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1715a707ab303d394132bb897d4c4c82 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_2 | For what value of $x$ does $10^{x}\cdot 100^{2x}=1000^{5}$
$\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5$ | We can rewrite $10^{x}\cdot 100^{2x}=1000^{5}$ as $10^{5x}=10^{15}$ \[\begin{split} 10^x\cdot100^{2x} & =10^x\cdot(10^2)^{2x} \\ 10^x\cdot10^{4x} & =(10^3)^5 \\ 10^{5x} & =10^{15} \end{split}\] Since the bases are equal, we can set the exponents equal, giving us $5x=15$ . Solving the equation gives us $x = \boxed{3}.$ | C | 3 |
1715a707ab303d394132bb897d4c4c82 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_2 | For what value of $x$ does $10^{x}\cdot 100^{2x}=1000^{5}$
$\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5$ | We can rewrite this expression as $\log(10^x \cdot 100^{2x})=\log(1000^5)$ , which can be simplified to $\log(10^{x}\cdot10^{4x})=5\log(1000)$ , and that can be further simplified to $\log(10^{5x})=5\log(10^3)$ . This leads to $5x=15$ . Solving this linear equation yields $x = \boxed{3}.$ | C | 3 |
f3137602561c1683be2c9192ef4b0f9b | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_4 | The mean, median, and mode of the $7$ data values $60, 100, x, 40, 50, 200, 90$ are all equal to $x$ . What is the value of $x$
$\textbf{(A)}\ 50 \qquad\textbf{(B)}\ 60 \qquad\textbf{(C)}\ 75 \qquad\textbf{(D)}\ 90 \qquad\textbf{(E)}\ 100$ | Since $x$ is the mean, \begin{align*} x&=\frac{60+100+x+40+50+200+90}{7}\\ &=\frac{540+x}{7}. \end{align*}
Therefore, $7x=540+x$ , so $x=\boxed{90}.$ | D | 90 |
f3137602561c1683be2c9192ef4b0f9b | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_4 | The mean, median, and mode of the $7$ data values $60, 100, x, 40, 50, 200, 90$ are all equal to $x$ . What is the value of $x$
$\textbf{(A)}\ 50 \qquad\textbf{(B)}\ 60 \qquad\textbf{(C)}\ 75 \qquad\textbf{(D)}\ 90 \qquad\textbf{(E)}\ 100$ | Note that $x$ must be the median so it must equal either $60$ or $90$ . You can see that the mean is also $x$ , and by intuition $x$ should be the greater one. $x=\boxed{90}.$ ~bjc | D | 90 |
89accbed968d3469afe551d15afb2e7e | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_6 | A triangular array of $2016$ coins has $1$ coin in the first row, $2$ coins in the second row, $3$ coins in the third row, and so on up to $N$ coins in the $N$ th row. What is the sum of the digits of $N$
$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 7\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 9\qquad\textbf{(E)}\ 10$ | We are trying to find the value of $N$ such that \[1+2+3\cdots+(N-1)+N=\frac{N(N+1)}{2}=2016.\] Noticing that $\frac{63\cdot 64}{2}=2016,$ we have $N=63,$ so our answer is $\boxed{9}.$ | D | 9 |
89accbed968d3469afe551d15afb2e7e | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_6 | A triangular array of $2016$ coins has $1$ coin in the first row, $2$ coins in the second row, $3$ coins in the third row, and so on up to $N$ coins in the $N$ th row. What is the sum of the digits of $N$
$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 7\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 9\qquad\textbf{(E)}\ 10$ | Knowing that each row number can stand for the number of coins there are in the row, we can just add until we get $2016$ .
Notice that $1 + 2 + 3 \cdots + 10 = 55.$ Knowing this, we can say that $11 + 12 \cdots + 20 = 155$ and $21 + \cdots +30 =255$ and so on. This is a quick way to get to the point that N is between 60 and 70. By subtracting from the sum of the number from 1 through 70, we learn that when we subtract $70, 69, 68, 67, 66, 65,$ and $64, N = 63.$ Adding those two digits, we get the answer $\boxed{9}.$ - CorgiARMY | D | 9 |
aee82ef56f7042c3ac61d14ca9f1de2d | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_9 | The five small shaded squares inside this unit square are congruent and have disjoint interiors. The midpoint of each side of the middle square coincides with one of the vertices of the other four small squares as shown. The common side length is $\tfrac{a-\sqrt{2}}{b}$ , where $a$ and $b$ are positive integers. What is $a+b$
[asy] real x=.369; draw((0,0)--(0,1)--(1,1)--(1,0)--cycle); filldraw((0,0)--(0,x)--(x,x)--(x,0)--cycle, gray); filldraw((0,1)--(0,1-x)--(x,1-x)--(x,1)--cycle, gray); filldraw((1,1)--(1,1-x)--(1-x,1-x)--(1-x,1)--cycle, gray); filldraw((1,0)--(1,x)--(1-x,x)--(1-x,0)--cycle, gray); filldraw((.5,.5-x*sqrt(2)/2)--(.5+x*sqrt(2)/2,.5)--(.5,.5+x*sqrt(2)/2)--(.5-x*sqrt(2)/2,.5)--cycle, gray); [/asy]
$\textbf{(A)}\ 7\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 11$ | Let $s$ be the side length of the small squares.
The diagonal of the big square can be written in two ways: $\sqrt{2}$ and $s \sqrt{2} + s + s \sqrt{2}$
Solving for $s$ , we get $s = \frac{4 - \sqrt{2}}{7}$ , so our answer is $4 + 7 \Rightarrow \boxed{11}$ | E | 11 |
aee82ef56f7042c3ac61d14ca9f1de2d | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_9 | The five small shaded squares inside this unit square are congruent and have disjoint interiors. The midpoint of each side of the middle square coincides with one of the vertices of the other four small squares as shown. The common side length is $\tfrac{a-\sqrt{2}}{b}$ , where $a$ and $b$ are positive integers. What is $a+b$
[asy] real x=.369; draw((0,0)--(0,1)--(1,1)--(1,0)--cycle); filldraw((0,0)--(0,x)--(x,x)--(x,0)--cycle, gray); filldraw((0,1)--(0,1-x)--(x,1-x)--(x,1)--cycle, gray); filldraw((1,1)--(1,1-x)--(1-x,1-x)--(1-x,1)--cycle, gray); filldraw((1,0)--(1,x)--(1-x,x)--(1-x,0)--cycle, gray); filldraw((.5,.5-x*sqrt(2)/2)--(.5+x*sqrt(2)/2,.5)--(.5,.5+x*sqrt(2)/2)--(.5-x*sqrt(2)/2,.5)--cycle, gray); [/asy]
$\textbf{(A)}\ 7\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 11$ | The diagonal of the small square can be written in two ways: $s \sqrt(2)$ and $2*(1-2s).$ Equating and simplifying gives $s = \frac{4 - \sqrt{2}}{7}$ . Hence our answer is $4 + 7 \Rightarrow \boxed{11}.$ | E | 11 |
25320ec9fece64ed8fcbb3c714aeb8ad | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_10 | Five friends sat in a movie theater in a row containing $5$ seats, numbered $1$ to $5$ from left to right. (The directions "left" and "right" are from the point of view of the people as they sit in the seats.) During the movie Ada went to the lobby to get some popcorn. When she returned, she found that Bea had moved two seats to the right, Ceci had moved one seat to the left, and Dee and Edie had switched seats, leaving an end seat for Ada. In which seat had Ada been sitting before she got up?
$\textbf{(A) }1 \qquad \textbf{(B) } 2 \qquad \textbf{(C) } 3 \qquad \textbf{(D) } 4\qquad \textbf{(E) } 5$ | Assume that Edie and Dee were originally in seats 3 and 4. If this were so, there is no possible position for which Bea can move 2 seats to the right. The same applies for seats 2 and 3. This means that either Edie or Dee was originally in an edge seat. If Edie and Dee were in seats 1 and 2, then Bea must have been in seat 3, which would mean that seat 5 would now be occupied and the positioning would not work. So, Edie and Dee are in seats 4 and 5. This means that Bea was originally in seat 1. Ceci must have been in seat 3 to keep seat 1 open, which leaves seat 2.
Thus, Ada was in seat $\boxed{2}$ | B | 2 |
25320ec9fece64ed8fcbb3c714aeb8ad | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_10 | Five friends sat in a movie theater in a row containing $5$ seats, numbered $1$ to $5$ from left to right. (The directions "left" and "right" are from the point of view of the people as they sit in the seats.) During the movie Ada went to the lobby to get some popcorn. When she returned, she found that Bea had moved two seats to the right, Ceci had moved one seat to the left, and Dee and Edie had switched seats, leaving an end seat for Ada. In which seat had Ada been sitting before she got up?
$\textbf{(A) }1 \qquad \textbf{(B) } 2 \qquad \textbf{(C) } 3 \qquad \textbf{(D) } 4\qquad \textbf{(E) } 5$ | Note that the person (out of A,B,C) that moves the most, moves the amount equal to the sum of what the other 2 move. They essentially make a cycle. D & E are seat fillers and can be ignored. A,B,C take up either seats 1,2,3 or 2,4,5. In each case you find A was originally in seat $\boxed{2}$ | null | 2 |
25320ec9fece64ed8fcbb3c714aeb8ad | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_10 | Five friends sat in a movie theater in a row containing $5$ seats, numbered $1$ to $5$ from left to right. (The directions "left" and "right" are from the point of view of the people as they sit in the seats.) During the movie Ada went to the lobby to get some popcorn. When she returned, she found that Bea had moved two seats to the right, Ceci had moved one seat to the left, and Dee and Edie had switched seats, leaving an end seat for Ada. In which seat had Ada been sitting before she got up?
$\textbf{(A) }1 \qquad \textbf{(B) } 2 \qquad \textbf{(C) } 3 \qquad \textbf{(D) } 4\qquad \textbf{(E) } 5$ | Note that the net displacements in the right direction sum up to $0$ . The sum of the net displacements of Bea, Ceci, Dee, Edie is $2-1 = 1$ , so Ada moved exactly $1$ place to the left. Since Ada ended on an end seat, she must have started on seat $\boxed{2}$ | null | 2 |
aa0bf97e09cfb906b9adb4f3d576633b | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_11 | Each of the $100$ students in a certain summer camp can either sing, dance, or act. Some students have more than one talent, but no student has all three talents. There are $42$ students who cannot sing, $65$ students who cannot dance, and $29$ students who cannot act. How many students have two of these talents?
$\textbf{(A)}\ 16\qquad\textbf{(B)}\ 25\qquad\textbf{(C)}\ 36\qquad\textbf{(D)}\ 49\qquad\textbf{(E)}\ 64$ | Let $a$ be the number of students that can only sing, $b$ can only dance, and $c$ can only act.
Let $ab$ be the number of students that can sing and dance, $ac$ can sing and act, and $bc$ can dance and act.
From the information given in the problem, $a + ab + b = 29, b + bc + c = 42,$ and $a + ac + c = 65$
Adding these equations together, we get $2(a + b + c) + ab + bc + ac = 136$
Since there are a total of $100$ students, $a + b + c + ab + bc + ac = 100$
Subtracting these equations, we get $a + b + c = 36$
Our answer is $ab + bc + ac = 100 - (a + b + c) = 100 - 36 = \boxed{64}$ | E | 64 |
aa0bf97e09cfb906b9adb4f3d576633b | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_11 | Each of the $100$ students in a certain summer camp can either sing, dance, or act. Some students have more than one talent, but no student has all three talents. There are $42$ students who cannot sing, $65$ students who cannot dance, and $29$ students who cannot act. How many students have two of these talents?
$\textbf{(A)}\ 16\qquad\textbf{(B)}\ 25\qquad\textbf{(C)}\ 36\qquad\textbf{(D)}\ 49\qquad\textbf{(E)}\ 64$ | An easier way to solve the problem:
Since $42$ students cannot sing, there are $100-42=58$ students who can.
Similarly $65$ students cannot dance, there are $100-65=35$ students who can.
And $29$ students cannot act, there are $100-29=71$ students who can.
Therefore, there are $58+35+71=164$ students in all ignoring the overlaps between $2$ of $3$ talent categories.
There are no students who have all $3$ talents, nor those who have none $(0)$ , so only $1$ or $2$ talents are viable.
Thus, there are $164-100=\boxed{64}$ students who have $2$ of $3$ talents. | E | 64 |
9c5b924f7e4a5c96dde0bfadef645041 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_13 | Let $N$ be a positive multiple of $5$ . One red ball and $N$ green balls are arranged in a line in random order. Let $P(N)$ be the probability that at least $\tfrac{3}{5}$ of the green balls are on the same side of the red ball. Observe that $P(5)=1$ and that $P(N)$ approaches $\tfrac{4}{5}$ as $N$ grows large. What is the sum of the digits of the least value of $N$ such that $P(N) < \tfrac{321}{400}$
$\textbf{(A) } 12 \qquad \textbf{(B) } 14 \qquad \textbf{(C) }16 \qquad \textbf{(D) } 18 \qquad \textbf{(E) } 20$ | Let $n = \frac{N}{5}$ . Then, consider $5$ blocks of $n$ green balls in a line, along with the red ball. Shuffling the line is equivalent to choosing one of the $N + 1$ positions between the green balls to insert the red ball. Less than $\frac{3}{5}$ of the green balls will be on the same side of the red ball if the red ball is inserted inside the middle block of $n$ balls, and there are $n - 1$ positions where this happens. Thus, $P(N) = 1 - \frac{n - 1}{N + 1} = \frac{4n + 2}{5n + 1}$ , so
\[P(N) = \frac{4n + 2}{5n + 1} < \frac{321}{400}.\]
Multiplying both sides of the inequality by $400(5n+1)$ , we have
\[400(4n+2)<321(5n+1),\]
and by the distributive property,
\[1600n+800<1605n+321.\]
Subtracting $1600n+321$ on both sides of the inequality gives us
\[479<5n.\]
Therefore, $N=5n>479$ , so the least possible value of $N$ is $480$ . The sum of the digits of $480$ is $\boxed{12}$ | A | 12 |
9c5b924f7e4a5c96dde0bfadef645041 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_13 | Let $N$ be a positive multiple of $5$ . One red ball and $N$ green balls are arranged in a line in random order. Let $P(N)$ be the probability that at least $\tfrac{3}{5}$ of the green balls are on the same side of the red ball. Observe that $P(5)=1$ and that $P(N)$ approaches $\tfrac{4}{5}$ as $N$ grows large. What is the sum of the digits of the least value of $N$ such that $P(N) < \tfrac{321}{400}$
$\textbf{(A) } 12 \qquad \textbf{(B) } 14 \qquad \textbf{(C) }16 \qquad \textbf{(D) } 18 \qquad \textbf{(E) } 20$ | Let $N=5$ $P(N)=1$ (Given)
Let $N=10$ $P(N)=\frac{10}{11}$
Let $N=15$ $P(N)=\frac{14}{16}$
Notice that the fraction can be written as $1-\frac{\frac{N}{5}-1}{N+1}$
Now it's quite simple to write the inequality as $1-\frac{\frac{N}{5}-1}{N+1}<\frac{321}{400}$
We can subtract $1$ on both sides to obtain $-\frac{\frac{N}{5}-1}{N+1}<-\frac{79}{400}$
Dividing both sides by $-1$ , we derive $\frac{\frac{N}{5}-1}{N+1}>\frac{79}{400}$ . (Switch the inequality sign when dividing by $-1$
We then cross multiply to get $80N - 400 > 79N + 79$
Finally we get $N > 479$
To achieve $N = 480$
So the sum of the digits of $N$ $\boxed{12}$ | A | 12 |
9c5b924f7e4a5c96dde0bfadef645041 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_13 | Let $N$ be a positive multiple of $5$ . One red ball and $N$ green balls are arranged in a line in random order. Let $P(N)$ be the probability that at least $\tfrac{3}{5}$ of the green balls are on the same side of the red ball. Observe that $P(5)=1$ and that $P(N)$ approaches $\tfrac{4}{5}$ as $N$ grows large. What is the sum of the digits of the least value of $N$ such that $P(N) < \tfrac{321}{400}$
$\textbf{(A) } 12 \qquad \textbf{(B) } 14 \qquad \textbf{(C) }16 \qquad \textbf{(D) } 18 \qquad \textbf{(E) } 20$ | We are trying to find the number of places to put the red ball, such that $\frac{3}{5}$ of the green balls or more are on one side of it. Notice that we can put the ball in a number of spaces describable with $N$ : Trying a few values, we see that the ball "works" in places $1$ to $\frac{2}{5}N + 1$ and spaces $\frac{3}{5}N+1$ to $N+1$ . This is a total of $\frac{4}{5}N + 2$ spaces, over a total possible $N + 1$ places to put the ball. So:
$\frac{\frac{4}{5}N + 2}{N+1} = \frac{321}{400} \rightarrow N = 479.$ And we know that the next value is what we are looking for, so $N+1 = 480$ , and the sum of its digits is $\boxed{12}$ | A | 12 |
678b5671dc839abe5ea18f5c4fb7fc61 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_14 | Each vertex of a cube is to be labeled with an integer $1$ through $8$ , with each integer being used once, in such a way that the sum of the four numbers on the vertices of a face is the same for each face. Arrangements that can be obtained from each other through rotations of the cube are considered to be the same. How many different arrangements are possible?
$\textbf{(A) } 1\qquad\textbf{(B) } 3\qquad\textbf{(C) }6 \qquad\textbf{(D) }12 \qquad\textbf{(E) }24$ | Note that the sum of the numbers on each face must be 18, because $\frac{1+2+\cdots+8}{2}=18$
So now consider the opposite edges (two edges which are parallel but not on same face of the cube);
they must have the same sum value too.
Now think about the points $1$ and $8$ . If they are not on the same edge, they must be endpoints of opposite edges, and we should have $1+X=8+Y$ . However, this scenario would yield no solution for $[7,2]$ , which is a contradiction. (Try drawing out the cube if it doesn't make sense to you.)
The points $1$ and $8$ are therefore on the same side and all edges parallel must also sum to $9$
Now we have $4$ parallel sides $1-8, 2-7, 3-6, 4-5$ .
Thinking about $4$ endpoints, we realize they need to sum to $18$ .
It is easy to notice only $1-7-6-4$ and $8-2-3-5$ would work.
So if we fix one direction $1-8 ($ or $8-1)$ all other $3$ parallel sides must lay in one particular direction. $(1-8,7-2,6-3,4-5)$ or $(8-1,2-7,3-6,5-4)$
Now, the problem is the same as arranging $4$ points in a two-dimensional square, which is $\frac{4!}{4}=\boxed{6.}$ | C | 6. |
678b5671dc839abe5ea18f5c4fb7fc61 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_14 | Each vertex of a cube is to be labeled with an integer $1$ through $8$ , with each integer being used once, in such a way that the sum of the four numbers on the vertices of a face is the same for each face. Arrangements that can be obtained from each other through rotations of the cube are considered to be the same. How many different arrangements are possible?
