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bf6f25525086c20ce831a91d88ba4f1d | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_23 | Ajay is standing at point $A$ near Pontianak, Indonesia, $0^\circ$ latitude and $110^\circ \text{ E}$ longitude. Billy is standing at point $B$ near Big Baldy Mountain, Idaho, USA, $45^\circ \text{ N}$ latitude and $115^\circ \text{ W}$ longitude. Assume that Earth is a perfect sphere with center $C.$ What is the degree measure of $\angle ACB?$
$\textbf{(A) }105 \qquad \textbf{(B) }112\frac{1}{2} \qquad \textbf{(C) }120 \qquad \textbf{(D) }135 \qquad \textbf{(E) }150 \qquad$ | This solution refers to the Diagram section.
Let $D$ be the orthogonal projection of $B$ onto the equator. Note that $\angle BDA = \angle BDC = 90^\circ$ and $\angle BCD = 45^\circ.$ Recall that $115^\circ \text{ W}$ longitude is the same as $245^\circ \text{ E}$ longitude, so $\angle ACD=135^\circ.$
Without the loss of generality, let $AC=BC=1.$ As shown below, we place Earth in the $xyz$ -plane with $C=(0,0,0)$ such that the positive $x$ -axis runs through $A,$ the positive $y$ -axis runs through $0^\circ$ latitude and $160^\circ \text{ W}$ longitude, and the positive $z$ -axis runs through the North Pole. [asy] /* Made by MRENTHUSIASM */ size(300); import graph3; import solids; currentprojection=orthographic((0.2,-0.5,0.2)); triple A, B, C, D; A = (1,0,0); B = (-1/2,1/2,sqrt(2)/2); C = (0,0,0); D = (-1/2,1/2,0); draw(unitsphere,white,light=White); dot(A^^B^^C^^D,linewidth(4.5)); draw(Circle(C,1,(0,0,1))^^B--C--D--cycle); label("$A$",A,5*dir((2.5,-3,0))); label("$B$",B,3*dir(B)); label("$C$",C,1.5*(1,0,-1)); label("$D$",D,3*(-1/2,-1/2,0)); draw((-1.5,0,0)--(1.5,0,0),linewidth(1.25),EndArrow3(10)); draw((0,-1.5,0)--(0,1.5,0),linewidth(1.25),EndArrow3(10)); draw((0,0,-1.5)--(0,0,1.5),linewidth(1.25),EndArrow3(10)); label("$x$",(1.5,0,0),2*dir((1.5,0,0))); label("$y$",(0,1.5,0),3*dir((0,1.5,0))); label("$z$",(0,0,1.5),2*dir((0,0,1.5))); [/asy] It follows that $A=(1,0,0)$ and $D=(-t,t,0)$ for some positive number $t.$ Since $\triangle BCD$ is an isosceles right triangle, we have $B=\left(-t,t,\sqrt{2}t\right).$ By the Distance Formula, we get $(-t)^2+t^2+\left(\sqrt{2}t\right)^2=1,$ from which $t=\frac12.$
As $\vec{A} = \begin{pmatrix}1 \\ 0 \\ 0 \end{pmatrix}$ and $\vec{B} = \begin{pmatrix}-1/2 \\ 1/2 \\ \sqrt2/2 \end{pmatrix},$ we obtain \[\cos\angle ACB=\frac{\vec{A}\bullet\vec{B}}{\left\lVert\vec{A}\right\rVert\left\lVert\vec{B}\right\rVert}=-\frac12\] by the dot product, so $\angle ACB=\boxed{120}$ degrees. | C | 120 |
bf6f25525086c20ce831a91d88ba4f1d | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_23 | Ajay is standing at point $A$ near Pontianak, Indonesia, $0^\circ$ latitude and $110^\circ \text{ E}$ longitude. Billy is standing at point $B$ near Big Baldy Mountain, Idaho, USA, $45^\circ \text{ N}$ latitude and $115^\circ \text{ W}$ longitude. Assume that Earth is a perfect sphere with center $C.$ What is the degree measure of $\angle ACB?$
$\textbf{(A) }105 \qquad \textbf{(B) }112\frac{1}{2} \qquad \textbf{(C) }120 \qquad \textbf{(D) }135 \qquad \textbf{(E) }150 \qquad$ | This solution refers to the diagram in Solution 2.
In spherical coordinates $(\rho,\theta,\phi),$ note that $\rho,\theta,$ and $\phi$ represent the radial distance, the polar angle, and the azimuthal angle, respectively.
Without the loss of generality, let $AC=BC=1.$ As shown in Solution 2, we place Earth in the $xyz$ -plane with origin $C$ such that the positive $x$ -axis runs through $A,$ the positive $y$ -axis runs through $0^\circ$ latitude and $160^\circ \text{ W}$ longitude, and the positive $z$ -axis runs through the North Pole.
In spherical coordinates, we have $A=(1,90^\circ,0^\circ)$ and $B=(1,45^\circ,135^\circ).$ Now, we express $\vec{A}$ and $\vec{B}$ in Cartesian coordinates: \[\vec{A} = \begin{pmatrix}\sin90^\circ \cos0^\circ \\ \sin90^\circ \sin0^\circ \\ \cos90^\circ \end{pmatrix} = \begin{pmatrix}1 \\ 0 \\ 0 \end{pmatrix} \text{ and } \vec{B} = \begin{pmatrix}\sin45^\circ \cos135^\circ \\ \sin45^\circ \sin135^\circ \\ \cos45^\circ \end{pmatrix} = \begin{pmatrix}-1/2 \\ 1/2 \\ \sqrt2/2 \end{pmatrix}.\] We continue with the last paragraph of Solution 2 to get the answer $\angle ACB=\boxed{120}$ degrees. | C | 120 |
23ab6d70a6c9c1da08c51991b78a9525 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_24 | Let $\lfloor x \rfloor$ denote the greatest integer less than or equal to $x$ . How many real numbers $x$ satisfy the equation $x^2 + 10,000\lfloor x \rfloor = 10,000x$
$\textbf{(A) } 197 \qquad \textbf{(B) } 198 \qquad \textbf{(C) } 199 \qquad \textbf{(D) } 200 \qquad \textbf{(E) } 201$ | This rewrites itself to $x^2=10,000\{x\}$ where $\lfloor x \rfloor + \{x\} = x$
Graphing $y=10,000\{x\}$ and $y=x^2$ we see that the former is a set of line segments with slope $10,000$ from $0$ to $1$ with a hole at $x=1$ , then $1$ to $2$ with a hole at $x=2$ etc.
Here is a graph of $y=x^2$ and $y=16\{x\}$ for visualization.
[asy] import graph; size(400); xaxis("$x$",Ticks(Label(fontsize(8pt)),new real[]{-5,-4,-3, -2, -1,0,1 2,3, 4,5})); yaxis("$y$",Ticks(Label(fontsize(8pt)),new real[]{0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18})); real y(real x) {return x^2;} draw(circle((-4,16), 0.1)); draw(circle((-3,16), 0.1)); draw(circle((-2,16), 0.1)); draw(circle((-1,16), 0.1)); draw(circle((0,16), 0.1)); draw(circle((1,16), 0.1)); draw(circle((2,16), 0.1)); draw(circle((3,16), 0.1)); draw(circle((4,16), 0.1)); draw((-5,0)--(-4,16), black); draw((-4,0)--(-3,16), black); draw((-3,0)--(-2,16), black); draw((-2,0)--(-1,16), black); draw((-1,0)--(-0,16), black); draw((0,0)--(1,16), black); draw((1,0)--(2,16), black); draw((2,0)--(3,16), black); draw((3,0)--(4,16), black); draw(graph(y,-4.2,4.2),green); [/asy]
Now notice that when $x=\pm 100$ the graph has a hole at $(\pm 100,10,000)$ which the equation $y=x^2$ passes through and then continues upwards. Thus our set of possible solutions is bounded by $(-100,100)$ . We can see that $y=x^2$ intersects each of the lines once and there are $99-(-99)+1=199$ lines for an answer of $\boxed{199}$ | C | 199 |
23ab6d70a6c9c1da08c51991b78a9525 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_24 | Let $\lfloor x \rfloor$ denote the greatest integer less than or equal to $x$ . How many real numbers $x$ satisfy the equation $x^2 + 10,000\lfloor x \rfloor = 10,000x$
$\textbf{(A) } 197 \qquad \textbf{(B) } 198 \qquad \textbf{(C) } 199 \qquad \textbf{(D) } 200 \qquad \textbf{(E) } 201$ | Same as the first solution, $x^2=10,000\{x\}$
We can write $x$ as $\lfloor x \rfloor+\{x\}$ . Expanding everything, we get a quadratic in $\{x\}$ in terms of $\lfloor x \rfloor$ \[\{x\}^2+ (2\lfloor x \rfloor -10,000)\{x\} + \lfloor x \rfloor ^2 = 0\]
We use the quadratic formula to solve for $\{x\}$ \[\{x\} = \frac {-2\lfloor x \rfloor + 10,000 \pm \sqrt{ ( 2\lfloor x \rfloor - 10,000 )^2 - 4\lfloor x \rfloor^2 }}{2} = \frac {-2\lfloor x \rfloor + 10,000 \pm \sqrt{ 4\lfloor x \rfloor^2 -40,000 \lfloor x \rfloor + 10,000^2- 4\lfloor x \rfloor^2 }}{2}\]
Since $0 \leq \{x\} < 1$ , we get an inequality which we can then solve. After simplifying a lot, we get that $\lfloor x \rfloor^2 + 2\lfloor x \rfloor - 9999 < 0$
Solving over the integers, $-101 < \lfloor x \rfloor < 99$ , and since $\lfloor x \rfloor$ is an integer, there are $\boxed{199}$ solutions. Each value of $\lfloor x \rfloor$ should correspond to one value of $x$ , so we are done. | C | 199 |
23ab6d70a6c9c1da08c51991b78a9525 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_24 | Let $\lfloor x \rfloor$ denote the greatest integer less than or equal to $x$ . How many real numbers $x$ satisfy the equation $x^2 + 10,000\lfloor x \rfloor = 10,000x$
$\textbf{(A) } 197 \qquad \textbf{(B) } 198 \qquad \textbf{(C) } 199 \qquad \textbf{(D) } 200 \qquad \textbf{(E) } 201$ | Let $x = a+k$ where $a$ is the integer part of $x$ and $k$ is the fractional part of $x$ .
