https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Mathmaster2012&feedformat=atomAoPS Wiki - User contributions [en]2024-03-29T12:59:00ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_12B_Problems&diff=840152017 AMC 12B Problems2017-02-17T05:30:47Z<p>Mathmaster2012: /* Problem 25 */</p>
<hr />
<div>WORK IN PROGRESS<br />
<br />
{{AMC12 Problems|year=2017|ab=B}}<br />
<br />
==Problem 1==<br />
<br />
Kymbrea's comic book collection currently has <math>30</math> comic books in it, and she is adding to her collection at the rate of <math>2</math> comic books per month. LaShawn's collection currently has <math>10</math> comic books in it, and he is adding to his collection at the rate of <math>6</math> comic books per month. After how many months will LaShawn's collection have twice as many comic books as Kymbrea's?<br />
<br />
<math>\textbf{(A)}\ 1\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 20\qquad\textbf{(E)}\ 25</math><br />
<br />
[[2017 AMC 10B Problems/Problem 2|Solution]]<br />
<br />
==Problem 2==<br />
<br />
Real numbers <math>x</math>, <math>y</math>, and <math>z</math> satify the inequalities<br />
<math>0<x<1</math>, <math>-1<y<0</math>, and <math>1<z<2</math>.<br />
Which of the following numbers is necessarily positive?<br />
<br />
<math>\textbf{(A)}\ y+x^2\qquad\textbf{(B)}\ y+xz\qquad\textbf{(C)}\ y+y^2\qquad\textbf{(D)}\ y+2y^2\qquad\textbf{(E)}\ y+z</math><br />
<br />
[[2017 AMC 10B Problems/Problem 3|Solution]]<br />
<br />
==Problem 3==<br />
<br />
Supposed that <math>x</math> and <math>y</math> are nonzero real numbers such that <math>\frac{3x+y}{x-3y}=-2</math>. What is the value of <math>\frac{x+3y}{3x-y}</math>?<br />
<br />
<math>\textbf{(A)}\ -3\qquad\textbf{(B)}\ -1\qquad\textbf{(C)}\ 1\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ 3</math><br />
<br />
[[2017 AMC 10B Problems/Problem 4|Solution]]<br />
<br />
==Problem 4==<br />
Samia set off on her bicycle to visit her friend, traveling at an average speed of <math>17</math> kilometers per hour. When she had gone half the distance to her friend's house, a tire went flat, and she walked the rest of the way at <math>5</math> kilometers per hour. In all it took her <math>44</math> minutes to reach her friend's house. In kilometers rounded to the nearest tenth, how far did Samia walk?<br />
<br />
<math>\textbf{(A)}\ 2.0\qquad\textbf{(B)}\ 2.2\qquad\textbf{(C)}\ 2.8\qquad\textbf{(D)}\ 3.4\qquad\textbf{(E)}\ 4.4</math><br />
<br />
[[2017 AMC 12B Problems/Problem 4|Solution]]<br />
<br />
==Problem 5==<br />
<br />
The data set <math>[6,19,33,33,39,41,41,43,51,57]</math> has median <math>Q_2 = 40</math>, first quartile <math>Q_1 = 33</math>, and third quartile <math>Q_3=43</math>. An outlier in a data set is a value that is more than <math>1.5</math> times the interquartile range below the first quartile <math>(Q_1)</math> or more than <math>1.5</math> times the interquartile range above the third quartile <math>(Q_3)</math>, where the interquartile range is defined as <math>Q_3 - Q_1</math>. How many outliers does this data set have?<br />
<br />
<math>\textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 4</math><br />
<br />
[[2017 AMC 12B Problems/Problem 5|Solution]]<br />
<br />
==Problem 6==<br />
The circle having <math>(0,0)</math> and <math>(8,6)</math> as the endpoints of a diameter intersects the <math>x</math>-axis at a second point. What is the <math>x</math>-coordinate of this point? <br />
<br />
<math>\textbf{(A)}\ 4\sqrt{2} \qquad \textbf{(B)}\ 6\qquad \textbf{(C)}\ 5\sqrt{2}\qquad \textbf{(D)}\ 8\qquad \textbf{(E)}\ 6\sqrt{2}</math><br />
<br />
[[2017 AMC 12B Problems/Problem 6|Solution]]<br />
<br />
==Problem 7==<br />
The functions <math>\sin(x)</math> and <math>\cos(x)</math> are periodic with least period <math>2\pi</math>. What is the least period of the function <math>\cos(\sin(x))</math>?<br />
<br />
<math>\textbf{(A)}\ \frac{\pi}{2}\qquad\textbf{(B)}\ \pi\qquad\textbf{(C)}\ 2\pi \qquad\textbf{(D)}\ 4\pi \qquad\textbf{(E)}</math> It's not periodic.<br />
<br />
[[2017 AMC 12B Problems/Problem 7|Solution]]<br />
<br />
==Problem 8==<br />
The ratio of the short side of a certain rectangle to the long side is equal to the ratio of the long side tho the diagonal. What is the square of the ratio of the short side to the long side of this rectangle?<br />
<br />
<math>\textbf{(A)}\ \frac{\sqrt{3}-1}{2}\qquad\textbf{(B)}\ \frac{1}{2}\qquad\textbf{(C)}\ \frac{\sqrt{5}-1}{2} \qquad\textbf{(D)}\ \frac{\sqrt{2}}{2} \qquad\textbf{(E)}\ \frac{\sqrt{6}-1}{2}</math><br />
<br />
[[2017 AMC 12B Problems/Problem 8|Solution]]<br />
<br />
==Problem 9==<br />
<br />
A circle has center <math>(-10,-4)</math> and radius <math>13</math>. Another circle has center <math>(3,9)</math> and radius <math>\sqrt{65}</math>. The line passing through the two points of intersection of the two circles has equation <math>x + y = c</math>. What is <math>c</math>?<br />
<br />
<math>\textbf{(A)}\ 3\qquad\textbf{(B)}\ 3\sqrt{3}\qquad\textbf{(C)}\ 4\sqrt{2}\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ \frac{13}{2}</math><br />
<br />
[[2017 AMC 12B Problems/Problem 9|Solution]]<br />
<br />
==Problem 10==<br />
At Typico High School, <math>60\%</math> of the students like dancing, and the rest dislike it. Of those who like dancing, <math>80\%</math> say that they like it, and the rest say that they dislike it. Of those who dislike dancing, <math>90\%</math> say that they dislike it, and the rest say that they like it. What fraction of students who say they dislike dancing actually like it?<br />
<br />
<math>\textbf{(A)}\ 10\%\qquad\textbf{(B)}\ 12\%\qquad\textbf{(C)}\ 20\%\qquad\textbf{(D)}\ 25\%\qquad\textbf{(E)}\ 33\frac{1}{3}\%</math><br />
<br />
[[2017 AMC 12B Problems/Problem 10|Solution]]<br />
<br />
==Problem 11==<br />
Call a positive integer <math>monotonous</math> if it is a one-digit number or its digits, when read from left to right, form either a strictly increasing or a strictly decreasing sequence. For example, <math>3</math>, <math>23578</math>, and <math>987620</math> are monotonous, but <math>88</math>, <math>7434</math>, and <math>23557</math> are not. How many monotonous positive integers are there?<br />
<br />
<math>\textbf{(A)}\ 1024\qquad\textbf{(B)}\ 1524\qquad\textbf{(C)}\ 1533\qquad\textbf{(D)}\ 1536\qquad\textbf{(E)}\ 2048</math><br />
<br />
[[2017 AMC 10B Problems/Problem 17|Solution]]<br />
<br />
==Problem 12==<br />
What is the sum of the roots of <math>z^{12}=64</math> that have a positive real part? <br />
<br />
<math>\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\ \sqrt{2}+2\sqrt{3} \qquad \textbf{(D)}\ 2\sqrt{2}+\sqrt{6} \qquad \textbf{(E)}\ (1+\sqrt{3}) + (1+\sqrt{3})i</math><br />
<br />
[[2017 AMC 12B Problems/Problem 12|Solution]]<br />
<br />
==Problem 13==<br />
<br />
In the figure below, <math>3</math> of the <math>6</math> disks are to be painted blue, <math>2</math> are to be painted red, and <math>1</math> is to be painted green. Two paintings that can be obtained from one another by a rotation or a reflection of the entire figure are considered the same. How many different paintings are possible?<br />
<br />
<asy><br />
size(100);<br />
pair A, B, C, D, E, F;<br />
A = (0,0);<br />
B = (1,0);<br />
C = (2,0);<br />
D = rotate(60, A)*B;<br />
E = B + D;<br />
F = rotate(60, A)*C;<br />
draw(Circle(A, 0.5));<br />
draw(Circle(B, 0.5));<br />
draw(Circle(C, 0.5));<br />
draw(Circle(D, 0.5));<br />
draw(Circle(E, 0.5));<br />
draw(Circle(F, 0.5));<br />
</asy><br />
<br />
<math>\textbf{(A) } 6 \qquad \textbf{(B) } 8 \qquad \textbf{(C) } 9 \qquad \textbf{(D) } 12 \qquad \textbf{(E) } 15</math><br />
<br />
==Problem 14==<br />
An ice-cream novelty item consists of a cup in the shape of a 4-inch-tall frustum of a right circular cone, with a 2-inch-diameter base at the bottom and a 4-inch-diameter base at the top, packed solid with ice cream, together with a solid cone of ice cream of height 4 inches, whose base, at the bottom, is the top base of the frustum. What is the total volume of the ice cream, in cubic inches? <br />
<br />
<math>\textbf{(A)}\ 8\pi \qquad \textbf{(B)}\ \frac{28\pi}{3} \qquad \textbf{(C)}\ 12\pi \qquad \textbf{(D)}\ 14\pi \qquad \textbf{(E)}\ \frac{44\pi}{3}</math><br />
<br />
[[2017 AMC 12B Problems/Problem 14|Solution]]<br />
<br />
==Problem 15==<br />
Let <math>ABC</math> be an equilateral triangle. Extend side <math>\overline{AB}</math> beyond <math>B</math> to a point <math>B'</math> so that <math>BB'=3AB</math>. Similarly, extend side <math>\overline{BC}</math> beyond <math>C</math> to a point <math>C'</math> so that <math>CC'=3BC</math>, and extend side <math>\overline{CA}</math> beyond <math>A</math> to a point <math>A'</math> so that <math>AA'=3CA</math>. What is the ratio of the area of <math>\triangle A'B'C'</math> to the area of <math>\triangle ABC</math>?<br />
<br />
<math>\textbf{(A)}\ 9:1\qquad\textbf{(B)}\ 16:1\qquad\textbf{(C)}\ 25:1\qquad\textbf{(D)}\ 36:1\qquad\textbf{(E)}\ 37:1</math><br />
<br />
[[2017 AMC 12B Problems/Problem 15|Solution]]<br />
<br />
==Problem 16==<br />
The number <math>21!=51,090,942,171,709,440,000</math> has over <math>60,000</math> positive integer divisors. One of them is chosen at random. What is the probability that it is odd?<br />
<br />
<math>\textbf{(A)}\ \frac{1}{21} \qquad \textbf{(B)}\ \frac{1}{19} \qquad \textbf{(C)}\ \frac{1}{18} \qquad \textbf{(D)}\ \frac{1}{2} \qquad \textbf{(E)}\ \frac{11}{21}</math><br />
<br />
[[2017 AMC 12B Problems/Problem 16|Solution]]<br />
<br />
==Problem 17==<br />
A coin is biased in such a way that on each toss the probability of heads is <math>\frac{2}{3}</math> and the probability of tails is <math>\frac{1}{3}</math>. The outcomes of the tosses are independent. A player has the choice of playing Game A or Game B. In Game A she tosses the coin three times and wins if all three outcomes are the same. In Game B she tosses the coin four times and wins if both the outcomes of the first and second tosses are the same and the outcomes of the third and fourth tosses are the same. How do the chances of winning Game A compare to the chances of winning Game B?<br />
<br />
<math>\textbf{(A)}</math> The probability of winning Game A is <math>\frac{4}{81}</math> less than the probability of winning Game B.<br />
<br />
<math>\textbf{(B)}</math> The probability of winning Game A is <math>\frac{2}{81}</math> less than the probability of winning Game B.<br />
<br />
<math>\textbf{(C)}</math> The probabilities are the same.<br />
<br />
<math>\textbf{(D)}</math> The probability of winning Game A is <math>\frac{2}{81}</math> greater than the probability of winning Game B.<br />
<br />
<math>\textbf{(E)}</math> The probability of winning Game A is <math>\frac{4}{81}</math> greater than the probability of winning Game B.<br />
<br />
[[2017 AMC 12B Problems/Problem 17|Solution]]<br />
<br />
==Problem 18==<br />
The diameter <math>AB</math> of a circle of radius <math>2</math> is extended to a point <math>D</math> outside the circle so that <math>BD=3</math>. Point <math>E</math> is chosen so that <math>ED=5</math> and line <math>ED</math> is perpendicular to line <math>AD</math>. Segment <math>AE</math> intersects the circle at a point <math>C</math> between <math>A</math> and <math>E</math>. What is the area of <math>\triangle <br />
ABC</math>?<br />
<br />
<math>\textbf{(A)}\ \frac{120}{37}\qquad\textbf{(B)}\ \frac{140}{39}\qquad\textbf{(C)}\ \frac{145}{39}\qquad\textbf{(D)}\ \frac{140}{37}\qquad\textbf{(E)}\ \frac{120}{31}</math><br />
<br />
[[2017 AMC 12B Problems/Problem 18|Solution]]<br />
<br />
==Problem 19==<br />
Let <math>N=123456789101112\dots4344</math> be the <math>79</math>-digit number that is formed by writing the integers from <math>1</math> to <math>44</math> in order, one after the other. What is the remainder when <math>N</math> is divided by <math>45</math>?<br />
<br />
<math>\textbf{(A)}\ 1\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 18\qquad\textbf{(E)}\ 44</math><br />
<br />
[[2017 AMC 12B Problems/Problem 19|Solution]]<br />
<br />
==Problem 20==<br />
Real numbers <math>x</math> and <math>y</math> are chosen independently and uniformly at random from the interval <math>(0,1)</math>. What is the probability that <math>\lfloor\log_2x\rfloor=\lfloor\log_2y\rfloor</math>, where <math>\lfloor r\rfloor</math> denotes the greatest integer less than or equal to the real number <math>r</math> ?<br />
<br />
<math>\textbf{(A)}\ \frac{1}{8}\qquad\textbf{(B)}\ \frac{1}{6}\qquad\textbf{(C)}\ \frac{1}{4}\qquad\textbf{(D)}\ \frac{1}{3}\qquad\textbf{(E)}\ \frac{1}{2}</math><br />
<br />
[[2017 AMC 12B Problems/Problem 20|Solution]]<br />
<br />
==Problem 21==<br />
Last year Isabella took <math>7</math> math tests and received <math>7</math> different scores, each an integer between <math>91</math> and <math>100</math>, inclusive. After each test she noticed that the average of her test scores was an integer. Her score on the seventh test was <math>95</math>. What was her score on the sixth test?<br />
<br />
<math>\textbf{(A)}\ 92\qquad\textbf{(B)}\ 94\qquad\textbf{(C)}\ 96\qquad\textbf{(D)}\ 98\qquad\textbf{(E)}\ 100</math><br />
<br />
[[2017 AMC 12B Problems/Problem 21|Solution]]<br />
<br />
==Problem 22==<br />
Abby, Bernardo, Carl, and Debra play a game in which each of them starts with four coins. The game consists of four rounds. In each round, four balls are placed in an urn---one green, one red, and two white. The players each draw a ball at random without replacement. Whoever gets the green ball gives one coin to whoever gets the red ball. What is the probability that, at the end of the fourth round, each of the players has four coins?<br />
<br />
<math>\textbf{(A)}\ \frac{7}{576} \qquad \textbf{(B)}\ \frac{5}{192} \qquad \textbf{(C)}\ \frac{1}{36} \qquad \textbf{(D)}\ \frac{5}{144} \qquad\textbf{(E)}\ \frac{7}{48}</math><br />
<br />
==Problem 23==<br />
The graph of <math>y=f(x)</math>, where <math>f(x)</math> is a polynomial of degree <math>3</math>, contains points <math>A(2,4)</math>, <math>B(3,9)</math>, and <math>C(4,16)</math>. Lines <math>AB</math>, <math>AC</math>, and <math>BC</math> intersect the graph again at points <math>D</math>, <math>E</math>, and <math>F</math>, respectively, and the sum of the <math>x</math>-coordinates of <math>D</math>, <math>E</math>, and <math>F</math> is 24. What is <math>f(0)</math>?<br />
<math>\textbf{(A)}\ -2 \qquad \textbf{(B)}\ 0 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ \frac{24}{5} \qquad\textbf{(E)}\ 8</math><br />
<br />
[[2017 AMC 12B Problems/Problem 23|Solution]]<br />
<br />
==Problem 24==<br />
<br />
Quadrilateral <math>ABCD</math> has right angles at <math>B</math> and <math>C</math>, <math>\triangle ABC \sim \triangle BCD</math>, and <math>AB > BC</math>. There is a point <math>E</math> in the interior of <math>ABCD</math> such that <math>\triangle ABC \sim \triangle CEB</math> and the area of <math>\triangle AED</math> is <math>17</math> times the area of <math>\triangle CEB</math>. What is <math>\frac{AB}{BC}</math>?<br />
<br />
<math>\textbf{(A)}\ 1 + \sqrt{2} \qquad \textbf{(B)}\ 2 + \sqrt{2} \qquad \textbf{(C)}\ \sqrt{17} \qquad \textbf{(D)}\ 2 + \sqrt{5} \qquad\textbf{(E)}\ 1 + 2\sqrt{3}</math><br />
<br />
[[2017 AMC 12B Problems/Problem 24|Solution]]<br />
<br />
==Problem 25==<br />
A set of <math>n</math> people participate in an online video basketball tournament. Each person may be a member of any number of <math>5</math>-player teams, but no teams may have exactly the same <math>5</math> members. The site statistics show a curious fact: The average, over all subsets of size <math>9</math> of the set of <math>n</math> participants, of the number of complete teams whose members are among those 9 people is equal to the reciprocal of the average, over all subsets of size <math>8</math> of the set of <math>n</math> participants, of the number of complete teams whose members are among those <math>8</math> people. How many values <math>n</math>, <math>9 \leq n \leq 2017</math>, can be the number of participants?<br />
<br />
<math>\textbf{(A)}\ 477 \qquad \textbf{(B)}\ 482 \qquad \textbf{(C)}\ 487 \qquad \textbf{(D)}\ 557 \qquad\textbf{(E)}\ 562</math><br />
<br />
[[2017 AMC 12B Problems/Problem 25|Solution]]<br />
<br />
==See also==<br />
{{AMC12 box|year=2017|ab=B|before=[[2017 AMC 12A Problems]]|after=[[2018 AMC 12A Problems]]}}<br />
{{MAA Notice}}</div>Mathmaster2012https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_12B_Problems&diff=840142017 AMC 12B Problems2017-02-17T05:29:10Z<p>Mathmaster2012: </p>
<hr />
<div>WORK IN PROGRESS<br />
<br />
{{AMC12 Problems|year=2017|ab=B}}<br />
<br />
==Problem 1==<br />
<br />
Kymbrea's comic book collection currently has <math>30</math> comic books in it, and she is adding to her collection at the rate of <math>2</math> comic books per month. LaShawn's collection currently has <math>10</math> comic books in it, and he is adding to his collection at the rate of <math>6</math> comic books per month. After how many months will LaShawn's collection have twice as many comic books as Kymbrea's?<br />
<br />
<math>\textbf{(A)}\ 1\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 20\qquad\textbf{(E)}\ 25</math><br />
<br />
[[2017 AMC 10B Problems/Problem 2|Solution]]<br />
<br />
==Problem 2==<br />
<br />
Real numbers <math>x</math>, <math>y</math>, and <math>z</math> satify the inequalities<br />
<math>0<x<1</math>, <math>-1<y<0</math>, and <math>1<z<2</math>.<br />
Which of the following numbers is necessarily positive?<br />
<br />
<math>\textbf{(A)}\ y+x^2\qquad\textbf{(B)}\ y+xz\qquad\textbf{(C)}\ y+y^2\qquad\textbf{(D)}\ y+2y^2\qquad\textbf{(E)}\ y+z</math><br />
<br />
[[2017 AMC 10B Problems/Problem 3|Solution]]<br />
<br />
==Problem 3==<br />
<br />
Supposed that <math>x</math> and <math>y</math> are nonzero real numbers such that <math>\frac{3x+y}{x-3y}=-2</math>. What is the value of <math>\frac{x+3y}{3x-y}</math>?<br />
<br />
<math>\textbf{(A)}\ -3\qquad\textbf{(B)}\ -1\qquad\textbf{(C)}\ 1\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ 3</math><br />
<br />
[[2017 AMC 10B Problems/Problem 4|Solution]]<br />
<br />
==Problem 4==<br />
Samia set off on her bicycle to visit her friend, traveling at an average speed of <math>17</math> kilometers per hour. When she had gone half the distance to her friend's house, a tire went flat, and she walked the rest of the way at <math>5</math> kilometers per hour. In all it took her <math>44</math> minutes to reach her friend's house. In kilometers rounded to the nearest tenth, how far did Samia walk?<br />
<br />
<math>\textbf{(A)}\ 2.0\qquad\textbf{(B)}\ 2.2\qquad\textbf{(C)}\ 2.8\qquad\textbf{(D)}\ 3.4\qquad\textbf{(E)}\ 4.4</math><br />
<br />
[[2017 AMC 12B Problems/Problem 4|Solution]]<br />
<br />
==Problem 5==<br />
<br />
The data set <math>[6,19,33,33,39,41,41,43,51,57]</math> has median <math>Q_2 = 40</math>, first quartile <math>Q_1 = 33</math>, and third quartile <math>Q_3=43</math>. An outlier in a data set is a value that is more than <math>1.5</math> times the interquartile range below the first quartile <math>(Q_1)</math> or more than <math>1.5</math> times the interquartile range above the third quartile <math>(Q_3)</math>, where the interquartile range is defined as <math>Q_3 - Q_1</math>. How many outliers does this data set have?<br />
<br />
<math>\textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 4</math><br />
<br />
[[2017 AMC 12B Problems/Problem 5|Solution]]<br />
<br />
==Problem 6==<br />
The circle having <math>(0,0)</math> and <math>(8,6)</math> as the endpoints of a diameter intersects the <math>x</math>-axis at a second point. What is the <math>x</math>-coordinate of this point? <br />
<br />
<math>\textbf{(A)}\ 4\sqrt{2} \qquad \textbf{(B)}\ 6\qquad \textbf{(C)}\ 5\sqrt{2}\qquad \textbf{(D)}\ 8\qquad \textbf{(E)}\ 6\sqrt{2}</math><br />
<br />
[[2017 AMC 12B Problems/Problem 6|Solution]]<br />
<br />
==Problem 7==<br />
The functions <math>\sin(x)</math> and <math>\cos(x)</math> are periodic with least period <math>2\pi</math>. What is the least period of the function <math>\cos(\sin(x))</math>?<br />
<br />
<math>\textbf{(A)}\ \frac{\pi}{2}\qquad\textbf{(B)}\ \pi\qquad\textbf{(C)}\ 2\pi \qquad\textbf{(D)}\ 4\pi \qquad\textbf{(E)}</math> It's not periodic.<br />
<br />
[[2017 AMC 12B Problems/Problem 7|Solution]]<br />
<br />
==Problem 8==<br />
The ratio of the short side of a certain rectangle to the long side is equal to the ratio of the long side tho the diagonal. What is the square of the ratio of the short side to the long side of this rectangle?<br />
<br />
<math>\textbf{(A)}\ \frac{\sqrt{3}-1}{2}\qquad\textbf{(B)}\ \frac{1}{2}\qquad\textbf{(C)}\ \frac{\sqrt{5}-1}{2} \qquad\textbf{(D)}\ \frac{\sqrt{2}}{2} \qquad\textbf{(E)}\ \frac{\sqrt{6}-1}{2}</math><br />
<br />
[[2017 AMC 12B Problems/Problem 8|Solution]]<br />
<br />
==Problem 9==<br />
<br />
A circle has center <math>(-10,-4)</math> and radius <math>13</math>. Another circle has center <math>(3,9)</math> and radius <math>\sqrt{65}</math>. The line passing through the two points of intersection of the two circles has equation <math>x + y = c</math>. What is <math>c</math>?<br />
<br />
<math>\textbf{(A)}\ 3\qquad\textbf{(B)}\ 3\sqrt{3}\qquad\textbf{(C)}\ 4\sqrt{2}\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ \frac{13}{2}</math><br />
<br />
[[2017 AMC 12B Problems/Problem 9|Solution]]<br />
<br />
==Problem 10==<br />
At Typico High School, <math>60\%</math> of the students like dancing, and the rest dislike it. Of those who like dancing, <math>80\%</math> say that they like it, and the rest say that they dislike it. Of those who dislike dancing, <math>90\%</math> say that they dislike it, and the rest say that they like it. What fraction of students who say they dislike dancing actually like it?<br />
<br />
<math>\textbf{(A)}\ 10\%\qquad\textbf{(B)}\ 12\%\qquad\textbf{(C)}\ 20\%\qquad\textbf{(D)}\ 25\%\qquad\textbf{(E)}\ 33\frac{1}{3}\%</math><br />
<br />
[[2017 AMC 12B Problems/Problem 10|Solution]]<br />
<br />
==Problem 11==<br />
Call a positive integer <math>monotonous</math> if it is a one-digit number or its digits, when read from left to right, form either a strictly increasing or a strictly decreasing sequence. For example, <math>3</math>, <math>23578</math>, and <math>987620</math> are monotonous, but <math>88</math>, <math>7434</math>, and <math>23557</math> are not. How many monotonous positive integers are there?<br />
<br />
<math>\textbf{(A)}\ 1024\qquad\textbf{(B)}\ 1524\qquad\textbf{(C)}\ 1533\qquad\textbf{(D)}\ 1536\qquad\textbf{(E)}\ 2048</math><br />
<br />
[[2017 AMC 10B Problems/Problem 17|Solution]]<br />
<br />
==Problem 12==<br />
What is the sum of the roots of <math>z^{12}=64</math> that have a positive real part? <br />
<br />
<math>\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\ \sqrt{2}+2\sqrt{3} \qquad \textbf{(D)}\ 2\sqrt{2}+\sqrt{6} \qquad \textbf{(E)}\ (1+\sqrt{3}) + (1+\sqrt{3})i</math><br />
<br />
[[2017 AMC 12B Problems/Problem 12|Solution]]<br />
<br />
==Problem 13==<br />
<br />
In the figure below, <math>3</math> of the <math>6</math> disks are to be painted blue, <math>2</math> are to be painted red, and <math>1</math> is to be painted green. Two paintings that can be obtained from one another by a rotation or a reflection of the entire figure are considered the same. How many different paintings are possible?<br />
<br />
<asy><br />
size(100);<br />
pair A, B, C, D, E, F;<br />
A = (0,0);<br />
B = (1,0);<br />
C = (2,0);<br />
D = rotate(60, A)*B;<br />
E = B + D;<br />
F = rotate(60, A)*C;<br />
draw(Circle(A, 0.5));<br />
draw(Circle(B, 0.5));<br />
draw(Circle(C, 0.5));<br />
draw(Circle(D, 0.5));<br />
draw(Circle(E, 0.5));<br />
draw(Circle(F, 0.5));<br />
</asy><br />
<br />
<math>\textbf{(A) } 6 \qquad \textbf{(B) } 8 \qquad \textbf{(C) } 9 \qquad \textbf{(D) } 12 \qquad \textbf{(E) } 15</math><br />
<br />
==Problem 14==<br />
An ice-cream novelty item consists of a cup in the shape of a 4-inch-tall frustum of a right circular cone, with a 2-inch-diameter base at the bottom and a 4-inch-diameter base at the top, packed solid with ice cream, together with a solid cone of ice cream of height 4 inches, whose base, at the bottom, is the top base of the frustum. What is the total volume of the ice cream, in cubic inches? <br />
<br />
<math>\textbf{(A)}\ 8\pi \qquad \textbf{(B)}\ \frac{28\pi}{3} \qquad \textbf{(C)}\ 12\pi \qquad \textbf{(D)}\ 14\pi \qquad \textbf{(E)}\ \frac{44\pi}{3}</math><br />
<br />
[[2017 AMC 12B Problems/Problem 14|Solution]]<br />
<br />
==Problem 15==<br />
Let <math>ABC</math> be an equilateral triangle. Extend side <math>\overline{AB}</math> beyond <math>B</math> to a point <math>B'</math> so that <math>BB'=3AB</math>. Similarly, extend side <math>\overline{BC}</math> beyond <math>C</math> to a point <math>C'</math> so that <math>CC'=3BC</math>, and extend side <math>\overline{CA}</math> beyond <math>A</math> to a point <math>A'</math> so that <math>AA'=3CA</math>. What is the ratio of the area of <math>\triangle A'B'C'</math> to the area of <math>\triangle ABC</math>?<br />
<br />
<math>\textbf{(A)}\ 9:1\qquad\textbf{(B)}\ 16:1\qquad\textbf{(C)}\ 25:1\qquad\textbf{(D)}\ 36:1\qquad\textbf{(E)}\ 37:1</math><br />
<br />
[[2017 AMC 12B Problems/Problem 15|Solution]]<br />
<br />
==Problem 16==<br />
The number <math>21!=51,090,942,171,709,440,000</math> has over <math>60,000</math> positive integer divisors. One of them is chosen at random. What is the probability that it is odd?<br />
<br />
<math>\textbf{(A)}\ \frac{1}{21} \qquad \textbf{(B)}\ \frac{1}{19} \qquad \textbf{(C)}\ \frac{1}{18} \qquad \textbf{(D)}\ \frac{1}{2} \qquad \textbf{(E)}\ \frac{11}{21}</math><br />
<br />
[[2017 AMC 12B Problems/Problem 16|Solution]]<br />
<br />
==Problem 17==<br />
A coin is biased in such a way that on each toss the probability of heads is <math>\frac{2}{3}</math> and the probability of tails is <math>\frac{1}{3}</math>. The outcomes of the tosses are independent. A player has the choice of playing Game A or Game B. In Game A she tosses the coin three times and wins if all three outcomes are the same. In Game B she tosses the coin four times and wins if both the outcomes of the first and second tosses are the same and the outcomes of the third and fourth tosses are the same. How do the chances of winning Game A compare to the chances of winning Game B?<br />
<br />
