https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Tehetrollr289&feedformat=atomAoPS Wiki - User contributions [en]2024-03-28T15:13:42ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2003_AMC_12A_Problems/Problem_19&diff=816762003 AMC 12A Problems/Problem 192016-11-27T17:27:53Z<p>Tehetrollr289: /* Problem */</p>
<hr />
<div>==Problem==<br />
<br />
A parabola with equation <math>y=ax^2+bx+c</math> is reflected about the <math>x</math>-axis. The parabola and its reflection are translated horizontally five units in opposite directions to become the graphs of <math>y=f(x)</math> and <math>y=g(x)</math>, respectively. Which of the following describes the graph of <math>y=(f+g)(x)?</math><br />
<br />
<math> \textbf{(A)}\ \text{a parabola tangent to the }x\text{-axis} </math><br />
<math> \textbf{(B)}\ \text{a parabola not tangent to the }x\text{-axis}\qquad\textbf{(C)}\ \text{a horizontal line} </math><br />
<math> \textbf{(D)}\ \text{a non-horizontal line}\qquad\textbf{(E)}\ \text{the graph of a cubic function} </math><br />
<br />
==Solution==<br />
<br />
If we take the parabola <math>ax^2 + bx + c</math> and reflect it over the x - axis, we have the parabola <math>-ax^2 - bx - c</math>. Without loss of generality, let us say that the parabola is translated 5 units to the left, and the reflection to the right. Then:<br />
<br />
<cmath> \begin{align*} f(x) = a(x+5)^2 + b(x+5) + c = ax^2 + (10a+b)x + 25a + 5b + c \\ g(x) = -a(x-5)^2 - b(x-5) - c = -ax^2 + 10ax -bx - 25a + 5b - c \end{align*} </cmath> <br />
<br />
Adding them up produces: <br />
<cmath> \begin{align*} (f + g)(x) = ax^2 + (10a+b)x + 25a + 5b + c - ax^2 + 10ax -bx - 25a + 5b - c = 20ax + 10b \end{align*}</cmath> <br />
<br />
This is a line with slope <math>20a</math>. Since <math>a</math> cannot be <math>0</math> (because <math>ax^2 + bx + c</math> would be a line) we end up with <math>\boxed{\textbf{(D)} \text{ a non-horizontal line }}</math><br />
<br />
==See Also==<br />
{{AMC12 box|year=2003|ab=A|num-b=18|num-a=20}}<br />
{{MAA Notice}}</div>Tehetrollr289https://artofproblemsolving.com/wiki/index.php?title=2000_AMC_12_Problems/Problem_17&diff=806442000 AMC 12 Problems/Problem 172016-10-14T11:38:46Z<p>Tehetrollr289: /* Solution */</p>
<hr />
<div>== Problem ==<br />
A circle centered at <math>O</math> has radius <math>1</math> and contains the point <math>A</math>. The segment <math>AB</math> is tangent to the circle at <math>A</math> and <math>\angle AOB = \theta</math>. If point <math>C</math> lies on <math>\overline{OA}</math> and <math>\overline{BC}</math> bisects <math>\angle ABO</math>, then <math>OC =</math><br />
<br />
<asy><br />
import olympiad;<br />
size(6cm);<br />
unitsize(1cm);<br />
defaultpen(fontsize(8pt)+linewidth(.8pt));<br />
labelmargin=0.2;<br />
dotfactor=3;<br />
pair O=(0,0);<br />
pair A=(1,0);<br />
pair B=(1,1.5);<br />
pair D=bisectorpoint(A,B,O);<br />
pair C=extension(B,D,O,A);<br />
draw(Circle(O,1));<br />
draw(O--A--B--cycle);<br />
draw(B--C);<br />
label("$O$",O,SW);<br />
dot(O);<br />
label("$\theta$",(0.1,0.05),ENE);<br />
dot(C);<br />
label("$C$",C,S);<br />
dot(A);<br />
label("$A$",A,E);<br />
dot(B);<br />
label("$B$",B,E);</asy><br />
<br />
<math>\text {(A)}\ \sec^2 \theta - \tan \theta \qquad \text {(B)}\ \frac 12 \qquad \text {(C)}\ \frac{\cos^2 \theta}{1 + \sin \theta}\qquad \text {(D)}\ \frac{1}{1+\sin\theta} \qquad \text {(E)}\ \frac{\sin \theta}{\cos^2 \theta}</math><br />
<br />
== Solution ==<br />
Since <math>\overline{AB}</math> is tangent to the circle, <math>\triangle OAB</math> is a right triangle. This means that <math>OA = 1</math>, <math>AB = \tan \theta</math> and <math>OB = \sec \theta</math>. By the [[Angle Bisector Theorem]], <cmath> \frac{OB}{OC} = \frac{AB}{AC} \Longrightarrow AC \sec \theta = OC \tan \theta </cmath> We multiply both sides by <math>\cos \theta</math> to simplify the trigonometric functions, <cmath> AC=OC \sin \theta </cmath> Since <math>AC + OC = 1</math>, <math>1 - OC = OC \sin \theta \Longrightarrow</math> <math>OC = \dfrac{1}{1+\sin \theta}</math>. Therefore, the answer is <math>\boxed{\textbf{(D)} \dfrac{1}{1+\sin \theta}}</math>.<br />
<br />
<br />
Alternatively, one could notice that OC approaches the value 1/2 as theta gets close to 90 degrees. The only choice that is consistent with this is (D).<br />
<br />
==Solution (with minimal trig)==<br />
Let's assign a value to <math>\theta</math> so we don't have to use trig functions to solve. <math>60</math> is a good value for <math>\theta</math>, because then we have a <math>30-60-90 \triangle</math> -- <math>\angle BAC=90</math> because <math>AB</math> is tangent to Circle <math>O</math>.<br />
<br />
Using our special right triangle, since <math>AO=1</math>, <math>OB=2</math>, and <math>AB=\sqrt{3}</math>.<br />
<br />
Let <math>OC=x</math>. Then <math>CA=1-x</math>. since <math>BC</math> bisects <math>\angle ABO</math>, we can use the angle bisector theorem:<br />
<br />
<math>\frac{2}{x}=\frac{\sqrt{3}}{1-x}</math><br />
<br />
<math>2-2x=\sqrt{3}x</math><br />
<br />
<math>2=(\sqrt{3}+2)x</math><br />
<br />
<math>x=\frac{2}{\sqrt{3}+2}</math>.<br />
<br />
Now, we only have to use a bit of trig to guess and check: the only trig facts we need to know to finish the problem is:<br />
<br />
<math>sin\theta =\frac{Opposite}{Hypotenuse}</math><br />
<br />
<math>cos\theta =\frac{Adjacent}{Hypotenuse}</math><br />
<br />
<math>tan\theta =\frac{Opposite}{Adjacent}</math>.<br />
<br />
With a bit of guess and check, we get that the answer is <math>\boxed{D}</math>.<br />
<br />
== See also ==<br />
{{AMC12 box|year=2000|num-b=16|num-a=18}}<br />
<br />
[[Category:Introductory Geometry Problems]]<br />
[[Category:Introductory Trigonometry Problems]]<br />
{{MAA Notice}}</div>Tehetrollr289https://artofproblemsolving.com/wiki/index.php?title=2000_AMC_12_Problems/Problem_16&diff=806432000 AMC 12 Problems/Problem 162016-10-14T11:33:00Z<p>Tehetrollr289: /* Solution */</p>
<hr />
<div>== Problem ==<br />
A checkerboard of <math>13</math> rows and <math>17</math> columns has a number written in each square, beginning in the upper left corner, so that the first row is numbered <math>1,2,\ldots,17</math>, the second row <math>18,19,\ldots,34</math>, and so on down the board. If the board is renumbered so that the left column, top to bottom, is <math>1,2,\ldots,13,</math>, the second column <math>14,15,\ldots,26</math> and so on across the board, some squares have the same numbers in both numbering systems. Find the sum of the numbers in these squares (under either system). <br />
<br />
<math>\text {(A)}\ 222 \qquad \text {(B)}\ 333\qquad \text {(C)}\ 444 \qquad \text {(D)}\ 555 \qquad \text {(E)}\ 666</math><br />
<br />
== Solution ==<br />
Index the rows with <math>i = 1, 2, 3, ..., 13</math><br />
Index the columns with <math>j = 1, 2, 3, ..., 17</math><br />
<br />
For the first row number the cells <math>1, 2, 3, ..., 17</math><br />
For the second, <math>18, 19, ..., 34</math><br />
and so on<br />
<br />
So the number in row = <math>i</math> and column = <math>j</math> is<br />
<math>f(i, j) = 17(i-1) + j = 17i + j - 17</math><br />
<br />
Similarly, numbering the same cells columnwise we<br />
find the number in row = <math>i</math> and column = <math>j</math> is<br />
<math>g(i, j) = i + 13j - 13</math><br />
<br />
So we need to solve<br />
<br />
<math>f(i, j) = g(i, j)<br />
17i + j - 17 = i + 13j - 13<br />
16i = 4 + 12j <br />
4i = 1 + 3j<br />
i = (1 + 3j)/4</math><br />
<br />
We get<br />
<math>(i, j) = (1, 1), f(i, j) = g(i, j) = 1<br />
(i, j) = (4, 5), f(i, j) = g(i, j) = 56<br />
(i, j) = (7, 9), f(i, j) = g(i, j) = 111<br />
(i, j) = (10, 13), f(i, j) = g(i, j) = 166<br />
(i, j) = (13, 17), f(i, j) = g(i, j) = 221</math><br />
<br />
<math>\boxed{D}</math> <math>555</math><br />
<br />
== See also ==<br />
{{AMC12 box|year=2000|num-b=15|num-a=17}}<br />
<br />
[[Category:Introductory Number Theory Problems]]<br />
{{MAA Notice}}</div>Tehetrollr289https://artofproblemsolving.com/wiki/index.php?title=2000_AMC_12_Problems/Problem_16&diff=806412000 AMC 12 Problems/Problem 162016-10-14T02:13:58Z<p>Tehetrollr289: /* Solution */</p>
<hr />
<div>== Problem ==<br />
A checkerboard of <math>13</math> rows and <math>17</math> columns has a number written in each square, beginning in the upper left corner, so that the first row is numbered <math>1,2,\ldots,17</math>, the second row <math>18,19,\ldots,34</math>, and so on down the board. If the board is renumbered so that the left column, top to bottom, is <math>1,2,\ldots,13,</math>, the second column <math>14,15,\ldots,26</math> and so on across the board, some squares have the same numbers in both numbering systems. Find the sum of the numbers in these squares (under either system). <br />
<br />
<math>\text {(A)}\ 222 \qquad \text {(B)}\ 333\qquad \text {(C)}\ 444 \qquad \text {(D)}\ 555 \qquad \text {(E)}\ 666</math><br />
<br />
== Solution ==<br />
Index the rows with <math>i = 1, 2, 3, ..., 13</math><br />
Index the columns with j = 1, 2, 3, ..., 17<br />