We can then rewrite the problem below:
$(a+k)^2 + 10000a = 10000(a+k)$
From here, we get
$(a+k)^2 + 10000a = 10000a + 10000k$
Solving for $a+k = x$
$(a+k)^2 = 10000k$
$x = a+k = \pm100\sqrt{k}$
Because $0 \leq k < 1$ , we know that $a+k$ cannot be less than or equal to $-100$ nor greater than or equal to $100$ . Therefore:
$-99 \leq x \leq 99$
There are $199$ elements in this range, so the answer is $\boxed{199}$ | C | 199 |
23ab6d70a6c9c1da08c51991b78a9525 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_24 | Let $\lfloor x \rfloor$ denote the greatest integer less than or equal to $x$ . How many real numbers $x$ satisfy the equation $x^2 + 10,000\lfloor x \rfloor = 10,000x$
$\textbf{(A) } 197 \qquad \textbf{(B) } 198 \qquad \textbf{(C) } 199 \qquad \textbf{(D) } 200 \qquad \textbf{(E) } 201$ | Notice the given equation is equivalent to $(\lfloor x \rfloor+\{x\})^2=10,000\{x\}$
Now we know that $\{x\} < 1$ so plugging in $1$ for $\{x\}$ we can find the upper and lower bounds for the values.
$(\lfloor x \rfloor +1)^2 = 10,000(1)$
$(\lfloor x \rfloor +1) = \pm 100$
$\lfloor x \rfloor = 99, -101$
And just like $\textbf{Solution 2}$ , we see that $-101 < \lfloor x \rfloor < 99$ , and since $\lfloor x \rfloor$ is an integer, there are $\boxed{199}$ solutions. Each value of $\lfloor x \rfloor$ should correspond to one value of $x$ , so we are done. | C | 199 |
23ab6d70a6c9c1da08c51991b78a9525 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_24 | Let $\lfloor x \rfloor$ denote the greatest integer less than or equal to $x$ . How many real numbers $x$ satisfy the equation $x^2 + 10,000\lfloor x \rfloor = 10,000x$
$\textbf{(A) } 197 \qquad \textbf{(B) } 198 \qquad \textbf{(C) } 199 \qquad \textbf{(D) } 200 \qquad \textbf{(E) } 201$ | Firstly, if $x$ is an integer, then $10,000\lfloor x \rfloor=10,000x$ , so $x$ must be $0$
If $0<x<1$ , then we know the following:
$0<x^2<1$
$10,000\lfloor x \rfloor =0$
$0<10,000x<10,000$
Therefore, $0<x^2+10,000\lfloor x \rfloor <1$ , which overlaps with $0<10,000x<10,000$ . This means that there is at least one real solution between $0$ and $1$ . Since $x^2+10,000\lfloor x \rfloor$ increases quadratically and $10,000x$ increases linearly, there is only one solution for this case.
Similarly, if $1<x<2$ , then we know the following:
$1<x^2<4$
$10,000\lfloor x \rfloor =10,000$
$<10,000<10,000x<20,000$
By following similar logic, we can find that there is one solution between $1$ ad $2$
We can also follow the same process to find that there are negative solutions for $x$ as well.
There are not an infinite amount of solutions, so at one point there will be no solutions when $n<x<n+1$ for some integer $n$ . For there to be no solutions in a given range means that the range of $10,000\lfloor x \rfloor + x^2$ does not intersect the range of $10,000x$ $x^2$ will always be positive, and $10,000\lfloor x \rfloor$ is less than $10,000$ less than $10,000x$ , so when $x^2 >= 10,000$ , the equation will have no solutions. This means that there are $99$ positive solutions, $99$ negative solutions, and $0$ for a total of $\boxed{199}$ solutions. | C | 199 |
23ab6d70a6c9c1da08c51991b78a9525 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_24 | Let $\lfloor x \rfloor$ denote the greatest integer less than or equal to $x$ . How many real numbers $x$ satisfy the equation $x^2 + 10,000\lfloor x \rfloor = 10,000x$
$\textbf{(A) } 197 \qquad \textbf{(B) } 198 \qquad \textbf{(C) } 199 \qquad \textbf{(D) } 200 \qquad \textbf{(E) } 201$ | General solution to this type of equation $f(x, \lfloor x \rfloor) = 0$
$x^2 - 10000x + 10000 \lfloor x \rfloor =0$
$x=5000 \pm 100 \sqrt{2500- \lfloor x \rfloor}$ $\lfloor x \rfloor \le 2500$
$\lfloor x \rfloor \le x < \lfloor x \rfloor + 1$
If $x= 5000 + 100 \sqrt{2500 - \lfloor x \rfloor}$ $x \ge 5000$ , it contradicts $x < \lfloor x \rfloor + 1 \le 2501$
So $x= 5000 - 100 \sqrt{2500 - \lfloor x \rfloor}$
Let $k = \lfloor x \rfloor$ $x= 5000 - 100 \sqrt{2500 - k}$
$k \le 5000 - 100 \sqrt{2500 - k} < k + 1$
$0 \le 5000 - k - 100 \sqrt{2500 - k} < 1$
$0 \le 2500 - k - 100 \sqrt{2500 - k} + 2500 < 1$
$0 \le (\sqrt{2500 - k} - 50)^2 < 1$
$-1 < \sqrt{2500 - k} - 50 < 1$
$49 < \sqrt{2500 - k} < 51$
$-101 < k < 99$
So the number of $k$ 's values is $99-(-101)-1=199$ . Because $x=5000-100\sqrt{2500-k}$ , for each value of $k$ , there is a value for $x$ . The answer is $\boxed{199}$ | C | 199 |
23ab6d70a6c9c1da08c51991b78a9525 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_24 | Let $\lfloor x \rfloor$ denote the greatest integer less than or equal to $x$ . How many real numbers $x$ satisfy the equation $x^2 + 10,000\lfloor x \rfloor = 10,000x$
$\textbf{(A) } 197 \qquad \textbf{(B) } 198 \qquad \textbf{(C) } 199 \qquad \textbf{(D) } 200 \qquad \textbf{(E) } 201$ | Subtracting $10000\lfloor x\rfloor$ from both sides gives $x^2=10000(x-\lfloor x\rfloor)=10000\{x\}$ . Dividing both sides by $10000$ gives $\left(\frac{x}{100}\right)^2=\{x\}<1$ $\left(\frac{x}{100}\right)^2<1$ when $-100<x<100$ so the answer is $\boxed{199}$ | null | 199 |
c17fd2d74a62e11ae945fa8c7c30ffab | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_25 | Circles $\omega_1$ $\omega_2$ , and $\omega_3$ each have radius $4$ and are placed in the plane so that each circle is externally tangent to the other two. Points $P_1$ $P_2$ , and $P_3$ lie on $\omega_1$ $\omega_2$ , and $\omega_3$ respectively such that $P_1P_2=P_2P_3=P_3P_1$ and line $P_iP_{i+1}$ is tangent to $\omega_i$ for each $i=1,2,3$ , where $P_4 = P_1$ . See the figure below. The area of $\triangle P_1P_2P_3$ can be written in the form $\sqrt{a}+\sqrt{b}$ for positive integers $a$ and $b$ . What is $a+b$
[asy] unitsize(12); pair A = (0, 8/sqrt(3)), B = rotate(-120)*A, C = rotate(120)*A; real theta = 41.5; pair P1 = rotate(theta)*(2+2*sqrt(7/3), 0), P2 = rotate(-120)*P1, P3 = rotate(120)*P1; filldraw(P1--P2--P3--cycle, gray(0.9)); draw(Circle(A, 4)); draw(Circle(B, 4)); draw(Circle(C, 4)); dot(P1); dot(P2); dot(P3); defaultpen(fontsize(10pt)); label("$P_1$", P1, E*1.5); label("$P_2$", P2, SW*1.5); label("$P_3$", P3, N); label("$\omega_1$", A, W*17); label("$\omega_2$", B, E*17); label("$\omega_3$", C, W*17); [/asy]
$\textbf{(A) }546\qquad\textbf{(B) }548\qquad\textbf{(C) }550\qquad\textbf{(D) }552\qquad\textbf{(E) }554$ | Let $O_1$ and $O_2$ be the centers of $\omega_1$ and $\omega_2$ respectively and draw $O_1O_2$ $O_1P_1$ , and $O_2P_2$ . Note that $\angle{O_1P_1P_2}$ and $\angle{O_2P_2P_3}$ are both right. Furthermore, since $\triangle{P_1P_2P_3}$ is equilateral, $m\angle{P_1P_2P_3} = 60^\circ$ and $m\angle{O_2P_2P_1} = 30^\circ$ . Mark $M$ as the base of the altitude from $O_2$ to $P_1P_2.$ Since $\triangle P_2O_2M$ is a 30-60-90 triangle, $O_2M = 2$ and $P_2M = 2\sqrt{3}$ . Also, since $O_1O_2 = 8$ and $O_1P_1 = 4$ , we can find $P_1M = \sqrt{8^2 - (4 + 2)^2} = 2\sqrt{7}$ . Thus, $P_1P_2 = P_1M + P_2M = 2\sqrt{3} + 2\sqrt{7}$ . This makes \[\left[P_1P_2P_3\right] = \frac{\sqrt 3}4\cdot\left(2\sqrt 3 + 2\sqrt 7\right)^2 = 10\sqrt 3 + 6\sqrt 7 = \sqrt{300} + \sqrt{252}.\] So, our answer is $252 + 300 = \boxed{552}$ | D | 552 |
