https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Thedrummer&feedformat=atomAoPS Wiki - User contributions [en]2024-03-29T15:42:53ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2008_AMC_12B_Problems/Problem_17&diff=504892008 AMC 12B Problems/Problem 172013-01-12T05:13:52Z<p>Thedrummer: /* Solution */</p>
<hr />
<div>==Problem==<br />
<br />
Let <math>A</math>, <math>B</math> and <math>C</math> be three distinct points on the graph of <math>y=x^2</math> such that line <math>AB</math> is parallel to the <math>x</math>-axis and <math>\triangle ABC</math> is a right triangle with area <math>2008</math>. What is the sum of the digits of the <math>y</math>-coordinate of <math>C</math>?<br />
<br />
<math>\textbf{(A)}\ 16\qquad\textbf{(B)}\ 17\qquad\textbf{(C)}\ 18\qquad\textbf{(D)}\ 19\qquad\textbf{(E)}\ 20</math><br />
<br />
==Solution==<br />
<br />
Supposing <math>\angle A=90^\circ</math>, <math>AC</math> is perpendicular to <math>AB</math> and, it follows, to the <math>x</math>-axis, making <math>AB</math> a segment of the line x=m. But that would mean that the coordinates of <math>C</math> are <math>(m, m^2)</math>, contradicting the given that points <math>A</math> and <math>C</math> are distinct. So <math>\angle A</math> is not <math>90^\circ</math>. By a similar logic, neither is <math>\angle B</math>. <br />
<br />
This means that <math>\angle C=90^\circ</math> and <math>AC</math> is perpendicular to <math>BC</math>. Let C be the point <math>(n, n^2)</math>. So the slope of <math>BC</math> is the negative reciprocal of the slope of <math>AC</math>, yielding <math>m+n=\frac{1}{m-n}</math> <math>\Rightarrow</math> <math>m^2-n^2=1</math>. <br />
<br />
Because <math>m^2-n^2</math> is the length of the altitude of triangle <math>ABC</math> from <math>AB</math>, and <math>2m</math> is the length of <math>AB</math>, the area of <math>\triangle ABC=m(m^2-n^2)=2008</math>. Since <math>m^2-n^2=1</math>, <math>m=2008</math>. <br />
Substituting, <math>2008^2-n^2=1</math> <math>\Rightarrow</math> <math>n^2=2008^2-1=4032063</math>, whose digits sum to <math>18 \Rightarrow \textbf{(C)}</math>.<br />
<br />
==See Also==<br />
{{AMC12 box|year=2008|num-b=16|num-a=18|ab=B}}</div>Thedrummerhttps://artofproblemsolving.com/wiki/index.php?title=2008_AMC_12A_Problems/Problem_19&diff=444422008 AMC 12A Problems/Problem 192012-02-03T18:03:53Z<p>Thedrummer: /* Solution */</p>
<hr />
<div>==Problem==<br />
In the expansion of<br />
<cmath>\left(1 + x + x^2 + \cdots + x^{27}\right)\left(1 + x + x^2 + \cdots + x^{14}\right)^2,</cmath><br />
what is the [[coefficient]] of <math>x^{28}</math>?<br />
<br />
<math>\mathrm{(A)}\ 195\qquad\mathrm{(B)}\ 196\qquad\mathrm{(C)}\ 224\qquad\mathrm{(D)}\ 378\qquad\mathrm{(E)}\ 405</math><br />
<br />
==Solution 1== <br />
Let <math>A = \left(1 + x + x^2 + \cdots + x^{14}\right)</math> and <math>B = \left(1 + x + x^2 + \cdots + x^{27}\right)</math>. We are expanding <math>A \cdot A \cdot B</math>. <br />
<br />