<br />
For the first row number the cells 1, 2, 3, ..., 17<br />
For the second, 18, 19, ..., 34<br />
and so on<br />
<br />
So the number in row = i and column = j is<br />
f(i, j) = 17(i-1) + j = 17i + j - 17<br />
<br />
Similarly, numbering the same cells columnwise we<br />
find the number in row = i and column = j is<br />
g(i, j) = i + 13j - 13<br />
<br />
So we need to solve<br />
<br />
f(i, j) = g(i, j)<br />
17i + j - 17 = i + 13j - 13<br />
16i = 4 + 12j<br />
4i = 1 + 3j<br />
i = (1 + 3j)/4<br />
<br />
We get<br />
(i, j) = (1, 1), f(i, j) = g(i, j) = 1<br />
(i, j) = (4, 5), f(i, j) = g(i, j) = 56<br />
(i, j) = (7, 9), f(i, j) = g(i, j) = 111<br />
(i, j) = (10, 13), f(i, j) = g(i, j) = 166<br />
(i, j) = (13, 17), f(i, j) = g(i, j) = 221 <br />
<br />
<math>\boxed{D}</math> 555<br />
<br />
== See also ==<br />
{{AMC12 box|year=2000|num-b=15|num-a=17}}<br />
<br />
[[Category:Introductory Number Theory Problems]]<br />
{{MAA Notice}}</div>Tehetrollr289https://artofproblemsolving.com/wiki/index.php?title=1997_AHSME_Problems/Problem_17&diff=803881997 AHSME Problems/Problem 172016-09-25T14:31:30Z<p>Tehetrollr289: /* Solution */ x=k is not horizontal</p>
<hr />
<div>==Problem==<br />
<br />
A line <math>x=k</math> intersects the graph of <math>y=\log_5 x</math> and the graph of <math>y=\log_5 (x + 4)</math>. The distance between the points of intersection is <math>0.5</math>. Given that <math>k = a + \sqrt{b}</math>, where <math>a</math> and <math>b</math> are integers, what is <math>a+b</math>?<br />
<br />
<math> \textbf{(A)}\ 6\qquad\textbf{(B)}\ 7\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 9\qquad\textbf{(E)}\ 10 </math><br />
<br />
==Solution==<br />
<br />
Since the line <math>x=k</math> is vertical, we are only concerned with vertical distance.<br />
<br />
In other words, we want to find the value of <math>k</math> for which the distance <math>|\log_5 x - \log_5 (x+4)| = \frac{1}{2}</math><br />
<br />
Since <math>\log_5 x</math> is a strictly increasing function, we have:<br />
<br />
<math>\log_5 (x + 4) - \log_5 x = \frac{1}{2}</math><br />
<br />
<math>\log_5 (\frac{x+4}{x}) = \frac{1}{2}</math><br />
<br />
<math>\frac{x+4}{x} = 5^\frac{1}{2}</math><br />
<br />
<math>x + 4 = x\sqrt{5}</math><br />
<br />
<math>x\sqrt{5} - x = 4</math><br />
<br />
<math>x = \frac{4}{\sqrt{5} - 1}</math><br />
<br />
<math>x = \frac{4(\sqrt{5} + 1)}{5 - 1^2}</math><br />
<br />
<math>x = 1 + \sqrt{5}</math><br />
<br />
The desired quantity is <math>1 + 5 = 6</math>, and the answer is <math>\boxed{A}</math>.<br />
<br />
== See also ==<br />
{{AHSME box|year=1997|num-b=16|num-a=18}}<br />
{{MAA Notice}}</div>Tehetrollr289https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_10A_Problems&diff=774482011 AMC 10A Problems2016-03-07T02:15:21Z<p>Tehetrollr289: /* Problem 1 */</p>
<hr />
<div>{{AMC10 Problems|year=2011|ab=A}}<br />
<br />
== Problem 1 ==<br />
<br />
A cell phone plan costs <math>\textdollar 20</math> each month, plus <math>5</math>¢ per text message sent, plus <math>10</math>¢ for each minute used over <math>30</math> hours. In January Michelle sent <math>100</math> text messages and talked for <math>30.5</math> hours. How much did she have to pay?<br />
<br />
<math> \textbf{(A)}\ \textdollar 24.00 \qquad\textbf{(B)}\ \textdollar 24.50 \qquad\textbf{(C)}\ \textdollar 25.50\qquad\textbf{(D)}\ \textdollar 28.00\qquad\textbf{(E)}\ \textdollar 30.00 </math><br />
<br />
[[2011 AMC 10A Problems/Problem 1|Solution]]<br />
<br />
== Problem 2 ==<br />
<br />
A small bottle of shampoo can hold 35 milliliters of shampoo, whereas a large bottle can hold 500 milliliters of shampoo. Jasmine wants to buy the minimum number of small bottles necessary to completely fill a large bottle. How many bottles must she buy?<br />
<br />
<math> \textbf{(A)}\ 11 \qquad\textbf{(B)}\ 12 \qquad\textbf{(C)}\ 13\qquad\textbf{(D)}\ 14\qquad\textbf{(E)}\ 15 </math><br />
<br />
[[2011 AMC 10A Problems/Problem 2|Solution]]<br />
<br />
== Problem 3 ==<br />
<br />
Suppose <math>[a\ b]</math> denotes the average of <math>a</math> and <math>b</math>, and <math>\{a\ b\ c\}</math> denotes the average of <math>a, b</math>, and <math>c</math>. What is <math>\{\{1\ 1\ 0\}\ [0\ 1]\ 0\}</math>?<br />
<br />
<math> \textbf{(A)}\ \frac{2}{9} \qquad\textbf{(B)}\ \frac{5}{18} \qquad\textbf{(C)}\ \frac{1}{3} \qquad\textbf{(D)}\ \frac{7}{18} \qquad\textbf{(E)}\ \frac{2}{3} </math><br />
<br />
[[2011 AMC 10A Problems/Problem 3|Solution]]<br />
<br />
== Problem 4 ==<br />
<br />
Let <math>X</math> and <math>Y</math> be the following sums of arithmetic sequences: <cmath> \begin{eqnarray*} X &=& 10 + 12 + 14 + \cdots + 100, \\ Y &=& 12 + 14 + 16 + \cdots + 102. \end{eqnarray*} </cmath> What is the value of <math>Y - X</math>?<br />
<br />
<math> \textbf{(A)}\ 92\qquad\textbf{(B)}\ 98\qquad\textbf{(C)}\ 100\qquad\textbf{(D)}\ 102\qquad\textbf{(E)}\ 112 </math><br />
<br />
<br />
[[2011 AMC 10A Problems/Problem 4|Solution]]<br />
<br />
== Problem 5 ==<br />
<br />
At an elementary school, the students in third grade, fourth grade, and fifth grade run an average of <math>12</math>, <math>15</math>, and <math>10</math> minutes per day, respectively. There are twice as many third graders as fourth graders, and twice as many fourth graders as fifth graders. What is the average number of minutes run per day by these students? <br />
<br />
<math> \textbf{(A)}\ 12 \qquad\textbf{(B)}\ \frac{37}{3} \qquad\textbf{(C)}\ \frac{88}{7} \qquad\textbf{(D)}\ 13\qquad\textbf{(E)}\ 14 </math><br />
<br />
[[2011 AMC 10A Problems/Problem 5|Solution]]<br />
<br />
== Problem 6 ==<br />
<br />
Set <math>A </math> has 20 elements, and set <math>B </math> has 15 elements. What is the smallest possible number of elements in <math>A \cup B </math>, the union of <math>A </math> and <math>B </math>?<br />
<br />
<math> \textbf{(A)}\ 5 \qquad\textbf{(B)}\ 15 \qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 35\qquad\textbf{(E)}\ 300 </math><br />
<br />
[[2011 AMC 10A Problems/Problem 6|Solution]]<br />
<br />
== Problem 7 ==<br />
Which of the following equations does NOT have a solution?<br />
<br />
<math>\text{(A)}\:(x+7)^2=0</math><br />
<br />
<math>\text{(B)}\:|-3x|+5=0</math><br />
<br />
<math>\text{(C)}\:\sqrt{-x}-2=0</math><br />
<br />
<math>\text{(D)}\:\sqrt{x}-8=0</math><br />
<br />
<math>\text{(E)}\:|-3x|-4=0 </math><br />
<br />
[[2011 AMC 10A Problems/Problem 7|Solution]]<br />
<br />
== Problem 8 ==<br />
Last summer 30% of the birds living on Town Lake were geese, 25% were swans, 10% were herons, and 35% were ducks. What percent of the birds that were not swans were geese?<br />
<br />
<br />
<math> \textbf{(A)}\ 20 \qquad\textbf{(B)}\ 30 \qquad\textbf{(C)}\ 40\qquad\textbf{(D)}\ 50\qquad\textbf{(E)}\ 60 </math><br />
<br />
[[2011 AMC 10A Problems/Problem 8|Solution]]<br />
<br />
== Problem 9 ==<br />
A rectangular region is bounded by the graphs of the equations <math>y=a, y=-b, x=-c,</math> and <math>x=d</math>, where <math>a,b,c,</math> and <math>d</math> are all positive numbers. Which of the following represents the area of this region?<br />
<br />
<math> \textbf{(A)}\ ac+ad+bc+bd\qquad\textbf{(B)}\ ac-ad</math> <math>+bc-bd\qquad\textbf{(C)}\ ac+ad</math> <math>-bc-bd \quad\quad\qquad\textbf{(D)}\ -ac-ad</math> <math>+bc+bd\qquad\textbf{(E)}\ ac-ad</math> <math>-bc+bd </math><br />
<br />
[[2011 AMC 10A Problems/Problem 9|Solution]]<br />
<br />
== Problem 10 ==<br />
A majority of the 30 students in Ms. Deameanor's class bought pencils at the school bookstore. Each of these students bought the same number of pencils, and this number was greater than 1. The cost of a pencil in cents was greater than the number of pencils each student bought, and the total cost of all the pencils was <math> \textdollar 17.71</math>. What was the cost of a pencil in cents?<br />
<br />
<math>\text{(A)}\,7 \qquad\text{(B)}\,11 \qquad\text{(C)}\,17 \qquad\text{(D)}\,23 \qquad\text{(E)}\,77</math><br />
<br />
[[2011 AMC 10A Problems/Problem 10|Solution]]<br />
<br />
== Problem 11 ==<br />
Square <math>EFGH</math> has one vertex on each side of square <math>ABCD</math>. Point <math>E</math> is on <math>\overline{AB}</math> with <math>AE=7\cdot EB</math>. What is the ratio of the area of <math>EFGH</math> to the area of <math>ABCD</math>?<br />
<br />
<math>\text{(A)}\,\frac{49}{64} \qquad\text{(B)}\,\frac{25}{32} \qquad\text{(C)}\,\frac78 \qquad\text{(D)}\,\frac{5\sqrt{2}}{8} \qquad\text{(E)}\,\frac{\sqrt{14}}{4} </math><br />
<br />
[[2011 AMC 10A Problems/Problem 11|Solution]]<br />
<br />
== Problem 12 ==<br />
<br />
The players on a basketball team made some three-point shots, some two-point shots, and some one-point free throws. They scored as many points with two-point shots as with three-point shots. Their number of successful free throws was one more than their number of successful two-point shots. The team's total score was 61 points. How many free throws did they make?<br />
<br />