c17fd2d74a62e11ae945fa8c7c30ffab | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_25 | Circles $\omega_1$ $\omega_2$ , and $\omega_3$ each have radius $4$ and are placed in the plane so that each circle is externally tangent to the other two. Points $P_1$ $P_2$ , and $P_3$ lie on $\omega_1$ $\omega_2$ , and $\omega_3$ respectively such that $P_1P_2=P_2P_3=P_3P_1$ and line $P_iP_{i+1}$ is tangent to $\omega_i$ for each $i=1,2,3$ , where $P_4 = P_1$ . See the figure below. The area of $\triangle P_1P_2P_3$ can be written in the form $\sqrt{a}+\sqrt{b}$ for positive integers $a$ and $b$ . What is $a+b$
[asy] unitsize(12); pair A = (0, 8/sqrt(3)), B = rotate(-120)*A, C = rotate(120)*A; real theta = 41.5; pair P1 = rotate(theta)*(2+2*sqrt(7/3), 0), P2 = rotate(-120)*P1, P3 = rotate(120)*P1; filldraw(P1--P2--P3--cycle, gray(0.9)); draw(Circle(A, 4)); draw(Circle(B, 4)); draw(Circle(C, 4)); dot(P1); dot(P2); dot(P3); defaultpen(fontsize(10pt)); label("$P_1$", P1, E*1.5); label("$P_2$", P2, SW*1.5); label("$P_3$", P3, N); label("$\omega_1$", A, W*17); label("$\omega_2$", B, E*17); label("$\omega_3$", C, W*17); [/asy]
$\textbf{(A) }546\qquad\textbf{(B) }548\qquad\textbf{(C) }550\qquad\textbf{(D) }552\qquad\textbf{(E) }554$ | Let $O_i$ be the center of circle $\omega_i$ for $i=1,2,3$ , and let $K$ be the intersection of lines $O_1P_1$ and $O_2P_2$ . Because $\angle P_1P_2P_3 = 60^\circ$ , it follows that $\triangle P_2KP_1$ is a $30-60-90$ triangle. Let $x=P_1K$ ; then $P_2K = 2x$ and $P_1P_2 = x \sqrt 3$ . The Law of Cosines in $\triangle O_1KO_2$ gives \[8^2 = (x+4)^2 + (2x-4)^2 - 2(x+4)(2x-4)\cos 60^\circ,\] which simplifies to $3x^2 - 12x - 16 = 0$ . The positive solution is $x = 2 + \tfrac23\sqrt{21}$ . Then $P_1P_2 = x\sqrt 3 = 2\sqrt 3 + 2\sqrt 7$ , and so the area of $\triangle P_1P_2P_3$ is \[\frac{\sqrt 3}4\cdot\left(2\sqrt 3 + 2\sqrt 7\right)^2 = 10\sqrt 3 + 6\sqrt 7 = \sqrt{300} + \sqrt{252}.\] The requested sum is $300 + 252 = \boxed{552}$ | null | 552 |
c17fd2d74a62e11ae945fa8c7c30ffab | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_25 | Circles $\omega_1$ $\omega_2$ , and $\omega_3$ each have radius $4$ and are placed in the plane so that each circle is externally tangent to the other two. Points $P_1$ $P_2$ , and $P_3$ lie on $\omega_1$ $\omega_2$ , and $\omega_3$ respectively such that $P_1P_2=P_2P_3=P_3P_1$ and line $P_iP_{i+1}$ is tangent to $\omega_i$ for each $i=1,2,3$ , where $P_4 = P_1$ . See the figure below. The area of $\triangle P_1P_2P_3$ can be written in the form $\sqrt{a}+\sqrt{b}$ for positive integers $a$ and $b$ . What is $a+b$
[asy] unitsize(12); pair A = (0, 8/sqrt(3)), B = rotate(-120)*A, C = rotate(120)*A; real theta = 41.5; pair P1 = rotate(theta)*(2+2*sqrt(7/3), 0), P2 = rotate(-120)*P1, P3 = rotate(120)*P1; filldraw(P1--P2--P3--cycle, gray(0.9)); draw(Circle(A, 4)); draw(Circle(B, 4)); draw(Circle(C, 4)); dot(P1); dot(P2); dot(P3); defaultpen(fontsize(10pt)); label("$P_1$", P1, E*1.5); label("$P_2$", P2, SW*1.5); label("$P_3$", P3, N); label("$\omega_1$", A, W*17); label("$\omega_2$", B, E*17); label("$\omega_3$", C, W*17); [/asy]
$\textbf{(A) }546\qquad\textbf{(B) }548\qquad\textbf{(C) }550\qquad\textbf{(D) }552\qquad\textbf{(E) }554$ | Let $O_i$ be the center of circle $\omega_i$ for $i=1,2,3$ . Let $X$ be the centroid of $\triangle{O_1O_2O_3}$ , which also happens to be the centroid of $\triangle{P_1P_2P_3}$ . Because $m\angle{O_1P_1P_2} = 90^\circ$ and $m\angle{O_1P_1X} = 30^\circ$ $m\angle{O_1P_1X} = 60^\circ$ $O_2M$ is $2/3$ the height of $\triangle{P_1P_2P_3}$ , thus $O_2M$ is $8*\sqrt{3}/3$
Applying cosine law on $\triangle{O_1P_1X}$ , one finds that $P_1X = 2 + 2*\sqrt{21}/3$ . Multiplying by $3/2$ to solve for the height of $\triangle{P_1P_2P_3}$ , one gets $3 + \sqrt{21}$ . Simply multiplying by $2/\sqrt{3}$ and then calculating the equilateral triangle's area, one would get the final result of $\sqrt{300} + \sqrt{252}$
This makes the answer $252 + 300 = 552$ $\boxed{552}$ | D | 552 |
c17fd2d74a62e11ae945fa8c7c30ffab | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_25 | Circles $\omega_1$ $\omega_2$ , and $\omega_3$ each have radius $4$ and are placed in the plane so that each circle is externally tangent to the other two. Points $P_1$ $P_2$ , and $P_3$ lie on $\omega_1$ $\omega_2$ , and $\omega_3$ respectively such that $P_1P_2=P_2P_3=P_3P_1$ and line $P_iP_{i+1}$ is tangent to $\omega_i$ for each $i=1,2,3$ , where $P_4 = P_1$ . See the figure below. The area of $\triangle P_1P_2P_3$ can be written in the form $\sqrt{a}+\sqrt{b}$ for positive integers $a$ and $b$ . What is $a+b$
[asy] unitsize(12); pair A = (0, 8/sqrt(3)), B = rotate(-120)*A, C = rotate(120)*A; real theta = 41.5; pair P1 = rotate(theta)*(2+2*sqrt(7/3), 0), P2 = rotate(-120)*P1, P3 = rotate(120)*P1; filldraw(P1--P2--P3--cycle, gray(0.9)); draw(Circle(A, 4)); draw(Circle(B, 4)); draw(Circle(C, 4)); dot(P1); dot(P2); dot(P3); defaultpen(fontsize(10pt)); label("$P_1$", P1, E*1.5); label("$P_2$", P2, SW*1.5); label("$P_3$", P3, N); label("$\omega_1$", A, W*17); label("$\omega_2$", B, E*17); label("$\omega_3$", C, W*17); [/asy]
$\textbf{(A) }546\qquad\textbf{(B) }548\qquad\textbf{(C) }550\qquad\textbf{(D) }552\qquad\textbf{(E) }554$ | [asy] unitsize(12); pair A = (0, 8/sqrt(3)), B = rotate(-120)*A, C = rotate(120)*A; real theta = 41.5; pair P1 = rotate(theta)*(2+2*sqrt(7/3), 0), P2 = rotate(-120)*P1, P3 = rotate(120)*P1; filldraw(P1--P2--P3--cycle, gray(0.9)); draw(Circle(A, 4)); draw(Circle(B, 4)); draw(Circle(C, 4)); dot(P1); dot(P2); dot(P3); defaultpen(fontsize(10pt)); label("$P_1$", P1, E*1.5); label("$P_2$", P2, SW*1.5); label("$P_3$", P3, N); label("$\omega_1$", A, W*30); label("$\omega_2$", B, E*17); label("$\omega_3$", C, W*17); label("$\Gamma$", A, W*15); draw(Circle(A,2),red); pair X=foot(A,P1,P3); dot(X,blue); draw(A--X,blue); label("$2\sqrt{7}$", X--P3); label("$2\sqrt{3}$",X--P1); label("$X$",X,dir(-80),blue); [/asy]
First, note that because the $\angle P_1=\angle P_2=\angle P_3=\pi/3$ , the arcs inside the shaded equilateral triangle are each $2\pi/3$ . Also, the distances between the centers of any two of the $3$ given circles are each $8$ .
Draw the circle $\Gamma$ concentric with $\omega_1$ with radius $2$ . Because the arc of $\omega_1$ inside said triangle is $2\pi/3$ $\Gamma$ touches $P_1P_3$ , say at a point $X$ . Thus, $P_1P_3$ is a common tangent of $\omega_3$ and $\Gamma$ , and it can be seen from inspection of the given diagram that the line is an common internal tangent.
The length of the common internal tangent segment $XP_3$ of $\Gamma$ and $\omega_3$ is then $\sqrt{8^2-(2+4)^2}=2\sqrt{7}$ , and it is easily seen that $XP_1=4\sin \pi/3=2\sqrt{3}$ .