$\textbf{(A) } 1\qquad\textbf{(B) } 3\qquad\textbf{(C) }6 \qquad\textbf{(D) }12 \qquad\textbf{(E) }24$ | Again, all faces sum to $18.$ If $x,y,z$ are the vertices next to $1$ , then the remaining vertices are $17-x-y, 17-y-z, 17-x-z, x+y+z-16.$ Now it remains to test possibilities. Note that we must have $x+y+z>17.$ Without loss of generality, let $x<y<z.$ \[3,7,8\text{: Does not work.}\] \[4,6,8\text{: Works.}\] \[4,7,8\text{: Works.}\] \[5,6,7\text{: Does not work.}\] \[5,6,8\text{: Does not work.}\] \[5,7,8\text{: Does not work.}\] \[6,7,8\text{: Works.}\]
Keeping in mind that a) solutions that can be obtained by rotating each other count as one solution and b) a cyclic sequence is always asymmetrical, we can see that there are two solutions (one with $[x, y, z]$ and one with $[z, y, x]$ ) for each combination of $x$ $y$ , and $z$ from above. So, our answer is $3\cdot 2=\boxed{6.}$ | C | 6. |
678b5671dc839abe5ea18f5c4fb7fc61 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_14 | Each vertex of a cube is to be labeled with an integer $1$ through $8$ , with each integer being used once, in such a way that the sum of the four numbers on the vertices of a face is the same for each face. Arrangements that can be obtained from each other through rotations of the cube are considered to be the same. How many different arrangements are possible?
$\textbf{(A) } 1\qquad\textbf{(B) } 3\qquad\textbf{(C) }6 \qquad\textbf{(D) }12 \qquad\textbf{(E) }24$ | We know the sum of each face is $18.$ If we look at an edge of the cube whose numbers sum to $x$ , it must be possible to achieve the sum $18-x$ in two distinct ways, looking at the two faces which contain the edge. If $8$ and $6$ were on the same edge, it is possible to achieve the desired sum only with the numbers $1$ and $3$ since the values must be distinct. Similarly, if $8$ and $7$ were on the same edge, the only way to get the sum is with $1$ and $2$ . This means that $6$ and $7$ are not on the same edge as $8$ , or in other words they are diagonally across from it on the same face, or on the other end of the cube.
Now we look at three cases, each yielding two solutions which are reflections of each other:
1) $6$ and $7$ are diagonally opposite $8$ on the same face.
2) $6$ is diagonally across the cube from $8$ , while $7$ is diagonally across from $8$ on the same face.
3) $7$ is diagonally across the cube from $8$ , while $6$ is diagonally across from $8$ on the same face.
This means the answer is $3\cdot 2=\boxed{6.}$ | C | 6. |
b4d0972c5fb5351b74e4648e973fe3a0 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_16 | The graphs of $y=\log_3 x, y=\log_x 3, y=\log_\frac{1}{3} x,$ and $y=\log_x \dfrac{1}{3}$ are plotted on the same set of axes. How many points in the plane with positive $x$ -coordinates lie on two or more of the graphs?
$\textbf{(A)}\ 2\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 5\qquad\textbf{(E)}\ 6$ | Setting the first two equations equal to each other, $\log_3 x = \log_x 3$
Solving this, we get $\left(3, 1\right)$ and $\left(\frac{1}{3}, -1\right)$
Similarly with the last two equations, we get $\left(3, -1\right)$ and $\left(\frac{1}{3}, 1\right)$
Now, by setting the first and third equations equal to each other, we get $\left(1, 0\right)$
Pairing the first and fourth or second and third equations won't work because then $\log x \leq 0$
Pairing the second and fourth equations will yield $x = 1$ , but since you can't divide by $\log 1 = 0$ , it doesn't work.
After trying all pairs, we have a total of $5$ solutions $\rightarrow \boxed{5}$ | D | 5 |
b4d0972c5fb5351b74e4648e973fe3a0 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_16 | The graphs of $y=\log_3 x, y=\log_x 3, y=\log_\frac{1}{3} x,$ and $y=\log_x \dfrac{1}{3}$ are plotted on the same set of axes. How many points in the plane with positive $x$ -coordinates lie on two or more of the graphs?
$\textbf{(A)}\ 2\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 5\qquad\textbf{(E)}\ 6$ | Note that $\log_b a =\log_c a / \log_c b$
Then $\log_b a = \log_a a / \log_a b = 1/ \log_a b$
$\log_\frac{1}{a} b = \log_a \frac{1}{a} / \log_a b = -1/ \log_a b$
$\log_\frac{1}{b} a = -\log_a b$
Therefore, the system of equations can be simplified to:
$y = t$
$y = -t$
$y = \frac{1}{t}$
$y = -\frac{1}{t}$
where $t = \log_3 x$ . Note that all values of $t$ correspond to exactly one positive $x$ value, so all $(t,y)$ intersections will correspond to exactly one $(x,y)$ intersection in the positive-x area.
Graphing this system of functions will generate a total of $5$ solutions $\rightarrow \boxed{5}$ | D | 5 |
1d01bd4418c390074d692ef401d636c5 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_18 | For some positive integer $n$ , the number $110n^3$ has $110$ positive integer divisors, including $1$ and the number $110n^3$ . How many positive integer divisors does the number $81n^4$ have?
$\textbf{(A) }110\qquad\textbf{(B) }191\qquad\textbf{(C) }261\qquad\textbf{(D) }325\qquad\textbf{(E) }425$ | Since the prime factorization of $110$ is $2 \cdot 5 \cdot 11$ , we have that the number is equal to $2 \cdot 5 \cdot 11 \cdot n^3$ . This has $2 \cdot 2 \cdot 2=8$ factors when $n=1$ . This needs a multiple of 11 factors, which we can achieve by setting $n=2^3$ , so we have $2^{10} \cdot 5 \cdot 11$ has $44$ factors. To achieve the desired $110$ factors, we need the number of factors to also be divisible by $5$ , so we can set $n=2^3 \cdot 5$ , so $2^{10} \cdot 5^4 \cdot 11$ has $110$ factors. Therefore, $n=2^3 \cdot 5$ . In order to find the number of factors of $81n^4$ , we raise this to the fourth power and multiply it by $81$ , and find the factors of that number. We have $3^4 \cdot 2^{12} \cdot 5^4$ , and this has $5 \cdot 13 \cdot 5=\boxed{325}$ factors. | D | 325 |
1d01bd4418c390074d692ef401d636c5 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_18 | For some positive integer $n$ , the number $110n^3$ has $110$ positive integer divisors, including $1$ and the number $110n^3$ . How many positive integer divisors does the number $81n^4$ have?
$\textbf{(A) }110\qquad\textbf{(B) }191\qquad\textbf{(C) }261\qquad\textbf{(D) }325\qquad\textbf{(E) }425$ | $110n^3$ clearly has at least three distinct prime factors, namely 2, 5, and 11.
The number of factors of $p_1^{n_1}\cdots p_k^{n_k}$ is $(n_1+1)\cdots(n_k+1)$ when the $p$ 's are distinct primes. This tells us that none of these factors can be 1. The number of factors is given as 110. The only way to write 110 as a product of at least three factors without $1$ s is $2\cdot 5\cdot 11$
We conclude that $110n^3$ has only the three prime factors 2, 5, and 11 and that the multiplicities are 1, 4, and 10 in some order. I.e., there are six different possible values of $n$ all of the form $n=p_1\cdot p_2^3$
$81n^4$ thus has prime factorization $81n^4=3^4\cdot p_1^4\cdot p_2^{12}$ and a factor count of $5\cdot5\cdot13=\boxed{325}$ | D | 325 |
03ba8fa9f71721051b46a01a8e620a94 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_19 | Jerry starts at $0$ on the real number line. He tosses a fair coin $8$ times. When he gets heads, he moves $1$ unit in the positive direction; when he gets tails, he moves $1$ unit in the negative direction. The probability that he reaches $4$ at some time during this process $\frac{a}{b},$ where $a$ and $b$ are relatively prime positive integers. What is $a + b?$ (For example, he succeeds if his sequence of tosses is $HTHHHHHH.$
$\textbf{(A)}\ 69\qquad\textbf{(B)}\ 151\qquad\textbf{(C)}\ 257\qquad\textbf{(D)}\ 293\qquad\textbf{(E)}\ 313$ | For $6$ to $8$ heads, we are guaranteed to hit $4$ heads, so the sum here is $\binom{8}{2}+\binom{8}{1}+\binom{8}{0}=28+8+1=37$
For $4$ heads, you have to hit the $4$ heads at the start so there's only one way, $1$
For $5$ heads, we either start off with $4$ heads, which gives us $_4\text{C}_1=4$ ways to arrange the other flips, or we start off with five heads and one tail, which has $6$ ways minus the $2$ overlapping cases, $\text{HHHHHTTT}$ and $\text{HHHHTHTT}$ . Total ways: $8$
Then we sum to get $46$ . There are a total of $2^8=256$ possible sequences of $8$ coin flips, so the probability is $\frac{46}{256}=\frac{23}{128}$ . Summing, we get $23+128=\boxed{151}$ | B | 151 |
03ba8fa9f71721051b46a01a8e620a94 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_19 | Jerry starts at $0$ on the real number line. He tosses a fair coin $8$ times. When he gets heads, he moves $1$ unit in the positive direction; when he gets tails, he moves $1$ unit in the negative direction. The probability that he reaches $4$ at some time during this process $\frac{a}{b},$ where $a$ and $b$ are relatively prime positive integers. What is $a + b?$ (For example, he succeeds if his sequence of tosses is $HTHHHHHH.$
$\textbf{(A)}\ 69\qquad\textbf{(B)}\ 151\qquad\textbf{(C)}\ 257\qquad\textbf{(D)}\ 293\qquad\textbf{(E)}\ 313$ | Reaching 4 will require either 4, 6, or 8 flips. Therefore we can split into 3 cases:
(Case 1): The first four flips are heads. Then, the last four flips can be anything so $2^4=16$ possibilities work.
(Case 2): It takes 6 flips to reach 4. There must be one tail in the first four flips so we don't repeat case 1. The tail can be in one of 4 positions. The next two flips must be heads. The last two flips can be anything so $2^2=4$ flips work. $4*4=16$
(Case 3): It takes 8 flips to reach 4. We can split this case into 2 sub-cases. There can either be 1 or 2 tails in the first 4 flips.
(1 tail in first four flips). In this case, the first tail can be in 4 positions. The second tail can be in either the 5th or 6th position so we don't repeat case 2. Thus, there are $4*2=8$ possibilities.
(2 tails in first four flips). In this case, the tails can be in $\binom{4}{2}=6$ positions.
Adding these cases up and taking the total out of $2^8=256$ yields $\frac{16+16+8+6}{256}=\frac{46}{256}=\frac{23}{128}$ . This means the answer is $23+128=\boxed{151}$ | B | 151 |
03ba8fa9f71721051b46a01a8e620a94 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_19 | Jerry starts at $0$ on the real number line. He tosses a fair coin $8$ times. When he gets heads, he moves $1$ unit in the positive direction; when he gets tails, he moves $1$ unit in the negative direction. The probability that he reaches $4$ at some time during this process $\frac{a}{b},$ where $a$ and $b$ are relatively prime positive integers. What is $a + b?$ (For example, he succeeds if his sequence of tosses is $HTHHHHHH.$
$\textbf{(A)}\ 69\qquad\textbf{(B)}\ 151\qquad\textbf{(C)}\ 257\qquad\textbf{(D)}\ 293\qquad\textbf{(E)}\ 313$ | Notice every $2$ flips, there is a $\dfrac{1}{4}$ chance to go $2$ steps left $(L),$ $\dfrac{1}{2}$ chance to stay put $(P),$ and $\dfrac{1}{4}$ chance to move $2$ steps right $(R).$ We have $4$ choices for how to get to $4:$ \[RR-1\text{ arrangement}-\dfrac{1}{4}\cdot\dfrac{1}{4}=\dfrac{1}{16}\] \[PRR-2\text{ arrangements}-2\cdot\dfrac{1}{2}\cdot\dfrac{1}{4}\cdot\dfrac{1}{4}=\dfrac{1}{16}\] \[LRRR-2\text{ arrangements}-2\cdot\dfrac{1}{4}\cdot\dfrac{1}{4}\cdot\dfrac{1}{4}\cdot\dfrac{1}{4}=\dfrac{1}{128}\] \[PPRR-3\text{ arrangements}-3\cdot\dfrac{1}{2}\cdot\dfrac{1}{2}\cdot\dfrac{1}{4}\cdot\dfrac{1}{4}=\dfrac{3}{64}\] Finally, our answer will be $\dfrac{1}{16}+\dfrac{1}{16}+\dfrac{1}{128}+\dfrac{3}{64}=\dfrac{23}{128}\implies a+b=\boxed{151}.$ | B | 151 |
93af4cac5a67157833d27feb7ade1fa9 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_20 | A binary operation $\diamondsuit$ has the properties that $a\,\diamondsuit\, (b\,\diamondsuit \,c) = (a\,\diamondsuit \,b)\cdot c$ and that $a\,\diamondsuit \,a=1$ for all nonzero real numbers $a, b,$ and $c$ . (Here $\cdot$ represents multiplication). The solution to the equation $2016 \,\diamondsuit\, (6\,\diamondsuit\, x)=100$ can be written as $\tfrac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. What is $p+q?$
$\textbf{(A) }109\qquad\textbf{(B) }201\qquad\textbf{(C) }301\qquad\textbf{(D) }3049\qquad\textbf{(E) }33,601$ | We see that $a \, \diamondsuit \, a = 1$ , and think of division. Testing, we see that the first condition $a \, \diamondsuit \, (b \, \diamondsuit \, c) = (a \, \diamondsuit \, b) \cdot c$ is satisfied, because $\frac{a}{\frac{b}{c}} = \frac{a}{b} \cdot c$ . Therefore, division can be the operation $\diamondsuit$ . Solving the equation, \[\frac{2016}{\frac{6}{x}} = \frac{2016}{6} \cdot x = 336x = 100\implies x=\frac{100}{336} = \frac{25}{84},\] so the answer is $25 + 84 = \boxed{109}$ | A | 109 |
93af4cac5a67157833d27feb7ade1fa9 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_20 | A binary operation $\diamondsuit$ has the properties that $a\,\diamondsuit\, (b\,\diamondsuit \,c) = (a\,\diamondsuit \,b)\cdot c$ and that $a\,\diamondsuit \,a=1$ for all nonzero real numbers $a, b,$ and $c$ . (Here $\cdot$ represents multiplication). The solution to the equation $2016 \,\diamondsuit\, (6\,\diamondsuit\, x)=100$ can be written as $\tfrac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. What is $p+q?$
$\textbf{(A) }109\qquad\textbf{(B) }201\qquad\textbf{(C) }301\qquad\textbf{(D) }3049\qquad\textbf{(E) }33,601$ | If the given conditions hold for all nonzero numbers $a, b,$ and $c$
Let $a=b=c.$ From the first two givens, this implies that
\[a\diamondsuit\, (a\diamondsuit\, {a})=(a\diamondsuit\, a)\cdot{a}.\]
From $a\diamondsuit\,{a}=1,$ this equation simply becomes $a\diamondsuit\,{1}=a.$
Let $c=b.$ Substituting this into the first two conditions, we see that
\[a\diamondsuit\, (b\diamondsuit\, {c})=(a\diamondsuit\, {b})\cdot{c} \implies a\diamondsuit\, (b\diamondsuit\, {b})=(a\diamondsuit\, {b})\cdot{b}.\]
Substituting $b\diamondsuit\, {b} =1$ , the second equation becomes
\[a\diamondsuit\, {1}=(a\diamondsuit\, {b})\cdot{b} \implies a=(a\diamondsuit\,{b})\cdot{b}.\]
Since $a, b$ and $c$ are nonzero, we can divide by $b$ which yields,
\[\frac{a}{b}=(a\diamondsuit\, {b}).\]
Now we can find the value of $x$ straightforwardly:
\[\frac{2016}{(\frac{6}{x})}=100 \implies 2016=\frac{600}{x} \implies x=\frac{600}{2016} = \frac{25}{84}.\]
Therefore, $a+b=25+84=\boxed{109}$ | A | 109 |
93af4cac5a67157833d27feb7ade1fa9 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_20 | A binary operation $\diamondsuit$ has the properties that $a\,\diamondsuit\, (b\,\diamondsuit \,c) = (a\,\diamondsuit \,b)\cdot c$ and that $a\,\diamondsuit \,a=1$ for all nonzero real numbers $a, b,$ and $c$ . (Here $\cdot$ represents multiplication). The solution to the equation $2016 \,\diamondsuit\, (6\,\diamondsuit\, x)=100$ can be written as $\tfrac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. What is $p+q?$
$\textbf{(A) }109\qquad\textbf{(B) }201\qquad\textbf{(C) }301\qquad\textbf{(D) }3049\qquad\textbf{(E) }33,601$ | One way to eliminate the $\diamondsuit$ in this equation is to make $a = b$ so that $a\,\diamondsuit\, (b\,\diamondsuit \,c) = c$ . In this case, we can make $b = 2016$
\[2016 \,\diamondsuit\, (6\,\diamondsuit\, x)=100\implies (2016\, \diamondsuit\, 6) \cdot x = 100\]
By multiplying both sides by $\frac{6}{x}$ , we get:
\[(2016\, \diamondsuit\, 6) \cdot 6 = \frac{600}{x}\implies 2016 \, \diamondsuit\, (6\, \diamondsuit\, 6) = \frac{600}{x}\]
Because $6\, \diamondsuit\, 6 = 2016\, \diamondsuit\, 2016 = 1:$
\[2016 \, \diamondsuit\, (2016\, \diamondsuit\, 2016) = \frac{600}{x}\implies (2016\, \diamondsuit\, 2016) \cdot 2016 = \frac{600}{x}\implies 2016 = \frac{600}{x}\]
Therefore, $x = \frac{600}{2016} = \frac{25}{84}$ , so the answer is $25 + 84 = \boxed{109.}$ | A | 109. |
93af4cac5a67157833d27feb7ade1fa9 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_20 | A binary operation $\diamondsuit$ has the properties that $a\,\diamondsuit\, (b\,\diamondsuit \,c) = (a\,\diamondsuit \,b)\cdot c$ and that $a\,\diamondsuit \,a=1$ for all nonzero real numbers $a, b,$ and $c$ . (Here $\cdot$ represents multiplication). The solution to the equation $2016 \,\diamondsuit\, (6\,\diamondsuit\, x)=100$ can be written as $\tfrac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. What is $p+q?$
$\textbf{(A) }109\qquad\textbf{(B) }201\qquad\textbf{(C) }301\qquad\textbf{(D) }3049\qquad\textbf{(E) }33,601$ | We can manipulate the given identities to arrive at a conclusion about the binary operator $\diamondsuit$ . Substituting $b = c$ into the first identity yields \[( a\ \diamondsuit\ b) \cdot b = a\ \diamondsuit\ (b\ \diamondsuit\ b) = a\ \diamondsuit\ 1 = a\ \diamondsuit\ ( a\ \diamondsuit\ a) = ( a\ \diamondsuit\ a) \cdot a = a.\] Hence, $( a\ \diamondsuit\ b) \cdot b = a,$ or, dividing both sides of the equation by $b,$ $( a\ \diamondsuit\ b) = \frac{a}{b}.$
Hence, the given equation becomes $\frac{2016}{\frac{6}{x}} = 100$ . Solving yields $x=\frac{100}{336} = \frac{25}{84},$ so the answer is $25 + 84 = \boxed{109.}$ | A | 109. |
93af4cac5a67157833d27feb7ade1fa9 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_20 | A binary operation $\diamondsuit$ has the properties that $a\,\diamondsuit\, (b\,\diamondsuit \,c) = (a\,\diamondsuit \,b)\cdot c$ and that $a\,\diamondsuit \,a=1$ for all nonzero real numbers $a, b,$ and $c$ . (Here $\cdot$ represents multiplication). The solution to the equation $2016 \,\diamondsuit\, (6\,\diamondsuit\, x)=100$ can be written as $\tfrac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. What is $p+q?$
$\textbf{(A) }109\qquad\textbf{(B) }201\qquad\textbf{(C) }301\qquad\textbf{(D) }3049\qquad\textbf{(E) }33,601$ | $2016 \diamondsuit (6 \diamondsuit x) = (2016 \diamondsuit 6) \cdot x = 100$
$2016 \diamondsuit (2016 \diamondsuit 1) = (2016 \diamondsuit 2016) \cdot 1 = 1 \cdot 1 = 1$
$2016 \diamondsuit 2016 = 1$ $2016 \diamondsuit (2016 \diamondsuit 1) = 1$ , so $2016 \diamondsuit 1 = 2016$
$2016 \diamondsuit 1 = (2016 \diamondsuit 6) \cdot 6$ $2016 \diamondsuit 6 = \frac{2016 \diamondsuit 1}{6} = 336$
$x = \frac{100}{2016 \diamondsuit 6} = \frac{100}{336} = \frac{25}{84}$ $24 + 85 = \boxed{109}$ | A | 109 |
93af4cac5a67157833d27feb7ade1fa9 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_20 | A binary operation $\diamondsuit$ has the properties that $a\,\diamondsuit\, (b\,\diamondsuit \,c) = (a\,\diamondsuit \,b)\cdot c$ and that $a\,\diamondsuit \,a=1$ for all nonzero real numbers $a, b,$ and $c$ . (Here $\cdot$ represents multiplication). The solution to the equation $2016 \,\diamondsuit\, (6\,\diamondsuit\, x)=100$ can be written as $\tfrac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. What is $p+q?$
$\textbf{(A) }109\qquad\textbf{(B) }201\qquad\textbf{(C) }301\qquad\textbf{(D) }3049\qquad\textbf{(E) }33,601$ | Notice that $2016 \diamondsuit (6 \diamondsuit 6)=(2016 \diamondsuit 6) \cdot 6 = 2016$ . Hence, $2016 \diamondsuit 6 = 336$ . Thus, $2016 \diamondsuit (6 \diamondsuit x)=100 \implies (2016 \diamondsuit 6) \cdot x = 100 \implies 336x=100 \implies x=\frac{25}{84}$ . Therefore, the answer is $\boxed{109}$ | A | 109 |
b7066cfe85c79dde1652e7df58145906 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_21 | A quadrilateral is inscribed in a circle of radius $200\sqrt{2}$ . Three of the sides of this quadrilateral have length $200$ . What is the length of the fourth side?