<math>\text{(A)}\,13 \qquad\text{(B)}\,14 \qquad\text{(C)}\,15 \qquad\text{(D)}\,16 \qquad\text{(E)}\,17</math><br />
<br />
<br />
[[2011 AMC 10A Problems/Problem 12|Solution]]<br />
<br />
== Problem 13 ==<br />
How many even integers are there between 200 and 700 whose digits are all different and come from the set {1,2,5,7,8,9}?<br />
<br />
<math>\text{(A)}\,12 \qquad\text{(B)}\,20 \qquad\text{(C)}\,72 \qquad\text{(D)}\,120 \qquad\text{(E)}\,200</math><br />
<br />
[[2011 AMC 10A Problems/Problem 13|Solution]]<br />
<br />
== Problem 14 ==<br />
A pair of standard 6-sided fair dice is rolled once. The sum of the numbers rolled determines the diameter of a circle. What is the probability that the numerical value of the area of the circle is less than the numerical value of the circle's circumference?<br />
<br />
<math>\text{(A)}\,\frac{1}{36} \qquad\text{(B)}\,\frac{1}{12} \qquad\text{(C)}\,\frac{1}{6} \qquad\text{(D)}\,\frac{1}{4} \qquad\text{(E)}\,\frac{5}{18}</math><br />
<br />
[[2011 AMC 10A Problems/Problem 14|Solution]]<br />
<br />
== Problem 15 ==<br />
Roy bought a new battery-gasoline hybrid car. On a trip the car ran exclusively on its battery for the first 40 miles, then ran exclusively on gasoline for the rest of the trip, using gasoline at a rate of 0.02 gallons per mile. On the whole trip he averaged 55 miles per gallon. How long was the trip in miles?<br />
<br />
<math>\text{(A)}\,140 \qquad\text{(B)}\,240 \qquad\text{(C)}\,440 \qquad\text{(D)}\,640 \qquad\text{(E)}\,840</math><br />
<br />
[[2011 AMC 10A Problems/Problem 15|Solution]]<br />
<br />
== Problem 16 ==<br />
Which of the following is equal to <math>\sqrt{9-6\sqrt{2}}+\sqrt{9+6\sqrt{2}}</math>?<br />
<br />
<math>\text{(A)}\,3\sqrt2 \qquad\text{(B)}\,2\sqrt6 \qquad\text{(C)}\,\frac{7\sqrt2}{2} \qquad\text{(D)}\,3\sqrt3 \qquad\text{(E)}\,6</math><br />
<br />
[[2011 AMC 10A Problems/Problem 16|Solution]]<br />
<br />
== Problem 17 ==<br />
In the eight-term sequence <math>A,B,C,D,E,F,G,H</math>, the value of <math>C</math> is 5 and the sum of any three consecutive terms is 30. What is <math>A+H</math>?<br />
<br />
<math>\text{(A)}\,17 \qquad\text{(B)}\,18 \qquad\text{(C)}\,25 \qquad\text{(D)}\,26 \qquad\text{(E)}\,43</math><br />
<br />
[[2011 AMC 10A Problems/Problem 17|Solution]]<br />
<br />
== Problem 18 ==<br />
<br />
Circles <math>A, B,</math> and <math>C</math> each have radius 1. Circles <math>A</math> and <math>B</math> share one point of tangency. Circle <math>C</math> has a point of tangency with the midpoint of <math>\overline{AB}</math>. What is the area inside Circle <math>C</math> but outside circle <math>A</math> and circle <math>B</math> ?<br />
<br />
<math><br />
\textbf{(A)}\ 3 - \frac{\pi}{2} \qquad<br />
\textbf{(B)}\ \frac{\pi}{2} \qquad<br />
\textbf{(C)}\ 2 \qquad<br />
\textbf{(D)}\ \frac{3\pi}{4} \qquad<br />
\textbf{(E)}\ 1+\frac{\pi}{2} </math><br />
<br />
[[2011 AMC 10A Problems/Problem 18|Solution]]<br />
<br />
== Problem 19 ==<br />
<br />
In 1991 the population of a town was a perfect square. Ten years later, after an increase of 150 people, the population was 9 more than a perfect square. Now, in 2011, with an increase of another 150 people, the population is once again a perfect square. Which of the following is closest to the percent growth of the town's population during this twenty-year period?<br />
<br />
<math> \textbf{(A)}\ 42 \qquad\textbf{(B)}\ 47 \qquad\textbf{(C)}\ 52\qquad\textbf{(D)}\ 57\qquad\textbf{(E)}\ 62 </math><br />
<br />
[[2011 AMC 10A Problems/Problem 19|Solution]]<br />
<br />
== Problem 20 ==<br />
Two points on the circumference of a circle of radius <math>r</math> are selected independently and at random. From each point a chord of length r is drawn in a clockwise direction. What is the probability that the two chords intersect?<br />
<br />
<math> \textbf{(A)}\ \frac{1}{6}\qquad\textbf{(B)}\ \frac{1}{5}\qquad\textbf{(C)}\ \frac{1}{4}\qquad\textbf{(D)}\ \frac{1}{3}\qquad\textbf{(E)}\ \frac{1}{2} </math><br />
<br />
[[2011 AMC 10A Problems/Problem 20|Solution]]<br />
<br />
== Problem 21 ==<br />
Two counterfeit coins of equal weight are mixed with 8 identical genuine coins. The weight of each of the counterfeit coins is different from the weight of each of the genuine coins. A pair of coins is selected at random without replacement from the 10 coins. A second pair is selected at random without replacement from the remaining 8 coins. The combined weight of the first pair is equal to the combined weight of the second pair. What is the probability that all 4 selected coins are genuine?<br />
<br />
<math> \textbf{(A)}\ \frac{7}{11}\qquad\textbf{(B)}\ \frac{9}{13}\qquad\textbf{(C)}\ \frac{11}{15}\qquad\textbf{(D)}\ \frac{15}{19}\qquad\textbf{(E)}\ \frac{15}{16} </math><br />
<br />
[[2011 AMC 10A Problems/Problem 21|Solution]]<br />
<br />
== Problem 22 ==<br />
Each vertex of convex pentagon <math>ABCDE</math> is to be assigned a color. There are <math>6</math> colors to choose from, and the ends of each diagonal must have different colors. How many different colorings are possible?<br />
<br />
<math> \textbf{(A)}\ 2520\qquad\textbf{(B)}\ 2880\qquad\textbf{(C)}\ 3120\qquad\textbf{(D)}\ 3250\qquad\textbf{(E)}\ 3750 </math><br />
<br />
[[2011 AMC 10A Problems/Problem 22|Solution]]<br />
<br />
== Problem 23 ==<br />
Seven students count from 1 to 1000 as follows:<br />
<br />
•Alice says all the numbers, except she skips the middle number in each consecutive group of three numbers. That is, Alice says 1, 3, 4, 6, 7, 9, . . ., 997, 999, 1000.<br />
<br />
•Barbara says all of the numbers that Alice doesn't say, except she also skips the middle number in each consecutive group of three numbers.<br />
<br />
•Candice says all of the numbers that neither Alice nor Barbara says, except she also skips the middle number in each consecutive group of three numbers.<br />
<br />
•Debbie, Eliza, and Fatima say all of the numbers that none of the students with the first names beginning before theirs in the alphabet say, except each also skips the middle number in each of her consecutive groups of three numbers.<br />
<br />
•Finally, George says the only number that no one else says.<br />
<br />
What number does George say?<br />
<br />
<math> \textbf{(A)}\ 37\qquad\textbf{(B)}\ 242\qquad\textbf{(C)}\ 365\qquad\textbf{(D)}\ 728\qquad\textbf{(E)}\ 998 </math><br />
<br />
[[2011 AMC 10A Problems/Problem 23|Solution]]<br />
<br />
== Problem 24 ==<br />
<br />
Two distinct regular tetrahedra have all their vertices among the vertices of the same unit cube. What is the volume of the region formed by the intersection of the tetrahedra?<br />
<br />
<math> \textbf{(A)}\ \frac{1}{12}\qquad\textbf{(B)}\ \frac{\sqrt{2}}{12}\qquad\textbf{(C)}\ \frac{\sqrt{3}}{12}\qquad\textbf{(D)}\ \frac{1}{6}\qquad\textbf{(E)}\ \frac{\sqrt{2}}{6} </math><br />
<br />
[[2011 AMC 10A Problems/Problem 24|Solution]]<br />
<br />
== Problem 25 ==<br />
Let <math>R</math> be a square region and <math>n\ge4</math> an integer. A point <math>X</math> in the interior of <math>R</math> is called <math>n\text{-}ray</math> partitional if there are <math>n</math> rays emanating from <math>X</math> that divide <math>R</math> into <math>n</math> triangles of equal area. How many points are 100-ray partitional but not 60-ray partitional?<br />
<br />
<math>\text{(A)}\,1500 \qquad\text{(B)}\,1560 \qquad\text{(C)}\,2320 \qquad\text{(D)}\,2480 \qquad\text{(E)}\,2500</math><br />
<br />
[[2011 AMC 10A Problems/Problem 25|Solution]]<br />
<br />
== See also ==<br />
{{AMC10 box|year=2011|ab=A|before=[[2010 AMC 10B Problems]]|after=[[2011 AMC 10B Problems]]}}<br />
* [[AMC 10]]<br />
* [[AMC 10 Problems and Solutions]]<br />
* [[Mathematics competition resources]]<br />
{{MAA Notice}}</div>Tehetrollr289https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_10B_Problems/Problem_24&diff=763332010 AMC 10B Problems/Problem 242016-02-16T21:07:07Z<p>Tehetrollr289: /* Solution */</p>
<hr />
<div>== Problem ==<br />
A high school basketball game between the Raiders and Wildcats was tied at the end of the first quarter. The number of points scored by the Raiders in each of the four quarters formed an increasing geometric sequence, and the number of points scored by the Wildcats in each of the four quarters formed an increasing arithmetic sequence. At the end of the fourth quarter, the Raiders had won by one point. Neither team scored more than <math>100</math> points. What was the total number of points scored by the two teams in the first half?<br />
<br />
<math>\textbf{(A)}\ 30 \qquad \textbf{(B)}\ 31 \qquad \textbf{(C)}\ 32 \qquad \textbf{(D)}\ 33 \qquad \textbf{(E)}\ 34</math><br />
<br />
== Solution ==<br />
Represent the teams' scores as: <math>(a, an, an^2, an^3)</math> and <math>(a, a+m, a+2m, a+3m)</math><br />
<br />
We have <math>a+an+an^2+an^3=4a+6m+1</math><br />
Factoring out the <math>a</math> from the left side of the equation, we can get <math>a(1+n+n^2+n^3)=4a+6m+1</math>, or <math>a(n^4-1)/(n-1)=4a+6m+1</math><br />
<br />
