}\]
[asy] draw((0,-1)--(1,-1)--(1,2)--(0,2)--cycle); draw((0,1)--(3,1)--(3,0)--(0,0... | [
"The $3$ -digit powers of $5$ are $125$ and $625$ , so space $2$ is filled with a $2$ .\nThe only $3$ -digit power of $2$ beginning with $2$ is $256$ , so the outlined block is filled with\na $\\boxed{6}$"
] |
https://artofproblemsolving.com/wiki/index.php/2007_AMC_8_Problems/Problem_9 | B | 2 | To complete the grid below, each of the digits 1 through 4 must occur once
in each row and once in each column. What number will occupy the lower
right-hand square?
\[\begin{tabular}{|c|c|c|c|}\hline 1 & & 2 &\\ \hline 2 & 3 & &\\ \hline & &&4\\ \hline & &&\\ \hline\end{tabular}\]
$\mathrm{(A)}\ 1 \qquad \mathrm{(B)}\ ... | [
"The number in the first row, last column must be a $3$ due to the fact if a $3$ was in the first row, second column, there would be two threes in that column. By the same reasoning, the number in the second row, last column has to be a $1$ . Therefore the number in the lower right-hand square is $\\boxed{2}$",
"... |
https://artofproblemsolving.com/wiki/index.php/1993_AJHSME_Problems/Problem_8 | D | 8 | To control her blood pressure, Jill's grandmother takes one half of a pill every other day. If one supply of medicine contains $60$ pills, then the supply of medicine would last approximately
$\text{(A)}\ 1\text{ month} \qquad \text{(B)}\ 4\text{ months} \qquad \text{(C)}\ 6\text{ months} \qquad \text{(D)}\ 8\text{ mo... | [
"If Jill's grandmother takes one half of a pill every other day, she takes a pill every $4$ days. Since she has $60$ pills, the supply will last $60 \\times 4=240$ days which is about $\\boxed{8}$"
] |
https://artofproblemsolving.com/wiki/index.php/2001_AMC_8_Problems/Problem_9 | D | 189 | To promote her school's annual Kite Olympics, Genevieve makes a small kite and a large kite for a bulletin board display. The kites look like the one in the diagram. For her small kite Genevieve draws the kite on a one-inch grid. For the large kite she triples both the height and width of the entire grid.
[asy] for (in... | [
"The large grid has dimensions three times that of the small grid, so its dimensions are $3(6)\\times3(7)$ , or $18\\times21$ , so the area is $(18)(21)=378$ . The area of the kite is half of the area of the rectangle as you can see, so the area of the waste material is also half the area of the rectangle. Thus, t... |
https://artofproblemsolving.com/wiki/index.php/2001_AMC_8_Problems/Problem_8 | E | 39 | To promote her school's annual Kite Olympics, Genevieve makes a small kite and a large kite for a bulletin board display. The kites look like the one in the diagram. For her small kite Genevieve draws the kite on a one-inch grid. For the large kite she triples both the height and width of the entire grid.
[asy] size(85... | [
"Each diagonal of the large kite is $3$ times the length of the corresponding diagonal of the short kite since it was made with a grid $3$ times as long in each direction. The diagonals of the small kite are $6$ and $7$ , so the diagonals of the large kite are $18$ and $21$ , and the amount of bracing Genevieve nee... |
https://artofproblemsolving.com/wiki/index.php/2001_AMC_8_Problems/Problem_7 | A | 21 | To promote her school's annual Kite Olympics, Genevieve makes a small kite and a large kite for a bulletin board display. The kites look like the one in the diagram. For her small kite Genevieve draws the kite on a one-inch grid. For the large kite she triples both the height and width of the entire grid. [asy] for (in... | [
"The area of a kite is half the product of its diagonals. The diagonals have lengths of $6$ and $7$ , so the area is $\\frac{(6)(7)}{2}=21, \\boxed{21}$",
"Drawing in the diagonals of the kite will form four right triangles on the \"inside\" part of the grid. Drawing in the border of the 7 by 6 grid will form fo... |
https://artofproblemsolving.com/wiki/index.php/2007_AMC_10B_Problems/Problem_12 | D | 5 | Tom's age is $T$ years, which is also the sum of the ages of his three children. His age $N$ years ago was twice the sum of their ages then. What is $T/N$
$\textbf{(A) } 2 \qquad\textbf{(B) } 3 \qquad\textbf{(C) } 4 \qquad\textbf{(D) } 5 \qquad\textbf{(E) } 6$ | [
"Tom's age $N$ years ago was $T-N$ . The sum of the ages of his three children $N$ years ago was $T-3N,$ since there are three children. If his age $N$ years ago was twice the sum of the children's ages then, \\begin{align*}T-N&=2(T-3N)\\\\ T-N&=2T-6N\\\\ T&=5N\\\\ T/N&=\\boxed{5} Note that actual values were not f... |
https://artofproblemsolving.com/wiki/index.php/2007_AMC_12B_Problems/Problem_8 | D | 5 | Tom's age is $T$ years, which is also the sum of the ages of his three children. His age $N$ years ago was twice the sum of their ages then. What is $T/N$