$\textbf{(A) }200\qquad \textbf{(B) }200\sqrt{2}\qquad\textbf{(C) }200\sqrt{3}\qquad\textbf{(D) }300\sqrt{2}\qquad\textbf{(E) } 500$ | [asy] size(250); defaultpen(linewidth(0.4)); //Variable Declarations real RADIUS; pair A, B, C, D, E, F, O; RADIUS=3; //Variable Definitions A=RADIUS*dir(148.414); B=RADIUS*dir(109.471); C=RADIUS*dir(70.529); D=RADIUS*dir(31.586); O=(0,0); //Path Definitions path quad= A -- B -- C -- D -- cycle; //Initial Diagram draw(Circle(O, RADIUS), linewidth(0.8)); draw(quad, linewidth(0.8)); label("$A$",A,W); label("$B$",B,NW); label("$C$",C,NE); label("$D$",D,ENE); label("$O$",O,S); label("$\theta$",O,3N); //Radii draw(O--A); draw(O--B); draw(O--C); draw(O--D); //Construction E=extension(B,O,A,D); label("$E$",E,NE); F=extension(C,O,A,D); label("$F$",F,NE); //Angle marks draw(anglemark(C,O,B)); [/asy]
Let $AD$ intersect $OB$ at $E$ and $OC$ at $F.$
$\overarc{AB}= \overarc{BC}= \overarc{CD}=\theta$
$\angle{BAD}=\frac{1}{2} \cdot \overarc{BCD}=\theta=\angle{AOB}$
From there, $\triangle{OAB} \sim \triangle{ABE}$ , thus:
$\frac{OA}{AB} = \frac{AB}{BE} = \frac{OB}{AE}$
$OA = OB$ because they are both radii of $\odot{O}$ . Since $\frac{OA}{AB} = \frac{OB}{AE}$ , we have that $AB = AE$ . Similarly, $CD = DF$
$OE = 100\sqrt{2} = \frac{OB}{2}$ and $EF=\frac{BC}{2}=100$ , so $AD=AE + EF + FD = 200 + 100 + 200 = \boxed{500}$ | E | 500 |
b7066cfe85c79dde1652e7df58145906 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_21 | A quadrilateral is inscribed in a circle of radius $200\sqrt{2}$ . Three of the sides of this quadrilateral have length $200$ . What is the length of the fourth side?
$\textbf{(A) }200\qquad \textbf{(B) }200\sqrt{2}\qquad\textbf{(C) }200\sqrt{3}\qquad\textbf{(D) }300\sqrt{2}\qquad\textbf{(E) } 500$ | To save us from getting big numbers with lots of zeros behind them, let's divide all side lengths by $200$ for now, then multiply it back at the end of our solution.
[asy] size(250); defaultpen(linewidth(0.4)); //Variable Declarations real RADIUS; pair A, B, C, D, E, O; RADIUS=3; //Variable Definitions A=RADIUS*dir(148.414); B=RADIUS*dir(109.471); C=RADIUS*dir(70.529); D=RADIUS*dir(31.586); E=extension(B,D,O,C); O=(0,0); //Path Definitions path quad= A -- B -- C -- D -- cycle; //Initial Diagram draw(Circle(O, RADIUS), linewidth(0.8)); draw(quad, linewidth(0.8)); label("$A$",A,W); label("$B$",B,NW); label("$C$",C,NE); label("$D$",D,E); label("$E$",E,WSW); label("$O$",O,S); //Radii draw(O--A); draw(O--B); draw(O--C); draw(O--D); //Construction draw(B--D); draw(rightanglemark(C,E,D)); [/asy]
Construct quadrilateral $ABCD$ on the circle with $AD$ being the missing side (Notice that since the side length is less than the radius, it will be very small on the top of the circle). Now, draw the radii from center $O$ to $A,B,C,$ and $D$ . Let the intersection of $BD$ and $OC$ be point $E$ . Notice that $BD$ and $OC$ are perpendicular because $BCDO$ is a kite.
We set lengths $BE=ED$ equal to $x$ (Solution 1.1 begins from here). By the Pythagorean Theorem, \[\sqrt{1^2-x^2}+\sqrt{(\sqrt{2})^2-x^2}=\sqrt{2}\]
We solve for $x$ \[1-x^2+2-x^2+2\sqrt{(1-x^2)(2-x^2)}=2\] \[2\sqrt{(1-x^2)(2-x^2)}=2x^2-1\] \[4(1-x^2)(2-x^2)=(2x^2-1)^2\] \[8-12x^2+4x^4=4x^4-4x^2+1\] \[8x^2=7\] \[x=\frac{\sqrt{14}}{4}\]
By Ptolemy's Theorem \[AB \cdot CD + BC \cdot AD = AC \cdot BD = BD^2 = (2 \cdot BE)^2\]
Substituting values, \[1^2+1 \cdot AD = 4{\left( \frac{\sqrt{14}}{4} \right)}^2\] \[1+AD=\frac{7}{2}\] \[AD=\frac{5}{2}\]
Finally, we multiply back the $200$ that we divided by at the beginning of the problem to get $AD=\boxed{500}$ | E | 500 |
b7066cfe85c79dde1652e7df58145906 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_21 | A quadrilateral is inscribed in a circle of radius $200\sqrt{2}$ . Three of the sides of this quadrilateral have length $200$ . What is the length of the fourth side?
$\textbf{(A) }200\qquad \textbf{(B) }200\sqrt{2}\qquad\textbf{(C) }200\sqrt{3}\qquad\textbf{(D) }300\sqrt{2}\qquad\textbf{(E) } 500$ | [asy] size(250); defaultpen(linewidth(0.4)); //Variable Declarations real RADIUS; pair A, B, C, D, O; RADIUS=3; //Variable Definitions A=RADIUS*dir(148.414); B=RADIUS*dir(109.471); C=RADIUS*dir(70.529); D=RADIUS*dir(31.586); O=(0,0); //Path Definitions path quad= A -- B -- C -- D -- cycle; //Initial Diagram draw(Circle(O, RADIUS), linewidth(0.8)); draw(quad, linewidth(0.8)); label("$A$",A,W); label("$B$",B,NW); label("$C$",C,NE); label("$D$",D,E); label("$O$",O,S); //Radii draw(O--A); draw(O--B); draw(O--C); draw(O--D); [/asy]
Let quadrilateral $ABCD$ be inscribed in circle $O$ , where $AD$ is the side of unknown length. Draw the radii from center $O$ to all four vertices of the quadrilateral, and draw the altitude of $\triangle BOC$ such that it passes through side $AD$ at the point $G$ and meets side $BC$ at point $H$
By the Pythagorean Theorem, the length of $OH$ is \begin{align*} \sqrt{CO^2 - HC^2} &= \sqrt{(200\sqrt{2})^2 - \left(\frac{200}{2}\right)^2} \\ &= \sqrt{80000 - 10000} \\ &= \sqrt{70000} \\ &= 100\sqrt{7}. \end{align*}
Note that $[ABCDO] = [AOB] + [BOC] + [COD] = [AOD] + [ABCD].$ Let the length of $OG$ be $h$ and the length of $AD$ be $x$ ; then we have that
$[AOB] + [BOC] + [COD] = \frac{200 \times 100\sqrt{7}}{2} + \frac{200 \times 100\sqrt{7}}{2} + \frac{200 \times 100\sqrt{7}}{2} = \frac{x \times h}{2} + \frac{(100\sqrt{7} - h)(200 + x)}{2} = [AOD] + [ABCD].$
Furthermore, \begin{align*} h &= \sqrt{OD^2 - GD^2} \\ &= \sqrt{(200\sqrt{2})^2 - \left(\frac{x}{2}\right)^2} \\ &= \sqrt{80000 - \frac{x^2}{4}} \end{align*}
Substituting this value of $h$ into the previous equation and evaluating for $x$ , we get: \[\frac{200 \times 100\sqrt{7}}{2} + \frac{200 \times 100\sqrt{7}}{2} + \frac{200 \times 100\sqrt{7}}{2} = \frac{x \times h}{2} + \frac{(100\sqrt{7} - h)(200 + x)}{2}\] \[\frac{3 \times 200 \times 100\sqrt{7}}{2} = \frac{x\sqrt{80000 - \frac{x^2}{4}}}{2} + \frac{\left(100\sqrt{7} - \sqrt{80000 - \frac{x^2}{4}}\right)(200 + x)}{2}\] \[60000\sqrt{7} = \left(x\sqrt{80000 - \frac{x^2}{4}}\right) + \left(20000\sqrt{7}\right) + \left(100x\sqrt{7}\right) - \left(200\sqrt{80000 - \frac{x^2}{4}}\right) - \left(x\sqrt{80000 - \frac{x^2}{4}}\right)\] \[40000\sqrt{7} = 100x\sqrt{7} - 200\sqrt{80000 - \frac{x^2}{4}}\] \[400\sqrt{7} = x\sqrt{7} - 2\sqrt{80000 - \frac{x^2}{4}}\] \[(x - 400)\sqrt{7} = 2\sqrt{80000 - \frac{x^2}{4}}\] \[7(x-400)^2 = 4\left(80000 - \frac{x^2}{4}\right)\] \[7x^2 - 5600x + 1120000 = 320000 - x^2\] \[8x^2 - 5600x + 800000 = 0\] \[x^2 - 700x + 100000 = 0\]
The roots of this quadratic are found by using the quadratic formula: \begin{align*} x &= \frac{-(-700) \pm \sqrt{(-700)^2 - 4 \times 1 \times 100000}}{2 \times 1} \\ &= \frac{700 \pm \sqrt{490000 - 400000}}{2} \\ &= \frac{700}{2} \pm \frac{\sqrt{90000}}{2} \\ &= 350 \pm \frac{300}{2} \\ &= 200, 500 \end{align*}
If the length of $AD$ is $200$ , then $ABCD$ would be a square. Thus, the radius of the circle would be \[\frac{\sqrt{200^2 + 200^2}}{2} = \frac{\sqrt{80000}}{2} = \frac{200\sqrt{2}}{2} = 100\sqrt{2}\] Which is a contradiction. Therefore, our answer is $\boxed{500}.$ | null | 500 |
b7066cfe85c79dde1652e7df58145906 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_21 | A quadrilateral is inscribed in a circle of radius $200\sqrt{2}$ . Three of the sides of this quadrilateral have length $200$ . What is the length of the fourth side?
$\textbf{(A) }200\qquad \textbf{(B) }200\sqrt{2}\qquad\textbf{(C) }200\sqrt{3}\qquad\textbf{(D) }300\sqrt{2}\qquad\textbf{(E) } 500$ | [asy] size(250); defaultpen(linewidth(0.4)); //Variable Declarations real RADIUS; pair A, B, C, D, O; RADIUS=3; //Variable Definitions A=RADIUS*dir(148.414); B=RADIUS*dir(109.471); C=RADIUS*dir(70.529); D=RADIUS*dir(31.586); O=(0,0); //Path Definitions path quad= A -- B -- C -- D -- cycle; //Initial Diagram draw(Circle(O, RADIUS), linewidth(0.8)); draw(quad, linewidth(0.8)); label("$A$",A,W); label("$B$",B,NW); label("$C$",C,NE); label("$D$",D,E); label("$O$",O,S); label("$\theta$",O,3N); //Radii draw(O--A); draw(O--B); draw(O--C); draw(O--D); //Angle mark for BOC draw(anglemark(C,O,B)); [/asy]
Construct quadrilateral $ABCD$ on the circle with $AD$ being the missing side (Notice that since the side length is less than the radius, it will be very small on the top of the circle). Now, draw the radii from center $O$ to $A,B,C,$ and $D$ . Apply the law of cosines on $\Delta BOC$ ; let $\theta = \angle BOC$ . We get the following equation: \[(BC)^{2}=(OB)^{2}+(OC)^{2}-2\cdot OB \cdot OC\cdot \cos\theta\] Substituting the values in, we get \[(200)^{2}=2\cdot (200)^{2}+ 2\cdot (200)^{2}- 2\cdot 2\cdot (200)^{2}\cdot \cos\theta\] Canceling out, we get \[\cos\theta=\frac{3}{4}\] Because $\angle AOB$ $\angle BOC$ , and $\angle COD$ are congruent, $\angle AOD = 3\theta$ . To find the remaining side ( $AD$ ), we simply have to apply the law of cosines to $\Delta AOD$ . Now, to find $\cos 3\theta$ , we can derive a formula that only uses $\cos\theta$ \[\cos 3\theta=\cos (2\theta+\theta)= \cos 2\theta \cos\theta- (2\sin\theta \cos\theta) \cdot \sin \theta\] \[\cos 3\theta= \cos\theta (\cos 2\theta-2\sin^{2}\theta)=\cos\theta (2\cos^{2}\theta-3+2\cos^{2}\theta)\] \[\Rightarrow \cos 3\theta=4\cos^{3}\theta-3\cos\theta\] It is useful to memorize the triple angle formulas ( $\cos 3\theta=4\cos^{3}\theta-3\cos\theta, \sin 3\theta=3\sin\theta-4\sin^{3}\theta$ ). Plugging in $\cos\theta=\frac{3}{4}$ , we get $\cos 3\theta= -\frac{9}{16}$ . Now, applying law of cosines on triangle $OAD$ , we get \[(AD)^{2}= 2\cdot (200)^{2}+ 2\cdot (200)^{2}+2\cdot 200\sqrt2 \cdot 200\sqrt2 \cdot \frac{9}{16}\] \[\Rightarrow 2\cdot (200)^{2} \cdot (1+1+ \frac{9}{8})=(200)^{2}\cdot \frac{25}{4}\] \[AD=200 \cdot \frac{5}{2}=\boxed{500}\] | null | 500 |
b7066cfe85c79dde1652e7df58145906 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_21 | A quadrilateral is inscribed in a circle of radius $200\sqrt{2}$ . Three of the sides of this quadrilateral have length $200$ . What is the length of the fourth side?
$\textbf{(A) }200\qquad \textbf{(B) }200\sqrt{2}\qquad\textbf{(C) }200\sqrt{3}\qquad\textbf{(D) }300\sqrt{2}\qquad\textbf{(E) } 500$ | [asy] size(250); defaultpen(linewidth(0.4)); //Variable Declarations real RADIUS; pair A, B, C, D, E, F, O; RADIUS=3; //Variable Definitions A=RADIUS*dir(148.414); B=RADIUS*dir(109.471); C=RADIUS*dir(70.529); D=RADIUS*dir(31.586); E=foot(A,B,C); F=foot(D,B,C); O=(0,0); //Path Definitions path quad= A -- B -- C -- D -- cycle; //Initial Diagram draw(Circle(O, RADIUS), linewidth(0.8)); draw(quad, linewidth(0.8)); label("$A$",A,W); label("$B$",B,NW); label("$C$",C,NE); label("$D$",D,ENE); label("$O$",O,S); label("$\theta$",O,3N); //Radii draw(O--A); draw(O--B); draw(O--C); draw(O--D); //Construction draw(A--E); draw(E--B); draw(C--F); draw(F--D); label("$E$",E,NW); label("$F$",F,NE); //Angle marks draw(anglemark(C,O,B)); draw(rightanglemark(A,E,B)); draw(rightanglemark(C,F,D)); [/asy]
Construct quadrilateral $ABCD$ on the circle $O$ with $AD$ being the desired side. Then, drop perpendiculars from $A$ and $D$ to the extended line of $\overline{BC}$ and let these points be $E$ and $F$ , respectively. Also, let $\theta = \angle BOC$ . From the Law of Cosines on $\triangle BOC$ , we have $\cos \theta = \frac{3}{4}$
Now, since $\triangle BOC$ is isosceles with $\overline{OB} \cong \overline{OC}$ , we have that $\angle BCO = \angle CBO = 90 - \frac{\theta}{2}$ . In addition, we know that $\overline{BC} \cong \overline{CD}$ as they are both equal to $200$ and $\overline{OB} \cong \overline{OC} \cong \overline{OD}$ as they are both radii of the same circle. By SSS Congruence, we have that $\triangle OBC \cong \triangle OCD$ , so we have that $\angle OCD = \angle BCO = 90 - \frac{\theta}{2}$ , so $\angle DCF = \theta$
Thus, we have $\frac{FC}{DC} = \cos \theta = \frac{3}{4}$ , so $FC = 150$ . Similarly, $BE = 150$ , and $AD = 150 + 200 + 150 = \boxed{500}$ | null | 500 |
b7066cfe85c79dde1652e7df58145906 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_21 | A quadrilateral is inscribed in a circle of radius $200\sqrt{2}$ . Three of the sides of this quadrilateral have length $200$ . What is the length of the fourth side?