Because $P_1P_3=2(\sqrt{3}+\sqrt{7})$ , the area of the shaded equilateral triangle is $\sqrt{3}(\sqrt{3}+\sqrt{7})^2=10\sqrt{3}+6\sqrt{7}$ . We get $\sqrt{300}+\sqrt{252}\Rightarrow\boxed{552}.$ | D | 552 |
c17fd2d74a62e11ae945fa8c7c30ffab | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_25 | Circles $\omega_1$ $\omega_2$ , and $\omega_3$ each have radius $4$ and are placed in the plane so that each circle is externally tangent to the other two. Points $P_1$ $P_2$ , and $P_3$ lie on $\omega_1$ $\omega_2$ , and $\omega_3$ respectively such that $P_1P_2=P_2P_3=P_3P_1$ and line $P_iP_{i+1}$ is tangent to $\omega_i$ for each $i=1,2,3$ , where $P_4 = P_1$ . See the figure below. The area of $\triangle P_1P_2P_3$ can be written in the form $\sqrt{a}+\sqrt{b}$ for positive integers $a$ and $b$ . What is $a+b$
[asy] unitsize(12); pair A = (0, 8/sqrt(3)), B = rotate(-120)*A, C = rotate(120)*A; real theta = 41.5; pair P1 = rotate(theta)*(2+2*sqrt(7/3), 0), P2 = rotate(-120)*P1, P3 = rotate(120)*P1; filldraw(P1--P2--P3--cycle, gray(0.9)); draw(Circle(A, 4)); draw(Circle(B, 4)); draw(Circle(C, 4)); dot(P1); dot(P2); dot(P3); defaultpen(fontsize(10pt)); label("$P_1$", P1, E*1.5); label("$P_2$", P2, SW*1.5); label("$P_3$", P3, N); label("$\omega_1$", A, W*17); label("$\omega_2$", B, E*17); label("$\omega_3$", C, W*17); [/asy]
$\textbf{(A) }546\qquad\textbf{(B) }548\qquad\textbf{(C) }550\qquad\textbf{(D) }552\qquad\textbf{(E) }554$ | This seems like a coordinate bashable problem. To do this, we notice that it is easier to graph the equilateral triangle first, then the circles, rather than the other way around. Let's forget the lengths of the radius for a moment and focus instead on the ratio of the circles' radius to $\triangle P_1P_2P_3$ 's side length.
WLOG let $P_1P_2 = 2$ . Place triangle $P_1P_2P_3$ on the coordinate plane with $P_1(0,\sqrt3)$ $P_2(1,0)$ , and $P_3(-1,0)$ . Let $r$ be the radius of the circles.
Now, we find the coordinates of the centers of circles $\omega_1$ and $\omega_2$ . Since $\omega_2$ is tangent to the x-axis at $(1,0)$ , the center of $\omega_2$ is $(1,r)$ . Draw a right triangle with legs parallel to the x and y axes, and with hypotenuse as the segment from the center of $\omega_1$ to $P_1$ . Since the slope of $\overline{P_1P_2}$ is $-\sqrt3$ , the slope of the hypotenuse is $\frac{1}{\sqrt{3}}$ , so the right triangle is $30-60-90$ . It's easy to see that the center of $\omega_1$ is $\left(-\frac{r\sqrt{3}}{2}, \sqrt{3} - \frac{r}{2}\right)$
Since $\omega_1$ and $\omega_2$ are tangent, the distance between the centers is $2r$ , so we have \[\sqrt{\left(1 + \frac{r\sqrt{3}}{2}\right)^2 + \left(\frac{3r}{2} - \sqrt{3}\right)^2} = 2r\] \[\left(1 + \frac{r\sqrt{3}}{2}\right)^2 + \left(\frac{3r}{2} - \sqrt{3}\right)^2 = 4r^2\] \[1 + r\sqrt{3} + \frac{3r^2}{4} + \frac{9r^2}{4} - 3r\sqrt{3} + 3 = 4r^2\] \[4 - 2r\sqrt{3} + 3r^2 = 4r^2\] \[r^2 + 2r\sqrt{3} - 4 = 0\] By the Quadratic Formula, $r = \frac{-2\sqrt{3} \pm \sqrt{12 + 16}}{2} = -\sqrt{3}\pm\sqrt{7}$ .
We take the positive value to get the radius of the circle is $\sqrt{7} - \sqrt{3}$
Therefore, the ratio of the radius to the side length of the equilateral triangle is $\sqrt{7} - \sqrt{3} : 2 = 4 : \frac{8}{\sqrt{7} - \sqrt{3}} = 4 : 2(\sqrt{7} + \sqrt{3})$
The side length of the equilateral triangle is $2(\sqrt{7} + \sqrt{3})$ , so its area is $(\sqrt{7} + \sqrt{3})^2\sqrt{3} = 10\sqrt{3} + 6\sqrt{7} = \sqrt{300} + \sqrt{252} \implies 300 + 252 = \boxed{552}$ | D | 552 |
c17fd2d74a62e11ae945fa8c7c30ffab | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_25 | Circles $\omega_1$ $\omega_2$ , and $\omega_3$ each have radius $4$ and are placed in the plane so that each circle is externally tangent to the other two. Points $P_1$ $P_2$ , and $P_3$ lie on $\omega_1$ $\omega_2$ , and $\omega_3$ respectively such that $P_1P_2=P_2P_3=P_3P_1$ and line $P_iP_{i+1}$ is tangent to $\omega_i$ for each $i=1,2,3$ , where $P_4 = P_1$ . See the figure below. The area of $\triangle P_1P_2P_3$ can be written in the form $\sqrt{a}+\sqrt{b}$ for positive integers $a$ and $b$ . What is $a+b$
[asy] unitsize(12); pair A = (0, 8/sqrt(3)), B = rotate(-120)*A, C = rotate(120)*A; real theta = 41.5; pair P1 = rotate(theta)*(2+2*sqrt(7/3), 0), P2 = rotate(-120)*P1, P3 = rotate(120)*P1; filldraw(P1--P2--P3--cycle, gray(0.9)); draw(Circle(A, 4)); draw(Circle(B, 4)); draw(Circle(C, 4)); dot(P1); dot(P2); dot(P3); defaultpen(fontsize(10pt)); label("$P_1$", P1, E*1.5); label("$P_2$", P2, SW*1.5); label("$P_3$", P3, N); label("$\omega_1$", A, W*17); label("$\omega_2$", B, E*17); label("$\omega_3$", C, W*17); [/asy]
$\textbf{(A) }546\qquad\textbf{(B) }548\qquad\textbf{(C) }550\qquad\textbf{(D) }552\qquad\textbf{(E) }554$ | Let $O_1$ $O_2$ , and $O_3$ be the centers of $\omega_1$ $\omega_2$ , and $\omega_3$
Let $P_1P_2 = a$ $\angle O_2O_1P_1= \alpha$
\[[O_1P_1O_2P_2O_3P_3] = [P_1P_2P_3] + 3 \cdot [O_1P_1P_3] = [O_1O_2O_3] + 3 \cdot [O_1P_1O_2]\]
\[[P_1P_2P_3] + 3 \cdot [O_1P_1P_3] = \frac12 \cdot a \cdot \frac{ \sqrt{3} }{2} a + 3 \cdot \frac12 \cdot 4 \cdot a \cdot \sin 30^{\circ} = \frac{ \sqrt{3} }{4} a^2 + 3a\]
\[[O_1O_2O_3] + 3 \cdot [O_1P_1O_2] = \frac12 \cdot 8 \cdot \frac{ \sqrt{3} }{2} \cdot 8 + 3 \cdot \frac12 \cdot 4 \cdot 8 \cdot \sin \alpha = 16 \sqrt{3} + 48 \sin \alpha\]
\[\frac{ \sqrt{3} }{4} a^2 + 3a = 16 \sqrt{3} + 48 \sin \alpha\]
\[\sin \alpha = \frac{ \sqrt{3} }{192} a^2 + \frac{a}{16} - \frac{ \sqrt{3} }{3}\]
By the Law of Cosine from $\triangle O_2P_1P_2$ $O_2P_1^2 = a^2 + 4^2 - 2 \cdot a \cdot 4 \cdot \cos 30^{\circ} = a^2 + 16 - 4a \sqrt{3}$
By the Law of Cosine from $\triangle O_1O_2P_1$ $O_2P_1^2 = 4^2 + 8 ^ 2 - 2 \cdot 4 \cdot 8 \cdot \cos \alpha = 80 - 64 \cos \alpha$
\[a^2 + 16 - 4a \sqrt{3} = 80 - 64 \cos \alpha\]
\[\cos \alpha = -(\frac{a^2}{64} - \frac{ \sqrt{3} }{16} a - 1)\]
\[\because \sin^2 \alpha + \cos^2 \alpha = 1\]
\[\therefore (\frac{ \sqrt{3} }{192} a^2 + \frac{a}{16} - \frac{ \sqrt{3} }{3})^2 + (\frac{a^2}{64} - \frac{ \sqrt{3} }{16} a - 1)^2 = 1\]
\[[ \frac{1}{\sqrt{3}} (\frac{ a^2 }{64} + \frac{ \sqrt{3} }{16} a - 1)]^2 + (\frac{a^2}{64} - \frac{ \sqrt{3} }{16} a - 1)^2 = 1\]
\[(\frac{ a^2 }{64} + \frac{ \sqrt{3} }{16} a - 1)^2 + 3 \cdot (\frac{a^2}{64} - \frac{ \sqrt{3} }{16} a - 1)^2 = 3\]
\[(a^2 + 4 \sqrt{3} a - 64)^2 + 3 \cdot (a^2 - 4 \sqrt{3} a - 64)^2 = 3 \cdot 64^2\]
\[(a^2 - 64)^2 + 2 (a^2 - 64) \cdot 4 \sqrt{3} a + (4 \sqrt{3} a)^2 + 3(a^2 - 64)^2 - 6 (a^2 - 64) \cdot 4 \sqrt{3} a + 3(4 \sqrt{3} a)^2 = 3 \cdot 64^2\]
\[4 (a^2 - 64)^2 - 4 (a^2 - 64) \cdot 4 \sqrt{3} a + 192 a^2 - 3 \cdot 64^2 = 0\]
\[4 (a^2 - 64)(a^2 - 64 - 4 \sqrt{3} a) + 192(a^2 - 64) = 0\]
\[4 (a^2 - 64)(a^2 - 64 - 4 \sqrt{3} a + 48) = 0\]
\[(a^2 - 64)(a^2 - 4 \sqrt{3} a - 16) = 0\]
\[a^2 -64 = 0 \quad or \quad a^2 - 4 \sqrt{3} a - 16 = 0\]
If $a^2 =64$ $a= 8$ $P_1O_2^2 = 8^2 + 4^2 - 2 \cdot 8 \cdot 4 \cdot \cos 30^{\circ} = 80 - 32 \sqrt{3} < 16$ $P_1O_2 < 4$ , meaning that $P_1$ is inside the circle.