Since both are increasing sequences, <math>n>1</math>. We can check cases up to <math>n=4</math> because when <math>n=5</math>, we get <math>156a>100</math>. When <br />
* <math>n=2, a=[1,6]</math><br />
*<math> n=3, a=[1,2]</math><br />
*<math> n=4, a=1</math><br />
Checking each of these cases individually back into the equation <math>a+an+an^2+an^3=4a+6m+1</math>, we see that only when <math>a=5</math> and <math>n=2</math>, we get an integer value for <math>m</math>, which is <math>9</math>. The original question asks for the first half scores summed, so we must find <math>(a)+(an)+(a)+(a+m)=(5)+(10)+(5)+(5+9)=\boxed{\textbf{(E)}\ 34}</math><br />
<br />
== See also ==<br />
{{AMC10 box|year=2010|ab=B|num-b=23|num-a=25}}<br />
{{MAA Notice}}</div>Tehetrollr289https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_10B_Problems/Problem_24&diff=763322010 AMC 10B Problems/Problem 242016-02-16T21:06:41Z<p>Tehetrollr289: /* Solution */</p>
<hr />
<div>== Problem ==<br />
A high school basketball game between the Raiders and Wildcats was tied at the end of the first quarter. The number of points scored by the Raiders in each of the four quarters formed an increasing geometric sequence, and the number of points scored by the Wildcats in each of the four quarters formed an increasing arithmetic sequence. At the end of the fourth quarter, the Raiders had won by one point. Neither team scored more than <math>100</math> points. What was the total number of points scored by the two teams in the first half?<br />
<br />
<math>\textbf{(A)}\ 30 \qquad \textbf{(B)}\ 31 \qquad \textbf{(C)}\ 32 \qquad \textbf{(D)}\ 33 \qquad \textbf{(E)}\ 34</math><br />
<br />
== Solution ==<br />
Represent the teams' scores as: <math>(a, an, an^2, an^3)</math> and <math>(a, a+m, a+2m, a+3m)</math><br />
<br />
We have <math>a+an+an^2+an^3=4a+6m+1</math><br />
Factoring out the <math>a</math>, we can get <math>a(1+n+n^2+n^3)=4a+6m+1</math>, or <math>a(n^4-1)/(n-1)=4a+6m+1</math><br />
<br />
Since both are increasing sequences, <math>n>1</math>. We can check cases up to <math>n=4</math> because when <math>n=5</math>, we get <math>156a>100</math>. When <br />
* <math>n=2, a=[1,6]</math><br />
*<math> n=3, a=[1,2]</math><br />
*<math> n=4, a=1</math><br />
Checking each of these cases individually back into the equation <math>a+an+an^2+an^3=4a+6m+1</math>, we see that only when <math>a=5</math> and <math>n=2</math>, we get an integer value for <math>m</math>, which is <math>9</math>. The original question asks for the first half scores summed, so we must find <math>(a)+(an)+(a)+(a+m)=(5)+(10)+(5)+(5+9)=\boxed{\textbf{(E)}\ 34}</math><br />
<br />
== See also ==<br />
{{AMC10 box|year=2010|ab=B|num-b=23|num-a=25}}<br />
{{MAA Notice}}</div>Tehetrollr289https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_10B_Problems/Problem_23&diff=763312010 AMC 10B Problems/Problem 232016-02-16T21:04:09Z<p>Tehetrollr289: /* Solution */</p>
<hr />
<div>==Problem==<br />
<br />
The entries in a <math>3 \times 3</math> array include all the digits from 1 through 9, arranged so that the entries in every row and column are in increasing order. How many such arrays are there? <br />
<br />
<math> \textbf{(A)}\ 18\qquad\textbf{(B)}\ 24 \qquad\textbf{(C)}\ 36\qquad\textbf{(D)}\ 42\qquad\textbf{(E)}\ 60 </math><br />
<br />
==Solution==<br />
The upper-left corner must contain the entry 1, and similarly the lower-right corner must contain the entry 9. Consider the entries 2 and 3 -- they may either both lie in the first row, both lie in the first column, or lie in the two squares neighboring 1. By symmetry (which we will take into account by a factor of 2 in the end), we may assume that 2 lies in the cell to the right of 1 and that 3 lies either in the cell to the right of 2 or in the cell below 1:<br />
<cmath><br />
\begin{array}{|c|c|c|}<br />
\hline<br />
1 & 2 & 3 \\\hline<br />
&& \\\hline<br />
&&\\\hline<br />
\end{array}<br />
\qquad<br />
\begin{array}{|c|c|c|}<br />
\hline<br />
1 & 2 & \\\hline<br />
3&& \\\hline<br />
&&\\\hline<br />
\end{array}<br />
</cmath><br />
Similarly, the entries 7 and 8 may either both lie in the last row, or lie in the two squares neighboring 9. This gives the following cases:<br />
<cmath><br />
\begin{array}{|c|c|c|}<br />
\hline<br />
1 & 2 & 3 \\\hline<br />
&& \\\hline<br />
7&8&9\\\hline<br />
\end{array}<br />
\qquad<br />
\begin{array}{|c|c|c|}<br />
\hline<br />
1 & 2 & \\\hline<br />
3&& \\\hline<br />
7&8&9\\\hline<br />
\end{array} \times 2<br />
\qquad<br />
\begin{array}{|c|c|c|}<br />
\hline<br />
1 & 2 & 3 \\\hline<br />
&&7 \\\hline<br />
&8&9\\\hline<br />
\end{array}\times 2<br />
\qquad<br />
\begin{array}{|c|c|c|}<br />
\hline<br />
1 & 2 & \\\hline<br />
3&&7 \\\hline<br />
&8&9\\\hline<br />
\end{array}\times 2,<br />
</cmath><br />
where the notation <math>\times 2</math> denotes two possible cases, either by switching a row and column or by switching the 7 and 8. Finally, there are respectively 1, 2, 2, 6 ways to complete these four cases. This gives a total of<br />
<cmath><br />
2\cdot\left(1+2\times2+2\times2+2\times6\right)=\boxed{\textbf{(D)}\ 42}<br />
</cmath><br />
possible ways to fill the diagram.<br />
<br />
==Notes==<br />
In fact, there is a general formula (coming from the fields of [[combinatorics]] and [[representation theory]]) to answer problems of this form; it is known as the [http://en.wikipedia.org/wiki/Young_tableau#Dimension_of_a_representation hook-length formula].<br />
<br />
== See also ==<br />
{{AMC10 box|year=2010|ab=B|num-b=22|num-a=24}}<br />
{{MAA Notice}}</div>Tehetrollr289https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_10B_Problems/Problem_24&diff=760472011 AMC 10B Problems/Problem 242016-02-16T02:41:22Z<p>Tehetrollr289: /* Solution */</p>
<hr />
<div>== Problem ==<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 \le 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 />
==Solution==<br />
We see that for the graph of <math>y=mx+2</math> to not pass through any lattice points, the denominator of <math>m</math> must be greater than <math>100</math>, or else it would be canceled by some <math>0<x\le100</math> which would make <math>y</math> an integer. By using common denominators, we find that the order of the fractions from smallest to largest is <math>(A), (B), (C), (D), (E)</math>. We can see that when <math>x=\frac{50}{99}</math>, <math>y</math> would be an integer, so therefore any fraction greater than <math>\frac{50}{99}</math> would not work, as substituting our fraction <math>\frac{50}{99}</math> for <math>m</math> would produce an integer for <math>y</math>. So now we are left with only <math>\frac{51}{101}</math> and <math>\frac{50}{99}</math>. But since <math>\frac{51}{101}=\frac{5049}{9999}</math> and <math>\frac{50}{99}=\frac{5050}{9999}</math>, we can be absolutely certain that there isn't a number between <math>\frac{51}{101}</math> and <math>\frac{50}{99}</math> that can reduce to a fraction whose denominator is less than or equal to <math>100</math>. Since we are looking for the maximum value of <math>a</math>, we take the larger of <math>\frac{51}{101}</math> and <math>\frac{50}{99}</math>, which is <math>\boxed{\textbf{(B)}\frac{50}{99}}</math>.<br />
<br />
==See Also==<br />
{{AMC10 box|year=2011|ab=B|num-a=25|num-b=23}}<br />
{{MAA Notice}}</div>Tehetrollr289https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_10B_Problems/Problem_24&diff=760462011 AMC 10B Problems/Problem 242016-02-16T02:40:42Z<p>Tehetrollr289: /* Solution */</p>
<hr />
<div>== Problem ==<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 \le 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 />
==Solution==<br />
We see that for the graph of <math>y=mx+2</math> to not pass through any lattice points, the denominator of <math>m</math> must be greater than <math>100</math>, or else it would be canceled by some <math>0<x\le100</math> which would make <math>y</math> an integer. By using common denominators, we find that the order of the fractions from smallest to largest is <math>(A), (B), (C), (D), (E)</math>. We can see that when <math>x=\frac{50}{99}</math>, <math>y</math> would be an integer, so therefore any fraction greater than <math>\frac{50}{99}</math> would not work, as substituting our fraction <math>\frac{50}{99}</math> for <math>m</math> would produce an integer for <math>y</math>. So now we are left with only <math>\frac{51}{101}</math> and <math>\frac{50}{99}</math>. But since <math>\frac{51}{101}=\frac{5049}{9999}</math> and <math>\frac{50}{99}=\frac{5050}{9999}</math>, we can be absolutely certain that there is no number between <math>\frac{51}{101}</math> and <math>\frac{50}{99}</math> that can reduce to a fraction whose denominator is less than or equal to <math>100</math>. Since we are looking for the maximum value of <math>a</math>, we take the larger of <math>\frac{51}{101}</math> and <math>\frac{50}{99}</math>, which is <math>\boxed{\textbf{(B)}\frac{50}{99}}</math>.<br />
<br />
==See Also==<br />
{{AMC10 box|year=2011|ab=B|num-a=25|num-b=23}}<br />
