https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Ccorder&feedformat=atomAoPS Wiki - User contributions [en]2024-03-28T16:36:56ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_12A_Problems/Problem_2&diff=949852018 AMC 12A Problems/Problem 22018-06-07T18:41:40Z<p>Ccorder: spacing</p>
<hr />
<div>== Problem ==<br />
<br />
While exploring a cave, Carl comes across a collection of <math>5</math>-pound rocks worth <math>\$14</math> each, <math>4</math>-pound rocks worth <math>\$11</math> each, and <math>1</math>-pound rocks worth <math>\$2</math> each. There are at least <math>20</math> of each size. He can carry at most <math>18</math> pounds. What is the maximum value, in dollars, of the rocks he can carry out of the cave?<br />
<br />
<math>\textbf{(A) } 48 \qquad \textbf{(B) } 49 \qquad \textbf{(C) } 50 \qquad \textbf{(D) } 51 \qquad \textbf{(E) } 52 </math><br />
<br />
== Solution ==<br />
<br />
The answer is just <math>3\cdot 18=54</math> minus the minimum number of rocks we need to make <math>18</math> pounds, or<br />
<cmath>54-4=\boxed{\textbf{(C)} 50.}</cmath><br />
<br />
== Solution 2 ==<br />
<br />
The ratio of dollar per pound is greatest for the <math>5</math> pound rock, then the <math>4</math> pound, lastly the <math>1</math> pound. So we should take two <math>5</math> pound rocks and two <math>4</math> pound rocks. <br />
Total weight: <cmath>2\cdot14+2\cdot11=\boxed{\textbf{(C)} 50.}</cmath><br />
~steakfails<br />
<br />
== See Also ==<br />
{{AMC12 box|year=2018|ab=A|num-b=1|num-a=3}}<br />
{{MAA Notice}}</div>Ccorderhttps://artofproblemsolving.com/wiki/index.php?title=2013_AMC_12A_Problems/Problem_15&diff=949842013 AMC 12A Problems/Problem 152018-06-07T18:38:15Z<p>Ccorder: spacing</p>
<hr />
<div>== Problem ==<br />
<br />
Rabbits Peter and Pauline have three offspring—Flopsie, Mopsie, and Cotton-tail. These five rabbits are to be distributed to four different pet stores so that no store gets both a parent and a child. It is not required that every store gets a rabbit. In how many different ways can this be done?<br />
<br />
<math>\textbf{(A)} \ 96 \qquad \textbf{(B)} \ 108 \qquad \textbf{(C)} \ 156 \qquad \textbf{(D)} \ 204 \qquad \textbf{(E)} \ 372 </math><br />
<br />
== Solution 1 ==<br />
<br />
There are two possibilities regarding the parents.<br />
<br />
1) Both are in the same store. In this case, we can treat them both as a single bunny, and they can go in any of the 4 stores. The 3 baby bunnies can go in any of the remaining 3 stores. There are <math>4 \cdot 3^3 = 108</math> combinations.<br />
<br />
<br />
2) The two are in different stores. In this case, one can go in any of the 4 stores, and the other can go in any of the 3 remaining stores. The 3 baby bunnies can each go in any of the remaining 2 stores. There are <math>4 \cdot 3 \cdot 2^3 = 96</math> combinations.<br />
<br />
Adding up, we get <math>108 + 96 = 204</math> combinations.<br />
<br />
== Solution 2 ==<br />
<br />
We tackle the problem by sorting it by how many stores are involved in the transaction.<br />
<br />
1) 2 stores are involved. There are <math>\binom{4}{2} = 6</math> ways to choose which of the stores are involved and 2 ways to choose which store recieves the parents. <math>6 \cdot 2 = 12</math> total arrangements.<br />
