https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=WolfOfAtlantis&feedformat=atomAoPS Wiki - User contributions [en]2024-03-28T12:18:41ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_12B_Problems&diff=631012011 AMC 12B Problems2014-08-16T02:28:16Z<p>WolfOfAtlantis: Fixing the link to Answer Key.</p>
<hr />
<div>{{AMC12 Problems|year=2011|ab=B}}<br />
<br />
==Problem 1==<br />
What is <center><math> \frac{2+4+6}{1+3+5}-\frac{1+3+5}{2+4+6}? </math></center><br />
<br />
<br />
<math>\textbf{(A)}\ -1 \qquad \textbf{(B)}\ \frac{5}{36} \qquad \textbf{(C)}\ \frac{7}{12} \qquad \textbf{(D)}\ \frac{147}{60} \qquad \textbf{(E)}\ \frac{43}{3}</math><br />
<br />
[[2011 AMC 12B Problems/Problem 1|Solution]]<br />
<br />
==Problem 2==<br />
Josanna's test scores to date are <math>90</math>, <math>80</math>, <math>70</math>, <math>60</math>, and <math>85</math>. Her goal is to raise her test average at least <math>3</math> points with her next test. What is the minimum test score she would need to accomplish this goal?<br />
<br />
<math>\textbf{(A)}\ 80 \qquad \textbf{(B)}\ 82 \qquad \textbf{(C)}\ 85 \qquad \textbf{(D)}\ 90 \qquad \textbf{(E)}\ 95</math><br />
<br />
[[2011 AMC 12B Problems/Problem 2|Solution]]<br />
<br />
==Problem 3==<br />
LeRoy and Bernardo went on a week-long trip together and agreed to share the costs equally. Over the week, each of them paid for various joint expenses such as gasoline and car rental. At the end of the trip it turned out that LeRoy had paid <math>A</math> dollars and Bernardo had paid <math>B</math> dollars, where <math>A<B</math>. How many dollars must LeRoy give to Bernardo so that they share the costs equally?<br />
<br />
<math>\textbf{(A)}\ \frac{A+B}{2} \qquad \textbf{(B)}\ \frac{A-B}{2} \qquad \textbf{(C)}\ \frac{B-A}{2} \qquad \textbf{(D)}\ B-A \qquad \textbf{(E)}\ A+B</math><br />
<br />
[[2011 AMC 12B Problems/Problem 3|Solution]]<br />
<br />
==Problem 4==<br />
In multiplying two positive integers <math>a</math> and <math>b</math>, Ron reversed the digits of the two-digit number <math>a</math>. His erroneous product was 161. What is the correct value of the product of <math>a</math> and <math>b</math>?<br />
<br />
<math>\textbf{(A)}\ 116 \qquad \textbf{(B)}\ 161 \qquad \textbf{(C)}\ 204 \qquad \textbf{(D)}\ 214 \qquad \textbf{(E)}\ 224</math><br />
<br />
[[2011 AMC 12B Problems/Problem 4|Solution]]<br />
<br />
==Problem 5==<br />
Let <math>N</math> be the second smallest positive integer that is divisible by every positive integer less than <math>7</math>. What is the sum of the digits of <math>N</math>?<br />
<br />
<math>\textbf{(A)}\ 3 \qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\ 5 \qquad \textbf{(D)}\ 6 \qquad \textbf{(E)}\ 9</math><br />
<br />
[[2011 AMC 12B Problems/Problem 5|Solution]]<br />
<br />
==Problem 6==<br />
Two tangents to a circle are drawn from a point <math>A</math>. The points of contact <math>B</math> and <math>C</math> divide the circle into arcs with lengths in the ratio <math>2 : 3</math>. What is the degree measure of <math>\angle{BAC}</math>?<br />
<br />
<math>\textbf{(A)}\ 24 \qquad \textbf{(B)}\ 30 \qquad \textbf{(C)}\ 36 \qquad \textbf{(D)}\ 48 \qquad \textbf{(E)}\ 60</math><br />
<br />
[[2011 AMC 12B Problems/Problem 6|Solution]]<br />
<br />
==Problem 7==<br />
Let <math>x</math> and <math>y</math> be two-digit positive integers with mean <math>60</math>. What is the maximum value of the ratio <math>\frac{x}{y}</math>?<br />