However, $P_1$ is not inside the circle.
\[So, \quad a^2 - 4 \sqrt{3} a - 16 = 0\]
\[a = \frac{ 4 \sqrt{3} + \sqrt{ (4 \sqrt{3})^2 - 4(-16) } }{2} = 2 \sqrt{3} + 2 \sqrt{7}\]
\[[P_1P_2P_3] = a^2 \cdot \frac{ \sqrt{3} }{4} = (2 \sqrt{3} + 2 \sqrt{7})^2 \cdot \frac{ \sqrt{3} }{4} = \sqrt{300} + \sqrt{252}\]
Therefore, the answer is $300 + 252 = \boxed{552}$ | D | 552 |
90eacb0e5f16eca5c500aaa9b4e26fb6 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12A_Problems/Problem_1 | Pablo buys popsicles for his friends. The store sells single popsicles for $$1$ each, 3-popsicle boxes for $$2$ , and 5-popsicle boxes for $$3$ . What is the greatest number of popsicles that Pablo can buy with $$8$
$\textbf{(A)}\ 8\qquad\textbf{(B)}\ 11\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 13\qquad\textbf{(E)}\ 15$ | We can take two 5-popsicle boxes and one 3-popsicle box with $$8$ . Note that it is optimal since one popsicle is at the rate of $$1$ per popsicle, three popsicles at $$\frac{2}{3}$ per popsicle and finally, five popsicles at $$\frac{3}{5}$ per popsicle, hence we want as many $$3$ sets as possible. It is clear that the above is the optimal method. $\boxed{13}$ | D | 13 |
be2553f5dbfcfc794f633e1c6fd6a5c3 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12A_Problems/Problem_2 | The sum of two nonzero real numbers is 4 times their product. What is the sum of the reciprocals of the two numbers?
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 12$ | Let $x, y$ be our two numbers. Then $x+y = 4xy$ . Thus,
$\frac{1}{x} + \frac{1}{y} = \frac{x+y}{xy} = 4$
$\boxed{4}$ | C | 4 |
be2553f5dbfcfc794f633e1c6fd6a5c3 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12A_Problems/Problem_2 | The sum of two nonzero real numbers is 4 times their product. What is the sum of the reciprocals of the two numbers?
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 12$ | We can let $x=y.$ Then, we have that $2x=4x^2$ making $x=\tfrac{1}{2}.$ The answer is $\dfrac{2}{x}=4=\boxed{4}.$ Solasky talk | C | 4 |
0ee849bcaedc3638d587e85b2f20e02e | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12A_Problems/Problem_4 | Jerry and Silvia wanted to go from the southwest corner of a square field to the northeast corner. Jerry walked due east and then due north to reach the goal, but Silvia headed northeast and reached the goal walking in a straight line. Which of the following is closest to how much shorter Silvia's trip was, compared to Jerry's trip?
$\textbf{(A)}\ 30\%\qquad\textbf{(B)}\ 40\%\qquad\textbf{(C)}\ 50\%\qquad\textbf{(D)}\ 60\%\qquad\textbf{(E)}\ 70\%$ | Let $j$ represent how far Jerry walked, and $s$ represent how far Sylvia walked. Since the field is a square, and Jerry walked two sides of it, while Silvia walked the diagonal, we can simply define the side of the square field to be one, and find the distances they walked. Since Jerry walked two sides, $j = 2$ Since Silvia walked the diagonal, she walked the hypotenuse of a $45$ $45$ $90$ triangle with leg length $1$ . Thus, $s = \sqrt{2} = 1.414...$ We can then take $\frac{j-s}{j} \approx \frac{2 - 1.4}{2} = 0.3 = 30\%$ $\boxed{30}$ | A | 30 |
740ac59ad788c55278e0795e3c58b96f | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12A_Problems/Problem_5 | At a gathering of $30$ people, there are $20$ people who all know each other and $10$ people who know no one. People who know each other hug, and people who do not know each other shake hands. How many handshakes occur?
$\textbf{(A)}\ 240\qquad\textbf{(B)}\ 245\qquad\textbf{(C)}\ 290\qquad\textbf{(D)}\ 480\qquad\textbf{(E)}\ 490$ | All of the handshakes will involve at least one person from the $10$ who knows no one. Label these ten people $A$ $B$ $C$ $D$ $E$ $F$ $G$ $H$ $I$ $J$
Person $A$ from the group of $10$ will initiate a handshake with everyone else ( $29$ people). Person $B$ initiates $28$ handshakes plus the one already counted from person $A$ . Person $C$ initiates $27$ new handshakes plus the two we already counted. This continues until person $J$ initiates $20$ handshakes plus the nine we already counted from $A$ ... $I$
$29+28+27+26+25+24+23+22+21+20 = \boxed{245}$ | B | 245 |
740ac59ad788c55278e0795e3c58b96f | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12A_Problems/Problem_5 | At a gathering of $30$ people, there are $20$ people who all know each other and $10$ people who know no one. People who know each other hug, and people who do not know each other shake hands. How many handshakes occur?
$\textbf{(A)}\ 240\qquad\textbf{(B)}\ 245\qquad\textbf{(C)}\ 290\qquad\textbf{(D)}\ 480\qquad\textbf{(E)}\ 490$ | Let the group of people who all know each other be $A$ , and let the group of people who know no one be $B$ . Handshakes occur between each pair $(a,b)$ such that $a\in A$ and $b\in B$ , and between each pair of members in $B$ . Thus, the answer is
$|A||B|+{|B|\choose 2} = 20\cdot 10+{10\choose 2} = 200+45 = \boxed{245}$ | B | 245 |
740ac59ad788c55278e0795e3c58b96f | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12A_Problems/Problem_5 | At a gathering of $30$ people, there are $20$ people who all know each other and $10$ people who know no one. People who know each other hug, and people who do not know each other shake hands. How many handshakes occur?
$\textbf{(A)}\ 240\qquad\textbf{(B)}\ 245\qquad\textbf{(C)}\ 290\qquad\textbf{(D)}\ 480\qquad\textbf{(E)}\ 490$ | The number of handshakes will be equivalent to the difference between the number of total interactions and the number of hugs, which are ${30\choose 2}$ and ${20\choose 2}$ , respectively. Thus, the total amount of handshakes is ${30\choose 2} - {20\choose 2} = 435 - 190= \boxed{245}$ | B | 245 |
740ac59ad788c55278e0795e3c58b96f | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12A_Problems/Problem_5 | At a gathering of $30$ people, there are $20$ people who all know each other and $10$ people who know no one. People who know each other hug, and people who do not know each other shake hands. How many handshakes occur?
$\textbf{(A)}\ 240\qquad\textbf{(B)}\ 245\qquad\textbf{(C)}\ 290\qquad\textbf{(D)}\ 480\qquad\textbf{(E)}\ 490$ | Each of the $10$ people who do not know anybody will shake hands with all $20$ of the people who do know each other. This means there will be at least $20 * 10 = 200$ handshakes. In addition, those $10$ people will also shake hands with each other, giving us another $9+8+7+6+5+4+3+2+1 = 45$ handshakes. Therefore, there is a total of $200+45 = \boxed{245}$ handshakes. | B | 245 |
740ac59ad788c55278e0795e3c58b96f | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12A_Problems/Problem_5 | At a gathering of $30$ people, there are $20$ people who all know each other and $10$ people who know no one. People who know each other hug, and people who do not know each other shake hands. How many handshakes occur?
$\textbf{(A)}\ 240\qquad\textbf{(B)}\ 245\qquad\textbf{(C)}\ 290\qquad\textbf{(D)}\ 480\qquad\textbf{(E)}\ 490$ | Every one of the $20$ people who know each other will shake hands with each of the $10$ people who know no one, so there are $20\cdot 10 = 200$ handshakes here. Each of the $10$ people will also shake hands with each other, so there will be ${10 \choose 2}$ $=45$ handshakes for this case. In total, there are $200+45 = \boxed{245}$ handshakes. | B | 245 |
682079b277f325c695c86ce8d5503633 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12A_Problems/Problem_6 | Joy has $30$ thin rods, one each of every integer length from $1 \text{ cm}$ through $30 \text{ cm}$ . She places the rods with lengths $3 \text{ cm}$ $7 \text{ cm}$ , and $15 \text{cm}$ on a table. She then wants to choose a fourth rod that she can put with these three to form a quadrilateral with positive area. How many of the remaining rods can she choose as the fourth rod?