<br />
<br />
2) 3 stores are involved. There are <math>\binom{4}{3} = 4</math> ways to choose which of the stores are involved. We then break the problem down to into two subsections - when the parents and grouped together or sold separately.<br />
<br />
<br />
'''Separately:'''<br />
All children must be in one store. There are <math>3!</math> ways to arrange this. <math>6</math> ways in total.<br />
<br />
'''Together:'''<br />
Both parents are in one store and the 3 children are split between the other two. There are <math>\binom{3}{2}</math> ways to split the children and <math>3!</math> ways to choose to which store each group will be sold. <math>3! \cdot \binom{3}{2} = 18</math>.<br />
<br />
<math>(6 + 18) \cdot 4 = 96</math> total arrangements.<br />
<br />
<br />
3) All 4 stores are involved. We break down the problem as previously shown.<br />
<br />
<br />
'''Separately:'''<br />
All children must be split between two stores. There are <math>\binom{3}{2} = 3</math> ways to arrange this. We can then arrange which group is sold to which store in <math>4!</math> ways. <math>4! \cdot 3 = 72</math>.<br />
<br />
'''Together:'''<br />
Both parents are in one store and the 3 children are each in another store. There are <math>4! = 24</math> ways to arrange this.<br />
<br />
<math>24 + 72 = 96</math> total arrangements.<br />
<br />
<br />
'''Final Answer:'''<br />
<math>12 + 96 + 96 = \boxed{\textbf{(D)} \: 204}</math>.<br />
<br />
== See also ==<br />
{{AMC12 box|year=2013|ab=A|num-b=14|num-a=16}}<br />
<br />
[[Category:Introductory Combinatorics Problems]]<br />
{{MAA Notice}}</div>Ccorderhttps://artofproblemsolving.com/wiki/index.php?title=2017_AMC_12A_Problems/Problem_9&diff=949832017 AMC 12A Problems/Problem 92018-06-07T18:36:36Z<p>Ccorder: spacing</p>
<hr />
<div>== Problem ==<br />
<br />
Let <math>S</math> be the set of points <math>(x,y)</math> in the coordinate plane such that two of the three quantities <math>3</math>, <math>x+2</math>, and <math>y-4</math> are equal and the third of the three quantities is no greater than the common value. Which of the following is a correct description of <math>S</math>?<br />
<br />
<math> \textbf{(A)}\ \text{a single point} \qquad\textbf{(B)}\ \text{two intersecting lines} \\ \qquad\textbf{(C)}\ \text{three lines whose pairwise intersections are three distinct points} \\ \qquad\textbf{(D)}\ \text{a triangle}\qquad\textbf{(E)}\ \text{three rays with a common point} </math><br />
<br />
== Solution ==<br />
If the two equal values are <math>3</math> and <math>x+2</math>, then <math>x=1</math>. Also, <math>y-4\leqslant 3</math> because 3 is the common value. Solving for <math>y</math>, we get <math>y\leqslant 7</math>. Therefore the portion of the line <math>x=1</math> where <math>y\leqslant 7</math> is part of <math>S</math>. This is a ray with an endpoint of <math>(1, 7)</math>.<br />
<br />
Similar to the process above, we assume that the two equal values are <math>3</math> and <math>y-4</math>. Solving the equation <math>3=y-4</math> then <math>y=7</math>. Also, <math>x+2\leqslant 3</math> because 3 is the common value. Solving for <math>x</math>, we get <math>x\leqslant 1</math>. Therefore the portion of the line <math>y=7</math> where <math>x\leqslant 1</math> is also part of <math>S</math>. This is another ray with the same endpoint as the above ray: <math>(1, 7)</math>.<br />