Since there are <math>15</math> terms in <math>A</math>, there are <math>15^2 = 225</math> ways to choose one term from each <math>A</math>. The product of the selected terms is <math>x^n</math> for some integer <math>n</math> between <math>0</math> and <math>28</math> inclusive. For each <math>n \neq 0</math>, there is one and only one <math>x^{28 - n}</math> in <math>B</math>. For example, if I choose <math>x^2</math> from <math>A</math> , then there is exactly one power of <math>x</math> in <math>B</math> that I can choose; in this case, it would be <math>x^{24}</math>. Since there is only one way to choose one term from each <math>A</math> to get a product of <math>x^0</math>, there are <math>225 - 1 = 224</math> ways to choose one term from each <math>A</math> and one term from <math>B</math> to get a product of <math>x^{28}</math>. Thus the coefficient of the <math>x^{28}</math> term is <math>224 \Rightarrow C</math>.<br />
<br />
== Solution 2 ==<br />
Let <math>P(x) = \left(1 + x + x^2 + \cdots + x^{14}\right)^2 = a_0 + a_1x + a_2x^2 + \cdots + a_{28}x^{28}</math>. Then the <math>x^{28}</math> term from the product in question <math>\left(1 + x + x^2 + \cdots + x^{27}\right)(a_0 + a_1x + a_2x^2 + \cdots + a_{28}x^{28})</math> is<br />
<br />
<math>1a_{28}x^{28} + xa_{27}x^{27} + x^2a_{26}x^{26} + \cdots + x^{27}a_1x = \left(a_1 + a_2 + \cdots a_{28}\right)x^{28}</math><br />
<br />
So we are trying to find the sum of the coefficients of <math>P(x)</math> minus <math>a_0</math>. Since the constant term <math>a_0</math> in <math>P(x)</math> (when expanded) is <math>1</math>, and the sum of the coefficients of <math>P(x)</math> is <math>P(1)</math>, we find the answer to be<br />
<math>P(1) - a_0<br />
= \left(1 + 1 + 1^2 + \cdots 1^{14}\right)^2 - 1<br />
= 15^2 - 1<br />
= 224 \rightarrow C</math><br />
<br />
<br />
==See Also==<br />
{{AMC12 box|year=2008|ab=A|num-b=18|num-a=20}}<br />
<br />
[[Category:Introductory Algebra Problems]]</div>Thedrummerhttps://artofproblemsolving.com/wiki/index.php?title=1993_AIME_Problems/Problem_14&diff=442221993 AIME Problems/Problem 142012-01-13T03:05:03Z<p>Thedrummer: /* Solution */</p>
<hr />
<div>== Problem ==<br />
A rectangle that is inscribed in a larger rectangle (with one vertex on each side) is called unstuck if it is possible to rotate (however slightly) the smaller rectangle about its center within the confines of the larger. Of all the rectangles that can be inscribed unstuck in a 6 by 8 rectangle, the smallest perimeter has the form <math>\sqrt{N}\,</math>, for a positive integer <math>N\,</math>. Find <math>N\,</math>.<br />
<br />
== Solution ==<br />
Answer: 448.<br />
<br />
Solution: Put the rectangle on the coordinate plane so its vertices are at <math>(\pm4,\pm3)</math>, for all four combinations of positive and negative. Then by symmetry, the other rectangle is also centered at the origin, <math>O</math>.<br />
<br />
Note that such a rectangle is unstuck if its four vertices are in or on the edge of all four quadrants, and it is not the same rectangle as the big one. Let the four vertices of this rectangle be <math>A(4,y)</math>, <math>B(-x,3)</math>, <math>C(-4,-y)</math> and <math>D(x,-3)</math> for nonnegative <math>x,y</math>. Then this is a rectangle, so <math>OA=OB</math>, or <math>16+y^2=9+x^2</math>, so <math>x^2=y^2+7</math>.<br />
<br />
[[Image:1993 AIME 14 Diagram.png|center]]<br />
<br />
Reflect <math>D</math> across the side of the rectangle containing <math>C</math> to <math>D'(-8-x,-3)</math>. Then <math>BD'=\sqrt{(-8-x-(-x))^2+(3-(-3))^2}=10</math> is constant, and the perimeter of the rectangle is equal to <math>2(BC+CD')</math>. The midpoint of <math>\overline{BD'}</math> is <math>(-4-x,0)</math>, and since <math>-4>-4-x</math> and <math>-y\le0</math>, <math>C</math> always lies below <math>\overline{BD'}</math>.<br />
<br />