{{MAA Notice}}</div>Tehetrollr289https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_10B_Problems/Problem_24&diff=760452011 AMC 10B Problems/Problem 242016-02-16T02:39:58Z<p>Tehetrollr289: /* Solution */</p>
<hr />
<div>== Problem ==<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 \le 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 />
==Solution==<br />
We see that for the graph of <math>y=mx+2</math> to not pass through any lattice points, the denominator of <math>m</math> must be greater than <math>100</math>, or else it would be canceled by some <math>0<x\le100</math> which would make <math>y</math> an integer. By using common denominators, we find that the order of the fractions from smallest to largest is <math>(A), (B), (C), (D), (E)</math>. We can see that when <math>x=\frac{50}{99}</math>, <math>y</math> would be an integer, so therefore any fraction greater than <math>\frac{50}{99}</math> would not work, as substituting our fraction <math>\frac{50}{99}</math> for <math>m</math> would produce an integer for <math>y</math>. So now we are left with only <math>\frac{51}{101}</math> and <math>\frac{50}{99}</math>. But since <math>\frac{51}{101}=\frac{5049}{9999}</math> and <math>\frac{50}{99}=\frac{5050}{9999}</math>, we can be certain that there is no number between <math>\frac{51}{101}</math> and <math>\frac{50}{99}</math> that can reduce to a fraction whose denominator is less than or equal to <math>100</math>. Since we are looking for the maximum value of <math>a</math>, we take the larger of <math>\frac{51}{101}</math> and <math>\frac{50}{99}</math>, which is <math>\boxed{\textbf{(B)}\frac{50}{99}}</math>.<br />
<br />
==See Also==<br />
{{AMC10 box|year=2011|ab=B|num-a=25|num-b=23}}<br />
{{MAA Notice}}</div>Tehetrollr289https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_10B_Problems/Problem_24&diff=760442011 AMC 10B Problems/Problem 242016-02-16T02:39:26Z<p>Tehetrollr289: /* Solution */</p>
<hr />
<div>== Problem ==<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 \le 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 />
==Solution==<br />
We see that for the graph of <math>y=mx+2</math> to not pass through any lattice points, the denominator of <math>m</math> must be greater than <math>100</math>, or else it would be canceled by some <math>0<x\le100</math> which would make <math>y</math> an integer. By using common denominators, we find that the order of the fractions from smallest to largest is <math>(A), (B), (C), (D), (E)</math>. We can see that when <math>x=\frac{50}{99}</math>, <math>y</math> would be an integer, so therefore any fraction greater than <math>\frac{50}{99}</math> would not work, as substituting our fraction <math>\frac{50}{99}</math> for <math>m</math> would produce an integer for <math>y</math>. So now we are left with only <math>\frac{51}{101}</math> and <math>\frac{50}{99}</math>. But because <math>\frac{51}{101}=\frac{5049}{9999}</math> and <math>\frac{50}{99}=\frac{5050}{9999}</math>, we can be certain that there is no number between <math>\frac{51}{101}</math> and <math>\frac{50}{99}</math> that can reduce to a fraction whose denominator is less than or equal to <math>100</math>. Since we are looking for the maximum value of <math>a</math>, we take the larger of <math>\frac{51}{101}</math> and <math>\frac{50}{99}</math>, which is <math>\boxed{\textbf{(B)}\frac{50}{99}}</math>.<br />
<br />
==See Also==<br />
{{AMC10 box|year=2011|ab=B|num-a=25|num-b=23}}<br />
{{MAA Notice}}</div>Tehetrollr289https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_10B_Problems/Problem_24&diff=760432011 AMC 10B Problems/Problem 242016-02-16T02:38:57Z<p>Tehetrollr289: /* Solution */</p>
<hr />
<div>== Problem ==<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 \le 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 />
==Solution==<br />
We see that for the graph of <math>y=mx+2</math> to not pass through any lattice points, the denominator of <math>m</math> must be greater than <math>100</math>, or else it would be canceled by some <math>0<x\le100</math> which would make <math>y</math> an integer. By using common denominators, we find that the order of the fractions from smallest to largest is <math>(A), (B), (C), (D), (E)</math>. We can see that when <math>x=\frac{50}{99}</math>, <math>y</math> would be an integer, so therefore any fraction greater than <math>\frac{50}{99}</math> would not work, as substituting <math>\frac{50}{99}</math> for <math>m</math> would produce an integer for <math>y</math>. So now we are left with only <math>\frac{51}{101}</math> and <math>\frac{50}{99}</math>. But because <math>\frac{51}{101}=\frac{5049}{9999}</math> and <math>\frac{50}{99}=\frac{5050}{9999}</math>, we can be certain that there is no number between <math>\frac{51}{101}</math> and <math>\frac{50}{99}</math> that can reduce to a fraction whose denominator is less than or equal to <math>100</math>. Since we are looking for the maximum value of <math>a</math>, we take the larger of <math>\frac{51}{101}</math> and <math>\frac{50}{99}</math>, which is <math>\boxed{\textbf{(B)}\frac{50}{99}}</math>.<br />
<br />
==See Also==<br />
{{AMC10 box|year=2011|ab=B|num-a=25|num-b=23}}<br />
{{MAA Notice}}</div>Tehetrollr289https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_10B_Problems/Problem_20&diff=760422011 AMC 10B Problems/Problem 202016-02-16T02:29:38Z<p>Tehetrollr289: /* Solution */</p>
<hr />
<div>{{duplicate|[[2011 AMC 12B Problems|2011 AMC 12B #16]] and [[2011 AMC 10B Problems|2011 AMC 10B #20]]}}<br />
<br />
== Problem==<br />
<br />
Rhombus <math>ABCD</math> has side length <math>2</math> and <math>\angle B = 120</math>°. Region <math>R</math> consists of all points inside 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 />
[[Category: Introductory Geometry Problems]]<br />
<br />
== Solution ==<br />
<br />
Suppose that <math>P</math> is a point in the rhombus <math>ABCD</math> and let <math>\ell_{BC}</math> be the [[perpendicular bisector]] of <math>\overline{BC}</math>. Then <math>PB < PC</math> if and only if <math>P</math> is on the same side of <math>\ell_{BC}</math> as <math>B</math>. The line <math>\ell_{BC}</math> divides the plane into two half-planes; let <math>S_{BC}</math> be the half-plane containing <math>B</math>. Let us define similarly <math>\ell_{BD},S_{BD}</math> and <math>\ell_{BA},S_{BA}</math>. Then <math>R</math> is equal to <math>ABCD \cap S_{BC} \cap S_{BD} \cap S_{BA}</math>. The region turns out to be an irregular pentagon. We can make it easier to find the area of this region by dividing it into four triangles:<br />
<br />
<asy><br />
unitsize(8mm);<br />
defaultpen(linewidth(0.8pt)+fontsize(10pt));<br />
dotfactor=4;<br />
<br />
pair A=(4,0), B=(2,2sqrt(3)), C=(-2,2sqrt(3)), D=(0,0), E=(B+C)/2, F=(B+C+D)/3, G=(A+C)/2, H=(A+B+D)/3, I=(A+B)/2;<br />
fill((0,2sqrt(3))--B--(3,sqrt(3))--(2,(2sqrt(3))/3)--(0,(4sqrt(3))/3)--cycle,lightgray);<br />
draw(A--B--C--D--cycle);<br />
draw(D--(0,2sqrt(3))); draw(D--(3,sqrt(3))); draw(A--C); draw(F--B--H); draw(B--G);<br />
<br />
label("$A$",A,SE);label("$B$",B,NE);label("$C$",C,NW);label("$D$",D,SW);<br />
label("$E$",E,N);label("$F$",F,SW);label("$G$",G,SW);label("$H$",H,S);label("$I$",I,NE);<br />
label("$2$",(D--C),SW);<br />
</asy><br />
Since <math>\triangle BCD</math> and <math>\triangle BAD</math> are equilateral, <math>\ell_{BC}</math> contains <math>D</math>, <math>\ell_{BD}</math> contains <math>A</math> and <math>C</math>, and <math>\ell_{BA}</math> contains <math>D</math>. Then <math>\triangle BEF \cong \triangle BGF \cong \triangle BGH \cong \triangle BIH</math> with <math>BE = 1</math> and <math>EF = \frac{1}{\sqrt{3}}</math> so <math>[BEF] = \frac{1}{2}\cdot 1 \cdot \frac{\sqrt{3}}{3}</math>. Multiply this by 4 and it turns out that the pentagon has area <math>\boxed{(C)\frac{2\sqrt{3}}{3}}</math>.<br />
<br />
== See Also==<br />
<br />
{{AMC10 box|year=2011|ab=B|num-b=19|num-a=21}}<br />
<br />
{{AMC12 box|year=2011|ab=B|num-b=15|num-a=17}}<br />
{{MAA Notice}}</div>Tehetrollr289https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_10B_Problems&diff=760412011 AMC 10B Problems2016-02-16T02:03:43Z<p>Tehetrollr289: /* Problem 1 */</p>
<hr />
<div>== Problem 1 ==<br />
<br />
What is <cmath>\dfrac{2+4+6}{1+3+5} - \dfrac{1+3+5}{2+4+6}</cmath><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 10B Problems/Problem 1|Solution]]<br />
<br />
== Problem 2 ==<br />
<br />
Josanna's test scores to date are <math>90, 80, 70, 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 10B Problems/Problem 2|Solution]]<br />
<br />
== Problem 3 ==<br />
<br />
At a store, when a length is reported as <math>x</math> inches that means the length is at least <math>x - 0.5</math> inches and at most <math>x + 0.5</math> inches. Suppose the dimensions of a rectangular tile are reported as <math>2</math> inches by <math>3</math> inches. In square inches, what is the minimum area for the rectangle?<br />
<br />
<math> \textbf{(A)}\ 3.75 \qquad\textbf{(B)}\ 4.5 \qquad\textbf{(C)}\ 5 \qquad\textbf{(D)}\ 6 \qquad\textbf{(E)}\ 8.75 </math><br />