<br />
If <math>x+2</math> and <math>y-4</math> are the two equal values, then <math>x+2=y-4</math>. Solving the equation for <math>y</math>, we get <math>y=x+6</math>. Also <math>3\leqslant y-4</math> because <math>y-4</math> is one way to express the common value (using <math>x-2</math> as the common value works as well). Solving for <math>y</math>, we get <math>y\geqslant 7</math>. Therefore the portion of the line <math>y=x+6</math> where <math>y\geqslant 7</math> is part of <math>S</math> like the other two rays. The lowest possible value that can be achieved is also <math>(1, 7)</math>.<br />
<br />
Since <math>S</math> is made up of three rays with common endpoint <math>(1, 7)</math>, the answer is <math>\boxed{E}</math>.<br />
<br />
Solution by TheMathematicsTiger7<br />
<br />
== See Also ==<br />
{{AMC10 box|year=2017|ab=A|num-b=11|num-a=13}}<br />
{{AMC12 box|year=2017|ab=A|num-b=8|num-a=10}}<br />
{{MAA Notice}}</div>Ccorderhttps://artofproblemsolving.com/wiki/index.php?title=2008_AMC_12A_Problems/Problem_24&diff=949822008 AMC 12A Problems/Problem 242018-06-07T18:35:49Z<p>Ccorder: spacing</p>
<hr />
<div>== Problem == <br />
[[Triangle]] <math>ABC</math> has <math>\angle C = 60^{\circ}</math> and <math>BC = 4</math>. Point <math>D</math> is the [[midpoint]] of <math>BC</math>. What is the largest possible value of <math>\tan{\angle BAD}</math>?<br />
<br />
<math>\mathrm{(A)}\ \frac{\sqrt{3}}{6}\qquad\mathrm{(B)}\ \frac{\sqrt{3}}{3}\qquad\mathrm{(C)}\ \frac{\sqrt{3}}{2\sqrt{2}}\qquad\mathrm{(D)}\ \frac{\sqrt{3}}{4\sqrt{2}-3}\qquad\mathrm{(E)}\ 1</math><br />
<br />
== Solution 1 ==<br />
<asy>unitsize(12mm);<br />
pair C=(0,0), B=(4 * dir(60)), A = (8,0), D=(2 * dir(60));<br />
pair E=(1,0), F=(2,0);<br />
draw(C--B--A--C);<br />
draw(A--D);draw(D--E);draw(B--F);<br />
dot(A);dot(B);dot(C);dot(D);dot(E);dot(F);<br />
label("\(C\)",C,SW);<br />
label("\(B\)",B,N);<br />
label("\(A\)",A,SE);<br />
label("\(D\)",D,NW);<br />
label("\(E\)",E,S);<br />
label("\(F\)",F,S);<br />
label("\(60^\circ\)",C+(.1,.1),ENE);<br />
label("\(2\)",1*dir(60),NW);<br />
label("\(2\)",3*dir(60),NW);<br />
label("\(\theta\)",(7,.4));<br />
label("\(1\)",(.5,0),S);<br />
label("\(1\)",(1.5,0),S);<br />
label("\(x-2\)",(5,0),S);</asy><br />
<br />
Let <math>x = CA</math>. Then <math>\tan\theta = \tan(\angle BAF - \angle DAE)</math>, and since <math>\tan\angle BAF = \frac{2\sqrt{3}}{x-2}</math> and <math>\tan\angle DAE = \frac{\sqrt{3}}{x-1}</math>, we have<br />
<br />
<cmath>\tan\theta = \frac{\frac{2\sqrt{3}}{x-2} - \frac{\sqrt{3}}{x-1}}{1 + \frac{2\sqrt{3}}{x-2}\cdot\frac{\sqrt{3}}{x-1}}= \frac{x\sqrt{3}}{x^2-3x+8}</cmath><br />
<br />
With calculus, taking the [[derivative]] and setting equal to zero will give the maximum value of <math>\tan \theta</math>. Otherwise, we can apply [[AM-GM]]:<br />
<br />
<cmath>\begin{align*}<br />
\frac{x^2 - 3x + 8}{x} = \left(x + \frac{8}{x}\right) -3 &\geq 2\sqrt{x \cdot \frac 8x} - 3 = 4\sqrt{2} - 3\\<br />