<math>\textbf{(A)}</math> The probability of winning Game A is <math>\frac{4}{81}</math> less than the probability of winning Game B.<br />
<br />
<math>\textbf{(B)}</math> The probability of winning Game A is <math>\frac{2}{81}</math> less than the probability of winning Game B.<br />
<br />
<math>\textbf{(C)}</math> The probabilities are the same.<br />
<br />
<math>\textbf{(D)}</math> The probability of winning Game A is <math>\frac{2}{81}</math> greater than the probability of winning Game B.<br />
<br />
<math>\textbf{(E)}</math> The probability of winning Game A is <math>\frac{4}{81}</math> greater than the probability of winning Game B.<br />
<br />
[[2017 AMC 12B Problems/Problem 17|Solution]]<br />
<br />
==Problem 18==<br />
The diameter <math>AB</math> of a circle of radius <math>2</math> is extended to a point <math>D</math> outside the circle so that <math>BD=3</math>. Point <math>E</math> is chosen so that <math>ED=5</math> and line <math>ED</math> is perpendicular to line <math>AD</math>. Segment <math>AE</math> intersects the circle at a point <math>C</math> between <math>A</math> and <math>E</math>. What is the area of <math>\triangle <br />
ABC</math>?<br />
<br />
<math>\textbf{(A)}\ \frac{120}{37}\qquad\textbf{(B)}\ \frac{140}{39}\qquad\textbf{(C)}\ \frac{145}{39}\qquad\textbf{(D)}\ \frac{140}{37}\qquad\textbf{(E)}\ \frac{120}{31}</math><br />
<br />
[[2017 AMC 12B Problems/Problem 18|Solution]]<br />
<br />
==Problem 19==<br />
Let <math>N=123456789101112\dots4344</math> be the <math>79</math>-digit number that is formed by writing the integers from <math>1</math> to <math>44</math> in order, one after the other. What is the remainder when <math>N</math> is divided by <math>45</math>?<br />
<br />
<math>\textbf{(A)}\ 1\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 18\qquad\textbf{(E)}\ 44</math><br />
<br />
[[2017 AMC 12B Problems/Problem 19|Solution]]<br />
<br />
==Problem 20==<br />
Real numbers <math>x</math> and <math>y</math> are chosen independently and uniformly at random from the interval <math>(0,1)</math>. What is the probability that <math>\lfloor\log_2x\rfloor=\lfloor\log_2y\rfloor</math>, where <math>\lfloor r\rfloor</math> denotes the greatest integer less than or equal to the real number <math>r</math> ?<br />
<br />
<math>\textbf{(A)}\ \frac{1}{8}\qquad\textbf{(B)}\ \frac{1}{6}\qquad\textbf{(C)}\ \frac{1}{4}\qquad\textbf{(D)}\ \frac{1}{3}\qquad\textbf{(E)}\ \frac{1}{2}</math><br />
<br />
[[2017 AMC 12B Problems/Problem 20|Solution]]<br />
<br />
==Problem 21==<br />
Last year Isabella took <math>7</math> math tests and received <math>7</math> different scores, each an integer between <math>91</math> and <math>100</math>, inclusive. After each test she noticed that the average of her test scores was an integer. Her score on the seventh test was <math>95</math>. What was her score on the sixth test?<br />
<br />
<math>\textbf{(A)}\ 92\qquad\textbf{(B)}\ 94\qquad\textbf{(C)}\ 96\qquad\textbf{(D)}\ 98\qquad\textbf{(E)}\ 100</math><br />
<br />
[[2017 AMC 12B Problems/Problem 21|Solution]]<br />
<br />
==Problem 22==<br />
Abby, Bernardo, Carl, and Debra play a game in which each of them starts with four coins. The game consists of four rounds. In each round, four balls are placed in an urn---one green, one red, and two white. The players each draw a ball at random without replacement. Whoever gets the green ball gives one coin to whoever gets the red ball. What is the probability that, at the end of the fourth round, each of the players has four coins?<br />
<br />
<math>\textbf{(A)}\ \frac{7}{576} \qquad \textbf{(B)}\ \frac{5}{192} \qquad \textbf{(C)}\ \frac{1}{36} \qquad \textbf{(D)}\ \frac{5}{144} \qquad\textbf{(E)}\ \frac{7}{48}</math><br />
<br />
==Problem 23==<br />
The graph of <math>y=f(x)</math>, where <math>f(x)</math> is a polynomial of degree <math>3</math>, contains points <math>A(2,4)</math>, <math>B(3,9)</math>, and <math>C(4,16)</math>. Lines <math>AB</math>, <math>AC</math>, and <math>BC</math> intersect the graph again at points <math>D</math>, <math>E</math>, and <math>F</math>, respectively, and the sum of the <math>x</math>-coordinates of <math>D</math>, <math>E</math>, and <math>F</math> is 24. What is <math>f(0)</math>?<br />
<math>\textbf{(A)}\ -2 \qquad \textbf{(B)}\ 0 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ \frac{24}{5} \qquad\textbf{(E)}\ 8</math><br />
<br />
[[2017 AMC 12B Problems/Problem 23|Solution]]<br />
<br />
==Problem 24==<br />
<br />
Quadrilateral <math>ABCD</math> has right angles at <math>B</math> and <math>C</math>, <math>\triangle ABC \sim \triangle BCD</math>, and <math>AB > BC</math>. There is a point <math>E</math> in the interior of <math>ABCD</math> such that <math>\triangle ABC \sim \triangle CEB</math> and the area of <math>\triangle AED</math> is <math>17</math> times the area of <math>\triangle CEB</math>. What is <math>\frac{AB}{BC}</math>?<br />
<br />
<math>\textbf{(A)}\ 1 + \sqrt{2} \qquad \textbf{(B)}\ 2 + \sqrt{2} \qquad \textbf{(C)}\ \sqrt{17} \qquad \textbf{(D)}\ 2 + \sqrt{5} \qquad\textbf{(E)}\ 1 + 2\sqrt{3}</math><br />
<br />
[[2017 AMC 12B Problems/Problem 24|Solution]]<br />
<br />
==Problem 25==<br />
A set of <math>n</math> people participate in an online video basketball tournament. Each person may be a member of any number of <math>5</math>-player teams, but no teams may have exactly the same <math>5</math> members. The site statistics show a curious fact: The average, over all subsets of size <math>9</math> of the set of <math>n</math> participants, of the number of complete teams whose members are among those 9 people is equal to the reciprocal of the average, over all subsets of size <math>8</math> of the set of <math>n</math> participants, of the number of complete teams whose members are among those <math>8</math> people. How many values <math>n</math>, <math>9 \leq n \leq 2017</math>, can be the number of participants?<br />
<br />
<math>\textbf{(A)}\quad {477} \qquad \textbf{(B)}\quad{482} \qquad \textbf{(C)}\quad{487} \qquad\textbf{(D)}\quad{557} \qquad\textbf{(E)}\quad{562}</math><br />
<br />
[[2017 AMC 12B Problems/Problem 25|Solution]]<br />
<br />
==See also==<br />
{{AMC12 box|year=2017|ab=B|before=[[2017 AMC 12A Problems]]|after=[[2018 AMC 12A Problems]]}}<br />
{{MAA Notice}}</div>Mathmaster2012https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_12B_Problems&diff=840132017 AMC 12B Problems2017-02-17T05:27:05Z<p>Mathmaster2012: /* Problem 23 */</p>
<hr />
<div>WORK IN PROGRESS<br />
<br />
{{AMC12 Problems|year=2017|ab=B}}<br />
<br />
==Problem 1==<br />
<br />
Kymbrea's comic book collection currently has <math>30</math> comic books in it, and she is adding to her collection at the rate of <math>2</math> comic books per month. LaShawn's collection currently has <math>10</math> comic books in it, and he is adding to his collection at the rate of <math>6</math> comic books per month. After how many months will LaShawn's collection have twice as many comic books as Kymbrea's?<br />
<br />
<math>\textbf{(A)}\ 1\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 20\qquad\textbf{(E)}\ 25</math><br />
<br />
[[2017 AMC 10B Problems/Problem 2|Solution]]<br />
<br />
==Problem 2==<br />
<br />
Real numbers <math>x</math>, <math>y</math>, and <math>z</math> satify the inequalities<br />
<math>0<x<1</math>, <math>-1<y<0</math>, and <math>1<z<2</math>.<br />
Which of the following numbers is necessarily positive?<br />
<br />
<math>\textbf{(A)}\ y+x^2\qquad\textbf{(B)}\ y+xz\qquad\textbf{(C)}\ y+y^2\qquad\textbf{(D)}\ y+2y^2\qquad\textbf{(E)}\ y+z</math><br />
<br />
[[2017 AMC 10B Problems/Problem 3|Solution]]<br />
<br />
==Problem 3==<br />
<br />
Supposed that <math>x</math> and <math>y</math> are nonzero real numbers such that <math>\frac{3x+y}{x-3y}=-2</math>. What is the value of <math>\frac{x+3y}{3x-y}</math>?<br />
<br />
<math>\textbf{(A)}\ -3\qquad\textbf{(B)}\ -1\qquad\textbf{(C)}\ 1\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ 3</math><br />
<br />
[[2017 AMC 10B Problems/Problem 4|Solution]]<br />
<br />
==Problem 4==<br />
Samia set off on her bicycle to visit her friend, traveling at an average speed of <math>17</math> kilometers per hour. When she had gone half the distance to her friend's house, a tire went flat, and she walked the rest of the way at <math>5</math> kilometers per hour. In all it took her <math>44</math> minutes to reach her friend's house. In kilometers rounded to the nearest tenth, how far did Samia walk?<br />
<br />
<math>\textbf{(A)}\ 2.0\qquad\textbf{(B)}\ 2.2\qquad\textbf{(C)}\ 2.8\qquad\textbf{(D)}\ 3.4\qquad\textbf{(E)}\ 4.4</math><br />
<br />
[[2017 AMC 12B Problems/Problem 4|Solution]]<br />
<br />
==Problem 5==<br />
<br />
The data set <math>[6,19,33,33,39,41,41,43,51,57]</math> has median <math>Q_2 = 40</math>, first quartile <math>Q_1 = 33</math>, and third quartile <math>Q_3=43</math>. An outlier in a data set is a value that is more than <math>1.5</math> times the interquartile range below the first quartile <math>(Q_1)</math> or more than <math>1.5</math> times the interquartile range above the third quartile <math>(Q_3)</math>, where the interquartile range is defined as <math>Q_3 - Q_1</math>. How many outliers does this data set have?<br />
<br />
<math>\textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 4</math><br />
<br />
[[2017 AMC 12B Problems/Problem 5|Solution]]<br />
<br />
==Problem 6==<br />
The circle having <math>(0,0)</math> and <math>(8,6)</math> as the endpoints of a diameter intersects the <math>x</math>-axis at a second point. What is the <math>x</math>-coordinate of this point? <br />
<br />
<math>\textbf{(A)}\ 4\sqrt{2} \qquad \textbf{(B)}\ 6\qquad \textbf{(C)}\ 5\sqrt{2}\qquad \textbf{(D)}\ 8\qquad \textbf{(E)}\ 6\sqrt{2}</math><br />
<br />
[[2017 AMC 12B Problems/Problem 6|Solution]]<br />
<br />
==Problem 7==<br />
The functions <math>\sin(x)</math> and <math>\cos(x)</math> are periodic with least period <math>2\pi</math>. What is the least period of the function <math>\cos(\sin(x))</math>?<br />
<br />
<math>\textbf{(A)}\ \frac{\pi}{2}\qquad\textbf{(B)}\ \pi\qquad\textbf{(C)}\ 2\pi \qquad\textbf{(D)}\ 4\pi \qquad\textbf{(E)}</math> It's not periodic.<br />
<br />
[[2017 AMC 12B Problems/Problem 7|Solution]]<br />
<br />
==Problem 8==<br />
The ratio of the short side of a certain rectangle to the long side is equal to the ratio of the long side tho the diagonal. What is the square of the ratio of the short side to the long side of this rectangle?<br />
<br />
<math>\textbf{(A)}\ \frac{\sqrt{3}-1}{2}\qquad\textbf{(B)}\ \frac{1}{2}\qquad\textbf{(C)}\ \frac{\sqrt{5}-1}{2} \qquad\textbf{(D)}\ \frac{\sqrt{2}}{2} \qquad\textbf{(E)}\ \frac{\sqrt{6}-1}{2}</math><br />
<br />
[[2017 AMC 12B Problems/Problem 8|Solution]]<br />
<br />
==Problem 9==<br />
<br />
A circle has center <math>(-10,-4)</math> and radius <math>13</math>. Another circle has center <math>(3,9)</math> and radius <math>\sqrt{65}</math>. The line passing through the two points of intersection of the two circles has equation <math>x + y = c</math>. What is <math>c</math>?<br />
<br />
<math>\textbf{(A)}\ 3\qquad\textbf{(B)}\ 3\sqrt{3}\qquad\textbf{(C)}\ 4\sqrt{2}\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ \frac{13}{2}</math><br />
<br />
[[2017 AMC 12B Problems/Problem 9|Solution]]<br />
<br />
==Problem 10==<br />
At Typico High School, <math>60\%</math> of the students like dancing, and the rest dislike it. Of those who like dancing, <math>80\%</math> say that they like it, and the rest say that they dislike it. Of those who dislike dancing, <math>90\%</math> say that they dislike it, and the rest say that they like it. What fraction of students who say they dislike dancing actually like it?<br />
<br />
<math>\textbf{(A)}\ 10\%\qquad\textbf{(B)}\ 12\%\qquad\textbf{(C)}\ 20\%\qquad\textbf{(D)}\ 25\%\qquad\textbf{(E)}\ 33\frac{1}{3}\%</math><br />
<br />
[[2017 AMC 12B Problems/Problem 10|Solution]]<br />
<br />
==Problem 11==<br />
Call a positive integer <math>monotonous</math> if it is a one-digit number or its digits, when read from left to right, form either a strictly increasing or a strictly decreasing sequence. For example, <math>3</math>, <math>23578</math>, and <math>987620</math> are monotonous, but <math>88</math>, <math>7434</math>, and <math>23557</math> are not. How many monotonous positive integers are there?<br />
<br />
<math>\textbf{(A)}\ 1024\qquad\textbf{(B)}\ 1524\qquad\textbf{(C)}\ 1533\qquad\textbf{(D)}\ 1536\qquad\textbf{(E)}\ 2048</math><br />
<br />
[[2017 AMC 10B Problems/Problem 17|Solution]]<br />
<br />
==Problem 12==<br />
What is the sum of the roots of <math>z^{12}=64</math> that have a positive real part? <br />
<br />
<math>\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\ \sqrt{2}+2\sqrt{3} \qquad \textbf{(D)}\ 2\sqrt{2}+\sqrt{6} \qquad \textbf{(E)}\ (1+\sqrt{3}) + (1+\sqrt{3})i</math><br />
<br />
[[2017 AMC 12B Problems/Problem 12|Solution]]<br />
<br />
==Problem 13==<br />
<br />
In the figure below, <math>3</math> of the <math>6</math> disks are to be painted blue, <math>2</math> are to be painted red, and <math>1</math> is to be painted green. Two paintings that can be obtained from one another by a rotation or a reflection of the entire figure are considered the same. How many different paintings are possible?<br />
<br />
<asy><br />
size(100);<br />
pair A, B, C, D, E, F;<br />
A = (0,0);<br />
B = (1,0);<br />
C = (2,0);<br />
D = rotate(60, A)*B;<br />
E = B + D;<br />
F = rotate(60, A)*C;<br />
draw(Circle(A, 0.5));<br />
draw(Circle(B, 0.5));<br />
draw(Circle(C, 0.5));<br />
draw(Circle(D, 0.5));<br />
draw(Circle(E, 0.5));<br />
draw(Circle(F, 0.5));<br />
</asy><br />
<br />
<math>\textbf{(A) } 6 \qquad \textbf{(B) } 8 \qquad \textbf{(C) } 9 \qquad \textbf{(D) } 12 \qquad \textbf{(E) } 15</math><br />
<br />
==Problem 14==<br />
An ice-cream novelty item consists of a cup in the shape of a 4-inch-tall frustum of a right circular cone, with a 2-inch-diameter base at the bottom and a 4-inch-diameter base at the top, packed solid with ice cream, together with a solid cone of ice cream of height 4 inches, whose base, at the bottom, is the top base of the frustum. What is the total volume of the ice cream, in cubic inches? <br />
<br />
<math>\textbf{(A)}\ 8\pi \qquad \textbf{(B)}\ \frac{28\pi}{3} \qquad \textbf{(C)}\ 12\pi \qquad \textbf{(D)}\ 14\pi \qquad \textbf{(E)}\ \frac{44\pi}{3}</math><br />
<br />
[[2017 AMC 12B Problems/Problem 14|Solution]]<br />
<br />
==Problem 15==<br />
Let <math>ABC</math> be an equilateral triangle. Extend side <math>\overline{AB}</math> beyond <math>B</math> to a point <math>B'</math> so that <math>BB'=3AB</math>. Similarly, extend side <math>\overline{BC}</math> beyond <math>C</math> to a point <math>C'</math> so that <math>CC'=3BC</math>, and extend side <math>\overline{CA}</math> beyond <math>A</math> to a point <math>A'</math> so that <math>AA'=3CA</math>. What is the ratio of the area of <math>\triangle A'B'C'</math> to the area of <math>\triangle ABC</math>?<br />
<br />
<math>\textbf{(A)}\ 9:1\qquad\textbf{(B)}\ 16:1\qquad\textbf{(C)}\ 25:1\qquad\textbf{(D)}\ 36:1\qquad\textbf{(E)}\ 37:1</math><br />
<br />
[[2017 AMC 12B Problems/Problem 15|Solution]]<br />
<br />
==Problem 16==<br />
The number <math>21!=51,090,942,171,709,440,000</math> has over <math>60,000</math> positive integer divisors. One of them is chosen at random. What is the probability that it is odd?<br />
<br />
<math>\textbf{(A)}\ \frac{1}{21} \qquad \textbf{(B)}\ \frac{1}{19} \qquad \textbf{(C)}\ \frac{1}{18} \qquad \textbf{(D)}\ \frac{1}{2} \qquad \textbf{(E)}\ \frac{11}{21}</math><br />
<br />
[[2017 AMC 12B Problems/Problem 16|Solution]]<br />
<br />
==Problem 17==<br />
A coin is biased in such a way that on each toss the probability of heads is <math>\frac{2}{3}</math> and the probability of tails is <math>\frac{1}{3}</math>. The outcomes of the tosses are independent. A player has the choice of playing Game A or Game B. In Game A she tosses the coin three times and wins if all three outcomes are the same. In Game B she tosses the coin four times and wins if both the outcomes of the first and second tosses are the same and the outcomes of the third and fourth tosses are the same. How do the chances of winning Game A compare to the chances of winning Game B?<br />
<br />
<math>\textbf{(A)}</math> The probability of winning Game A is <math>\frac{4}{81}</math> less than the probability of winning Game B.<br />
<br />
<math>\textbf{(B)}</math> The probability of winning Game A is <math>\frac{2}{81}</math> less than the probability of winning Game B.<br />
<br />
<math>\textbf{(C)}</math> The probabilities are the same.<br />
<br />
<math>\textbf{(D)}</math> The probability of winning Game A is <math>\frac{2}{81}</math> greater than the probability of winning Game B.<br />
<br />
<math>\textbf{(E)}</math> The probability of winning Game A is <math>\frac{4}{81}</math> greater than the probability of winning Game B.<br />
<br />
[[2017 AMC 12B Problems/Problem 17|Solution]]<br />
<br />
==Problem 18==<br />
The diameter <math>AB</math> of a circle of radius <math>2</math> is extended to a point <math>D</math> outside the circle so that <math>BD=3</math>. Point <math>E</math> is chosen so that <math>ED=5</math> and line <math>ED</math> is perpendicular to line <math>AD</math>. Segment <math>AE</math> intersects the circle at a point <math>C</math> between <math>A</math> and <math>E</math>. What is the area of <math>\triangle <br />
ABC</math>?<br />
<br />
<math>\textbf{(A)}\ \frac{120}{37}\qquad\textbf{(B)}\ \frac{140}{39}\qquad\textbf{(C)}\ \frac{145}{39}\qquad\textbf{(D)}\ \frac{140}{37}\qquad\textbf{(E)}\ \frac{120}{31}</math><br />
<br />
[[2017 AMC 12B Problems/Problem 18|Solution]]<br />
<br />
==Problem 19==<br />
Let <math>N=123456789101112\dots4344</math> be the <math>79</math>-digit number that is formed by writing the integers from <math>1</math> to <math>44</math> in order, one after the other. What is the remainder when <math>N</math> is divided by <math>45</math>?<br />
<br />
<math>\textbf{(A)}\ 1\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 18\qquad\textbf{(E)}\ 44</math><br />
<br />
[[2017 AMC 12B Problems/Problem 19|Solution]]<br />
<br />
==Problem 20==<br />
Real numbers <math>x</math> and <math>y</math> are chosen independently and uniformly at random from the interval <math>(0,1)</math>. What is the probability that <math>\lfloor\log_2x\rfloor=\lfloor\log_2y\rfloor</math>, where <math>\lfloor r\rfloor</math> denotes the greatest integer less than or equal to the real number <math>r</math> ?<br />
<br />
<math>\textbf{(A)}\ \frac{1}{8}\qquad\textbf{(B)}\ \frac{1}{6}\qquad\textbf{(C)}\ \frac{1}{4}\qquad\textbf{(D)}\ \frac{1}{3}\qquad\textbf{(E)}\ \frac{1}{2}</math><br />
<br />
[[2017 AMC 12B Problems/Problem 20|Solution]]<br />
<br />
==Problem 21==<br />
Last year Isabella took <math>7</math> math tests and received <math>7</math> different scores, each an integer between <math>91</math> and <math>100</math>, inclusive. After each test she noticed that the average of her test scores was an integer. Her score on the seventh test was <math>95</math>. What was her score on the sixth test?<br />
<br />
<math>\textbf{(A)}\ 92\qquad\textbf{(B)}\ 94\qquad\textbf{(C)}\ 96\qquad\textbf{(D)}\ 98\qquad\textbf{(E)}\ 100</math><br />
<br />
[[2017 AMC 12B Problems/Problem 21|Solution]]<br />
<br />
==Problem 22==<br />
Abby, Bernardo, Carl, and Debra play a game in which each of them starts with four coins. The game consists of four rounds. In each round, four balls are placed in an urn---one green, one red, and two white. The players each draw a ball at random without replacement. Whoever gets the green ball gives one coin to whoever gets the red ball. What is the probability that, at the end of the fourth round, each of the players has four coins?<br />
<br />
<math>\textbf{(A)}\ \frac{7}{576} \qquad \textbf{(B)}\ \frac{5}{192} \qquad \textbf{(C)}\ \frac{1}{36} \qquad \textbf{(D)}\ \frac{5}{144} \qquad\textbf{(E)}\ \frac{7}{48}</math><br />
<br />
==Problem 23==<br />
The graph of <math>y=f(x)</math>, where <math>f(x)</math> is a polynomial of degree <math>3</math>, contains points <math>A(2,4)</math>, <math>B(3,9)</math>, and <math>C(4,16)</math>. Lines <math>AB</math>, <math>AC</math>, and <math>BC</math> intersect the graph again at points <math>D</math>, <math>E</math>, and <math>F</math>, respectively, and the sum of the <math>x</math>-coordinates of <math>D</math>, <math>E</math>, and <math>F</math> is 24. What is <math>f(0)</math>?<br />
<math>\textbf{(A)}\ -2 \qquad \textbf{(B)}\ 0 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ \frac{24}{5} \qquad\textbf{(E)}\ 8</math><br />
<br />
[[2017 AMC 12B Problems/Problem 23|Solution]]<br />
<br />
==Problem 24==<br />
<br />
Quadrilateral <math>ABCD</math> has right angles at <math>B</math> and <math>C</math>, <math>\triangle ABC \sim \triangle BCD</math>, and <math>AB > BC</math>. There is a point <math>E</math> in the interior of <math>ABCD</math> such that <math>\triangle ABC \sim \triangle CEB</math> and the area of <math>\triangle AED</math> is <math>17</math> times the area of <math>\triangle CEB</math>. What is <math>\frac{AB}{BC}</math>?<br />
<br />
<math>\textbf{(A)}\quad {1 + \sqrt{2}} \qquad \textbf{(B)}\quad{2 + \sqrt{2}} \qquad \textbf{(C)}\quad{\sqrt{17}} \qquad\textbf{(D)}\quad{2+\sqrt{5}} \qquad\textbf{(E)}\quad{1 + 2\sqrt{3}}</math><br />
<br />
[[2017 AMC 12B Problems/Problem 24|Solution]]<br />
<br />
==Problem 25==<br />
A set of <math>n</math> people participate in an online video basketball tournament. Each person may be a member of any number of <math>5</math>-player teams, but no teams may have exactly the same <math>5</math> members. The site statistics show a curious fact: The average, over all subsets of size <math>9</math> of the set of <math>n</math> participants, of the number of complete teams whose members are among those 9 people is equal to the reciprocal of the average, over all subsets of size <math>8</math> of the set of <math>n</math> participants, of the number of complete teams whose members are among those <math>8</math> people. How many values <math>n</math>, <math>9 \leq n \leq 2017</math>, can be the number of participants?<br />
<br />
<math>\textbf{(A)}\quad {477} \qquad \textbf{(B)}\quad{482} \qquad \textbf{(C)}\quad{487} \qquad\textbf{(D)}\quad{557} \qquad\textbf{(E)}\quad{562}</math><br />
<br />
[[2017 AMC 12B Problems/Problem 25|Solution]]<br />
<br />
==See also==<br />
{{AMC12 box|year=2017|ab=B|before=[[2017 AMC 12A Problems]]|after=[[2018 AMC 12A Problems]]}}<br />
{{MAA Notice}}</div>Mathmaster2012https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_12B_Problems&diff=840122017 AMC 12B Problems2017-02-17T05:26:32Z<p>Mathmaster2012: /* Problem 22 */</p>
<hr />
<div>WORK IN PROGRESS<br />
<br />
{{AMC12 Problems|year=2017|ab=B}}<br />
<br />
==Problem 1==<br />
<br />
Kymbrea's comic book collection currently has <math>30</math> comic books in it, and she is adding to her collection at the rate of <math>2</math> comic books per month. LaShawn's collection currently has <math>10</math> comic books in it, and he is adding to his collection at the rate of <math>6</math> comic books per month. After how many months will LaShawn's collection have twice as many comic books as Kymbrea's?<br />
<br />
<math>\textbf{(A)}\ 1\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 20\qquad\textbf{(E)}\ 25</math><br />
<br />
[[2017 AMC 10B Problems/Problem 2|Solution]]<br />
<br />
==Problem 2==<br />
<br />
Real numbers <math>x</math>, <math>y</math>, and <math>z</math> satify the inequalities<br />
<math>0<x<1</math>, <math>-1<y<0</math>, and <math>1<z<2</math>.<br />
Which of the following numbers is necessarily positive?<br />
<br />
<math>\textbf{(A)}\ y+x^2\qquad\textbf{(B)}\ y+xz\qquad\textbf{(C)}\ y+y^2\qquad\textbf{(D)}\ y+2y^2\qquad\textbf{(E)}\ y+z</math><br />
<br />
[[2017 AMC 10B Problems/Problem 3|Solution]]<br />
<br />
==Problem 3==<br />
<br />
Supposed that <math>x</math> and <math>y</math> are nonzero real numbers such that <math>\frac{3x+y}{x-3y}=-2</math>. What is the value of <math>\frac{x+3y}{3x-y}</math>?<br />
<br />
<math>\textbf{(A)}\ -3\qquad\textbf{(B)}\ -1\qquad\textbf{(C)}\ 1\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ 3</math><br />
<br />
[[2017 AMC 10B Problems/Problem 4|Solution]]<br />
<br />
==Problem 4==<br />
Samia set off on her bicycle to visit her friend, traveling at an average speed of <math>17</math> kilometers per hour. When she had gone half the distance to her friend's house, a tire went flat, and she walked the rest of the way at <math>5</math> kilometers per hour. In all it took her <math>44</math> minutes to reach her friend's house. In kilometers rounded to the nearest tenth, how far did Samia walk?<br />
<br />
<math>\textbf{(A)}\ 2.0\qquad\textbf{(B)}\ 2.2\qquad\textbf{(C)}\ 2.8\qquad\textbf{(D)}\ 3.4\qquad\textbf{(E)}\ 4.4</math><br />
<br />
[[2017 AMC 12B Problems/Problem 4|Solution]]<br />
<br />
==Problem 5==<br />
<br />
The data set <math>[6,19,33,33,39,41,41,43,51,57]</math> has median <math>Q_2 = 40</math>, first quartile <math>Q_1 = 33</math>, and third quartile <math>Q_3=43</math>. An outlier in a data set is a value that is more than <math>1.5</math> times the interquartile range below the first quartile <math>(Q_1)</math> or more than <math>1.5</math> times the interquartile range above the third quartile <math>(Q_3)</math>, where the interquartile range is defined as <math>Q_3 - Q_1</math>. How many outliers does this data set have?<br />
<br />
<math>\textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 4</math><br />
<br />
[[2017 AMC 12B Problems/Problem 5|Solution]]<br />
<br />
==Problem 6==<br />
The circle having <math>(0,0)</math> and <math>(8,6)</math> as the endpoints of a diameter intersects the <math>x</math>-axis at a second point. What is the <math>x</math>-coordinate of this point? <br />
<br />
<math>\textbf{(A)}\ 4\sqrt{2} \qquad \textbf{(B)}\ 6\qquad \textbf{(C)}\ 5\sqrt{2}\qquad \textbf{(D)}\ 8\qquad \textbf{(E)}\ 6\sqrt{2}</math><br />
<br />
[[2017 AMC 12B Problems/Problem 6|Solution]]<br />
<br />
==Problem 7==<br />
The functions <math>\sin(x)</math> and <math>\cos(x)</math> are periodic with least period <math>2\pi</math>. What is the least period of the function <math>\cos(\sin(x))</math>?<br />
<br />
<math>\textbf{(A)}\ \frac{\pi}{2}\qquad\textbf{(B)}\ \pi\qquad\textbf{(C)}\ 2\pi \qquad\textbf{(D)}\ 4\pi \qquad\textbf{(E)}</math> It's not periodic.<br />
<br />