<br />
[[2011 AMC 10B Problems/Problem 3|Solution]]<br />
<br />
== Problem 4 ==<br />
<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)}\ \dfrac{A - B}{2}\qquad\textbf{(C)}\ \dfrac{B - A}{2}\qquad\textbf{(D)}\ B - A \qquad\textbf{(E)}\ A + B </math><br />
<br />
[[2011 AMC 10B Problems/Problem 4|Solution]]<br />
<br />
== Problem 5 ==<br />
<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 <math>161</math>. 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 10B Problems/Problem 5|Solution]]<br />
<br />
== Problem 6 ==<br />
<br />
On Halloween Casper ate <math>1/3</math> of his candies and then gave <math>2</math> candies to his brother. The next day he ate <math>1/3</math> of his remaining candies and then gave <math>4</math> candies to his sister. On the third day he ate his final <math>8</math> candies. How many candies did Casper have at the beginning?<br />
<br />
<math> \textbf{(A)}\ 30 \qquad\textbf{(B)}\ 39 \qquad\textbf{(C)}\ 48 \qquad\textbf{(D)}\ 57 \qquad\textbf{(E)}\ 66 </math><br />
<br />
[[2011 AMC 10B Problems/Problem 6|Solution]]<br />
<br />
== Problem 7 ==<br />
<br />
The sum of two angles of a triangle is <math>6/5</math> of a right angle, and one of these two angles is <math>30^{\circ}</math> larger than the other. What is the degree measure of the largest angle in the triangle?<br />
<br />
<math> \textbf{(A)}\ 69 \qquad\textbf{(B)}\ 72 \qquad\textbf{(C)}\ 90 \qquad\textbf{(D)}\ 102 \qquad\textbf{(E)}\ 108 </math><br />
<br />
[[2011 AMC 10B Problems/Problem 7|Solution]]<br />
<br />
== Problem 8 ==<br />
<br />
At a certain beach if it is at least <math>80^{\circ} F</math> and sunny, then the beach will be crowded. On June 10 the beach was not crowded. What can be concluded about the weather conditions on June 10?<br />
<br />
<math> \textbf{(A)}\ \text{The temperature was cooler than } 80^{\circ} \text{F and it was not sunny.}</math><br />
<br />
<math> \textbf{(B)}\ \text{The temperature was cooler than } 80^{\circ} \text{F or it was not sunny.}</math><br />
<br />
<math> \textbf{(C)}\ \text{If the temperature was at least } 80^{\circ} \text{F, then it was sunny.}</math><br />
<br />
<math> \textbf{(D)}\ \text{If the temperature was cooler than } 80^{\circ} \text{F, then it was sunny.}</math><br />
<br />
<math> \textbf{(E)}\ \text{If the temperature was cooler than } 80^{\circ} \text{F, then it was not sunny.}</math><br />
<br />
[[2011 AMC 10B Problems/Problem 8|Solution]]<br />
<br />
== Problem 9 ==<br />
<br />
The area of <math>\triangle</math><math>EBD</math> is one third of the area of <math>3-4-5</math> <math>\triangle</math><math>ABC</math>. Segment <math>DE</math> is perpendicular to segment <math>AB</math>. What is <math>BD</math>? <p><br />
<center><asy><br />
unitsize(10mm);<br />
defaultpen(linewidth(.8pt)+fontsize(10pt));<br />
dotfactor=4;<br />
<br />
pair A=(0,0), B=(5,0), C=(1.8,2.4), D=(5-4sqrt(3)/3,0), E=(5-4sqrt(3)/3,sqrt(3));<br />
pair[] ps={A,B,C,D,E};<br />
<br />
draw(A--B--C--cycle);<br />
draw(E--D);<br />
draw(rightanglemark(E,D,B));<br />
<br />
dot(ps);<br />
label("$A$",A,SW);<br />
label("$B$",B,SE);<br />
label("$C$",C,N);<br />
label("$D$",D,S);<br />
label("$E$",E,NE);<br />
label("$3$",midpoint(A--C),NW);<br />
label("$4$",midpoint(C--B),NE);<br />
label("$5$",midpoint(A--B),SW);<br />
</asy><br />
</center><br />
<br />
<math> \textbf{(A)}\ \frac{4}{3} \qquad\textbf{(B)}\ \sqrt{5} \qquad\textbf{(C)}\ \frac{9}{4} \qquad\textbf{(D)}\ \frac{4\sqrt{3}}{3} \qquad\textbf{(E)}\ \frac{5}{2} </math><br />
<br />
[[2011 AMC 10B Problems/Problem 9|Solution]]<br />
<br />
== Problem 10 ==<br />
<br />
Consider the set of numbers <math>\{1, 10, 10^2, 10^3, \ldots, 10^{10}\}</math>. The ratio of the largest element of the set to the sum of the other ten elements of the set is closest to which integer?<br />
<br />
<math> \textbf{(A)}\ 1 \qquad\textbf{(B)}\ 9 \qquad\textbf{(C)}\ 10 \qquad\textbf{(D)}\ 11 \qquad\textbf{(E)} 101 </math><br />
<br />
[[2011 AMC 10B Problems/Problem 10|Solution]]<br />
<br />
== Problem 11 ==<br />
<br />
There are <math>52</math> people in a room. What is the largest value of <math>n</math> such that the statement "At least <math>n</math> people in this room have birthdays falling in the same month" is always true?<br />
<br />
<math> \textbf{(A)}\ 2 \qquad\textbf{(B)}\ 3 \qquad\textbf{(C)}\ 4 \qquad\textbf{(D)}\ 5 \qquad\textbf{(E)}\ 12 </math><br />
<br />
[[2011 AMC 10B Problems/Problem 11|Solution]]<br />
<br />
== Problem 12 ==<br />
<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 a width of <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 10B Problems/Problem 12|Solution]]<br />
<br />
== Problem 13 ==<br />
<br />
Two real numbers are selected independently 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 10B Problems/Problem 13|Solution]]<br />
<br />
== Problem 14 ==<br />
<br />
A rectangular parking lot has a diagonal of <math>25</math> meters and an area of <math>168</math> square meters. In meters, what is the perimeter of the parking lot?<br />
<br />
<math> \textbf{(A)}\ 52 \qquad\textbf{(B)}\ 58 \qquad\textbf{(C)}\ 62 \qquad\textbf{(D)}\ 68 \qquad\textbf{(E)}\ 70</math><br />
<br />
[[2011 AMC 10B Problems/Problem 14|Solution]]<br />
<br />
== Problem 15 ==<br />
<br />
Let <math>@</math> denote the "averaged with" operation: <math>a @ b = (a+b)/2</math>. Which of the following distributive laws hold for all numbers <math>x, y,</math> and <math>z</math>? <cmath>\text{I. x @ (y + z) = (x @ y) + (x @ z)}</cmath> <cmath>\text{II. x + (y @ z) = (x + y) @ (x + z)}</cmath> <cmath>\text{III. x @ (y @ z) = (x @ y) @ (x @ z)}</cmath><br />
<br />
<math> \textbf{(A)}\ \text{I only} \qquad\textbf{(B)}\ \text{II only} \qquad\textbf{(C)}\ \text{III only} \qquad\textbf{(D)}\ \text{I and III only} \qquad\textbf{(E)}\ \text{II and III only}</math><br />
<br />
[[2011 AMC 10B Problems/Problem 15|Solution]]<br />
<br />
== Problem 16 ==<br />
<br />
A dart board is a regular octagon divided into regions as shown. Suppose that a dart thrown at the board is equally likely to land anywhere on the board. What is probability that the dart lands within the center square?<br />
<br />
<center><asy><br />
unitsize(10mm);<br />
defaultpen(linewidth(.8pt)+fontsize(10pt));<br />
dotfactor=4;<br />
<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 />
<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);<br />
<br />
</asy><br />
</center><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 10B Problems/Problem 16|Solution]]<br />
<br />
== Problem 17 ==<br />
<br />
In the given circle, the diameter <math>\overline{EB}</math> is parallel to <math>\overline{DC}</math>, and <math>\overline{AB}</math> is parallel to <math>\overline{ED}</math>. The angles <math>AEB</math> and <math>ABE</math> are in the ratio <math>4 : 5</math>. What is the degree measure of angle <math>BCD</math>?<br />
<br />
<center><asy><br />
unitsize(7mm);<br />
defaultpen(linewidth(.8pt)+fontsize(10pt));<br />
dotfactor=4;<br />
<br />
real r=3;<br />
pair A=(-3cos(80),-3sin(80));<br />
pair D=(3cos(80),3sin(80)), C=(-3cos(80),3sin(80));<br />
pair O=(0,0), E=(-3,0), B=(3,0);<br />
path outer=Circle(O,r);<br />
draw(outer);<br />
draw(E--B);<br />
draw(E--A);<br />
draw(B--A);<br />
draw(E--D);<br />
draw(C--D);<br />
draw(B--C);<br />
<br />
pair[] ps={A,B,C,D,E,O};<br />
dot(ps);<br />
<br />
label("$A$",A,N);<br />
label("$B$",B,NE);<br />
label("$C$",C,S);<br />
label("$D$",D,S);<br />
label("$E$",E,NW);<br />
label("$$",O,N);<br />
</asy></center><br />
<br />
<math> \textbf{(A)}\ 120 \qquad\textbf{(B)}\ 125 \qquad\textbf{(C)}\ 130 \qquad\textbf{(D)}\ 135 \qquad\textbf{(E)}\ 140</math><br />
<br />
[[2011 AMC 10B Problems/Problem 17|Solution]]<br />
<br />
== Problem 18 ==<br />
<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 10B Problems/Problem 18|Solution]]<br />
<br />
== Problem 19 ==<br />
<br />
What is the product of all the roots of the equation <cmath>\sqrt{5 | x | + 8} = \sqrt{x^2 - 16}.</cmath><br />
<br />
<math> \textbf{(A)}\ -64 \qquad\textbf{(B)}\ -24 \qquad\textbf{(C)}\ -9 \qquad\textbf{(D)}\ 24 \qquad\textbf{(E)}\ 576</math><br />
<br />
[[2011 AMC 10B Problems/Problem 19|Solution]]<br />
<br />
== Problem 20 ==<br />
<br />
Rhombus <math>ABCD</math> has side length <math>2</math> and <math>\angle B = 120</math>°. Region <math>R</math> consists of all points inside 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 10B Problems/Problem 20|Solution]]<br />
<br />
== Problem 21 ==<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 for <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 10B Problems/Problem 21|Solution]]<br />
<br />
== Problem 22 ==<br />
<br />
A pyramid has a square base with sides of length <math>1</math> 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 10B Problems/Problem 22|Solution]]<br />
<br />
== Problem 23 ==<br />
<br />
What is the hundreds digit of <math>2011^{2011}</math>?<br />
<br />