$\textbf{(A) }200\qquad \textbf{(B) }200\sqrt{2}\qquad\textbf{(C) }200\sqrt{3}\qquad\textbf{(D) }300\sqrt{2}\qquad\textbf{(E) } 500$ | [asy] pathpen = black; pointpen = black; size(6cm); draw(unitcircle); pair A = D("A", dir(50), dir(50)); pair B = D("B", dir(90), dir(90)); pair C = D("C", dir(130), dir(130)); pair D = D("D", dir(170), dir(170)); pair O = D("O", (0,0), dir(-90)); draw(A--C, red); draw(B--D, blue+dashed); draw(A--B--C--D--cycle); draw(A--O--C); draw(O--B); [/asy]
Let $s = 200$ . Let $O$ be the center of the circle. Then $AC$ is twice the altitude of $\triangle OBC$ to $\overline{OB}$ . Since $\triangle OBC$ is isosceles we can compute its area to be $\frac{s^2 \sqrt{7}}{4}$ , hence $CA = 2 \cdot \tfrac{2 \cdot s^2\sqrt7/4}{s\sqrt2} = s\sqrt{\frac{7}{2}}$
Now by Ptolemy's Theorem we have $CA^2 = s^2 + AD \cdot s \implies AD = \left(\frac{7}{2}-1\right)s.$ This gives us: \[\boxed{500.}\] | E | 500. |
b7066cfe85c79dde1652e7df58145906 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_21 | A quadrilateral is inscribed in a circle of radius $200\sqrt{2}$ . Three of the sides of this quadrilateral have length $200$ . What is the length of the fourth side?
$\textbf{(A) }200\qquad \textbf{(B) }200\sqrt{2}\qquad\textbf{(C) }200\sqrt{3}\qquad\textbf{(D) }300\sqrt{2}\qquad\textbf{(E) } 500$ | Since all three sides equal $200$ , they subtend three equal angles from the center. The right triangle between the center of the circle, a vertex, and the midpoint between two vertices has side lengths $100,100\sqrt{7},200\sqrt{2}$ by the Pythagorean Theorem. Thus, the sine of half of the subtended angle is $\frac{100}{200\sqrt{2}}=\frac{\sqrt{2}}{4}$ . Similarly, the cosine is $\frac{100\sqrt{7}}{200\sqrt{2}}=\frac{\sqrt{14}}{4}$ .
Since there are three sides, and since $\sin\theta=\sin\left(180-\theta\right)$ ,we seek to find $2r\sin 3\theta$ .
First, $\sin 2\theta=2\sin\theta\cos\theta=2\cdot\left(\frac{\sqrt{2}}{4}\right)\left(\frac{\sqrt{14}}{4}\right)=\frac{2\sqrt{2}\sqrt{14}}{16}=\frac{\sqrt{7}}{4}$ and $\cos 2\theta=\frac{3}{4}$ by Pythagorean. \[\sin 3\theta=\sin(2\theta+\theta)=\sin 2\theta\cos\theta+\sin \theta\cos 2\theta=\frac{\sqrt{7}}{4}\left(\frac{\sqrt{14}}{4}\right)+\frac{\sqrt{2}}{4}\left(\frac{3}{4}\right)=\frac{7\sqrt{2}+3\sqrt{2}}{16}=\frac{5\sqrt{2}}{8}\] \[2r\sin 3\theta=2\left(200\sqrt{2}\right)\left(\frac{5\sqrt{2}}{8}\right)=400\sqrt{2}\left(\frac{5\sqrt{2}}{8}\right)=\frac{800\cdot 5}{8}=\boxed{500}\] | E | 500 |
b7066cfe85c79dde1652e7df58145906 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_21 | A quadrilateral is inscribed in a circle of radius $200\sqrt{2}$ . Three of the sides of this quadrilateral have length $200$ . What is the length of the fourth side?
$\textbf{(A) }200\qquad \textbf{(B) }200\sqrt{2}\qquad\textbf{(C) }200\sqrt{3}\qquad\textbf{(D) }300\sqrt{2}\qquad\textbf{(E) } 500$ | For simplicity, scale everything down by a factor of 100. Let the inscribed trapezoid be $ABCD$ , where $AB=BC=CD=2$ and $DA$ is the missing side length. Let $DA=2x$ . If $M$ and $N$ are the midpoints of $BC$ and $AD$ , respectively, the height of the trapezoid is $OM-ON$ . By the pythagorean theorem, $OM=\sqrt{OB^2-BM^2}=\sqrt7$ and $ON=\sqrt{OA^2-AN^2}=\sqrt{8-x^2}$ . Thus the height of the trapezoid is $\sqrt7-\sqrt{8-x^2}$ , so the area is $\frac{(2+2x)(\sqrt7-\sqrt{8-x^2})}{2}=(x+1)(\sqrt7-\sqrt{8-x^2})$ . By Brahmagupta's formula , the area is $\sqrt{(x+1)(x+1)(x+1)(3-x)}$ . Setting these two equal, we get $(x+1)(\sqrt7-\sqrt{8-x^2})=\sqrt{(x+1)(x+1)(x+1)(3-x)}$ . Dividing both sides by $x+1$ and then squaring, we get $7-2(\sqrt7)(\sqrt{8-x^2})+8-x^2=(x+1)(3-x)$ . Expanding the right hand side and canceling the $x^2$ terms gives us $15-2(\sqrt7)(\sqrt{8-x^2})=2x+3$ . Rearranging and dividing by two, we get $(\sqrt7)(\sqrt{8-x^2})=6-x$ . Squaring both sides, we get $56-7x^2=x^2-12x+36$ . Rearranging, we get $8x^2-12x-20=0$ . Dividing by 4 we get $2x^2-3x-5=0$ . Factoring we get, $(2x-5)(x+1)=0$ , and since $x$ cannot be negative, we get $x=2.5$ . Since $DA=2x$ $DA=5$ . Scaling up by 100, we get $\boxed{500}$ | E | 500 |
b7066cfe85c79dde1652e7df58145906 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_21 | A quadrilateral is inscribed in a circle of radius $200\sqrt{2}$ . Three of the sides of this quadrilateral have length $200$ . What is the length of the fourth side?
$\textbf{(A) }200\qquad \textbf{(B) }200\sqrt{2}\qquad\textbf{(C) }200\sqrt{3}\qquad\textbf{(D) }300\sqrt{2}\qquad\textbf{(E) } 500$ | [asy] size(250); defaultpen(linewidth(0.4)); //Variable Declarations, L is used to write alpha= statement real RADIUS; pair A, B, C, D, E, F, O, L; RADIUS=3; //Variable Definitions A=RADIUS*dir(148.414); B=RADIUS*dir(109.471); C=RADIUS*dir(70.529); D=RADIUS*dir(31.586); E=extension(A,D,O,B); F=extension(A,D,O,C); L=midpoint(C--D); O=(0,0); //Path Definitions path quad = A -- B -- C -- D -- cycle; //Initial Diagram draw(circle(O, RADIUS), linewidth(0.8)); draw(quad, linewidth(0.8)); label("$A$",A,NW); label("$B$",B,NW); label("$C$",C,NE); label("$D$",D,NE); label("$E$",E,SW); label("$F$",F,SE); label("$O$",O,SE); dot(O,linewidth(5)); //Radii draw(O--A); draw(O--B); draw(O--C); draw(O--D); //Construction label("$\alpha = 90-\frac{\theta}{2}$",L,5NE,rgb(128, 0, 0)); draw(anglemark(C,O,B)); label("$\theta$",O,3N); draw(anglemark(E,F,O)); label("$\alpha$",F,3SW); draw(anglemark(D,F,C)); label("$\alpha$",F,3NE); draw(anglemark(F,C,D)); label("$\alpha$",C,3SSE); draw(anglemark(C,D,F)); label("$\theta$",(RADIUS-0.04)*dir(31.586),3WNW); [/asy] Label the points as shown, and let $\angle{EOF} = \theta$ . Since $\overline{OB} = \overline{OC}$ , and $\triangle{OFE} \sim \triangle{OCB}$ , we get that $\angle{EFO} = 90-\frac{\theta}{2}$ . We assign $\alpha$ to $90-\frac{\theta}{2}$ for simplicity.
From here, by vertical angles $\angle{CFD} = \alpha$ . Also, since $\triangle{OCB} \cong \triangle{ODC}$ $\angle{OCD} = \alpha$ . This means that $\angle{CDF} = 180-2\alpha = \theta$ , which leads to $\triangle{OCB} \sim \triangle{DCF}$ .
Since we know that $\overline{CD} = 200$ $\overline{DF} = 200$ , and by similar reasoning $\overline{AE} = 200$ .
Finally, again using similar triangles, we get that $\overline{CF} = 100\sqrt{2}$ , which means that $\overline{OF} = \overline{OC} - \overline{CF} = 200\sqrt{2} - 100\sqrt{2} = 100\sqrt{2}$ . We can again apply similar triangles (or use Power of a Point) to get $\overline{EF} = 100$ , and finally $\overline{AD} = \overline{AE}+\overline{EF}+\overline{FD} = 200+100+200=\boxed{500}$ - ColtsFan10 | E | 500 |
b7066cfe85c79dde1652e7df58145906 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_21 | A quadrilateral is inscribed in a circle of radius $200\sqrt{2}$ . Three of the sides of this quadrilateral have length $200$ . What is the length of the fourth side?
$\textbf{(A) }200\qquad \textbf{(B) }200\sqrt{2}\qquad\textbf{(C) }200\sqrt{3}\qquad\textbf{(D) }300\sqrt{2}\qquad\textbf{(E) } 500$ | We first scale down by a factor of $200\sqrt{2}$ . Let the vertices of the quadrilateral be $A$ $B$ $C$ , and $D$ , so that $AD$ is the length of the fourth side. We draw this in the complex plane so that $D$ corresponds to the complex number $1$ , and we let $C$ correspond to the complex number $z$ . Then, $A$ corresponds to $z^3$ and $B$ corresponds to $z^2$ . We are given that $\lvert z \rvert = 1$ and $\lvert z-1 \rvert = 1/\sqrt{2}$ , and we wish to find $\lvert z^3 - 1 \rvert=\lvert z^2+z+1\rvert \cdot \lvert z-1 \rvert=\lvert (z^2+z+1)/\sqrt{2} \rvert$ . Let $z=a+bi$ , where $a$ and $b$ are real numbers. Then, $a^2+b^2=1$ and $a^2-2a+1+b^2=1/2$ ; solving for $a$ and $b$ yields $a=3/4$ and $b=\sqrt{7}/4$ . Thus, $AD = \lvert z^3 - 1 \rvert = \lvert (z^2+z+1)/\sqrt{2} \rvert = \lvert (15/8 + 5\sqrt{7}/8 \cdot i)/\sqrt{2} \rvert = \frac{5\sqrt{2}}{4}$ . Scaling back up gives us a final answer of $\frac{5\sqrt{2}}{4} \cdot 200\sqrt{2} = \boxed{500}$ | E | 500 |
b7066cfe85c79dde1652e7df58145906 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_21 | A quadrilateral is inscribed in a circle of radius $200\sqrt{2}$ . Three of the sides of this quadrilateral have length $200$ . What is the length of the fourth side?
$\textbf{(A) }200\qquad \textbf{(B) }200\sqrt{2}\qquad\textbf{(C) }200\sqrt{3}\qquad\textbf{(D) }300\sqrt{2}\qquad\textbf{(E) } 500$ | Let angle $C$ be $2a$ . This way $BD$ will be $400sin(a)$ . Now we can trig bash. As the circumradius of triangle $BCD$ is $200\sqrt{2}$ , we can use the formula \[R=\frac{abc}{4A}\] and \[A=\frac{absin(C)}{2}\] and plug in all the values we got to get \[200\sqrt{2}=\frac{200^2 \cdot 400sin(a)}{4 \cdot (\frac{200^2 sin(2a)}{2})}\] . This boils down to \[\sqrt{2}=\frac{sin(a)}{sin{2a}}\] . This expression can further be simplified by the trig identity \[sin(2a)=2sin(a)cos(a)\] . This leads to the final simplified form \[2\sqrt{2}=\frac{1}{cos(a)}\] . Solving this expression gives us \[cos(a)=\frac{\sqrt{2}}{4}\] . However, as we want $sin(a)$ , we use the identity $sin^2+cos^2=1$ , and substitute to get that $sin(a)=\frac{\sqrt{14}}{4}$ , and therefore BD is $100\sqrt{14}$
Then, as $ABCD$ is a cyclic quadrilateral, we can use Ptolemy’s Theorem (with $AD=x$ ) to get \[14 \cdot 100^2=200x+200^2\] . Finally, we solve to get $\boxed{500}$ | E | 500 |
b7066cfe85c79dde1652e7df58145906 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_21 | A quadrilateral is inscribed in a circle of radius $200\sqrt{2}$ . Three of the sides of this quadrilateral have length $200$ . What is the length of the fourth side?
$\textbf{(A) }200\qquad \textbf{(B) }200\sqrt{2}\qquad\textbf{(C) }200\sqrt{3}\qquad\textbf{(D) }300\sqrt{2}\qquad\textbf{(E) } 500$ | [asy] size(250); defaultpen(linewidth(0.4)); //Variable Declarations real RADIUS; pair A, B, C, D, O; RADIUS=3; //Variable Definitions A=RADIUS*dir(148.414); B=RADIUS*dir(109.471); C=RADIUS*dir(70.529); D=RADIUS*dir(31.586); O=(0,0); //Path Definitions path quad= A -- B -- C -- D -- cycle; //Initial Diagram draw(Circle(O, RADIUS), linewidth(0.8)); draw(quad, linewidth(0.8)); label("$A$",A,W); label("$B$",B,NW); label("$C$",C,NE); label("$D$",D,E); label("$O$",O,S); //Radii draw(O--A); draw(O--B); draw(O--C); draw(O--D); [/asy] Claim: $[ABCD]$ is an isosceles trapezoid.
Proof: Notice that $[ABCD]$ is cyclic, triangle $BOC$ is isosceles, and triangle $AOB$ is congruent to $DOC$ by SSS congruence. Therefore, $\angle BAD = 180 - \angle BCD = 180-(\angle BCO + \angle DCO)=180-(\angle CBO+\angle ABO) = 180 - \angle ABC = \angle CDA$ . Hence, $[ABCD]$ is an isosceles trapezoid.
Let $\angle CDA=\alpha$ . Notice that the length of the altitude from $C$ to $AD$ is $200sin(\alpha)$ . Furthermore, the length of the altitude from $O$ to $BC$ is $100\sqrt{7}$ by the Pythagorean theorem. Therefore, the length of the altitude from $O$ to $AD$ is $100\sqrt{7}-200sin\alpha$ . Let $F$ the feet of the altitude from $O$ to $AD$ . Then, $FD=(200+400cos(\alpha))/2=100+200cos(\alpha)$ , because $AOD$ is isosceles.
Therefore, by the Pythagorean theorem, $(100+200cos(\alpha))^2+(100\sqrt{7}-200sin(\alpha))^2=80000$ . Simplifying, we have $1+cos(\alpha)=sin(\alpha) \cdot sqrt{7} \implies cos^2(\alpha)+2cos(\alpha)+1=sin^2(\alpha) \cdot 7 = 7-7cos^2(\alpha) \implies 8cos^2(\alpha)+2cos(\alpha) - 6 =0$ . Solving this quadratic, we have $cos(\alpha)=\frac{3}{4}, -1$ , but $0<\alpha<180 \implies cos(\alpha)=3/4$ . Therefore, $AD=200cos(\alpha)+200cos(\alpha)+200=\boxed{500}$ | null | 500 |
a96e5075aa015913eae871d1e7af66a9 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_22 | How many ordered triples $(x,y,z)$ of positive integers satisfy $\text{lcm}(x,y) = 72, \text{lcm}(x,z) = 600 \text{ and lcm}(y,z)=900$
$\textbf{(A)}\ 15\qquad\textbf{(B)}\ 16\qquad\textbf{(C)}\ 24\qquad\textbf{(D)}\ 27\qquad\textbf{(E)}\ 64$ | We prime factorize $72,600,$ and $900$ . The prime factorizations are $2^3\times 3^2$ $2^3\times 3\times 5^2$ and $2^2\times 3^2\times 5^2$ , respectively. Let $x=2^a\times 3^b\times 5^c$ $y=2^d\times 3^e\times 5^f$ and $z=2^g\times 3^h\times 5^i$ . We know that \[\max(a,d)=3\] \[\max(b,e)=2\] \[\max(a,g)=3\] \[\max(b,h)=1\] \[\max(c,i)=2\] \[\max(d,g)=2\] \[\max(e,h)=2\] and $c=f=0$ since $\text{lcm}(x,y)$ isn't a multiple of 5. Since $\max(d,g)=2$ we know that $a=3$ . We also know that since $\max(b,h)=1$ that $e=2$ . So now some equations have become useless to us...let's take them out. \[\max(b,h)=1\] \[\max(d,g)=2\] are the only two important ones left. We do casework on each now. If $\max(b,h)=1$ then $(b,h)=(1,0),(0,1)$ or $(1,1)$ . Similarly if $\max(d,g)=2$ then $(d,g)=(2,0),(2,1),(2,2),(1,2),(0,2)$ . Thus our answer is $5\times 3=\boxed{15}.$ | A | 15 |
a96e5075aa015913eae871d1e7af66a9 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_22 | How many ordered triples $(x,y,z)$ of positive integers satisfy $\text{lcm}(x,y) = 72, \text{lcm}(x,z) = 600 \text{ and lcm}(y,z)=900$
$\textbf{(A)}\ 15\qquad\textbf{(B)}\ 16\qquad\textbf{(C)}\ 24\qquad\textbf{(D)}\ 27\qquad\textbf{(E)}\ 64$ | It is well known that if the $\text{lcm}(a,b)=c$ and $c$ can be written as $p_1^ap_2^bp_3^c\dots$ , then the highest power of all prime numbers $p_1,p_2,p_3\dots$ must divide into either $a$ and/or $b$ . Or else a lower $c_0=p_1^{a-\epsilon}p_2^{b-\epsilon}p_3^{c-\epsilon}\dots$ is the $\text{lcm}$
Start from $x$ $\text{lcm}(x,y)=72$ so $8\mid x$ or $9\mid x$ or both. But $9\nmid x$ because $\text{lcm}(x,z)=600$ and $9\nmid 600$ .
So $x=8,24$
$y$ can be $9,18,36$ in both cases of $x$ but NOT $72$ because $\text{lcm}(y,z)=900$ and $72\nmid 900$
So there are six sets of $x,y$ and we will list all possible values of $z$ based on those.
