https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Qszwdxefc&feedformat=atomAoPS Wiki - User contributions [en]2024-03-29T05:41:35ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2005_AMC_12A_Problems/Problem_9&diff=333382005 AMC 12A Problems/Problem 92010-02-08T05:04:48Z<p>Qszwdxefc: /* Solution */</p>
<hr />
<div>== Problem ==<br />
There are two values of <math>a</math> for which the equation <math>4x^2 + ax + 8x + 9 = 0</math> has only one solution for <math>x</math>. What is the sum of these values of <math>a</math>?<br />
<br />
<math>(\mathrm {A}) \ -16 \qquad (\mathrm {B}) \ -8 \qquad (\mathrm {C})\ 0 \qquad (\mathrm {D}) \ 8 \qquad (\mathrm {E})\ 20</math><br />
<br />
== Solution ==<br />
=== Solution 1 ===<br />
A [[quadratic equation]] always has two roots, unless it has a double root. That means we can write the quadratic as a square, and the coefficients 4 and 9 suggest this. [[Completing the square]], <math>0 = (2x \pm 3)^2 = 4x^2 \pm 12x + 9</math>, so <math>\pm 12 = a + 8 \Longrightarrow a = 4, -20</math>. The sum of these is <math>-20 + 4 = -16 \Rightarrow \mathrm{(A)}</math>.<br />
<br />
=== Solution 2 ===<br />
Another method would be to use the quadratic formula, since our <math>x^2</math> coefficient is given as 4, the <math>x</math> coefficient is <math>a+8</math> and the constant term is <math>9</math>. Hence, <math>x = \frac{-(a+8) \pm \sqrt {(a+8)^2-4(4)(9)}}{2(4)}</math> Because we want only a single solution for <math>x</math>, the determinant must equal 0. Therefore, we can write <math>(a+8)^2 - 144 = 0</math> which factors to <math>a^2 + 16a - 80 = 0</math>; using [[Vieta's formulas]] we see that the sum of the solutions for <math>a</math> is the opposite of the coefficient of <math>a</math>, or <math>-16 \Rightarrow \mathrm{ (A)}</math>.<br />
<br />
=== Solution 3 ===<br />
Using the [[discriminant]], the result must equal <math>0</math>.<br />
<math>D = b^2 - 4ac</math><br />
<math> = (a+8)^2 - 4(4)(9)</math><br />
<math> = a^2 + 16a + 64 - 144</math><br />
<math> = a^2 + 16a - 80 = 0 \Rightarrow</math><br />
<math> (a + 20)(a - 4) = 0</math><br />
Therefore, <math>a = -20</math> or <math>a = 4</math>, giving a sum of <math>-16 \Rightarrow \mathrm{ (A)}</math>.<br />
<br />
== See also ==<br />
{{AMC12 box|year=2005|num-b=8|num-a=10|ab=A}}<br />
<br />
[[Category:Introductory Algebra Problems]]</div>Qszwdxefchttps://artofproblemsolving.com/wiki/index.php?title=2006_AMC_12A_Problems/Problem_19&diff=333212006 AMC 12A Problems/Problem 192010-02-06T03:41:25Z<p>Qszwdxefc: /* Solution 1 */</p>
<hr />
<div>== Problem ==<br />
[[Circle]]s with [[center]]s <math>(2,4)</math> and <math>(14,9)</math> have [[radius | radii]] <math>4</math> and <math>9</math>, respectively. The equation of a common external [[tangent line|tangent]] to the circles can be written in the form <math>y=mx+b</math> with <math>m>0</math>. What is <math>b</math>?<br />
<center>[[Image:AMC12_2006A_19.png]]</center><br />
<br />
<math> \mathrm{(A) \ } \frac{908}{199}\qquad \mathrm{(B) \ } \frac{909}{119}\qquad \mathrm{(C) \ } \frac{130}{17}\qquad \mathrm{(D) \ } \frac{911}{119}\qquad \mathrm{(E) \ } \frac{912}{119}</math><br />
<br />
== Solutions ==<br />
<br />
=== Solution 1 ===<br />
:''This solution needs a clearer explanation and a diagram.''<br />
Notice that both [[circle]]s are [[tangent]] to the [[x-axis]] and each other. Call the circles (respectively) A and B; the [[distance]] between the two centers is <math>4 + 9 = 13</math>. If we draw the parallel radii that lead to the common [[external tangent]], a line can be extended parallel to the tangent from A to the radius of circle B. This creates a 5-12-13 triangle. To find the slope of that line (which is parallel to the tangent), note that another 5-12-13 triangle can be drawn below the first one such that the side with length 12 is parallel to the x-axis. The slope can be found by using the [[double tangent identity]],<br />
<br />
:<math>\tan (2 \tan ^{-1} \left(\frac{5}{12}\right)) = \frac{\frac{5}{12} + \frac{5}{12}}{1 - \frac{5}{12}\frac{5}{12}}</math><br />
:<math>= \frac{120}{119}</math><br />
<br />
To find the x and y coordinates of the point of tangency of circle A, we can set up a ratio (the slope will be –119/120 because it is the negative reciprocal):<br />
<br />
:<math>\frac{119}{\sqrt{119^2 + 120^2}}</math> <math>=</math> <math>\frac{119}{169} = \frac{y - 4}{4}</math><br />
<br />
:<math>\frac{-120}{\sqrt{119^2 + 120^2}} = \frac{-120}{169} = \frac{x - 2}{4}</math><br />
<br />
:<math>x = \frac{-142}{169}, y = \frac{1152}{169}</math><br />
<br />
We can plug this into the equation of the line for the tangent to get:<br />
<br />
:<math>\frac{1152}{169} = \frac{120}{119}\frac{-142}{169} + b</math><br />
:<math>b = \frac{912}{119}</math> <math>\Rightarrow \mathrm{E}</math><br />
<br />
=== Solution 2 ===<br />
<br />
By [[User:skiron|skiron]].<br />
<br />
Let <math>L_1</math> be the line that goes through <math>(2,4)</math> and <math>(14,9)</math>, and let <math>L_2</math> be the line <math>y=mx+b</math>. If we let <math>\theta</math> be the measure of the acute angle formed by <math>L_1</math> and the x-axis, then <math>\tan\theta=\frac{5}{12}</math>. <math>L_1</math> clearly bisects the angle formed by <math>L_2</math> and the x-axis, so <math>m=\tan{2\theta}=\frac{2\tan\theta}{1-\tan^2{\theta}}=\frac{120}{119}</math>. We also know that <math>L_1</math> and <math>L_2</math> intersect at a point on the x-axis. The equation of <math>L_1</math> is <math>y=\frac{5}{12}x+\frac{19}{6}</math>, so the coordinate of this point is <math>\left(-\frac{38}{5},0\right)</math>. Hence the equation of <math>L_2</math> is <math>y=\frac{120}{119}x+\frac{912}{119}</math>, so <math>b=\frac{912}{119}</math>, and our answer choice is <math>\boxed{\mathrm{E}}</math>. <br />
<br />
== See also ==<br />
{{AMC12 box|year=2006|ab=A|num-b=18|num-a=20}}<br />
<br />
[[Category:Intermediate Geometry Problems]]</div>Qszwdxefchttps://artofproblemsolving.com/wiki/index.php?title=2006_AMC_12B_Problems/Problem_16&diff=333042006 AMC 12B Problems/Problem 162010-02-04T05:12:30Z<p>Qszwdxefc: /* Solution */</p>
<hr />
<div>{{empty}}<br />
<br />
== Problem ==<br />
Regular hexagon <math>ABCDEF</math> has vertices <math>A</math> and <math>C</math> at <math>(0,0)</math> and <math>(7,1)</math>, respectively. What is its area?<br />
<br />
<math><br />
\mathrm{(A)}\ 20\sqrt {3}<br />
\qquad<br />
\mathrm{(B)}\ 22\sqrt {3}<br />
\qquad<br />
\mathrm{(C)}\ 25\sqrt {3}<br />
\qquad<br />
\mathrm{(D)}\ 27\sqrt {3}<br />
\qquad<br />
\mathrm{(E)}\ 50<br />
</math><br />
<br />
== Solution ==<br />
http://www.unl.edu/amc/mathclub/5-0,problems/H-problems/H-pdfs/2006/HB2006-16.pdf<br />
<br />
== See also ==<br />
{{AMC12 box|year=2006|ab=B|num-b=15|num-a=17}}</div>Qszwdxefchttps://artofproblemsolving.com/wiki/index.php?title=2006_AMC_12B_Problems/Problem_16&diff=333032006 AMC 12B Problems/Problem 162010-02-04T05:12:08Z<p>Qszwdxefc: /* Problem */</p>
<hr />
<div>{{empty}}<br />
<br />
== Problem ==<br />
Regular hexagon <math>ABCDEF</math> has vertices <math>A</math> and <math>C</math> at <math>(0,0)</math> and <math>(7,1)</math>, respectively. What is its area?<br />
<br />
<math><br />
\mathrm{(A)}\ 20\sqrt {3}<br />