$\textbf{(A)}\ 16 \qquad\textbf{(B)}\ 17 \qquad\textbf{(C)}\ 18 \qquad\textbf{(D)}\ 19 \qquad\textbf{(E)}\ 20$ | The quadrilateral cannot be a straight line. Thus, the fourth side must be longer than $15 - (3 + 7) = 5$ and shorter than $15 + 3 + 7 = 25$ . This means Joy can use the $19$ possible integer rod lengths that fall into $[6, 24]$ . However, she has already used the rods of length $7$ cm and $15$ cm so the answer is $19 - 2 = 17$ $\boxed{17}$ | B | 17 |
5692898ef55e50f396ff802921cfc91e | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12A_Problems/Problem_7 | Define a function on the positive integers recursively by $f(1) = 2$ $f(n) = f(n-1) + 1$ if $n$ is even, and $f(n) = f(n-2) + 2$ if $n$ is odd and greater than $1$ . What is $f(2017)$
$\textbf{(A)}\ 2017 \qquad\textbf{(B)}\ 2018 \qquad\textbf{(C)}\ 4034 \qquad\textbf{(D)}\ 4035 \qquad\textbf{(E)}\ 4036$ | This is a recursive function, which means the function refers back to itself to calculate subsequent terms. To solve this, we must identify the base case, $f(1)=2$ . We also know that when $n$ is odd, $f(n)=f(n-2)+2$ . Thus we know that $f(2017)=f(2015)+2$ . Thus we know that n will always be odd in the recursion of $f(2017)$ , and we add $2$ each recursive cycle, which there are $1008$ of. Thus the answer is $1008*2+2=2018$ , which is answer $\boxed{2018}$ .
Note that when you write out a few numbers, you find that $f(n)=n+1$ for any $n$ , so $f(2017)=2018$ | B | 2018 |
5e0e2c2f600767a848a6cd25582dae8b | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12A_Problems/Problem_11 | Claire adds the degree measures of the interior angles of a convex polygon and arrives at a sum of $2017$ . She then discovers that she forgot to include one angle. What is the degree measure of the forgotten angle?
$\textbf{(A)}\ 37\qquad\textbf{(B)}\ 63\qquad\textbf{(C)}\ 117\qquad\textbf{(D)}\ 143\qquad\textbf{(E)}\ 163$ | We know that the sum of the interior angles of the polygon is a multiple of $180$ . Note that $\left\lceil\frac{2017}{180}\right\rceil = 12$ and $180\cdot 12 = 2160$ , so the angle Claire forgot is $\equiv 2160-2017=143\mod 180$ . Since the polygon is convex, the angle is $\leq 180$ , so the answer is $\boxed{143}$ | D | 143 |
d159bd59c67d733932ba26809b415919 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12A_Problems/Problem_12 | There are $10$ horses, named Horse 1, Horse 2, $\ldots$ , Horse 10. They get their names from how many minutes it takes them to run one lap around a circular race track: Horse $k$ runs one lap in exactly $k$ minutes. At time 0 all the horses are together at the starting point on the track. The horses start running in the same direction, and they keep running around the circular track at their constant speeds. The least time $S > 0$ , in minutes, at which all $10$ horses will again simultaneously be at the starting point is $S = 2520$ . Let $T>0$ be the least time, in minutes, such that at least $5$ of the horses are again at the starting point. What is the sum of the digits of $T$
$\textbf{(A)}\ 2\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 5\qquad\textbf{(E)}\ 6$ | We know that Horse $k$ will be at the starting point after $n$ minutes if $k|n$ . Thus, we are looking for the smallest $n$ such that at least $5$ of the numbers $\{1,2,\cdots,10\}$ divide $n$ . Thus, $n$ has at least $5$ positive integer divisors.
We quickly see that $12$ is the smallest number with at least $5$ positive integer divisors and that $1,2,3,4,6$ are each numbers of horses. Thus, our answer is $1+2=\boxed{3}$ | B | 3 |
d159bd59c67d733932ba26809b415919 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12A_Problems/Problem_12 | There are $10$ horses, named Horse 1, Horse 2, $\ldots$ , Horse 10. They get their names from how many minutes it takes them to run one lap around a circular race track: Horse $k$ runs one lap in exactly $k$ minutes. At time 0 all the horses are together at the starting point on the track. The horses start running in the same direction, and they keep running around the circular track at their constant speeds. The least time $S > 0$ , in minutes, at which all $10$ horses will again simultaneously be at the starting point is $S = 2520$ . Let $T>0$ be the least time, in minutes, such that at least $5$ of the horses are again at the starting point. What is the sum of the digits of $T$
$\textbf{(A)}\ 2\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 5\qquad\textbf{(E)}\ 6$ | In order for at least $5$ horses to finish simultaneously, the current time needs to have at least $5$ divisors. Thus the number must have a form of either $p^4$ or $p^2*q$ , which have $5$ and $6$ factors respectively. The smallest number of the first form is $16$ , and the smallest number of the second form is $12.$ Thus, our answer is $1+2=\boxed{3}$ | B | 3 |
b2db102ee633b3bda598c923e8cc9c76 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12A_Problems/Problem_13 | Driving at a constant speed, Sharon usually takes $180$ minutes to drive from her house to her mother's house. One day Sharon begins the drive at her usual speed, but after driving $\frac{1}{3}$ of the way, she hits a bad snowstorm and reduces her speed by $20$ miles per hour. This time the trip takes her a total of $276$ minutes. How many miles is the drive from Sharon's house to her mother's house?
$\textbf{(A)}\ 132 \qquad\textbf{(B)}\ 135 \qquad\textbf{(C)}\ 138 \qquad\textbf{(D)}\ 141 \qquad\textbf{(E)}\ 144$ | Let total distance be $x$ . Her speed in miles per minute is $\tfrac{x}{180}$ . Then, the distance that she drove before hitting the snowstorm is $\tfrac{x}{3}$ . Her speed in snowstorm is reduced $20$ miles per hour, or $\tfrac{1}{3}$ miles per minute. Knowing it took her $276$ minutes in total, we create equation: \[\text{Time before Storm}\, + \, \text{Time after Storm} = \text{Total Time} \Longrightarrow\] \[\frac{\text{Distance before Storm}}{\text{Speed before Storm}} + \frac{\text{Distance in Storm}}{\text{Speed in Storm}} = \text{Total Time} \Longrightarrow \frac{\tfrac{x}{3}}{\tfrac{x}{180}} + \frac{\tfrac{2x}{3}}{\tfrac{x}{180} - \tfrac{1}{3}} = 276\]
Solving equation, we get $x=135$ $\Longrightarrow \boxed{135}$ | B | 135 |
0c89fe769516cc2eea1cb8f9ae9424aa | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12A_Problems/Problem_14 | Alice refuses to sit next to either Bob or Carla. Derek refuses to sit next to Eric. How many ways are there for the five of them to sit in a row of $5$ chairs under these conditions?
$\textbf{(A)}\ 12 \qquad \textbf{(B)}\ 16 \qquad\textbf{(C)}\ 28 \qquad\textbf{(D)}\ 32 \qquad\textbf{(E)}\ 40$ | Alice may sit in the center chair, in an end chair, or in a next-to-end chair. Suppose she sits in the center chair. The 2nd and 4th chairs (next to her) must be occupied by Derek and Eric, in either order, leaving the end chairs for Bob and Carla in either order; this yields $2! * 2! = 4$ ways to seat the group.
Next, suppose Alice sits in one of the end chairs. Then the chair beside her will be occupied by either Derek or Eric. The center chair must be occupied by Bob or Carla, leaving the last two people to fill the last two chairs in either order. $2$ ways to seat Alice times $2$ ways to fill the next chair times $2$ ways to fill the center chair times $2$ ways to fill the last two chairs yields $2 * 2 * 2 * 2 = 16$ ways to fill the chairs.
Finally, suppose Alice sits in the second or fourth chair. Then the chairs next to her must be occupied by Derek and Eric in either order, and the other two chairs must be occupied by Bob and Carla in either order. This yields $2 * 2 * 2 = 8$ ways to fill the chairs.
In total, there are $4 + 8 + 16$ ways to fill the chairs, so the answer is $\boxed{28}$ | C | 28 |
ce4c3c1f04a62cce08b508228ccdafba | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12A_Problems/Problem_17 | There are $24$ different complex numbers $z$ such that $z^{24}=1$ . For how many of these is $z^6$ a real number?
$\textbf{(A)}\ 0 \qquad\textbf{(B)}\ 4 \qquad\textbf{(C)}\ 6 \qquad\textbf{(D)}\ 12 \qquad\textbf{(E)}\ 24$ | Note that these $z$ such that $z^{24}=1$ are $e^{\frac{ni\pi}{12}}$ for integer $0\leq n<24$ . So
$z^6=e^{\frac{ni\pi}{2}}$
This is real if $\frac{n}{2}\in \mathbb{Z} \Leftrightarrow (n$ is even $)$ . Thus, the answer is the number of even $0\leq n<24$ which is $\boxed{12}$ | D | 12 |
ce4c3c1f04a62cce08b508228ccdafba | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12A_Problems/Problem_17 | There are $24$ different complex numbers $z$ such that $z^{24}=1$ . For how many of these is $z^6$ a real number?
$\textbf{(A)}\ 0 \qquad\textbf{(B)}\ 4 \qquad\textbf{(C)}\ 6 \qquad\textbf{(D)}\ 12 \qquad\textbf{(E)}\ 24$ | $z = \sqrt[24]{1} = 1^{\frac{1}{24}}$
By Euler's identity $1 = e^{0 \times i} = \cos (0+2k\pi) + i \sin(0+2k\pi)$ , where $k$ is an integer.
Using De Moivre's Theorem , we have $z = 1^{\frac{1}{24}} = {\cos (\frac{k\pi}{12}) + i \sin (\frac{k\pi}{12})}$ , where $0 \leq k<24$ that produce $24$ unique results.