$25\mid z$ because $z$ must source all powers of $5$ $z\in\{25,50,75,100,150,300\}$ $z\ne\{200,225\}$ because of $\text{lcm}$ restrictions.
By different sourcing of powers of $2$ and $3$
\[(8,9):z=300\] \[(8,18):z=300\] \[(8,36):z=75,150,300\] \[(24,9):z=100,300\] \[(24,18):z=100,300\] \[(24,36):z=25,50,75,100,150,300\]
$z=100$ is "enabled" by $x$ sourcing the power of $3$ $z=75,150$ is uncovered by $y$ sourcing all powers of $2$ . And $z=25,50$ is uncovered by $x$ and $y$ both at full power capacity.
Counting the cases, $1+1+3+2+2+6=\boxed{15}.$ | A | 15 |
a96e5075aa015913eae871d1e7af66a9 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_22 | How many ordered triples $(x,y,z)$ of positive integers satisfy $\text{lcm}(x,y) = 72, \text{lcm}(x,z) = 600 \text{ and lcm}(y,z)=900$
$\textbf{(A)}\ 15\qquad\textbf{(B)}\ 16\qquad\textbf{(C)}\ 24\qquad\textbf{(D)}\ 27\qquad\textbf{(E)}\ 64$ | As said in previous solutions, start by factoring $72, 600,$ and $900$ . The prime factorizations are as follows: \[72=2^3\cdot 3^2,\] \[600=2^3\cdot 3\cdot 5^2,\] \[\text{and } 900=2^2\cdot 3^2\cdot 5^2\] To organize $x,y, \text{ and } z$ and their respective LCMs in a simpler way, we can draw a triangle as follows such that $x,y, \text{and } z$ are the vertices and the LCMs are on the edges.
[asy] //Variable Declarations defaultpen(0.45); size(200pt); fontsize(15pt); pair X, Y, Z; real R; path tri; //Variable Definitions R = 1; X = R*dir(90); Y = R*dir(210); Z = R*dir(-30); tri = X--Y--Z--cycle; //Diagram draw(tri); label("$x$",X,N); label("$y$",Y,SW); label("$z$",Z,SE); label("$2^33^25^0$",X--Y,2W); label("$2^33^15^2$",X--Z,2E); label("$2^23^25^2$",Y--Z,2S); [/asy]
Now we can split this triangle into three separate ones for each of the three different prime factors $2,3, \text{and } 5$
[asy] //Variable Declarations defaultpen(0.45); size(200pt); fontsize(15pt); pair X, Y, Z; real R; path tri; //Variable Definitions R = 1; X = R*dir(90); Y = R*dir(210); Z = R*dir(-30); tri = X--Y--Z--cycle; //Diagram draw(tri); label("$x$",X,N); label("$y$",Y,SW); label("$z$",Z,SE); label("$2^3$",X--Y,2W); label("$2^3$",X--Z,2E); label("$2^2$",Y--Z,2S); [/asy]
Analyzing for powers of $2$ , it is quite obvious that $x$ must have $2^3$ as one of its factors since neither $y \text{ nor } z$ can have a power of $2$ exceeding $2$ . Turning towards the vertices $y$ and $z$ , we know at least one of them must have $2^2$ as its factors. Therefore, we have $5$ ways for the powers of $2$ for $y \text{ and } z$ since the only ones that satisfy the previous conditions are for ordered pairs $(y,z) \{(2,0)(2,1)(0,2)(1,2)(2,2)\}$
[asy] //Variable Declarations defaultpen(0.45); size(200pt); fontsize(15pt); pair X, Y, Z; real R; path tri; //Variable Definitions R = 1; X = R*dir(90); Y = R*dir(210); Z = R*dir(-30); tri = X--Y--Z--cycle; //Diagram draw(tri); label("$x$",X,N); label("$y$",Y,SW); label("$z$",Z,SE); label("$3^2$",X--Y,2W); label("$3^1$",X--Z,2E); label("$3^2$",Y--Z,2S); [/asy]
Using the same logic as we did for powers of $2$ , it becomes quite easy to note that $y$ must have $3^2$ as one of its factors. Moving onto $x \text{ and } z$ , we can use the same logic to find the only ordered pairs $(x,z)$ that will work are $\{(1,0)(0,1)(1,1)\}$
The final and last case is the powers of $5$
[asy] //Variable Declarations defaultpen(0.45); size(200pt); fontsize(15pt); pair X, Y, Z; real R; path tri; //Variable Definitions R = 1; X = R*dir(90); Y = R*dir(210); Z = R*dir(-30); tri = X--Y--Z--cycle; //Diagram draw(tri); label("$x$",X,N); label("$y$",Y,SW); label("$z$",Z,SE); label("$5^0$",X--Y,2W); label("$5^2$",X--Z,2E); label("$5^2$",Y--Z,2S); [/asy]
This is actually quite a simple case since we know $z$ must have $5^2$ as part of its factorization while $x \text{ and } y$ cannot have a factor of $5$ in their prime factorization.
Multiplying all the possible arrangements for prime factors $2,3, \text{ and } 5$ , we get the answer: \[5\cdot3\cdot1=\boxed{15}\] | A | 15 |
06b49b706d32fe0b7d75989de1d68365 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_23 | Three numbers in the interval $\left[0,1\right]$ are chosen independently and at random. What is the probability that the chosen numbers are the side lengths of a triangle with positive area?
$\textbf{(A) }\frac16\qquad\textbf{(B) }\frac13\qquad\textbf{(C) }\frac12\qquad\textbf{(D) }\frac23\qquad\textbf{(E) }\frac56$ | Because we can let the the sides of the triangle be any variable we want, to make it easier for us when solving, let’s let the side lengths be $x,y,$ and $a$ . WLOG assume $a$ is the largest. Then, $x+y>a$ , meaning the solution is $\boxed{12}$ , as shown in the graph below. [asy] pair A = (0,0); pair B = (1,0); pair C = (1,1); pair D = (0,1); pair E = (0,0); draw(A--B--C--D--cycle); draw(B--D,dashed); fill(B--D--C--cycle,gray); label("$0$",A,SW); label("$a$",B,S); label("$a$",D,W); label("$y$",(0,.5),W); label("$x$",(.5,0),S); label("$x+y>a$",(5/7,5/7)); [/asy] | C | 12 |
06b49b706d32fe0b7d75989de1d68365 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_23 | Three numbers in the interval $\left[0,1\right]$ are chosen independently and at random. What is the probability that the chosen numbers are the side lengths of a triangle with positive area?
$\textbf{(A) }\frac16\qquad\textbf{(B) }\frac13\qquad\textbf{(C) }\frac12\qquad\textbf{(D) }\frac23\qquad\textbf{(E) }\frac56$ | WLOG, let the largest of the three numbers drawn be $a>0$ . Then the other two numbers are drawn uniformly and independently from the interval $[0,a]$ . The probability that their sum is greater than $a$ is $\boxed{12}$ | C | 12 |
06b49b706d32fe0b7d75989de1d68365 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_23 | Three numbers in the interval $\left[0,1\right]$ are chosen independently and at random. What is the probability that the chosen numbers are the side lengths of a triangle with positive area?
$\textbf{(A) }\frac16\qquad\textbf{(B) }\frac13\qquad\textbf{(C) }\frac12\qquad\textbf{(D) }\frac23\qquad\textbf{(E) }\frac56$ | This problem is going to require some geometric probability, so let's dive right in.
Take three integers $x$ $y$ , and $z$ . Then for the triangle inequality to hold, the following $3$ inequalities must be true
\[x + y > z\] \[y + z > x\] \[x + z > y\]
Now, it would be really easy if these equations only had two variables instead of $3$ , because then we could graph it in a $2$ -dimensional plane instead of a $3$ -dimensional cube. So, we assume $z$ is a constant. We will deal with it later.
Now, since we are graphing, we should probably write these equations in terms of $y$ so they are in slope-intercept form and are easier to graph.
\[y > -x + z\] \[y > x - z\] \[y < x + z\]
Now, note that all solutions $(x,y)$ are in a $1$ $x$ $1$ square in the $xy$ -plane because $x$ $y \in [0, 1]$
I recommend drawing the following figure to get an idea of what is going on.
The first line is a line with a negative slope that cuts off a $45-45-90$ triangle with side length $z$ of the bottom left corner of the square. The second line is a line with a positive slope that cuts off a $45-45-90$ triangle with side length $1-z$ off the top left corner of the square. The third line also has a positive slope and cuts a $45-45-90$ triangle with side length $1 - z$ off the bottom right corner of the square.
Note: All triangles are $45-45-90$ because the lines have slopes of $1$ or $-1$
Using the $>$ and $<$ signs in the lines, we see that the area that satisfies all three inequalities is the area not enclosed in the three triangles. So, our plan of attack will be to
Find the area of the triangles -> Subtract that from the area of the square -> Use probability to get the answer.
Except, now, we have one problem. $z$ is still a variable. But, we want $z$ to be a constant. Well, what if we just took the area over every possible value of $z$ ? Well, that would be a bit hard, if not impossible to do by hand, but there is a handy math tool that will let us do that: the integral!
To find the area of the triangles, our plan of attack will be
Find the area in terms of $z$ -> take the integral from $0$ to $1$ of the expression for the area (this will cover every possible value of $z$
The area of the triangles is
\[\frac{z^2}{2} + (1-z)^2 = \frac{1}{2}\left(-3x^2 + 4x\right)\]
The integral from $0$ to $1$ is
\[\frac{1}{2}\int_0^1\left(-3x^2 + 4x\right)dx = \frac{1}{2}\]
The total area of all the possible unit squares is quite obviously
\[\int_0^11dx = 1\]
Thus, the area not enclosed by the triangles is $1 - \dfrac{1}{2} = \dfrac{1}{2}$ , and the total area of the square is $1$ . Thus, the desired probability is
\[\frac{\frac{1}{2}}{1} = \boxed{12}\] | C | 12 |
06b49b706d32fe0b7d75989de1d68365 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_23 | Three numbers in the interval $\left[0,1\right]$ are chosen independently and at random. What is the probability that the chosen numbers are the side lengths of a triangle with positive area?
$\textbf{(A) }\frac16\qquad\textbf{(B) }\frac13\qquad\textbf{(C) }\frac12\qquad\textbf{(D) }\frac23\qquad\textbf{(E) }\frac56$ | Consider a stick of length $1$ . Cutting the stick at two random points gives a triangle from the three new segments. These two random points must be on opposite sides of the halfway mark. Thus, after the first cut is made, there is $\boxed{12}$ probability that the second cut is on the opposite side. | C | 12 |
9bde57e2cf1efee03997fd42c145ba88 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_24 | There is a smallest positive real number $a$ such that there exists a positive real number $b$ such that all the roots of the polynomial $x^3-ax^2+bx-a$ are real. In fact, for this value of $a$ the value of $b$ is unique. What is this value of $b$
$\textbf{(A)}\ 8\qquad\textbf{(B)}\ 9\qquad\textbf{(C)}\ 10\qquad\textbf{(D)}\ 11\qquad\textbf{(E)}\ 12$ | The acceleration must be zero at the $x$ -intercept; this intercept must be an inflection point for the minimum $a$ value.
Derive $f(x)$ so that the acceleration $f''(x)=0$ . Using the power rule, \begin{align*} f(x) &= x^3-ax^2+bx-a \\ f’(x) &= 3x^2-2ax+b \\ f’’(x) &= 6x-2a \end{align*} So $x=\frac{a}{3}$ for the inflection point/root. Furthermore, the slope of the function must be zero - maximum - at the intercept, thus having a triple root at $x=a/3$ (if the slope is greater than zero, there will be two complex roots and we do not want that).
The function with the minimum $a$
\[f(x)=\left(x-\frac{a}{3}\right)^3\] \[x^3-ax^2+\left(\frac{a^2}{3}\right)x-\frac{a^3}{27}\] Since this is equal to the original equation $x^3-ax^2+bx-a$ , equating the coefficients, we get that
\[\frac{a^3}{27}=a\rightarrow a^2=27\rightarrow a=3\sqrt{3}\] \[b=\frac{a^2}{3}=\frac{27}{3}=\boxed{9}\] | B | 9 |
9bde57e2cf1efee03997fd42c145ba88 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_24 | There is a smallest positive real number $a$ such that there exists a positive real number $b$ such that all the roots of the polynomial $x^3-ax^2+bx-a$ are real. In fact, for this value of $a$ the value of $b$ is unique. What is this value of $b$
$\textbf{(A)}\ 8\qquad\textbf{(B)}\ 9\qquad\textbf{(C)}\ 10\qquad\textbf{(D)}\ 11\qquad\textbf{(E)}\ 12$ | Let the roots of the polynomial be $r, s, t$ . By Vieta's formulas we have $r+s+t=a$ $rs+st+rt=b$ , and $rst=a$ . Since both $a$ and $b$ are positive, it follows that all 3 roots $r, s, t$ are positive as well, and so we can apply AM-GM to get \[\tfrac 13 (r+s+t) \ge \sqrt[3]{rst} \quad \Rightarrow \quad a \ge 3\sqrt[3]{a}.\] Cubing both sides and then dividing by $a$ (since $a$ is positive we can divide by $a$ and not change the sign of the inequality) yields \[a^2 \ge 27 \quad \Rightarrow \quad a \ge 3\sqrt{3}.\] Thus, the smallest possible value of $a$ is $3\sqrt{3}$ which is achieved when all the roots are equal to $\sqrt{3}$ . For this value of $a$ , we can use Vieta's to get $b=\boxed{9}$ | B | 9 |
9bde57e2cf1efee03997fd42c145ba88 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_24 | There is a smallest positive real number $a$ such that there exists a positive real number $b$ such that all the roots of the polynomial $x^3-ax^2+bx-a$ are real. In fact, for this value of $a$ the value of $b$ is unique. What is this value of $b$
$\textbf{(A)}\ 8\qquad\textbf{(B)}\ 9\qquad\textbf{(C)}\ 10\qquad\textbf{(D)}\ 11\qquad\textbf{(E)}\ 12$ | We see that with cubics, the number $3$ comes up a lot, and as $9=3\cdot3$ has the most relation to $3$ , we can assume $b=\boxed{9}$ | B | 9 |
9820da393dde1cc3355ee5d83ead68da | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_25 | Let $k$ be a positive integer. Bernardo and Silvia take turns writing and erasing numbers on a blackboard as follows: Bernardo starts by writing the smallest perfect square with $k+1$ digits. Every time Bernardo writes a number, Silvia erases the last $k$ digits of it. Bernardo then writes the next perfect square, Silvia erases the last $k$ digits of it, and this process continues until the last two numbers that remain on the board differ by at least 2. Let $f(k)$ be the smallest positive integer not written on the board. For example, if $k = 1$ , then the numbers that Bernardo writes are $16, 25, 36, 49, 64$ , and the numbers showing on the board after Silvia erases are $1, 2, 3, 4,$ and $6$ , and thus $f(1) = 5$ . What is the sum of the digits of $f(2) + f(4)+ f(6) + \dots + f(2016)$
$\textbf{(A)}\ 7986\qquad\textbf{(B)}\ 8002\qquad\textbf{(C)}\ 8030\qquad\textbf{(D)}\ 8048\qquad\textbf{(E)}\ 8064$ | Consider $f(2)$ . The numbers left on the blackboard will show the hundreds place at the end. In order for the hundreds place to differ by 2, the difference between two perfect squares needs to be at least $100$ . Since $100\le (x+1)^2-x^2=2x+1$ , this first happens at $x\ge \lfloor 99/2\rfloor = 50$ . The perfect squares from here go: $2500, 2601, 2704, 2809\dots$ . Note that the ones and tens also make the perfect squares, $1^2,2^2,3^2\dots$ . After the ones and tens make $100$ , the hundreds place will go up by $2$ , thus reaching our goal. Since $10^2=100$ , the last perfect square to be written will be $\left(50+10\right)^2=60^2=3600$ . The missing number is one less than the number of hundreds $(k=2)$ of $3600$ , or $35$
Now consider $f(4)$ . Instead of the difference between two squares needing to be $100$ , the difference must now be $10000$ . This first happens at $x\ge 5000$ . After this point, similarly, $\sqrt{10000}=100$ more numbers are needed to make the $10^4$ th's place go up by $2$ . This will take place at $\left(5000+100\right)^2=5100^2= 26010000$ . Removing the last four digits (the zeros) and subtracting one yields $2600$ for the skipped value.
In general, each new value of $f(k+2)$ will add two digits to the " $5$ " and one digit to the " $1$ ". This means that the last number Bernardo writes for $k=6$ is $\left(500000+1000\right)^2$ , the last for $k = 8$ will be $\left(50000000+10000\right)^2$ , and so on until $k=2016$ . Removing the last $k$ digits as Silvia does will be the same as removing $k/2$ trailing zeroes on the number to be squared. This means that the last number on the board for $k=6$ is $5001^2$ $k=8$ is $50001^2$ , and so on. So the first missing number is $5001^2-1,50001^2-1\text{ etc.}$ The squaring will make a " $25$ " with two more digits than the last number, a " $10$ " with one more digit, and a " $1$ ". The missing number is one less than that, so the "1" will be subtracted from $f(k)$ . In other words, $f(k) = 25\cdot 10^{k-2}+1\cdot 10^{k/2}$
Therefore:
\[f(2) =35 =25 +10\] \[f(4) =2600 =2500 +100\] \[f(6) =251000 =250000 +1000\] \[f(8) = 25010000 = 25000000 + 10000\]
And so on. The sum $f(2) + f(4) + f(6) +\dots + f(2016)$ is:
$2.52525252525\dots 2525\cdot 10^{2015}$ $1.11111\dots 110\cdot 10^{1008}$ , with $2016$ repetitions each of " $25$ " and " $1$ ".
There is no carrying in this addition. Therefore each $f(k)$ adds $2 + 5 + 1 = 8$ to the sum of the digits.
Since $2n = 2016$ $n = 1008$ , and $8n = 8064$ , or $\boxed{8064}$ | E | 8064 |
9820da393dde1cc3355ee5d83ead68da | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_25 | Let $k$ be a positive integer. Bernardo and Silvia take turns writing and erasing numbers on a blackboard as follows: Bernardo starts by writing the smallest perfect square with $k+1$ digits. Every time Bernardo writes a number, Silvia erases the last $k$ digits of it. Bernardo then writes the next perfect square, Silvia erases the last $k$ digits of it, and this process continues until the last two numbers that remain on the board differ by at least 2. Let $f(k)$ be the smallest positive integer not written on the board. For example, if $k = 1$ , then the numbers that Bernardo writes are $16, 25, 36, 49, 64$ , and the numbers showing on the board after Silvia erases are $1, 2, 3, 4,$ and $6$ , and thus $f(1) = 5$ . What is the sum of the digits of $f(2) + f(4)+ f(6) + \dots + f(2016)$
$\textbf{(A)}\ 7986\qquad\textbf{(B)}\ 8002\qquad\textbf{(C)}\ 8030\qquad\textbf{(D)}\ 8048\qquad\textbf{(E)}\ 8064$ | We assume $n \geq 1$ for all claims.