\qquad<br />
\mathrm{(B)}\ 22\sqrt {3}<br />
\qquad<br />
\mathrm{(C)}\ 25\sqrt {3}<br />
\qquad<br />
\mathrm{(D)}\ 27\sqrt {3}<br />
\qquad<br />
\mathrm{(E)}\ 50<br />
</math><br />
<br />
== Solution ==<br />
<br />
== See also ==<br />
{{AMC12 box|year=2006|ab=B|num-b=15|num-a=17}}</div>Qszwdxefchttps://artofproblemsolving.com/wiki/index.php?title=2006_AMC_12B_Problems/Problem_25&diff=333022006 AMC 12B Problems/Problem 252010-02-04T05:10:47Z<p>Qszwdxefc: /* Problem */</p>
<hr />
<div><br />
{{empty}}<br />
== Problem ==<br />
A sequence <math>a_1,a_2,\dots</math> of non-negative integers is defined by the rule <math>a_{n+2}=|a_{n+1}-a_n|</math> for <math>n\geq 1</math>. If <math>a_1=999</math>, <math>a_2<999</math> and <math>a_{2006}=1</math>, how many different values of <math>a_2</math> are possible?<br />
<br />
<math><br />
\mathrm{(A)}\ 165<br />
\qquad<br />
\mathrm{(B)}\ 324<br />
\qquad<br />
\mathrm{(C)}\ 495<br />
\qquad<br />
\mathrm{(D)}\ 499<br />
\qquad<br />
\mathrm{(E)}\ 660<br />
</math><br />
<br />
== Solution ==<br />
http://www.unl.edu/amc/mathclub/5-0,problems/H-problems/H-pdfs/2006/HB2006-25.pdf<br />
<br />
== See also ==<br />
{{AMC12 box|year=2006|ab=B|num-b=24|after=Last Question}}</div>Qszwdxefchttps://artofproblemsolving.com/wiki/index.php?title=2006_AMC_12B_Problems/Problem_25&diff=333012006 AMC 12B Problems/Problem 252010-02-04T05:10:18Z<p>Qszwdxefc: /* Solution */</p>
<hr />
<div><br />
{{empty}}<br />
== Problem ==<br />
{{problem}}<br />
<br />
== Solution ==<br />
http://www.unl.edu/amc/mathclub/5-0,problems/H-problems/H-pdfs/2006/HB2006-25.pdf<br />
<br />
== See also ==<br />
{{AMC12 box|year=2006|ab=B|num-b=24|after=Last Question}}</div>Qszwdxefchttps://artofproblemsolving.com/wiki/index.php?title=2006_AMC_12B_Problems&diff=333002006 AMC 12B Problems2010-02-04T03:57:39Z<p>Qszwdxefc: /* Problem */</p>
<hr />
<div>== Problem 1 ==<br />
What is <math>( - 1)^1 + ( - 1)^2 + \cdots + ( - 1)^{2006}</math>?<br />
<br />
<math><br />
\text {(A) } - 2006 \qquad \text {(B) } - 1 \qquad \text {(C) } 0 \qquad \text {(D) } 1 \qquad \text {(E) } 2006<br />
</math><br />
<br />
[[2006 AMC 12B Problems/Problem 1|Solution]]<br />
<br />
== Problem 2 ==<br />
For real numbers <math>x</math> and <math>y</math>, define <math>x\spadesuit y = (x + y)(x - y)</math>. What is <math>3\spadesuit(4\spadesuit 5)</math>?<br />
<br />
<math>\text {(A) } - 72 \qquad \text {(B) } - 27 \qquad \text {(C) } - 24 \qquad \text {(D) } 24 \qquad \text {(E) } 72</math><br />
<br />
[[2006 AMC 12B Problems/Problem 2|Solution]]<br />
<br />
== Problem 3 ==<br />
A football game was played between two teams, the Cougars and the Panthers. The two teams scored a total of 34 points, and the Cougars won by a margin of 14 points. How many points did the Panthers score?<br />
<br />
<math><br />
\text {(A) } 10 \qquad \text {(B) } 14 \qquad \text {(C) } 17 \qquad \text {(D) } 20 \qquad \text {(E) } 24<br />
</math><br />
<br />
[[2006 AMC 12B Problems/Problem 3|Solution]]<br />
<br />
== Problem 4 ==<br />
Mary is about to pay for five items at the grocery store. The prices of the items are <math>\</math> <math>7.99</math>, <math>\</math> <math>4.99</math>, <math>\</math> <math>2.99</math>, <math>\</math> <math>1.99</math>, and <math>\</math> <math>0.99</math>. Mary will pay with a twenty-dollar bill. Which of the following is closest to the percentage of the <math>\</math> <math>20.00</math> that she will receive in change?<br />
<br />
<math><br />
\text {(A) } 5 \qquad \text {(B) } 10 \qquad \text {(C) } 15 \qquad \text {(D) } 20 \qquad \text {(E) } 25<br />
</math><br />
<br />
[[2006 AMC 12B Problems/Problem 4|Solution]]<br />
<br />
== Problem 5 ==<br />
John is walking east at a speed of 3 miles per hour, while Bob is also walking east, but at a speed of 5 miles per hour. If Bob is now 1 mile west of John, how many minutes will it take for Bob to catch up to John?<br />
<br />
<math><br />
\text {(A) } 30 \qquad \text {(B) } 50 \qquad \text {(C) } 60 \qquad \text {(D) } 90 \qquad \text {(E) } 120<br />
</math><br />
<br />
[[2006 AMC 12B Problems/Problem 5|Solution]]<br />
<br />
== Problem 6 ==<br />
Francesca uses 100 grams of lemon juice, 100 grams of sugar, and 400 grams of water to make lemonade. There are 25 calories in 100 grams of lemon juice and 386 calories in 100 grams of sugar. Water contains no calories. How many calories are in 200 grams of her lemonade.<br />
<br />
<math><br />
\text {(A) } 129 \qquad \text {(B) } 137 \qquad \text {(C) } 174 \qquad \text {(D) } 223 \qquad \text {(E) } 411<br />
</math><br />
<br />
[[2006 AMC 12B Problems/Problem 6|Solution]]<br />
<br />
== Problem 7 ==<br />
Mr. and Mrs. Lopez have two children. When they get into their family car, two people sit in the front, and the other two sit in the back. Either Mr. Lopez or Mrs. Lopez must sit in the driver's seat. How many seating arrangements are possible?<br />
<br />
<math><br />
\text {(A) } 4 \qquad \text {(B) } 12 \qquad \text {(C) } 16 \qquad \text {(D) } 24 \qquad \text {(E) } 48<br />
</math><br />
<br />
[[2006 AMC 12B Problems/Problem 7|Solution]]<br />
<br />
== Problem 8 ==<br />
The lines <math>x = \frac 14y + a</math> and <math>y = \frac 14x + b</math> intersect at the point <math>(1,2)</math>. What is <math>a + b</math>?<br />
<br />
<math><br />
\text {(A) } 0 \qquad \text {(B) } \frac 34 \qquad \text {(C) } 1 \qquad \text {(D) } 2 \qquad \text {(E) } \frac 94<br />
</math><br />
<br />
[[2006 AMC 12B Problems/Problem 8|Solution]]<br />
<br />
== Problem 9 ==<br />
How many even three-digit integers have the property that their digits, read left to right, are in strictly increasing order? <br />
<br />
<math><br />
\text {(A) } 21 \qquad \text {(B) } 34 \qquad \text {(C) } 51 \qquad \text {(D) } 72 \qquad \text {(E) } 150<br />
</math><br />
<br />
[[2006 AMC 12B Problems/Problem 9|Solution]]<br />
<br />
== Problem 10 ==<br />
Joe and JoAnn each bought 12 ounces of coffee in a 16-ounce cup. Joe drank 2 ounces of his coffee and then added 2 ounces of cream. JoAnn added 2 ounces of cream, stirred the coffee well, and then drank 2 ounces. What is the resulting ratio of the amount of cream in Joe's coffee to that in JoAnn's coffee?<br />
<br />
<math><br />
\text {(A) } \frac 67 \qquad \text {(B) } \frac{13}{14} \qquad \text {(C) } 1 \qquad \text {(D) } \frac{14}{13} \qquad \text {(E) } \frac 76<br />
</math><br />
<br />
[[2006 AMC 12B Problems/Problem 10|Solution]]<br />
<br />
== Problem 11 ==<br />
Joe and JoAnn each bought 12 ounces of coffee in a 16-ounce cup. Joe drank 2 ounces of his coffee and then added 2 ounces of cream. JoAnn added 2 ounces of cream, stirred the coffee well, and then drank 2 ounces. What is the resulting ratio of the amount of cream in Joe's coffee to that in JoAnn's coffee?<br />
<br />
<math><br />
\text {(A) } \frac 67 \qquad \text {(B) } \frac {13}{14} \qquad \text {(C) } 1 \qquad \text {(D) } \frac {14}{13} \qquad \text {(E) } \frac 76<br />
</math><br />
<br />
[[2006 AMC 12B Problems/Problem 11|Solution]]<br />
<br />
== Problem 12 ==<br />
The parabola <math>y=ax^2+bx+c</math> has vertex <math>(p,p)</math> and <math>y</math>-intercept <math>(0,-p)</math>, where <math>p\ne 0</math>. What is <math>b</math>?<br />