[[2017 AMC 12B Problems/Problem 7|Solution]]<br />
<br />
==Problem 8==<br />
The ratio of the short side of a certain rectangle to the long side is equal to the ratio of the long side tho the diagonal. What is the square of the ratio of the short side to the long side of this rectangle?<br />
<br />
<math>\textbf{(A)}\ \frac{\sqrt{3}-1}{2}\qquad\textbf{(B)}\ \frac{1}{2}\qquad\textbf{(C)}\ \frac{\sqrt{5}-1}{2} \qquad\textbf{(D)}\ \frac{\sqrt{2}}{2} \qquad\textbf{(E)}\ \frac{\sqrt{6}-1}{2}</math><br />
<br />
[[2017 AMC 12B Problems/Problem 8|Solution]]<br />
<br />
==Problem 9==<br />
<br />
A circle has center <math>(-10,-4)</math> and radius <math>13</math>. Another circle has center <math>(3,9)</math> and radius <math>\sqrt{65}</math>. The line passing through the two points of intersection of the two circles has equation <math>x + y = c</math>. What is <math>c</math>?<br />
<br />
<math>\textbf{(A)}\ 3\qquad\textbf{(B)}\ 3\sqrt{3}\qquad\textbf{(C)}\ 4\sqrt{2}\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ \frac{13}{2}</math><br />
<br />
[[2017 AMC 12B Problems/Problem 9|Solution]]<br />
<br />
==Problem 10==<br />
At Typico High School, <math>60\%</math> of the students like dancing, and the rest dislike it. Of those who like dancing, <math>80\%</math> say that they like it, and the rest say that they dislike it. Of those who dislike dancing, <math>90\%</math> say that they dislike it, and the rest say that they like it. What fraction of students who say they dislike dancing actually like it?<br />
<br />
<math>\textbf{(A)}\ 10\%\qquad\textbf{(B)}\ 12\%\qquad\textbf{(C)}\ 20\%\qquad\textbf{(D)}\ 25\%\qquad\textbf{(E)}\ 33\frac{1}{3}\%</math><br />
<br />
[[2017 AMC 12B Problems/Problem 10|Solution]]<br />
<br />
==Problem 11==<br />
Call a positive integer <math>monotonous</math> if it is a one-digit number or its digits, when read from left to right, form either a strictly increasing or a strictly decreasing sequence. For example, <math>3</math>, <math>23578</math>, and <math>987620</math> are monotonous, but <math>88</math>, <math>7434</math>, and <math>23557</math> are not. How many monotonous positive integers are there?<br />
<br />
<math>\textbf{(A)}\ 1024\qquad\textbf{(B)}\ 1524\qquad\textbf{(C)}\ 1533\qquad\textbf{(D)}\ 1536\qquad\textbf{(E)}\ 2048</math><br />
<br />
[[2017 AMC 10B Problems/Problem 17|Solution]]<br />
<br />
==Problem 12==<br />
What is the sum of the roots of <math>z^{12}=64</math> that have a positive real part? <br />
<br />
<math>\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\ \sqrt{2}+2\sqrt{3} \qquad \textbf{(D)}\ 2\sqrt{2}+\sqrt{6} \qquad \textbf{(E)}\ (1+\sqrt{3}) + (1+\sqrt{3})i</math><br />
<br />
[[2017 AMC 12B Problems/Problem 12|Solution]]<br />
<br />
==Problem 13==<br />
<br />
In the figure below, <math>3</math> of the <math>6</math> disks are to be painted blue, <math>2</math> are to be painted red, and <math>1</math> is to be painted green. Two paintings that can be obtained from one another by a rotation or a reflection of the entire figure are considered the same. How many different paintings are possible?<br />
<br />
<asy><br />
size(100);<br />
pair A, B, C, D, E, F;<br />
A = (0,0);<br />
B = (1,0);<br />
C = (2,0);<br />
D = rotate(60, A)*B;<br />
E = B + D;<br />
F = rotate(60, A)*C;<br />
draw(Circle(A, 0.5));<br />
draw(Circle(B, 0.5));<br />
draw(Circle(C, 0.5));<br />
draw(Circle(D, 0.5));<br />
draw(Circle(E, 0.5));<br />
draw(Circle(F, 0.5));<br />
</asy><br />
<br />
<math>\textbf{(A) } 6 \qquad \textbf{(B) } 8 \qquad \textbf{(C) } 9 \qquad \textbf{(D) } 12 \qquad \textbf{(E) } 15</math><br />
<br />
==Problem 14==<br />
An ice-cream novelty item consists of a cup in the shape of a 4-inch-tall frustum of a right circular cone, with a 2-inch-diameter base at the bottom and a 4-inch-diameter base at the top, packed solid with ice cream, together with a solid cone of ice cream of height 4 inches, whose base, at the bottom, is the top base of the frustum. What is the total volume of the ice cream, in cubic inches? <br />
<br />
<math>\textbf{(A)}\ 8\pi \qquad \textbf{(B)}\ \frac{28\pi}{3} \qquad \textbf{(C)}\ 12\pi \qquad \textbf{(D)}\ 14\pi \qquad \textbf{(E)}\ \frac{44\pi}{3}</math><br />
<br />
[[2017 AMC 12B Problems/Problem 14|Solution]]<br />
<br />
==Problem 15==<br />
Let <math>ABC</math> be an equilateral triangle. Extend side <math>\overline{AB}</math> beyond <math>B</math> to a point <math>B'</math> so that <math>BB'=3AB</math>. Similarly, extend side <math>\overline{BC}</math> beyond <math>C</math> to a point <math>C'</math> so that <math>CC'=3BC</math>, and extend side <math>\overline{CA}</math> beyond <math>A</math> to a point <math>A'</math> so that <math>AA'=3CA</math>. What is the ratio of the area of <math>\triangle A'B'C'</math> to the area of <math>\triangle ABC</math>?<br />
<br />
<math>\textbf{(A)}\ 9:1\qquad\textbf{(B)}\ 16:1\qquad\textbf{(C)}\ 25:1\qquad\textbf{(D)}\ 36:1\qquad\textbf{(E)}\ 37:1</math><br />
<br />
[[2017 AMC 12B Problems/Problem 15|Solution]]<br />
<br />
==Problem 16==<br />
The number <math>21!=51,090,942,171,709,440,000</math> has over <math>60,000</math> positive integer divisors. One of them is chosen at random. What is the probability that it is odd?<br />
<br />
<math>\textbf{(A)}\ \frac{1}{21} \qquad \textbf{(B)}\ \frac{1}{19} \qquad \textbf{(C)}\ \frac{1}{18} \qquad \textbf{(D)}\ \frac{1}{2} \qquad \textbf{(E)}\ \frac{11}{21}</math><br />
<br />
[[2017 AMC 12B Problems/Problem 16|Solution]]<br />
<br />
==Problem 17==<br />
A coin is biased in such a way that on each toss the probability of heads is <math>\frac{2}{3}</math> and the probability of tails is <math>\frac{1}{3}</math>. The outcomes of the tosses are independent. A player has the choice of playing Game A or Game B. In Game A she tosses the coin three times and wins if all three outcomes are the same. In Game B she tosses the coin four times and wins if both the outcomes of the first and second tosses are the same and the outcomes of the third and fourth tosses are the same. How do the chances of winning Game A compare to the chances of winning Game B?<br />
<br />
<math>\textbf{(A)}</math> The probability of winning Game A is <math>\frac{4}{81}</math> less than the probability of winning Game B.<br />
<br />
<math>\textbf{(B)}</math> The probability of winning Game A is <math>\frac{2}{81}</math> less than the probability of winning Game B.<br />
<br />
<math>\textbf{(C)}</math> The probabilities are the same.<br />
<br />
<math>\textbf{(D)}</math> The probability of winning Game A is <math>\frac{2}{81}</math> greater than the probability of winning Game B.<br />
<br />
<math>\textbf{(E)}</math> The probability of winning Game A is <math>\frac{4}{81}</math> greater than the probability of winning Game B.<br />
<br />
[[2017 AMC 12B Problems/Problem 17|Solution]]<br />
<br />
==Problem 18==<br />
The diameter <math>AB</math> of a circle of radius <math>2</math> is extended to a point <math>D</math> outside the circle so that <math>BD=3</math>. Point <math>E</math> is chosen so that <math>ED=5</math> and line <math>ED</math> is perpendicular to line <math>AD</math>. Segment <math>AE</math> intersects the circle at a point <math>C</math> between <math>A</math> and <math>E</math>. What is the area of <math>\triangle <br />
ABC</math>?<br />
<br />
<math>\textbf{(A)}\ \frac{120}{37}\qquad\textbf{(B)}\ \frac{140}{39}\qquad\textbf{(C)}\ \frac{145}{39}\qquad\textbf{(D)}\ \frac{140}{37}\qquad\textbf{(E)}\ \frac{120}{31}</math><br />
<br />
[[2017 AMC 12B Problems/Problem 18|Solution]]<br />
<br />
==Problem 19==<br />
Let <math>N=123456789101112\dots4344</math> be the <math>79</math>-digit number that is formed by writing the integers from <math>1</math> to <math>44</math> in order, one after the other. What is the remainder when <math>N</math> is divided by <math>45</math>?<br />
<br />
<math>\textbf{(A)}\ 1\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 18\qquad\textbf{(E)}\ 44</math><br />
<br />
[[2017 AMC 12B Problems/Problem 19|Solution]]<br />
<br />
==Problem 20==<br />
Real numbers <math>x</math> and <math>y</math> are chosen independently and uniformly at random from the interval <math>(0,1)</math>. What is the probability that <math>\lfloor\log_2x\rfloor=\lfloor\log_2y\rfloor</math>, where <math>\lfloor r\rfloor</math> denotes the greatest integer less than or equal to the real number <math>r</math> ?<br />
<br />
<math>\textbf{(A)}\ \frac{1}{8}\qquad\textbf{(B)}\ \frac{1}{6}\qquad\textbf{(C)}\ \frac{1}{4}\qquad\textbf{(D)}\ \frac{1}{3}\qquad\textbf{(E)}\ \frac{1}{2}</math><br />
<br />
[[2017 AMC 12B Problems/Problem 20|Solution]]<br />
<br />
==Problem 21==<br />
Last year Isabella took <math>7</math> math tests and received <math>7</math> different scores, each an integer between <math>91</math> and <math>100</math>, inclusive. After each test she noticed that the average of her test scores was an integer. Her score on the seventh test was <math>95</math>. What was her score on the sixth test?<br />
<br />
<math>\textbf{(A)}\ 92\qquad\textbf{(B)}\ 94\qquad\textbf{(C)}\ 96\qquad\textbf{(D)}\ 98\qquad\textbf{(E)}\ 100</math><br />
<br />
[[2017 AMC 12B Problems/Problem 21|Solution]]<br />
<br />
==Problem 22==<br />
Abby, Bernardo, Carl, and Debra play a game in which each of them starts with four coins. The game consists of four rounds. In each round, four balls are placed in an urn---one green, one red, and two white. The players each draw a ball at random without replacement. Whoever gets the green ball gives one coin to whoever gets the red ball. What is the probability that, at the end of the fourth round, each of the players has four coins?<br />
<br />
<math>\textbf{(A)}\ \frac{7}{576} \qquad \textbf{(B)}\ \frac{5}{192} \qquad \textbf{(C)}\ \frac{1}{36} \qquad \textbf{(D)}\ \frac{5}{144} \qquad\textbf{(E)}\ \frac{7}{48}</math><br />
<br />
==Problem 23==<br />
The graph of <math>y=f(x)</math>, where <math>f(x)</math> is a polynomial of degree <math>3</math>, contains points <math>A(2,4)</math>, <math>B(3,9)</math>, and <math>C(4,16)</math>. Lines <math>AB</math>, <math>AC</math>, and <math>BC</math> intersect the graph again at points <math>D</math>, <math>E</math>, and <math>F</math>, respectively, and the sum of the <math>x</math>-coordinates of <math>D</math>, <math>E</math>, and <math>F</math> is 24. What is <math>f(0)</math>?<br />
<br />
<math>\textbf{(A)}\quad {-2} \qquad \qquad \textbf{(B)}\quad 0 \qquad\qquad \textbf{(C)}\quad 2 \qquad\qquad \textbf{(D)}\quad \dfrac{24}5 \qquad\qquad\textbf{(E)}\quad 8</math><br />
<br />
[[2017 AMC 12B Problems/Problem 23|Solution]]<br />
<br />
==Problem 24==<br />
<br />
Quadrilateral <math>ABCD</math> has right angles at <math>B</math> and <math>C</math>, <math>\triangle ABC \sim \triangle BCD</math>, and <math>AB > BC</math>. There is a point <math>E</math> in the interior of <math>ABCD</math> such that <math>\triangle ABC \sim \triangle CEB</math> and the area of <math>\triangle AED</math> is <math>17</math> times the area of <math>\triangle CEB</math>. What is <math>\frac{AB}{BC}</math>?<br />
<br />
<math>\textbf{(A)}\quad {1 + \sqrt{2}} \qquad \textbf{(B)}\quad{2 + \sqrt{2}} \qquad \textbf{(C)}\quad{\sqrt{17}} \qquad\textbf{(D)}\quad{2+\sqrt{5}} \qquad\textbf{(E)}\quad{1 + 2\sqrt{3}}</math><br />
<br />
[[2017 AMC 12B Problems/Problem 24|Solution]]<br />
<br />
==Problem 25==<br />
A set of <math>n</math> people participate in an online video basketball tournament. Each person may be a member of any number of <math>5</math>-player teams, but no teams may have exactly the same <math>5</math> members. The site statistics show a curious fact: The average, over all subsets of size <math>9</math> of the set of <math>n</math> participants, of the number of complete teams whose members are among those 9 people is equal to the reciprocal of the average, over all subsets of size <math>8</math> of the set of <math>n</math> participants, of the number of complete teams whose members are among those <math>8</math> people. How many values <math>n</math>, <math>9 \leq n \leq 2017</math>, can be the number of participants?<br />
<br />
<math>\textbf{(A)}\quad {477} \qquad \textbf{(B)}\quad{482} \qquad \textbf{(C)}\quad{487} \qquad\textbf{(D)}\quad{557} \qquad\textbf{(E)}\quad{562}</math><br />
<br />
[[2017 AMC 12B Problems/Problem 25|Solution]]<br />
<br />
==See also==<br />
{{AMC12 box|year=2017|ab=B|before=[[2017 AMC 12A Problems]]|after=[[2018 AMC 12A Problems]]}}<br />
{{MAA Notice}}</div>Mathmaster2012https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_12B_Problems&diff=840112017 AMC 12B Problems2017-02-17T05:20:03Z<p>Mathmaster2012: </p>
<hr />
<div>WORK IN PROGRESS<br />
<br />
{{AMC12 Problems|year=2017|ab=B}}<br />
<br />
==Problem 1==<br />
<br />
Kymbrea's comic book collection currently has <math>30</math> comic books in it, and she is adding to her collection at the rate of <math>2</math> comic books per month. LaShawn's collection currently has <math>10</math> comic books in it, and he is adding to his collection at the rate of <math>6</math> comic books per month. After how many months will LaShawn's collection have twice as many comic books as Kymbrea's?<br />
<br />
<math>\textbf{(A)}\ 1\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 20\qquad\textbf{(E)}\ 25</math><br />
<br />
[[2017 AMC 10B Problems/Problem 2|Solution]]<br />
<br />
==Problem 2==<br />
<br />
Real numbers <math>x</math>, <math>y</math>, and <math>z</math> satify the inequalities<br />
<math>0<x<1</math>, <math>-1<y<0</math>, and <math>1<z<2</math>.<br />
Which of the following numbers is necessarily positive?<br />
<br />
<math>\textbf{(A)}\ y+x^2\qquad\textbf{(B)}\ y+xz\qquad\textbf{(C)}\ y+y^2\qquad\textbf{(D)}\ y+2y^2\qquad\textbf{(E)}\ y+z</math><br />
<br />
[[2017 AMC 10B Problems/Problem 3|Solution]]<br />
<br />
==Problem 3==<br />
<br />
Supposed that <math>x</math> and <math>y</math> are nonzero real numbers such that <math>\frac{3x+y}{x-3y}=-2</math>. What is the value of <math>\frac{x+3y}{3x-y}</math>?<br />
<br />
<math>\textbf{(A)}\ -3\qquad\textbf{(B)}\ -1\qquad\textbf{(C)}\ 1\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ 3</math><br />
<br />
[[2017 AMC 10B Problems/Problem 4|Solution]]<br />
<br />
==Problem 4==<br />
Samia set off on her bicycle to visit her friend, traveling at an average speed of <math>17</math> kilometers per hour. When she had gone half the distance to her friend's house, a tire went flat, and she walked the rest of the way at <math>5</math> kilometers per hour. In all it took her <math>44</math> minutes to reach her friend's house. In kilometers rounded to the nearest tenth, how far did Samia walk?<br />
<br />
<math>\textbf{(A)}\ 2.0\qquad\textbf{(B)}\ 2.2\qquad\textbf{(C)}\ 2.8\qquad\textbf{(D)}\ 3.4\qquad\textbf{(E)}\ 4.4</math><br />
<br />
[[2017 AMC 12B Problems/Problem 4|Solution]]<br />
<br />
==Problem 5==<br />
<br />
The data set <math>[6,19,33,33,39,41,41,43,51,57]</math> has median <math>Q_2 = 40</math>, first quartile <math>Q_1 = 33</math>, and third quartile <math>Q_3=43</math>. An outlier in a data set is a value that is more than <math>1.5</math> times the interquartile range below the first quartile <math>(Q_1)</math> or more than <math>1.5</math> times the interquartile range above the third quartile <math>(Q_3)</math>, where the interquartile range is defined as <math>Q_3 - Q_1</math>. How many outliers does this data set have?<br />
<br />
<math>\textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 4</math><br />
<br />
[[2017 AMC 12B Problems/Problem 5|Solution]]<br />
<br />
==Problem 6==<br />
The circle having <math>(0,0)</math> and <math>(8,6)</math> as the endpoints of a diameter intersects the <math>x</math>-axis at a second point. What is the <math>x</math>-coordinate of this point? <br />
<br />
<math>\textbf{(A)}\ 4\sqrt{2} \qquad \textbf{(B)}\ 6\qquad \textbf{(C)}\ 5\sqrt{2}\qquad \textbf{(D)}\ 8\qquad \textbf{(E)}\ 6\sqrt{2}</math><br />
<br />
[[2017 AMC 12B Problems/Problem 6|Solution]]<br />
<br />
==Problem 7==<br />
The functions <math>\sin(x)</math> and <math>\cos(x)</math> are periodic with least period <math>2\pi</math>. What is the least period of the function <math>\cos(\sin(x))</math>?<br />
<br />
<math>\textbf{(A)}\ \frac{\pi}{2}\qquad\textbf{(B)}\ \pi\qquad\textbf{(C)}\ 2\pi \qquad\textbf{(D)}\ 4\pi \qquad\textbf{(E)}</math> It's not periodic.<br />
<br />
[[2017 AMC 12B Problems/Problem 7|Solution]]<br />
<br />
==Problem 8==<br />
The ratio of the short side of a certain rectangle to the long side is equal to the ratio of the long side tho the diagonal. What is the square of the ratio of the short side to the long side of this rectangle?<br />
<br />
<math>\textbf{(A)}\ \frac{\sqrt{3}-1}{2}\qquad\textbf{(B)}\ \frac{1}{2}\qquad\textbf{(C)}\ \frac{\sqrt{5}-1}{2} \qquad\textbf{(D)}\ \frac{\sqrt{2}}{2} \qquad\textbf{(E)}\ \frac{\sqrt{6}-1}{2}</math><br />
<br />
[[2017 AMC 12B Problems/Problem 8|Solution]]<br />
<br />
==Problem 9==<br />
<br />
A circle has center <math>(-10,-4)</math> and radius <math>13</math>. Another circle has center <math>(3,9)</math> and radius <math>\sqrt{65}</math>. The line passing through the two points of intersection of the two circles has equation <math>x + y = c</math>. What is <math>c</math>?<br />
<br />
<math>\textbf{(A)}\ 3\qquad\textbf{(B)}\ 3\sqrt{3}\qquad\textbf{(C)}\ 4\sqrt{2}\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ \frac{13}{2}</math><br />
<br />
[[2017 AMC 12B Problems/Problem 9|Solution]]<br />
<br />
==Problem 10==<br />
At Typico High School, <math>60\%</math> of the students like dancing, and the rest dislike it. Of those who like dancing, <math>80\%</math> say that they like it, and the rest say that they dislike it. Of those who dislike dancing, <math>90\%</math> say that they dislike it, and the rest say that they like it. What fraction of students who say they dislike dancing actually like it?<br />
<br />
<math>\textbf{(A)}\ 10\%\qquad\textbf{(B)}\ 12\%\qquad\textbf{(C)}\ 20\%\qquad\textbf{(D)}\ 25\%\qquad\textbf{(E)}\ 33\frac{1}{3}\%</math><br />
<br />
[[2017 AMC 12B Problems/Problem 10|Solution]]<br />
<br />
==Problem 11==<br />
Call a positive integer <math>monotonous</math> if it is a one-digit number or its digits, when read from left to right, form either a strictly increasing or a strictly decreasing sequence. For example, <math>3</math>, <math>23578</math>, and <math>987620</math> are monotonous, but <math>88</math>, <math>7434</math>, and <math>23557</math> are not. How many monotonous positive integers are there?<br />
<br />
<math>\textbf{(A)}\ 1024\qquad\textbf{(B)}\ 1524\qquad\textbf{(C)}\ 1533\qquad\textbf{(D)}\ 1536\qquad\textbf{(E)}\ 2048</math><br />
<br />
[[2017 AMC 10B Problems/Problem 17|Solution]]<br />
<br />
==Problem 12==<br />
What is the sum of the roots of <math>z^{12}=64</math> that have a positive real part? <br />
<br />
<math>\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\ \sqrt{2}+2\sqrt{3} \qquad \textbf{(D)}\ 2\sqrt{2}+\sqrt{6} \qquad \textbf{(E)}\ (1+\sqrt{3}) + (1+\sqrt{3})i</math><br />
<br />
[[2017 AMC 12B Problems/Problem 12|Solution]]<br />
<br />
==Problem 13==<br />
<br />
In the figure below, <math>3</math> of the <math>6</math> disks are to be painted blue, <math>2</math> are to be painted red, and <math>1</math> is to be painted green. Two paintings that can be obtained from one another by a rotation or a reflection of the entire figure are considered the same. How many different paintings are possible?<br />
<br />
<asy><br />
size(100);<br />
pair A, B, C, D, E, F;<br />
A = (0,0);<br />
B = (1,0);<br />
C = (2,0);<br />
D = rotate(60, A)*B;<br />
E = B + D;<br />
F = rotate(60, A)*C;<br />
draw(Circle(A, 0.5));<br />
draw(Circle(B, 0.5));<br />
draw(Circle(C, 0.5));<br />
draw(Circle(D, 0.5));<br />
draw(Circle(E, 0.5));<br />
draw(Circle(F, 0.5));<br />
</asy><br />
<br />
<math>\textbf{(A) } 6 \qquad \textbf{(B) } 8 \qquad \textbf{(C) } 9 \qquad \textbf{(D) } 12 \qquad \textbf{(E) } 15</math><br />
<br />
==Problem 14==<br />
An ice-cream novelty item consists of a cup in the shape of a 4-inch-tall frustum of a right circular cone, with a 2-inch-diameter base at the bottom and a 4-inch-diameter base at the top, packed solid with ice cream, together with a solid cone of ice cream of height 4 inches, whose base, at the bottom, is the top base of the frustum. What is the total volume of the ice cream, in cubic inches? <br />
<br />
<math>\textbf{(A)}\ 8\pi \qquad \textbf{(B)}\ \frac{28\pi}{3} \qquad \textbf{(C)}\ 12\pi \qquad \textbf{(D)}\ 14\pi \qquad \textbf{(E)}\ \frac{44\pi}{3}</math><br />
<br />
[[2017 AMC 12B Problems/Problem 14|Solution]]<br />
<br />
==Problem 15==<br />
Let <math>ABC</math> be an equilateral triangle. Extend side <math>\overline{AB}</math> beyond <math>B</math> to a point <math>B'</math> so that <math>BB'=3AB</math>. Similarly, extend side <math>\overline{BC}</math> beyond <math>C</math> to a point <math>C'</math> so that <math>CC'=3BC</math>, and extend side <math>\overline{CA}</math> beyond <math>A</math> to a point <math>A'</math> so that <math>AA'=3CA</math>. What is the ratio of the area of <math>\triangle A'B'C'</math> to the area of <math>\triangle ABC</math>?<br />
<br />
<math>\textbf{(A)}\ 9:1\qquad\textbf{(B)}\ 16:1\qquad\textbf{(C)}\ 25:1\qquad\textbf{(D)}\ 36:1\qquad\textbf{(E)}\ 37:1</math><br />
<br />
[[2017 AMC 12B Problems/Problem 15|Solution]]<br />
<br />
==Problem 16==<br />
The number <math>21!=51,090,942,171,709,440,000</math> has over <math>60,000</math> positive integer divisors. One of them is chosen at random. What is the probability that it is odd?<br />
<br />
<math>\textbf{(A)}\ \frac{1}{21} \qquad \textbf{(B)}\ \frac{1}{19} \qquad \textbf{(C)}\ \frac{1}{18} \qquad \textbf{(D)}\ \frac{1}{2} \qquad \textbf{(E)}\ \frac{11}{21}</math><br />
<br />
[[2017 AMC 12B Problems/Problem 16|Solution]]<br />
<br />
==Problem 17==<br />
A coin is biased in such a way that on each toss the probability of heads is <math>\frac{2}{3}</math> and the probability of tails is <math>\frac{1}{3}</math>. The outcomes of the tosses are independent. A player has the choice of playing Game A or Game B. In Game A she tosses the coin three times and wins if all three outcomes are the same. In Game B she tosses the coin four times and wins if both the outcomes of the first and second tosses are the same and the outcomes of the third and fourth tosses are the same. How do the chances of winning Game A compare to the chances of winning Game B?<br />
<br />
<math>\textbf{(A)}</math> The probability of winning Game A is <math>\frac{4}{81}</math> less than the probability of winning Game B.<br />
<br />
<math>\textbf{(B)}</math> The probability of winning Game A is <math>\frac{2}{81}</math> less than the probability of winning Game B.<br />
<br />
<math>\textbf{(C)}</math> The probabilities are the same.<br />
<br />
<math>\textbf{(D)}</math> The probability of winning Game A is <math>\frac{2}{81}</math> greater than the probability of winning Game B.<br />
<br />
<math>\textbf{(E)}</math> The probability of winning Game A is <math>\frac{4}{81}</math> greater than the probability of winning Game B.<br />
<br />
[[2017 AMC 12B Problems/Problem 17|Solution]]<br />
<br />
==Problem 18==<br />
The diameter <math>AB</math> of a circle of radius <math>2</math> is extended to a point <math>D</math> outside the circle so that <math>BD=3</math>. Point <math>E</math> is chosen so that <math>ED=5</math> and line <math>ED</math> is perpendicular to line <math>AD</math>. Segment <math>AE</math> intersects the circle at a point <math>C</math> between <math>A</math> and <math>E</math>. What is the area of <math>\triangle <br />
ABC</math>?<br />
<br />
<math>\textbf{(A)}\ \frac{120}{37}\qquad\textbf{(B)}\ \frac{140}{39}\qquad\textbf{(C)}\ \frac{145}{39}\qquad\textbf{(D)}\ \frac{140}{37}\qquad\textbf{(E)}\ \frac{120}{31}</math><br />
<br />
[[2017 AMC 12B Problems/Problem 18|Solution]]<br />
<br />
==Problem 19==<br />
Let <math>N=123456789101112\dots4344</math> be the <math>79</math>-digit number that is formed by writing the integers from <math>1</math> to <math>44</math> in order, one after the other. What is the remainder when <math>N</math> is divided by <math>45</math>?<br />
<br />
<math>\textbf{(A)}\ 1\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 18\qquad\textbf{(E)}\ 44</math><br />
<br />
[[2017 AMC 12B Problems/Problem 19|Solution]]<br />
<br />
==Problem 20==<br />
Real numbers <math>x</math> and <math>y</math> are chosen independently and uniformly at random from the interval <math>(0,1)</math>. What is the probability that <math>\lfloor\log_2x\rfloor=\lfloor\log_2y\rfloor</math>, where <math>\lfloor r\rfloor</math> denotes the greatest integer less than or equal to the real number <math>r</math> ?<br />
<br />
<math>\textbf{(A)}\ \frac{1}{8}\qquad\textbf{(B)}\ \frac{1}{6}\qquad\textbf{(C)}\ \frac{1}{4}\qquad\textbf{(D)}\ \frac{1}{3}\qquad\textbf{(E)}\ \frac{1}{2}</math><br />
<br />
[[2017 AMC 12B Problems/Problem 20|Solution]]<br />
<br />
==Problem 21==<br />
Last year Isabella took <math>7</math> math tests and received <math>7</math> different scores, each an integer between <math>91</math> and <math>100</math>, inclusive. After each test she noticed that the average of her test scores was an integer. Her score on the seventh test was <math>95</math>. What was her score on the sixth test?<br />
<br />
<math>\textbf{(A)}\ 92\qquad\textbf{(B)}\ 94\qquad\textbf{(C)}\ 96\qquad\textbf{(D)}\ 98\qquad\textbf{(E)}\ 100</math><br />
<br />
[[2017 AMC 12B Problems/Problem 21|Solution]]<br />
<br />
==Problem 22==<br />
Abby, Bernardo, Carl, and Debra play a game in which each of them starts with four coins. The game consists of four rounds. In each round, four balls are placed in an urn---one green, one red, and two white. The players each draw a ball at random without replacement. Whoever gets the green ball gives one coin to whoever gets the red ball. What is the probability that, at the end of the fourth round, each of the players has four coins?<br />
<br />
<math>\textbf{(A)}\quad \dfrac{7}{576} \qquad \qquad \textbf{(B)}\quad \dfrac{5}{192} \qquad\qquad \textbf{(C)}\quad \dfrac{1}{36} \qquad\qquad \textbf{(D)}\quad \dfrac{5}{144} \qquad\qquad\textbf{(E)}\quad \dfrac{7}{48}</math><br />
<br />
==Problem 23==<br />