<br />
<math>\textbf{(A)}\ 3 \qquad \textbf{(B)}\ \frac{33}{7} \qquad \textbf{(C)}\ \frac{39}{7} \qquad \textbf{(D)}\ 9 \qquad \textbf{(E)}\ \frac{99}{10}</math><br />
<br />
[[2011 AMC 12B Problems/Problem 7|Solution]]<br />
<br />
==Problem 8==<br />
Keiko walks once around a track at exactly the same constant speed every day. The sides of the track are straight, and the ends are semicircles. The track has width <math>6</math> meters, and it takes her <math>36</math> seconds longer to walk around the outside edge of the track than around the inside edge. What is Keiko's speed in meters per second?<br />
<br />
<math>\textbf{(A)}\ \frac{\pi}{3} \qquad \textbf{(B)}\ \frac{2\pi}{3} \qquad \textbf{(C)}\ \pi \qquad \textbf{(D)}\ \frac{4\pi}{3} \qquad \textbf{(E)}\ \frac{5\pi}{3}</math><br />
<br />
[[2011 AMC 12B Problems/Problem 8|Solution]]<br />
<br />
==Problem 9==<br />
Two real numbers are selected independently and at random from the interval <math>[-20,10]</math>. What is the probability that the product of those numbers is greater than zero?<br />
<br />
<math>\textbf{(A)}\ \frac{1}{9} \qquad \textbf{(B)}\ \frac{1}{3} \qquad \textbf{(C)}\ \frac{4}{9} \qquad \textbf{(D)}\ \frac{5}{9} \qquad \textbf{(E)}\ \frac{2}{3}</math><br />
<br />
[[2011 AMC 12B Problems/Problem 9|Solution]]<br />
<br />
==Problem 10==<br />
Rectangle <math>ABCD</math> has <math>AB=6</math> and <math>BC=3</math>. Point <math>M</math> is chosen on side <math>AB</math> so that <math>\angle AMD=\angle CMD</math>. What is the degree measure of <math>\angle AMD</math>?<br />
<br />
<math>\textbf{(A)}\ 15 \qquad \textbf{(B)}\ 30 \qquad \textbf{(C)}\ 45 \qquad \textbf{(D)}\ 60 \qquad \textbf{(E)}\ 75</math><br />
<br />
[[2011 AMC 12B Problems/Problem 10|Solution]]<br />
<br />
==Problem 11==<br />
A frog located at <math>(x,y)</math>, with both <math>x</math> and <math>y</math> integers, makes successive jumps of length <math>5</math> and always lands on points with integer coordinates. Suppose that the frog starts at <math>(0,0)</math> and ends at <math>(1,0)</math>. What is the smallest possible number of jumps the frog makes?<br />
<br />
<math>\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 3 \qquad \textbf{(C)}\ 4 \qquad \textbf{(D)}\ 5 \qquad \textbf{(E)}\ 6</math><br />
<br />
[[2011 AMC 12B Problems/Problem 11|Solution]]<br />
<br />
==Problem 12==<br />
<br />
A dart board is a regular octagon divided into regions as shown below. Suppose that a dart thrown at the board is equally likely to land anywhere on the board. What is the probability that the dart lands within the center square?<br />
<br />
<asy><br />
unitsize(10mm);<br />
defaultpen(linewidth(.8pt)+fontsize(10pt));<br />
dotfactor=4;<br />
pair A=(0,1), B=(1,0), C=(1+sqrt(2),0), D=(2+sqrt(2),1), E=(2+sqrt(2),1+sqrt(2)), F=(1+sqrt(2),2+sqrt(2)), G=(1,2+sqrt(2)), H=(0,1+sqrt(2));<br />
draw(A--B--C--D--E--F--G--H--cycle);<br />
draw(A--D);<br />
draw(B--G);<br />
draw(C--F);<br />
draw(E--H);</asy><br />
<br />
<math>\textbf{(A)}\ \frac{\sqrt{2} - 1}{2} \qquad \textbf{(B)}\ \frac{1}{4} \qquad \textbf{(C)}\ \frac{2 - \sqrt{2}}{2} \qquad \textbf{(D)}\ \frac{\sqrt{2}}{4} \qquad \textbf{(E)}\ 2 - \sqrt{2}</math><br />
<br />
[[2011 AMC 12B Problems/Problem 12|Solution]]<br />
<br />
==Problem 13==<br />
<br />