<br />
<math><br />
\text {(A) } -p \qquad \text {(B) } 0 \qquad \text {(C) } 2 \qquad \text {(D) } 4 \qquad \text {(E) } p<br />
</math><br />
<br />
[[2006 AMC 12B Problems/Problem 12|Solution]]<br />
<br />
== Problem 13 ==<br />
{{problem}}<br />
<br />
[[2006 AMC 12B Problems/Problem 13|Solution]]<br />
<br />
== Problem 14 ==<br />
Elmo makes <math>N</math> sandwiches for a fundraiser. For each sandwich he uses <math>B</math> globs of peanut butter at <math>4</math> cents per glob and <math>J</math> blobs of jam at <math>5</math> cents per glob. The cost of the peanut butter and jam to make all the sandwiches is <math>\</math><math> 2.53</math>. Assume that <math>B</math>, <math>J</math> and <math>N</math> are all positive integers with <math>N>1</math>. What is the cost of the jam Elmo uses to make the sandwiches? <br />
<br />
<math><br />
\mathrm{(A)}\ 1.05<br />
\qquad<br />
\mathrm{(B)}\ 1.25<br />
\qquad<br />
\mathrm{(C)}\ 1.45<br />
\qquad<br />
\mathrm{(D)}\ 1.65<br />
\qquad<br />
\mathrm{(E)}\ 1.85<br />
</math><br />
<br />
[[2006 AMC 12B Problems/Problem 14|Solution]]<br />
<br />
== Problem 15 ==<br />
{{problem}}<br />
<br />
[[2006 AMC 12B Problems/Problem 15|Solution]]<br />
<br />
== Problem 16 ==<br />
<br />
Regular hexagon <math>ABCDEF</math> has vertices <math>A</math> and <math>C</math> at <math>(0,0)</math> and <math>(7,1)</math>, respectively. What is its area?<br />
<br />
<math><br />
\mathrm{(A)}\ 20\sqrt {3}<br />
\qquad<br />
\mathrm{(B)}\ 22\sqrt {3}<br />
\qquad<br />
\mathrm{(C)}\ 25\sqrt {3}<br />
\qquad<br />
\mathrm{(D)}\ 27\sqrt {3}<br />
\qquad<br />
\mathrm{(E)}\ 50<br />
</math><br />
<br />
[[2006 AMC 12B Problems/Problem 16|Solution]]<br />
<br />
== Problem 17 ==<br />
<br />
For a particular peculiar pair of dice, the probabilities of rolling <math>1</math>, <math>2</math>, <math>3</math>, <math>4</math>, <math>5</math> and <math>6</math> on each die are in the ratio <math>1:2:3:4:5:6</math>. What is the probability of rolling a total of <math>7</math> on the two dice?<br />
<br />
<math><br />
\mathrm{(A)}\ \frac 4{63}<br />
\qquad<br />
\mathrm{(B)}\ \frac 18 <br />
\qquad<br />
\mathrm{(C)}\ \frac 8{63}<br />
\qquad<br />
\mathrm{(D)}\ \frac 16<br />
\qquad<br />
\mathrm{(E)}\ \frac 27<br />
</math><br />
<br />
[[2006 AMC 12B Problems/Problem 17|Solution]]<br />
<br />
== Problem 18 ==<br />
An object in the plane moves from one lattice point to another. At each step, the object may move one unit to the right, one unit to the left, one unit up, or one unit down. If the object starts at the origin and takes a ten-step path, how many different points could be the final point?<br />
<br />
<math><br />
\mathrm{(A)}\ 120<br />
\qquad<br />
\mathrm{(B)}\ 121<br />
\qquad<br />
\mathrm{(C)}\ 221<br />
\qquad<br />
\mathrm{(D)}\ 230<br />
\qquad<br />
\mathrm{(E)}\ 231<br />
</math><br />
<br />
[[2006 AMC 12B Problems/Problem 18|Solution]]<br />
<br />
== Problem 19 ==<br />
Mr. Jones has eight children of different ages. On a family trip his olderst child, who is 9, spots a license plate with a 4-digit number in which each of two digits appears two times. "Look, daddy!" she exclaims. "That number is evenly divisible by the age of each of us kids!" "That's right," replies Mr. Jones, "and the last two digits just happen to be my age." Which of the following is not the age of one of Mr. Jones's children?<br />
<br />
<math><br />
\mathrm{(A)}\ 4<br />
\qquad<br />
\mathrm{(B)}\ 5<br />
\qquad<br />
\mathrm{(C)}\ 6<br />
\qquad<br />
\mathrm{(D)}\ 7<br />
\qquad<br />
\mathrm{(E)}\ 8<br />
</math><br />
<br />
[[2006 AMC 12B Problems/Problem 19|Solution]]<br />
<br />
== Problem 20 ==<br />
Let <math>x</math> be chosen at random from the interval <math>(0,1)</math>. What is the probability that<br />
<math>\lfloor\log_{10}4x\rfloor - \lfloor\log_{10}x\rfloor = 0</math>?<br />
Here <math>\lfloor x\rfloor</math> denotes the greatest integer that is less than or equal to <math>x</math>.<br />
<br />
<br />
<math><br />
\mathrm{(A)}\ \frac 18<br />
\qquad<br />
\mathrm{(B)}\ \frac 3{20}<br />
\qquad<br />
\mathrm{(C)}\ \frac 16<br />
\qquad<br />
\mathrm{(D)}\ \frac 15 <br />
\qquad<br />
\mathrm{(E)}\ \frac 14<br />
</math><br />
<br />
[[2006 AMC 12B Problems/Problem 20|Solution]]<br />
<br />
== Problem 21 ==<br />
Rectange <math>ABCD</math> has area <math>2006</math>. An ellipse with area <math>2006\pi</math> passes through <math>A</math> and <math>C</math> and has foci at <math>B</math> and <math>D</math>. What is the perimeter of the rectangle? (The area of an ellipse is <math>ab\pi</math> where <math>2a</math> and <math>2b</math> are the lengths of the axes.)<br />
<br />
<math><br />
\mathrm{(A)}\ \frac {16\sqrt {2006}}{\pi}<br />
\qquad<br />
\mathrm{(B)}\ \frac {1003}4<br />
\qquad<br />
\mathrm{(C)}\ 8\sqrt {1003}<br />
\qquad<br />
\mathrm{(D)}\ 6\sqrt {2006}<br />
\qquad<br />
\mathrm{(E)}\ \frac {32\sqrt {1003}}\pi<br />
</math><br />
<br />
[[2006 AMC 12B Problems/Problem 21|Solution]]<br />
<br />
== Problem 22 ==<br />
Suppose <math>a</math>, <math>b</math> and <math>c</math> are positive integers with <math>a+b+c=2006</math>, and <math>a!b!c!=m\cdot 10^n</math>, where <math>m</math> and <math>n</math> are integers and <math>m</math> is not divisible by <math>10</math>. What is the smallest possible value of <math>n</math>?<br />
<br />
<math><br />
\mathrm{(A)}\ 489<br />
\qquad<br />
\mathrm{(B)}\ 492 <br />
\qquad<br />
\mathrm{(C)}\ 495<br />
\qquad<br />
\mathrm{(D)}\ 498<br />
\qquad<br />
\mathrm{(E)}\ 501<br />
</math><br />
<br />
[[2006 AMC 12B Problems/Problem 22|Solution]]<br />
<br />
== Problem 23 ==<br />
Isosceles <math>\triangle ABC</math> has a right angle at <math>C</math>. Point <math>P</math> is inside <math>\triangle ABC</math>, such that <math>PA=11</math>, <math>PB=7</math>, and <math>PC=6</math>. Legs <math>\overline{AC}</math> and <math>\overline{BC}</math> have length <math>s=\sqrt{a+b\sqrt{2}{</math>, where <math>a</math> and <math>b</math> are positive integers. What is <math>a+b</math>?<br />
<br />
<asy><br />
pathpen = linewidth(0.7);<br />
pen f = fontsize(10);<br />
size(5cm);<br />
pair B = (0,sqrt(85+42*sqrt(2)));<br />
pair A = (B.y,0);<br />
pair C = (0,0);<br />
pair P = IP(arc(B,7,180,360),arc(C,6,0,90));<br />
D(A--B--C--cycle);<br />
D(P--A);<br />
D(P--B);<br />
D(P--C);<br />
MP("A",D(A),plain.E,f);<br />
MP("B",D(B),plain.N,f);<br />
MP("C",D(C),plain.SW,f);<br />
MP("P",D(P),plain.NE,f);<br />
</asy><br />
<br />
<math><br />
\mathrm{(A)}\ 85<br />
\qquad<br />
\mathrm{(B)}\ 91<br />
\qquad<br />
\mathrm{(C)}\ 108<br />
\qquad<br />
\mathrm{(D)}\ 121<br />
\qquad<br />
\mathrm{(E)}\ 127<br />
</math><br />
<br />
<br />
[[2006 AMC 12B Problems/Problem 23|Solution]]<br />
<br />
== Problem 24 ==<br />
<br />
Let <math>S</math> be the set of all points <math>(x,y)</math> in the coordinate plane such that <math>0\leq x\leq \frac\pi 2</math> and <math>0\leq y\leq \frac\pi 2</math>. What is the area of the subset of <math>S</math> for which <math>\sin^2 x - \sin x\sin y + \sin^2 y\le \frac 34</math>?<br />
<br />
<math><br />
\mathrm{(A)}\ \frac {\pi^2}9<br />
\qquad<br />
\mathrm{(B)}\ \frac {\pi^2}8<br />
\qquad<br />