The graph of <math>y=f(x)</math>, where <math>f(x)</math> is a polynomial of degree <math>3</math>, contains points <math>A(2,4)</math>, <math>B(3,9)</math>, and <math>C(4,16)</math>. Lines <math>AB</math>, <math>AC</math>, and <math>BC</math> intersect the graph again at points <math>D</math>, <math>E</math>, and <math>F</math>, respectively, and the sum of the <math>x</math>-coordinates of <math>D</math>, <math>E</math>, and <math>F</math> is 24. What is <math>f(0)</math>?<br />
<br />
<math>\textbf{(A)}\quad {-2} \qquad \qquad \textbf{(B)}\quad 0 \qquad\qquad \textbf{(C)}\quad 2 \qquad\qquad \textbf{(D)}\quad \dfrac{24}5 \qquad\qquad\textbf{(E)}\quad 8</math><br />
<br />
[[2017 AMC 12B Problems/Problem 23|Solution]]<br />
<br />
==Problem 24==<br />
<br />
Quadrilateral <math>ABCD</math> has right angles at <math>B</math> and <math>C</math>, <math>\triangle ABC \sim \triangle BCD</math>, and <math>AB > BC</math>. There is a point <math>E</math> in the interior of <math>ABCD</math> such that <math>\triangle ABC \sim \triangle CEB</math> and the area of <math>\triangle AED</math> is <math>17</math> times the area of <math>\triangle CEB</math>. What is <math>\frac{AB}{BC}</math>?<br />
<br />
<math>\textbf{(A)}\quad {1 + \sqrt{2}} \qquad \textbf{(B)}\quad{2 + \sqrt{2}} \qquad \textbf{(C)}\quad{\sqrt{17}} \qquad\textbf{(D)}\quad{2+\sqrt{5}} \qquad\textbf{(E)}\quad{1 + 2\sqrt{3}}</math><br />
<br />
[[2017 AMC 12B Problems/Problem 24|Solution]]<br />
<br />
==Problem 25==<br />
A set of <math>n</math> people participate in an online video basketball tournament. Each person may be a member of any number of <math>5</math>-player teams, but no teams may have exactly the same <math>5</math> members. The site statistics show a curious fact: The average, over all subsets of size <math>9</math> of the set of <math>n</math> participants, of the number of complete teams whose members are among those 9 people is equal to the reciprocal of the average, over all subsets of size <math>8</math> of the set of <math>n</math> participants, of the number of complete teams whose members are among those <math>8</math> people. How many values <math>n</math>, <math>9 \leq n \leq 2017</math>, can be the number of participants?<br />
<br />
<math>\textbf{(A)}\quad {477} \qquad \textbf{(B)}\quad{482} \qquad \textbf{(C)}\quad{487} \qquad\textbf{(D)}\quad{557} \qquad\textbf{(E)}\quad{562}</math><br />
<br />
[[2017 AMC 12B Problems/Problem 25|Solution]]<br />
<br />
==See also==<br />
{{AMC12 box|year=2017|ab=B|before=[[2017 AMC 12A Problems]]|after=[[2018 AMC 12A Problems]]}}<br />
{{MAA Notice}}</div>Mathmaster2012https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_12B_Problems&diff=840102017 AMC 12B Problems2017-02-17T05:15:01Z<p>Mathmaster2012: /* Problem 7 */</p>
<hr />
<div>WORK IN PROGRESS<br />
<br />
{{AMC12 Problems|year=2017|ab=B}}<br />
<br />
==Problem 1==<br />
<br />
Kymbrea's comic book collection currently has <math>30</math> comic books in it, and she is adding to her collection at the rate of <math>2</math> comic books per month. LaShawn's collection currently has <math>10</math> comic books in it, and he is adding to his collection at the rate of <math>6</math> comic books per month. After how many months will LaShawn's collection have twice as many comic books as Kymbrea's?<br />
<br />
<math>\textbf{(A)}\ 1\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 20\qquad\textbf{(E)}\ 25</math><br />
<br />
[[2017 AMC 10B Problems/Problem 2|Solution]]<br />
<br />
==Problem 2==<br />
<br />
Real numbers <math>x</math>, <math>y</math>, and <math>z</math> satify the inequalities<br />
<math>0<x<1</math>, <math>-1<y<0</math>, and <math>1<z<2</math>.<br />
Which of the following numbers is necessarily positive?<br />
<br />
<math>\textbf{(A)}\ y+x^2\qquad\textbf{(B)}\ y+xz\qquad\textbf{(C)}\ y+y^2\qquad\textbf{(D)}\ y+2y^2\qquad\textbf{(E)}\ y+z</math><br />
<br />
[[2017 AMC 10B Problems/Problem 3|Solution]]<br />
<br />
==Problem 3==<br />
<br />
Supposed that <math>x</math> and <math>y</math> are nonzero real numbers such that <math>\frac{3x+y}{x-3y}=-2</math>. What is the value of <math>\frac{x+3y}{3x-y}</math>?<br />
<br />
<math>\textbf{(A)}\ -3\qquad\textbf{(B)}\ -1\qquad\textbf{(C)}\ 1\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ 3</math><br />
<br />
[[2017 AMC 10B Problems/Problem 4|Solution]]<br />
<br />
==Problem 4==<br />
Samia set off on her bicycle to visit her friend, traveling at an average speed of <math>17</math> kilometers per hour. When she had gone half the distance to her friend's house, a tire went flat, and she walked the rest of the way at <math>5</math> kilometers per hour. In all it took her <math>44</math> minutes to reach her friend's house. In kilometers rounded to the nearest tenth, how far did Samia walk?<br />
<br />
<math>\textbf{(A)}\ 2.0\qquad\textbf{(B)}\ 2.2\qquad\textbf{(C)}\ 2.8\qquad\textbf{(D)}\ 3.4\qquad\textbf{(E)}\ 4.4</math><br />
<br />
[[2017 AMC 12B Problems/Problem 4|Solution]]<br />
<br />
==Problem 5==<br />
<br />
The data set <math>[6,19,33,33,39,41,41,43,51,57]</math> has median <math>Q_2 = 40</math>, first quartile <math>Q_1 = 33</math>, and third quartile <math>Q_3=43</math>. An outlier in a data set is a value that is more than <math>1.5</math> times the interquartile range below the first quartile <math>(Q_1)</math> or more than <math>1.5</math> times the interquartile range above the third quartile <math>(Q_3)</math>, where the interquartile range is defined as <math>Q_3 - Q_1</math>. How many outliers does this data set have?<br />
<br />
<math>\textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 4</math><br />
<br />
[[2017 AMC 12B Problems/Problem 5|Solution]]<br />
<br />
==Problem 6==<br />
The circle having <math>(0,0)</math> and <math>(8,6)</math> as the endpoints of a diameter intersects the <math>x</math>-axis at a second point. What is the <math>x</math>-coordinate of this point? <br />
<br />
<math>\textbf{(A)}\ 4\sqrt{2} \qquad \textbf{(B)}\ 6\qquad \textbf{(C)}\ 5\sqrt{2}\qquad \textbf{(D)}\ 8\qquad \textbf{(E)}\ 6\sqrt{2}</math><br />
<br />
[[2017 AMC 12B Problems/Problem 6|Solution]]<br />
<br />
==Problem 7==<br />
The functions <math>\sin(x)</math> and <math>\cos(x)</math> are periodic with least period <math>2\pi</math>. What is the least period of the function <math>\cos(\sin(x))</math>?<br />
<br />
<math>\textbf{(A)}\ \frac{\pi}{2}\qquad\textbf{(B)}\ \pi\qquad\textbf{(C)}\ 2\pi \qquad\textbf{(D)}\ 4\pi \qquad\textbf{(E)}</math> It's not periodic.<br />
<br />
[[2017 AMC 12B Problems/Problem 7|Solution]]<br />
<br />
==Problem 8==<br />
The ratio of the short side of a certain rectangle to the long side is equal to the ratio of the long side tho the diagonal. What is the square of the ratio of the short side to the long side of this rectangle?<br />
<br />
<math>\textbf{(A)}\ \frac{\sqrt{3}-1}{2}\qquad\textbf{(B)}\ \frac{1}{2}\qquad\textbf{(C)}\ \frac{\sqrt{5}-1}{2} \qquad\textbf{(D)}\ \frac{\sqrt{2}}{2} \qquad\textbf{(E)}\ \frac{\sqrt{6}-1}{2}</math><br />
<br />
[[2017 AMC 12B Problems/Problem 8|Solution]]<br />
<br />
==Problem 9==<br />
<br />
A circle has center <math>(-10,-4)</math> and radius <math>13</math>. Another circle has center <math>(3,9)</math> and radius <math>\sqrt{65}</math>. The line passing through the two points of intersection of the two circles has equation <math>x + y = c</math>. What is <math>c</math>?<br />
<br />
<math>\textbf{(A)}\ 3\qquad\textbf{(B)}\ 3\sqrt{3}\qquad\textbf{(C)}\ 4\sqrt{2}\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ \frac{13}{2}</math><br />
<br />
[[2017 AMC 12B Problems/Problem 9|Solution]]<br />
<br />
==Problem 10==<br />
At Typico High School, <math>60\%</math> of the students like dancing, and the rest dislike it. Of those who like dancing, <math>80\%</math> say that they like it, and the rest say that they dislike it. Of those who dislike dancing, <math>90\%</math> say that they dislike it, and the rest say that they like it. What fraction of students who say they dislike dancing actually like it?<br />
<br />
<math>\textbf{(A)}\ 10\%\qquad\textbf{(B)}\ 12\%\qquad\textbf{(C)}\ 20\%\qquad\textbf{(D)}\ 25\%\qquad\textbf{(E)}\ 33\frac{1}{3}\%</math><br />
<br />
[[2017 AMC 12B Problems/Problem 10|Solution]]<br />
<br />
==Problem 11==<br />
Call a positive integer <math>monotonous</math> if it is a one-digit number or its digits, when read from left to right, form either a strictly increasing or a strictly decreasing sequence. For example, <math>3</math>, <math>23578</math>, and <math>987620</math> are monotonous, but <math>88</math>, <math>7434</math>, and <math>23557</math> are not. How many monotonous positive integers are there?<br />
<br />
<math>\textbf{(A)}\ 1024\qquad\textbf{(B)}\ 1524\qquad\textbf{(C)}\ 1533\qquad\textbf{(D)}\ 1536\qquad\textbf{(E)}\ 2048</math><br />
<br />
[[2017 AMC 10B Problems/Problem 17|Solution]]<br />
<br />
==Problem 12==<br />
What is the sum of the roots of <math>z^{12}=64</math> that have a positive real part? <br />
<br />
<math>\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\ \sqrt{2}+2\sqrt{3} \qquad \textbf{(D)}\ 2\sqrt{2}+\sqrt{6} \qquad \textbf{(E)}\ (1+\sqrt{3}) + (1+\sqrt{3})i</math><br />
<br />
[[2017 AMC 12B Problems/Problem 12|Solution]]<br />
<br />
==Problem 13==<br />
<br />
==Problem 14==<br />
An ice-cream novelty item consists of a cup in the shape of a 4-inch-tall frustum of a right circular cone, with a 2-inch-diameter base at the bottom and a 4-inch-diameter base at the top, packed solid with ice cream, together with a solid cone of ice cream of height 4 inches, whose base, at the bottom, is the top base of the frustum. What is the total volume of the ice cream, in cubic inches? <br />
<br />
<math>\textbf{(A)}\ 8\pi \qquad \textbf{(B)}\ \frac{28\pi}{3} \qquad \textbf{(C)}\ 12\pi \qquad \textbf{(D)}\ 14\pi \qquad \textbf{(E)}\ \frac{44\pi}{3}</math><br />
<br />
[[2017 AMC 12B Problems/Problem 14|Solution]]<br />
<br />
==Problem 15==<br />
Let <math>ABC</math> be an equilateral triangle. Extend side <math>\overline{AB}</math> beyond <math>B</math> to a point <math>B'</math> so that <math>BB'=3AB</math>. Similarly, extend side <math>\overline{BC}</math> beyond <math>C</math> to a point <math>C'</math> so that <math>CC'=3BC</math>, and extend side <math>\overline{CA}</math> beyond <math>A</math> to a point <math>A'</math> so that <math>AA'=3CA</math>. What is the ratio of the area of <math>\triangle A'B'C'</math> to the area of <math>\triangle ABC</math>?<br />
<br />
<math>\textbf{(A)}\ 9:1\qquad\textbf{(B)}\ 16:1\qquad\textbf{(C)}\ 25:1\qquad\textbf{(D)}\ 36:1\qquad\textbf{(E)}\ 37:1</math><br />
<br />
[[2017 AMC 12B Problems/Problem 15|Solution]]<br />
<br />
==Problem 16==<br />
The number <math>21!=51,090,942,171,709,440,000</math> has over <math>60,000</math> positive integer divisors. One of them is chosen at random. What is the probability that it is odd?<br />
<br />
<math>\textbf{(A)}\ \frac{1}{21} \qquad \textbf{(B)}\ \frac{1}{19} \qquad \textbf{(C)}\ \frac{1}{18} \qquad \textbf{(D)}\ \frac{1}{2} \qquad \textbf{(E)}\ \frac{11}{21}</math><br />
<br />
[[2017 AMC 12B Problems/Problem 16|Solution]]<br />
<br />
==Problem 17==<br />
A coin is biased in such a way that on each toss the probability of heads is <math>\frac{2}{3}</math> and the probability of tails is <math>\frac{1}{3}</math>. The outcomes of the tosses are independent. A player has the choice of playing Game A or Game B. In Game A she tosses the coin three times and wins if all three outcomes are the same. In Game B she tosses the coin four times and wins if both the outcomes of the first and second tosses are the same and the outcomes of the third and fourth tosses are the same. How do the chances of winning Game A compare to the chances of winning Game B?<br />
<br />
<math>\textbf{(A)}</math> The probability of winning Game A is <math>\frac{4}{81}</math> less than the probability of winning Game B.<br />
<br />
<math>\textbf{(B)}</math> The probability of winning Game A is <math>\frac{2}{81}</math> less than the probability of winning Game B.<br />
<br />
<math>\textbf{(C)}</math> The probabilities are the same.<br />
<br />
<math>\textbf{(D)}</math> The probability of winning Game A is <math>\frac{2}{81}</math> greater than the probability of winning Game B.<br />
<br />
<math>\textbf{(E)}</math> The probability of winning Game A is <math>\frac{4}{81}</math> greater than the probability of winning Game B.<br />
<br />
[[2017 AMC 12B Problems/Problem 17|Solution]]<br />
<br />
==Problem 18==<br />
The diameter <math>AB</math> of a circle of radius <math>2</math> is extended to a point <math>D</math> outside the circle so that <math>BD=3</math>. Point <math>E</math> is chosen so that <math>ED=5</math> and line <math>ED</math> is perpendicular to line <math>AD</math>. Segment <math>AE</math> intersects the circle at a point <math>C</math> between <math>A</math> and <math>E</math>. What is the area of <math>\triangle <br />
ABC</math>?<br />
<br />
<math>\textbf{(A)}\ \frac{120}{37}\qquad\textbf{(B)}\ \frac{140}{39}\qquad\textbf{(C)}\ \frac{145}{39}\qquad\textbf{(D)}\ \frac{140}{37}\qquad\textbf{(E)}\ \frac{120}{31}</math><br />
<br />
[[2017 AMC 12B Problems/Problem 18|Solution]]<br />
<br />
==Problem 19==<br />
Let <math>N=123456789101112\dots4344</math> be the <math>79</math>-digit number that is formed by writing the integers from <math>1</math> to <math>44</math> in order, one after the other. What is the remainder when <math>N</math> is divided by <math>45</math>?<br />
<br />
<math>\textbf{(A)}\ 1\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 18\qquad\textbf{(E)}\ 44</math><br />
<br />
[[2017 AMC 12B Problems/Problem 19|Solution]]<br />
<br />
==Problem 20==<br />
Real numbers <math>x</math> and <math>y</math> are chosen independently and uniformly at random from the interval <math>(0,1)</math>. What is the probability that <math>\lfloor\log_2x\rfloor=\lfloor\log_2y\rfloor</math>, where <math>\lfloor r\rfloor</math> denotes the greatest integer less than or equal to the real number <math>r</math> ?<br />
<br />
<math>\textbf{(A)}\ \frac{1}{8}\qquad\textbf{(B)}\ \frac{1}{6}\qquad\textbf{(C)}\ \frac{1}{4}\qquad\textbf{(D)}\ \frac{1}{3}\qquad\textbf{(E)}\ \frac{1}{2}</math><br />
<br />
[[2017 AMC 12B Problems/Problem 20|Solution]]<br />
<br />
==Problem 21==<br />
Last year Isabella took <math>7</math> math tests and received <math>7</math> different scores, each an integer between <math>91</math> and <math>100</math>, inclusive. After each test she noticed that the average of her test scores was an integer. Her score on the seventh test was <math>95</math>. What was her score on the sixth test?<br />
<br />
<math>\textbf{(A)}\ 92\qquad\textbf{(B)}\ 94\qquad\textbf{(C)}\ 96\qquad\textbf{(D)}\ 98\qquad\textbf{(E)}\ 100</math><br />
<br />
[[2017 AMC 12B Problems/Problem 21|Solution]]<br />
<br />
==Problem 22==<br />
Abby, Bernardo, Carl, and Debra play a game in which each of them starts with four coins. The game consists of four rounds. In each round, four balls are placed in an urn---one green, one red, and two white. The players each draw a ball at random without replacement. Whoever gets the green ball gives one coin to whoever gets the red ball. What is the probability that, at the end of the fourth round, each of the players has four coins?<br />
<br />
<math>\textbf{(A)}\quad \dfrac{7}{576} \qquad \qquad \textbf{(B)}\quad \dfrac{5}{192} \qquad\qquad \textbf{(C)}\quad \dfrac{1}{36} \qquad\qquad \textbf{(D)}\quad \dfrac{5}{144} \qquad\qquad\textbf{(E)}\quad \dfrac{7}{48}</math><br />
<br />
==Problem 23==<br />
The graph of <math>y=f(x)</math>, where <math>f(x)</math> is a polynomial of degree <math>3</math>, contains points <math>A(2,4)</math>, <math>B(3,9)</math>, and <math>C(4,16)</math>. Lines <math>AB</math>, <math>AC</math>, and <math>BC</math> intersect the graph again at points <math>D</math>, <math>E</math>, and <math>F</math>, respectively, and the sum of the <math>x</math>-coordinates of <math>D</math>, <math>E</math>, and <math>F</math> is 24. What is <math>f(0)</math>?<br />
<br />
<math>\textbf{(A)}\quad {-2} \qquad \qquad \textbf{(B)}\quad 0 \qquad\qquad \textbf{(C)}\quad 2 \qquad\qquad \textbf{(D)}\quad \dfrac{24}5 \qquad\qquad\textbf{(E)}\quad 8</math><br />
<br />
[[2017 AMC 12B Problems/Problem 23|Solution]]<br />
<br />
==Problem 24==<br />
<br />
Quadrilateral <math>ABCD</math> has right angles at <math>B</math> and <math>C</math>, <math>\triangle ABC \sim \triangle BCD</math>, and <math>AB > BC</math>. There is a point <math>E</math> in the interior of <math>ABCD</math> such that <math>\triangle ABC \sim \triangle CEB</math> and the area of <math>\triangle AED</math> is <math>17</math> times the area of <math>\triangle CEB</math>. What is <math>\frac{AB}{BC}</math>?<br />
<br />
<math>\textbf{(A)}\quad {1 + \sqrt{2}} \qquad \textbf{(B)}\quad{2 + \sqrt{2}} \qquad \textbf{(C)}\quad{\sqrt{17}} \qquad\textbf{(D)}\quad{2+\sqrt{5}} \qquad\textbf{(E)}\quad{1 + 2\sqrt{3}}</math><br />
<br />
[[2017 AMC 12B Problems/Problem 24|Solution]]<br />
<br />
==Problem 25==<br />
A set of <math>n</math> people participate in an online video basketball tournament. Each person may be a member of any number of <math>5</math>-player teams, but no teams may have exactly the same <math>5</math> members. The site statistics show a curious fact: The average, over all subsets of size <math>9</math> of the set of <math>n</math> participants, of the number of complete teams whose members are among those 9 people is equal to the reciprocal of the average, over all subsets of size <math>8</math> of the set of <math>n</math> participants, of the number of complete teams whose members are among those <math>8</math> people. How many values <math>n</math>, <math>9 \leq n \leq 2017</math>, can be the number of participants?<br />
<br />
<math>\textbf{(A)}\quad {477} \qquad \textbf{(B)}\quad{482} \qquad \textbf{(C)}\quad{487} \qquad\textbf{(D)}\quad{557} \qquad\textbf{(E)}\quad{562}</math><br />
<br />
[[2017 AMC 12B Problems/Problem 25|Solution]]<br />
<br />
==See also==<br />
{{AMC12 box|year=2017|ab=B|before=[[2017 AMC 12A Problems]]|after=[[2018 AMC 12A Problems]]}}<br />
{{MAA Notice}}</div>Mathmaster2012https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_12B_Problems&diff=840092017 AMC 12B Problems2017-02-17T05:13:08Z<p>Mathmaster2012: /* Problem 5 */</p>
<hr />
<div>WORK IN PROGRESS<br />
<br />
{{AMC12 Problems|year=2017|ab=B}}<br />
<br />
==Problem 1==<br />
<br />
Kymbrea's comic book collection currently has <math>30</math> comic books in it, and she is adding to her collection at the rate of <math>2</math> comic books per month. LaShawn's collection currently has <math>10</math> comic books in it, and he is adding to his collection at the rate of <math>6</math> comic books per month. After how many months will LaShawn's collection have twice as many comic books as Kymbrea's?<br />
<br />
<math>\textbf{(A)}\ 1\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 20\qquad\textbf{(E)}\ 25</math><br />
<br />
[[2017 AMC 10B Problems/Problem 2|Solution]]<br />
<br />
==Problem 2==<br />
<br />
Real numbers <math>x</math>, <math>y</math>, and <math>z</math> satify the inequalities<br />
<math>0<x<1</math>, <math>-1<y<0</math>, and <math>1<z<2</math>.<br />
Which of the following numbers is necessarily positive?<br />
<br />
<math>\textbf{(A)}\ y+x^2\qquad\textbf{(B)}\ y+xz\qquad\textbf{(C)}\ y+y^2\qquad\textbf{(D)}\ y+2y^2\qquad\textbf{(E)}\ y+z</math><br />
<br />
[[2017 AMC 10B Problems/Problem 3|Solution]]<br />
<br />
==Problem 3==<br />
<br />
Supposed that <math>x</math> and <math>y</math> are nonzero real numbers such that <math>\frac{3x+y}{x-3y}=-2</math>. What is the value of <math>\frac{x+3y}{3x-y}</math>?<br />
<br />
<math>\textbf{(A)}\ -3\qquad\textbf{(B)}\ -1\qquad\textbf{(C)}\ 1\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ 3</math><br />
<br />
[[2017 AMC 10B Problems/Problem 4|Solution]]<br />
<br />
==Problem 4==<br />
Samia set off on her bicycle to visit her friend, traveling at an average speed of <math>17</math> kilometers per hour. When she had gone half the distance to her friend's house, a tire went flat, and she walked the rest of the way at <math>5</math> kilometers per hour. In all it took her <math>44</math> minutes to reach her friend's house. In kilometers rounded to the nearest tenth, how far did Samia walk?<br />
<br />
<math>\textbf{(A)}\ 2.0\qquad\textbf{(B)}\ 2.2\qquad\textbf{(C)}\ 2.8\qquad\textbf{(D)}\ 3.4\qquad\textbf{(E)}\ 4.4</math><br />
<br />
[[2017 AMC 12B Problems/Problem 4|Solution]]<br />
<br />
==Problem 5==<br />
<br />
The data set <math>[6,19,33,33,39,41,41,43,51,57]</math> has median <math>Q_2 = 40</math>, first quartile <math>Q_1 = 33</math>, and third quartile <math>Q_3=43</math>. An outlier in a data set is a value that is more than <math>1.5</math> times the interquartile range below the first quartile <math>(Q_1)</math> or more than <math>1.5</math> times the interquartile range above the third quartile <math>(Q_3)</math>, where the interquartile range is defined as <math>Q_3 - Q_1</math>. How many outliers does this data set have?<br />
<br />
<math>\textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 4</math><br />
<br />
[[2017 AMC 12B Problems/Problem 5|Solution]]<br />
<br />
==Problem 6==<br />
The circle having <math>(0,0)</math> and <math>(8,6)</math> as the endpoints of a diameter intersects the <math>x</math>-axis at a second point. What is the <math>x</math>-coordinate of this point? <br />
<br />
<math>\textbf{(A)}\ 4\sqrt{2} \qquad \textbf{(B)}\ 6\qquad \textbf{(C)}\ 5\sqrt{2}\qquad \textbf{(D)}\ 8\qquad \textbf{(E)}\ 6\sqrt{2}</math><br />
<br />
[[2017 AMC 12B Problems/Problem 6|Solution]]<br />
<br />
==Problem 7==<br />
The functions <math>\sin(x)</math> and <math>\cos(x)</math> are periodic with least period <math>2\pi</math>. What is the least period of the function <math>\cos(\sin(x))</math>?<br />
<br />
<math>\textbf{(A)}\ \frac{\sqrt{3}-1}{2}\qquad\textbf{(B)}\ \frac{1}{2}\qquad\textbf{(C)}\ \frac{\sqrt{5}-1}{2} \qquad\textbf{(D)}\ \frac{\sqrt{2}}{2} \qquad\textbf{(E)}\ \frac{\sqrt{6}-1}{2}</math><br />
<br />
[[2017 AMC 12B Problems/Problem 7|Solution]]<br />
<br />
==Problem 8==<br />
The ratio of the short side of a certain rectangle to the long side is equal to the ratio of the long side tho the diagonal. What is the square of the ratio of the short side to the long side of this rectangle?<br />
<br />
<math>\textbf{(A)}\ \frac{\sqrt{3}-1}{2}\qquad\textbf{(B)}\ \frac{1}{2}\qquad\textbf{(C)}\ \frac{\sqrt{5}-1}{2} \qquad\textbf{(D)}\ \frac{\sqrt{2}}{2} \qquad\textbf{(E)}\ \frac{\sqrt{6}-1}{2}</math><br />
<br />
[[2017 AMC 12B Problems/Problem 8|Solution]]<br />
<br />
==Problem 9==<br />
<br />
A circle has center <math>(-10,-4)</math> and radius <math>13</math>. Another circle has center <math>(3,9)</math> and radius <math>\sqrt{65}</math>. The line passing through the two points of intersection of the two circles has equation <math>x + y = c</math>. What is <math>c</math>?<br />
<br />
<math>\textbf{(A)}\ 3\qquad\textbf{(B)}\ 3\sqrt{3}\qquad\textbf{(C)}\ 4\sqrt{2}\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ \frac{13}{2}</math><br />
<br />
[[2017 AMC 12B Problems/Problem 9|Solution]]<br />
<br />
==Problem 10==<br />
At Typico High School, <math>60\%</math> of the students like dancing, and the rest dislike it. Of those who like dancing, <math>80\%</math> say that they like it, and the rest say that they dislike it. Of those who dislike dancing, <math>90\%</math> say that they dislike it, and the rest say that they like it. What fraction of students who say they dislike dancing actually like it?<br />
<br />
<math>\textbf{(A)}\ 10\%\qquad\textbf{(B)}\ 12\%\qquad\textbf{(C)}\ 20\%\qquad\textbf{(D)}\ 25\%\qquad\textbf{(E)}\ 33\frac{1}{3}\%</math><br />
<br />
[[2017 AMC 12B Problems/Problem 10|Solution]]<br />
<br />
==Problem 11==<br />
Call a positive integer <math>monotonous</math> if it is a one-digit number or its digits, when read from left to right, form either a strictly increasing or a strictly decreasing sequence. For example, <math>3</math>, <math>23578</math>, and <math>987620</math> are monotonous, but <math>88</math>, <math>7434</math>, and <math>23557</math> are not. How many monotonous positive integers are there?<br />
<br />
<math>\textbf{(A)}\ 1024\qquad\textbf{(B)}\ 1524\qquad\textbf{(C)}\ 1533\qquad\textbf{(D)}\ 1536\qquad\textbf{(E)}\ 2048</math><br />
<br />
[[2017 AMC 10B Problems/Problem 17|Solution]]<br />
<br />
==Problem 12==<br />
What is the sum of the roots of <math>z^{12}=64</math> that have a positive real part? <br />
<br />
<math>\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\ \sqrt{2}+2\sqrt{3} \qquad \textbf{(D)}\ 2\sqrt{2}+\sqrt{6} \qquad \textbf{(E)}\ (1+\sqrt{3}) + (1+\sqrt{3})i</math><br />
<br />
[[2017 AMC 12B Problems/Problem 12|Solution]]<br />
<br />
==Problem 13==<br />
<br />
==Problem 14==<br />
An ice-cream novelty item consists of a cup in the shape of a 4-inch-tall frustum of a right circular cone, with a 2-inch-diameter base at the bottom and a 4-inch-diameter base at the top, packed solid with ice cream, together with a solid cone of ice cream of height 4 inches, whose base, at the bottom, is the top base of the frustum. What is the total volume of the ice cream, in cubic inches? <br />
<br />
<math>\textbf{(A)}\ 8\pi \qquad \textbf{(B)}\ \frac{28\pi}{3} \qquad \textbf{(C)}\ 12\pi \qquad \textbf{(D)}\ 14\pi \qquad \textbf{(E)}\ \frac{44\pi}{3}</math><br />
<br />
[[2017 AMC 12B Problems/Problem 14|Solution]]<br />
<br />
==Problem 15==<br />
Let <math>ABC</math> be an equilateral triangle. Extend side <math>\overline{AB}</math> beyond <math>B</math> to a point <math>B'</math> so that <math>BB'=3AB</math>. Similarly, extend side <math>\overline{BC}</math> beyond <math>C</math> to a point <math>C'</math> so that <math>CC'=3BC</math>, and extend side <math>\overline{CA}</math> beyond <math>A</math> to a point <math>A'</math> so that <math>AA'=3CA</math>. What is the ratio of the area of <math>\triangle A'B'C'</math> to the area of <math>\triangle ABC</math>?<br />
<br />
<math>\textbf{(A)}\ 9:1\qquad\textbf{(B)}\ 16:1\qquad\textbf{(C)}\ 25:1\qquad\textbf{(D)}\ 36:1\qquad\textbf{(E)}\ 37:1</math><br />
<br />
[[2017 AMC 12B Problems/Problem 15|Solution]]<br />
<br />
==Problem 16==<br />
The number <math>21!=51,090,942,171,709,440,000</math> has over <math>60,000</math> positive integer divisors. One of them is chosen at random. What is the probability that it is odd?<br />
<br />
<math>\textbf{(A)}\ \frac{1}{21} \qquad \textbf{(B)}\ \frac{1}{19} \qquad \textbf{(C)}\ \frac{1}{18} \qquad \textbf{(D)}\ \frac{1}{2} \qquad \textbf{(E)}\ \frac{11}{21}</math><br />