We will let $g_k(x) = \left \lfloor \frac{x^2}{10^k}\right \rfloor$ . This is the result when the last k digits are truncated off $x^2$
Let $x_n$ = the smallest a, such that $g_{2n}(a) - g_{2n}(a-1) \geq 2$
We then have $\left \lfloor \frac{a^2}{10^{2n}} \right \rfloor - \left \lfloor \frac{(a-1)^2}{10^{2n}} \right \rfloor \geq 2$
Claim 1: $x_n > 5\cdot10^{2n-1}$
Proof of Claim 1:
Assume for the sake of contradiction, we have $x_n \leq 5 \cdot 10^{2n-1}$
Let $c = \frac{2x_n - 1}{10^{2n}}$ $d = \frac{(x_n-1)^2}{10^{2n}}$
Note that since $x_n \leq 5 \cdot 10^{2n-1}$ , we have $c < 1$
It is well known that $\lfloor i+j \rfloor \leq \lfloor i \rfloor + \lfloor j \rfloor + 1$ for any $i$ and $j$
Since $x_n$ satisfies the condition, we have:
$1 = \lfloor c \rfloor + 1 \geq \lfloor{c+d} \rfloor - \lfloor d \rfloor \geq 2$
This is a contradiction. $\blacksquare$
Claim 2: $x_n = 5 \cdot 10^{2n-1} + 10^n$
Proof of Claim 2:
We will show our choice of $x_n = 5 \cdot 10^{2n-1} + 10^n$ satisfies the criteria above.
$\left \lfloor \frac{{x_n}^2}{10^{2n}} \right \rfloor - \left \lfloor \frac{(x_n-1)^2}{10^{2n}} \right \rfloor = \left \lfloor \frac{25 \cdot 10^{4n - 2} + 10^{3n} + 10^{2n}}{10^{2n}} \right \rfloor - \left \lfloor \frac{25 \cdot 10^{4n - 2} + 10^{3n} - 2 \cdot 10^n + 1}{10^{2n}} \right \rfloor$
$= (25 \cdot 10^{2n-2} + 10^{n} + 1) - (25 \cdot 10^{2n-2} + 10^{n} - 1) = 2$
Now, we will show all values smaller than $5 \cdot 10^{2n-1} + 10^n$ don’t satisfy the criteria.
We will assume for the sake of contradiction, there exists an $x'_n$ which satisfies the criteria.
We will write $x'_n$ as $5 \cdot 10^{2n - 1} + k$ . By our hypothesis, we have $k < 10^n$ . We also have $k > 0$ by claim 1.
We have $\left \lfloor \frac{{x'}_n^2}{10^{2n}} \right \rfloor - \left \lfloor \frac{(x'_n-1)^2}{10^{2n}} \right \rfloor = \left \lfloor \frac{25 \cdot 10^{4n - 2} + k \cdot 10^{2n} + k^2}{10^{2n}} \right \rfloor - \left \lfloor \frac{25 \cdot 10^{4n - 2} + (k-1) \cdot 10^{2n} + (k-1)^2 }{10^{2n}} \right \rfloor$
$= (25 \cdot 10^{2n-2} + k) - (25 \cdot 10^{2n-2} + k - 1) = 1 \geq 2.$ This gets a contradiction. $\blacksquare$
Claim 3: $f(2n) = 25 \cdot10^{2n - 2} + 10^{n}$
Proof of Claim 3: Because of claim 2, $x_n$ is $5 \cdot 10^{2n-1} + 10^n$ . Note that $g_n(x_n)$ and $g_n(x_n - 1)$ are the first numbers written that will differ by at least 2. Thus, $f(2n) = g_n(x_n - 1) + 1 = 25 \cdot 10^{2n-2} + 10^n$ $\blacksquare$
Thus, $f(2) + f(4) \dots f(2016) = \underbrace{(2525 \dots 25)}_{1008 \ 25s} + \underbrace{(111111 \dots 1111)}_{1008 \ 1s}$ . This addition has no regroups/carry overs, so we can just take the sums of the digits of each of the addends, and sum them together to get 8064. $\boxed{8064}$ | E | 8064 |
7cea25e26d83b9afa50c7c42dc6e0803 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12B_Problems/Problem_1 | What is the value of $\frac{2a^{-1}+\frac{a^{-1}}{2}}{a}$ when $a= \frac{1}{2}$
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ \frac{5}{2}\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 20$ | By: Dragonfly
We find that $a^{-1}$ is the same as $2$ , since a number to the power of $-1$ is just the reciprocal of that number. We then get the equation to be
\[\frac{2\times2+\frac{2}{2}}{\frac{1}{2}}\]
We can then simplify the equation to get $\boxed{10}$ | D | 10 |
3db863d9220758835d21960c23b3f53e | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12B_Problems/Problem_2 | The harmonic mean of two numbers can be calculated as twice their product divided by their sum. The harmonic mean of $1$ and $2016$ is closest to which integer?
$\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 45 \qquad \textbf{(C)}\ 504 \qquad \textbf{(D)}\ 1008 \qquad \textbf{(E)}\ 2015$ | Since the harmonic mean is $2$ times their product divided by their sum, we get the equation
$\frac{2\times1\times2016}{1+2016}$
which is then
$\frac{4032}{2017}$
which is finally closest to $\boxed{2}$ .
-dragonfly | A | 2 |
e5e922acf502a1832f6b4f6693da6149 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12B_Problems/Problem_3 | Let $x=-2016$ . What is the value of $\bigg|$ $||x|-x|-|x|$ $\bigg|$ $-x$
$\textbf{(A)}\ -2016\qquad\textbf{(B)}\ 0\qquad\textbf{(C)}\ 2016\qquad\textbf{(D)}\ 4032\qquad\textbf{(E)}\ 6048$ | By: dragonfly
First of all, lets plug in all of the $x$ 's into the equation.
$\bigg|$ $||-2016|-(-2016)|-|-2016|$ $\bigg|$ $-(-2016)$
Then we simplify to get
$\bigg|$ $|2016+2016|-2016$ $\bigg|$ $+2016$
which simplifies into
$\bigg|$ $2016$ $\bigg|$ $+2016$
and finally we get $\boxed{4032}$ | D | 4032 |
e5e922acf502a1832f6b4f6693da6149 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12B_Problems/Problem_3 | Let $x=-2016$ . What is the value of $\bigg|$ $||x|-x|-|x|$ $\bigg|$ $-x$
$\textbf{(A)}\ -2016\qquad\textbf{(B)}\ 0\qquad\textbf{(C)}\ 2016\qquad\textbf{(D)}\ 4032\qquad\textbf{(E)}\ 6048$ | Consider $x$ is negative.
We replace all instances of $x$ with $|x|$
$\bigg|$ $||x|+|x||-|x|$ $\bigg|$ $+|x|$
$=$ $\bigg|$ $|2x|-|x|$ $\bigg|$ $+|x|$
$=$ $|x|$ $+|x|$
$=|2x|$
$=4032 \implies \boxed{4032}$ | D | 4032 |
11f60b8c3e87d6f508548c98e9c48b0a | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12B_Problems/Problem_4 | The ratio of the measures of two acute angles is $5:4$ , and the complement of one of these two angles is twice as large as the complement of the other. What is the sum of the degree measures of the two angles?
$\textbf{(A)}\ 75\qquad\textbf{(B)}\ 90\qquad\textbf{(C)}\ 135\qquad\textbf{(D)}\ 150\qquad\textbf{(E)}\ 270$ | By: dragonfly
We set up equations to find each angle. The larger angle will be represented as $x$ and the smaller angle will be represented as $y$ , in degrees. This implies that
$4x=5y$
and
$2\times(90-x)=90-y$
since the larger the original angle, the smaller the complement.
We then find that $x=75$ and $y=60$ , and their sum is $\boxed{135}$ | C | 135 |
11f60b8c3e87d6f508548c98e9c48b0a | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12B_Problems/Problem_4 | The ratio of the measures of two acute angles is $5:4$ , and the complement of one of these two angles is twice as large as the complement of the other. What is the sum of the degree measures of the two angles?
$\textbf{(A)}\ 75\qquad\textbf{(B)}\ 90\qquad\textbf{(C)}\ 135\qquad\textbf{(D)}\ 150\qquad\textbf{(E)}\ 270$ | We can visualize the problem like so: [asy] path b = brace((0,10),(90,10),5); draw(b); label("$90^\circ$",b,N); draw("$5x$",(0,0)--(75,0),N); draw((75,2.5)--(75,-2.5)); draw("$1y$",(75,0)--(90,0),N); draw("$4x$",(0,-10)--(60,-10),S); draw((60,-7.5)--(60,-12.5)); draw("$2y$",(60,-10)--(90,-10),S); draw((0,5)--(0,-15)); draw((90,5)--(90,-15)); [/asy] \[5x+1y = 90^\circ = 4x+2y\] Moving like terms to the same side gets $x = y$ , and substituting this back gets $6x = 90^\circ \implies x = \frac{90^\circ}{6} = 15^\circ$ , so the sum of the degree measures is $5x + 4x = 9x = 9(15) = \boxed{135}$ . ~ emerald_block | C | 135 |
b46ce22b01c7e2b5682b64bb8e2cfb2a | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12B_Problems/Problem_6 | All three vertices of $\bigtriangleup ABC$ lie on the parabola defined by $y=x^2$ , with $A$ at the origin and $\overline{BC}$ parallel to the $x$ -axis. The area of the triangle is $64$ . What is the length of $BC$
$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 16$ | By: Albert471
Plotting points $B$ and $C$ on the graph shows that they are at $\left( -x,x^2\right)$ and $\left( x,x^2\right)$ , which is isosceles. By setting up the triangle area formula you get: $64=\frac{1}{2}*2x*x^2 = 64=x^3$ Making x=4, and the length of $BC$ is $2x$ , so the answer is $\boxed{8}$ | C | 8 |
d52dfb9cc52fc517b40836e568456d9e | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12B_Problems/Problem_7 | Josh writes the numbers $1,2,3,\dots,99,100$ . He marks out $1$ , skips the next number $(2)$ , marks out $3$ , and continues skipping and marking out the next number to the end of the list. Then he goes back to the start of his list, marks out the first remaining number $(2)$ , skips the next number $(4)$ , marks out $6$ , skips $8$ , marks out $10$ , and so on to the end. Josh continues in this manner until only one number remains. What is that number?
$\textbf{(A)}\ 13 \qquad \textbf{(B)}\ 32 \qquad \textbf{(C)}\ 56 \qquad \textbf{(D)}\ 64 \qquad \textbf{(E)}\ 96$ | Following the pattern, you are crossing out...
Time 1: Every non-multiple of $2$
Time 2: Every non-multiple of $4$
Time 3: Every non-multiple of $8$
Following this pattern, you are left with every multiple of $64$ which is only $\boxed{64}$ | D | 64 |
f0a19114deef7af9c3ee4cf9afc075ab | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12B_Problems/Problem_8 | A thin piece of wood of uniform density in the shape of an equilateral triangle with side length $3$ inches weighs $12$ ounces. A second piece of the same type of wood, with the same thickness, also in the shape of an equilateral triangle, has side length of $5$ inches. Which of the following is closest to the weight, in ounces, of the second piece?
$\textbf{(A)}\ 14.0\qquad\textbf{(B)}\ 16.0\qquad\textbf{(C)}\ 20.0\qquad\textbf{(D)}\ 33.3\qquad\textbf{(E)}\ 55.6$ | By: dragonfly
We can solve this problem by using similar triangles, since two equilateral triangles are always similar. We can then use
$\left(\frac{3}{5}\right)^2=\frac{12}{x}$
We can then solve the equation to get $x=\frac{100}{3}$ which is closest to $\boxed{33.3}$ | D | 33.3 |
f0a19114deef7af9c3ee4cf9afc075ab | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12B_Problems/Problem_8 | A thin piece of wood of uniform density in the shape of an equilateral triangle with side length $3$ inches weighs $12$ ounces. A second piece of the same type of wood, with the same thickness, also in the shape of an equilateral triangle, has side length of $5$ inches. Which of the following is closest to the weight, in ounces, of the second piece?
$\textbf{(A)}\ 14.0\qquad\textbf{(B)}\ 16.0\qquad\textbf{(C)}\ 20.0\qquad\textbf{(D)}\ 33.3\qquad\textbf{(E)}\ 55.6$ | Another approach to this problem, very similar to the previous one but perhaps explained more thoroughly, is to use proportions. First, since the thickness and density are the same, we can set up a proportion based on the principle that $d=\frac{m}{V}$ , thus $dV=m$
However, since density and thickness are the same and $A$ is proportional to $s^2$ (recognizing that the area of an equilateral triangle is $\frac{(s)^2\sqrt{3}}{4}$ ), we can say that $m$ is proportional to $s^2$
Then, by increasing s by a factor of $\frac{5}{3}$ $s^2$ is increased by a factor of $\frac{25}{9}$ , thus $m=12*\frac{25}{9}$ or $\boxed{33.3}$ | D | 33.3 |
45c438d90e75e3eea1c40e38a106d5ef | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12B_Problems/Problem_9 | Carl decided to fence in his rectangular garden. He bought $20$ fence posts, placed one on each of the four corners, and spaced out the rest evenly along the edges of the garden, leaving exactly $4$ yards between neighboring posts. The longer side of his garden, including the corners, has twice as many posts as the shorter side, including the corners. What is the area, in square yards, of Carl’s garden?
$\textbf{(A)}\ 256\qquad\textbf{(B)}\ 336\qquad\textbf{(C)}\ 384\qquad\textbf{(D)}\ 448\qquad\textbf{(E)}\ 512$ | By Albert471
To start, use algebra to determine the number of posts on each side. You have (the long sides count for $2$ because there are twice as many) $6x = 20 + 4$ (each corner is double counted so you must add $4$ ) Making the shorter end have $4$ , and the longer end have $8$ $((8-1)*4)*((4-1)*4) = 28*12 = 336$ . Therefore, the answer is $\boxed{336}$ | B | 336 |
a0fcbe800fc7653b7a55b5e1d462cd1a | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12B_Problems/Problem_10 | A quadrilateral has vertices $P(a,b)$ $Q(b,a)$ $R(-a, -b)$ , and $S(-b, -a)$ , where $a$ and $b$ are integers with $a>b>0$ . The area of $PQRS$ is $16$ . What is $a+b$
$\textbf{(A)}\ 4 \qquad\textbf{(B)}\ 5 \qquad\textbf{(C)}\ 6 \qquad\textbf{(D)}\ 12 \qquad\textbf{(E)}\ 13$ | Note that the slope of $PQ$ is $\frac{a-b}{b-a}=-1$ and the slope of $PS$ is $\frac{b+a}{a+b}=1$ . Hence, $PQ\perp PS$ and we can similarly prove that the other angles are right angles. This means that $PQRS$ is a rectangle. By distance formula we have $(a-b)^2+(b-a)^2*2*(a+b)^2 = 256$ . Simplifying we get $(a-b)(a+b) = 8$ . Thus $a+b$ and $a-b$ have to be a factor of 8. The only way for them to be factors of $8$ and remain integers is if $a+b = 4$ and $a-b = 2$ . So the answer is $\boxed{4}$ | A | 4 |
a0fcbe800fc7653b7a55b5e1d462cd1a | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12B_Problems/Problem_10 | A quadrilateral has vertices $P(a,b)$ $Q(b,a)$ $R(-a, -b)$ , and $S(-b, -a)$ , where $a$ and $b$ are integers with $a>b>0$ . The area of $PQRS$ is $16$ . What is $a+b$
$\textbf{(A)}\ 4 \qquad\textbf{(B)}\ 5 \qquad\textbf{(C)}\ 6 \qquad\textbf{(D)}\ 12 \qquad\textbf{(E)}\ 13$ | Solution by e_power_pi_times_i
By the Shoelace Theorem, the area of the quadrilateral is $2a^2 - 2b^2$ , so $a^2 - b^2 = 8$ . Since $a$ and $b$ are integers, $a = 3$ and $b = 1$ , so $a + b = \boxed{4}$ | A | 4 |
f11905d9fb7b8aef70eac5feb4a2ae46 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12B_Problems/Problem_11 | How many squares whose sides are parallel to the axes and whose vertices have coordinates that are integers lie entirely within the region bounded by the line $y=\pi x$ , the line $y=-0.1$ and the line $x=5.1?$
$\textbf{(A)}\ 30 \qquad \textbf{(B)}\ 41 \qquad \textbf{(C)}\ 45 \qquad \textbf{(D)}\ 50 \qquad \textbf{(E)}\ 57$ | Solution by e_power_pi_times_i
Revised by Kinglogic and RJ5303707
[asy] Label l; l.p=fontsize(8); xaxis(-1,8,Ticks(l, 1.0)); yaxis(-1,16,Ticks(l, 1.0)); real f(real x) { return x * pi; } D(graph(f,-1/pi,5.1)); D((5.1,-1)--(5.1,16)); D((-1,-0.1)--(8,-0.1)); for(int x = 0; x < 5.1; ++x) { for(int y = 0; y < 16; ++y) { if(x * pi > y) { D((x,y)); } } } [/asy] (red shows lattice points within the triangle)
If we draw a picture showing the triangle, we see that it would be easier to count the squares vertically and not horizontally. The upper bound is $16$ squares $(y=5.1*\pi)$ , and the limit for the $x$ -value is $5$ squares. First we count the $1*1$ squares. In the rightmost column, there are $12$ squares with length $1$ because $y=4*\pi$ generates squares from $(4,0)$ to $(4,4\pi)$ , and continuing on we have $9$ $6$ , and $3$ for $x$ -values for $1$ $2$ , and $3$ in the equation $y=\pi x$ . So there are $12+9+6+3 = 30$ squares with length $1$ in the figure.
For $2*2$ squares, each square takes up $2$ units left and $2$ units up. Squares can also overlap. For $2*2$ squares, the rightmost column stretches from $(3,0)$ to $(3,3\pi)$ , so there are $8$ squares with length $2$ in a $2$ by $9$ box. Repeating the process, the next column stretches from $(2,0)$ to $(2,2\pi)$ , so there are $5$ squares. Continuing and adding up in the end, there are $8+5+2=15$ squares with length $2$ in the figure.
Squares with length $3$ in the rightmost column start at $(2,0)$ and end at $(2,2\pi)$ , so there are $4$ such squares in the right column. As the left row starts at $(1,0)$ and ends at $(1,\pi)$ there are $4+1=5$ squares with length $3$ . As squares with length $4$ would not fit in the triangle, the answer is $30+15+5$ which is $\boxed{50}$ | D | 50 |
3899010a341759a057dd346753c42009 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12B_Problems/Problem_13 | Alice and Bob live $10$ miles apart. One day Alice looks due north from her house and sees an airplane. At the same time Bob looks due west from his house and sees the same airplane. The angle of elevation of the airplane is $30^\circ$ from Alice's position and $60^\circ$ from Bob's position. Which of the following is closest to the airplane's altitude, in miles?