\mathrm{(C)}\ \frac {\pi^2}6<br />
\qquad<br />
\mathrm{(D)}\ \frac {3\pi^2}{16} <br />
\qquad<br />
\mathrm{(E)}\ \frac {2\pi^2}9<br />
</math><br />
<br />
[[2006 AMC 12B Problems/Problem 24|Solution]]<br />
<br />
== Problem 25 ==<br />
<br />
A sequence <math>a_1,a_2,\dots</math> of non-negative integers is defined by the rule <math>a_{n+2}=|a_{n+1}-a_n|</math> for <math>n\geq 1</math>. If <math>a_1=999</math>, <math>a_2<999</math> and <math>a_{2006}=1</math>, how many different values of <math>a_2</math> are possible?<br />
<br />
<math><br />
\mathrm{(A)}\ 165<br />
\qquad<br />
\mathrm{(B)}\ 324<br />
\qquad<br />
\mathrm{(C)}\ 495<br />
\qquad<br />
\mathrm{(D)}\ 499<br />
\qquad<br />
\mathrm{(E)}\ 660<br />
</math><br />
<br />
[[2006 AMC 12B Problems/Problem 25|Solution]]<br />
<br />
== See also ==<br />
* [[AMC 12]]<br />
* [[AMC 12 Problems and Solutions]]<br />
* [[2006 AMC 12B]]<br />
* [http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php?mj_id=143 2006 AMC B Math Jam Transcript]<br />
* [[Mathematics competition resources]]</div>Qszwdxefchttps://artofproblemsolving.com/wiki/index.php?title=2006_AMC_12B_Problems&diff=332992006 AMC 12B Problems2010-02-04T03:57:14Z<p>Qszwdxefc: /* Problem 23 */</p>
<hr />
<div>== Problem 1 ==<br />
What is <math>( - 1)^1 + ( - 1)^2 + \cdots + ( - 1)^{2006}</math>?<br />
<br />
<math><br />
\text {(A) } - 2006 \qquad \text {(B) } - 1 \qquad \text {(C) } 0 \qquad \text {(D) } 1 \qquad \text {(E) } 2006<br />
</math><br />
<br />
[[2006 AMC 12B Problems/Problem 1|Solution]]<br />
<br />
== Problem 2 ==<br />
For real numbers <math>x</math> and <math>y</math>, define <math>x\spadesuit y = (x + y)(x - y)</math>. What is <math>3\spadesuit(4\spadesuit 5)</math>?<br />
<br />
<math>\text {(A) } - 72 \qquad \text {(B) } - 27 \qquad \text {(C) } - 24 \qquad \text {(D) } 24 \qquad \text {(E) } 72</math><br />
<br />
[[2006 AMC 12B Problems/Problem 2|Solution]]<br />
<br />
== Problem 3 ==<br />
A football game was played between two teams, the Cougars and the Panthers. The two teams scored a total of 34 points, and the Cougars won by a margin of 14 points. How many points did the Panthers score?<br />
<br />
<math><br />
\text {(A) } 10 \qquad \text {(B) } 14 \qquad \text {(C) } 17 \qquad \text {(D) } 20 \qquad \text {(E) } 24<br />
</math><br />
<br />
[[2006 AMC 12B Problems/Problem 3|Solution]]<br />
<br />
== Problem 4 ==<br />
Mary is about to pay for five items at the grocery store. The prices of the items are <math>\</math> <math>7.99</math>, <math>\</math> <math>4.99</math>, <math>\</math> <math>2.99</math>, <math>\</math> <math>1.99</math>, and <math>\</math> <math>0.99</math>. Mary will pay with a twenty-dollar bill. Which of the following is closest to the percentage of the <math>\</math> <math>20.00</math> that she will receive in change?<br />
<br />
<math><br />
\text {(A) } 5 \qquad \text {(B) } 10 \qquad \text {(C) } 15 \qquad \text {(D) } 20 \qquad \text {(E) } 25<br />
</math><br />
<br />
[[2006 AMC 12B Problems/Problem 4|Solution]]<br />
<br />
== Problem 5 ==<br />
John is walking east at a speed of 3 miles per hour, while Bob is also walking east, but at a speed of 5 miles per hour. If Bob is now 1 mile west of John, how many minutes will it take for Bob to catch up to John?<br />
<br />
<math><br />
\text {(A) } 30 \qquad \text {(B) } 50 \qquad \text {(C) } 60 \qquad \text {(D) } 90 \qquad \text {(E) } 120<br />
</math><br />
<br />
[[2006 AMC 12B Problems/Problem 5|Solution]]<br />
<br />
== Problem 6 ==<br />
Francesca uses 100 grams of lemon juice, 100 grams of sugar, and 400 grams of water to make lemonade. There are 25 calories in 100 grams of lemon juice and 386 calories in 100 grams of sugar. Water contains no calories. How many calories are in 200 grams of her lemonade.<br />
<br />
<math><br />
\text {(A) } 129 \qquad \text {(B) } 137 \qquad \text {(C) } 174 \qquad \text {(D) } 223 \qquad \text {(E) } 411<br />
</math><br />
<br />
[[2006 AMC 12B Problems/Problem 6|Solution]]<br />
<br />
== Problem 7 ==<br />
Mr. and Mrs. Lopez have two children. When they get into their family car, two people sit in the front, and the other two sit in the back. Either Mr. Lopez or Mrs. Lopez must sit in the driver's seat. How many seating arrangements are possible?<br />
<br />
<math><br />
\text {(A) } 4 \qquad \text {(B) } 12 \qquad \text {(C) } 16 \qquad \text {(D) } 24 \qquad \text {(E) } 48<br />
</math><br />
<br />
[[2006 AMC 12B Problems/Problem 7|Solution]]<br />
<br />
== Problem 8 ==<br />
The lines <math>x = \frac 14y + a</math> and <math>y = \frac 14x + b</math> intersect at the point <math>(1,2)</math>. What is <math>a + b</math>?<br />
<br />
<math><br />
\text {(A) } 0 \qquad \text {(B) } \frac 34 \qquad \text {(C) } 1 \qquad \text {(D) } 2 \qquad \text {(E) } \frac 94<br />
</math><br />
<br />
[[2006 AMC 12B Problems/Problem 8|Solution]]<br />
<br />
== Problem 9 ==<br />
How many even three-digit integers have the property that their digits, read left to right, are in strictly increasing order? <br />
<br />
<math><br />
\text {(A) } 21 \qquad \text {(B) } 34 \qquad \text {(C) } 51 \qquad \text {(D) } 72 \qquad \text {(E) } 150<br />
</math><br />
<br />
[[2006 AMC 12B Problems/Problem 9|Solution]]<br />
<br />
== Problem 10 ==<br />
Joe and JoAnn each bought 12 ounces of coffee in a 16-ounce cup. Joe drank 2 ounces of his coffee and then added 2 ounces of cream. JoAnn added 2 ounces of cream, stirred the coffee well, and then drank 2 ounces. What is the resulting ratio of the amount of cream in Joe's coffee to that in JoAnn's coffee?<br />
<br />
<math><br />
\text {(A) } \frac 67 \qquad \text {(B) } \frac{13}{14} \qquad \text {(C) } 1 \qquad \text {(D) } \frac{14}{13} \qquad \text {(E) } \frac 76<br />
</math><br />
<br />
[[2006 AMC 12B Problems/Problem 10|Solution]]<br />
<br />
== Problem 11 ==<br />
Joe and JoAnn each bought 12 ounces of coffee in a 16-ounce cup. Joe drank 2 ounces of his coffee and then added 2 ounces of cream. JoAnn added 2 ounces of cream, stirred the coffee well, and then drank 2 ounces. What is the resulting ratio of the amount of cream in Joe's coffee to that in JoAnn's coffee?<br />
<br />
<math><br />
\text {(A) } \frac 67 \qquad \text {(B) } \frac {13}{14} \qquad \text {(C) } 1 \qquad \text {(D) } \frac {14}{13} \qquad \text {(E) } \frac 76<br />
</math><br />
<br />
[[2006 AMC 12B Problems/Problem 11|Solution]]<br />
<br />
== Problem 12 ==<br />
The parabola <math>y=ax^2+bx+c</math> has vertex <math>(p,p)</math> and <math>y</math>-intercept <math>(0,-p)</math>, where <math>p\ne 0</math>. What is <math>b</math>?<br />
<br />
<math><br />
\text {(A) } -p \qquad \text {(B) } 0 \qquad \text {(C) } 2 \qquad \text {(D) } 4 \qquad \text {(E) } p<br />
</math><br />
<br />
[[2006 AMC 12B Problems/Problem 12|Solution]]<br />
<br />
== Problem 13 ==<br />
{{problem}}<br />
<br />
[[2006 AMC 12B Problems/Problem 13|Solution]]<br />
<br />
== Problem 14 ==<br />
Elmo makes <math>N</math> sandwiches for a fundraiser. For each sandwich he uses <math>B</math> globs of peanut butter at <math>4</math> cents per glob and <math>J</math> blobs of jam at <math>5</math> cents per glob. The cost of the peanut butter and jam to make all the sandwiches is <math>\</math><math> 2.53</math>. Assume that <math>B</math>, <math>J</math> and <math>N</math> are all positive integers with <math>N>1</math>. What is the cost of the jam Elmo uses to make the sandwiches? <br />