$\textbf{(A) } 2 \qquad\textbf{(B) } 3 \qquad\textbf{(C) } 4 \qquad\textbf{(D) } 5 \qquad\textbf{(E) } 6$ | [
"Tom's age $N$ years ago was $T-N$ . The sum of the ages of his three children $N$ years ago was $T-3N,$ since there are three children. If his age $N$ years ago was twice the sum of the children's ages then, \\begin{align*}T-N&=2(T-3N)\\\\ T-N&=2T-6N\\\\ T&=5N\\\\ T/N&=\\boxed{5} Note that actual values were not f... |
https://artofproblemsolving.com/wiki/index.php/2013_AMC_10A_Problems/Problem_5 | B | 20 | Tom, Dorothy, and Sammy went on a vacation and agreed to split the costs evenly. During their trip Tom paid $105, Dorothy paid $125, and Sammy paid $175. In order to share costs equally, Tom gave Sammy $t$ dollars, and Dorothy gave Sammy $d$ dollars. What is $t-d$
$\textbf{(A)}\ 15\qquad\textbf{(B)}\ 20\qquad\textbf... | [
"The total amount paid is $105 + 125 + 175 = 405$ . To get how much each should have paid, we do $405/3 = 135$\nThus, we know that Tom needs to give Sammy 30 dollars, and Dorothy 10 dollars. This means that $t-d = 30 - 10 = \\boxed{20}$",
"The difference in the money that Tom paid and Dorothy paid is $20$ . In... |
https://artofproblemsolving.com/wiki/index.php/2013_AMC_12A_Problems/Problem_5 | B | 20 | Tom, Dorothy, and Sammy went on a vacation and agreed to split the costs evenly. During their trip Tom paid $ $105$ , Dorothy paid $ $125$ , and Sammy paid $ $175$ . In order to share the costs equally, Tom gave Sammy $t$ dollars, and Dorothy gave Sammy $d$ dollars. What is $t-d$
$\textbf{(A)}\ 15\qquad\textbf{(B)}\ 20... | [
"Simply write down two algebraic equations. We know that Tom gave $t$ dollars and Dorothy gave $d$ dollars. In addition, Tom originally paid $105$ dollars and Dorothy paid $125$ dollars originally. Since they all pay the same amount, we have: \\[105 + t = 125 + d.\\] Rearranging, we have \\[t-d = \\boxed{20}.\\]",
... |
https://artofproblemsolving.com/wiki/index.php/2010_AMC_10A_Problems/Problem_8 | D | 13 | Tony works $2$ hours a day and is paid $ $0.50$ per hour for each full year of his age. During a six month period Tony worked $50$ days and earned $ $630$ . How old was Tony at the end of the six month period?
$\mathrm{(A)}\ 9 \qquad \mathrm{(B)}\ 11 \qquad \mathrm{(C)}\ 12 \qquad \mathrm{(D)}\ 13 \qquad \mathrm{(E)}\ ... | [
"Tony works $2$ hours a day and is paid $0.50$ dollars per hour for each full year of his age. This basically says that he gets a dollar for each year of his age. So if he is $12$ years old, he gets $12$ dollars a day. We also know that he worked $50$ days and earned $630$ dollars. If he was $12$ years old at the b... |
https://artofproblemsolving.com/wiki/index.php/1989_AHSME_Problems/Problem_5 | null | 430 | Toothpicks of equal length are used to build a rectangular grid as shown. If the grid is 20 toothpicks high and 10 toothpicks wide, then the number of toothpicks used is
[asy] real xscl = 1.2; int[] x = {0,1,2,4,5},y={0,1,3,4,5}; for(int a:x){ for(int b:y) { dot((a*xscl,b)); } } for(int a:x) { pair prev = (a,y[0]); for... | [
"There are 21 horizontal lines made of 10 matches, and 11 vertical lines made of 20 matches, and $21\\cdot10+11\\cdot20=\\boxed{430}$"
] |
https://artofproblemsolving.com/wiki/index.php/1999_AMC_8_Problems/Problem_16 | B | 5 | Tori's mathematics test had 75 problems: 10 arithmetic, 30 algebra, and 35 geometry problems. Although she answered 70% of the arithmetic, 40% of the algebra, and 60% of the geometry problems correctly, she did not pass the test because she got less than 60% of the problems right. How many more problems would she have ... | [
"First, calculate how many of each type of problem she got right:\nArithmetic: $70\\% \\cdot 10 = 0.70 \\cdot 10 = 7$\nAlgebra: $40\\% \\cdot 30 = 0.40 \\cdot 30 = 12$\nGeometry: $60\\% \\cdot 35 = 0.60 \\cdot 35 = 21$\nAltogether, Tori answered $7 + 12 + 21 = 40$ questions correct.\nTo get a $60\\%$ on her test o... |
https://artofproblemsolving.com/wiki/index.php/2024_AIME_II_Problems/Problem_8 | null | 127 | Torus $T$ is the surface produced by revolving a circle with radius $3$ around an axis in the plane of the circle that is a distance $6$ from the center of the circle (so like a donut). Let $S$ be a sphere with a radius $11$ . When $T$ rests on the inside of $S$ , it is internally tangent to $S$ along a circle with rad... | [
"First, let's consider a section $\\mathcal{P}$ of the solids, along the axis.\nBy some 3D-Geomerty thinking, we can simply know that the axis crosses the sphere center. So, that is saying, the $\\mathcal{P}$ we took crosses one of the equator of the sphere.\nHere I drew two graphs, the first one is the case when $... |