Using De Moivre's Theorem again, we have $z^6 = {\cos (\frac{k\pi}{2}) + i \sin (\frac{k\pi}{2})}$
For $z^6$ to be real, $\sin(\frac{k\pi}{2})$ has to equal $0$ to negate the imaginary component. This occurs whenever $\frac{k\pi}{2}$ is an integer multiple of $\pi$ , requiring that $k$ is even. There are exactly $\boxed{12}$ | D | 12 |
ce4c3c1f04a62cce08b508228ccdafba | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12A_Problems/Problem_17 | There are $24$ different complex numbers $z$ such that $z^{24}=1$ . For how many of these is $z^6$ a real number?
$\textbf{(A)}\ 0 \qquad\textbf{(B)}\ 4 \qquad\textbf{(C)}\ 6 \qquad\textbf{(D)}\ 12 \qquad\textbf{(E)}\ 24$ | From the start, recall from the Fundamental Theorem of Algebra that $z^{24} = 1$ must have $24$ solutions (and these must be distinct since the equation factors into $0 = (z-1)(z^{23} + z^{22} + z^{21}... + z + 1)$ ), or notice that the question is simply referring to the 24th roots of unity, of which we know there must be $24$ . Notice that $1 = z^{24} = (z^6)^4$ , so for any solution $z$ $z^6$ will be one of the 4th roots of unity ( $1$ $i$ $-1$ , or $-i$ ). Then $6$ solutions $z$ will satisfy $z^6 = 1$ $6$ will satisfy $z^6 = -1$ (and this is further justified by knowledge of the 6th roots of unity), so there must be $\boxed{12}$ such $z$ | D | 12 |
ce4c3c1f04a62cce08b508228ccdafba | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12A_Problems/Problem_17 | There are $24$ different complex numbers $z$ such that $z^{24}=1$ . For how many of these is $z^6$ a real number?
$\textbf{(A)}\ 0 \qquad\textbf{(B)}\ 4 \qquad\textbf{(C)}\ 6 \qquad\textbf{(D)}\ 12 \qquad\textbf{(E)}\ 24$ | Let $a\in\mathbb{R}$ and $a = z^6.$ We have \[a^4 = 1 \implies a = 1,-1.\] $z^6 = \pm 1$ has 6 solutions for $1$ and $-1$ respectively, so $6+6=\boxed{12}.$ \[\] -svyn | D | 12 |
8b0e8e3477f3b15ce59be11e5609e756 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12A_Problems/Problem_18 | Let $S(n)$ equal the sum of the digits of positive integer $n$ . For example, $S(1507) = 13$ . For a particular positive integer $n$ $S(n) = 1274$ . Which of the following could be the value of $S(n+1)$
$\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 1239\qquad\textbf{(E)}\ 1265$ | Note that $n\equiv S(n)\bmod 9$ , so $S(n+1)-S(n)\equiv n+1-n = 1\bmod 9$ . So, since $S(n)=1274\equiv 5\bmod 9$ , we have that $S(n+1)\equiv 6\bmod 9$ . The only one of the answer choices $\equiv 6\bmod 9$ is $\boxed{1239}$ | D | 1239 |
8b0e8e3477f3b15ce59be11e5609e756 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12A_Problems/Problem_18 | Let $S(n)$ equal the sum of the digits of positive integer $n$ . For example, $S(1507) = 13$ . For a particular positive integer $n$ $S(n) = 1274$ . Which of the following could be the value of $S(n+1)$
$\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 1239\qquad\textbf{(E)}\ 1265$ | One possible value of $S(n)$ would be $1275$ , but this is not any of the choices. Therefore, we know that $n$ ends in $9$ , and after adding $1$ , the last digit $9$ carries over, turning the last digit into $0$ . If the next digit is also a $9$ , this process repeats until we get to a non- $9$ digit. By the end, the sum of digits would decrease by $9$ multiplied by the number of carry-overs but increase by $1$ as a result of the final carrying over. Therefore, the result must be $9x-1$ less than original value of $S(n)$ $1274$ , where $x$ is a positive integer. The only choice that satisfies this condition is $\boxed{1239}$ | D | 1239 |
8b0e8e3477f3b15ce59be11e5609e756 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12A_Problems/Problem_18 | Let $S(n)$ equal the sum of the digits of positive integer $n$ . For example, $S(1507) = 13$ . For a particular positive integer $n$ $S(n) = 1274$ . Which of the following could be the value of $S(n+1)$
$\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 1239\qquad\textbf{(E)}\ 1265$ | Another way to solve this is to realize that if you continuously add the digits of the number $1274 (1 + 2 + 7 + 4 = 14, 1 + 4 = 5)$ , we get $5$ . Adding one to that, we get $6$ . So, if we assess each option to see which one attains $6$ , we would discover that $1239$ satisfies the requirement, because $1 + 2 + 3 + 9 = 15$ $1 + 5 = 6$ . The answer is $\boxed{1239}$ | D | 1239 |
8b0e8e3477f3b15ce59be11e5609e756 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12A_Problems/Problem_18 | Let $S(n)$ equal the sum of the digits of positive integer $n$ . For example, $S(1507) = 13$ . For a particular positive integer $n$ $S(n) = 1274$ . Which of the following could be the value of $S(n+1)$
$\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 1239\qquad\textbf{(E)}\ 1265$ | Note that a lot of numbers can have a sum of $1274$ , but what we use wishful thinking and want is some simple number $n$ where it is easy to compute the sum of the digits of $n+1$ . This number would consists of basically all digits $9$ , since when you add $1$ a lot of stuff will cancel out and end up at $0$ (ex: $399+1=400$ ). We see that the maximum number of $9$ s that can be in $1274$ is $141$ and we are left with a remainder of $5$ , so $n$ is in the form $99...9599...9$ . If we add $1$ to this number we will get $99...9600...0$ so this the sum of the digits of $n+1$ is congruent to $6 \mod 9$ . The only answer choice that is equivalent to $6 \mod 9$ is $1239$ , so our answer is $\boxed{1239}$ -srisainandan6 | D | 1239 |
7d8dda36869170c50035e3c0d478ab5b | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12A_Problems/Problem_20 | How many ordered pairs $(a,b)$ such that $a$ is a positive real number and $b$ is an integer between $2$ and $200$ , inclusive, satisfy the equation $(\log_b a)^{2017}=\log_b(a^{2017})?$
$\textbf{(A)}\ 198\qquad\textbf{(B)}\ 199\qquad\textbf{(C)}\ 398\qquad\textbf{(D)}\ 399\qquad\textbf{(E)}\ 597$ | By the properties of logarithms, we can rearrange the equation to read $x^{2017}=2017x$ with $x=\log_b a$ . If $x\neq 0$ , we may divide by it and get $x^{2016}=2017$ , which implies $x=\pm \root{2016}\of{2017}$ . Hence, we have $3$ possible values $x$ , namely \[x=0,\qquad x=2017^{\frac1{2016}},\, \text{and}\quad x=-2017^{\frac1{2016}}.\]
Since $\log_b a=x$ is equivalent to $a=b^x$ , each possible value $x$ yields exactly $199$ solutions $(b,a)$ , as we can assign $a=b^x$ to each $b=2,3,\dots,200$ . In total, we have $3\cdot 199=\boxed{597}$ solutions. | E | 597 |
0de955d0a36eff5e826d196861b5f847 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12A_Problems/Problem_21 | A set $S$ is constructed as follows. To begin, $S = \{0,10\}$ . Repeatedly, as long as possible, if $x$ is an integer root of some polynomial $a_{n}x^n + a_{n-1}x^{n-1} + \dots + a_{1}x + a_0$ for some $n\geq{1}$ , all of whose coefficients $a_i$ are elements of $S$ , then $x$ is put into $S$ . When no more elements can be added to $S$ , how many elements does $S$ have?
$\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 5 \qquad\textbf{(C)}\ 7 \qquad\textbf{(D)}\ 9 \qquad\textbf{(E)}\ 11$ | At first, $S=\{0,10\}$
\[\begin{tabular}{r c l c l} \(10x+10\) & has root & \(x=-1\) & so now & \(S=\{-1,0,10\}\) \\ \(-x^{10}-x^9-x^8-x^7-x^6-x^5-x^4-x^3-x^2-x+10\) & has root & \(x=1\) & so now & \(S=\{-1,0,1,10\}\) \\ \(x+10\) & has root & \(x=-10\) & so now & \(S=\{-10,-1,0,1,10\}\) \\ \(x^3+x-10\) & has root & \(x=2\) & so now & \(S=\{-10,-1,0,1,2,10\}\) \\ \(x+2\) & has root & \(x=-2\) & so now & \(S=\{-10,-2,-1,0,1,2,10\}\) \\ \(2x-10\) & has root & \(x=5\) & so now & \(S=\{-10,-2,-1,0,1,2,5,10\}\) \\ \(x+5\) & has root & \(x=-5\) & so now & \(S=\{-10,-5,-2,-1,0,1,2,5,10\}\) \end{tabular}\]
At this point, no more elements can be added to $S$ . To see this, let
\begin{align*} a_{n}x^n + a_{n-1}x^{n-1} + ... + a_{2}x^2 + a_{1}x + a_0 &= 0 \\ x(a_{n}x^{n-1} + a_{n-1}x^{n-2} + ... + a_{2}x + a_{1}) + a_0 &= 0 \\ x(a_{n}x^{n-1} + a_{n-1}x^{n-2} + ... + a_{2}x + a_{1}) &= -a_0 \end{align*}
with each $a_i$ in $S$ $x$ is a factor of $a_0$ , and $a_0$ is in $S$ , so $x$ has to be a factor of some element in $S$ . There are no such integers left, so there can be no more additional elements. $\{-10,-5,-2,-1,0,1,2,5,10\}$ has $9$ elements $\to \boxed{9}$ | D | 9 |
1ff64818e741a3ec60d426c57af671a2 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12A_Problems/Problem_22 | A square is drawn in the Cartesian coordinate plane with vertices at $(2, 2)$ $(-2, 2)$ $(-2, -2)$ $(2, -2)$ . A particle starts at $(0,0)$ . Every second it moves with equal probability to one of the eight lattice points (points with integer coordinates) closest to its current position, independently of its previous moves. In other words, the probability is $1/8$ that the particle will move from $(x, y)$ to each of $(x, y + 1)$ $(x + 1, y + 1)$ $(x + 1, y)$ $(x + 1, y - 1)$ $(x, y - 1)$ $(x - 1, y - 1)$ $(x - 1, y)$ , or $(x - 1, y + 1)$ . The particle will eventually hit the square for the first time, either at one of the 4 corners of the square or at one of the 12 lattice points in the interior of one of the sides of the square. The probability that it will hit at a corner rather than at an interior point of a side is $m/n$ , where $m$ and $n$ are relatively prime positive integers. What is $m + n$
$\textbf{(A) } 4 \qquad\textbf{(B) } 5 \qquad\textbf{(C) } 7 \qquad\textbf{(D) } 15 \qquad\textbf{(E) } 39$ | We let $c, e,$ and $m$ be the probability of reaching a corner before an edge when starting at an "inside corner" (e.g. $(1, 1)$ ), an "inside edge" (e.g. $(1, 0)$ ), and the middle respectively.