<br />
<math><br />
\mathrm{(A)}\ 1.05<br />
\qquad<br />
\mathrm{(B)}\ 1.25<br />
\qquad<br />
\mathrm{(C)}\ 1.45<br />
\qquad<br />
\mathrm{(D)}\ 1.65<br />
\qquad<br />
\mathrm{(E)}\ 1.85<br />
</math><br />
<br />
[[2006 AMC 12B Problems/Problem 14|Solution]]<br />
<br />
== Problem 15 ==<br />
{{problem}}<br />
<br />
[[2006 AMC 12B Problems/Problem 15|Solution]]<br />
<br />
== Problem 16 ==<br />
<br />
Regular hexagon <math>ABCDEF</math> has vertices <math>A</math> and <math>C</math> at <math>(0,0)</math> and <math>(7,1)</math>, respectively. What is its area?<br />
<br />
<math><br />
\mathrm{(A)}\ 20\sqrt {3}<br />
\qquad<br />
\mathrm{(B)}\ 22\sqrt {3}<br />
\qquad<br />
\mathrm{(C)}\ 25\sqrt {3}<br />
\qquad<br />
\mathrm{(D)}\ 27\sqrt {3}<br />
\qquad<br />
\mathrm{(E)}\ 50<br />
</math><br />
<br />
[[2006 AMC 12B Problems/Problem 16|Solution]]<br />
<br />
== Problem 17 ==<br />
<br />
For a particular peculiar pair of dice, the probabilities of rolling <math>1</math>, <math>2</math>, <math>3</math>, <math>4</math>, <math>5</math> and <math>6</math> on each die are in the ratio <math>1:2:3:4:5:6</math>. What is the probability of rolling a total of <math>7</math> on the two dice?<br />
<br />
<math><br />
\mathrm{(A)}\ \frac 4{63}<br />
\qquad<br />
\mathrm{(B)}\ \frac 18 <br />
\qquad<br />
\mathrm{(C)}\ \frac 8{63}<br />
\qquad<br />
\mathrm{(D)}\ \frac 16<br />
\qquad<br />
\mathrm{(E)}\ \frac 27<br />
</math><br />
<br />
[[2006 AMC 12B Problems/Problem 17|Solution]]<br />
<br />
== Problem 18 ==<br />
An object in the plane moves from one lattice point to another. At each step, the object may move one unit to the right, one unit to the left, one unit up, or one unit down. If the object starts at the origin and takes a ten-step path, how many different points could be the final point?<br />
<br />
<math><br />
\mathrm{(A)}\ 120<br />
\qquad<br />
\mathrm{(B)}\ 121<br />
\qquad<br />
\mathrm{(C)}\ 221<br />
\qquad<br />
\mathrm{(D)}\ 230<br />
\qquad<br />
\mathrm{(E)}\ 231<br />
</math><br />
<br />
[[2006 AMC 12B Problems/Problem 18|Solution]]<br />
<br />
== Problem 19 ==<br />
Mr. Jones has eight children of different ages. On a family trip his olderst child, who is 9, spots a license plate with a 4-digit number in which each of two digits appears two times. "Look, daddy!" she exclaims. "That number is evenly divisible by the age of each of us kids!" "That's right," replies Mr. Jones, "and the last two digits just happen to be my age." Which of the following is not the age of one of Mr. Jones's children?<br />
<br />
<math><br />
\mathrm{(A)}\ 4<br />
\qquad<br />
\mathrm{(B)}\ 5<br />
\qquad<br />
\mathrm{(C)}\ 6<br />
\qquad<br />
\mathrm{(D)}\ 7<br />
\qquad<br />
\mathrm{(E)}\ 8<br />
</math><br />
<br />
[[2006 AMC 12B Problems/Problem 19|Solution]]<br />
<br />
== Problem 20 ==<br />
Let <math>x</math> be chosen at random from the interval <math>(0,1)</math>. What is the probability that<br />
<math>\lfloor\log_{10}4x\rfloor - \lfloor\log_{10}x\rfloor = 0</math>?<br />
Here <math>\lfloor x\rfloor</math> denotes the greatest integer that is less than or equal to <math>x</math>.<br />
<br />
<br />
<math><br />
\mathrm{(A)}\ \frac 18<br />
\qquad<br />
\mathrm{(B)}\ \frac 3{20}<br />
\qquad<br />
\mathrm{(C)}\ \frac 16<br />
\qquad<br />
\mathrm{(D)}\ \frac 15 <br />
\qquad<br />
\mathrm{(E)}\ \frac 14<br />
</math><br />
<br />
[[2006 AMC 12B Problems/Problem 20|Solution]]<br />
<br />
== Problem 21 ==<br />
Rectange <math>ABCD</math> has area <math>2006</math>. An ellipse with area <math>2006\pi</math> passes through <math>A</math> and <math>C</math> and has foci at <math>B</math> and <math>D</math>. What is the perimeter of the rectangle? (The area of an ellipse is <math>ab\pi</math> where <math>2a</math> and <math>2b</math> are the lengths of the axes.)<br />
<br />
<math><br />
\mathrm{(A)}\ \frac {16\sqrt {2006}}{\pi}<br />
\qquad<br />
\mathrm{(B)}\ \frac {1003}4<br />
\qquad<br />
\mathrm{(C)}\ 8\sqrt {1003}<br />
\qquad<br />
\mathrm{(D)}\ 6\sqrt {2006}<br />
\qquad<br />
\mathrm{(E)}\ \frac {32\sqrt {1003}}\pi<br />
</math><br />
<br />
[[2006 AMC 12B Problems/Problem 21|Solution]]<br />
<br />
== Problem 22 ==<br />
Suppose <math>a</math>, <math>b</math> and <math>c</math> are positive integers with <math>a+b+c=2006</math>, and <math>a!b!c!=m\cdot 10^n</math>, where <math>m</math> and <math>n</math> are integers and <math>m</math> is not divisible by <math>10</math>. What is the smallest possible value of <math>n</math>?<br />
<br />
<math><br />
\mathrm{(A)}\ 489<br />
\qquad<br />
\mathrm{(B)}\ 492 <br />
\qquad<br />
\mathrm{(C)}\ 495<br />
\qquad<br />
\mathrm{(D)}\ 498<br />
\qquad<br />
\mathrm{(E)}\ 501<br />
</math><br />
<br />
[[2006 AMC 12B Problems/Problem 22|Solution]]<br />
<br />
== Problem 23 ==<br />
== Problem ==<br />
Isosceles <math>\triangle ABC</math> has a right angle at <math>C</math>. Point <math>P</math> is inside <math>\triangle ABC</math>, such that <math>PA=11</math>, <math>PB=7</math>, and <math>PC=6</math>. Legs <math>\overline{AC}</math> and <math>\overline{BC}</math> have length <math>s=\sqrt{a+b\sqrt{2}{</math>, where <math>a</math> and <math>b</math> are positive integers. What is <math>a+b</math>?<br />
<br />
<asy><br />
pathpen = linewidth(0.7);<br />
pen f = fontsize(10);<br />
size(5cm);<br />
pair B = (0,sqrt(85+42*sqrt(2)));<br />
pair A = (B.y,0);<br />
pair C = (0,0);<br />
pair P = IP(arc(B,7,180,360),arc(C,6,0,90));<br />
D(A--B--C--cycle);<br />
D(P--A);<br />
D(P--B);<br />
D(P--C);<br />
MP("A",D(A),plain.E,f);<br />
MP("B",D(B),plain.N,f);<br />
MP("C",D(C),plain.SW,f);<br />
MP("P",D(P),plain.NE,f);<br />
</asy><br />
<br />
<math><br />
\mathrm{(A)}\ 85<br />
\qquad<br />
\mathrm{(B)}\ 91<br />
\qquad<br />
\mathrm{(C)}\ 108<br />
\qquad<br />
\mathrm{(D)}\ 121<br />
\qquad<br />
\mathrm{(E)}\ 127<br />
</math><br />
<br />
<br />
[[2006 AMC 12B Problems/Problem 23|Solution]]<br />
<br />
== Problem 24 ==<br />
<br />
Let <math>S</math> be the set of all points <math>(x,y)</math> in the coordinate plane such that <math>0\leq x\leq \frac\pi 2</math> and <math>0\leq y\leq \frac\pi 2</math>. What is the area of the subset of <math>S</math> for which <math>\sin^2 x - \sin x\sin y + \sin^2 y\le \frac 34</math>?<br />
<br />
<math><br />
\mathrm{(A)}\ \frac {\pi^2}9<br />
\qquad<br />
\mathrm{(B)}\ \frac {\pi^2}8<br />
\qquad<br />
\mathrm{(C)}\ \frac {\pi^2}6<br />
\qquad<br />
\mathrm{(D)}\ \frac {3\pi^2}{16} <br />
\qquad<br />
\mathrm{(E)}\ \frac {2\pi^2}9<br />
</math><br />
<br />
[[2006 AMC 12B Problems/Problem 24|Solution]]<br />
<br />
== Problem 25 ==<br />
<br />
A sequence <math>a_1,a_2,\dots</math> of non-negative integers is defined by the rule <math>a_{n+2}=|a_{n+1}-a_n|</math> for <math>n\geq 1</math>. If <math>a_1=999</math>, <math>a_2<999</math> and <math>a_{2006}=1</math>, how many different values of <math>a_2</math> are possible?<br />
<br />
<math><br />
\mathrm{(A)}\ 165<br />
\qquad<br />
\mathrm{(B)}\ 324<br />
\qquad<br />
\mathrm{(C)}\ 495<br />