If <math>y</math> is positive, it can be decreased to <math>y'<y</math>. This causes <math>x</math> to decrease as well, to <math>x'</math>, where <math>x'^2=y'^2+7</math> and <math>x'</math> is still positive. If <math>B</math> and <math>D'</math> are held in place as everything else moves, then <math>C</math> moves <math>(y-y')</math> units up and <math>(x-x')</math> units left to <math>C'</math>, which must lie within <math>\triangle BCD'</math>. Then we must have <math>BC'+C'D'<BC+CD'</math>, and the perimeter of the rectangle is decreased. Therefore, the minimum perimeter must occur with <math>y=0</math>, so <math>x=\sqrt7</math>.<br />
<br />
By the distance formula, this minimum perimeter is <cmath>2\left(\sqrt{(4-\sqrt7)^2+3^2}+\sqrt{(4+\sqrt7)^2+3^2}\right)=4\left(\sqrt{8-2\sqrt7}+\sqrt{8+2\sqrt7}\right)</cmath> <cmath>=4(\sqrt7-1+\sqrt7+1)=8\sqrt7=\sqrt{448}.</cmath><br />
<br />
== Solution 2 ==<br />
Note that the diagonal of the rectangle with minimum perimeter must have the diagonal along the middle segment of length 8 of the rectangle (any other inscribed rectangle can be rotated a bit, then made smaller; this one can't because then the rectangle cannot be inscribed since its longest diagonal is less than 8 in length). Then since a rectangle must have right angles, we draw a circle of radius 4 around the center of the rectangle. Picking the two midpoints on the sides of length 6 and opposite intersection points on the segments of length 8, we form a rectangle. Let <math>a</math> and <math>b</math> be the sides of the rectangle. Then <math>ab = 3(8) = 24</math> since both are twice the area of the same right triangle, and <math>a^2+b^2 = 64</math>. So <math>(a+b)^2 = 64+2(24) = 112</math>, so <math>2(a+b) = \boxed{\sqrt{448}}</math>.<br />
<br />
== See also ==<br />
{{AIME box|year=1993|num-b=13|num-a=15}}</div>Thedrummerhttps://artofproblemsolving.com/wiki/index.php?title=2011_AMC_10A_Problems/Problem_25&diff=369162011 AMC 10A Problems/Problem 252011-02-13T18:04:33Z<p>Thedrummer: Created page with '==Problem 25== 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</ma…'</p>
<hr />
<div>==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></div>Thedrummerhttps://artofproblemsolving.com/wiki/index.php?title=2011_AMC_10A_Problems/Problem_24&diff=369152011 AMC 10A Problems/Problem 242011-02-13T06:52:59Z<p>Thedrummer: Created page with '==Problem 24== 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 t…'</p>
<hr />
<div>==Problem 24==<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>\text{(A)}\,\frac12 \qquad\text{(B)}\,\frac{\sqrt2}{12} \qquad\text{(C)}\,\frac{\sqrt3}{12} \qquad\text{(D)}\,\frac{1}{6} \qquad\text{(E)}\,\frac{\sqrt2}{6}</math></div>Thedrummerhttps://artofproblemsolving.com/wiki/index.php?title=2011_AMC_10A_Problems/Problem_17&diff=369032011 AMC 10A Problems/Problem 172011-02-13T00:25:40Z<p>Thedrummer: Created page with '==Problem 17== 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</mat…'</p>
<hr />
<div>==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></div>Thedrummerhttps://artofproblemsolving.com/wiki/index.php?title=2011_AMC_10A_Problems/Problem_16&diff=369002011 AMC 10A Problems/Problem 162011-02-13T00:02:48Z<p>Thedrummer: Created page with '==Problem 16== Which of the following is equal to <math>\sqrt{9-6\sqrt{2}}+\sqrt{9+6\sqrt{2}}</math>? <math>\text{(A)}\,3\sqrt2 \qquad\text{(B)}\,2\sqrt6 \qquad\text{(C)}\,\frac…'</p>
<hr />