https://artofproblemsolving.com/wiki/index.php/2009_AMC_12B_Problems/Problem_16 | null | 45 | Trapezoid $ABCD$ has $AD||BC$ $BD = 1$ $\angle DBA = 23^{\circ}$ , and $\angle BDC = 46^{\circ}$ . The ratio $BC: AD$ is $9: 5$ . What is $CD$
$\mathrm{(A)}\ \frac 79\qquad \mathrm{(B)}\ \frac 45\qquad \mathrm{(C)}\ \frac {13}{15}\qquad \mathrm{(D)}\ \frac 89\qquad \mathrm{(E)}\ \frac {14}{15}$ | [
"Extend $\\overline {AB}$ and $\\overline {DC}$ to meet at $E$ . Then\n\\begin{align*} \\angle BED &= 180^{\\circ} - \\angle EDB - \\angle DBE\\\\ &= 180^{\\circ} - 134^{\\circ} -23^{\\circ} = 23^{\\circ}. \\end{align*}\nThus $\\triangle BDE$ is isosceles with $DE = BD$ . Because $\\overline {AD} \\parallel \\ove... |
https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_17 | D | 194 | Trapezoid $ABCD$ has $\overline{AB}\parallel\overline{CD},BC=CD=43$ , and $\overline{AD}\perp\overline{BD}$ . Let $O$ be the intersection of the diagonals $\overline{AC}$ and $\overline{BD}$ , and let $P$ be the midpoint of $\overline{BD}$ . Given that $OP=11$ , the length of $AD$ can be written in the form $m\sqrt{n}$... | [
"Angle chasing* reveals that $\\triangle BPC\\sim\\triangle BDA$ , therefore \\[2=\\frac{BD}{BP}=\\frac{AB}{BC}=\\frac{AB}{43},\\] or $AB=86$\nAdditional angle chasing shows that $\\triangle ABO\\sim\\triangle CDO$ , therefore \\[2=\\frac{AB}{CD}=\\frac{BO}{OD}=\\frac{BP+11}{BP-11},\\] or $BP=33$ and $BD=66$\nSince... |
https://artofproblemsolving.com/wiki/index.php/2021_AMC_12A_Problems/Problem_17 | D | 194 | Trapezoid $ABCD$ has $\overline{AB}\parallel\overline{CD},BC=CD=43$ , and $\overline{AD}\perp\overline{BD}$ . Let $O$ be the intersection of the diagonals $\overline{AC}$ and $\overline{BD}$ , and let $P$ be the midpoint of $\overline{BD}$ . Given that $OP=11$ , the length of $AD$ can be written in the form $m\sqrt{n}$... | [
"Angle chasing* reveals that $\\triangle BPC\\sim\\triangle BDA$ , therefore \\[2=\\frac{BD}{BP}=\\frac{AB}{BC}=\\frac{AB}{43},\\] or $AB=86$\nAdditional angle chasing shows that $\\triangle ABO\\sim\\triangle CDO$ , therefore \\[2=\\frac{AB}{CD}=\\frac{BO}{OD}=\\frac{BP+11}{BP-11},\\] or $BP=33$ and $BD=66$\nSince... |
https://artofproblemsolving.com/wiki/index.php/2014_AMC_10B_Problems/Problem_21 | B | 25 | Trapezoid $ABCD$ has parallel sides $\overline{AB}$ of length $33$ and $\overline {CD}$ of length $21$ . The other two sides are of lengths $10$ and $14$ . The angles $A$ and $B$ are acute. What is the length of the shorter diagonal of $ABCD$
$\textbf{(A) }10\sqrt{6}\qquad\textbf{(B) }25\qquad\textbf{(C) }8\sqrt{10}\qq... | [
"\nIn the diagram, $\\overline{DE} \\perp \\overline{AB}, \\overline{FC} \\perp \\overline{AB}$ . \nDenote $\\overline{AE} = x$ and $\\overline{DE} = h$ . In right triangle $AED$ , we have from the Pythagorean theorem: $x^2+h^2=100$ . Note that since $EF = DC$ , we have $BF = 33-DC-x = 12-x$ . Using the Pythagorean... |
https://artofproblemsolving.com/wiki/index.php/1992_AIME_Problems/Problem_9 | null | 164 | Trapezoid $ABCD^{}_{}$ has sides $AB=92^{}_{}$ $BC=50^{}_{}$ $CD=19^{}_{}$ , and $AD=70^{}_{}$ , with $AB^{}_{}$ parallel to $CD^{}_{}$ . A circle with center $P^{}_{}$ on $AB^{}_{}$ is drawn tangent to $BC^{}_{}$ and $AD^{}_{}$ . Given that $AP^{}_{}=\frac mn$ , where $m^{}_{}$ and $n^{}_{}$ are relatively prime posit... | [
"Let $AP=x$ so that $PB=92-x.$ Extend $AD, BC$ to meet at $X,$ and note that $XP$ bisects $\\angle AXB;$ let it meet $CD$ at $E.$ Using the angle bisector theorem, we let $XB=y(92-x), XA=xy$ for some $y.$\nThen $XD=xy-70, XC=y(92-x)-50,$ thus \\[\\frac{xy-70}{y(92-x)-50} = \\frac{XD}{XC} = \\frac{ED}{EC}=\\frac{AP}... |
https://artofproblemsolving.com/wiki/index.php/2019_AMC_10A_Problems/Problem_23 | C | 5,979 | Travis has to babysit the terrible Thompson triplets. Knowing that they love big numbers, Travis devises a counting game for them. First Tadd will say the number $1$ , then Todd must say the next two numbers ( $2$ and $3$ ), then Tucker must say the next three numbers ( $4$ $5$ $6$ ), then Tadd must say the next four ... | [
"Define a round as one complete rotation through each of the three children, and define a turn as the portion when one child says his numbers (similar to how a game is played).\nWe create a table to keep track of what numbers each child says for each round.\n$\\begin{tabular}{||c c c c||} \\hline Round & Tadd & ... |
https://artofproblemsolving.com/wiki/index.php/2013_AMC_8_Problems/Problem_7 | C | 100 | Trey and his mom stopped at a railroad crossing to let a train pass. As the train began to pass, Trey counted 6 cars in the first 10 seconds. It took the train 2 minutes and 45 seconds to clear the crossing at a constant speed. Which of the following was the most likely number of cars in the train?