\frac{x}{x^2 - 3x + 8} &\leq \frac{1}{4\sqrt{2}-3}\\<br />
\frac{x\sqrt{3}}{x^2 - 3x + 8} = \tan \theta &\leq \frac{\sqrt{3}}{4\sqrt{2}-3}\end{align*}</cmath><br />
<br />
Thus, the maximum is at <br />
<math>\frac{\sqrt{3}}{4\sqrt{2}-3} \Rightarrow \mathbf{(D)}</math>.<br />
<br />
== Solution 2 ==<br />
We notice that <math>\tan(x)</math> is strictly increasing on the interval <math>[0, \frac{\pi}{2})</math> (if <math>\angle BAD\ge 90^\circ</math>, then it is impossible for <math>\angle C=60^\circ</math>), so we want to maximize <math>\angle BAD</math>. <br />
<br />
Consider the circumcircle of <math>BAD</math> and let it meet <math>AC</math> again at <math>F</math>. Any point <math>P</math> between <math>A</math> and <math>F</math> on line <math>AC</math> is inside this circle, so it follows that <math>\angle BPD>\angle BAD</math>. Therefore to maximize <math>\angle BAD</math>, the circumcircle of <math>BAD</math> must be tangent to <math>AC</math> at <math>A</math>. By PoP we find that <math>CA^2=CD\cdot CB \Rightarrow AC = 2\sqrt{2}</math>.<br />
<br />
Now our computations are straightforward:<br />
<cmath>\tan\angle BAD = \frac{\sin \angle BAD}{\cos \angle BAD} = \frac{\frac{2\sin\angle ABD}{AD}}{\frac{AB^2+AD^2-BD^2}{2AB\cdot AD}}</cmath><br />
<cmath>=\frac{4\sin \angle ABD\cdot AB}{AB^2+AD^2-4} = \frac{4 AC \sin \angle ACB}{AB^2 + AD^2 - 4}</cmath><br />
<cmath>=\frac{4\sqrt{6}}{(4^2+(2\sqrt{2})^2-4\cdot 2\sqrt{2}) + (2^2+(2\sqrt{2})^2 - 2\cdot 2\sqrt{2}) - 4} = \frac{4\sqrt{6}}{32-12\sqrt{2}}</cmath><br />
<cmath>=\boxed{\frac{\sqrt{3}}{4\sqrt{2}-3}}</cmath><br />
<br />
== See also ==<br />
{{AMC12 box|year=2008|ab=A|num-b=23|num-a=25}}<br />
<br />
[[Category:Intermediate Geometry Problems]]<br />
{{MAA Notice}}</div>Ccorderhttps://artofproblemsolving.com/wiki/index.php?title=2018_AMC_12A_Problems/Problem_1&diff=949812018 AMC 12A Problems/Problem 12018-06-07T18:24:14Z<p>Ccorder: making spacing consistent for problem and solution headings.</p>
<hr />
<div>== Problem ==<br />
<br />
A large urn contains <math>100</math> balls, of which <math>36 \%</math> are red and the rest are blue. How many of the blue balls must be removed so that the percentage of red balls in the urn will be <math>72 \%</math>? (No red balls are to be removed.)<br />
<br />
<math> \textbf{(A)}\ 28 \qquad\textbf{(B)}\ 32 \qquad\textbf{(C)}\ 36 \qquad\textbf{(D)}\ <br />
50 \qquad\textbf{(E)}\ 64 </math><br />
<br />
== Solution 1 ==<br />
<br />
There are <math>36</math> red balls; for these red balls to comprise <math>72 \%</math> of the urn, there must be only <math>14</math> blue balls. Since there are currently <math>64</math> blue balls, this means we must remove <math>50 = \boxed{ \textbf{(D)}.}</math><br />
<br />
== Solution 2 ==<br />
<br />
There are <math>36</math> red balls and <math>64</math> blue balls. For the percentage of the red balls to double from <math>36 \%</math> to <math>72 \%</math> of the urn, half of the total number of balls must be removed. Therefore, the number of blue balls that need to be removed is <math>50 = \boxed{ \textbf{(D)}}</math><br />
<br />
== See Also ==<br />
{{AMC12 box|year=2018|ab=A|before = First Problem|num-a=2}}<br />
{{MAA Notice}}</div>Ccorder