<div>==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></div>Thedrummerhttps://artofproblemsolving.com/wiki/index.php?title=2011_AMC_10A_Problems/Problem_14&diff=368992011 AMC 10A Problems/Problem 142011-02-12T23:27:42Z<p>Thedrummer: Created page with '==Problem 14== 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 numerica…'</p>
<hr />
<div>==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></div>Thedrummerhttps://artofproblemsolving.com/wiki/index.php?title=2011_AMC_10A_Problems/Problem_13&diff=368942011 AMC 10A Problems/Problem 132011-02-12T18:02:56Z<p>Thedrummer: Created page with '==Problem 13== 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}? <math>\text{(A)}\,12 \qquad\text{(B)}\,20…'</p>
<hr />
<div>==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></div>Thedrummerhttps://artofproblemsolving.com/wiki/index.php?title=2011_AMC_10A_Problems/Problem_12&diff=368932011 AMC 10A Problems/Problem 122011-02-12T18:00:19Z<p>Thedrummer: /* Problem 12 */</p>
<hr />
<div>==Problem 12==<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></div>Thedrummerhttps://artofproblemsolving.com/wiki/index.php?title=2011_AMC_10A_Problems/Problem_12&diff=368602011 AMC 10A Problems/Problem 122011-02-11T20:15:39Z<p>Thedrummer: Created page with '==Problem 12== 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 shot…'</p>
<hr />
<div>==Problem 12==<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></div>Thedrummerhttps://artofproblemsolving.com/wiki/index.php?title=2011_AMC_10A_Problems/Problem_11&diff=368482011 AMC 10A Problems/Problem 112011-02-11T06:07:54Z<p>Thedrummer: Created page with '==Problem 11== Square <math>EFGH</math> has one vertex on each side of square <math>ABCD</math>. Point <math>E</math> is on <math>AB</math> with <math>AE=7\cdot EB</math>. What…'</p>
<hr />
<div>==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>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></div>Thedrummerhttps://artofproblemsolving.com/wiki/index.php?title=2011_AMC_10A_Problems/Problem_10&diff=368462011 AMC 10A Problems/Problem 102011-02-11T05:59:24Z<p>Thedrummer: Created page with '==Problem 10== 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 …'</p>
<hr />
<div>==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>\</math><math>17.71</math>. What was the cost of a pencil in cents?<br />
<br />
<math>\text{(A)}\,13 \qquad\text{(B)}\,14 \qquad\text{(C)}\,15 \qquad\text{(D)}\,16 \qquad\text{(E)}\,17</math></div>Thedrummerhttps://artofproblemsolving.com/wiki/index.php?title=2011_AMC_10A_Problems/Problem_7&diff=368422011 AMC 10A Problems/Problem 72011-02-11T05:36:46Z<p>Thedrummer: Created page with '==Problem 7== Which of the following equations does NOT have a solution? <math>\text{(A)}\:(x+7)^2=0</math> <math>\text{(B)}\:|-3x|+5=0</math> <math>\text{(C)}\:\sqrt{-x}-2=0<…'</p>
<hr />
<div>==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></div>Thedrummerhttps://artofproblemsolving.com/wiki/index.php?title=2011_AMC_10A_Problems/Problem_6&diff=368402011 AMC 10A Problems/Problem 62011-02-11T05:29:12Z<p>Thedrummer: Created page with '==Problem 6== Set <math>A</math> has <math>20</math> elements, and set <math>B</math> has <math>15</math> elements. What is the smallest possible number of elements in <math>A \…'</p>
<hr />
<div>==Problem 6==<br />