$\textbf{(A)}\ 60 \q... | [
"Clearly, for every $5$ seconds, $3$ cars pass. It's more convenient to have everything in seconds: $2$ minutes and $45$ seconds $=2\\cdot60 + 45 = 165$ seconds. We then set up a ratio: \\[\\frac{3}{5}=\\frac{x}{165}\\] \\[3(165)=5x\\] \\[x=3(33)=99\\approx\\boxed{100}.\\]",
"$2$ minutes and $45$ seconds is equal... |
https://artofproblemsolving.com/wiki/index.php/2008_AMC_12A_Problems/Problem_18 | C | 95 | Triangle $ABC$ , with sides of length $5$ $6$ , and $7$ , has one vertex on the positive $x$ -axis, one on the positive $y$ -axis, and one on the positive $z$ -axis. Let $O$ be the origin . What is the volume of tetrahedron $OABC$
$\mathrm{(A)}\ \sqrt{85}\qquad\mathrm{(B)}\ \sqrt{90}\qquad\mathrm{(C)}\ \sqrt{95}\qquad\... | [
"Without loss of generality, let $A$ be on the $x$ axis, $B$ be on the $y$ axis, and $C$ be on the $z$ axis, and let $AB, BC, CA$ have respective lengths of 5, 6, and 7. Let $a,b,c$ denote the lengths of segments $OA,OB,OC,$ respectively. Then by the Pythagorean Theorem \\begin{align*} a^2+b^2 &=5^2 , \\\\ b^2+c... |
https://artofproblemsolving.com/wiki/index.php/1997_AHSME_Problems/Problem_26 | A | 5 | Triangle $ABC$ and point $P$ in the same plane are given. Point $P$ is equidistant from $A$ and $B$ , angle $APB$ is twice angle $ACB$ , and $\overline{AC}$ intersects $\overline{BP}$ at point $D$ . If $PB = 3$ and $PD= 2$ , then $AD\cdot CD =$
[asy] defaultpen(linewidth(.8pt)); dotfactor=4; pair A = origin; pair B = (... | [
"The product of two lengths with a common point brings to mind the Power of a Point Theorem\nSince $PA = PB$ , we can make a circle with radius $PA$ that is centered on $P$ , and both $A$ and $B$ will be on that circle. Since $\\angle APB = \\widehat {AB} = 2 \\angle ACB$ , we can see that point $C$ will also lie ... |
https://artofproblemsolving.com/wiki/index.php/2009_AMC_12B_Problems/Problem_13 | D | 18 | Triangle $ABC$ has $AB = 13$ and $AC = 15$ , and the altitude to $\overline{BC}$ has length $12$ . What is the sum of the two possible values of $BC$
$\mathrm{(A)}\ 15\qquad \mathrm{(B)}\ 16\qquad \mathrm{(C)}\ 17\qquad \mathrm{(D)}\ 18\qquad \mathrm{(E)}\ 19$ | [
"Let $D$ be the foot of the altitude to $\\overline BC$ . Then $BD = \\sqrt {13^2 - 12^2} = 5$ and $DC = \\sqrt {15^2 - 12^2} = 9$ . Thus $BC = BD + BC = 5 + 9 = 14$ or assume that the triangle is obtuse at angle $B$ then $BC = DC - BD = 9 -5 = 4$ . The sum of the two possible values is $14 + 4 = \\boxed{18}$"
... |
https://artofproblemsolving.com/wiki/index.php/2010_AMC_10A_Problems/Problem_14 | C | 90 | Triangle $ABC$ has $AB=2 \cdot AC$ . Let $D$ and $E$ be on $\overline{AB}$ and $\overline{BC}$ , respectively, such that $\angle BAE = \angle ACD$ . Let $F$ be the intersection of segments $AE$ and $CD$ , and suppose that $\triangle CFE$ is equilateral. What is $\angle ACB$
$\textbf{(A)}\ 60^\circ \qquad \textbf{(B)}\ ... | [
"Let $\\angle BAE = \\angle ACD = x$\n\\begin{align*}\\angle BCD &= \\angle AEC = 60^\\circ\\\\ \\angle EAC + \\angle FCA + \\angle ECF + \\angle AEC &= \\angle EAC + x + 60^\\circ + 60^\\circ = 180^\\circ\\\\ \\angle EAC &= 60^\\circ - x\\\\ \\angle BAC &= \\angle EAC + \\angle BAE = 60^\\circ - x + x = 60^\\ci... |
https://artofproblemsolving.com/wiki/index.php/2010_AMC_12A_Problems/Problem_8 | C | 90 | Triangle $ABC$ has $AB=2 \cdot AC$ . Let $D$ and $E$ be on $\overline{AB}$ and $\overline{BC}$ , respectively, such that $\angle BAE = \angle ACD$ . Let $F$ be the intersection of segments $AE$ and $CD$ , and suppose that $\triangle CFE$ is equilateral. What is $\angle ACB$
$\textbf{(A)}\ 60^\circ \qquad \textbf{(B)}\ ... | [
"Let $\\angle BAE = \\angle ACD = x$\n\\begin{align*}\\angle BCD &= \\angle AEC = 60^\\circ\\\\ \\angle EAC + \\angle FCA + \\angle ECF + \\angle AEC &= \\angle EAC + x + 60^\\circ + 60^\\circ = 180^\\circ\\\\ \\angle EAC &= 60^\\circ - x\\\\ \\angle BAC &= \\angle EAC + \\angle BAE = 60^\\circ - x + x = 60^\\ci... |
https://artofproblemsolving.com/wiki/index.php/2001_AIME_I_Problems/Problem_7 | null | 923 | Triangle $ABC$ has $AB=21$ $AC=22$ and $BC=20$ . Points $D$ and $E$ are located on $\overline{AB}$ and $\overline{AC}$ , respectively, such that $\overline{DE}$ is parallel to $\overline{BC}$ and contains the center of the inscribed circle of triangle $ABC$ . Then $DE=m/n$ , where $m$ and $n$ are relatively prime posit... | [