<br />
[[2017 AMC 12B Problems/Problem 16|Solution]]<br />
<br />
==Problem 17==<br />
A coin is biased in such a way that on each toss the probability of heads is <math>\frac{2}{3}</math> and the probability of tails is <math>\frac{1}{3}</math>. The outcomes of the tosses are independent. A player has the choice of playing Game A or Game B. In Game A she tosses the coin three times and wins if all three outcomes are the same. In Game B she tosses the coin four times and wins if both the outcomes of the first and second tosses are the same and the outcomes of the third and fourth tosses are the same. How do the chances of winning Game A compare to the chances of winning Game B?<br />
<br />
<math>\textbf{(A)}</math> The probability of winning Game A is <math>\frac{4}{81}</math> less than the probability of winning Game B.<br />
<br />
<math>\textbf{(B)}</math> The probability of winning Game A is <math>\frac{2}{81}</math> less than the probability of winning Game B.<br />
<br />
<math>\textbf{(C)}</math> The probabilities are the same.<br />
<br />
<math>\textbf{(D)}</math> The probability of winning Game A is <math>\frac{2}{81}</math> greater than the probability of winning Game B.<br />
<br />
<math>\textbf{(E)}</math> The probability of winning Game A is <math>\frac{4}{81}</math> greater than the probability of winning Game B.<br />
<br />
[[2017 AMC 12B Problems/Problem 17|Solution]]<br />
<br />
==Problem 18==<br />
The diameter <math>AB</math> of a circle of radius <math>2</math> is extended to a point <math>D</math> outside the circle so that <math>BD=3</math>. Point <math>E</math> is chosen so that <math>ED=5</math> and line <math>ED</math> is perpendicular to line <math>AD</math>. Segment <math>AE</math> intersects the circle at a point <math>C</math> between <math>A</math> and <math>E</math>. What is the area of <math>\triangle <br />
ABC</math>?<br />
<br />
<math>\textbf{(A)}\ \frac{120}{37}\qquad\textbf{(B)}\ \frac{140}{39}\qquad\textbf{(C)}\ \frac{145}{39}\qquad\textbf{(D)}\ \frac{140}{37}\qquad\textbf{(E)}\ \frac{120}{31}</math><br />
<br />
[[2017 AMC 12B Problems/Problem 18|Solution]]<br />
<br />
==Problem 19==<br />
Let <math>N=123456789101112\dots4344</math> be the <math>79</math>-digit number that is formed by writing the integers from <math>1</math> to <math>44</math> in order, one after the other. What is the remainder when <math>N</math> is divided by <math>45</math>?<br />
<br />
<math>\textbf{(A)}\ 1\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 18\qquad\textbf{(E)}\ 44</math><br />
<br />
[[2017 AMC 12B Problems/Problem 19|Solution]]<br />
<br />
==Problem 20==<br />
Real numbers <math>x</math> and <math>y</math> are chosen independently and uniformly at random from the interval <math>(0,1)</math>. What is the probability that <math>\lfloor\log_2x\rfloor=\lfloor\log_2y\rfloor</math>, where <math>\lfloor r\rfloor</math> denotes the greatest integer less than or equal to the real number <math>r</math> ?<br />
<br />
<math>\textbf{(A)}\ \frac{1}{8}\qquad\textbf{(B)}\ \frac{1}{6}\qquad\textbf{(C)}\ \frac{1}{4}\qquad\textbf{(D)}\ \frac{1}{3}\qquad\textbf{(E)}\ \frac{1}{2}</math><br />
<br />
[[2017 AMC 12B Problems/Problem 20|Solution]]<br />
<br />
==Problem 21==<br />
Last year Isabella took <math>7</math> math tests and received <math>7</math> different scores, each an integer between <math>91</math> and <math>100</math>, inclusive. After each test she noticed that the average of her test scores was an integer. Her score on the seventh test was <math>95</math>. What was her score on the sixth test?<br />
<br />
<math>\textbf{(A)}\ 92\qquad\textbf{(B)}\ 94\qquad\textbf{(C)}\ 96\qquad\textbf{(D)}\ 98\qquad\textbf{(E)}\ 100</math><br />
<br />
[[2017 AMC 12B Problems/Problem 21|Solution]]<br />
<br />
==Problem 22==<br />
Abby, Bernardo, Carl, and Debra play a game in which each of them starts with four coins. The game consists of four rounds. In each round, four balls are placed in an urn---one green, one red, and two white. The players each draw a ball at random without replacement. Whoever gets the green ball gives one coin to whoever gets the red ball. What is the probability that, at the end of the fourth round, each of the players has four coins?<br />
<br />
<math>\textbf{(A)}\quad \dfrac{7}{576} \qquad \qquad \textbf{(B)}\quad \dfrac{5}{192} \qquad\qquad \textbf{(C)}\quad \dfrac{1}{36} \qquad\qquad \textbf{(D)}\quad \dfrac{5}{144} \qquad\qquad\textbf{(E)}\quad \dfrac{7}{48}</math><br />
<br />
==Problem 23==<br />
The graph of <math>y=f(x)</math>, where <math>f(x)</math> is a polynomial of degree <math>3</math>, contains points <math>A(2,4)</math>, <math>B(3,9)</math>, and <math>C(4,16)</math>. Lines <math>AB</math>, <math>AC</math>, and <math>BC</math> intersect the graph again at points <math>D</math>, <math>E</math>, and <math>F</math>, respectively, and the sum of the <math>x</math>-coordinates of <math>D</math>, <math>E</math>, and <math>F</math> is 24. What is <math>f(0)</math>?<br />
<br />
<math>\textbf{(A)}\quad {-2} \qquad \qquad \textbf{(B)}\quad 0 \qquad\qquad \textbf{(C)}\quad 2 \qquad\qquad \textbf{(D)}\quad \dfrac{24}5 \qquad\qquad\textbf{(E)}\quad 8</math><br />
<br />
[[2017 AMC 12B Problems/Problem 23|Solution]]<br />
<br />
==Problem 24==<br />
<br />
Quadrilateral <math>ABCD</math> has right angles at <math>B</math> and <math>C</math>, <math>\triangle ABC \sim \triangle BCD</math>, and <math>AB > BC</math>. There is a point <math>E</math> in the interior of <math>ABCD</math> such that <math>\triangle ABC \sim \triangle CEB</math> and the area of <math>\triangle AED</math> is <math>17</math> times the area of <math>\triangle CEB</math>. What is <math>\frac{AB}{BC}</math>?<br />
<br />
<math>\textbf{(A)}\quad {1 + \sqrt{2}} \qquad \textbf{(B)}\quad{2 + \sqrt{2}} \qquad \textbf{(C)}\quad{\sqrt{17}} \qquad\textbf{(D)}\quad{2+\sqrt{5}} \qquad\textbf{(E)}\quad{1 + 2\sqrt{3}}</math><br />
<br />
[[2017 AMC 12B Problems/Problem 24|Solution]]<br />
<br />
==Problem 25==<br />
A set of <math>n</math> people participate in an online video basketball tournament. Each person may be a member of any number of <math>5</math>-player teams, but no teams may have exactly the same <math>5</math> members. The site statistics show a curious fact: The average, over all subsets of size <math>9</math> of the set of <math>n</math> participants, of the number of complete teams whose members are among those 9 people is equal to the reciprocal of the average, over all subsets of size <math>8</math> of the set of <math>n</math> participants, of the number of complete teams whose members are among those <math>8</math> people. How many values <math>n</math>, <math>9 \leq n \leq 2017</math>, can be the number of participants?<br />
<br />
<math>\textbf{(A)}\quad {477} \qquad \textbf{(B)}\quad{482} \qquad \textbf{(C)}\quad{487} \qquad\textbf{(D)}\quad{557} \qquad\textbf{(E)}\quad{562}</math><br />
<br />
[[2017 AMC 12B Problems/Problem 25|Solution]]<br />
<br />
==See also==<br />
{{AMC12 box|year=2017|ab=B|before=[[2017 AMC 12A Problems]]|after=[[2018 AMC 12A Problems]]}}<br />
{{MAA Notice}}</div>Mathmaster2012https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_12B_Problems/Problem_18&diff=840082017 AMC 12B Problems/Problem 182017-02-17T04:12:50Z<p>Mathmaster2012: </p>
<hr />
<div>==Problem==<br />
The diameter <math>AB</math> of a circle of radius <math>2</math> is extended to a point <math>D</math> outside the circle so that <math>BD=3</math>. Point <math>E</math> is chosen so that <math>ED=5</math> and line <math>ED</math> is perpendicular to line <math>AD</math>. Segment <math>AE</math> intersects the circle at a point <math>C</math> between <math>A</math> and <math>E</math>. What is the area of <math>\triangle ABC</math>?<br />
<br />
<math>\textbf{(A)}\ \frac{120}{37}\qquad\textbf{(B)}\ \frac{140}{39}\qquad\textbf{(C)}\ \frac{145}{39}\qquad\textbf{(D)}\ \frac{140}{37}\qquad\textbf{(E)}\ \frac{120}{31}</math><br />
<br />
==Solution==<br />
<br />
<asy><br />
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */<br />
import graph; size(8.865514650638614cm); <br />
real labelscalefactor = 0.5; /* changes label-to-point distance */<br />
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ <br />
pen dotstyle = black; /* point style */ <br />
real xmin = -6.36927122464312, xmax = 11.361758076634109, ymin = -3.789601803155515, ymax = 7.420015026296013; /* image dimensions */<br />
<br />
<br />
draw((-2.,0.)--(0.6486486486486486,1.8918918918918919)--(2.,0.)--cycle); <br />
/* draw figures */<br />
draw(circle((0.,0.), 2.)); <br />
draw((-2.,0.)--(5.,5.)); <br />
draw((5.,5.)--(5.,0.)); <br />
draw((5.,0.)--(-2.,0.)); <br />
draw((-2.,0.)--(0.6486486486486486,1.8918918918918919)); <br />
draw((0.6486486486486486,1.8918918918918919)--(2.,0.)); <br />
draw((2.,0.)--(-2.,0.)); <br />
draw((2.,0.)--(5.,5.)); <br />
draw((0.,0.)--(5.,5.)); <br />
/* dots and labels */<br />
dot((0.,0.),dotstyle); <br />
label("$O$", (-0.10330578512396349,-0.39365890308038826), NE * labelscalefactor); <br />
dot((-2.,0.),dotstyle); <br />
label("$A$", (-2.2370398196844437,-0.42371149511645134), NE * labelscalefactor); <br />
dot((2.,0.),dotstyle); <br />
label("$B$", (2.045454545454548,-0.36360631104432517), NE * labelscalefactor); <br />
dot((5.,0.),dotstyle); <br />
label("$D$", (4.900450788880542,-0.42371149511645134), NE * labelscalefactor); <br />
dot((5.,5.),dotstyle); <br />
label("$E$", (5.06574004507889,5.15104432757325), NE * labelscalefactor); <br />
dot((0.6486486486486486,1.8918918918918919),linewidth(3.pt) + dotstyle); <br />
label("$C$", (0.48271975957926694,2.100706235912847), NE * labelscalefactor); <br />
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); <br />
/* end of picture */<br />
</asy><br />
<br />
Let <math>O</math> be the center of the circle. Note that <math>EC + CA = EA = \sqrt{AD^2 + DE^2} = \sqrt{(2+2+3)^2 + 5^2} = \sqrt{74}</math>. However, by Power of a Point, <math>(EC)(EC + CA) = EO^2 - R^2 = (2+3)^2 + 5^2 - 2^2 = 25 + 25 - 4 = 46 \implies EC = \frac{46}{\sqrt{74}}</math>, so <math>AC = \sqrt{74} - \frac{46}{\sqrt{74}} = \frac{28}{\sqrt{74}}</math>. Now <math>BC = \sqrt{AB^2 - AC^2} = \sqrt{4^2 - \frac{28^2}{74}} = \sqrt{\frac{16 \cdot 74 - 28^2}{74}} = \sqrt{\frac{1184 - 784}{74}} = \frac{20}{\sqrt{74}}</math>. Since <math>\angle ACB = 90^{\circ}, [ABC] = \frac{1}{2} \cdot BC \cdot AC = \frac{1}{2} \cdot \frac{20}{\sqrt{74}} \cdot \frac{28}{\sqrt{74}} = \boxed{\textbf{(D)}\ \frac{140}{37}}</math>.<br />
<br />
==See Also==<br />
{{AMC12 box|year=2017|ab=B|num-b=17|num-a=19}}<br />
{{MAA Notice}}</div>Mathmaster2012https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_12B&diff=840072017 AMC 12B2017-02-17T04:12:28Z<p>Mathmaster2012: </p>
<hr />
<div>'''2017 AMC 12B''' problems and solutions. The test was held on February 15, 2017. <br />
<br />
*[[2017 AMC 12B Problems]]<br />
*[[2017 AMC 12B Answer Key]]<br />
**[[2017 AMC 12B Problems/Problem 1|Problem 1]]<br />
**[[2017 AMC 12B Problems/Problem 2|Problem 2]]<br />
**[[2017 AMC 12B Problems/Problem 3|Problem 3]]<br />
**[[2017 AMC 12B Problems/Problem 4|Problem 4]]<br />
**[[2017 AMC 12B Problems/Problem 5|Problem 5]]<br />
**[[2017 AMC 12B Problems/Problem 6|Problem 6]]<br />
**[[2017 AMC 12B Problems/Problem 7|Problem 7]]<br />
**[[2017 AMC 12B Problems/Problem 8|Problem 8]]<br />
**[[2017 AMC 12B Problems/Problem 9|Problem 9]]<br />
**[[2017 AMC 12B Problems/Problem 10|Problem 10]]<br />
**[[2017 AMC 12B Problems/Problem 11|Problem 11]]<br />
**[[2017 AMC 12B Problems/Problem 12|Problem 12]]<br />
**[[2017 AMC 12B Problems/Problem 13|Problem 13]]<br />
**[[2017 AMC 12B Problems/Problem 14|Problem 14]]<br />
**[[2017 AMC 12B Problems/Problem 15|Problem 15]]<br />
**[[2017 AMC 12B Problems/Problem 16|Problem 16]]<br />
**[[2017 AMC 12B Problems/Problem 17|Problem 17]]<br />
**[[2017 AMC 12B Problems/Problem 18|Problem 18]]<br />
**[[2017 AMC 12B Problems/Problem 19|Problem 19]]<br />
**[[2017 AMC 12B Problems/Problem 20|Problem 20]]<br />
**[[2017 AMC 12B Problems/Problem 21|Problem 21]]<br />
**[[2017 AMC 12B Problems/Problem 22|Problem 22]]<br />
**[[2017 AMC 12B Problems/Problem 23|Problem 23]]<br />
**[[2017 AMC 12B Problems/Problem 24|Problem 24]]<br />
**[[2017 AMC 12B Problems/Problem 25|Problem 25]]<br />
== See Also ==<br />
{{AMC12 box|year=2017|before=[[2016 AMC 12A|2016 AMC 12A]], [[2016 AMC 12B|B]]|after=[[2017 AMC 12A|2017 AMC 12A]], [[2017 AMC 12B|B]]|ab=B}}<br />
* [[Mathematics competitions]]<br />
* [[AHSME Problems and Solutions]]<br />
* [[Math books]]<br />
* [[Mathematics competition resources]]</div>Mathmaster2012https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_12B_Problems/Problem_18&diff=840062017 AMC 12B Problems/Problem 182017-02-17T03:48:27Z<p>Mathmaster2012: </p>
<hr />
<div>==Problem==<br />
The diameter <math>AB</math> of a circle of radius <math>2</math> is extended to a point <math>D</math> outside the circle so that <math>BD=3</math>. Point <math>E</math> is chosen so that <math>ED=5</math> and line <math>ED</math> is perpendicular to line <math>AD</math>. Segment <math>AE</math> intersects the circle at a point <math>C</math> between <math>A</math> and <math>E</math>. What is the area of <math>\triangle ABC</math>?<br />
<br />
<math>\textbf{(A)}\ \frac{120}{37}\qquad\textbf{(B)}\ \frac{140}{39}\qquad\textbf{(C)}\ \frac{145}{39}\qquad\textbf{(D)}\ \frac{140}{37}\qquad\textbf{(E)}\ \frac{120}{31}</math><br />
<br />
==Solution==<br />
<br />
[asy]<br />
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */<br />
import graph; size(8.865514650638614cm); <br />
real labelscalefactor = 0.5; /* changes label-to-point distance */<br />
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ <br />
pen dotstyle = black; /* point style */ <br />
real xmin = -6.36927122464312, xmax = 11.361758076634109, ymin = -3.789601803155515, ymax = 7.420015026296013; /* image dimensions */<br />
<br />
<br />
draw((-2.,0.)--(0.6486486486486486,1.8918918918918919)--(2.,0.)--cycle); <br />
/* draw figures */<br />
draw(circle((0.,0.), 2.)); <br />
draw((-2.,0.)--(5.,5.)); <br />
draw((5.,5.)--(5.,0.)); <br />
draw((5.,0.)--(-2.,0.)); <br />
draw((-2.,0.)--(0.6486486486486486,1.8918918918918919)); <br />
draw((0.6486486486486486,1.8918918918918919)--(2.,0.)); <br />
draw((2.,0.)--(-2.,0.)); <br />
draw((2.,0.)--(5.,5.)); <br />
draw((0.,0.)--(5.,5.)); <br />
/* dots and labels */<br />
dot((0.,0.),dotstyle); <br />
label("<math>O</math>", (-0.10330578512396349,-0.39365890308038826), NE * labelscalefactor); <br />
dot((-2.,0.),dotstyle); <br />
label("<math>A</math>", (-2.2370398196844437,-0.42371149511645134), NE * labelscalefactor); <br />
dot((2.,0.),dotstyle); <br />
label("<math>B</math>", (2.045454545454548,-0.36360631104432517), NE * labelscalefactor); <br />
dot((5.,0.),dotstyle); <br />
label("<math>D</math>", (4.900450788880542,-0.42371149511645134), NE * labelscalefactor); <br />
dot((5.,5.),dotstyle); <br />
label("<math>E</math>", (5.06574004507889,5.15104432757325), NE * labelscalefactor); <br />
dot((0.6486486486486486,1.8918918918918919),linewidth(3.pt) + dotstyle); <br />
label("<math>C</math>", (0.48271975957926694,2.100706235912847), NE * labelscalefactor); <br />
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); <br />
/* end of picture */<br />
[/asy]<br />
<br />
Let <math>O</math> be the center of the circle. Note that <math>EC + CA = EA = \sqrt{AD^2 + DE^2} = \sqrt{(2+2+3)^2 + 5^2} = \sqrt{74}</math>. However, by Power of a Point, <math>(EC)(EC + CA) = EO^2 - R^2 = (2+3)^2 + 5^2 - 2^2 = 25 + 25 - 4 = 46 \implies EC = \frac{46}{\sqrt{74}}</math>, so <math>AC = \sqrt{74} - \frac{46}{\sqrt{74}} = \frac{28}{\sqrt{74}}</math>. Now <math>BC = \sqrt{AB^2 - AC^2} = \sqrt{4^2 - \frac{28^2}{74}} = \sqrt{\frac{16 \cdot 74 - 28^2}{74}} = \sqrt{\frac{1184 - 784}{74}} = \frac{20}{\sqrt{74}}</math>. Since <math>\angle ACB = 90^{\circ}, [ABC] = \frac{1}{2} \cdot BC \cdot AC = \frac{1}{2} \cdot \frac{20}{\sqrt{74}} \cdot \frac{28}{\sqrt{74}} = \boxed{\textbf{(D)}\ \frac{140}{37}}</math>.<br />
<br />
==See Also==<br />
{{AMC12 box|year=2017|ab=B|num-b=17|num-a=19}}<br />
{{MAA Notice}}</div>Mathmaster2012https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10A_Problems/Problem_11&diff=831662017 AMC 10A Problems/Problem 112017-02-09T00:34:50Z<p>Mathmaster2012: </p>
<hr />
<div>==Problem==<br />
<br />
The region consisting of all point in three-dimensional space within 3 units of line segment <math>\overline{AB}</math> has volume 216<math>\pi</math>. What is the length <math>\textit{AB}</math>?<br />
<br />
<math>\textbf{(A)}\ 6\qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 18\qquad\textbf{(D)}\ 20\qquad\textbf{(E)}\ 24</math><br />
<br />
==Solution 1==<br />
In order to solve this problem, we must first visualize what the region contained looks like. We know that, in a three dimensional plane, the region consisting of all points within <math>3</math> units of a point would be a sphere with radius <math>3</math>. However, we need to find the region containing all points within 3 units of a segment. It can be seen that our region is a cylinder with two hemispheres on either end. We know the volume of our region, so we set up the following equation (the volume of our cylinder + the volume of our two hemispheres will equal <math>216 \pi</math>):<br />
<br />
<math>\frac{4 \pi }{3}3^3+9 \pi x=216</math><br />
<br />
Where <math>x</math> is equal to the length of our line segment.<br />
<br />
We isolate <math>x</math>. This comes out to be <math>\boxed{\textbf{(D)}\ 20}</math><br />
<br />
==Solution 2==<br />
To envision what the region must look like, we simplify the problem to finding all points within <math>3</math> units from a point. This is a sphere. To account for the line, we drag the sphere's center across the line, sweeping out the desired volume. As stated above, this is a cylinder with two hemispheres on both ends.<br />
<br />
==Diagram==<br />
<br />
http://i.imgur.com/cwNt293.png<br />
<br />
==See Also==<br />
{{AMC10 box|year=2017|ab=A|num-b=10|num-a=12}}<br />
{{MAA Notice}}</div>Mathmaster2012https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10A_Problems/Problem_11&diff=831622017 AMC 10A Problems/Problem 112017-02-09T00:32:53Z<p>Mathmaster2012: </p>
<hr />
<div>==Problem==<br />
<br />
The region consisting of all point in three-dimensional space within 3 units of line segment <math>\overline{AB}</math> has volume 216<math>\pi</math>. What is the length <math>\textit{AB}</math>?<br />
<br />
<math>\textbf{(A)}\ 6\qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 18\qquad\textbf{(D)}\ 20\qquad\textbf{(E)}\ 24</math><br />
<br />
http://i.imgur.com/cwNt293.png<br />
<br />
==Solution 1==<br />
In order to solve this problem, we must first visualize what the region contained looks like. We know that, in a three dimensional plane, the region consisting of all points within <math>3</math> units of a point would be a sphere with radius <math>3</math>. However, we need to find the region containing all points within 3 units of a segment. It can be seen that our region is a cylinder with two hemispheres on either end. We know the volume of our region, so we set up the following equation (the volume of our cylinder + the volume of our two hemispheres will equal <math>216 \pi</math>):<br />
<br />
<math>\frac{4 \pi }{3}3^3+9 \pi x=216</math><br />
<br />
Where <math>x</math> is equal to the length of our line segment.<br />
<br />
We isolate <math>x</math>. This comes out to be <math>\boxed{\textbf{(D)}\ 20}</math><br />
<br />
==Solution 2==<br />
To envision what the region must look like, we simplify the problem to finding all points within <math>3</math> units from a point. This is a sphere. To account for the line, we drag the sphere's center across the line, sweeping out the desired volume. As stated above, this is a cylinder with two hemispheres on both ends.<br />
<br />
==See Also==<br />
{{AMC10 box|year=2017|ab=A|num-b=10|num-a=12}}<br />
{{MAA Notice}}</div>Mathmaster2012https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10A_Problems/Problem_11&diff=831602017 AMC 10A Problems/Problem 112017-02-09T00:32:16Z<p>Mathmaster2012: </p>
<hr />
<div>==Problem==<br />
<br />
The region consisting of all point in three-dimensional space within 3 units of line segment <math>\overline{AB}</math> has volume 216<math>\pi</math>. What is the length <math>\textit{AB}</math>?<br />
<br />
<math>\textbf{(A)}\ 6\qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 18\qquad\textbf{(D)}\ 20\qquad\textbf{(E)}\ 24</math><br />
<br />
[img]<br />
<br />
http://i.imgur.com/cwNt293.png<br />
<br />
[/img]<br />
<br />
==Solution 1==<br />
In order to solve this problem, we must first visualize what the region contained looks like. We know that, in a three dimensional plane, the region consisting of all points within <math>3</math> units of a point would be a sphere with radius <math>3</math>. However, we need to find the region containing all points within 3 units of a segment. It can be seen that our region is a cylinder with two hemispheres on either end. We know the volume of our region, so we set up the following equation (the volume of our cylinder + the volume of our two hemispheres will equal <math>216 \pi</math>):<br />
<br />
<math>\frac{4 \pi }{3}3^3+9 \pi x=216</math><br />
<br />
Where <math>x</math> is equal to the length of our line segment.<br />
<br />
We isolate <math>x</math>. This comes out to be <math>\boxed{\textbf{(D)}\ 20}</math><br />
<br />
==Solution 2==<br />
To envision what the region must look like, we simplify the problem to finding all points within <math>3</math> units from a point. This is a sphere. To account for the line, we drag the sphere's center across the line, sweeping out the desired volume. As stated above, this is a cylinder with two hemispheres on both ends.<br />
<br />
==See Also==<br />
{{AMC10 box|year=2017|ab=A|num-b=10|num-a=12}}<br />
{{MAA Notice}}</div>Mathmaster2012https://artofproblemsolving.com/wiki/index.php?title=AMC_historical_results&diff=77923AMC historical results2016-04-03T20:45:42Z<p>Mathmaster2012: /* AIME II */</p>
<hr />
<div><!-- Post AMC statistics and lists of high scorers here so that the AMC page doesn't get cluttered. --><br />
This is the '''AMC historical results''' page. This page should include results for the [[AIME]] as well. For [[USAMO]] results, see [[USAMO historical results]].<br />
<br />
==2016==<br />
===AMC 10A===<br />
*Average score: 64.73<br />
*AIME floor: 110<br />
<br />
===AMC 10B===<br />
*Average score: 64.54<br />
*AIME floor: 110<br />
<br />
===AMC 12A===<br />
*Average score: 59.06<br />
*AIME floor: 92<br />
<br />
===AMC 12B===<br />
*Average score: 67.96<br />
*AIME floor: 100.0 (Top 5% 106.5)<br />
<br />
===AIME I===<br />
*Average score: <br />
*Median score: <br />
*USAMO cutoff: UNDETERMINED <br />
*USAJMO cutoff: UNDETERMINED<br />
<br />
===AIME II===<br />
*Average score: <br />
*Median score: <br />
*USAMO cutoff: UNDETERMINED<br />
*USAJMO cutoff: UNDETERMINED<br />
<br />
==2015==<br />
===AMC 10A===<br />
*Average score: 73.39<br />
*AIME floor: 106.5<br />
<br />
===AMC 10B===<br />
*Average score: 76.10<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 69.90<br />
*AIME floor: 99<br />
<br />
===AMC 12B===<br />
*Average score: 66.92<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score: 5.29<br />
*Median score: 5<br />
*USAMO cutoff: 219.0<br />
*USAJMO cutoff: 213.0<br />
<br />
===AIME II===<br />
*Average score: 6.63<br />
*Median score: 6<br />
*USAMO cutoff: 229.0<br />
*USAJMO cutoff: 223.5<br />
<br />
==2014==<br />
===AMC 10A===<br />
*Average score: 63.83<br />
*AIME floor: 120<br />
<br />
===AMC 10B===<br />
*Average score: 71.44<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 64.01<br />
*AIME floor: 93<br />
<br />
===AMC 12B===<br />
*Average score: 68.11<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score: 4.88<br />
*Median score: 5<br />
*USAMO cutoff: 211.5<br />
*USAJMO cutoff: 211<br />
<br />
===AIME II===<br />
*Average score: 5.49<br />
*Median score: 5<br />
*USAMO cutoff: 211.5<br />
*USAJMO cutoff: 211<br />
<br />
==2013==<br />
===AMC 10A===<br />
*Average score: 72.50<br />
*AIME floor: 108<br />
<br />
===AMC 10B===<br />
*Average score: 72.62<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 65.06<br />
*AIME floor:88.5<br />
<br />
===AMC 12B===<br />
*Average score: 64.21<br />
*AIME floor: 93<br />
<br />
===AIME I===<br />
*Average score: 4.69<br />
*Median score: <br />
*USAMO cutoff: 209<br />
*USAJMO cutoff: 210.5<br />
<br />
===AIME II===<br />
*Average score: 6.56<br />
*Median score: <br />
*USAMO cutoff: 209<br />
*USAJMO cutoff: 210.5<br />
<br />
==2012==<br />
===AMC 10A===<br />
*Average score: 72.51<br />
*AIME floor: 115.5<br />
<br />
===AMC 10B===<br />
*Average score: 76.59<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 64.62<br />
*AIME floor: 94.5<br />
<br />
===AMC 12B===<br />
*Average score: 70.08<br />
*AIME floor: 99<br />
<br />
===AIME I===<br />
*Average score: 5.13<br />
*Median score: <br />
*USAMO cutoff: 204.5<br />
*USAJMO cutoff: 204<br />
<br />
===AIME II===<br />
*Average score: 4.94<br />
*Median score: <br />
*USAMO cutoff: 204.5<br />
*USAJMO cutoff: 204<br />
<br />
==2011==<br />
===AMC 10A===<br />
*Average score: 67.61<br />
*AIME floor: 117<br />
<br />
===AMC 10B===<br />
*Average score: 71.78<br />
*AIME floor: 117<br />
<br />
===AMC 12A===<br />
*Average score: 66.77<br />
*AIME floor: 93<br />
<br />
===AMC 12B===<br />
*Average score: 64.71<br />
*AIME floor: 97.5<br />
<br />
===AIME I===<br />
*Average score: 5.47<br />
*Median score: <br />
*USAMO cutoff: 188<br />
*USAJMO cutoff: 179<br />
<br />
===AIME II===<br />
*Average score: 2.23<br />
*Median score: <br />
*USAMO cutoff: 215.5<br />
*USAJMO cutoff: 196.5<br />
<br />
==2010==<br />
===AMC 10A===<br />
*Average score: 68.11<br />
*AIME floor: 115.5<br />
<br />
===AMC 10B===<br />
*Average score: 68.57<br />
*AIME floor: 118.5<br />
<br />
===AMC 12A===<br />
*Average score: 61.02<br />
*AIME floor: 88.5<br />
<br />
===AMC 12B===<br />
*Average score: 59.58<br />
*AIME floor: 88.5<br />
<br />
===AIME I===<br />
*Average score: 5.90<br />
*Median score: <br />
*USAMO cutoff: 208.5 (204.5 for non juniors and seniors)<br />
*USAJMO cutoff: 188.5<br />
<br />
===AIME II===<br />
*Average score: 3.39<br />
*Median score: <br />
*USAMO cutoff: 208.5 (204.5 for non juniors and seniors)<br />
*USAJMO cutoff: 188.5<br />