Set <math>A</math> has <math>20</math> elements, and set <math>B</math> has <math>15</math> elements. What is the smallest possible number of elements in <math>A \cup 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></div>Thedrummerhttps://artofproblemsolving.com/wiki/index.php?title=2010_AMC_10B_Problems/Problem_24&diff=364862010 AMC 10B Problems/Problem 242011-01-28T20:56:57Z<p>Thedrummer: </p>
<hr />
<div>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 />
Manipulating this, 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 />
n=2, a=[1,6]<br />
n=3, a=[1,2]<br />
n=4, a=1<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 a=5 and n=2, we get an integer value for m, which is 9. 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)=34</math></div>Thedrummerhttps://artofproblemsolving.com/wiki/index.php?title=2010_AMC_10B_Problems&diff=364852010 AMC 10B Problems2011-01-28T20:48:20Z<p>Thedrummer: /* Problem 7 */</p>
<hr />
<div>== Problem 1 ==<br />
1. What is <math>100(100-3)-(100\cdot100-3)</math>?<br />
<br />
<math><br />
\mathrm{(A)}\ -20,000<br />
\qquad<br />
\mathrm{(B)}\ -10,000<br />
\qquad<br />
\mathrm{(C)}\ -297<br />
\qquad<br />
\mathrm{(D)}\ -6<br />
\qquad<br />
\mathrm{(E)}\ 0<br />
</math><br />
<br />
<br />
[[2010 AMC 10B Problems/Problem 1|Solution]]<br />
<br />
== Problem 2 ==<br />
<br />
Makarla attended two meetings during her <math>9</math>-hour work day. The first meeting took <math>45</math> minutes and the second meeting took twice as long. What percent of her work day was spent attending meetings?<br />
<br />
<math>\textbf{(A)}\ 15 \qquad \textbf{(B)}\ 20 \qquad \textbf{(C)}\ 25 \qquad \textbf{(D)}\ 30 \qquad \textbf{(E)}\ 35</math><br />
<br />
<br />
[[2010 AMC 10B Problems/Problem 2|Solution]]<br />
<br />
== Problem 3 ==<br />
A drawer contains red, green, blue, and white socks with at least 2 of each color. What is<br />
the minimum number of socks that must be pulled from the drawer to guarantee a matching<br />
pair?<br />
<br />
<math><br />
\mathrm{(A)}\ 3<br />
\qquad<br />
\mathrm{(B)}\ 4<br />
\qquad<br />
\mathrm{(C)}\ 5<br />
\qquad<br />
\mathrm{(D)}\ 8<br />
\qquad<br />
\mathrm{(E)}\ 9<br />
</math><br />
<br />
[[2010 AMC 10B Problems/Problem 3|Solution]]<br />
<br />
== Problem 4 ==<br />
For a real number <math>x</math>, define <math>\heartsuit(x)</math> to be the average of <math>x</math> and <math>x^2</math>. What is <math>\heartsuit(1)+\heartsuit(2)+\heartsuit(3)</math>?<br />
<br />
<math><br />
\mathrm{(A)}\ 3<br />
\qquad<br />
\mathrm{(B)}\ 6<br />
\qquad<br />
\mathrm{(C)}\ 10<br />
\qquad<br />
\mathrm{(D)}\ 12<br />
\qquad<br />
\mathrm{(E)}\ 20<br />
</math><br />
<br />
[[2010 AMC 10B Problems/Problem 4|Solution]]<br />
<br />
== Problem 5 ==<br />
<br />
<br />
A month with <math>31</math> days has the same number of Mondays and Wednesdays.How many of the seven days of the week could be the first day of this month?<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 />
[[2010 AMC 10B Problems/Problem 5|Solution]]<br />
<br />
== Problem 6 ==<br />
<br />
A circle is centered at <math>O</math>, <math>\overbar{AB}</math> is a diameter and <math>C</math> is a point on the circle with <math>\angle COB = 50^\circ</math>.<br />
What is the degree measure of <math>\angle CAB</math>?<br />
<br />
<math><br />
\mathrm{(A)}\ 20<br />
\qquad<br />
\mathrm{(B)}\ 25<br />
\qquad<br />
\mathrm{(C)}\ 45<br />
\qquad<br />
\mathrm{(D)}\ 50<br />
\qquad<br />
\mathrm{(E)}\ 65<br />
</math><br />
<br />
[[2010 AMC 10B Problems/Problem 6|Solution]]<br />
<br />
== Problem 7 ==<br />
<br />