Brian writes down four integers <math>w > x > y > z</math> whose sum is <math>44</math>. The pairwise positive differences of these numbers are <math>1, 3, 4, 5, 6</math> and <math>9</math>. What is the sum of the possible values of <math>w</math>?<br />
<br />
<math>\textbf{(A)}\ 16 \qquad \textbf{(B)}\ 31 \qquad \textbf{(C)}\ 48 \qquad \textbf{(D)}\ 62 \qquad \textbf{(E)}\ 93</math><br />
<br />
[[2011 AMC 12B Problems/Problem 13|Solution]]<br />
<br />
==Problem 14==<br />
<br />
A segment through the focus <math>F</math> of a parabola with vertex <math>V</math> is perpendicular to <math>\overline{FV}</math> and intersects the parabola in points <math>A</math> and <math>B</math>. What is <math>\cos\left(\angle AVB\right)</math>?<br />
<br />
<math>\textbf{(A)}\ -\frac{3\sqrt{5}}{7} \qquad \textbf{(B)}\ -\frac{2\sqrt{5}}{5} \qquad \textbf{(C)}\ -\frac{4}{5} \qquad \textbf{(D)}\ -\frac{3}{5} \qquad \textbf{(E)}\ -\frac{1}{2}</math><br />
<br />
[[2011 AMC 12B Problems/Problem 14|Solution]]<br />
<br />
==Problem 15==<br />
<br />
How many positive two-digit integers are factors of <math>2^{24}-1</math>?<br />
<br />
<math>\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 8 \qquad \textbf{(C)}\ 10 \qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ 14</math><br />
<br />
[[2011 AMC 12B Problems/Problem 15|Solution]]<br />
<br />
==Problem 16==<br />
<br />
Rhombus <math>ABCD</math> has side length <math>2</math> and <math>\angle B = 120^{\circ}</math>. Region <math>R</math> consists of all points inside of the rhombus that are closer to vertex <math>B</math> than any of the other three vertices. What is the area of <math>R</math>?<br />
<br />
<math>\textbf{(A)}\ \frac{\sqrt{3}}{3} \qquad \textbf{(B)}\ \frac{\sqrt{3}}{2} \qquad \textbf{(C)}\ \frac{2\sqrt{3}}{3} \qquad \textbf{(D)}\ 1 + \frac{\sqrt{3}}{3} \qquad \textbf{(E)}\ 2</math><br />
<br />
[[2011 AMC 12B Problems/Problem 16|Solution]]<br />
<br />
==Problem 17==<br />
<br />
Let <math>f(x) = 10^{10x}, g(x) = \log_{10}\left(\frac{x}{10}\right), h_1(x) = g(f(x))</math>, and <math>h_n(x) = h_1(h_{n-1}(x))</math> for integers <math>n \geq 2</math>. What is the sum of the digits of <math>h_{2011}(1)</math>?<br />
<br />
<math>\textbf{(A)}\ 16081 \qquad \textbf{(B)}\ 16089 \qquad \textbf{(C)}\ 18089 \qquad \textbf{(D)}\ 18098 \qquad \textbf{(E)}\ 18099</math><br />
<br />
[[2011 AMC 12B Problems/Problem 17|Solution]]<br />
<br />
==Problem 18==<br />
<br />
A pyramid has a square base with side of length 1 and has lateral faces that are equilateral triangles. A cube is placed within the pyramid so that one face is on the base of the pyramid and its opposite face has all its edges on the lateral faces of the pyramid. What is the volume of this cube?<br />
<br />
<math>\textbf{(A)}\ 5\sqrt{2} - 7 \qquad \textbf{(B)}\ 7 - 4\sqrt{3} \qquad \textbf{(C)}\ \frac{2\sqrt{2}}{27} \qquad \textbf{(D)}\ \frac{\sqrt{2}}{9} \qquad \textbf{(E)}\ \frac{\sqrt{3}}{9}</math><br />
<br />
[[2011 AMC 12B Problems/Problem 18|Solution]]<br />
<br />
==Problem 19==<br />
<br />
A lattice point in an <math>xy</math>-coordinate system is any point <math>(x, y)</math> where both <math>x</math> and <math>y</math> are integers. The graph of <math>y = mx + 2</math> passes through no lattice point with <math>0 < x \leq 100</math> for all <math>m</math> such that <math>\frac{1}{2} < m < a</math>. What is the maximum possible value of <math>a</math>?<br />
<br />
<math>\textbf{(A)}\ \frac{51}{101} \qquad \textbf{(B)}\ \frac{50}{99} \qquad \textbf{(C)}\ \frac{51}{100} \qquad \textbf{(D)}\ \frac{52}{101} \qquad \textbf{(E)}\ \frac{13}{25}</math><br />