"Let $I$ be the incenter of $\\triangle ABC$ , so that $BI$ and $CI$ are angle bisectors of $\\angle ABC$ and $\\angle ACB$ respectively. Then, $\\angle BID = \\angle CBI = \\angle DBI,$ so $\\triangle BDI$ is isosceles , and similarly $\\triangle CEI$ is isosceles. It follows that $DE = DB + EC$ , so the perimeter... |
https://artofproblemsolving.com/wiki/index.php/2012_AMC_12A_Problems/Problem_18 | A | 15 | Triangle $ABC$ has $AB=27$ $AC=26$ , and $BC=25$ . Let $I$ be the intersection of the internal angle bisectors of $\triangle ABC$ . What is $BI$
$\textbf{(A)}\ 15\qquad\textbf{(B)}\ 5+\sqrt{26}+3\sqrt{3}\qquad\textbf{(C)}\ 3\sqrt{26}\qquad\textbf{(D)}\ \frac{2}{3}\sqrt{546}\qquad\textbf{(E)}\ 9\sqrt{3}$ | [
"Inscribe circle $C$ of radius $r$ inside triangle $ABC$ so that it meets $AB$ at $Q$ $BC$ at $R$ , and $AC$ at $S$ . Note that angle bisectors of triangle $ABC$ are concurrent at the center $O$ (also $I$ ) of circle $C$ . Let $x=QB$ $y=RC$ and $z=AS$ . Note that $BR=x$ $SC=y$ and $AQ=z$ . Hence $x+z=27$ $x+y=25$ ,... |
https://artofproblemsolving.com/wiki/index.php/1992_AIME_Problems/Problem_13 | null | 820 | Triangle $ABC$ has $AB=9$ and $BC: AC=40: 41$ . What's the largest area that this triangle can have? | [
"First, consider the triangle in a coordinate system with vertices at $(0,0)$ $(9,0)$ , and $(a,b)$ . Applying the distance formula , we see that $\\frac{ \\sqrt{a^2 + b^2} }{ \\sqrt{ (a-9)^2 + b^2 } } = \\frac{40}{41}$\nWe want to maximize $b$ , the height, with $9$ being the base.\nSimplifying gives $-a^2 -\\frac... |
https://artofproblemsolving.com/wiki/index.php/2009_AIME_I_Problems/Problem_5 | null | 72 | Triangle $ABC$ has $AC = 450$ and $BC = 300$ . Points $K$ and $L$ are located on $\overline{AC}$ and $\overline{AB}$ respectively so that $AK = CK$ , and $\overline{CL}$ is the angle bisector of angle $C$ . Let $P$ be the point of intersection of $\overline{BK}$ and $\overline{CL}$ , and let $M$ be the point on line $B... | [
"Since $K$ is the midpoint of $\\overline{PM}$ and $\\overline{AC}$ , quadrilateral $AMCP$ is a parallelogram, which implies $AM||LP$ and $\\bigtriangleup{AMB}$ is similar to $\\bigtriangleup{LPB}$\nThus,\n\\[\\frac {AM}{LP}=\\frac {AB}{LB}=\\frac {AL+LB}{LB}=\\frac {AL}{LB}+1\\]\nNow let's apply the angle bisector... |
https://artofproblemsolving.com/wiki/index.php/2005_AIME_I_Problems/Problem_15 | null | 38 | Triangle $ABC$ has $BC=20.$ The incircle of the triangle evenly trisects the median $AD.$ If the area of the triangle is $m \sqrt{n}$ where $m$ and $n$ are integers and $n$ is not divisible by the square of a prime, find $m+n.$ | [
"Let $E$ $F$ and $G$ be the points of tangency of the incircle with $BC$ $AC$ and $AB$ , respectively. Without loss of generality, let $AC < AB$ , so that $E$ is between $D$ and $C$ . Let the length of the median be $3m$ . Then by two applications of the Power of a Point Theorem $DE^2 = 2m \\cdot m = AF^2$ , so ... |
https://artofproblemsolving.com/wiki/index.php/2011_AMC_12A_Problems/Problem_25 | D | 80 | Triangle $ABC$ has $\angle BAC = 60^{\circ}$ $\angle CBA \leq 90^{\circ}$ $BC=1$ , and $AC \geq AB$ . Let $H$ $I$ , and $O$ be the orthocenter, incenter, and circumcenter of $\triangle ABC$ , respectively. Assume that the area of pentagon $BCOIH$ is the maximum possible. What is $\angle CBA$
$\textbf{(A)}\ 60^{\circ} \... | [
"Let $\\angle CAB=A$ $\\angle ABC=B$ $\\angle BCA=C$ for convenience.\nIt's well-known that $\\angle BOC=2A$ $\\angle BIC=90+\\frac{A}{2}$ , and $\\angle BHC=180-A$ (verifiable by angle chasing). Then, as $A=60$ , it follows that $\\angle BOC=\\angle BIC=\\angle BHC=120$ and consequently pentagon $BCOIH$ is cyclic.... |
https://artofproblemsolving.com/wiki/index.php/2015_AIME_I_Problems/Problem_11 | null | 108 | Triangle $ABC$ has positive integer side lengths with $AB=AC$ . Let $I$ be the intersection of the bisectors of $\angle B$ and $\angle C$ . Suppose $BI=8$ . Find the smallest possible perimeter of $\triangle ABC$ | [
"Let $D$ be the midpoint of $\\overline{BC}$ . Then by SAS Congruence, $\\triangle ABD \\cong \\triangle ACD$ , so $\\angle ADB = \\angle ADC = 90^o$\nNow let $BD=y$ $AB=x$ , and $\\angle IBD = \\dfrac{\\angle ABD}{2} = \\theta$\nThen $\\mathrm{cos}{(\\theta)} = \\dfrac{y}{8}$\nand $\\mathrm{cos}{(2\\theta)} = \\df... |