A triangle has side lengths <math>10</math>, <math>10</math>, and <math>12</math>. A rectangle has width <math>4</math> and area equal to the<br />
area of the triangle. What is the perimeter of this rectangle?<br />
<br />
<math><br />
\mathrm{(A)}\ 16<br />
\qquad<br />
\mathrm{(B)}\ 24<br />
\qquad<br />
\mathrm{(C)}\ 28<br />
\qquad<br />
\mathrm{(D)}\ 32<br />
\qquad<br />
\mathrm{(E)}\ 36<br />
</math><br />
<br />
[[2010 AMC 10B Problems/Problem 7|Solution]]<br />
<br />
== Problem 8 ==<br />
<br />
A ticket to a school play cost <math>x</math> dollars, where <math>x</math> is a whole number. A group of 9<sup>th</sup> graders buys tickets costing a total of &#36;<math>48</math>, and a group of 10<sup>th</sup> graders buys tickets costing a total of &#36;<math>64</math>. How many values for <math>x</math> are possible?<br />
<br />
<math>\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 5</math><br />
<br />
[[2010 AMC 10B Problems/Problem 8|Solution]]<br />
<br />
== Problem 9 ==<br />
<br />
Lucky Larry's teacher asked him to substitute numbers for <math>a</math>, <math>b</math>, <math>c</math>, <math>d</math>, and <math>e</math> in the expression <math>a-(b-(c-(d+e)))</math> and evaluate the result. Larry ignored the parenthese but added and subtracted correctly and obtained the correct result by coincidence. The number Larry sustitued for <math>a</math>, <math>b</math>, <math>c</math>, and <math>d</math> were <math>1</math>, <math>2</math>, <math>3</math>, and <math>4</math>, respectively. What number did Larry substitude for <math>e</math>?<br />
<br />
<math>\textbf{(A)}\ -5 \qquad \textbf{(B)}\ -3 \qquad \textbf{(C)}\ 0 \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 5</math><br />
<br />
[[2010 AMC 10B Problems/Problem 9|Solution]]<br />
<br />
== Problem 10 ==<br />
<br />
Shelby drives her scooter at a speed of <math>30</math> miles per hour if it is not raining, and <math>20</math> miles per hour if it is raining. Today she drove in the sun in the morning and in the rain in the evening, for a total of <math>16</math> miles in <math>40</math> minutes. How many minutes did she drive in the rain?<br />
<br />
<math>\textbf{(A)}\ 18 \qquad \textbf{(B)}\ 21 \qquad \textbf{(C)}\ 24 \qquad \textbf{(D)}\ 27 \qquad \textbf{(E)}\ 30</math><br />
<br />
[[2010 AMC 10B Problems/Problem 10|Solution]]<br />
<br />
== Problem 11 ==<br />
A shopper plans to purchase an item that has a listed price greater than &#36;<math>100</math> and can use any one of the three coupns. Coupon A gives <math>15\%</math> off the listed price, Coupon B gives &#36;<math>30</math> off the listed price, and Coupon C gives <math>25\%</math> off the amount by which the listed price exceeds<br />
&#36;<math>100</math>. <br/><br />
Let <math>x</math> and <math>y</math> be the smallest and largest prices, respectively, for which Coupon A saves at least as many dollars as Coupon B or C. What is <math>y</math> − <math>x</math>?<br />
<br />
<math><br />
\mathrm{(A)}\ 50<br />
\qquad<br />
\mathrm{(B)}\ 60<br />
\qquad<br />
\mathrm{(C)}\ 75<br />
\qquad<br />
\mathrm{(D)}\ 80 <br />
\qquad<br />
\mathrm{(E)}\ 100<br />
</math><br />
<br />
[[2010 AMC 10B Problems/Problem 11|Solution]]<br />
<br />
== Problem 12 ==<br />
<br />