<br />
==2009==<br />
===AMC 10A===<br />
*Average score: 67.41<br />
*AIME floor: <br />
<br />
===AMC 10B===<br />
*Average score: 74.73<br />
*AIME floor: <br />
<br />
===AMC 12A===<br />
*Average score: 66.37<br />
*AIME floor: <br />
<br />
===AMC 12B===<br />
*Average score: 71.88<br />
*AIME floor: <br />
<br />
===AIME I===<br />
*Average score: <br />
*Median score: <br />
*USAMO floor: <br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2008==<br />
===AMC 10A===<br />
*Average score: <br />
*AIME floor: <br />
<br />
===AMC 10B===<br />
*Average score: <br />
*AIME floor: <br />
<br />
===AMC 12A===<br />
*Average score: 65.6<br />
*AIME floor: 97.5<br />
<br />
===AMC 12B===<br />
*Average score: 68.9<br />
*AIME floor: 97.5<br />
<br />
===AIME I===<br />
*Average score: <br />
*Median score: <br />
*USAMO floor: <br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2007==<br />
<br />
===AMC 10A===<br />
*Average score: 67.9<br />
*AIME floor: 117<br />
<br />
===AMC 10B===<br />
*Average score: 61.5<br />
*AIME floor: 115.5<br />
<br />
===AMC 12A===<br />
*Average score: 66.8<br />
*AIME floor: 97.5<br />
<br />
===AMC 12B===<br />
*Average score: 73.1<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score: 5<br />
*Median score: 3<br />
*USAMO floor: 6<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2006==<br />
===AMC 10A===<br />
*Average score: 79.0<br />
*AIME floor: 120<br />
<br />
===AMC 10B===<br />
*Average score: 68.5<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 85.7<br />
*AIME floor: 100<br />
<br />
===AMC 12B===<br />
*Average score: 85.5<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2005==<br />
===AMC 10A===<br />
*Average score: 74.0<br />
*AIME floor: 120<br />
<br />
===AMC 10B===<br />
*Average score: 79.0<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 78.7<br />
*AIME floor: 100<br />
<br />
===AMC 12B===<br />
*Average score: 83.4<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2004==<br />
===AMC 10A===<br />
*Average score: 69.1<br />
*AIME floor: 110<br />
<br />
===AMC 10B===<br />
*Average score: 80.4<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 73.9<br />
*AIME floor: 100<br />
<br />
===AMC 12B===<br />
*Average score: 84.5<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2003==<br />
===AMC 10A===<br />
*Average score: 74.4<br />
*AIME floor: 119<br />
<br />
===AMC 10B===<br />
*Average score: 79.6<br />
*AIME floor: 121<br />
<br />
===AMC 12A===<br />
*Average score: 77.8<br />
*AIME floor: 100<br />
<br />
===AMC 12B===<br />
*Average score: 76.6<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2002==<br />
===AMC 10A===<br />
*Average score: 68.5<br />
*AIME floor: 115<br />
<br />
===AMC 10B===<br />
*Average score: 74.9<br />
*AIME floor: 118<br />
<br />
===AMC 12A===<br />
*Average score: 72.7<br />
*AIME floor: 100<br />
<br />
===AMC 12B===<br />
*Average score: 80.8<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2001==<br />
===AMC 10===<br />
*Average score: 67.8<br />
*AIME floor: 116<br />
<br />
===AMC 12===<br />
*Average score: 56.6<br />
*AIME floor: 84<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2000==<br />
===AMC 10===<br />
*Average score: 64.2<br />
*AIME floor: <br />
<br />
===AMC 12===<br />
*Average score: 64.9<br />
*AIME floor: <br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==1999==<br />
===AHSME===<br />
*Average score: 68.8<br />
*AIME floor:<br />
<br />
===AIME===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==1998==<br />
==1997==<br />
==1996==<br />
==1995==<br />
==1994==<br />
==1993==<br />
==1992==<br />
==1991==<br />
==1990==<br />
==1989==<br />
==1988==<br />
==1987==<br />
==1986==<br />
==1985==<br />
==1984==<br />
==1983==<br />
==1982==<br />
==1981==<br />
==1980==<br />
==1979==<br />
==1978==<br />
==1977==<br />
==1976==<br />
==1975==<br />
==1974==<br />
==1973==<br />
==1972==<br />
==1971==<br />
==1970==<br />
==1969==<br />
==1968==<br />
==1967==<br />
==1966==<br />
==1965==<br />
==1964==<br />
==1963==<br />
==1962==<br />
==1961==<br />
==1960==<br />
==1959==<br />
==1958==<br />
==1957==<br />
==1956==<br />
==1955==<br />
==1954==<br />
==1953==<br />
==1952==<br />
==1951==<br />
==1950==<br />
==1949==<br />
<br />
[[Category:Historical results]]</div>Mathmaster2012https://artofproblemsolving.com/wiki/index.php?title=AMC_historical_results&diff=77922AMC historical results2016-04-03T20:45:08Z<p>Mathmaster2012: /* AIME I */</p>
<hr />
<div><!-- Post AMC statistics and lists of high scorers here so that the AMC page doesn't get cluttered. --><br />
This is the '''AMC historical results''' page. This page should include results for the [[AIME]] as well. For [[USAMO]] results, see [[USAMO historical results]].<br />
<br />
==2016==<br />
===AMC 10A===<br />
*Average score: 64.73<br />
*AIME floor: 110<br />
<br />
===AMC 10B===<br />
*Average score: 64.54<br />
*AIME floor: 110<br />
<br />
===AMC 12A===<br />
*Average score: 59.06<br />
*AIME floor: 92<br />
<br />
===AMC 12B===<br />
*Average score: 67.96<br />
*AIME floor: 100.0 (Top 5% 106.5)<br />
<br />
===AIME I===<br />
*Average score: <br />
*Median score: <br />
*USAMO cutoff: UNDETERMINED <br />
*USAJMO cutoff: UNDETERMINED<br />
<br />
===AIME II===<br />
*Average score: <br />
*Median score: <br />
*USAMO cutoff: 209 <br />
*USAJMO cutoff: 204<br />
<br />
==2015==<br />
===AMC 10A===<br />
*Average score: 73.39<br />
*AIME floor: 106.5<br />
<br />
===AMC 10B===<br />
*Average score: 76.10<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 69.90<br />
*AIME floor: 99<br />
<br />
===AMC 12B===<br />
*Average score: 66.92<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score: 5.29<br />
*Median score: 5<br />
*USAMO cutoff: 219.0<br />
*USAJMO cutoff: 213.0<br />
<br />
===AIME II===<br />
*Average score: 6.63<br />
*Median score: 6<br />
*USAMO cutoff: 229.0<br />
*USAJMO cutoff: 223.5<br />
<br />
==2014==<br />
===AMC 10A===<br />
*Average score: 63.83<br />
*AIME floor: 120<br />
<br />
===AMC 10B===<br />
*Average score: 71.44<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 64.01<br />
*AIME floor: 93<br />
<br />
===AMC 12B===<br />
*Average score: 68.11<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score: 4.88<br />
*Median score: 5<br />
*USAMO cutoff: 211.5<br />
*USAJMO cutoff: 211<br />
<br />
===AIME II===<br />
*Average score: 5.49<br />
*Median score: 5<br />
*USAMO cutoff: 211.5<br />
*USAJMO cutoff: 211<br />
<br />
==2013==<br />
===AMC 10A===<br />
*Average score: 72.50<br />
*AIME floor: 108<br />
<br />
===AMC 10B===<br />
*Average score: 72.62<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 65.06<br />
*AIME floor:88.5<br />
<br />
===AMC 12B===<br />
*Average score: 64.21<br />
*AIME floor: 93<br />
<br />
===AIME I===<br />
*Average score: 4.69<br />
*Median score: <br />
*USAMO cutoff: 209<br />
*USAJMO cutoff: 210.5<br />
<br />
===AIME II===<br />
*Average score: 6.56<br />
*Median score: <br />
*USAMO cutoff: 209<br />
*USAJMO cutoff: 210.5<br />
<br />
==2012==<br />
===AMC 10A===<br />
*Average score: 72.51<br />
*AIME floor: 115.5<br />
<br />
===AMC 10B===<br />
*Average score: 76.59<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 64.62<br />
*AIME floor: 94.5<br />
<br />
===AMC 12B===<br />
*Average score: 70.08<br />
*AIME floor: 99<br />
<br />
===AIME I===<br />
*Average score: 5.13<br />
*Median score: <br />
*USAMO cutoff: 204.5<br />
*USAJMO cutoff: 204<br />
<br />
===AIME II===<br />
*Average score: 4.94<br />
*Median score: <br />
*USAMO cutoff: 204.5<br />
*USAJMO cutoff: 204<br />
<br />
==2011==<br />
===AMC 10A===<br />
*Average score: 67.61<br />
*AIME floor: 117<br />
<br />
===AMC 10B===<br />
*Average score: 71.78<br />
*AIME floor: 117<br />
<br />
===AMC 12A===<br />
*Average score: 66.77<br />
*AIME floor: 93<br />
<br />
===AMC 12B===<br />
*Average score: 64.71<br />
*AIME floor: 97.5<br />
<br />
===AIME I===<br />
*Average score: 5.47<br />
*Median score: <br />
*USAMO cutoff: 188<br />
*USAJMO cutoff: 179<br />
<br />
===AIME II===<br />
*Average score: 2.23<br />
*Median score: <br />
*USAMO cutoff: 215.5<br />
*USAJMO cutoff: 196.5<br />
<br />
==2010==<br />
===AMC 10A===<br />
*Average score: 68.11<br />
*AIME floor: 115.5<br />
<br />
===AMC 10B===<br />
*Average score: 68.57<br />
*AIME floor: 118.5<br />
<br />
===AMC 12A===<br />
*Average score: 61.02<br />
*AIME floor: 88.5<br />
<br />
===AMC 12B===<br />
*Average score: 59.58<br />
*AIME floor: 88.5<br />
<br />
===AIME I===<br />
*Average score: 5.90<br />
*Median score: <br />
*USAMO cutoff: 208.5 (204.5 for non juniors and seniors)<br />
*USAJMO cutoff: 188.5<br />
<br />
===AIME II===<br />
*Average score: 3.39<br />
*Median score: <br />
*USAMO cutoff: 208.5 (204.5 for non juniors and seniors)<br />
*USAJMO cutoff: 188.5<br />
<br />
==2009==<br />
===AMC 10A===<br />
*Average score: 67.41<br />
*AIME floor: <br />
<br />
===AMC 10B===<br />
*Average score: 74.73<br />
*AIME floor: <br />
<br />
===AMC 12A===<br />
*Average score: 66.37<br />
*AIME floor: <br />
<br />
===AMC 12B===<br />
*Average score: 71.88<br />
*AIME floor: <br />
<br />
===AIME I===<br />
*Average score: <br />
*Median score: <br />
*USAMO floor: <br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2008==<br />
===AMC 10A===<br />
*Average score: <br />
*AIME floor: <br />
<br />
===AMC 10B===<br />
*Average score: <br />
*AIME floor: <br />
<br />
===AMC 12A===<br />
*Average score: 65.6<br />
*AIME floor: 97.5<br />
<br />
===AMC 12B===<br />
*Average score: 68.9<br />
*AIME floor: 97.5<br />
<br />
===AIME I===<br />
*Average score: <br />
*Median score: <br />
*USAMO floor: <br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2007==<br />
<br />
===AMC 10A===<br />
*Average score: 67.9<br />
*AIME floor: 117<br />
<br />
===AMC 10B===<br />
*Average score: 61.5<br />
*AIME floor: 115.5<br />
<br />
===AMC 12A===<br />
*Average score: 66.8<br />
*AIME floor: 97.5<br />
<br />
===AMC 12B===<br />
*Average score: 73.1<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score: 5<br />
*Median score: 3<br />
*USAMO floor: 6<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2006==<br />
===AMC 10A===<br />
*Average score: 79.0<br />
*AIME floor: 120<br />
<br />
===AMC 10B===<br />
*Average score: 68.5<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 85.7<br />
*AIME floor: 100<br />
<br />
===AMC 12B===<br />
*Average score: 85.5<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2005==<br />
===AMC 10A===<br />
*Average score: 74.0<br />
*AIME floor: 120<br />
<br />
===AMC 10B===<br />
*Average score: 79.0<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 78.7<br />
*AIME floor: 100<br />
<br />
===AMC 12B===<br />
*Average score: 83.4<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2004==<br />
===AMC 10A===<br />
*Average score: 69.1<br />
*AIME floor: 110<br />
<br />
===AMC 10B===<br />
*Average score: 80.4<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 73.9<br />
*AIME floor: 100<br />
<br />
===AMC 12B===<br />
*Average score: 84.5<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2003==<br />
===AMC 10A===<br />
*Average score: 74.4<br />
*AIME floor: 119<br />
<br />
===AMC 10B===<br />
*Average score: 79.6<br />
*AIME floor: 121<br />
<br />
===AMC 12A===<br />
*Average score: 77.8<br />
*AIME floor: 100<br />
<br />
===AMC 12B===<br />
*Average score: 76.6<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2002==<br />
===AMC 10A===<br />
*Average score: 68.5<br />
*AIME floor: 115<br />
<br />
===AMC 10B===<br />
*Average score: 74.9<br />
*AIME floor: 118<br />
<br />
===AMC 12A===<br />
*Average score: 72.7<br />
*AIME floor: 100<br />
<br />
===AMC 12B===<br />
*Average score: 80.8<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2001==<br />
===AMC 10===<br />
*Average score: 67.8<br />
*AIME floor: 116<br />
<br />
===AMC 12===<br />
*Average score: 56.6<br />
*AIME floor: 84<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2000==<br />
===AMC 10===<br />
*Average score: 64.2<br />
*AIME floor: <br />
<br />
===AMC 12===<br />
*Average score: 64.9<br />
*AIME floor: <br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==1999==<br />
===AHSME===<br />
*Average score: 68.8<br />
*AIME floor:<br />
<br />
===AIME===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==1998==<br />
==1997==<br />
==1996==<br />
==1995==<br />
==1994==<br />
==1993==<br />
==1992==<br />
==1991==<br />
==1990==<br />
==1989==<br />
==1988==<br />
==1987==<br />
==1986==<br />
==1985==<br />
==1984==<br />
==1983==<br />
==1982==<br />
==1981==<br />
==1980==<br />
==1979==<br />
==1978==<br />
==1977==<br />
==1976==<br />
==1975==<br />
==1974==<br />
==1973==<br />
==1972==<br />
==1971==<br />
==1970==<br />
==1969==<br />
==1968==<br />
==1967==<br />
==1966==<br />
==1965==<br />
==1964==<br />
==1963==<br />
==1962==<br />
==1961==<br />
==1960==<br />
==1959==<br />
==1958==<br />
==1957==<br />
==1956==<br />
==1955==<br />
==1954==<br />
==1953==<br />
==1952==<br />
==1951==<br />
==1950==<br />
==1949==<br />
<br />
[[Category:Historical results]]</div>Mathmaster2012https://artofproblemsolving.com/wiki/index.php?title=AMC_historical_results&diff=77898AMC historical results2016-04-03T00:15:42Z<p>Mathmaster2012: </p>
<hr />
<div><!-- Post AMC statistics and lists of high scorers here so that the AMC page doesn't get cluttered. --><br />
This is the '''AMC historical results''' page. This page should include results for the [[AIME]] as well. For [[USAMO]] results, see [[USAMO historical results]].<br />
<br />
==2016==<br />
===AMC 10A===<br />
*Average score: 64.73<br />
*AIME floor: 110<br />
<br />
===AMC 10B===<br />
*Average score: 64.54<br />
*AIME floor: 110<br />
<br />
===AMC 12A===<br />
*Average score: 59.06<br />
*AIME floor: 92<br />
<br />
===AMC 12B===<br />
*Average score: 67.96<br />
*AIME floor: 100.0 (Top 5% 106.5)<br />
<br />
===AIME I===<br />
*Average score: <br />
*Median score: <br />
*USAMO cutoff: UNDETERMINED<br />
*USAJMO cutoff: UNDETERMINED<br />
<br />
===AIME II===<br />
*Average score: <br />
*Median score: <br />
*USAMO cutoff: UNDETERMINED <br />
*USAJMO cutoff: UNDETERMINED<br />
<br />
==2015==<br />
===AMC 10A===<br />
*Average score: 73.39<br />
*AIME floor: 106.5<br />
<br />
===AMC 10B===<br />
*Average score: 76.10<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 69.90<br />
*AIME floor: 99<br />
<br />
===AMC 12B===<br />
*Average score: 66.92<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score: 5.29<br />
*Median score: 5<br />
*USAMO cutoff: 219.0<br />
*USAJMO cutoff: 213.0<br />
<br />
===AIME II===<br />
*Average score: 6.63<br />
*Median score: 6<br />
*USAMO cutoff: 229.0<br />
*USAJMO cutoff: 223.5<br />
<br />
==2014==<br />
===AMC 10A===<br />
*Average score: 63.83<br />
*AIME floor: 120<br />
<br />
===AMC 10B===<br />
*Average score: 71.44<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 64.01<br />
*AIME floor: 93<br />
<br />
===AMC 12B===<br />
*Average score: 68.11<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score: 4.88<br />
*Median score: 5<br />
*USAMO cutoff: 211.5<br />
*USAJMO cutoff: 211<br />
<br />
===AIME II===<br />
*Average score: 5.49<br />
*Median score: 5<br />
*USAMO cutoff: 211.5<br />
*USAJMO cutoff: 211<br />
<br />
==2013==<br />
===AMC 10A===<br />
*Average score: 72.50<br />
*AIME floor: 108<br />
<br />
===AMC 10B===<br />
*Average score: 72.62<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 65.06<br />
*AIME floor:88.5<br />
<br />
===AMC 12B===<br />
*Average score: 64.21<br />
*AIME floor: 93<br />
<br />
===AIME I===<br />
*Average score: 4.69<br />
*Median score: <br />
*USAMO cutoff: 209<br />
*USAJMO cutoff: 210.5<br />
<br />
===AIME II===<br />
*Average score: 6.56<br />
*Median score: <br />
*USAMO cutoff: 209<br />
*USAJMO cutoff: 210.5<br />
<br />
==2012==<br />
===AMC 10A===<br />
*Average score: 72.51<br />
*AIME floor: 115.5<br />
<br />
===AMC 10B===<br />
*Average score: 76.59<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 64.62<br />
*AIME floor: 94.5<br />
<br />
===AMC 12B===<br />
*Average score: 70.08<br />
*AIME floor: 99<br />
<br />
===AIME I===<br />
*Average score: 5.13<br />
*Median score: <br />
*USAMO cutoff: 204.5<br />
*USAJMO cutoff: 204<br />
<br />
===AIME II===<br />
*Average score: 4.94<br />
*Median score: <br />
*USAMO cutoff: 204.5<br />
*USAJMO cutoff: 204<br />
<br />
==2011==<br />
===AMC 10A===<br />
*Average score: 67.61<br />
*AIME floor: 117<br />
<br />
===AMC 10B===<br />
*Average score: 71.78<br />
*AIME floor: 117<br />
<br />
===AMC 12A===<br />
*Average score: 66.77<br />
*AIME floor: 93<br />
<br />
===AMC 12B===<br />
*Average score: 64.71<br />
*AIME floor: 97.5<br />
<br />
===AIME I===<br />
*Average score: 5.47<br />
*Median score: <br />
*USAMO cutoff: 188<br />
*USAJMO cutoff: 179<br />
<br />
===AIME II===<br />
*Average score: 2.23<br />
*Median score: <br />
*USAMO cutoff: 215.5<br />
*USAJMO cutoff: 196.5<br />
<br />
==2010==<br />
===AMC 10A===<br />
*Average score: 68.11<br />
*AIME floor: 115.5<br />
<br />
===AMC 10B===<br />
*Average score: 68.57<br />
*AIME floor: 118.5<br />
<br />
===AMC 12A===<br />
*Average score: 61.02<br />
*AIME floor: 88.5<br />
<br />
===AMC 12B===<br />
*Average score: 59.58<br />
*AIME floor: 88.5<br />
<br />
===AIME I===<br />
*Average score: 5.90<br />
*Median score: <br />
*USAMO cutoff: 208.5 (204.5 for non juniors and seniors)<br />
*USAJMO cutoff: 188.5<br />
<br />
===AIME II===<br />
*Average score: 3.39<br />
*Median score: <br />
*USAMO cutoff: 208.5 (204.5 for non juniors and seniors)<br />
*USAJMO cutoff: 188.5<br />
<br />
==2009==<br />
===AMC 10A===<br />
*Average score: 67.41<br />
*AIME floor: <br />
<br />
===AMC 10B===<br />
*Average score: 74.73<br />
*AIME floor: <br />
<br />
===AMC 12A===<br />
*Average score: 66.37<br />
*AIME floor: <br />
<br />
===AMC 12B===<br />
*Average score: 71.88<br />
*AIME floor: <br />
<br />
===AIME I===<br />
*Average score: <br />
*Median score: <br />
*USAMO floor: <br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2008==<br />
===AMC 10A===<br />
*Average score: <br />
*AIME floor: <br />
<br />
===AMC 10B===<br />
*Average score: <br />
*AIME floor: <br />
<br />
===AMC 12A===<br />
*Average score: 65.6<br />
*AIME floor: 97.5<br />
<br />
===AMC 12B===<br />
*Average score: 68.9<br />
*AIME floor: 97.5<br />
<br />
===AIME I===<br />
*Average score: <br />
*Median score: <br />
*USAMO floor: <br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2007==<br />
<br />
===AMC 10A===<br />
*Average score: 67.9<br />
*AIME floor: 117<br />
<br />
===AMC 10B===<br />
*Average score: 61.5<br />
*AIME floor: 115.5<br />
<br />
===AMC 12A===<br />
*Average score: 66.8<br />
*AIME floor: 97.5<br />
<br />
===AMC 12B===<br />
*Average score: 73.1<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score: 5<br />
*Median score: 3<br />
*USAMO floor: 6<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2006==<br />
===AMC 10A===<br />
*Average score: 79.0<br />
*AIME floor: 120<br />
<br />
===AMC 10B===<br />
*Average score: 68.5<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 85.7<br />
*AIME floor: 100<br />
<br />
===AMC 12B===<br />
*Average score: 85.5<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2005==<br />
===AMC 10A===<br />
*Average score: 74.0<br />
*AIME floor: 120<br />
<br />
===AMC 10B===<br />
*Average score: 79.0<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 78.7<br />
*AIME floor: 100<br />
<br />
===AMC 12B===<br />
*Average score: 83.4<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2004==<br />
===AMC 10A===<br />
*Average score: 69.1<br />
*AIME floor: 110<br />
<br />
===AMC 10B===<br />
*Average score: 80.4<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 73.9<br />
*AIME floor: 100<br />
<br />
===AMC 12B===<br />
*Average score: 84.5<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2003==<br />
===AMC 10A===<br />
*Average score: 74.4<br />
*AIME floor: 119<br />
<br />
===AMC 10B===<br />
*Average score: 79.6<br />
*AIME floor: 121<br />
<br />
===AMC 12A===<br />
*Average score: 77.8<br />
*AIME floor: 100<br />
<br />
===AMC 12B===<br />
*Average score: 76.6<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2002==<br />
===AMC 10A===<br />
*Average score: 68.5<br />
*AIME floor: 115<br />
<br />
===AMC 10B===<br />
*Average score: 74.9<br />
*AIME floor: 118<br />
<br />
===AMC 12A===<br />
*Average score: 72.7<br />
*AIME floor: 100<br />
<br />
===AMC 12B===<br />
*Average score: 80.8<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2001==<br />
===AMC 10===<br />
*Average score: 67.8<br />
*AIME floor: 116<br />
<br />
===AMC 12===<br />
*Average score: 56.6<br />
*AIME floor: 84<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2000==<br />
===AMC 10===<br />
*Average score: 64.2<br />
*AIME floor: <br />
<br />
===AMC 12===<br />
*Average score: 64.9<br />
*AIME floor: <br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==1999==<br />
===AHSME===<br />
*Average score: 68.8<br />
*AIME floor:<br />
<br />
===AIME===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==1998==<br />
==1997==<br />
==1996==<br />
==1995==<br />
==1994==<br />
==1993==<br />
==1992==<br />
==1991==<br />
==1990==<br />
==1989==<br />
==1988==<br />
==1987==<br />
==1986==<br />
==1985==<br />
==1984==<br />
==1983==<br />
==1982==<br />
==1981==<br />
==1980==<br />
==1979==<br />
==1978==<br />
==1977==<br />
==1976==<br />
==1975==<br />
==1974==<br />
==1973==<br />
==1972==<br />
==1971==<br />
==1970==<br />
==1969==<br />
==1968==<br />
==1967==<br />
==1966==<br />
==1965==<br />
==1964==<br />
==1963==<br />
==1962==<br />
==1961==<br />
==1960==<br />
==1959==<br />
==1958==<br />
==1957==<br />
==1956==<br />
==1955==<br />
==1954==<br />
==1953==<br />
==1952==<br />
==1951==<br />
==1950==<br />
<br />
[[Category:Historical results]]</div>Mathmaster2012https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_12B_Problems/Problem_21&diff=766372016 AMC 12B Problems/Problem 212016-02-21T16:17:46Z<p>Mathmaster2012: Created page with "Let <math>ABCD</math> be a unit square. Let <math>Q_1</math> be the midpoint of <math>\overline{CD}</math>. For <math>i=1,2,\dots,</math> let <math>P_i</math> be the intersect..."</p>
<hr />
<div>Let <math>ABCD</math> be a unit square. Let <math>Q_1</math> be the midpoint of <math>\overline{CD}</math>. For <math>i=1,2,\dots,</math> let <math>P_i</math> be the intersection of <math>\overline{AQ_i}</math> and <math>\overline{BD}</math>, and let <math>Q_{i+1}</math> be the foot of the perpendicular from <math>P_i</math> to <math>\overline{CD}</math>. What is <br />
<cmath>\sum_{i=1}^{\infty} \text{Area of } \triangle DQ_i P_i \, ?</cmath><br />
<br />
<math>\textbf{(A)}\ \frac{1}{6} \qquad<br />
\textbf{(B)}\ \frac{1}{4} \qquad<br />
\textbf{(C)}\ \frac{1}{3} \qquad<br />
\textbf{(D)}\ \frac{1}{2} \qquad<br />
\textbf{(E)}\ 1</math><br />
<br />
==Solution==<br />
<br />
==See Also==<br />
{{AMC12 box|year=2016|ab=B|num-b=20|num-a=22}}<br />
{{MAA Notice}}</div>Mathmaster2012https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_12B_Problems/Problem_20&diff=766362016 AMC 12B Problems/Problem 202016-02-21T16:16:47Z<p>Mathmaster2012: Created page with "==Problem== A set of teams held a round-robin tournament in which every team played every other team exactly once. Every team won <math>10</math> games and lost <math>10</mat..."</p>
<hr />
<div>==Problem==<br />
<br />
A set of teams held a round-robin tournament in which every team played every other team exactly once. Every team won <math>10</math> games and lost <math>10</math> games; there were no ties. How many sets of three teams <math>\{A, B, C\}</math> were there in which <math>A</math> beat <math>B</math>, <math>B</math> beat <math>C</math>, and <math>C</math> beat <math>A?</math><br />
<br />
<math>\textbf{(A)}\ 385 \qquad<br />
\textbf{(B)}\ 665 \qquad<br />
\textbf{(C)}\ 945 \qquad<br />
\textbf{(D)}\ 1140 \qquad<br />
\textbf{(E)}\ 1330</math><br />
<br />
==Solution==<br />
<br />
==See Also==<br />
{{AMC12 box|year=2016|ab=B|num-b=7|num-a=9}}<br />
{{MAA Notice}}</div>Mathmaster2012https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_12B_Problems/Problem_19&diff=766322016 AMC 12B Problems/Problem 192016-02-21T16:15:56Z<p>Mathmaster2012: Created page with "==Problem== Tom, Dick, and Harry are playing a game. Starting at the same time, each of them flips a fair coin repeatedly until he gets his first head, at which point he stop..."</p>
<hr />
<div>==Problem==<br />
<br />
Tom, Dick, and Harry are playing a game. Starting at the same time, each of them flips a fair coin repeatedly until he gets his first head, at which point he stops. What is the probability that all three flip their coins the same number of times?<br />
<br />
<math>\textbf{(A)}\ \frac{1}{8} \qquad<br />
\textbf{(B)}\ \frac{1}{7} \qquad<br />
\textbf{(C)}\ \frac{1}{6} \qquad<br />
\textbf{(D)}\ \frac{1}{4} \qquad<br />
\textbf{(E)}\ \frac{1}{3}</math><br />
<br />
==Solution==<br />
<br />
==See Also==<br />
{{AMC12 box|year=2016|ab=B|num-b=18|num-a=20}}<br />
{{MAA Notice}}</div>Mathmaster2012https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_12B_Problems/Problem_18&diff=766292016 AMC 12B Problems/Problem 182016-02-21T16:15:04Z<p>Mathmaster2012: Created page with "==Problem== What is the area of the region enclosed by the graph of the equation <math>x^2+y^2=|x|+|y|?</math> <math>\textbf{(A)}\ \pi+\sqrt{2} \qquad\textbf{(B)}\ \pi+2 \qq..."</p>
<hr />
<div>==Problem==<br />
<br />
What is the area of the region enclosed by the graph of the equation <math>x^2+y^2=|x|+|y|?</math><br />
<br />
<math>\textbf{(A)}\ \pi+\sqrt{2} \qquad\textbf{(B)}\ \pi+2 \qquad\textbf{(C)}\ \pi+2\sqrt{2} \qquad\textbf{(D)}\ 2\pi+\sqrt{2} \qquad\textbf{(E)}\ 2\pi+2\sqrt{2}</math><br />
<br />
==Solution==<br />
<br />
==See Also==<br />
{{AMC12 box|year=2016|ab=B|num-b=17|num-a=19}}<br />
{{MAA Notice}}</div>Mathmaster2012https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_12B_Problems/Problem_17&diff=766272016 AMC 12B Problems/Problem 172016-02-21T16:13:53Z<p>Mathmaster2012: </p>
<hr />
<div>==Problem==<br />
<br />
In <math>\triangle ABC</math> shown in the figure, <math>AB=7</math>, <math>BC=8</math>, <math>CA=9</math>, and <math>\overline{AH}</math> is an altitude. Points <math>D</math> and <math>E</math> lie on sides <math>\overline{AC}</math> and <math>\overline{AB}</math>, respectively, so that <math>\overline{BD}</math> and <math>\overline{CE}</math> are angle bisectors, intersecting <math>\overline{AH}</math> at <math>Q</math> and <math>P</math>, respectively. What is <math>PQ</math>?<br />