https://artofproblemsolving.com/wiki/index.php/1987_AIME_Problems/Problem_9 | null | 33 | Triangle $ABC$ has right angle at $B$ , and contains a point $P$ for which $PA = 10$ $PB = 6$ , and $\angle APB = \angle BPC = \angle CPA$ . Find $PC$
[asy] unitsize(0.2 cm); pair A, B, C, P; A = (0,14); B = (0,0); C = (21*sqrt(3),0); P = intersectionpoint(arc(B,6,0,180),arc(C,33,0,180)); draw(A--B--C--cycle); draw... | [
"Let $PC = x$ . Since $\\angle APB = \\angle BPC = \\angle CPA$ , each of them is equal to $120^\\circ$ . By the Law of Cosines applied to triangles $\\triangle APB$ $\\triangle BPC$ and $\\triangle CPA$ at their respective angles $P$ , remembering that $\\cos 120^\\circ = -\\frac12$ , we have\n\\[AB^2 = 36 + 100 ... |
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12B_Problems/Problem_24 | C | 30 | Triangle $ABC$ has side lengths $AB = 11, BC=24$ , and $CA = 20$ . The bisector of $\angle{BAC}$ intersects $\overline{BC}$ in point $D$ , and intersects the circumcircle of $\triangle{ABC}$ in point $E \ne A$ . The circumcircle of $\triangle{BED}$ intersects the line $AB$ in points $B$ and $F \ne B$ . What is $CF$
$\t... | [
"By the Law of Cosine $\\cos A = \\frac{AC^2 + AB^2 - BC^2}{ 2 \\cdot AC \\cdot AB} = \\frac{20^2 + 11^2 - 24^2}{2\\cdot20\\cdot11} = -\\frac18$\nAs $ABEC$ is a cyclic quadrilateral, $\\angle CEA = \\angle CBA$ . As $BDEF$ is a cyclic quadrilateral, $\\angle CBA = \\angle FEA$\n$\\because \\quad \\angle CEA = \\ang... |
https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_7 | null | 161 | Triangle $ABC$ has side lengths $AB = 12$ $BC = 25$ , and $CA = 17$ . Rectangle $PQRS$ has vertex $P$ on $\overline{AB}$ , vertex $Q$ on $\overline{AC}$ , and vertices $R$ and $S$ on $\overline{BC}$ . In terms of the side length $PQ = \omega$ , the area of $PQRS$ can be expressed as the quadratic polynomial \[Area(PQRS... | [
"If $\\omega = 25$ , the area of rectangle $PQRS$ is $0$ , so\n\\[\\alpha\\omega - \\beta\\omega^2 = 25\\alpha - 625\\beta = 0\\]\nand $\\alpha = 25\\beta$ . If $\\omega = \\frac{25}{2}$ , we can reflect $APQ$ over $PQ$ $PBS$ over $PS$ , and $QCR$ over $QR$ to completely cover rectangle $PQRS$ , so the area of $P... |
https://artofproblemsolving.com/wiki/index.php/2018_AIME_II_Problems/Problem_7 | null | 20 | Triangle $ABC$ has side lengths $AB = 9$ $BC =$ $5\sqrt{3}$ , and $AC = 12$ . Points $A = P_{0}, P_{1}, P_{2}, ... , P_{2450} = B$ are on segment $\overline{AB}$ with $P_{k}$ between $P_{k-1}$ and $P_{k+1}$ for $k = 1, 2, ..., 2449$ , and points $A = Q_{0}, Q_{1}, Q_{2}, ... , Q_{2450} = C$ are on segment $\overline{AC... | [
"For each $k$ between $2$ and $2450$ , the area of the trapezoid with $\\overline{P_kQ_k}$ as its bottom base is the difference between the areas of two triangles, both similar to $\\triangle{ABC}$ . Let $d_k$ be the length of segment $\\overline{P_kQ_k}$ . The area of the trapezoid with bases $\\overline{P_{k-1}Q_... |
https://artofproblemsolving.com/wiki/index.php/2019_AIME_II_Problems/Problem_7 | null | 715 | Triangle $ABC$ has side lengths $AB=120,BC=220$ , and $AC=180$ . Lines $\ell_A,\ell_B$ , and $\ell_C$ are drawn parallel to $\overline{BC},\overline{AC}$ , and $\overline{AB}$ , respectively, such that the intersections of $\ell_A,\ell_B$ , and $\ell_C$ with the interior of $\triangle ABC$ are segments of lengths $55,4... | [
"Let the points of intersection of $\\ell_A, \\ell_B,\\ell_C$ with $\\triangle ABC$ divide the sides into consecutive segments $BD,DE,EC,CF,FG,GA,AH,HI,IB$ . Furthermore, let the desired triangle be $\\triangle XYZ$ , with $X$ closest to side $BC$ $Y$ closest to side $AC$ , and $Z$ closest to side $AB$ . Hence, the... |
https://artofproblemsolving.com/wiki/index.php/2019_AIME_I_Problems/Problem_13 | null | 32 | Triangle $ABC$ has side lengths $AB=4$ $BC=5$ , and $CA=6$ . Points $D$ and $E$ are on ray $AB$ with $AB<AD<AE$ . The point $F \neq C$ is a point of intersection of the circumcircles of $\triangle ACD$ and $\triangle EBC$ satisfying $DF=2$ and $EF=7$ . Then $BE$ can be expressed as $\tfrac{a+b\sqrt{c}}{d}$ , where $a$ ... | [
"\nNotice that \\[\\angle DFE=\\angle CFE-\\angle CFD=\\angle CBE-\\angle CAD=180-B-A=C.\\] By the Law of Cosines, \\[\\cos C=\\frac{AC^2+BC^2-AB^2}{2\\cdot AC\\cdot BC}=\\frac34.\\] Then, \\[DE^2=DF^2+EF^2-2\\cdot DF\\cdot EF\\cos C=32\\implies DE=4\\sqrt2.\\] Let $X=\\overline{AB}\\cap\\overline{CF}$ $a=XB$ , and... |