$\textbf{(A)}\ 3.5 \qquad\textbf{(B)}\ 4 \qquad\textbf{(C)}\ 4.5 \qquad\textbf{(D)}\ 5 \qquad\textbf{(E)}\ 5.5$ | We have two 30-60-90 triangles $ABC$ and $DBC$ that are perpendicular and share leg $BC$ (the altitude of the plane $p$ ). $AD=10$ The shared leg is the shortest leg of one triangle and the longest leg of the other. $A$ and $B$ are Bob and Alice respectively.
Find $AC$ and $DC$ in terms of $p$ . Use Pythagorean Theorem on triangle $ADC$ to produce $p=\sqrt{30}\implies\boxed{5.5}$ | E | 5.5 |
3899010a341759a057dd346753c42009 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12B_Problems/Problem_13 | Alice and Bob live $10$ miles apart. One day Alice looks due north from her house and sees an airplane. At the same time Bob looks due west from his house and sees the same airplane. The angle of elevation of the airplane is $30^\circ$ from Alice's position and $60^\circ$ from Bob's position. Which of the following is closest to the airplane's altitude, in miles?
$\textbf{(A)}\ 3.5 \qquad\textbf{(B)}\ 4 \qquad\textbf{(C)}\ 4.5 \qquad\textbf{(D)}\ 5 \qquad\textbf{(E)}\ 5.5$ | Non-trig solution by e_power_pi_times_i
Set the distance from Alice's and Bob's position to the point directly below the airplane to be $x$ and $y$ , respectively. From the Pythagorean Theorem, $x^2 + y^2 = 100$ . As both are $30-60-90$ triangles, the altitude of the airplane can be expressed as $\dfrac{x\sqrt{3}}{3}$ or $y\sqrt{3}$ . Solving the equation $\dfrac{x\sqrt{3}}{3} = y\sqrt{3}$ , we get $x = 3y$ . Plugging this into the equation $x^2 + y^2 = 100$ , we get $10y^2 = 100$ , or $y = \sqrt{10}$ $y$ cannot be negative), so the altitude is $\sqrt{3*10} = \sqrt{30}$ , which is closest to $\boxed{5.5}$ | E | 5.5 |
3899010a341759a057dd346753c42009 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12B_Problems/Problem_13 | Alice and Bob live $10$ miles apart. One day Alice looks due north from her house and sees an airplane. At the same time Bob looks due west from his house and sees the same airplane. The angle of elevation of the airplane is $30^\circ$ from Alice's position and $60^\circ$ from Bob's position. Which of the following is closest to the airplane's altitude, in miles?
$\textbf{(A)}\ 3.5 \qquad\textbf{(B)}\ 4 \qquad\textbf{(C)}\ 4.5 \qquad\textbf{(D)}\ 5 \qquad\textbf{(E)}\ 5.5$ | Let Alice be at point $A$ , Bob be at point $B$ . Let the plane be at point $P$ and $X$ be the projection of $P$ onto the ground (the plane with contains $A$ and $B$ ). Let the height of the plane, or $PX$ , be $h$ . So, because of the $30-60-90$ triangles, \[AX = \dfrac{h}{\sqrt{3}}, BX = h\sqrt{3}\] By Pythagorean Theorem on $\triangle ABX$ \[\dfrac{h^2}{3} + 3h^2 = 100 \implies \dfrac{10h^2}{3} = 100 \implies h = \sqrt{30},\] which is clossest to $\boxed{5.5}.$ | E | 5.5 |
3c180d92f3cd6e1f58dc450d4ac342a0 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12B_Problems/Problem_14 | The sum of an infinite geometric series is a positive number $S$ , and the second term in the series is $1$ . What is the smallest possible value of $S?$
$\textbf{(A)}\ \frac{1+\sqrt{5}}{2} \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ \sqrt{5} \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 4$ | The sum of the geometric sequence is $\frac{a}{1 - r}$ where $a$ is the first term and $r$ is the common ratio. We know the second term, $ar,$ is equal to $1.$ Thus $ar = 1 \Rightarrow a = \frac{1}{r}.$ This means, \[S = \frac{a}{1 - r} = \frac{1/r}{1 - r} = \frac{1}{r(1 - r)}.\] In order to minimize $S,$ we maximize the denominator. By AM-GM, \[\frac{(r) + (1 - r)}{2} \ge \sqrt{r(1-r)} \Rightarrow \frac{1}{4} \ge r(1-r).\] Equality occurs at $r = 1-r \Rightarrow r = \frac{1}{2}.$ This gives the minimum value of $S$ as $\frac{\frac{1}{1/2}}{1 - \frac{1}{2}} = \boxed{4}.$ | E | 4 |
3c180d92f3cd6e1f58dc450d4ac342a0 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12B_Problems/Problem_14 | The sum of an infinite geometric series is a positive number $S$ , and the second term in the series is $1$ . What is the smallest possible value of $S?$
$\textbf{(A)}\ \frac{1+\sqrt{5}}{2} \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ \sqrt{5} \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 4$ | A geometric sequence always looks like
\[a,ar,ar^2,ar^3,\dots\]
and they say that the second term $ar=1$ . You should know that the sum of an infinite geometric series (denoted by $S$ here) is $\frac{a}{1-r}$ . We now have a system of equations which allows us to find $S$ in one variable.
\begin{align*} ar&=1 \\ S&=\frac{a}{1-r} \end{align*}
$\textbf{Solving in terms of \textit{a} then graphing}$
\[S=\frac{a^2}{a-1}\]
We seek the smallest positive value of $S$ . We proceed by graphing in the $aS$ plane (if you want to be rigorous, only stop graphing when you know all the rest you didn't graph is just the approaching of asymptotes and infinities) and find the answer is $\boxed{4}.$ | E | 4 |
3c180d92f3cd6e1f58dc450d4ac342a0 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12B_Problems/Problem_14 | The sum of an infinite geometric series is a positive number $S$ , and the second term in the series is $1$ . What is the smallest possible value of $S?$
$\textbf{(A)}\ \frac{1+\sqrt{5}}{2} \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ \sqrt{5} \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 4$ | \[\textbf{Completing the Square and Quadratics}\] Let $r$ be the common ratio. If the second term is $1$ , the first must be $\frac{1}{r}$ . By the infinite geometric series formula, the sum must be \[S=\frac{\frac{1}{r}}{1-r}\] This equals $\frac{1}{r(1-r)}$ . To find the minimum value of S, we must find the maximum value of the denominator, $r(1-r)$ , which is $\frac{1}{4}$ , completing the square. Thus, the minimum value of $S$ is $\boxed{4}$ | E | 4 |
3c180d92f3cd6e1f58dc450d4ac342a0 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12B_Problems/Problem_14 | The sum of an infinite geometric series is a positive number $S$ , and the second term in the series is $1$ . What is the smallest possible value of $S?$
$\textbf{(A)}\ \frac{1+\sqrt{5}}{2} \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ \sqrt{5} \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 4$ | Our sequence is $a_1,1,...$ . Since we know this is a converging series, our ratio is in $(0,1)$ . Because the 2nd term in the sequence is a 1, the ratio must be $\frac{1}{a_1}$ , so we can write $S$ as $\frac{a_1}{1-\frac{1}{a_1}}$ . With some manipulation we get $S=\frac{1}{a_1-1}$ . Since S has to be a "positive number," we come to think $a_1$ is $2$ (makes S positive & we know a sequence/series of a ratio $1/2$ is definitely convergent). So our sequence is $2,1,1/2,1,4,...$ $2+1+1=\boxed{4}$ .
-thedodecagon | E | 4 |
da97c32daee19ef6f973a98bae425e09 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12B_Problems/Problem_15 | All the numbers $2, 3, 4, 5, 6, 7$ are assigned to the six faces of a cube, one number to each face. For each of the eight vertices of the cube, a product of three numbers is computed, where the three numbers are the numbers assigned to the three faces that include that vertex. What is the greatest possible value of the sum of these eight products?
$\textbf{(A)}\ 312 \qquad \textbf{(B)}\ 343 \qquad \textbf{(C)}\ 625 \qquad \textbf{(D)}\ 729 \qquad \textbf{(E)}\ 1680$ | First assign each face the letters $a,b,c,d,e,f$ . The sum of the product of the faces is $abc+acd+ade+aeb+fbc+fcd+fde+feb$ . We can factor this into $(a+f)(b+c)(d+e)$ which is the product of the sum of each pair of opposite faces. In order to maximize $(a+f)(b+c)(d+e)$ we use the numbers $(7+2)(6+3)(5+4)$ or $\boxed{729}$ | D | 729 |
da97c32daee19ef6f973a98bae425e09 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12B_Problems/Problem_15 | All the numbers $2, 3, 4, 5, 6, 7$ are assigned to the six faces of a cube, one number to each face. For each of the eight vertices of the cube, a product of three numbers is computed, where the three numbers are the numbers assigned to the three faces that include that vertex. What is the greatest possible value of the sum of these eight products?
$\textbf{(A)}\ 312 \qquad \textbf{(B)}\ 343 \qquad \textbf{(C)}\ 625 \qquad \textbf{(D)}\ 729 \qquad \textbf{(E)}\ 1680$ | We'll proceed from the factoring process above.
By the AM-GM inequality,
\[\frac{a_1+a_2+a_3}{3}\geq\sqrt[3]{a_1a_2a_3}\]
Cubing both sides,
\[\left(\frac{a_1+a_2+a_3}{3}\right)^3\geq{a_1a_2a_3}\]
Let $a_1=(a+f)$ $a_2=(b+c)$ , and $a_3=(d+e)$ . Let's substitute in these values.
\[\left(\frac{a+b+c+d+e+f}{3}\right)^3\geq{(a+f)(b+c)(d+e)}\]
$a+b+c+d+e+f$ is fixed at 27.
\[\left(\frac{27}{3}\right)^3\geq{(a+f)(b+c)(d+e)}\]
\[\boxed{729}\] | D | 729 |
da97c32daee19ef6f973a98bae425e09 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12B_Problems/Problem_15 | All the numbers $2, 3, 4, 5, 6, 7$ are assigned to the six faces of a cube, one number to each face. For each of the eight vertices of the cube, a product of three numbers is computed, where the three numbers are the numbers assigned to the three faces that include that vertex. What is the greatest possible value of the sum of these eight products?
$\textbf{(A)}\ 312 \qquad \textbf{(B)}\ 343 \qquad \textbf{(C)}\ 625 \qquad \textbf{(D)}\ 729 \qquad \textbf{(E)}\ 1680$ | First, we see that we want to try and maximize each vertex. Since the multiplication of each vertex is the product of three values, we want to maximize those three values. Doing so, we see that we want them to be as close as possible, giving $4.5^3$ (the average of all the values). This gives us the maximum for each vertex, multiplied by the 8 vertices, yields our answer $\boxed{729}$ Also note that if you cannot evaluate $4.5^3$ quickly, a rough approximation of $5*4.5*4*8$ will yield 720, very close to our answer. -rayprati | D | 729 |
da97c32daee19ef6f973a98bae425e09 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12B_Problems/Problem_15 | All the numbers $2, 3, 4, 5, 6, 7$ are assigned to the six faces of a cube, one number to each face. For each of the eight vertices of the cube, a product of three numbers is computed, where the three numbers are the numbers assigned to the three faces that include that vertex. What is the greatest possible value of the sum of these eight products?
$\textbf{(A)}\ 312 \qquad \textbf{(B)}\ 343 \qquad \textbf{(C)}\ 625 \qquad \textbf{(D)}\ 729 \qquad \textbf{(E)}\ 1680$ | It is obvious to put $5$ $6$ , and $7$ on the faces that share the same vertex. As $4$ is the next biggest number, the face with $4$ has to be next to the faces with $6$ and $7$ . As $4$ is the next biggest number, the face with $3$ has to be next to the faces with $5$ and $7$ . making the face with $2$ next to the faces with $6$ and $5$
Therefore the answer is $7 \cdot 5 \cdot 6 + 7 \cdot 4 \cdot 6 + 7 \cdot 3 \cdot 5 + 7 \cdot 3 \cdot 4 + 2 \cdot 5 \cdot 6 + 2 \cdot 4 \cdot 6 + 2 \cdot 3 \cdot 5 + 2 \cdot 3 \cdot 4 = \boxed{729}$ | D | 729 |
21cb2702105d88e2f4a4bb31f1485960 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12B_Problems/Problem_16 | In how many ways can $345$ be written as the sum of an increasing sequence of two or more consecutive positive integers?
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 7$ | We proceed with this problem by considering two cases, when: 1) There are an odd number of consecutive numbers, 2) There are an even number of consecutive numbers.
For the first case, we can cleverly choose the convenient form of our sequence to be \[a-n,\cdots, a-1, a, a+1, \cdots, a+n\]
because then our sum will just be $(2n+1)a$ . We now have \[(2n+1)a = 345\] and $a$ will have a solution when $\frac{345}{2n+1}$ is an integer, namely when $2n+1$ is a divisor of 345. We check that \[2n+1 = 3, 5, 15, 23\] work, and no more, because $2n+1=1$ does not satisfy the requirements of two or more consecutive integers, and when $2n+1$ equals the next biggest factor, $69$ , there must be negative integers in the sequence. Our solutions are $\{114,115, 116\}, \{67, \cdots, 71\}, \{16, \cdots, 30\}, \{4, \cdots, 26\}$
For the even cases, we choose our sequence to be of the form: \[a-(n-1), \cdots, a, a+1, \cdots, a+n\] so the sum is $\frac{(2n)(2a+1)}{2} = n(2a+1)$ . In this case, we find our solutions to be $\{172, 173\}, \{55,\cdots, 60\}, \{30,\cdots, 39\}$
We have found all 7 solutions and our answer is $\boxed{7}$ | E | 7 |
21cb2702105d88e2f4a4bb31f1485960 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12B_Problems/Problem_16 | In how many ways can $345$ be written as the sum of an increasing sequence of two or more consecutive positive integers?
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 7$ | The sum from $a$ to $b$ where $a$ and $b$ are integers and $a>b$ is
$S=\dfrac{(a-b+1)(a+b)}{2}$
$345=\dfrac{(a-b+1)(a+b)}{2}$
$2\cdot 3\cdot 5\cdot 23=(a-b+1)(a+b)$
Let $c=a-b+1$ and $d=a+b$
$2\cdot 3\cdot 5\cdot 23=c\cdot d$
If we factor $690$ into all of its factor groups $(\text{exg}~ (10,69) ~\text{or} ~(15,46))$ we will have several ordered pairs $(c,d)$ where $c<d$
The number of possible values for $c$ is half the number of factors of $690$ which is $\frac{1}{2}\cdot2\cdot2\cdot2\cdot2=8$
However, we have one extraneous case of $(1,690)$ because here, $a=b$ and we have the sum of one consecutive number which is not allowed by the question.
Thus the answer is $8-1=7$
$\boxed{7}$ | E | 7 |
21cb2702105d88e2f4a4bb31f1485960 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12B_Problems/Problem_16 | In how many ways can $345$ be written as the sum of an increasing sequence of two or more consecutive positive integers?
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 7$ | The consecutive sums can be written as $345=kn+\sum_{i=1}^{k-1}{i}$ where $k$ is the number of terms in a sequence, and $n$ is the first term. Then, $\{k,n\}\in \mathbb{N}$ and $k\geq2$ . Evaluating the sum and rearranging yields $n=\frac{345}{k}-\frac{k-1}{2}$
The prime factorization of $345$ is $1\cdot3\cdot5\cdot23$ . Then, $3\cdot5$ $3\cdot23$ , and $5\cdot23$ are also divisors. As odd divisors of $345$ , note that they all produce integer solutions to $n$ as $k$ . Only $k!=1$ is not valid, as $k\geq2$ . Similarly, quickly notice that $k=2$ is a solution. Multiplying an odd divisor by $2$ always yields an integer solution (see below). As such, the even integer solutions are $k=2, 2\cdot3, 2\cdot5, 2\cdot15 ... 2\cdot345$
Note that the function is decreasing for increasing values of $k$ and that $\frac{345}{k}-\frac{k-1}{2}=4$ for $k=23$ . Thus, when $n$ is negative, $k$ is only slightly more than $k=23$ . Recall $\{k,n\}\in \mathbb{N}$ . Since the next highest solution, $k=2\cdot23=46$ is twice $k=23$ $k\leq23$ . Thus, the remaining solutions are when $k=2,6,10,3,5,15,23$ $\implies \boxed{7}$ | E | 7 |
21cb2702105d88e2f4a4bb31f1485960 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12B_Problems/Problem_16 | In how many ways can $345$ be written as the sum of an increasing sequence of two or more consecutive positive integers?
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 7$ | We're dealing with an increasing arithmetic progression of common difference 1. Let $x$ be the number of terms in a summation. Let $y$ be the first term in a summation. The sum of an arithmetic progression is the average of the first term and the last term multiplied by the number of terms. The problem tells us that the sum must be 345.
\begin{align*} x \cdot \frac{y+y+[(x-1)1]}{2}&=345 \\ 2xy+x^2-x&=690 \end{align*}
In order to satisfy the constraints of the problem, x and y must be positive integers. Maybe we can make this into a Diophantine thing! In fact, if we just factor out that $x$ ... voilà!
\[(x)(x+2y-1)=690\]
There are 16 possible factor pairs to try (for brevity, I will not enumerate them here). Notice that the expression in the right parenthesis is $2y-1$ more than the expression in the parenthesis on the left. $y$ is at least 1. Thus, the expression in the right parenthesis will always be greater than the expression on the left. This eliminates 8 factor pairs. The problem also says the "increasing sequence" has to have "two or more" terms, so $x \geqslant 2$ . This eliminates the factor pair $1 \cdot 690$ . With brief testing, we find that the the other 7 factor pairs produce 7 viable ordered pairs. This means we have found $\boxed{7}$ ways to write 345 in the silly way outlined by the problem. | E | 7 |
21cb2702105d88e2f4a4bb31f1485960 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12B_Problems/Problem_16 | In how many ways can $345$ be written as the sum of an increasing sequence of two or more consecutive positive integers?
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 7$ | By the sum of an arithmetic sequence... this ultimately comes to $n+n+1+n+2....+n+p=345=(2n+p)(p+1)=690=23\cdot3\cdot5\cdot2$
Quick testing (would take you roughly a minute)
We see that the first 7 values of $p$ that work are
$p=1,2,4,5,9,14,22$
We see that each one of them works.
Hence, the answer is $\boxed{7}$ | null | 7 |
21cb2702105d88e2f4a4bb31f1485960 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12B_Problems/Problem_16 | In how many ways can $345$ be written as the sum of an increasing sequence of two or more consecutive positive integers?
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 7$ | The sum of all integers from $x$ to $y$ inclusive is equal to
\[\frac{y(y+1)}{2}-\left(\frac{x(x+1)}{2}\right)+x.\]
In words, $x$ added to sum of the first $x$ positive inegers subtracted from the sum of the first positive $y$ integers.
Setting this equal to $345$ and multiplying by $2$ to clear fractions, we can see that \[y(y+1)-x(x+1)+2x\] \[=y^2-x^2+y+x\] \[=(y+x)(y-x+1)=690.\] Now we know that, $y+x$ and $y-x+1$ multiply together to $690$ , and they're both integers. Now, we can set $y+x$ and $y-x+1$ equal to factors of $690$ . Clearly, $x>\frac{1}{2},$ so $y+x$ will take the larger of the two factors.