<math> \textbf{(A)}\ 1 \qquad\textbf{(B)}\ 4 \qquad\textbf{(C)}\ 5 \qquad\textbf{(D)}\ 6 \qquad\textbf{(E)}\ 9</math><br />
<br />
[[2011 AMC 10B Problems/Problem 23|Solution]]<br />
<br />
== Problem 24 ==<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 \le 100</math> for all <math>m</math> such that <math>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 10B Problems/Problem 24|Solution]]<br />
<br />
== Problem 25 ==<br />
<br />
Let <math>T_1</math> be a triangle with sides <math>2011, 2012,</math> and <math>2013</math>. For <math>n \ge 1</math>, if <math>T_n = \triangle ABC</math> and <math>D, E,</math> and <math>F</math> are the points of tangency of the incircle of <math>\triangle 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>( T_n )</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 10B Problems/Problem 25|Solution]]<br />
==See also==<br />
{{AMC10 box|year=2011|ab=B|before=[[2011 AMC 10A Problems]]|after=[[2012 AMC 10A Problems]]}}<br />
* [[AMC 10]]<br />
* [[AMC 10 Problems and Solutions]]<br />
* [[2011 AMC 10B]]<br />
* [[Mathematics competition resources]]<br />
{{MAA Notice}}</div>Tehetrollr289https://artofproblemsolving.com/wiki/index.php?title=Talk:2012_AMC_10B_Problems/Problem_21&diff=75987Talk:2012 AMC 10B Problems/Problem 212016-02-15T18:23:00Z<p>Tehetrollr289: Discussion for 2012 AMC 10B problem #21</p>
<hr />
<div>Wait, so the points can be colinear?</div>Tehetrollr289https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_10B_Problems/Problem_25&diff=758292013 AMC 10B Problems/Problem 252016-02-14T00:57:54Z<p>Tehetrollr289: /* Solution */</p>
<hr />
<div>{{duplicate|[[2013 AMC 12B Problems|2013 AMC 12B #23]] and [[2013 AMC 10B Problems|2013 AMC 10B #25]]}}<br />
<br />
==Problem==<br />
<br />
Bernardo chooses a three-digit positive integer <math>N</math> and writes both its base-5 and base-6 representations on a blackboard. Later LeRoy sees the two numbers Bernardo has written. Treating the two numbers as base-10 integers, he adds them to obtain an integer <math>S</math>. For example, if <math>N = 749</math>, Bernardo writes the numbers <math>10,\!444</math> and <math>3,\!245</math>, and LeRoy obtains the sum <math>S = 13,\!689</math>. For how many choices of <math>N</math> are the two rightmost digits of <math>S</math>, in order, the same as those of <math>2N</math>?<br />
<br />
<math> \textbf{(A)}\ 5 \qquad\textbf{(B)}\ 10 \qquad\textbf{(C)}\ 15 \qquad\textbf{(D)}\ 20 \qquad\textbf{(E)}\ 25</math><br />
<br />
==Solution==<br />
First, we can examine the units digits of the number base 5 and base 6 and eliminate some possibilities.<br />
<br />
Say that <math>N \equiv a \pmod{6}</math><br />
<br />
also that <math>N \equiv b \pmod{5}</math><br />
<br />
Substituting these equations into the question and setting the units digits of 2N and S equal to each other, it can be seen that <math>a=b</math>, and <math>b < 5</math>, (otherwise <math>a</math> and <math>b</math> always have different parities) so<br />
<math>N \equiv a \pmod{6}</math>,<br />
<math>N \equiv a \pmod{5}</math>,<br />
<math>\implies N=a \pmod{30}</math>,<br />
<math>0 \le a \le 4 </math><br />
<br />
Therefore, <math>N</math> can be written as <math>30x+y</math><br />
and <math>2N</math> can be written as <math>60x+2y</math> <br />
<br />
Keep in mind that <math>y</math> can be one of five choices: <math>0, 1, 2, 3,</math> or <math>4</math>, ;<br />
Also, we have already found which digits of <math>y</math> will add up into the units digits of <math>2N</math>.<br />
<br />
Now, examine the tens digit, <math>x</math> by using <math>\mod{25}</math> and <math>\mod{36}</math> to find the tens digit (units digits can be disregarded because <math>y=0,1,2,3,4</math> will always work)<br />
Then we see that <math>N=30x+y</math> and take it <math>\mod{25}</math> and <math>\mod{36}</math> to find the last two digits in the base <math>5</math> and <math>6</math> representation.<br />
<cmath>N \equiv 30x \pmod{36}</cmath><br />
<cmath>N \equiv 30x \equiv 5x \pmod{25}</cmath> <br />
Both of those must add up to <br />
<cmath>2N\equiv60x \pmod{100}</cmath><br />
<br />
(<math>33 \ge x \ge 4</math>)<br />
<br />
Now, since <math>y=0,1,2,3,4</math> will always work if <math>x</math> works, then we can treat <math>x</math> as a units digit instead of a tens digit in the respective bases and decrease the mods so that <math>x</math> is now the units digit.<br />
<cmath>N \equiv 6x \equiv x \pmod{5}</cmath> <br />
<cmath>N \equiv 5x \pmod{6}</cmath> <br />
<cmath>2N\equiv 6x \pmod{10}</cmath><br />
<br />
Say that <math>x=5m+n</math> (m is between 0-6, n is 0-4 because of constraints on x)<br />
Then <br />
<br />
<cmath>N \equiv 5m+n \pmod{5}</cmath> <br />
<cmath>N \equiv 25m+5n \pmod{6}</cmath> <br />
<cmath>2N\equiv30m + 6n \pmod{10}</cmath><br />
<br />
and this simplifies to <br />
<br />
<cmath>N \equiv n \pmod{5}</cmath> <br />
<cmath>N \equiv m+5n \pmod{6}</cmath><br />
<cmath>2N\equiv 6n \pmod{10}</cmath><br />
<br />
From inspection, when<br />
<br />
<math>n=0, m=6</math><br />
<br />
<math>n=1, m=6</math><br />
<br />
<math>n=2, m=2</math><br />
<br />
<math>n=3, m=2</math><br />
<br />
<math>n=4, m=4</math><br />
<br />
This gives you <math>5</math> choices for <math>x</math>, and <math>5</math> choices for <math>y</math>, so the answer is <br />
<math>5* 5 = \boxed{\textbf{(E) }25}</math><br />
<br />
== See also ==<br />
{{AMC12 box|year=2013|ab=B|num-b=22|num-a=24}}<br />
<br />
{{AMC10 box|year=2013|ab=B|num-b=24|after=Last Question}}<br />
<br />
{{MAA Notice}}</div>Tehetrollr289https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_10B_Problems/Problem_19&diff=758242013 AMC 10B Problems/Problem 192016-02-14T00:28:06Z<p>Tehetrollr289: /* Solution 1 */</p>
<hr />
<div>==Problem==<br />
The real numbers <math>c,b,a</math> form an arithmetic sequence with <math>a\ge b\ge c\ge 0</math>. The quadratic <math>ax^2+bx+c</math> has exactly one root. What is this root?<br />
<br />
<math> \textbf{(A)}\ -7-4\sqrt{3}\qquad\textbf{(B)}\ -2-\sqrt{3}\qquad\textbf{(C)}\ -1\qquad\textbf{(D)}\ -2+\sqrt{3}\qquad\textbf{(E)}\ -7+4\sqrt{3} </math><br />
<br />
==Solutions==<br />
===Solution 1===<br />
It is given that <math>ax^2+bx+c=0</math> has 1 real root, so the discriminant is zero, or <math>b^2=4ac</math>. Because a, b, c are in arithmetic progression, <math>b-a=c-b</math>, or <math>b=\frac {a+c} {2} </math>. We need to find the unique root, or <math>-\frac {b} {2a} </math> (discriminant is 0). From <math>b^2=4ac</math>, we can get <math>-\frac {b} {2a} =-\frac {2c} {b}</math>. Ignoring the negatives, we have <br />
<math>\frac {2c} {b} = \frac {2c} {\frac {a+c} {2}} = \frac {4c} {a+c} = \frac {1} {\frac {1} {\frac {4c} {a+c}}} = \frac {1} {\frac {a+c} {4c}} = \frac {1} {\frac {a} {4c} + \frac {1} {4} }</math>. Fortunately, finding <math>\frac {a} {c} </math> is not very hard. Plug in <math>b=\frac {a+c} {2}</math> to <math>b^2=4ac</math>, we have <math>a^2+2ac+c^2=16ac</math>, or <math>a^2-14ac+c^2=0</math>, and dividing by <math>c^2</math> gives <math>(\frac {a} {c} ) ^2-14( \frac {a} {c} ) +1 = 0</math>, so <math>\frac {a} {c} = \frac {14 \pm \sqrt {192} } {2} = 7 \pm 4 \sqrt {3} </math>. But <math>7-4\sqrt {3} <1</math>, violating the assumption that <math>a \ge c</math>. Therefore, <math>\frac {a} {c} = 7 +4\sqrt {3} </math>. Plugging this in, we have <math>\frac {1} {\frac {a} {4c} + \frac {1} {4} } = \frac {1} {2+ \sqrt {3} } = 2- \sqrt {3} </math>. But we need the negative of this, so the answer is <math>\boxed {\textbf{(D)}}.</math><br />
<br />
===Solution 2===<br />
Note that we can divide the polynomial by <math>a</math> to make the leading coefficient 1 since dividing does not change the roots or the fact that the coefficients are in an arithmetic sequence. Also, we know that there is exactly one root so this equation must be of the form <br />
<math>(x-r)^2 = x^2 - 2rx + r^2</math> where <math>1 \ge -2r \ge r^2 \ge 0</math>.<br />
We now use the fact that the coefficients are in an arithmetic sequence. Note that in any arithmetic sequence, the average is equal to the median. Thus,<br />
<math>r^2 + 1 = -4r</math> and <math>r = -2 \pm \sqrt{3}</math>. Since <math>1 > r^2</math>, we easily see that <math>|r|</math> has to be between 1 and 0. Thus, we can eliminate <math>-2 - \sqrt{3}</math> and are left with <math>\boxed{\textbf{(D)} -2 + \sqrt{3}}</math> as the answer.<br />
<br />
===Solution 3===<br />
Given that <math>ax^2+bx+c=0</math> has only 1 real root, we know that the discriminant must equal 0, or that <math>b^2=4ac</math>. Because the discriminant equals 0, we have that the root of the quadratic is <math>r=\frac {-b} {2a}</math>. We are also given that the coefficients of the quadratic are in arithmetic progression, where <math>a \ge b \ge c \ge 0</math>. Letting the arbitrary difference equal variable <math>d</math>, we have that <math>a=b+d</math> and that <math>c=b-d</math>. Plugging those two equations into <math>b^2=4ac</math>, we have <math>b^2=4(b^2-d^2)=4b^2-4d^2</math> which yields <math>3b^2=4d^2</math>. Isolating <math>d</math>, we have <math>d=\frac {b \sqrt{3}} {2}</math>. Substituting that in for <math>d</math> in <math>a=b+d</math>, we get <math>a=b+\frac {b \sqrt{3}} {2}=b(1+\frac {\sqrt{3}} {2})</math>. Once again, substituting that in for <math>a</math> in <math>r=\frac {-b} {2a}</math>, we have <math>r=\frac {-b} {2b(1+\frac {\sqrt{3}} {2})}=\frac {-1} {2+\sqrt {3}}=-2+\sqrt {3}</math>. The answer is: <cmath>\boxed {\textbf{(D)}}.</cmath><br />