https://artofproblemsolving.com/wiki/index.php/2019_AIME_II_Problems/Problem_11 | null | 11 | Triangle $ABC$ has side lengths $AB=7, BC=8,$ and $CA=9.$ Circle $\omega_1$ passes through $B$ and is tangent to line $AC$ at $A.$ Circle $\omega_2$ passes through $C$ and is tangent to line $AB$ at $A.$ Let $K$ be the intersection of circles $\omega_1$ and $\omega_2$ not equal to $A.$ Then $AK=\tfrac mn,$ where $m$ an... | [
" -Diagram by Brendanb4321\nNote that from the tangency condition that the supplement of $\\angle CAB$ with respects to lines $AB$ and $AC$ are equal to $\\angle AKB$ and $\\angle AKC$ , respectively, so from tangent-chord, \\[\\angle AKC=\\angle AKB=180^{\\circ}-\\angle BAC\\] Also note that $\\angle ABK=\\angle K... |
https://artofproblemsolving.com/wiki/index.php/2011_AMC_12A_Problems/Problem_13 | B | 30 | Triangle $ABC$ has side-lengths $AB = 12, BC = 24,$ and $AC = 18.$ The line through the incenter of $\triangle ABC$ parallel to $\overline{BC}$ intersects $\overline{AB}$ at $M$ and $\overline{AC}$ at $N.$ What is the perimeter of $\triangle AMN?$
$\textbf{(A)}\ 27 \qquad \textbf{(B)}\ 30 \qquad \textbf{(C)}\ 33 \qqua... | [
"Let $O$ be the incenter of $\\triangle{ABC}$ . Because $\\overline{MO} \\parallel \\overline{BC}$ and $\\overline{BO}$ is the angle bisector of $\\angle{ABC}$ , we have\n\\[\\angle{MBO} = \\angle{CBO} = \\angle{MOB} = \\frac{1}{2}\\angle{MBC}\\]\nIt then follows due to alternate interior angles and base angles of ... |
https://artofproblemsolving.com/wiki/index.php/2009_AMC_12B_Problems/Problem_9 | A | 6 | Triangle $ABC$ has vertices $A = (3,0)$ $B = (0,3)$ , and $C$ , where $C$ is on the line $x + y = 7$ . What is the area of $\triangle ABC$
$\mathrm{(A)}\ 6\qquad \mathrm{(B)}\ 8\qquad \mathrm{(C)}\ 10\qquad \mathrm{(D)}\ 12\qquad \mathrm{(E)}\ 14$ | [
"The base of the triangle is $AB = \\sqrt{3^2 + 3^2} = 3\\sqrt 2$ . Its altitude is the distance between the point $A$ and the parallel line $x + y = 7$ , which is $\\frac 4{\\sqrt 2} = 2\\sqrt 2$ . Therefore its area is $\\frac 12 \\cdot 3\\sqrt 2 \\cdot 2\\sqrt 2 = \\boxed{6}$"
] |
https://artofproblemsolving.com/wiki/index.php/2009_AMC_12B_Problems/Problem_9 | null | 6 | Triangle $ABC$ has vertices $A = (3,0)$ $B = (0,3)$ , and $C$ , where $C$ is on the line $x + y = 7$ . What is the area of $\triangle ABC$
$\mathrm{(A)}\ 6\qquad \mathrm{(B)}\ 8\qquad \mathrm{(C)}\ 10\qquad \mathrm{(D)}\ 12\qquad \mathrm{(E)}\ 14$ | [
"Because the line $x + y = 7$ is parallel to $\\overline {AB}$ , the area of $\\triangle ABC$ is independent of the location of $C$ on that line. Therefore it may be assumed that $C$ is $(7,0)$ . In that case the triangle has base $AC = 4$ and altitude $3$ , so its area is $\\frac 12 \\cdot 4 \\cdot 3 = \\boxed{6... |
https://artofproblemsolving.com/wiki/index.php/2003_AIME_II_Problems/Problem_11 | null | 578 | Triangle $ABC$ is a right triangle with $AC = 7,$ $BC = 24,$ and right angle at $C.$ Point $M$ is the midpoint of $AB,$ and $D$ is on the same side of line $AB$ as $C$ so that $AD = BD = 15.$ Given that the area of triangle $CDM$ may be expressed as $\frac {m\sqrt {n}}{p},$ where $m,$ $n,$ and $p$ are positive integers... | [
"We use the Pythagorean Theorem on $ABC$ to determine that $AB=25.$\nLet $N$ be the orthogonal projection from $C$ to $AB.$ Thus, $[CDM]=\\frac{(DM)(MN)} {2}$ $MN=BN-BM$ , and $[ABC]=\\frac{24 \\cdot 7} {2} =\\frac{25 \\cdot (CN)} {2}.$\nFrom the third equation, we get $CN=\\frac{168} {25}.$\nBy the Pythagorean The... |
https://artofproblemsolving.com/wiki/index.php/2018_AMC_12A_Problems/Problem_20 | D | 12 | Triangle $ABC$ is an isosceles right triangle with $AB=AC=3$ . Let $M$ be the midpoint of hypotenuse $\overline{BC}$ . Points $I$ and $E$ lie on sides $\overline{AC}$ and $\overline{AB}$ , respectively, so that $AI>AE$ and $AIME$ is a cyclic quadrilateral. Given that triangle $EMI$ has area $2$ , the length $CI$ can be... | [
"Observe that $\\triangle{EMI}$ is isosceles right ( $M$ is the midpoint of diameter arc $EI$ since $m\\angle MEI = m\\angle MAI = 45^\\circ$ ), so $MI=2,MC=\\frac{3}{\\sqrt{2}}$ . With $\\angle{MCI}=45^\\circ$ , we can use Law of Cosines to determine that $CI=\\frac{3\\pm\\sqrt{7}}{2}$ . The same calculations hold... |