<br />
[[2011 AMC 12B Problems/Problem 19|Solution]]<br />
<br />
==Problem 20==<br />
<br />
Triangle <math>ABC</math> has <math>AB = 13, BC = 14</math>, and <math>AC = 15</math>. The points <math>D, E</math>, and <math>F</math> are the midpoints of <math>\overline{AB}, \overline{BC}</math>, and <math>\overline{AC}</math> respectively. Let <math>X \not= E</math> be the intersection of the circumcircles of <math>\Delta BDE</math> and <math>\Delta CEF</math>. What is <math>XA + XB + XC</math>?<br />
<br />
<math>\textbf{(A)}\ 24 \qquad \textbf{(B)}\ 14\sqrt{3} \qquad \textbf{(C)}\ \frac{195}{8} \qquad \textbf{(D)}\ \frac{129\sqrt{7}}{14} \qquad \textbf{(E)}\ \frac{69\sqrt{2}}{4}</math><br />
<br />
[[2011 AMC 12B Problems/Problem 20|Solution]]<br />
<br />
==Problem 21==<br />
<br />
The arithmetic mean of two distinct positive integers <math>x</math> and <math>y</math> is a two-digit integer. The geometric mean of <math>x</math> and <math>y</math> is obtained by reversing the digits of the arithmetic mean. What is <math>|x - y|</math>?<br />
<br />
<math>\textbf{(A)}\ 24 \qquad \textbf{(B)}\ 48 \qquad \textbf{(C)}\ 54 \qquad \textbf{(D)}\ 66 \qquad \textbf{(E)}\ 70</math><br />
<br />
[[2011 AMC 12B Problems/Problem 21|Solution]]<br />
<br />
==Problem 22==<br />
<br />
Let <math>T_1</math> be a triangle with sides <math>2011, 2012</math>, and <math>2013</math>. For <math>n \geq 1</math>, if <math>T_n = \Delta ABC</math> and <math>D, E</math>, and <math>F</math> are the points of tangency of the incircle of <math>\Delta ABC</math> to the sides <math>AB, BC</math>, and <math>AC</math>, respectively, then <math>T_{n+1}</math> is a triangle with side lengths <math>AD, BE</math>, and <math>CF</math>, if it exists. What is the perimeter of the last triangle in the sequence <math>\left(T_n\right)</math>?<br />
<br />
<math>\textbf{(A)}\ \frac{1509}{8} \qquad \textbf{(B)}\ \frac{1509}{32} \qquad \textbf{(C)}\ \frac{1509}{64} \qquad \textbf{(D)}\ \frac{1509}{128} \qquad \textbf{(E)}\ \frac{1509}{256}</math><br />
<br />
[[2011 AMC 12B Problems/Problem 22|Solution]]<br />
<br />
==Problem 23==<br />
<br />
A bug travels in the coordinate plane, moving only along the lines that are parallel to the <math>x</math>-axis or <math>y</math>-axis. Let <math>A = (-3, 2)</math> and <math>B = (3, -2)</math>. Consider all possible paths of the bug from <math>A</math> to <math>B</math> of length at most <math>20</math>. How many points with integer coordinates lie on at least one of these paths?<br />
<br />
<math>\textbf{(A)}\ 161 \qquad \textbf{(B)}\ 185 \qquad \textbf{(C)}\ 195 \qquad \textbf{(D)}\ 227 \qquad \textbf{(E)}\ 255</math><br />
<br />
[[2011 AMC 12B Problems/Problem 23|Solution]]<br />
<br />
==Problem 24==<br />
<br />
Let <math>P(z) = z^8 + \left(4\sqrt{3} + 6\right)z^4 - \left(4\sqrt{3} + 7\right)</math>. What is the minimum perimeter among all the <math>8</math>-sided polygons in the complex plane whose vertices are precisely the zeros of <math>P(z)</math>?<br />
<br />
<math>\textbf{(A)}\ 4\sqrt{3} + 4 \qquad \textbf{(B)}\ 8\sqrt{2} \qquad \textbf{(C)}\ 3\sqrt{2} + 3\sqrt{6} \qquad \textbf{(D)}\ 4\sqrt{2} + 4\sqrt{3} \qquad \textbf{(E)}\ 4\sqrt{3} + 6</math><br />
<br />
[[2011 AMC 12B Problems/Problem 24|Solution]]<br />
<br />
==Problem 25==<br />
<br />