At the beginning of the school year, <math>50\%</math> of all students in Mr. Wells' math class answered "Yes" to the question "Do you love math", and <math>50\%</math> answered "No." At the end of the school year, <math>70\%</math> answered "Yes" and <math>30\%</math> answerws "No." Altogether, <math>x\%</math> of the students gave a different answer at the beginning and end of the school year. What is the difference between the maximum and the minimum possible values of <math>x</math>?<br />
<br />
<math>\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 20 \qquad \textbf{(C)}\ 40 \qquad \textbf{(D)}\ 60 \qquad \textbf{(E)}\ 80</math><br />
<br />
[[2010 AMC 10B Problems/Problem 12|Solution]]<br />
<br />
== Problem 13 ==<br />
What is the sum of all the solutions of <math>x = \left|2x-|60-2x|\right|</math>?<br />
<br />
<math><br />
\mathrm{(A)}\ 32<br />
\qquad<br />
\mathrm{(B)}\ 60<br />
\qquad<br />
\mathrm{(C)}\ 92<br />
\qquad<br />
\mathrm{(D)}\ 120<br />
\qquad<br />
\mathrm{(E)}\ 124<br />
</math><br />
<br />
[[2010 AMC 10B Problems/Problem 13|Solution]]<br />
<br />
== Problem 14 ==<br />
<br />
The average of the numbers <math>1, 2, 3,\cdots, 98, 99,</math> and <math>x</math> is <math>100x</math>. What is <math>x</math>?<br />
<br />
<math>\textbf{(A)}\ \dfrac{49}{101} \qquad \textbf{(B)}\ \dfrac{50}{101} \qquad \textbf{(C)}\ \dfrac{1}{2} \qquad \textbf{(D)}\ \dfrac{51}{101} \qquad \textbf{(E)}\ \dfrac{50}{99}</math><br />
<br />
[[2010 AMC 10B Problems/Problem 14|Solution]]<br />
<br />
== Problem 15 ==<br />
On a <math>50</math>-question multiple choice math contest, students receive <math>4</math> points for a correct answer, <math>0</math> points for an answer left blank, and <math>-1</math> point for an incorrect answer. Jesse’s total score on the contest was <math>99</math>. What is the maximum number of questions that Jesse could have answered correctly?<br />
<br />
<math><br />
\mathrm{(A)}\ 25<br />
\qquad<br />
\mathrm{(B)}\ 27<br />
\qquad<br />
\mathrm{(C)}\ 29<br />
\qquad<br />
\mathrm{(D)}\ 31<br />
\qquad<br />
\mathrm{(E)}\ 33<br />
</math><br />
<br />
[[2010 AMC 10B Problems/Problem 15|Solution]]<br />
<br />
== Problem 16 ==<br />
A square of side length <math>1</math> and a circle of radius <math>\dfrac{\sqrt{3}}{3}</math> share the same center. What is the area inside the circle, but outside the square?<br />
<br />
<math><br />
\mathrm{(A)}\ \dfrac{\pi}{3}-1<br />
\qquad<br />
\mathrm{(B)}\ \dfrac{2\pi}{9}-\dfrac{\sqrt{3}}{3}<br />
\qquad<br />
\mathrm{(C)}\ \dfrac{\pi}{18}<br />
\qquad<br />
\mathrm{(D)}\ \dfrac{1}{4}<br />
\qquad<br />
\mathrm{(E)}\ \dfrac{2\pi}{9}<br />
</math><br />
<br />
[[2010 AMC 10B Problems/Problem 16|Solution]]<br />
<br />
== Problem 17 ==<br />
Every high school in the city of Euclid sent a team of <math>3</math> students to a math contest. Each participant in the contest received a different score. Andrea's score was the median among all students, and hers was the highest score on her team. Andrea's teammates Beth and Carla placed <math>37</math><sup>th</sup> and <math>64</math><sup>th</sup>, respectively. How many schools are in the city?<br />
<br />
<math>\textbf{(A)}\ 22 \qquad \textbf{(B)}\ 23 \qquad \textbf{(C)}\ 24 \qquad \textbf{(D)}\ 25 \qquad \textbf{(E)}\ 26</math><br />
<br />
[[2010 AMC 10B Problems/Problem 17|Solution]]<br />
<br />
== Problem 18 ==<br />
<br />
Positive integers <math>a</math>, <math>b</math>, and <math>c</math> are randomly and independently selected with replacement from the set <math>\{1, 2, 3,\dots, 2010\}</math>. What is the probability that <math>abc + ab + a</math> is divisible by <math>3</math>?<br />