<br />
<asy><br />
import graph; size(9cm); <br />
real labelscalefactor = 0.5; /* changes label-to-point distance */<br />
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ <br />
pen dotstyle = black; /* point style */ <br />
real xmin = -4.381056062031275, xmax = 15.020004395092375, ymin = -4.051697595316909, ymax = 10.663513514111651; /* image dimensions */<br />
<br />
<br />
draw((0.,0.)--(4.714285714285714,7.666518779999279)--(7.,0.)--cycle); <br />
/* draw figures */<br />
draw((0.,0.)--(4.714285714285714,7.666518779999279)); <br />
draw((4.714285714285714,7.666518779999279)--(7.,0.)); <br />
draw((7.,0.)--(0.,0.)); <br />
label("7",(3.2916797119724284,-0.07831656949355523),SE*labelscalefactor); <br />
label("9",(2.0037562070503783,4.196493361737088),SE*labelscalefactor); <br />
label("8",(6.114150371695219,3.785453945272603),SE*labelscalefactor); <br />
draw((0.,0.)--(6.428571428571427,1.9166296949998194)); <br />
draw((7.,0.)--(2.2,3.5777087639996634)); <br />
draw((4.714285714285714,7.666518779999279)--(3.7058823529411766,0.)); <br />
/* dots and labels */<br />
dot((0.,0.),dotstyle); <br />
label("$A$", (-0.2432592696221352,-0.5715638692509372), NE * labelscalefactor); <br />
dot((7.,0.),dotstyle); <br />
label("$B$", (7.0458397156813835,-0.48935598595804014), NE * labelscalefactor); <br />
dot((3.7058823529411766,0.),dotstyle); <br />
label("$E$", (3.8123296394941084,0.16830708038513573), NE * labelscalefactor); <br />
dot((4.714285714285714,7.666518779999279),dotstyle); <br />
label("$C$", (4.579603216894479,7.895848109917452), NE * labelscalefactor); <br />
dot((2.2,3.5777087639996634),linewidth(3.pt) + dotstyle); <br />
label("$D$", (2.1407693458718726,3.127790878929427), NE * labelscalefactor); <br />
dot((6.428571428571427,1.9166296949998194),linewidth(3.pt) + dotstyle); <br />
label("$H$", (6.004539860638023,1.9494778850645704), NE * labelscalefactor); <br />
dot((5.,1.49071198499986),linewidth(3.pt) + dotstyle); <br />
label("$Q$", (4.935837377830365,1.7302568629501784), NE * labelscalefactor); <br />
dot((3.857142857142857,1.1499778169998918),linewidth(3.pt) + dotstyle); <br />
label("$P$", (3.538303361851119,1.2370095631927964), NE * labelscalefactor); <br />
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); <br />
/* end of picture */<br />
</asy><br />
<br />
<math>\textbf{(A)}\ 1 \qquad<br />
\textbf{(B)}\ \frac{5}{8}\sqrt{3} \qquad<br />
\textbf{(C)}\ \frac{4}{5}\sqrt{2} \qquad<br />
\textbf{(D)}\ \frac{8}{15}\sqrt{5} \qquad<br />
\textbf{(E)}\ \frac{6}{5}</math><br />
<br />
==Solution==<br />
<br />
==See Also==<br />
{{AMC12 box|year=2016|ab=B|num-b=16|num-a=18}}<br />
{{MAA Notice}}</div>Mathmaster2012https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_12B_Problems/Problem_17&diff=766212016 AMC 12B Problems/Problem 172016-02-21T16:12:12Z<p>Mathmaster2012: Created page with "==Problem== In <math>\triangle ABC</math> shown in the figure, <math>AB=7</math>, <math>BC=8</math>, <math>CA=9</math>, and <math>\overline{AH}</math> is an altitude. Points ..."</p>
<hr />
<div>==Problem==<br />
<br />
In <math>\triangle ABC</math> shown in the figure, <math>AB=7</math>, <math>BC=8</math>, <math>CA=9</math>, and <math>\overline{AH}</math> is an altitude. Points <math>D</math> and <math>E</math> lie on sides <math>\overline{AC}</math> and <math>\overline{AB}</math>, respectively, so that <math>\overline{BD}</math> and <math>\overline{CE}</math> are angle bisectors, intersecting <math>\overline{AH}</math> at <math>Q</math> and <math>P</math>, respectively. What is <math>PQ</math>?<br />
<br />
[asy]<br />
import graph; size(9cm); <br />
real labelscalefactor = 0.5; /* changes label-to-point distance */<br />
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ <br />
pen dotstyle = black; /* point style */ <br />
real xmin = -4.381056062031275, xmax = 15.020004395092375, ymin = -4.051697595316909, ymax = 10.663513514111651; /* image dimensions */<br />
<br />
<br />
draw((0.,0.)--(4.714285714285714,7.666518779999279)--(7.,0.)--cycle); <br />
/* draw figures */<br />
draw((0.,0.)--(4.714285714285714,7.666518779999279)); <br />
draw((4.714285714285714,7.666518779999279)--(7.,0.)); <br />
draw((7.,0.)--(0.,0.)); <br />
label("7",(3.2916797119724284,-0.07831656949355523),SE*labelscalefactor); <br />
label("9",(2.0037562070503783,4.196493361737088),SE*labelscalefactor); <br />
label("8",(6.114150371695219,3.785453945272603),SE*labelscalefactor); <br />
draw((0.,0.)--(6.428571428571427,1.9166296949998194)); <br />
draw((7.,0.)--(2.2,3.5777087639996634)); <br />
draw((4.714285714285714,7.666518779999279)--(3.7058823529411766,0.)); <br />
/* dots and labels */<br />
dot((0.,0.),dotstyle); <br />
label("<math>A</math>", (-0.2432592696221352,-0.5715638692509372), NE * labelscalefactor); <br />
dot((7.,0.),dotstyle); <br />
label("<math>B</math>", (7.0458397156813835,-0.48935598595804014), NE * labelscalefactor); <br />
dot((3.7058823529411766,0.),dotstyle); <br />
label("<math>E</math>", (3.8123296394941084,0.16830708038513573), NE * labelscalefactor); <br />
dot((4.714285714285714,7.666518779999279),dotstyle); <br />
label("<math>C</math>", (4.579603216894479,7.895848109917452), NE * labelscalefactor); <br />
dot((2.2,3.5777087639996634),linewidth(3.pt) + dotstyle); <br />
label("<math>D</math>", (2.1407693458718726,3.127790878929427), NE * labelscalefactor); <br />
dot((6.428571428571427,1.9166296949998194),linewidth(3.pt) + dotstyle); <br />
label("<math>H</math>", (6.004539860638023,1.9494778850645704), NE * labelscalefactor); <br />
dot((5.,1.49071198499986),linewidth(3.pt) + dotstyle); <br />
label("<math>Q</math>", (4.935837377830365,1.7302568629501784), NE * labelscalefactor); <br />
dot((3.857142857142857,1.1499778169998918),linewidth(3.pt) + dotstyle); <br />
label("<math>P</math>", (3.538303361851119,1.2370095631927964), NE * labelscalefactor); <br />
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); <br />
/* end of picture */<br />
[/asy]<br />
<br />
<math>\textbf{(A)}\ 1 \qquad<br />
\textbf{(B)}\ \frac{5}{8}\sqrt{3} \qquad<br />
\textbf{(C)}\ \frac{4}{5}\sqrt{2} \qquad<br />
\textbf{(D)}\ \frac{8}{15}\sqrt{5} \qquad<br />
\textbf{(E)}\ \frac{6}{5}</math><br />
<br />
==Solution==<br />
<br />
==See Also==<br />
{{AMC12 box|year=2016|ab=B|num-b=16|num-a=18}}<br />
{{MAA Notice}}</div>Mathmaster2012https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_12B_Problems/Problem_16&diff=766192016 AMC 12B Problems/Problem 162016-02-21T16:10:40Z<p>Mathmaster2012: Created page with "==Problem== In how many ways can <math>345</math> be written as the sum of an increasing sequence of two or more consecutive positive integers? <math>\textbf{(A)}\ 1\qquad\t..."</p>
<hr />
<div>==Problem==<br />
<br />
In how many ways can <math>345</math> be written as the sum of an increasing sequence of two or more consecutive positive integers?<br />
<br />
<math>\textbf{(A)}\ 1\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 7</math><br />
<br />
==Solution==<br />
<br />
==See Also==<br />
{{AMC12 box|year=2016|ab=B|num-b=15|num-a=17}}<br />
{{MAA Notice}}</div>Mathmaster2012https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_12B_Problems/Problem_15&diff=766182016 AMC 12B Problems/Problem 152016-02-21T16:09:15Z<p>Mathmaster2012: Created page with "==Problem== All the numbers <math>2, 3, 4, 5, 6, 7</math> are assigned to the six faces of a cube, one number to each face. For each of the eight vertices of the cube, a prod..."</p>
<hr />
<div>==Problem==<br />
<br />
All the numbers <math>2, 3, 4, 5, 6, 7</math> are assigned to the six faces of a cube, one number to each face. For each of the eight vertices of the cube, a product of three numbers is computed, where the three numbers are the numbers assigned to the three faces that include that vertex. What is the greatest possible value of the sum of these eight products?<br />
<br />
<math>\textbf{(A)}\ 312 \qquad<br />
\textbf{(B)}\ 343 \qquad<br />
\textbf{(C)}\ 625 \qquad<br />
\textbf{(D)}\ 729 \qquad<br />
\textbf{(E)}\ 1680</math><br />
<br />
==Solution==<br />
<br />
==See Also==<br />
{{AMC12 box|year=2016|ab=B|num-b=14|num-a=16}}<br />
{{MAA Notice}}</div>Mathmaster2012https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_12B_Problems/Problem_14&diff=766122016 AMC 12B Problems/Problem 142016-02-21T16:03:11Z<p>Mathmaster2012: Created page with "==Problem== The sum of an infinite geometric series is a positive number <math>S</math>, and the second term in the series is <math>1</math>. What is the smallest possible va..."</p>
<hr />
<div>==Problem==<br />
<br />
The sum of an infinite geometric series is a positive number <math>S</math>, and the second term in the series is <math>1</math>. What is the smallest possible value of <math>S?</math><br />
<br />
<math>\textbf{(A)}\ \frac{1+\sqrt{5}}{2} \qquad<br />
\textbf{(B)}\ 2 \qquad<br />
\textbf{(C)}\ \sqrt{5} \qquad<br />
\textbf{(D)}\ 3 \qquad<br />
\textbf{(E)}\ 4</math><br />
<br />
==Solution==<br />
<br />
==See Also==<br />
{{AMC12 box|year=2016|ab=B|num-b=13|num-a=15}}<br />
{{MAA Notice}}</div>Mathmaster2012https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_12B_Problems/Problem_13&diff=766062016 AMC 12B Problems/Problem 132016-02-21T15:59:04Z<p>Mathmaster2012: Created page with "==Problem== Alice and Bob live <math>10</math> miles apart. One day Alice looks due north from her house and sees an airplane. At the same time Bob looks due west from his ho..."</p>
<hr />
<div>==Problem==<br />
<br />
Alice and Bob live <math>10</math> miles apart. One day Alice looks due north from her house and sees an airplane. At the same time Bob looks due west from his house and sees the same airplane. The angle of elevation of the airplane is <math>30^\circ</math> from Alice's position and <math>60^\circ</math> from Bob's position. Which of the following is closest to the airplane's altitude, in miles?<br />
<br />
<math>\textbf{(A)}\ 3.5 \qquad\textbf{(B)}\ 4 \qquad\textbf{(C)}\ 4.5 \qquad\textbf{(D)}\ 5 \qquad\textbf{(E)}\ 5.5</math><br />
<br />
==Solution==<br />
<br />
==See Also==<br />
{{AMC12 box|year=2016|ab=B|num-b=12|num-a=14}}<br />
{{MAA Notice}}</div>Mathmaster2012https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_12B_Problems/Problem_12&diff=766042016 AMC 12B Problems/Problem 122016-02-21T15:57:37Z<p>Mathmaster2012: </p>
<hr />
<div>==Problem==<br />
<br />
All the numbers <math>1, 2, 3, 4, 5, 6, 7, 8, 9</math> are written in a <math>3\times3</math> array of squares, one number in each square, in such a way that if two numbers of consecutive then they occupy squares that share an edge. The numbers in the four corners add up to <math>18</math>. What is the number in the center?<br />
<br />
<math>\textbf{(A)}\ 5\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 7\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 9</math><br />
<br />
==Solution==<br />
<br />
==See Also==<br />
{{AMC12 box|year=2016|ab=B|num-b=11|num-a=13}}<br />
{{MAA Notice}}</div>Mathmaster2012https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_12B_Problems/Problem_12&diff=766032016 AMC 12B Problems/Problem 122016-02-21T15:57:11Z<p>Mathmaster2012: Created page with "==Problem== All the numbers <math>1, 2, 3, 4, 5, 6, 7, 8, 9</math> are written in a <math>3\times3</math> array of squares, one number in each square, in such a way that if t..."</p>
<hr />
<div>==Problem==<br />
<br />
All the numbers <math>1, 2, 3, 4, 5, 6, 7, 8, 9</math> are written in a <math>3\times3</math> array of squares, one number in each square, in such a way that if two numbers of consecutive then they occupy squares that share an edge. The numbers in the four corners add up to <math>18</math>. What is the number in the center?<br />
<br />
<math>\textbf{(A)}\ 5\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 7\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 9</math><br />
<br />
==Solution==<br />
<br />
==See Also==<br />
{{AMC12 box|year=2016|ab=B|num-b=7|num-a=9}}<br />
{{MAA Notice}}</div>Mathmaster2012https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_12B_Problems/Problem_11&diff=766012016 AMC 12B Problems/Problem 112016-02-21T15:56:02Z<p>Mathmaster2012: Created page with "==Problem== How many squares whose sides are parallel to the axes and whose vertices have coordinates that are integers lie entirely within the region bounded by the line <ma..."</p>
<hr />
<div>==Problem==<br />
<br />
How many squares whose sides are parallel to the axes and whose vertices have coordinates that are integers lie entirely within the region bounded by the line <math>y=\pi x</math>, the line <math>y=-0.1</math> and the line <math>x=5.1?</math><br />
<br />
<math>\textbf{(A)}\ 30 \qquad<br />
\textbf{(B)}\ 41 \qquad<br />
\textbf{(C)}\ 45 \qquad<br />
\textbf{(D)}\ 50 \qquad<br />
\textbf{(E)}\ 57</math><br />
<br />
==Solution==<br />
<br />
==See Also==<br />
{{AMC12 box|year=2016|ab=B|num-b=10|num-a=12}}<br />
{{MAA Notice}}</div>Mathmaster2012https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_12B_Problems/Problem_10&diff=766002016 AMC 12B Problems/Problem 102016-02-21T15:54:51Z<p>Mathmaster2012: Created page with "==Problem== A quadrilateral has vertices <math>P(a,b)</math>, <math>Q(b,a)</math>, <math>R(-a, -b)</math>, and <math>S(-b, -a)</math>, where <math>a</math> and <math>b</math>..."</p>
<hr />
<div>==Problem==<br />
<br />
A quadrilateral has vertices <math>P(a,b)</math>, <math>Q(b,a)</math>, <math>R(-a, -b)</math>, and <math>S(-b, -a)</math>, where <math>a</math> and <math>b</math> are integers with <math>a>b>0</math>. The area of <math>PQRS</math> is <math>16</math>. What is <math>a+b</math>?<br />
<br />
<math>\textbf{(A)}\ 4 \qquad\textbf{(B)}\ 5 \qquad\textbf{(C)}\ 6 \qquad\textbf{(D)}\ 12 \qquad\textbf{(E)}\ 13</math><br />
<br />
==Solution==<br />
<br />
==See Also==<br />
{{AMC12 box|year=2016|ab=B|num-b=9|num-a=11}}<br />
{{MAA Notice}}</div>Mathmaster2012https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_12B_Problems/Problem_9&diff=765992016 AMC 12B Problems/Problem 92016-02-21T15:53:58Z<p>Mathmaster2012: Created page with "==Problem== Carl decided to in his rectangular garden. He bought <math>20</math> fence posts, placed one on each of the four corners, and spaced out the rest evenly along the..."</p>
<hr />
<div>==Problem==<br />
<br />
Carl decided to in his rectangular garden. He bought <math>20</math> fence posts, placed one on each of the four corners, and spaced out the rest evenly along the edges of the garden, leaving exactly <math>4</math> yards between neighboring posts. The longer side of his garden, including the corners, has twice as many posts as the shorter side, including the corners. What is the area, in square yards, of Carl’s garden?<br />
<br />
<math>\textbf{(A)}\ 256\qquad\textbf{(B)}\ 336\qquad\textbf{(C)}\ 384\qquad\textbf{(D)}\ 448\qquad\textbf{(E)}\ 512</math><br />
<br />
==Solution==<br />
<br />
==See Also==<br />
{{AMC12 box|year=2016|ab=B|num-b=8|num-a=10}}<br />
{{MAA Notice}}</div>Mathmaster2012https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_12B_Problems/Problem_8&diff=765982016 AMC 12B Problems/Problem 82016-02-21T15:50:21Z<p>Mathmaster2012: Created page with "==Problem== A thin piece of wood of uniform density in the shape of an equilateral triangle with side length <math>3</math> inches weighs <math>12</math> ounces. A second pie..."</p>
<hr />
<div>==Problem==<br />
<br />
A thin piece of wood of uniform density in the shape of an equilateral triangle with side length <math>3</math> inches weighs <math>12</math> ounces. A second piece of the same type of wood, with the same thickness, also in the shape of an equilateral triangle, has side length of <math>5</math> inches. Which of the following is closest to the weight, in ounces, of the second piece?<br />
<br />
<math>\textbf{(A)}\ 14.0\qquad\textbf{(B)}\ 16.0\qquad\textbf{(C)}\ 20.0\qquad\textbf{(D)}\ 33.3\qquad\textbf{(E)}\ 55.6</math><br />
==Solution==<br />
<br />
==See Also==<br />
{{AMC12 box|year=2016|ab=B|num-b=7|num-a=9}}<br />
{{MAA Notice}}</div>Mathmaster2012https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_12B_Problems/Problem_7&diff=765962016 AMC 12B Problems/Problem 72016-02-21T15:48:32Z<p>Mathmaster2012: Created page with "==Problem== Josh writes the numbers <math>1,2,3,\dots,99,100</math>. He marks out <math>1</math>, skips the next number <math>(2)</math>, marks out <math>3</math>, and contin..."</p>
<hr />
<div>==Problem==<br />
<br />
Josh writes the numbers <math>1,2,3,\dots,99,100</math>. He marks out <math>1</math>, skips the next number <math>(2)</math>, marks out <math>3</math>, and continues skipping and marking out the next number to the end of the list. Then he goes back to the start of his list, marks out the first remaining number <math>(2)</math>, skips the next number <math>(4)</math>, marks out <math>6</math>, skips <math>8</math>, marks out <math>10</math>, and so on to the end. Josh continues in this manner until only one number remains. What is that number?<br />
<br />
<math>\textbf{(A)}\ 13 \qquad<br />
\textbf{(B)}\ 32 \qquad<br />
\textbf{(C)}\ 56 \qquad<br />
\textbf{(D)}\ 64 \qquad<br />
\textbf{(E)}\ 96</math><br />
<br />
==Solution==<br />
<br />
==See Also==<br />
{{AMC12 box|year=2016|ab=B|num-b=6|num-a=8}}<br />
{{MAA Notice}}</div>Mathmaster2012https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_12B_Problems/Problem_6&diff=765952016 AMC 12B Problems/Problem 62016-02-21T15:47:16Z<p>Mathmaster2012: Created page with "==Problem== All three vertices of <math>\bigtriangleup ABC</math> lie on the parabola defined by <math>y=x^2</math>, with <math>A</math> at the origin and <math>\overline{BC}..."</p>
<hr />
<div>==Problem==<br />
<br />
All three vertices of <math>\bigtriangleup ABC</math> lie on the parabola defined by <math>y=x^2</math>, with <math>A</math> at the origin and <math>\overline{BC}</math> parallel to the <math>x</math>-axis. The area of the triangle is <math>64</math>. What is the length of <math>BC</math>? <br />
<br />
<math>\textbf{(A)}\ 4\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 16</math><br />
<br />
==Solution==<br />
<br />
==See Also==<br />
{{AMC12 box|year=2016|ab=B|num-b=5|num-a=7}}<br />
{{MAA Notice}}</div>Mathmaster2012https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_12B_Problems/Problem_5&diff=765942016 AMC 12B Problems/Problem 52016-02-21T15:45:23Z<p>Mathmaster2012: Created page with "==Problem== The War of <math>1812</math> started with a declaration of war on Thursday, June <math>18</math>, <math>1812</math>. The peace treaty to end the war was signed <m..."</p>
<hr />
<div>==Problem==<br />
<br />
The War of <math>1812</math> started with a declaration of war on Thursday, June <math>18</math>, <math>1812</math>. The peace treaty to end the war was signed <math>919</math> days later, on December <math>24</math>, <math>1814</math>. On what day of the week was the treaty signed? <br />
<br />
<math>\textbf{(A)}\ \text{Friday} \qquad<br />
\textbf{(B)}\ \text{Saturday} \qquad<br />
\textbf{(C)}\ \text{Sunday} \qquad<br />
\textbf{(D)}\ \text{Monday} \qquad<br />
\textbf{(E)}\ \text{Tuesday} </math><br />
<br />
==Solution==<br />
<br />
==See Also==<br />
{{AMC12 box|year=2016|ab=B|num-b=4|num-a=6}}<br />
{{MAA Notice}}</div>Mathmaster2012https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_12B_Problems/Problem_4&diff=765922016 AMC 12B Problems/Problem 42016-02-21T15:44:12Z<p>Mathmaster2012: </p>
<hr />
<div>==Problem==<br />
<br />
The ratio of the measures of two acute angles is <math>5:4</math>, and the complement of one of these two angles is twice as large as the complement of the other. What is the sum of the degree measures of the two angles?<br />
<br />
<math>\textbf{(A)}\ 75\qquad\textbf{(B)}\ 90\qquad\textbf{(C)}\ 135\qquad\textbf{(D)}\ 150\qquad\textbf{(E)}\ 270</math><br />
<br />
==Solution==<br />
<br />
==See Also==<br />
{{AMC12 box|year=2016|ab=B|num-b=3|num-a=5}}<br />
{{MAA Notice}}</div>Mathmaster2012https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_12B_Problems/Problem_4&diff=765912016 AMC 12B Problems/Problem 42016-02-21T15:43:43Z<p>Mathmaster2012: Created page with "==Problem== The ratio of the measures of two acute angles is <math>5:4</math>, and the complement of one of these two angles is twice as large as the complement of the other...."</p>
<hr />
<div>==Problem==<br />
<br />
The ratio of the measures of two acute angles is <math>5:4</math>, and the complement of one of these two angles is twice as large as the complement of the other. What is the sum of the degree measures of the two angles?<br />
<br />
<math>\textbf{(A)}\ 75\qquad\textbf{(B)}\ 90\qquad\textbf{(C)}\ 135\qquad\textbf{(D)}\ 150\qquad\textbf{(E)}\ 270</math><br />
<br />
==Solution==<br />
<br />
==See Also==<br />
{{AMC12 box|year=2016|ab=B|num-b=2|num-a=4}}<br />
{{MAA Notice}}</div>Mathmaster2012https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_12B_Problems/Problem_3&diff=765902016 AMC 12B Problems/Problem 32016-02-21T15:42:52Z<p>Mathmaster2012: </p>
<hr />
<div>==Problem==<br />
<br />
Let <math>x=-2016</math>. What is the value of <math>\bigg|</math> <math>||x|-x|-|x|</math> <math>\bigg|</math> <math>-x</math>?<br />
<br />
==Solution==<br />
<br />
==See Also==<br />
{{AMC12 box|year=2016|ab=B|num-b=2|num-a=4}}<br />
{{MAA Notice}}</div>Mathmaster2012https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_12B_Problems/Problem_3&diff=765892016 AMC 12B Problems/Problem 32016-02-21T15:40:45Z<p>Mathmaster2012: Created page with "==Problem== Let <math>x=-2016</math>. What is the value of <math>\bigg|</math> <math>||x|-x|-|x|</math> <math>\bigg|</math> <math>-x</math>? ==Solution== ==See Also== {{AMC..."</p>
<hr />
<div>==Problem==<br />
<br />
Let <math>x=-2016</math>. What is the value of <math>\bigg|</math> <math>||x|-x|-|x|</math> <math>\bigg|</math> <math>-x</math>?<br />
<br />
==Solution==<br />
<br />
==See Also==<br />
{{AMC12 box|year=2016|ab=B|num-b=1|num-a=3}}<br />
{{MAA Notice}}</div>Mathmaster2012https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_12B_Problems/Problem_2&diff=765862016 AMC 12B Problems/Problem 22016-02-21T15:39:01Z<p>Mathmaster2012: /* See Also */</p>
<hr />
<div>==Problem==<br />
<br />
The harmonic mean of two numbers can be calculated as twice their product divided by their sum. The harmonic mean of <math>1</math> and <math>2016</math> is closest to which integer?<br />
<br />
<math>\textbf{(A)}\ 2 \qquad<br />
\textbf{(B)}\ 45 \qquad<br />
\textbf{(C)}\ 504 \qquad<br />
\textbf{(D)}\ 1008 \qquad<br />
\textbf{(E)}\ 2015 </math><br />
<br />
==Solution==<br />
<br />
==See Also==<br />
{{AMC12 box|year=2016|ab=B|num-b=1|num-a=3}}<br />
{{MAA Notice}}</div>Mathmaster2012https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_12B_Problems/Problem_2&diff=765842016 AMC 12B Problems/Problem 22016-02-21T15:38:28Z<p>Mathmaster2012: /* See Also */</p>
<hr />
<div>==Problem==<br />
<br />
The harmonic mean of two numbers can be calculated as twice their product divided by their sum. The harmonic mean of <math>1</math> and <math>2016</math> is closest to which integer?<br />
<br />
<math>\textbf{(A)}\ 2 \qquad<br />
\textbf{(B)}\ 45 \qquad<br />
\textbf{(C)}\ 504 \qquad<br />
\textbf{(D)}\ 1008 \qquad<br />
\textbf{(E)}\ 2015 </math><br />
<br />
==Solution==<br />
<br />
==See Also==<br />
{{AMC12 box|year=2016|ab=A|num-b=1|num-a=3}}<br />
{{MAA Notice}}</div>Mathmaster2012https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_12B_Problems/Problem_2&diff=765832016 AMC 12B Problems/Problem 22016-02-21T15:37:20Z<p>Mathmaster2012: Created page with "==Problem== The harmonic mean of two numbers can be calculated as twice their product divided by their sum. The harmonic mean of <math>1</math> and <math>2016</math> is close..."</p>
<hr />
<div>==Problem==<br />
<br />
The harmonic mean of two numbers can be calculated as twice their product divided by their sum. The harmonic mean of <math>1</math> and <math>2016</math> is closest to which integer?<br />
<br />
<math>\textbf{(A)}\ 2 \qquad<br />
\textbf{(B)}\ 45 \qquad<br />
\textbf{(C)}\ 504 \qquad<br />
\textbf{(D)}\ 1008 \qquad<br />
\textbf{(E)}\ 2015 </math><br />
<br />
==Solution==<br />
<br />
==See Also==<br />
{{AMC12 box|year=2016|ab=B|before=2|num-a=3}}<br />
{{MAA Notice}}</div>Mathmaster2012https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_12B_Problems/Problem_1&diff=765812016 AMC 12B Problems/Problem 12016-02-21T15:36:17Z<p>Mathmaster2012: </p>
<hr />
<div>==Problem==<br />
<br />
What is the value of <math>\frac{2a^{-1}+\frac{a^{-1}}{2}}{a}</math> when <math>a= \frac{1}{2}</math>?<br />
<br />
<math>\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ \frac{5}{2}\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 20</math><br />
<br />
==Solution==<br />
<br />
==See Also==<br />
{{AMC12 box|year=2016|ab=B|before=First Problem|num-a=2}}<br />
{{MAA Notice}}</div>Mathmaster2012https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_12B_Problems/Problem_1&diff=765802016 AMC 12B Problems/Problem 12016-02-21T15:35:47Z<p>Mathmaster2012: </p>
<hr />
<div>==Problem==<br />
<br />
What is the value of <math>\frac{2a^{-1}+\frac{a^{-1}}{2}}{a}</math> when <math>a= \frac{1}{2}</math>?<br />
<br />
<math>\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ \frac{5}{2}\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 20</math><br />
<br />
==Solution==<br />
<br />
==See Also==<br />
{{AMC10 box|year=2016|ab=A|before=First Problem|num-a=2}}<br />
{{AMC12 box|year=2016|ab=A|before=First Problem|num-a=2}}<br />
{{MAA Notice}}</div>Mathmaster2012https://artofproblemsolving.com/wiki/index.php?title=User:Mathmaster2012&diff=75681User:Mathmaster20122016-02-08T06:17:56Z<p>Mathmaster2012: </p>
<hr />
<div>MathStudent2002 Is Very Smart</div>Mathmaster2012https://artofproblemsolving.com/wiki/index.php?title=User:MathStudent2002&diff=74704User:MathStudent20022016-01-19T03:38:27Z<p>Mathmaster2012: </p>
<hr />
<div>i eat you</div>Mathmaster2012https://artofproblemsolving.com/wiki/index.php?title=2014_IMO_Problems/Problem_4&diff=737052014 IMO Problems/Problem 42015-12-14T05:07:55Z<p>Mathmaster2012: </p>
<hr />
<div>==Problem==<br />