For every <math>m</math> and <math>k</math> integers with <math>k</math> odd, denote by <math>\left[\frac{m}{k}\right]</math> the integer closest to <math>\frac{m}{k}</math>. For every odd integer <math>k</math>, let <math>P(k)</math> be the probability that<br />
<br />
<cmath> \left[\frac{n}{k}\right] + \left[\frac{100 - n}{k}\right] = \left[\frac{100}{k}\right] </cmath><br />
<br />
for an integer <math>n</math> randomly chosen from the interval <math>1 \leq n \leq 99!</math>. What is the minimum possible value of <math>P(k)</math> over the odd integers <math>k</math> in the interval <math>1 \leq k \leq 99</math>?<br />
<br />
<math>\textbf{(A)}\ \frac{1}{2} \qquad \textbf{(B)}\ \frac{50}{99} \qquad \textbf{(C)}\ \frac{44}{87} \qquad \textbf{(D)}\ \frac{34}{67} \qquad \textbf{(E)}\ \frac{7}{13}</math><br />
<br />
[[2011 AMC 12B Problems/Problem 25|Solution]]<br />
<br />
== See also ==<br />
{{AMC12 box|year=2011|ab=B|before=[[2011 AMC 12A Problems]]|after=[[2012 AMC 12A Problems]]}}{{MAA Notice}}</div>WolfOfAtlantishttps://artofproblemsolving.com/wiki/index.php?title=2010_AMC_8_Problems/Problem_23&diff=490742010 AMC 8 Problems/Problem 232012-11-04T01:32:24Z<p>WolfOfAtlantis: </p>
<hr />
<div>Semicircles <math>POQ</math> and <math>ROS</math> pass through the center <math>O</math>. What is the ratio of the combined areas of the two semicircles to the area of circle <math>O</math>? <br />
<asy><br />
import graph; size(7.5cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-6.27,xmax=10.01,ymin=-5.65,ymax=10.98; draw(circle((0,0),2)); draw((-3,0)--(3,0),EndArrow(6)); draw((0,-3)--(0,3),EndArrow(6)); draw(shift((0.01,1.42))*xscale(1.41)*yscale(1.41)*arc((0,0),1,179.76,359.76)); draw(shift((-0.01,-1.42))*xscale(1.41)*yscale(1.41)*arc((0,0),1,-0.38,179.62)); draw((-1.4,1.43)--(1.41,1.41)); draw((-1.42,-1.41)--(1.4,-1.42)); label("$ P(-1,1) $",(-2.57,2.17),SE*lsf); label("$ Q(1,1) $",(1.55,2.21),SE*lsf); label("$ R(-1,1) $",(-2.72,-1.45),SE*lsf); label("$S(1,-1)$",(1.59,-1.49),SE*lsf); <br />
dot((0,0),ds); label("$O$",(-0.24,-0.35),NE*lsf); dot((1.41,1.41),ds); dot((-1.4,1.43),ds); dot((1.4,-1.42),ds); dot((-1.42,-1.41),ds); <br />
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);</asy><br />
<br />
<b>Solution</b><br />
According to the pythagorean theorem, The radius of the larger circle is:<br />
<br />
<math>1^2 + 1^2 = \sqrt{2}</math><br />
<br />
Therefore the area of the larger circle is:<br />
<br />
<math>(\sqrt{2})^2\pi = 2\pi </math><br />
<br />
Using the coordinate plane given we find that the radius of the two semicircles to be 1. Therefore the area of the two semicircles is:<br />
<br />
<math>1^2\pi=\pi</math><br />
<br />
Finally the ratio of the combined areas of the two semicircles to the area of circle <math>O</math> is <math>\frac{1}{2}</math>.</div>WolfOfAtlantishttps://artofproblemsolving.com/wiki/index.php?title=Principle_of_Inclusion-Exclusion&diff=47867Principle of Inclusion-Exclusion2012-08-10T21:05:45Z<p>WolfOfAtlantis: /* Solution */</p>
<hr />
<div>The '''Principle of Inclusion-Exclusion''' (abbreviated PIE) provides an organized method/formula to find the number of [[element]]s in the [[union]] of a given group of [[set]]s, the size of each set, and the size of all possible [[intersection]]s among the sets.<br />
<br />
== Application ==<br />
Here we will illustrate how PIE is applied with various numbers of sets.<br />
<br />
=== Two Set Example ===<br />