<br />
<math>\textbf{(A)}\ \dfrac{1}{3} \qquad \textbf{(B)}\ \dfrac{29}{81} \qquad \textbf{(C)}\ \dfrac{31}{81} \qquad \textbf{(D)}\ \dfrac{11}{27} \qquad \textbf{(E)}\ \dfrac{13}{27}</math><br />
<br />
[[2010 AMC 10B Problems/Problem 18|Solution]]<br />
<br />
== Problem 19 ==<br />
<br />
A circle with center <math>O</math> has area <math>156\pi</math>. Triangle <math>ABC</math> is equilateral, <math>\overbar{BC}</math> is a chord on the circle, <math>OA = 4\sqrt{3}</math>, and point <math>O</math> is outside <math>\triangle ABC</math>. What is the side length of <math>\triangle ABC</math>?<br />
<br />
<math><br />
\mathrm{(A)}\ 2\sqrt{3}<br />
\qquad<br />
\mathrm{(B)}\ 6<br />
\qquad<br />
\mathrm{(C)}\ 4\sqrt{3}<br />
\qquad<br />
\mathrm{(D)}\ 12<br />
\qquad<br />
\mathrm{(E)}\ 18<br />
</math><br />
<br />
[[2010 AMC 10B Problems/Problem 19|Solution]]<br />
<br />
== Problem 20 ==<br />
<br />
Two circles lie outside regular hexagon <math>ABCDEF</math>. The first is tangent to <math>\overbar{AB}</math>, and the second is tangent to <math>\overbar{DE}</math>. Both are tangent to lines <math>BC</math> and <math>FA</math>. What is the ratio of the area of the second circle to that of the first circle?<br />
<br />
<math><br />
\mathrm{(A)}\ 18<br />
\qquad<br />
\mathrm{(B)}\ 27<br />
\qquad<br />
\mathrm{(C)}\ 36<br />
\qquad<br />
\mathrm{(D)}\ 81<br />
\qquad<br />
\mathrm{(E)}\ 108<br />
</math><br />
<br />
[[2010 AMC 10B Problems/Problem 20|Solution]]<br />
<br />
== Problem 21 ==<br />
<br />
<br />
A palindrome between <math>1000</math> and <math>10,000</math> is chosen at random. What is the probability that it is divisible by <math>7</math>?<br />
<br />
<math>\textbf{(A)}\ \dfrac{1}{10} \qquad \textbf{(B)}\ \dfrac{1}{9} \qquad \textbf{(C)}\ \dfrac{1}{7} \qquad \textbf{(D)}\ \dfrac{1}{6} \qquad \textbf{(E)}\ \dfrac{1}{5}</math><br />
<br />
[[2010 AMC 10B Problems/Problem 21|Solution]]<br />
<br />
== Problem 22 ==<br />
<br />
Seven distinct pieces of candy are to be distributed among three bags. The red bag and the blue bag must each receive at least one piece of candy; the white bag may remain empty. How many arrangements are possible?<br />
<br />
<math><br />
\mathrm{(A)}\ 1930<br />
\qquad<br />
\mathrm{(B)}\ 1931<br />
\qquad<br />
\mathrm{(C)}\ 1932<br />
\qquad<br />
\mathrm{(D)}\ 1933<br />
\qquad<br />
\mathrm{(E)}\ 1934<br />
</math><br />
<br />
[[2010 AMC 10B Problems/Problem 22|Solution]]<br />
<br />
== Problem 23 ==<br />
The entries in a <math>3 \times 3</math> array include all the digits from <math>1</math> through <math>9</math>, 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 />
[[2010 AMC 10B Problems/Problem 23|Solution]]<br />
<br />
== Problem 24 ==<br />
<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 />
[[2010 AMC 10B Problems/Problem 24|Solution]]<br />
<br />
== Problem 25 ==<br />
<br />
Let <math>a > 0</math>, and let <math>P(x)</math> be a polynomial with integer coefficients such that<br />
<br />
<center><br />
<math>P(1) = P(3) = P(5) = P(7) = a</math>, and<br/><br />
<math>P(2) = P(4) = P(6) = P(8) = -a</math>.<br />
</center><br />
<br />
What is the smallest possible value of <math>a</math>?<br />
<br />
<math>\textbf{(A)}\ 105 \qquad \textbf{(B)}\ 315 \qquad \textbf{(C)}\ 945 \qquad \textbf{(D)}\ 7! \qquad \textbf{(E)}\ 8!</math><br />
<br />
[[2010 AMC 10B Problems/Problem 25|Solution]]</div>Thedrummer