<br />
===Solution 4===<br />
Let the double root be <math>r</math>. Then by the arithmetic progression and Vieta's,<cmath>\begin{align*}a-b & =b-c\\<br />
1-\frac{b}a & =\frac{b}a-\frac{c}a\\<br />
1+2r & =-2r-r^2\\<br />
r^2+4r+1 & =0\\<br />
r & =-2\pm\sqrt{3}\end{align*}</cmath><br />
<br />
We see <math>0\le b\le a\Rightarrow 0\le \frac{b}{a}\le 1</math>, and so we want <math>0\le -2r\le 1</math> . Note that since <math>0\le -2(-2-\sqrt{3})=4+2\sqrt{3}\ge 1</math> and <math>0 \le -2(-2+\sqrt{3})=4-2\sqrt{3}\le 1</math>, we can conclude that <math>r=-2+\sqrt{3}</math>, so the answer is: <cmath>\boxed{\textbf{(D)}}.</cmath><br />
<br />
== See also ==<br />
{{AMC10 box|year=2013|ab=B|num-b=18|num-a=20}}<br />
{{MAA Notice}}</div>Tehetrollr289https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_10B_Problems/Problem_19&diff=758232013 AMC 10B Problems/Problem 192016-02-14T00:27:20Z<p>Tehetrollr289: /* Solutions */</p>
<hr />
<div>==Problem==<br />
The real numbers <math>c,b,a</math> form an arithmetic sequence with <math>a\ge b\ge c\ge 0</math>. The quadratic <math>ax^2+bx+c</math> has exactly one root. What is this root?<br />
<br />
<math> \textbf{(A)}\ -7-4\sqrt{3}\qquad\textbf{(B)}\ -2-\sqrt{3}\qquad\textbf{(C)}\ -1\qquad\textbf{(D)}\ -2+\sqrt{3}\qquad\textbf{(E)}\ -7+4\sqrt{3} </math><br />
<br />
==Solutions==<br />
===Solution 1===<br />
It is given that <math>ax^2+bx+c=0</math> has 1 real root, so the discriminant is zero, or <math>b^2=4ac</math>. Because a, b, c are in arithmetic progression, <math>b-a=c-b</math>, or <math>b=\frac {a+c} {2} </math>. We need to find the unique root, or <math>-\frac {b} {2a} </math> (discriminant is 0). From <math>b^2=4ac</math>, we get <math>-\frac {b} {2a} =-\frac {2c} {b}</math>. Ignoring the negatives, we have <br />
<math>\frac {2c} {b} = \frac {2c} {\frac {a+c} {2}} = \frac {4c} {a+c} = \frac {1} {\frac {1} {\frac {4c} {a+c}}} = \frac {1} {\frac {a+c} {4c}} = \frac {1} {\frac {a} {4c} + \frac {1} {4} }</math>. Fortunately, finding <math>\frac {a} {c} </math> is not very hard. Plug in <math>b=\frac {a+c} {2}</math> to <math>b^2=4ac</math>, we have <math>a^2+2ac+c^2=16ac</math>, or <math>a^2-14ac+c^2=0</math>, and dividing by <math>c^2</math> gives <math>(\frac {a} {c} ) ^2-14( \frac {a} {c} ) +1 = 0</math>, so <math>\frac {a} {c} = \frac {14 \pm \sqrt {192} } {2} = 7 \pm 4 \sqrt {3} </math>. But <math>7-4\sqrt {3} <1</math>, violating the assumption that <math>a \ge c</math>. Therefore, <math>\frac {a} {c} = 7 +4\sqrt {3} </math>. Plugging this in, we have <math>\frac {1} {\frac {a} {4c} + \frac {1} {4} } = \frac {1} {2+ \sqrt {3} } = 2- \sqrt {3} </math>. But we need the negative of this, so the answer is <math>\boxed {\textbf{(D)}}.</math><br />
<br />
===Solution 2===<br />
Note that we can divide the polynomial by <math>a</math> to make the leading coefficient 1 since dividing does not change the roots or the fact that the coefficients are in an arithmetic sequence. Also, we know that there is exactly one root so this equation must be of the form <br />
<math>(x-r)^2 = x^2 - 2rx + r^2</math> where <math>1 \ge -2r \ge r^2 \ge 0</math>.<br />
We now use the fact that the coefficients are in an arithmetic sequence. Note that in any arithmetic sequence, the average is equal to the median. Thus,<br />
<math>r^2 + 1 = -4r</math> and <math>r = -2 \pm \sqrt{3}</math>. Since <math>1 > r^2</math>, we easily see that <math>|r|</math> has to be between 1 and 0. Thus, we can eliminate <math>-2 - \sqrt{3}</math> and are left with <math>\boxed{\textbf{(D)} -2 + \sqrt{3}}</math> as the answer.<br />
<br />
===Solution 3===<br />
Given that <math>ax^2+bx+c=0</math> has only 1 real root, we know that the discriminant must equal 0, or that <math>b^2=4ac</math>. Because the discriminant equals 0, we have that the root of the quadratic is <math>r=\frac {-b} {2a}</math>. We are also given that the coefficients of the quadratic are in arithmetic progression, where <math>a \ge b \ge c \ge 0</math>. Letting the arbitrary difference equal variable <math>d</math>, we have that <math>a=b+d</math> and that <math>c=b-d</math>. Plugging those two equations into <math>b^2=4ac</math>, we have <math>b^2=4(b^2-d^2)=4b^2-4d^2</math> which yields <math>3b^2=4d^2</math>. Isolating <math>d</math>, we have <math>d=\frac {b \sqrt{3}} {2}</math>. Substituting that in for <math>d</math> in <math>a=b+d</math>, we get <math>a=b+\frac {b \sqrt{3}} {2}=b(1+\frac {\sqrt{3}} {2})</math>. Once again, substituting that in for <math>a</math> in <math>r=\frac {-b} {2a}</math>, we have <math>r=\frac {-b} {2b(1+\frac {\sqrt{3}} {2})}=\frac {-1} {2+\sqrt {3}}=-2+\sqrt {3}</math>. The answer is: <cmath>\boxed {\textbf{(D)}}.</cmath><br />
<br />
===Solution 4===<br />
Let the double root be <math>r</math>. Then by the arithmetic progression and Vieta's,<cmath>\begin{align*}a-b & =b-c\\<br />
1-\frac{b}a & =\frac{b}a-\frac{c}a\\<br />
1+2r & =-2r-r^2\\<br />
r^2+4r+1 & =0\\<br />
r & =-2\pm\sqrt{3}\end{align*}</cmath><br />
<br />
We see <math>0\le b\le a\Rightarrow 0\le \frac{b}{a}\le 1</math>, and so we want <math>0\le -2r\le 1</math> . Note that since <math>0\le -2(-2-\sqrt{3})=4+2\sqrt{3}\ge 1</math> and <math>0 \le -2(-2+\sqrt{3})=4-2\sqrt{3}\le 1</math>, we can conclude that <math>r=-2+\sqrt{3}</math>, so the answer is: <cmath>\boxed{\textbf{(D)}}.</cmath><br />
<br />
== See also ==<br />
{{AMC10 box|year=2013|ab=B|num-b=18|num-a=20}}<br />
{{MAA Notice}}</div>Tehetrollr289https://artofproblemsolving.com/wiki/index.php?title=2009_AMC_10A_Problems/Problem_24&diff=757792009 AMC 10A Problems/Problem 242016-02-12T16:40:24Z<p>Tehetrollr289: /* Solution 1 */</p>
<hr />
<div>== Problem ==<br />
<br />
Three distinct vertices of a cube are chosen at random. What is the probability that the plane determined by these three vertices contains points inside the cube?<br />
<br />
<math><br />
\mathrm{(A)}\ \frac{1}{4}<br />
\qquad<br />
\mathrm{(B)}\ \frac{3}{8}<br />
\qquad<br />
\mathrm{(C)}\ \frac{4}{7}<br />
\qquad<br />
\mathrm{(D)}\ \frac{5}{7}<br />
\qquad<br />
\mathrm{(E)}\ \frac{3}{4}<br />
</math><br />
<br />
== Solution ==<br />
=== Solution 1 ===<br />
<br />
We will try to use symmetry as much as possible.<br />
<br />
Pick the first vertex <math>A</math>, its choice clearly does not influence anything.<br />
<br />
Pick the second vertex <math>B</math>. With probability <math>3/7</math> vertices <math>A</math> and <math>B</math> have a common edge, with probability <math>3/7</math> they are in opposite corners of the same face, and with probability <math>1/7</math> they are in opposite corners of the cube. We will handle each of the cases separately.<br />
<br />
In the first case, there are <math>2</math> faces that contain the edge <math>AB</math>. In each of these faces there are <math>2</math> other vertices. If one of these <math>4</math> vertices is the third vertex <math>C</math>, the entire triangle <math>ABC</math> will be on a face. On the other hand, if <math>C</math> is one of the two remaining vertices, the triangle will contain points inside the cube. Hence in this case the probability of choosing a good <math>C</math> is <math>2/6 = 1/3</math>.<br />
<br />
In the second case, the triangle <math>ABC</math> will not intersect the cube if point <math>C</math> is one of the two points on the side that contains <math>AB</math>. Hence the probability of <math>ABC</math> intersecting the inside of the cube is <math>2/3</math>.<br />
<br />
In the third case, already the diagonal <math>AB</math> contains points inside the cube, hence this case will be good regardless of the choice of <math>C</math>.<br />
<br />
Summing up all cases, the resulting probability is:<br />
<cmath><br />
\frac 37\cdot\frac 13 + \frac 37\cdot \frac 23 + \frac 17\cdot 1 = \boxed{\frac 47}<br />
</cmath><br />
<br />
=== Solution 2 ===<br />
<br />
There are <math>\binom{8}{3}=56</math> ways to pick three vertices from eight total vertices; this is our denominator. In order to have three points inside the cube, they cannot be on the surface. Thus, we can use complementary probability.<br />
<br />
There are four ways to choose three points from the vertices of a single face. Since there are six faces, <math>4 \times 6 = 24</math>.<br />
<br />
Thus, the probability of what we don't want is <math>\frac{24}{56} = \frac{3}{7}</math>. Using complementary, <br />
<br />
<cmath><br />
1- \frac 37 = \boxed{\frac 47}<br />
</cmath><br />
<br />
== See Also ==<br />
<br />
{{AMC10 box|year=2009|ab=A|num-b=23|num-a=25}}<br />
<br />
<br />
[[Category:Introductory Combinatorics Problems]]<br />
{{MAA Notice}}</div>Tehetrollr289