Assume we are given the sizes of two sets, <math>|A_1|</math> and <math>|A_2|</math>, and the size of their intersection, <math>|A_1\cap A_2|</math>. We wish to find the size of their union, <math>|A_1\cup A_2|</math>.<br />
<br />
To find the union, we can add <math>|A_1|</math> and <math>|A_2|</math>. In doing so, we know we have counted everything in <math>|A_1\cup A_2|</math> at least once. However, some things were counted twice. The elements that were counted twice are precisely those in <math> {}A_1\cap A_2 </math>. Thus, we have that:<br />
<br />
<center><math> |A_1 \cup A_2| = |A_1| + |A_2| - |A_1\cap A_2|</math>.</center><br />
<br />
=== Three Set Example ===<br />
Assume we are given the sizes of three sets, <math>|A_1|, |A_2|,{}</math> and <math>|A_3|</math>, the size of their pairwise intersections, <math>|A_1\cap A_2|, |A_2\cap A_3|</math>, and <math>|A_3\cap A_1|</math>, and the size their overall intersection, <math>|A_1\cap A_2\cap A_3|</math>. We wish to find the size of their union, <math>|A_1\cup A_2\cup A_3|</math>.<br />
<br />
Just like in the Two Set Example, we start with the sum of the sizes of the individual sets <math>|A_1|+|A_2|+|A_3|</math>. We have counted the elements which are in exactly one of the original three sets once, but we've obviously counted other things twice, and even other things thrice! To account for the elements that are in two of the three sets, we first subtract out <math>|A_1\cap A_2|+|A_2\cap A_3| + |A_3\cap A_1|</math>. Now we have correctly accounted for them since we counted them twice originally, and just subtracted them out once. However, the elements that are in all three sets were originally counted three times, and then subtracted out three times. We have to add back in <math>|A_1\cap A_2\cap A_3|</math>. Putting this all together gives:<br />
<br />
<center><math> |A_1\cup A_2\cup A_3| = |A_1| + |A_2| + |A_3| -|A_1\cap A_2| - |A_2\cap A_3| - |A_3\cap A_1| +|A_1\cap A_2\cap A_3|</math>.</center><br />
<br />
=== Five Set Example ===<br />
==== Problem ====<br />
<br />
There are five courses at my school. Students take the classes as follows:<br />
243 take algebra.<br />
323 take language arts.<br />
143 take social studies.<br />
241 take biology.<br />
300 take history.<br />
213 take algebra and language arts.<br />
264 take algebra and social studies.<br />
144 take algebra and biology.<br />
121 take algebra and history.<br />
111 take language arts and social studies.<br />
90 take language arts and biology.<br />
80 take language arts and history.<br />
60 take social studies and biology.<br />
70 take social studies and history.<br />
60 take biology and history.<br />
50 take algebra, language arts, and social studies.<br />
50 take algebra, language arts, and biology.<br />
50 take algebra, language arts, and history.<br />
50 take algebra, social studies, and biology.<br />
50 take algebra, social studies, and history.<br />
50 take algebra, biology, and history.<br />
50 take language arts, social studies, and biology.<br />
50 take language arts, social studies, and history.<br />
50 take language arts, biology, and history.<br />
50 take social studies, biology, and history.<br />
20 take algebra, language arts, social studies, and biology.<br />
15 take algebra, language arts, social studies, and history.<br />
15 take algebra, language arts, biology, and history.<br />
10 take algebra, social studies, biology, and history.<br />
10 take language arts, social studies, biology, and history.<br />
5 take all five.<br />