\qquad<br />
\mathrm{(D)}\ 499<br />
\qquad<br />
\mathrm{(E)}\ 660<br />
</math><br />
<br />
[[2006 AMC 12B Problems/Problem 25|Solution]]<br />
<br />
== See also ==<br />
* [[AMC 12]]<br />
* [[AMC 12 Problems and Solutions]]<br />
* [[2006 AMC 12B]]<br />
* [http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php?mj_id=143 2006 AMC B Math Jam Transcript]<br />
* [[Mathematics competition resources]]</div>Qszwdxefchttps://artofproblemsolving.com/wiki/index.php?title=2007_AMC_12B_Problems/Problem_20&diff=332932007 AMC 12B Problems/Problem 202010-02-02T05:36:14Z<p>Qszwdxefc: See also misdirect</p>
<hr />
<div>==Problem==<br />
The parallelogram bounded by the lines <math>y=ax+c</math>, <math>y=ax+d</math>, <math>y=bx+c</math>, and <math>y=bx+d</math> has area <math>18</math>. The parallelogram bounded by the lines <math>y=ax+c</math>, <math>y=ax-d</math>, <math>y=bx+c</math>, and <math>y=bx-d</math> has area <math>72</math>. Given that <math>a</math>, <math>b</math>, <math>c</math>, and <math>d</math> are positive integers, what is the smallest possible value of <math>a+b+c+d</math>?<br />
<br />
<math>\mathrm {(A)} 13\qquad \mathrm {(B)} 14\qquad \mathrm {(C)} 15\qquad \mathrm {(D)} 16\qquad \mathrm {(E)} 17</math><br />
<br />
==Solution==<br />
{{incomplete|solution}}<br />
<!-- <center><asy><br />
pathpen = linewidth(0.7);<br />
real a = 3, b = 1, c = 9, d = 3; <br />
D((0,c) -- ((d-c)/(a-b),(a*d-b*c)/(a-b)) -- (0,d) -- ((c-d)/(a-b),(b*c-a*d)/(a-b)) -- cycle);<br />
D((0,c) -- ((-d-c)/(a-b),(-a*d-b*c)/(a-b)) -- (0,-d) -- ((c+d)/(a-b),-(-a*d-b*c)/(a-b)) -- cycle);<br />
</asy></center> --><br />
Plotting the parallelogram on the coordinate plane, the 4 corners are at <math>(0,c),(0,d),\left(\frac{d-c}{a-b},\frac{ad-bc}{a-b}\right),\left(\frac{c-d}{a-b},\frac{bc-ad}{a-b}\right)</math>. Because <math>72= 4\cdot 18</math>, we have that <math>4(c-d)\left(\frac{c-d}{a-b}\right) = (c+d)\left(\frac{c+d}{a-b}\right)</math> or that <math>2(c-d)=c+d</math>, which gives <math>c=3d</math> (consider a [[homothety]], or dilation, that carries the first parallelogram to the second parallelogram; because the area increases by <math>4\times</math>, it follows that the stretch along the diagonal is <math>2\times</math>). The area of triangular half of the parallelogram on the right side of the y-axis is given by <math>9 = \frac{1}{2} (c-d)\left(\frac{d-c}{a-b}\right)</math>, so substituting <math>c = 3d</math>:<br />
<center><cmath><br />
\frac{1}{2} (c-d)\left(\frac{c-d}{a-b}\right) &= 9 \quad \Longrightarrow \quad 2d^2 &= 9(a-b)</cmath></center><br />
Thus <math>3|d</math>, and we verify that <math>d = 3</math>, <math>a-b = 2 \Longrightarrow a = 3, b = 1</math> will give us a minimum value for <math>a+b+c+d</math>. Then <math>a+b+c+d = 3 + 1 + 9 + 3 = 16\ \mathbf{(D)}</math>.<br />
==Solution 2==<br />
{{incomplete|solution}}<br />
The key to this solution is that area is invariant under translation. By suitably shifting the plane, the problem is mapped to the lines <math>c,d,(b-a)x+c,(b-a)x+d</math> and <math>c,-d,(b-a)x+c,(b-a)x-d</math>. Now, the area of the parallelogram contained by is the former is equal to the area of a rectangle with sides <math>d-c</math> and <math>\frac{d-c}{b-a}</math>, <math>\frac{(d-c)^2}{b-a}=18</math>, and the area contained by the latter is <math>\frac{(c+d)^2}{b-a}=72</math>. Thus, <math>d=3c</math> and <math>b-a</math> must be even if the former quantity is to equal <math>18</math>. <math>c^2=18(b-a)</math> so <math>c</math> is a multiple of <math>3</math>. Putting this all together, the minimal solution is <math>(a,b,c,d)=(3,1,3,9)</math><br />
<br />
==See also==<br />
{{AMC12 box|year=2007|ab=B|num-b=19|num-a=21}}<br />
<br />
[[Category:Introductory Geometry Problems]]</div>Qszwdxefchttps://artofproblemsolving.com/wiki/index.php?title=2007_AMC_12B_Problems/Problem_20&diff=332922007 AMC 12B Problems/Problem 202010-02-02T05:35:06Z<p>Qszwdxefc: /* Solution 2 */</p>
<hr />
<div>==Problem==<br />
The parallelogram bounded by the lines <math>y=ax+c</math>, <math>y=ax+d</math>, <math>y=bx+c</math>, and <math>y=bx+d</math> has area <math>18</math>. The parallelogram bounded by the lines <math>y=ax+c</math>, <math>y=ax-d</math>, <math>y=bx+c</math>, and <math>y=bx-d</math> has area <math>72</math>. Given that <math>a</math>, <math>b</math>, <math>c</math>, and <math>d</math> are positive integers, what is the smallest possible value of <math>a+b+c+d</math>?<br />
<br />
<math>\mathrm {(A)} 13\qquad \mathrm {(B)} 14\qquad \mathrm {(C)} 15\qquad \mathrm {(D)} 16\qquad \mathrm {(E)} 17</math><br />
<br />
==Solution==<br />
{{incomplete|solution}}<br />
<!-- <center><asy><br />
pathpen = linewidth(0.7);<br />
real a = 3, b = 1, c = 9, d = 3; <br />
D((0,c) -- ((d-c)/(a-b),(a*d-b*c)/(a-b)) -- (0,d) -- ((c-d)/(a-b),(b*c-a*d)/(a-b)) -- cycle);<br />
D((0,c) -- ((-d-c)/(a-b),(-a*d-b*c)/(a-b)) -- (0,-d) -- ((c+d)/(a-b),-(-a*d-b*c)/(a-b)) -- cycle);<br />
</asy></center> --><br />
Plotting the parallelogram on the coordinate plane, the 4 corners are at <math>(0,c),(0,d),\left(\frac{d-c}{a-b},\frac{ad-bc}{a-b}\right),\left(\frac{c-d}{a-b},\frac{bc-ad}{a-b}\right)</math>. Because <math>72= 4\cdot 18</math>, we have that <math>4(c-d)\left(\frac{c-d}{a-b}\right) = (c+d)\left(\frac{c+d}{a-b}\right)</math> or that <math>2(c-d)=c+d</math>, which gives <math>c=3d</math> (consider a [[homothety]], or dilation, that carries the first parallelogram to the second parallelogram; because the area increases by <math>4\times</math>, it follows that the stretch along the diagonal is <math>2\times</math>). The area of triangular half of the parallelogram on the right side of the y-axis is given by <math>9 = \frac{1}{2} (c-d)\left(\frac{d-c}{a-b}\right)</math>, so substituting <math>c = 3d</math>:<br />
<center><cmath><br />
\frac{1}{2} (c-d)\left(\frac{c-d}{a-b}\right) &= 9 \quad \Longrightarrow \quad 2d^2 &= 9(a-b)</cmath></center><br />
Thus <math>3|d</math>, and we verify that <math>d = 3</math>, <math>a-b = 2 \Longrightarrow a = 3, b = 1</math> will give us a minimum value for <math>a+b+c+d</math>. Then <math>a+b+c+d = 3 + 1 + 9 + 3 = 16\ \mathbf{(D)}</math>.<br />
==Solution 2==<br />
{{incomplete|solution}}<br />
The key to this solution is that area is invariant under translation. By suitably shifting the plane, the problem is mapped to the lines <math>c,d,(b-a)x+c,(b-a)x+d</math> and <math>c,-d,(b-a)x+c,(b-a)x-d</math>. Now, the area of the parallelogram contained by is the former is equal to the area of a rectangle with sides <math>d-c</math> and <math>\frac{d-c}{b-a}</math>, <math>\frac{(d-c)^2}{b-a}=18</math>, and the area contained by the latter is <math>\frac{(c+d)^2}{b-a}=72</math>. Thus, <math>d=3c</math> and <math>b-a</math> must be even if the former quantity is to equal <math>18</math>. <math>c^2=18(b-a)</math> so <math>c</math> is a multiple of <math>3</math>. Putting this all together, the minimal solution is <math>(a,b,c,d)=(3,1,3,9)</math><br />
<br />
==See also==<br />
{{AMC12 box|year=2007|ab=B|num-b=21|num-a=23}}<br />
<br />
[[Category:Introductory Geometry Problems]]</div>Qszwdxefchttps://artofproblemsolving.com/wiki/index.php?title=2009_AMC_12B_Problems&diff=332652009 AMC 12B Problems2010-02-01T05:22:55Z<p>Qszwdxefc: Q2 typo</p>