Points <math>P</math> and <math>Q</math> lie on side <math>BC</math> of acute-angled <math>\triangle{ABC}</math> so that <math>\angle{PAB}=\angle{BCA}</math> and <math>\angle{CAQ}=\angle{ABC}</math>. Points <math>M</math> and <math>N</math> lie on lines <math>AP</math> and <math>AQ</math>, respectively, such that <math>P</math> is the midpoint of <math>AM</math>, and <math>Q</math> is the midpoint of <math>AN</math>. Prove that lines <math>BM</math> and <math>CN</math> intersect on the circumcircle of <math>\triangle{ABC}</math>.<br />
<br />
==Solution==<br />
<asy><br />
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */<br />
import graph; size(10.60000000000002cm); <br />
real labelscalefactor = 0.5; /* changes label-to-point distance */<br />
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pen dotstyle = black; /* point style */ <br />
real xmin = -4.740000000000007, xmax = 16.46000000000002, ymin = -7.520000000000004, ymax = 4.140000000000004; /* image dimensions */<br />
pen zzttqq = rgb(0.6000000000000006,0.2000000000000002,0.000000000000000); pen qqwuqq = rgb(0.000000000000000,0.3921568627450985,0.000000000000000); <br />
<br />
draw((1.800000000000002,3.640000000000004)--(-0.2200000000000002,-1.200000000000001)--(7.660000000000009,-1.140000000000001)--cycle, zzttqq); <br />
draw(arc((7.660000000000009,-1.140000000000001),0.6000000000000009,140.7958863822920,180.4362538499006)--(7.660000000000009,-1.140000000000001)--cycle, red); <br />
draw(arc((-0.2200000000000002,-1.200000000000001),0.6000000000000009,0.4362538499004549,67.34652063545073)--(-0.2200000000000002,-1.200000000000001)--cycle, qqwuqq); <br />
draw(arc((1.800000000000002,3.640000000000004),0.6000000000000009,-106.1141136177082,-39.20411361770813)--(1.800000000000002,3.640000000000004)--cycle, qqwuqq); <br />
draw(arc((1.800000000000002,3.640000000000004),0.6000000000000009,-112.6534793645495,-73.01347936454944)--(1.800000000000002,3.640000000000004)--cycle, red); <br />
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draw((7.660000000000009,-1.140000000000001)--(1.800000000000002,3.640000000000004), zzttqq); <br />
draw(arc((7.660000000000009,-1.140000000000001),0.6000000000000009,140.7958863822920,180.4362538499006), red); <br />
draw(arc((7.660000000000009,-1.140000000000001),0.5000000000000008,140.7958863822920,180.4362538499006), red); <br />
draw(arc((-0.2200000000000002,-1.200000000000001),0.6000000000000009,0.4362538499004549,67.34652063545073), qqwuqq); <br />
draw(arc((-0.2200000000000002,-1.200000000000001),0.5000000000000008,0.4362538499004549,67.34652063545073), qqwuqq); <br />
draw(arc((-0.2200000000000002,-1.200000000000001),0.4000000000000006,0.4362538499004549,67.34652063545073), qqwuqq); <br />
draw(arc((1.800000000000002,3.640000000000004),0.6000000000000009,-106.1141136177082,-39.20411361770813), qqwuqq); <br />
draw(arc((1.800000000000002,3.640000000000004),0.5000000000000008,-106.1141136177082,-39.20411361770813), qqwuqq); <br />
draw(arc((1.800000000000002,3.640000000000004),0.4000000000000006,-106.1141136177082,-39.20411361770813), qqwuqq); <br />
draw(arc((1.800000000000002,3.640000000000004),0.6000000000000009,-112.6534793645495,-73.01347936454944), red); <br />
draw(arc((1.800000000000002,3.640000000000004),0.5000000000000008,-112.6534793645495,-73.01347936454944), red); <br />
draw((-1.022670636276736,-6.130338243877306)--(7.660000000000009,-1.140000000000001)); <br />
draw((4.740746205921980,-5.986847110107199)--(-0.2200000000000002,-1.200000000000001)); <br />
draw((1.800000000000002,3.640000000000004)--(4.740746205921980,-5.986847110107199)); <br />
draw((1.800000000000002,3.640000000000004)--(-1.022670636276736,-6.130338243877306)); <br />
draw(circle((3.711084749329270,0.0008695880898521494), 4.110415438128883)); <br />
/* dots and labels */<br />
dot((1.800000000000002,3.640000000000004),dotstyle); <br />
label("$A$", (1.880000000000002,3.760000000000004), NE * labelscalefactor); <br />
dot((-0.2200000000000002,-1.200000000000001),dotstyle); <br />
label("$B$", (-0.1400000000000008,-1.080000000000000), NE * labelscalefactor); <br />
dot((7.660000000000009,-1.140000000000001),dotstyle); <br />
label("$C$", (7.740000000000012,-1.020000000000000), NE * labelscalefactor); <br />
dot((0.3886646818616330,-1.245169121938651),dotstyle); <br />
label("$Q$", (0.4600000000000001,-1.120000000000000), NE * labelscalefactor); <br />
dot((3.270373102960991,-1.173423555053598),dotstyle); <br />
label("$P$", (3.360000000000004,-1.060000000000000), NE * labelscalefactor); <br />
dot((4.740746205921980,-5.986847110107199),dotstyle); <br />
label("$M$", (4.820000000000006,-5.860000000000003), NE * labelscalefactor); <br />
dot((-1.022670636276736,-6.130338243877306),dotstyle); <br />
label("$N$", (-0.9400000000000020,-6.020000000000004), NE * labelscalefactor); <br />
dot((2.709057008802497,-3.985539257126989),dotstyle); <br />
label("$D$", (2.780000000000003,-3.860000000000002), NE * labelscalefactor); <br />
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); <br />
/* end of picture */<br />
</asy><br />
<br />
We are trying to prove that the intersection of BM and CN, call it point D, is on the circumcircle of triangle ABC. In other words, we are trying to prove angle BAC plus angle BDC is 180 degrees. <br />
Let the intersection of BM and AN be point E, and the intersection of AM and CN be point F. <br />
Let us assume (angle BDC) + (angle BAC) = 180. ''Note: This is circular reasoning.'' If angle BDC plus angle BAC is 180, then angle BAC should be equal to angles BDN and CDM. We can quickly prove that the triangles ABC, APB, and AQC are similar, so angles BAC = AQC = APB. We also see that angles AQC = BQN = APB = CPF. Also because angles BEQ and NED, MFD and CFP are equal, the triangles BEQ and NED, MDF and FCP must be two pairs of similar triangles. Therefore we must prove angles CBM and ANC, AMB and BCN are equal. <br />
We have angles BQA = APC = NQC = BPM. We also have AQ = QN, AP = PM. Because the triangles ABP and ACQ are similar, we have EC/EN = BF/FM, so triangles BFM and NEC are similar. So the angles CBM and ANC, BCN and AMB are equal and we are done.<br />
<br />
==Solution 2==<br />
Let <math>L</math> be the midpoint of <math>BC</math>. Easy angle chasing gives <math>\angle{AQP} = \angle{APQ} = \angle{BAC}</math>. Because <math>P</math> is the midpoint of <math>AM</math>, the cotangent rule applied on triangle <math>MBA</math> gives us<br />
<cmath>\cot \angle{MBC} - \cot \angle{ABC} = 2\cot \angle{BAC}.</cmath><br />
Hence, by the cotangent rule on <math>ABC</math>, we have<br />
<cmath>\cot \angle{BAL} = 2\cot \angle{BAC} + \cot \angle{ABC} = \cot \angle{MBC}.</cmath><br />
Because the period of cotangent is <math>180^\circ</math>, but angles are less than <math>180^\circ</math>, we have <math>\angle{BAL} = \angle{MBC}.</math><br />
<br />
Similarly, we have <math>\angle{LAC} = \angle{NCB}.</math> Hence, if <math>BM</math> and <math>CN</math> intersect at <math>Z</math>, then <math>\angle{BZC} = 180^\circ - \angle{BAC}</math> by the Angle Sum in a Triangle Theorem. Hence, <math>BACZ</math> is cyclic, which is equivalent to the desired result.<br />
<br />
--[[User:Suli|Suli]] 23:27, 7 February 2015 (EST)<br />
<br />
==Solution 3==<br />
Let <math>L</math> be the midpoint of <math>BC</math>. By AA Similarity, triangles <math>BAP</math> and <math>BCA</math> are similar, so <math>\dfrac{BA}{AP} = \dfrac{BC}{CA}</math> and <math>\angle{BPA} = \angle{BAC}</math>. Similarly, <math>\angle{CQA} = \angle{BAC}</math>, and so triangle <math>AQP</math> is isosceles. Thus, <math>AQ = AP</math>, and so <math>\dfrac{BA}{AQ} = \dfrac{BC}{CA}</math>. Dividing both sides by 2, we have <math>\dfrac{BA}{AN} = \dfrac{BL}{AC}</math>, or<br />
<cmath>\frac{BA}{BL} = \frac{AN}{AC}.</cmath><br />
But we also have <math>\angle{ABL} = \angle{CAQ}</math>, so triangles <math>ABL</math> and <math>NAC</math> are similar by <math>SAS</math> similarity. In particular, <math>\angle{ANC} = \angle{BAL}</math>. Similarly, <math>\angle{BMA} = \angle{CAL}</math>, so <math>\angle{ANC} + \angle{BMA} = \angle{BAC}</math>. In addition, angle sum in triangle <math>AQP</math> gives <math>\angle{QAP} = 180^\circ - 2\angle{A}</math>. Therefore, if we let lines <math>BM</math> and <math>CN</math> intersect at <math>T</math>, by Angle Sum in quadrilateral <math>AMTN</math> concave <math>\angle{NTM} = 180^\circ + \angle{A}</math>, and so convex <math>\angle{BTC} = 180^\circ - \angle{A}</math>, which is enough to prove that <math>BACT</math> is cyclic. This completes the proof.<br />
<br />
--[[User:Suli|Suli]] 10:38, 8 February 2015 (EST)<br />
<br />
{{alternate solutions}}<br />
<br />
==See Also==<br />
<br />
{{IMO box|year=2014|num-b=3|num-a=5}}<br />
<br />
<br />
[[Category:Olympiad Geometry Problems]]</div>Mathmaster2012https://artofproblemsolving.com/wiki/index.php?title=2014_IMO_Problems/Problem_4&diff=737042014 IMO Problems/Problem 42015-12-14T05:05:53Z<p>Mathmaster2012: </p>
<hr />
<div>==Problem==<br />
Points <math>P</math> and <math>Q</math> lie on side <math>BC</math> of acute-angled <math>\triangle{ABC}</math> so that <math>\angle{PAB}=\angle{BCA}</math> and <math>\angle{CAQ}=\angle{ABC}</math>. Points <math>M</math> and <math>N</math> lie on lines <math>AP</math> and <math>AQ</math>, respectively, such that <math>P</math> is the midpoint of <math>AM</math>, and <math>Q</math> is the midpoint of <math>AN</math>. Prove that lines <math>BM</math> and <math>CN</math> intersect on the circumcircle of <math>\triangle{ABC}</math>.<br />
<br />
==Solution==<br />
<math><br />
\begin{asy}<br />
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */<br />
import graph; size(10.60000000000002cm); <br />
real labelscalefactor = 0.5; /* changes label-to-point distance */<br />
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ <br />
pen dotstyle = black; /* point style */ <br />
real xmin = -4.740000000000007, xmax = 16.46000000000002, ymin = -7.520000000000004, ymax = 4.140000000000004; /* image dimensions */<br />
pen zzttqq = rgb(0.6000000000000006,0.2000000000000002,0.000000000000000); pen qqwuqq = rgb(0.000000000000000,0.3921568627450985,0.000000000000000); <br />
<br />
draw((1.800000000000002,3.640000000000004)--(-0.2200000000000002,-1.200000000000001)--(7.660000000000009,-1.140000000000001)--cycle, zzttqq); <br />
draw(arc((7.660000000000009,-1.140000000000001),0.6000000000000009,140.7958863822920,180.4362538499006)--(7.660000000000009,-1.140000000000001)--cycle, red); <br />
draw(arc((-0.2200000000000002,-1.200000000000001),0.6000000000000009,0.4362538499004549,67.34652063545073)--(-0.2200000000000002,-1.200000000000001)--cycle, qqwuqq); <br />
draw(arc((1.800000000000002,3.640000000000004),0.6000000000000009,-106.1141136177082,-39.20411361770813)--(1.800000000000002,3.640000000000004)--cycle, qqwuqq); <br />
draw(arc((1.800000000000002,3.640000000000004),0.6000000000000009,-112.6534793645495,-73.01347936454944)--(1.800000000000002,3.640000000000004)--cycle, red); <br />
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draw((1.800000000000002,3.640000000000004)--(-0.2200000000000002,-1.200000000000001), zzttqq); <br />
draw((-0.2200000000000002,-1.200000000000001)--(7.660000000000009,-1.140000000000001), zzttqq); <br />
draw((7.660000000000009,-1.140000000000001)--(1.800000000000002,3.640000000000004), zzttqq); <br />
draw(arc((7.660000000000009,-1.140000000000001),0.6000000000000009,140.7958863822920,180.4362538499006), red); <br />
draw(arc((7.660000000000009,-1.140000000000001),0.5000000000000008,140.7958863822920,180.4362538499006), red); <br />
draw(arc((-0.2200000000000002,-1.200000000000001),0.6000000000000009,0.4362538499004549,67.34652063545073), qqwuqq); <br />
draw(arc((-0.2200000000000002,-1.200000000000001),0.5000000000000008,0.4362538499004549,67.34652063545073), qqwuqq); <br />
draw(arc((-0.2200000000000002,-1.200000000000001),0.4000000000000006,0.4362538499004549,67.34652063545073), qqwuqq); <br />
draw(arc((1.800000000000002,3.640000000000004),0.6000000000000009,-106.1141136177082,-39.20411361770813), qqwuqq); <br />
draw(arc((1.800000000000002,3.640000000000004),0.5000000000000008,-106.1141136177082,-39.20411361770813), qqwuqq); <br />
draw(arc((1.800000000000002,3.640000000000004),0.4000000000000006,-106.1141136177082,-39.20411361770813), qqwuqq); <br />
draw(arc((1.800000000000002,3.640000000000004),0.6000000000000009,-112.6534793645495,-73.01347936454944), red); <br />
draw(arc((1.800000000000002,3.640000000000004),0.5000000000000008,-112.6534793645495,-73.01347936454944), red); <br />
draw((-1.022670636276736,-6.130338243877306)--(7.660000000000009,-1.140000000000001)); <br />
draw((4.740746205921980,-5.986847110107199)--(-0.2200000000000002,-1.200000000000001)); <br />
draw((1.800000000000002,3.640000000000004)--(4.740746205921980,-5.986847110107199)); <br />
draw((1.800000000000002,3.640000000000004)--(-1.022670636276736,-6.130338243877306)); <br />
draw(circle((3.711084749329270,0.0008695880898521494), 4.110415438128883)); <br />
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dot((1.800000000000002,3.640000000000004),dotstyle); <br />
label("</math>A<math>", (1.880000000000002,3.760000000000004), NE * labelscalefactor); <br />
dot((-0.2200000000000002,-1.200000000000001),dotstyle); <br />
label("</math>B<math>", (-0.1400000000000008,-1.080000000000000), NE * labelscalefactor); <br />
dot((7.660000000000009,-1.140000000000001),dotstyle); <br />
label("</math>C<math>", (7.740000000000012,-1.020000000000000), NE * labelscalefactor); <br />
dot((0.3886646818616330,-1.245169121938651),dotstyle); <br />
label("</math>Q<math>", (0.4600000000000001,-1.120000000000000), NE * labelscalefactor); <br />
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\end{asy}<br />
</math><br />
<br />
We are trying to prove that the intersection of BM and CN, call it point D, is on the circumcircle of triangle ABC. In other words, we are trying to prove angle BAC plus angle BDC is 180 degrees. <br />
Let the intersection of BM and AN be point E, and the intersection of AM and CN be point F. <br />
Let us assume (angle BDC) + (angle BAC) = 180. ''Note: This is circular reasoning.'' If angle BDC plus angle BAC is 180, then angle BAC should be equal to angles BDN and CDM. We can quickly prove that the triangles ABC, APB, and AQC are similar, so angles BAC = AQC = APB. We also see that angles AQC = BQN = APB = CPF. Also because angles BEQ and NED, MFD and CFP are equal, the triangles BEQ and NED, MDF and FCP must be two pairs of similar triangles. Therefore we must prove angles CBM and ANC, AMB and BCN are equal. <br />
We have angles BQA = APC = NQC = BPM. We also have AQ = QN, AP = PM. Because the triangles ABP and ACQ are similar, we have EC/EN = BF/FM, so triangles BFM and NEC are similar. So the angles CBM and ANC, BCN and AMB are equal and we are done.<br />
<br />
==Solution 2==<br />
Let <math>L</math> be the midpoint of <math>BC</math>. Easy angle chasing gives <math>\angle{AQP} = \angle{APQ} = \angle{BAC}</math>. Because <math>P</math> is the midpoint of <math>AM</math>, the cotangent rule applied on triangle <math>MBA</math> gives us<br />
<cmath>\cot \angle{MBC} - \cot \angle{ABC} = 2\cot \angle{BAC}.</cmath><br />
Hence, by the cotangent rule on <math>ABC</math>, we have<br />
<cmath>\cot \angle{BAL} = 2\cot \angle{BAC} + \cot \angle{ABC} = \cot \angle{MBC}.</cmath><br />
Because the period of cotangent is <math>180^\circ</math>, but angles are less than <math>180^\circ</math>, we have <math>\angle{BAL} = \angle{MBC}.</math><br />
<br />
Similarly, we have <math>\angle{LAC} = \angle{NCB}.</math> Hence, if <math>BM</math> and <math>CN</math> intersect at <math>Z</math>, then <math>\angle{BZC} = 180^\circ - \angle{BAC}</math> by the Angle Sum in a Triangle Theorem. Hence, <math>BACZ</math> is cyclic, which is equivalent to the desired result.<br />
<br />
--[[User:Suli|Suli]] 23:27, 7 February 2015 (EST)<br />
<br />
==Solution 3==<br />
Let <math>L</math> be the midpoint of <math>BC</math>. By AA Similarity, triangles <math>BAP</math> and <math>BCA</math> are similar, so <math>\dfrac{BA}{AP} = \dfrac{BC}{CA}</math> and <math>\angle{BPA} = \angle{BAC}</math>. Similarly, <math>\angle{CQA} = \angle{BAC}</math>, and so triangle <math>AQP</math> is isosceles. Thus, <math>AQ = AP</math>, and so <math>\dfrac{BA}{AQ} = \dfrac{BC}{CA}</math>. Dividing both sides by 2, we have <math>\dfrac{BA}{AN} = \dfrac{BL}{AC}</math>, or<br />
<cmath>\frac{BA}{BL} = \frac{AN}{AC}.</cmath><br />
But we also have <math>\angle{ABL} = \angle{CAQ}</math>, so triangles <math>ABL</math> and <math>NAC</math> are similar by <math>SAS</math> similarity. In particular, <math>\angle{ANC} = \angle{BAL}</math>. Similarly, <math>\angle{BMA} = \angle{CAL}</math>, so <math>\angle{ANC} + \angle{BMA} = \angle{BAC}</math>. In addition, angle sum in triangle <math>AQP</math> gives <math>\angle{QAP} = 180^\circ - 2\angle{A}</math>. Therefore, if we let lines <math>BM</math> and <math>CN</math> intersect at <math>T</math>, by Angle Sum in quadrilateral <math>AMTN</math> concave <math>\angle{NTM} = 180^\circ + \angle{A}</math>, and so convex <math>\angle{BTC} = 180^\circ - \angle{A}</math>, which is enough to prove that <math>BACT</math> is cyclic. This completes the proof.<br />
<br />
--[[User:Suli|Suli]] 10:38, 8 February 2015 (EST)<br />
<br />
{{alternate solutions}}<br />
<br />
==See Also==<br />
<br />
{{IMO box|year=2014|num-b=3|num-a=5}}<br />
<br />
<br />
[[Category:Olympiad Geometry Problems]]</div>Mathmaster2012https://artofproblemsolving.com/wiki/index.php?title=2014_IMO_Problems/Problem_4&diff=737032014 IMO Problems/Problem 42015-12-14T05:05:15Z<p>Mathmaster2012: </p>
<hr />
<div>==Problem==<br />
Points <math>P</math> and <math>Q</math> lie on side <math>BC</math> of acute-angled <math>\triangle{ABC}</math> so that <math>\angle{PAB}=\angle{BCA}</math> and <math>\angle{CAQ}=\angle{ABC}</math>. Points <math>M</math> and <math>N</math> lie on lines <math>AP</math> and <math>AQ</math>, respectively, such that <math>P</math> is the midpoint of <math>AM</math>, and <math>Q</math> is the midpoint of <math>AN</math>. Prove that lines <math>BM</math> and <math>CN</math> intersect on the circumcircle of <math>\triangle{ABC}</math>.<br />
<br />
==Solution==<br />
<br />
\begin{asy}<br />
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\end{asy}<br />
<br />
We are trying to prove that the intersection of BM and CN, call it point D, is on the circumcircle of triangle ABC. In other words, we are trying to prove angle BAC plus angle BDC is 180 degrees. <br />
Let the intersection of BM and AN be point E, and the intersection of AM and CN be point F. <br />
Let us assume (angle BDC) + (angle BAC) = 180. ''Note: This is circular reasoning.'' If angle BDC plus angle BAC is 180, then angle BAC should be equal to angles BDN and CDM. We can quickly prove that the triangles ABC, APB, and AQC are similar, so angles BAC = AQC = APB. We also see that angles AQC = BQN = APB = CPF. Also because angles BEQ and NED, MFD and CFP are equal, the triangles BEQ and NED, MDF and FCP must be two pairs of similar triangles. Therefore we must prove angles CBM and ANC, AMB and BCN are equal. <br />
We have angles BQA = APC = NQC = BPM. We also have AQ = QN, AP = PM. Because the triangles ABP and ACQ are similar, we have EC/EN = BF/FM, so triangles BFM and NEC are similar. So the angles CBM and ANC, BCN and AMB are equal and we are done.<br />
<br />
==Solution 2==<br />
Let <math>L</math> be the midpoint of <math>BC</math>. Easy angle chasing gives <math>\angle{AQP} = \angle{APQ} = \angle{BAC}</math>. Because <math>P</math> is the midpoint of <math>AM</math>, the cotangent rule applied on triangle <math>MBA</math> gives us<br />
<cmath>\cot \angle{MBC} - \cot \angle{ABC} = 2\cot \angle{BAC}.</cmath><br />
Hence, by the cotangent rule on <math>ABC</math>, we have<br />
<cmath>\cot \angle{BAL} = 2\cot \angle{BAC} + \cot \angle{ABC} = \cot \angle{MBC}.</cmath><br />
Because the period of cotangent is <math>180^\circ</math>, but angles are less than <math>180^\circ</math>, we have <math>\angle{BAL} = \angle{MBC}.</math><br />
<br />
Similarly, we have <math>\angle{LAC} = \angle{NCB}.</math> Hence, if <math>BM</math> and <math>CN</math> intersect at <math>Z</math>, then <math>\angle{BZC} = 180^\circ - \angle{BAC}</math> by the Angle Sum in a Triangle Theorem. Hence, <math>BACZ</math> is cyclic, which is equivalent to the desired result.<br />
<br />
--[[User:Suli|Suli]] 23:27, 7 February 2015 (EST)<br />
<br />
==Solution 3==<br />
Let <math>L</math> be the midpoint of <math>BC</math>. By AA Similarity, triangles <math>BAP</math> and <math>BCA</math> are similar, so <math>\dfrac{BA}{AP} = \dfrac{BC}{CA}</math> and <math>\angle{BPA} = \angle{BAC}</math>. Similarly, <math>\angle{CQA} = \angle{BAC}</math>, and so triangle <math>AQP</math> is isosceles. Thus, <math>AQ = AP</math>, and so <math>\dfrac{BA}{AQ} = \dfrac{BC}{CA}</math>. Dividing both sides by 2, we have <math>\dfrac{BA}{AN} = \dfrac{BL}{AC}</math>, or<br />
<cmath>\frac{BA}{BL} = \frac{AN}{AC}.</cmath><br />
But we also have <math>\angle{ABL} = \angle{CAQ}</math>, so triangles <math>ABL</math> and <math>NAC</math> are similar by <math>SAS</math> similarity. In particular, <math>\angle{ANC} = \angle{BAL}</math>. Similarly, <math>\angle{BMA} = \angle{CAL}</math>, so <math>\angle{ANC} + \angle{BMA} = \angle{BAC}</math>. In addition, angle sum in triangle <math>AQP</math> gives <math>\angle{QAP} = 180^\circ - 2\angle{A}</math>. Therefore, if we let lines <math>BM</math> and <math>CN</math> intersect at <math>T</math>, by Angle Sum in quadrilateral <math>AMTN</math> concave <math>\angle{NTM} = 180^\circ + \angle{A}</math>, and so convex <math>\angle{BTC} = 180^\circ - \angle{A}</math>, which is enough to prove that <math>BACT</math> is cyclic. This completes the proof.<br />
<br />
--[[User:Suli|Suli]] 10:38, 8 February 2015 (EST)<br />
<br />
{{alternate solutions}}<br />
<br />
==See Also==<br />
<br />
{{IMO box|year=2014|num-b=3|num-a=5}}<br />
<br />
<br />
[[Category:Olympiad Geometry Problems]]</div>Mathmaster2012https://artofproblemsolving.com/wiki/index.php?title=2014_IMO_Problems/Problem_4&diff=737022014 IMO Problems/Problem 42015-12-14T05:04:47Z<p>Mathmaster2012: </p>
<hr />
<div>==Problem==<br />
Points <math>P</math> and <math>Q</math> lie on side <math>BC</math> of acute-angled <math>\triangle{ABC}</math> so that <math>\angle{PAB}=\angle{BCA}</math> and <math>\angle{CAQ}=\angle{ABC}</math>. Points <math>M</math> and <math>N</math> lie on lines <math>AP</math> and <math>AQ</math>, respectively, such that <math>P</math> is the midpoint of <math>AM</math>, and <math>Q</math> is the midpoint of <math>AN</math>. Prove that lines <math>BM</math> and <math>CN</math> intersect on the circumcircle of <math>\triangle{ABC}</math>.<br />
<br />
==Solution==<br />
<br />
[asy]<br />
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[/asy]<br />
<br />
We are trying to prove that the intersection of BM and CN, call it point D, is on the circumcircle of triangle ABC. In other words, we are trying to prove angle BAC plus angle BDC is 180 degrees. <br />
Let the intersection of BM and AN be point E, and the intersection of AM and CN be point F. <br />
Let us assume (angle BDC) + (angle BAC) = 180. ''Note: This is circular reasoning.'' If angle BDC plus angle BAC is 180, then angle BAC should be equal to angles BDN and CDM. We can quickly prove that the triangles ABC, APB, and AQC are similar, so angles BAC = AQC = APB. We also see that angles AQC = BQN = APB = CPF. Also because angles BEQ and NED, MFD and CFP are equal, the triangles BEQ and NED, MDF and FCP must be two pairs of similar triangles. Therefore we must prove angles CBM and ANC, AMB and BCN are equal. <br />
We have angles BQA = APC = NQC = BPM. We also have AQ = QN, AP = PM. Because the triangles ABP and ACQ are similar, we have EC/EN = BF/FM, so triangles BFM and NEC are similar. So the angles CBM and ANC, BCN and AMB are equal and we are done.<br />
<br />
==Solution 2==<br />
Let <math>L</math> be the midpoint of <math>BC</math>. Easy angle chasing gives <math>\angle{AQP} = \angle{APQ} = \angle{BAC}</math>. Because <math>P</math> is the midpoint of <math>AM</math>, the cotangent rule applied on triangle <math>MBA</math> gives us<br />
<cmath>\cot \angle{MBC} - \cot \angle{ABC} = 2\cot \angle{BAC}.</cmath><br />
Hence, by the cotangent rule on <math>ABC</math>, we have<br />
<cmath>\cot \angle{BAL} = 2\cot \angle{BAC} + \cot \angle{ABC} = \cot \angle{MBC}.</cmath><br />
Because the period of cotangent is <math>180^\circ</math>, but angles are less than <math>180^\circ</math>, we have <math>\angle{BAL} = \angle{MBC}.</math><br />
<br />
Similarly, we have <math>\angle{LAC} = \angle{NCB}.</math> Hence, if <math>BM</math> and <math>CN</math> intersect at <math>Z</math>, then <math>\angle{BZC} = 180^\circ - \angle{BAC}</math> by the Angle Sum in a Triangle Theorem. Hence, <math>BACZ</math> is cyclic, which is equivalent to the desired result.<br />
<br />
--[[User:Suli|Suli]] 23:27, 7 February 2015 (EST)<br />
<br />
==Solution 3==<br />
Let <math>L</math> be the midpoint of <math>BC</math>. By AA Similarity, triangles <math>BAP</math> and <math>BCA</math> are similar, so <math>\dfrac{BA}{AP} = \dfrac{BC}{CA}</math> and <math>\angle{BPA} = \angle{BAC}</math>. Similarly, <math>\angle{CQA} = \angle{BAC}</math>, and so triangle <math>AQP</math> is isosceles. Thus, <math>AQ = AP</math>, and so <math>\dfrac{BA}{AQ} = \dfrac{BC}{CA}</math>. Dividing both sides by 2, we have <math>\dfrac{BA}{AN} = \dfrac{BL}{AC}</math>, or<br />
<cmath>\frac{BA}{BL} = \frac{AN}{AC}.</cmath><br />
But we also have <math>\angle{ABL} = \angle{CAQ}</math>, so triangles <math>ABL</math> and <math>NAC</math> are similar by <math>SAS</math> similarity. In particular, <math>\angle{ANC} = \angle{BAL}</math>. Similarly, <math>\angle{BMA} = \angle{CAL}</math>, so <math>\angle{ANC} + \angle{BMA} = \angle{BAC}</math>. In addition, angle sum in triangle <math>AQP</math> gives <math>\angle{QAP} = 180^\circ - 2\angle{A}</math>. Therefore, if we let lines <math>BM</math> and <math>CN</math> intersect at <math>T</math>, by Angle Sum in quadrilateral <math>AMTN</math> concave <math>\angle{NTM} = 180^\circ + \angle{A}</math>, and so convex <math>\angle{BTC} = 180^\circ - \angle{A}</math>, which is enough to prove that <math>BACT</math> is cyclic. This completes the proof.<br />
<br />
--[[User:Suli|Suli]] 10:38, 8 February 2015 (EST)<br />
<br />
{{alternate solutions}}<br />
<br />
==See Also==<br />
<br />
{{IMO box|year=2014|num-b=3|num-a=5}}<br />
<br />
<br />
[[Category:Olympiad Geometry Problems]]</div>Mathmaster2012