We can factorize $690$ as $1\cdot2\cdot3\cdot5\cdot23$
We also know that when solving for $y$ , we can add the two systems in $y+x$ and $y-x+1$ to get a new equation in terms of $2y+1.$ In order for $y$ to be an integer, the sum of the two factors must be odd.
We also know that there is only one factor of $2$ , so either $y+x$ or $y-x+1$ will be even. The number of odd factors multiplied together can either be $1, 2$ or $3$ (0 doesn't work, since $(690, 1)$ doesn't). There are a total of $3$ odd numbers present in our new prime factorization, now excluding $1$
Therefore, the number of combinations that yield integral values for both $x$ and $y$ are \[{3\choose1}+{3\choose2}+{3\choose3}= \boxed{7}\] | E | 7 |
e0402cc1fed1c942211c64b7a0e97350 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12B_Problems/Problem_18 | What is the area of the region enclosed by the graph of the equation $x^2+y^2=|x|+|y|?$
$\textbf{(A)}\ \pi+\sqrt{2} \qquad\textbf{(B)}\ \pi+2 \qquad\textbf{(C)}\ \pi+2\sqrt{2} \qquad\textbf{(D)}\ 2\pi+\sqrt{2} \qquad\textbf{(E)}\ 2\pi+2\sqrt{2}$ | Consider the case when $x \geq 0$ $y \geq 0$ \[x^2+y^2=x+y\] \[(x - \frac{1}{2})^2+(y - \frac{1}{2})^2=\frac{1}{2}\] Notice the circle intersect the axes at points $(0, 1)$ and $(1, 0)$ . Find the area of this circle in the first quadrant. The area is made of a semi-circle with radius of $\frac{\sqrt{2}}{2}$ and a triangle: \[A = \frac{\pi}{4} +\frac{1}{2}\] [asy]draw((0,-1.5)--(0,1.5),EndArrow);draw((-1.5,0)--(1.5,0),EndArrow);draw((0,1)--(1,0)--(0,-1)--(-1,0)--cycle,dotted);draw(arc((1/2,1/2),sqrt(2)/2,135, 315),dotted);for(int i=0;i<4;++i){draw(rotate(i*90,(0,0))*arc((1/2,1/2),sqrt(2)/2,-45,135)); dot(rotate(i*90,(0,0))*(1/2,1/2));}[/asy] Because of symmetry, the area is the same in all four quadrants.
The answer is $\boxed{2}$ | B | 2 |
431abc337c350c25806ed4ebe14d245f | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12B_Problems/Problem_20 | A set of teams held a round-robin tournament in which every team played every other team exactly once. Every team won $10$ games and lost $10$ games; there were no ties. How many sets of three teams $\{A, B, C\}$ were there in which $A$ beat $B$ $B$ beat $C$ , and $C$ beat $A?$
$\textbf{(A)}\ 385 \qquad \textbf{(B)}\ 665 \qquad \textbf{(C)}\ 945 \qquad \textbf{(D)}\ 1140 \qquad \textbf{(E)}\ 1330$ | We use complementary counting. First, because each team played $20$ other teams, there are $21$ teams total. All sets that do not have $A$ beat $B$ $B$ beat $C$ , and $C$ beat $A$ have one team that beats both the other teams. Thus we must count the number of sets of three teams such that one team beats the two other teams and subtract that number from the total number of ways to choose three teams.
There are $21$ ways to choose the team that beat the two other teams, and $\binom{10}{2} = 45$ to choose two teams that the first team both beat. This is $21 * 45 = 945$ sets. There are $\binom{21}{3} = 1330$ sets of three teams total. Subtracting, we obtain $1330 - 945 = \boxed{385})$ is our answer. | A | 385 |
431abc337c350c25806ed4ebe14d245f | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12B_Problems/Problem_20 | A set of teams held a round-robin tournament in which every team played every other team exactly once. Every team won $10$ games and lost $10$ games; there were no ties. How many sets of three teams $\{A, B, C\}$ were there in which $A$ beat $B$ $B$ beat $C$ , and $C$ beat $A?$
$\textbf{(A)}\ 385 \qquad \textbf{(B)}\ 665 \qquad \textbf{(C)}\ 945 \qquad \textbf{(D)}\ 1140 \qquad \textbf{(E)}\ 1330$ | As above, note that there are 21 teams, and call them A, B, C, ... T, U. WLOG, assume that A beat teams B-L and lost to teams M-U. We will count the number of sets satisfying the “cycle-win” condition—e.g. here, A beats a team in X which beats a team in Y which beats A. The first and third part of the condition are already met by our wlog, so we just need to count of number of ways the second condition is true (a team in X beats a team in Y). These are the number of cycle-wins that include A, then multiply by 21 (for each team) and divide by 3 (since every set will be counted by each of the 3 teams that are a part of that set).
To do this, let X $=\{B, ..., L\}$ and Y $=\{M, ..., U\}$ . Since a total of $10*10=100$ losses total were suffered by teams in Y and $\binom{10}{2}=45^{*}$ losses were suffered by teams in Y from teams in Y, we have $100-45=55$ losses suffered by teams in Y from teams in X. Hence, for each of these $55$ losses, there is exactly one set of three teams that includes A that satisfies the problem conditions. Thus, the answer is $\frac{55\cdot 21}{3}=\boxed{385}$ | null | 385 |
431abc337c350c25806ed4ebe14d245f | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12B_Problems/Problem_20 | A set of teams held a round-robin tournament in which every team played every other team exactly once. Every team won $10$ games and lost $10$ games; there were no ties. How many sets of three teams $\{A, B, C\}$ were there in which $A$ beat $B$ $B$ beat $C$ , and $C$ beat $A?$
$\textbf{(A)}\ 385 \qquad \textbf{(B)}\ 665 \qquad \textbf{(C)}\ 945 \qquad \textbf{(D)}\ 1140 \qquad \textbf{(E)}\ 1330$ | Note that there are $21$ teams total and $\binom{21}{3}=1330$ ways to pick ${A,B,C}.$ The possible arrangements are one team beats the other two or they each win/lose equally (we want the second case). Approximately $\frac{1}{4}$ of all the arrangements satisfy the second case, and $\frac{1330}{4}=332.5,$ which is by far the closest to $\boxed{385}.$ | A | 385 |
841ab39d1d1a8feee96736635a704133 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12B_Problems/Problem_22 | For a certain positive integer $n$ less than $1000$ , the decimal equivalent of $\frac{1}{n}$ is $0.\overline{abcdef}$ , a repeating decimal of period of $6$ , and the decimal equivalent of $\frac{1}{n+6}$ is $0.\overline{wxyz}$ , a repeating decimal of period $4$ . In which interval does $n$ lie?
$\textbf{(A)}\ [1,200]\qquad\textbf{(B)}\ [201,400]\qquad\textbf{(C)}\ [401,600]\qquad\textbf{(D)}\ [601,800]\qquad\textbf{(E)}\ [801,999]$ | Solution by e_power_pi_times_i
If $\frac{1}{n} = 0.\overline{abcdef}$ $n$ must be a factor of $999999$ . Also, by the same procedure, $n+6$ must be a factor of $9999$ . Checking through all the factors of $999999$ and $9999$ that are less than $1000$ , we see that $n = 297$ is a solution, so the answer is $\boxed{201,400}$ | B | 201,400 |
841ab39d1d1a8feee96736635a704133 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12B_Problems/Problem_22 | For a certain positive integer $n$ less than $1000$ , the decimal equivalent of $\frac{1}{n}$ is $0.\overline{abcdef}$ , a repeating decimal of period of $6$ , and the decimal equivalent of $\frac{1}{n+6}$ is $0.\overline{wxyz}$ , a repeating decimal of period $4$ . In which interval does $n$ lie?
$\textbf{(A)}\ [1,200]\qquad\textbf{(B)}\ [201,400]\qquad\textbf{(C)}\ [401,600]\qquad\textbf{(D)}\ [601,800]\qquad\textbf{(E)}\ [801,999]$ | Notice that the repeating fraction $0.\overline{abcdef}$ can be represented as $\frac{abcdef}{999999},$ and thereby, $n|999999.$ Also, notice that $0.\overline{wxyz} = \frac{wxyz}{9999},$ so $(n+6)|9999.$ However, we have to make some restrictions here. For instance, if $n|99999,$ then $\frac{1}{n}$ could be expressed as $\frac{a’b’c’d’e’}{99999} = .\overline{a’b’c’d’e’}$ which cannot happen. Therefore, from this, we see that the smallest $m$ such that $n|\underbrace{999\cdots999}_{m \text{ nines}}$ is $m = 6.$ Also, the smallest number $m$ such that $(n+6)|\underbrace{999\cdots999}_{m \text{ nines}}$ is $m = 4$ by similar reasoning.
Proceeding, we can factorize $9999 = 99 \times 101,$ after which we see that $n+6$ must contain a prime factor of $101$ as it cannot divide $99$ but must divide $9999.$ However, $101$ is prime, so $101|(n+6)$ ! Looking at the answer choices, all of the intervals are less than $1000,$ so we know that (the minimum value of) $n+6$ is thereby either $101, 101 \times 3,$ or $101 \times 9.$ Testing, we see that $n+6 = 303$ gives $n = 297 = 3^3 \times 11,$ which in fact is a divisor of $999 \times 1001$ while not being a divisor of $999.$ Therefore, the answer is $\boxed{201,400}.$ | B | 201,400 |
342e3a8c9f252ac800c2e0321b7cd7a2 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12B_Problems/Problem_24 | There are exactly $77,000$ ordered quadruplets $(a, b, c, d)$ such that $\gcd(a, b, c, d) = 77$ and $\operatorname{lcm}(a, b, c, d) = n$ . What is the smallest possible value for $n$
$\textbf{(A)}\ 13,860\qquad\textbf{(B)}\ 20,790\qquad\textbf{(C)}\ 21,560 \qquad\textbf{(D)}\ 27,720 \qquad\textbf{(E)}\ 41,580$ | Let $A=\frac{a}{77},\ B=\frac{b}{77}$ , etc., so that $\gcd(A,B,C,D)=1$ . Then for each prime power $p^k$ in the prime factorization of $N=\frac{n}{77}$ , at least one of the prime factorizations of $(A,B,C,D)$ has $p^k$ , at least one has $p^0$ , and all must have $p^m$ with $0\le m\le k$
Let $f(k)$ be the number of ordered quadruplets of integers $(m_1,m_2,m_3,m_4)$ such that $0\le m_i\le k$ for all $i$ , the largest is $k$ , and the smallest is $0$ . Then for the prime factorization $N=2^{k_2}3^{k_3}5^{k_5}\ldots$ we must have $77000=f(k_2)f(k_3)f(k_5)\ldots$ So let's take a look at the function $f(k)$ by counting the quadruplets we just mentioned.
There are $14$ quadruplets which consist only of $0$ and $k$ . Then there are $36(k-1)$ quadruplets which include three different values, and $12(k-1)(k-2)$ with four. Thus $f(k)=14+12(k-1)(3+k-2)=14+12(k^2-1)$ and the first few values from $k=1$ onwards are \[14,50,110,194,302,434,590,770,\ldots\] Straight away we notice that $14\cdot 50\cdot 110=77000$ , so the prime factorization of $N$ can use the exponents $1,2,3$ . To make it as small as possible, assign the larger exponents to smaller primes. The result is $N=2^33^25^1=360$ , so $n=360\cdot 77=27720$ which is answer $\boxed{27,720}$ | D | 27,720 |
d87f677f4af119431ff81402e87f5aaf | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12B_Problems/Problem_25 | The sequence $(a_n)$ is defined recursively by $a_0=1$ $a_1=\sqrt[19]{2}$ , and $a_n=a_{n-1}a_{n-2}^2$ for $n\geq 2$ . What is the smallest positive integer $k$ such that the product $a_1a_2\cdots a_k$ is an integer?
$\textbf{(A)}\ 17\qquad\textbf{(B)}\ 18\qquad\textbf{(C)}\ 19\qquad\textbf{(D)}\ 20\qquad\textbf{(E)}\ 21$ | Let $b_i=19\text{log}_2a_i$ . Then $b_0=0, b_1=1,$ and $b_n=b_{n-1}+2b_{n-2}$ for all $n\geq 2$ . The characteristic polynomial of this linear recurrence is $x^2-x-2=0$ , which has roots $2$ and $-1$
Therefore, $b_n=k_12^{n}+k_2(-1)^n$ for constants to be determined $k_1, k_2$ . Using the fact that $b_0=0, b_1=1,$ we can solve a pair of linear equations for $k_1, k_2$
$k_1+k_2=0$ $2k_1-k_2=1$
Thus $k_1=\frac{1}{3}$ $k_2=-\frac{1}{3}$ , and $b_n=\frac{2^n-(-1)^n}{3}$
Now, $a_1a_2\cdots a_k=2^{\frac{(b_1+b_2+\cdots+b_k)}{19}}$ , so we are looking for the least value of $k$ so that
$b_1+b_2+\cdots+b_k \equiv 0 \pmod{19}$
Note that we can multiply all $b_i$ by three for convenience, as the $b_i$ are always integers, and it does not affect divisibility by $19$
Now, for all even $k$ the sum (adjusted by a factor of three) is $2^1+2^2+\cdots+2^k=2^{k+1}-2$ . The smallest $k$ for which this is a multiple of $19$ is $k=18$ by Fermat's Little Theorem, as it is seen with further testing that $2$ is a primitive root $\pmod{19}$
Now, assume $k$ is odd. Then the sum (again adjusted by a factor of three) is $2^1+2^2+\cdots+2^k+1=2^{k+1}-1$ . The smallest $k$ for which this is a multiple of $19$ is $k=17$ , by the same reasons. Thus, the minimal value of $k$ is $\boxed{17}$ | A | 17 |
d87f677f4af119431ff81402e87f5aaf | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12B_Problems/Problem_25 | The sequence $(a_n)$ is defined recursively by $a_0=1$ $a_1=\sqrt[19]{2}$ , and $a_n=a_{n-1}a_{n-2}^2$ for $n\geq 2$ . What is the smallest positive integer $k$ such that the product $a_1a_2\cdots a_k$ is an integer?
$\textbf{(A)}\ 17\qquad\textbf{(B)}\ 18\qquad\textbf{(C)}\ 19\qquad\textbf{(D)}\ 20\qquad\textbf{(E)}\ 21$ | Since the product $a_1a_2\cdots a_k$ is an integer, it must be a power of $2$ , so the sum of the base- $2$ logarithms must be an integer. Multiply all of these logarithms by $19$ (to make them integers), so the sum must be a multiple of $19$
The logarithms are $b_n = 19\log_2 a_n$ . Using the recursion $b_0 = 0, b_1 = 1, b_n = b_{n-1}+2b_{n-2}$ (modulo $19$ to save calculation time), we get the sequence \[0,1,1,3,5,11,2,5,9,0,-1,-1,-3,-5,-11,-2,-5,-9,0,\dots\] Listing the numbers out is expedited if you notice $b_{n+1}=2b_n+(-1)^n$
The cycle repeats every $9+9=18$ terms. Notice that since $b_n+b_{n+9} \equiv 0 \pmod{19}$ , the first $18$ terms sum up to a multiple of $19$ . Since $b_{18}=0$ , we only need at most the first $\boxed{17}$ terms to sum up to a multiple of $19$ , and this is the lowest answer choice. | A | 17 |
d87f677f4af119431ff81402e87f5aaf | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12B_Problems/Problem_25 | The sequence $(a_n)$ is defined recursively by $a_0=1$ $a_1=\sqrt[19]{2}$ , and $a_n=a_{n-1}a_{n-2}^2$ for $n\geq 2$ . What is the smallest positive integer $k$ such that the product $a_1a_2\cdots a_k$ is an integer?
$\textbf{(A)}\ 17\qquad\textbf{(B)}\ 18\qquad\textbf{(C)}\ 19\qquad\textbf{(D)}\ 20\qquad\textbf{(E)}\ 21$ | Like in Solution 2 , calculate the first few terms of the sequence, but also keep a running sum $c_n$ of the logarithms (not modulo $19$ here): \[0,1,2,5,10,21,42,\dots\] Notice that $c_n=2c_{n-1}+1$ for odd $n$ and $c_n=2c_{n-1}$ for even $n$ . Since $2$ is relatively prime to $19$ , we can ignore even $n$ and calculate odd $n$ using $c_1 = 1, c_{n} = 4c_{n-2}+1$ (modulo $19$ ): \[,1,,5,,2,,9,,-1,,-3,,8,,-5,,0,\dots\] $c_n$ is first a multiple of $19$ at $n = \boxed{17}$ . ~ emerald_block | A | 17 |
844a686676a431172327f38cf209d205 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_12A_Problems/Problem_2 | Two of the three sides of a triangle are 20 and 15. Which of the following numbers is not a possible perimeter of the triangle?
$\textbf{(A)}\ 52\qquad\textbf{(B)}\ 57\qquad\textbf{(C)}\ 62\qquad\textbf{(D)}\ 67\qquad\textbf{(E)}\ 72$ | Letting $x$ be the third side, then by the triangle inequality, $20-15 < x < 20+15$ , or $5 < x < 35$ . Therefore the perimeter must be greater than 40 but less than 70. 72 is not in this range, so $\boxed{72}$ is our answer. | E | 72 |
8e2a2516d416f1a341693d2b0dea3b9c | https://artofproblemsolving.com/wiki/index.php/2015_AMC_12A_Problems/Problem_3 | Mr. Patrick teaches math to $15$ students. He was grading tests and found that when he graded everyone's test except Payton's, the average grade for the class was $80$ . After he graded Payton's test, the test average became $81$ . What was Payton's score on the test?
$\textbf{(A)}\ 81\qquad\textbf{(B)}\ 85\qquad\textbf{(C)}\ 91\qquad\textbf{(D)}\ 94\qquad\textbf{(E)}\ 95$ | If the average of the first $14$ peoples' scores was $80$ , then the sum of all of their tests is $14 \cdot 80 = 1120$ . When Payton's score was added, the sum of all of the scores became $15 \cdot 81 = 1215$ . So, Payton's score must be $1215-1120 = \boxed{95}$ | E | 95 |
8e2a2516d416f1a341693d2b0dea3b9c | https://artofproblemsolving.com/wiki/index.php/2015_AMC_12A_Problems/Problem_3 | Mr. Patrick teaches math to $15$ students. He was grading tests and found that when he graded everyone's test except Payton's, the average grade for the class was $80$ . After he graded Payton's test, the test average became $81$ . What was Payton's score on the test?
$\textbf{(A)}\ 81\qquad\textbf{(B)}\ 85\qquad\textbf{(C)}\ 91\qquad\textbf{(D)}\ 94\qquad\textbf{(E)}\ 95$ | The average of a set of numbers is the value we get if we evenly distribute the total across all entries. So assume that the first $14$ students each scored $80$ . If Payton also scored an $80$ , the average would still be $80$ . In order to increase the overall average to $81$ , we need to add one more point to all of the scores, including Payton's. This means we need to add a total of $15$ more points, so Payton needs $80+15 = \boxed{95}$ | E | 95 |
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