<hr />
<div>== Problem 1 ==<br />
Each morning of her five-day workweek, Jane bought either a <math>50</math>-cent muffin or a <math>75</math>-cent bagel. Her total cost for the week was a whole number of dollars. How many bagels did she buy?<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 />
[[2009 AMC 12B Problems/Problem 1|Solution]]<br />
<br />
== Problem 2 ==<br />
Paula the painter had just enough paint for <math>30</math> identically sized rooms. Unfortunately, on the way to work, three cans of paint fell off her truck, so she had only enough paint for <math>25</math> rooms. How many cans of paint did she use for the <math>25</math> rooms?<br />
<br />
<math>\textbf{(A)}\ 10 \qquad \textbf{(B)}\ 12 \qquad \textbf{(C)}\ 15 \qquad \textbf{(D)}\ 18 \qquad \textbf{(E)}\ 25</math><br />
<br />
[[2009 AMC 12B Problems/Problem 2|Solution]]<br />
<br />
== Problem 3 ==<br />
Twenty percent less than <math>60</math> is one-third more than what number?<br />
<br />
<math>\textbf{(A)}\ 16\qquad \textbf{(B)}\ 30\qquad \textbf{(C)}\ 32\qquad \textbf{(D)}\ 36\qquad \textbf{(E)}\ 48</math><br />
<br />
[[2009 AMC 12B Problems/Problem 3|Solution]]<br />
<br />
== Problem 4 ==<br />
A rectangular yard contains two flower beds in the shape of congruent isosceles right triangles. The remainder of the yard has a trapezoidal shape, as shown. The parallel sides of the trapezoid have lengths <math>15</math> and <math>25</math> meters. What fraction of the yard is occupied by the flower beds?<br />
<center><asy><br />
unitsize(2mm);<br />
defaultpen(linewidth(.8pt));<br />
<br />
fill((0,0)--(0,5)--(5,5)--cycle,gray);<br />
fill((25,0)--(25,5)--(20,5)--cycle,gray);<br />
draw((0,0)--(0,5)--(25,5)--(25,0)--cycle);<br />
draw((0,0)--(5,5));<br />
draw((20,5)--(25,0));<br />
</asy></center><math>\textbf{(A)}\ \frac18\qquad \textbf{(B)}\ \frac16\qquad \textbf{(C)}\ \frac15\qquad \textbf{(D)}\ \frac14\qquad \textbf{(E)}\ \frac13</math><br />
<br />
[[2009 AMC 12B Problems/Problem 4|Solution]]<br />
<br />
== Problem 5 ==<br />
Kiana has two older twin brothers. The product of their ages is <math>128</math>. What is the sum of their three ages?<br />
<br />
<math>\textbf{(A)}\ 10\qquad \textbf{(B)}\ 12\qquad \textbf{(C)}\ 16\qquad \textbf{(D)}\ 18\qquad \textbf{(E)}\ 24</math><br />
<br />
[[2009 AMC 12B Problems/Problem 5|Solution]]<br />
<br />
== Problem 6 ==<br />
By inserting parentheses, it is possible to give the expression<br />
<cmath>2\times3 + 4\times5</cmath><br />
several values. How many different values can be obtained?<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 />
[[2009 AMC 12B Problems/Problem 6|Solution]]<br />
<br />
== Problem 7 ==<br />
In a certain year the price of gasoline rose by <math>20\%</math> during January, fell by <math>20\%</math> during February, rose by <math>25\%</math> during March, and fell by <math>x\%</math> during April. The price of gasoline at the end of April was the same as it had been at the beginning of January. To the nearest integer, what is <math>x</math>?<br />
<br />
<math>\textbf{(A)}\ 12\qquad \textbf{(B)}\ 17\qquad \textbf{(C)}\ 20\qquad \textbf{(D)}\ 25\qquad \textbf{(E)}\ 35</math><br />
<br />
[[2009 AMC 12B Problems/Problem 7|Solution]]<br />
<br />
== Problem 8 ==<br />
When a bucket is two-thirds full of water, the bucket and water weigh <math>a</math> kilograms. When the bucket is one-half full of water the total weight is <math>b</math> kilograms. In terms of <math>a</math> and <math>b</math>, what is the total weight in kilograms when the bucket is full of water?<br />
<br />
<math>\textbf{(A)}\ \frac23a + \frac13b\qquad \textbf{(B)}\ \frac32a - \frac12b\qquad \textbf{(C)}\ \frac32a + b</math><br />
<math>\textbf{(D)}\ \frac32a + 2b\qquad \textbf{(E)}\ 3a - 2b</math><br />
<br />
[[2009 AMC 12B Problems/Problem 8|Solution]]<br />
<br />
== Problem 9 ==<br />
Triangle <math>ABC</math> has vertices <math>A = (3,0)</math>, <math>B = (0,3)</math>, and <math>C</math>, where <math>C</math> is on the line <math>x + y = 7</math>. What is the area of <math>\triangle ABC</math>?<br />
<br />
<math>\textbf{(A)}\ 6\qquad \textbf{(B)}\ 8\qquad \textbf{(C)}\ 10\qquad \textbf{(D)}\ 12\qquad \textbf{(E)}\ 14</math><br />
<br />
[[2009 AMC 12B Problems/Problem 9|Solution]]<br />
<br />
== Problem 10 ==<br />
A particular <math>12</math>-hour digital clock displays the hour and minute of a day. Unfortunately, whenever it is supposed to display a <math>1</math>, it mistakenly displays a <math>9</math>. For example, when it is 1:16 PM the clock incorrectly shows 9:96 PM. What fraction of the day will the clock show the correct time?<br />
<br />
<math>\textbf{(A)}\ \frac12\qquad \textbf{(B)}\ \frac58\qquad \textbf{(C)}\ \frac34\qquad \textbf{(D)}\ \frac56\qquad \textbf{(E)}\ \frac {9}{10}</math><br />
<br />
[[2009 AMC 12B Problems/Problem 10|Solution]]<br />
<br />
== Problem 11 ==<br />
On Monday, Millie puts a quart of seeds, <math>25\%</math> of which are millet, into a bird feeder. On each successive day she adds another quart of the same mix of seeds without removing any seeds that are left. Each day the birds eat only <math>25\%</math> of the millet in the feeder, but they eat all of the other seeds. On which day, just after Millie has placed the seeds, will the birds find that more than half the seeds in the feeder are millet?<br />
<br />
<math>\textbf{(A)}\ \text{Tuesday}\qquad \textbf{(B)}\ \text{Wednesday}\qquad \textbf{(C)}\ \text{Thursday} \qquad \textbf{(D)}\ \text{Friday}\qquad \textbf{(E)}\ \text{Saturday}</math><br />
<br />
[[2009 AMC 12B Problems/Problem 11|Solution]]<br />
<br />
== Problem 12 ==<br />
The fifth and eighth terms of a geometric sequence of real numbers are <math>7!</math> and <math>8!</math> respectively. What is the first term?<br />
<br />
<math>\textbf{(A)}\ 60\qquad \textbf{(B)}\ 75\qquad \textbf{(C)}\ 120\qquad \textbf{(D)}\ 225\qquad \textbf{(E)}\ 315</math><br />
<br />
[[2009 AMC 12B Problems/Problem 12|Solution]]<br />
<br />
== Problem 13 ==<br />
Triangle <math>ABC</math> has <math>AB = 13</math> and <math>AC = 15</math>, and the altitude to <math>\overline{BC}</math> has length <math>12</math>. What is the sum of the two possible values of <math>BC</math>?<br />
<br />
<math>\textbf{(A)}\ 15\qquad \textbf{(B)}\ 16\qquad \textbf{(C)}\ 17\qquad \textbf{(D)}\ 18\qquad \textbf{(E)}\ 19</math><br />
<br />
[[2009 AMC 12B Problems/Problem 13|Solution]]<br />
<br />
== Problem 14 ==<br />
Five unit squares are arranged in the coordinate plane as shown, with the lower left corner at the origin. The slanted line, extending from <math>(a,0)</math> to <math>(3,3)</math>, divides the entire region into two regions of equal area. What is <math>a</math>?<br />
<asy><br />
unitsize(1cm);<br />
defaultpen(linewidth(.8pt)+fontsize(8pt));<br />
<br />
fill((2/3,0)--(3,3)--(3,1)--(2,1)--(2,0)--cycle,gray);<br />
<br />
xaxis("$x$",-0.5,4,EndArrow(HookHead,4));<br />
yaxis("$y$",-0.5,4,EndArrow(4));<br />
<br />
draw((0,1)--(3,1)--(3,3)--(2,3)--(2,0));<br />
draw((1,0)--(1,2)--(3,2));<br />
draw((2/3,0)--(3,3));<br />
<br />
label("$(a,0)$",(2/3,0),S);<br />
label("$(3,3)$",(3,3),NE);<br />