None take none. <br />
<br />
How many people are in my school?<br />
<br />
==== Solution ====<br />
Let A be the subset of students who take algebra, L-languages, S-social s., B-biology, H-history, M the set of all students. We have:<br />
<br />
<math>|M|=|A|+|L|+|S|+|B|+|H|-|A\cap L|-|A\cap S|-|A\cap B|-|A\cap H|-|L\cap S|-|L\cap B|</math><br />
<br />
<math>-|L\cap H|-|S\cap B|-|S\cap H|-|B\cap H|+|A\cap L\cap S|+|A\cap L\cap B|+|A\cap L\cap H|+|A\cap S\cap B|+|A\cap S\cap H|</math><br />
<br />
<math>+|A\cap B\cap H|+|L\cap S\cap H|+|L\cap S \cap B|+|S\cap B\cap H|+|L \cap B\cap H|-|A\cap L\cap S\cap B|-|A\cap L\cap S\cap H|</math><br />
<br />
<math>-|A\cap L\cap B\cap H|-|A\cap S\cap B\cap H|-|L\cap S\cap B\cap H|+|A\cap L\cap S\cap B\cap H|</math><br />
<br />
<math>=243+323+143+241+300-213-264-144-121-111-90-80-60-70-60</math><br />
<br />
<math>+50+50+50+50+50+50+50+50+50+50-20-15-15-10-10+5=472</math><br />
<br />
Thus, there are <math> \boxed{472} </math> people in my school.<br />
<br />
== Statement ==<br />
If <math>(A_i)_{1\leq i\leq n}</math> are finite sets, then:<br />
<center><math> \left|\bigcup_{i=1}^n A_i\right|=\sum_{i=1}^n\left|A_i\right|<br />
-\sum_{i < j}\left|A_i\cap A_j\right| +\sum_{i<j<k}\left|A_i\cap A_j\cap A_k\right|-\cdots\ +(-1)^n \left|A_1\cap\cdots\cap A_n\right|{}</math>.</center><br />
== Remark ==<br />
Sometimes it is also useful to know that, if you take into account only the first <math>m\le n</math> sums on the right, then you will get an overestimate if <math>m</math> is [[odd integer | odd]] and an underestimate if <math>m</math> is [[even integer | even]].<br />
So, <br />
<br />
<math>\left|\bigcup_{i=1}^n A_i\right|\le \sum_{i=1}^n\left|A_i\right|</math>,<br />
<br />
<math>\left|\bigcup_{i=1}^n A_i\right|\ge \sum_{i=1}^n\left|A_i\right|-\sum_{i < j}\left|A_i\cap A_j\right|</math>,<br />
<br />
<math>\left|\bigcup_{i=1}^n A_i\right|\le \sum_{i=1}^n\left|A_i\right|-\sum_{i < j}\left|A_i\cap A_j\right| +\sum_{i<j<k}\left|A_i\cap A_j\cap A_k\right|</math>,<br />
<br />
and so on.<br />
<br />
== Examples ==<br />
* http://www.artofproblemsolving.com/Forum/viewtopic?t=83102<br />
* http://www.artofproblemsolving.com/Forum/viewtopic?t=61283<br />
<br />
* [http://mathideas.org/problems/2006/6/5.pdf Counting Divisors with PIE!]<br />
<br />
== See also ==<br />
* [[Combinatorics]]<br />
* [[Overcounting]]<br />
<br />
[[Category:Combinatorics]]</div>WolfOfAtlantishttps://artofproblemsolving.com/wiki/index.php?title=2010_AMC_8_Problems/Problem_18&diff=469932010 AMC 8 Problems/Problem 182012-05-16T03:32:26Z<p>WolfOfAtlantis: Created page with "The ratio of AD to AB is 3:2 and AB=30 inches. <math>If 30=2, then x=3. x=45. </math> The area of the rectangle is <math>45*30= 1350</math>. The semicircles, when combined, have..."</p>
<hr />
<div>The ratio of AD to AB is 3:2 and AB=30 inches.<br />
<math>If 30=2, then x=3. x=45.<br />
</math><br />
The area of the rectangle is <math>45*30= 1350</math>.<br />
<br />
The semicircles, when combined, have the area of a regular circle.<br />
Radius is 15 (<math>30/2=15</math>, for the record). Area of a circle is <math>\pi r^2</math> , so it's equal to 225<math>\pi</math><br />
<br />
<math>1350/225 \pi</math> is equivalent to <math>6/ \pi </math><br />
[http://i.imgur.com/5qEb2.png]<br />
The image given in the test is given above for reference.<br />
<br />
Answer is <math> \box{C: 6:\pi} </math></div>WolfOfAtlantis