https://artofproblemsolving.com/wiki/index.php/2006_AMC_8_Problems/Problem_19 | D | 5.5 | Triangle $ABC$ is an isosceles triangle with $\overline{AB}=\overline{BC}$ . Point $D$ is the midpoint of both $\overline{BC}$ and $\overline{AE}$ , and $\overline{CE}$ is 11 units long. Triangle $ABD$ is congruent to triangle $ECD$ . What is the length of $\overline{BD}$
[asy] size(100); draw((0,0)--(2,4)--(4,0)--(6,4... | [
"Since triangle $ABD$ is congruent to triangle $ECD$ and $\\overline{CE} =11$ $\\overline{AB}=11$ . Since $\\overline{AB}=\\overline{BC}$ $\\overline{BC}=11$ . Because point $D$ is the midpoint of $\\overline{BC}$ $\\overline{BD}=\\frac{\\overline{BC}}{2}=\\frac{11}{2}=\\boxed{5.5}$"
] |
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12B_Problems/Problem_9 | B | 12 | Triangle $ABC$ is equilateral with side length $6$ . Suppose that $O$ is the center of the inscribed
circle of this triangle. What is the area of the circle passing through $A$ $O$ , and $C$
$\textbf{(A)} \: 9\pi \qquad\textbf{(B)} \: 12\pi \qquad\textbf{(C)} \: 18\pi \qquad\textbf{(D)} \: 24\pi \qquad\textbf{(E)} \: 2... | [
"Construct the circle that passes through $A$ $O$ , and $C$ , centered at $X$\nAlso notice that $\\overline{OA}$ and $\\overline{OC}$ are the angle bisectors of angle $\\angle BAC$ and $\\angle BCA$ respectively. We then deduce $\\angle AOC=120^\\circ$\nConsider another point $M$ on Circle $X$ opposite to point $O$... |
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12B_Problems/Problem_9 | null | 12 | Triangle $ABC$ is equilateral with side length $6$ . Suppose that $O$ is the center of the inscribed
circle of this triangle. What is the area of the circle passing through $A$ $O$ , and $C$
$\textbf{(A)} \: 9\pi \qquad\textbf{(B)} \: 12\pi \qquad\textbf{(C)} \: 18\pi \qquad\textbf{(D)} \: 24\pi \qquad\textbf{(E)} \: 2... | [
"Call the diameter of the circle $d$ . If we extend points $A$ and $C$ to meet at a point on the circle and call it $E$ , then $\\bigtriangleup OAE=\\bigtriangleup OCE$ . Note that both triangles are right, since their hypotenuse is the diameter of the circle. Therefore, $CE=AE=\\sqrt{d^2-12}$ . We know this since ... |
https://artofproblemsolving.com/wiki/index.php/1994_AHSME_Problems/Problem_18 | C | 9 | Triangle $ABC$ is inscribed in a circle, and $\angle B = \angle C = 4\angle A$ . If $B$ and $C$ are adjacent vertices of a regular polygon of $n$ sides inscribed in this circle, then $n=$ [asy] draw(Circle((0,0), 5)); draw((0,5)--(3,-4)--(-3,-4)--cycle); label("A", (0,5), N); label("B", (-3,-4), SW); label("C", (3,-4),... | [
"We solve for $\\angle A$ as follows: \\[4\\angle A+4\\angle A+\\angle A=180\\implies 9\\angle A=180\\implies \\angle A=20.\\] That means that minor arc $\\widehat{BC}$ has measure $40^\\circ$ . We can fit a maximum of $\\frac{360}{40}=\\boxed{9}$ of these arcs in the circle."
] |
https://artofproblemsolving.com/wiki/index.php/1969_AHSME_Problems/Problem_8 | D | 61 | Triangle $ABC$ is inscribed in a circle. The measure of the non-overlapping minor arcs $AB$ $BC$ and $CA$ are, respectively, $x+75^{\circ} , 2x+25^{\circ},3x-22^{\circ}$ . Then one interior angle of the triangle is:
$\text{(A) } 57\tfrac{1}{2}^{\circ}\quad \text{(B) } 59^{\circ}\quad \text{(C) } 60^{\circ}\quad \text{(... | [
" Because the triangle is inscribed, the sum of the minor arcs equals $360^\\circ$ . Thus, \\[x+75+2x+25+3x-22=360\\] \\[6x+78=360\\] Solving this yields $x = 47$ , so the inscribed angles are $122^\\circ$ $99^\\circ$ , and $119^\\circ$ . Noting that an angle of $\\triangle ABC$ is half of its corresponding inscr... |
https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_10 | null | 43 | Triangle $ABC$ is inscribed in circle $\omega$ . Points $P$ and $Q$ are on side $\overline{AB}$ with $AP<AQ$ . Rays $CP$ and $CQ$ meet $\omega$ again at $S$ and $T$ (other than $C$ ), respectively. If $AP=4,PQ=3,QB=6,BT=5,$ and $AS=7$ , then $ST=\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. F... | [
" Let $\\angle ACP=\\alpha$ $\\angle PCQ=\\beta$ , and $\\angle QCB=\\gamma$ . Note that since $\\triangle ACQ\\sim\\triangle TBQ$ we have $\\tfrac{AC}{CQ}=\\tfrac56$ , so by the Ratio Lemma \\[\\dfrac{AP}{PQ}=\\dfrac{AC}{CQ}\\cdot\\dfrac{\\sin\\alpha}{\\sin\\beta}\\quad\\implies\\quad \\dfrac{\\sin\\alpha}{\\sin\\... |
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