</asy><math>\textbf{(A)}\ \frac12\qquad \textbf{(B)}\ \frac35\qquad \textbf{(C)}\ \frac23\qquad \textbf{(D)}\ \frac34\qquad \textbf{(E)}\ \frac45</math><br />
<br />
[[2009 AMC 12B Problems/Problem 14|Solution]]<br />
<br />
== Problem 15 ==<br />
Assume <math>0 < r < 3</math>. Below are five equations for <math>x</math>. Which equation has the largest solution <math>x</math>?<br />
<br />
<math>\textbf{(A)}\ 3(1 + r)^x = 7\qquad \textbf{(B)}\ 3(1 + r/10)^x = 7\qquad \textbf{(C)}\ 3(1 + 2r)^x = 7</math><br />
<math>\textbf{(D)}\ 3(1 + \sqrt {r})^x = 7\qquad \textbf{(E)}\ 3(1 + 1/r)^x = 7</math><br />
<br />
[[2009 AMC 12B Problems/Problem 15|Solution]]<br />
<br />
== Problem 16 ==<br />
Trapezoid <math>ABCD</math> has <math>AD||BC</math>, <math>BD = 1</math>, <math>\angle DBA = 23^{\circ}</math>, and <math>\angle BDC = 46^{\circ}</math>. The ratio <math>BC: AD</math> is <math>9: 5</math>. What is <math>CD</math>?<br />
<br />
<math>\textbf{(A)}\ \frac {7}{9}\qquad \textbf{(B)}\ \frac {4}{5}\qquad \textbf{(C)}\ \frac {13}{15} \qquad \textbf{(D)}\ \frac {8}{9}\qquad \textbf{(E)}\ \frac {14}{15}</math><br />
<br />
[[2009 AMC 12B Problems/Problem 16|Solution]]<br />
<br />
== Problem 17 ==<br />
Each face of a cube is given a single narrow stripe painted from the center of one edge to the center of its opposite edge. The choice of the edge pairing is made at random and independently for each face. What is the probability that there is a continuous stripe encircling the cube?<br />
<br />
<math>\textbf{(A)}\ \frac {1}{8}\qquad \textbf{(B)}\ \frac {3}{16}\qquad \textbf{(C)}\ \frac {1}{4} \qquad \textbf{(D)}\ \frac {3}{8}\qquad \textbf{(E)}\ \frac {1}{2}</math><br />
<br />
[[2009 AMC 12B Problems/Problem 17|Solution]]<br />
<br />
== Problem 18 ==<br />
Rachel and Robert run on a circular track. Rachel runs counterclockwise and completes a lap every <math>90</math> seconds, and Robert runs clockwise and completes a lap every <math>80</math> seconds. Both start from the start line at the same time. At some random time between <math>10</math> minutes and <math>11</math> minutes after they begin to run, a photographer standing inside the track takes a picture that shows one-fourth of the track, centered on the starting line. What is the probability that both Rachel and Robert are in the picture?<br />
<br />
<math>\textbf{(A)}\ \frac {1}{16}\qquad \textbf{(B)}\ \frac18\qquad \textbf{(C)}\ \frac {3}{16} \qquad \textbf{(D)}\ \frac14\qquad \textbf{(E)}\ \frac {5}{16}</math><br />
<br />
[[2009 AMC 12B Problems/Problem 18|Solution]]<br />
<br />
== Problem 19 ==<br />
For each positive integer <math>n</math>, let <math>f(n) = n^4 - 360n^2 + 400</math>. What is the sum of all values of <math>f(n)</math> that are prime numbers?<br />
<br />
<math>\textbf{(A)}\ 794\qquad \textbf{(B)}\ 796\qquad \textbf{(C)}\ 798\qquad \textbf{(D)}\ 800\qquad \textbf{(E)}\ 802</math><br />
<br />
[[2009 AMC 12B Problems/Problem 19|Solution]]<br />
<br />
== Problem 20 ==<br />
A convex polyhedron <math>Q</math> has vertices <math>V_1,V_2,\ldots,V_n</math>, and <math>100</math> edges. The polyhedron is cut by planes <math>P_1,P_2,\ldots,P_n</math> in such a way that plane <math>P_k</math> cuts only those edges that meet at vertex <math>V_k</math>. In addition, no two planes intersect inside or on <math>Q</math>. The cuts produce <math>n</math> pyramids and a new polyhedron <math>R</math>. How many edges does <math>R</math> have?<br />
<br />
<math>\textbf{(A)}\ 200\qquad \textbf{(B)}\ 2n\qquad \textbf{(C)}\ 300\qquad \textbf{(D)}\ 400\qquad \textbf{(E)}\ 4n</math><br />
<br />
[[2009 AMC 12B Problems/Problem 20|Solution]]<br />
<br />
== Problem 21 ==<br />
Ten women sit in <math>10</math> seats in a line. All of the <math>10</math> get up and then reseat themselves using all <math>10</math> seats, each sitting in the seat she was in before or a seat next to the one she occupied before. In how many ways can the women be reseated?<br />
<br />
<math>\textbf{(A)}\ 89\qquad \textbf{(B)}\ 90\qquad \textbf{(C)}\ 120\qquad \textbf{(D)}\ 2^{10}\qquad \textbf{(E)}\ 2^2 3^8</math><br />
<br />
[[2009 AMC 12B Problems/Problem 21|Solution]]<br />
<br />
== Problem 22 ==<br />
Parallelogram <math>ABCD</math> has area <math>1,\!000,\!000</math>. Vertex <math>A</math> is at <math>(0,0)</math> and all other vertices are in the first quadrant. Vertices <math>B</math> and <math>D</math> are lattice points on the lines <math>y = x</math> and <math>y = kx</math> for some integer <math>k > 1</math>, respectively. How many such parallelograms are there?<br />
<br />
<math>\textbf{(A)}\ 49\qquad \textbf{(B)}\ 720\qquad \textbf{(C)}\ 784\qquad \textbf{(D)}\ 2009\qquad \textbf{(E)}\ 2048</math><br />
<br />
[[2009 AMC 12B Problems/Problem 22|Solution]]<br />
<br />
== Problem 23 ==<br />
A region <math>S</math> in the complex plane is defined by<br />
<cmath><br />
S = \{x + iy: - 1\le x\le1, - 1\le y\le1\}.<br />
</cmath><br />
A complex number <math>z = x + iy</math> is chosen uniformly at random from <math>S</math>. What is the probability that <math>\left(\frac34 + \frac34i\right)z</math> is also in <math>S</math>?<br />
<br />
<math>\textbf{(A)}\ \frac12\qquad \textbf{(B)}\ \frac23\qquad \textbf{(C)}\ \frac34\qquad \textbf{(D)}\ \frac79\qquad \textbf{(E)}\ \frac78</math><br />
<br />
[[2009 AMC 12B Problems/Problem 23|Solution]]<br />
<br />
== Problem 24 ==<br />
For how many values of <math>x</math> in <math>[0,\pi]</math> is <math>\sin^{ - 1}(\sin 6x) = \cos^{ - 1}(\cos x)</math>?<br />
Note: The functions <math>\sin^{ - 1} = \arcsin</math> and <math>\cos^{ - 1} = \arccos</math> denote inverse trigonometric functions.<br />
<br />
<math>\textbf{(A)}\ 3\qquad \textbf{(B)}\ 4\qquad \textbf{(C)}\ 5\qquad \textbf{(D)}\ 6\qquad \textbf{(E)}\ 7</math><br />
<br />
[[2009 AMC 12B Problems/Problem 24|Solution]]<br />
<br />
== Problem 25 ==<br />
The set <math>G</math> is defined by the points <math>(x,y)</math> with integer coordinates, <math>3\le|x|\le7</math>, <math>3\le|y|\le7</math>. How many squares of side at least <math>6</math> have their four vertices in <math>G</math>?<br />
<asy><br />
defaultpen(black+0.75bp+fontsize(8pt));<br />
size(5cm);<br />
path p = scale(.15)*unitcircle;<br />
draw((-8,0)--(8.5,0),Arrow(HookHead,1mm));<br />
draw((0,-8)--(0,8.5),Arrow(HookHead,1mm));<br />
int i,j;<br />
for (i=-7;i<8;++i) {<br />
for (j=-7;j<8;++j) {<br />
if (((-7 <= i) && (i <= -3)) || ((3 <= i) && (i<= 7))) { if (((-7 <= j) && (j <= -3)) || ((3 <= j) && (j<= 7))) { fill(shift(i,j)*p,black); }}}} draw((-7,-.2)--(-7,.2),black+0.5bp);<br />
draw((-3,-.2)--(-3,.2),black+0.5bp);<br />
draw((3,-.2)--(3,.2),black+0.5bp);<br />
draw((7,-.2)--(7,.2),black+0.5bp);<br />
draw((-.2,-7)--(.2,-7),black+0.5bp);<br />
draw((-.2,-3)--(.2,-3),black+0.5bp);<br />
draw((-.2,3)--(.2,3),black+0.5bp);<br />
draw((-.2,7)--(.2,7),black+0.5bp);<br />
label("$-7$",(-7,0),S);<br />
label("$-3$",(-3,0),S);<br />
label("$3$",(3,0),S);<br />
label("$7$",(7,0),S);<br />
label("$-7$",(0,-7),W);<br />
label("$-3$",(0,-3),W);<br />
label("$3$",(0,3),W);<br />
label("$7$",(0,7),W);<br />
</asy><math>\textbf{(A)}\ 125\qquad \textbf{(B)}\ 150\qquad \textbf{(C)}\ 175\qquad \textbf{(D)}\ 200\qquad \textbf{(E)}\ 225</math><br />
<br />
[[2009 AMC 12B Problems/Problem 25|Solution]]</div>Qszwdxefc