Starting in the middle, there is a $\frac{4}{8}$ chance of moving to an inside edge and a $\frac{4}{8}$ chance of moving to an inside corner, so
\[m = \frac{1}{2}e + \frac{1}{2}c.\]
Starting at an inside edge, there is a $\frac{2}{8}$ chance of moving to another inside edge, a $\frac{2}{8}$ chance of moving to an inside corner, a $\frac{1}{8}$ chance of moving into the middle, and a $\frac{3}{8}$ chance of reaching an outside edge and stopping. Therefore,
\[e = \frac{1}{4}e + \frac{1}{4}c + \frac{1}{8}m + \frac{3}{8}\cdot 0 = \frac{1}{4}e + \frac{1}{4}c + \frac{1}{8}m.\]
Starting at an inside corner, there is a $\frac{2}{8}$ chance of moving to an inside edge, a $\frac{1}{8}$ chance of moving into the middle, a $\frac{4}{8}$ chance of moving to an outside edge and stopping, and finally a $\frac{1}{8}$ chance of reaching that elusive outside corner. This gives
\[c = \frac{1}{4}e + \frac{1}{8}m + \frac{1}{2}0 + \frac{1}{8}\cdot 1 = \frac{1}{4}e + \frac{1}{8}m + \frac{1}{8}.\]
Solving this system of equations gives
\[m = \frac{4}{35},\] \[e = \frac{1}{14},\] \[c = \frac{11}{70}.\]
Since the particle starts at $(0, 0),$ it is $m$ we are looking for, so the final answer is
\[4 + 35 = \boxed{39}.\] | E | 39 |
a91a4d5f4ab36ee43189c02f60aa3ba2 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12A_Problems/Problem_23 | For certain real numbers $a$ $b$ , and $c$ , the polynomial \[g(x) = x^3 + ax^2 + x + 10\] has three distinct roots, and each root of $g(x)$ is also a root of the polynomial \[f(x) = x^4 + x^3 + bx^2 + 100x + c.\] What is $f(1)$
$\textbf{(A)}\ -9009 \qquad\textbf{(B)}\ -8008 \qquad\textbf{(C)}\ -7007 \qquad\textbf{(D)}\ -6006 \qquad\textbf{(E)}\ -5005$ | Let $r_1,r_2,$ and $r_3$ be the roots of $g(x)$ . Let $r_4$ be the additional root of $f(x)$ . Then from Vieta's formulas on the quadratic term of $g(x)$ and the cubic term of $f(x)$ , we obtain the following:
\begin{align*} r_1+r_2+r_3&=-a \\ r_1+r_2+r_3+r_4&=-1 \end{align*}
Thus $r_4=a-1$
Now applying Vieta's formulas on the constant term of $g(x)$ , the linear term of $g(x)$ , and the linear term of $f(x)$ , we obtain:
\begin{align*} r_1r_2r_3 & = -10\\ r_1r_2+r_2r_3+r_3r_1 &= 1\\ r_1r_2r_3+r_2r_3r_4+r_3r_4r_1+r_4r_1r_2 & = -100\\ \end{align*}
Substituting for $r_1r_2r_3$ in the bottom equation and factoring the remainder of the expression, we obtain:
\[-10+(r_1r_2+r_2r_3+r_3r_1)r_4=-10+r_4=-100\]
It follows that $r_4=-90$ . But $r_4=a-1$ so $a=-89$
Now we can factor $f(x)$ in terms of $g(x)$ as
\[f(x)=(x-r_4)g(x)=(x+90)g(x)\]
Then $f(1)=91g(1)$ and
\[g(1)=1^3-89\cdot 1^2+1+10=-77\]
Hence $f(1)=91\cdot(-77)=\boxed{7007}$ | C | 7007 |
a91a4d5f4ab36ee43189c02f60aa3ba2 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12A_Problems/Problem_23 | For certain real numbers $a$ $b$ , and $c$ , the polynomial \[g(x) = x^3 + ax^2 + x + 10\] has three distinct roots, and each root of $g(x)$ is also a root of the polynomial \[f(x) = x^4 + x^3 + bx^2 + 100x + c.\] What is $f(1)$
$\textbf{(A)}\ -9009 \qquad\textbf{(B)}\ -8008 \qquad\textbf{(C)}\ -7007 \qquad\textbf{(D)}\ -6006 \qquad\textbf{(E)}\ -5005$ | Since all of the roots of $g(x)$ are distinct and are roots of $f(x)$ , and the degree of $f$ is one more than the degree of $g$ , we have that
\[f(x) = C(x-k)g(x)\]
for some number $k$ . By comparing $x^4$ coefficients, we see that $C=1$ . Thus,
\[x^4+x^3+bx^2+100x+c=(x-k)(x^3+ax^2+x+10)\]
Expanding and equating coefficients we get that
\[a-k=1,1-ak=b,10-k=100,-10k=c\]
The third equation yields $k=-90$ , and the first equation yields $a=-89$ . So we have that
$f(1)=(1+90)g(1)=91(1-89+1+10)=(91)(-77)=\boxed{7007}$ | C | 7007 |
a91a4d5f4ab36ee43189c02f60aa3ba2 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12A_Problems/Problem_23 | For certain real numbers $a$ $b$ , and $c$ , the polynomial \[g(x) = x^3 + ax^2 + x + 10\] has three distinct roots, and each root of $g(x)$ is also a root of the polynomial \[f(x) = x^4 + x^3 + bx^2 + 100x + c.\] What is $f(1)$
$\textbf{(A)}\ -9009 \qquad\textbf{(B)}\ -8008 \qquad\textbf{(C)}\ -7007 \qquad\textbf{(D)}\ -6006 \qquad\textbf{(E)}\ -5005$ | $f(x)$ must have four roots, three of which are roots of $g(x)$ . Using the fact that every polynomial has a unique factorization into its roots, and since the leading coefficient of $f(x)$ and $g(x)$ are the same, we know that
\[f(x)=g(x)(x-r)\]
where $r\in\mathbb{C}$ is the fourth root of $f(x)$ . Substituting $g(x)$ and expanding, we find that
\begin{align*}f(x)&=(x^3+ax^2+x+10)(x-r)\\ &=x^4+(a-r)x^3+(1-ar)x^2+(10-r)x-10r.\end{align*}
Comparing coefficients with $f(x)$ , we see that
\begin{align*} a-r&=1\\ 1-ar&=b\\ 10-r&=100\\ -10r&=c.\\ \end{align*}
(Solution 1.1 picks up here.)
Let's solve for $a,b,c,$ and $r$ . Since $10-r=100$ $r=-90$ , so $c=(-10)(-90)=900$ . Since $a-r=1$ $a=-89$ , and $b=1-ar=-8009$ . Thus, we know that
\[f(x)=x^4+x^3-8009x^2+100x+900.\]
Taking $f(1)$ , we find that
\begin{align*} f(1)&=1^4+1^3-8009(1)^2+100(1)+900\\ &=1+1-8009+100+900\\ &=\boxed{7007} | C | 7007 |
a91a4d5f4ab36ee43189c02f60aa3ba2 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12A_Problems/Problem_23 | For certain real numbers $a$ $b$ , and $c$ , the polynomial \[g(x) = x^3 + ax^2 + x + 10\] has three distinct roots, and each root of $g(x)$ is also a root of the polynomial \[f(x) = x^4 + x^3 + bx^2 + 100x + c.\] What is $f(1)$
$\textbf{(A)}\ -9009 \qquad\textbf{(B)}\ -8008 \qquad\textbf{(C)}\ -7007 \qquad\textbf{(D)}\ -6006 \qquad\textbf{(E)}\ -5005$ | A faster ending to Solution 1 is as follows.
We shall solve for only $a$ and $r$ . Since $10-r=100$ $r=-90$ , and since $a-r=1$ $a=-89$ . Then, \begin{align*} f(1)&=(1-r)(1^3+a\cdot1^2+1+10)\\ &=(91)(-77)\\ &=\boxed{7007} | C | 7007 |
a91a4d5f4ab36ee43189c02f60aa3ba2 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12A_Problems/Problem_23 | For certain real numbers $a$ $b$ , and $c$ , the polynomial \[g(x) = x^3 + ax^2 + x + 10\] has three distinct roots, and each root of $g(x)$ is also a root of the polynomial \[f(x) = x^4 + x^3 + bx^2 + 100x + c.\] What is $f(1)$
$\textbf{(A)}\ -9009 \qquad\textbf{(B)}\ -8008 \qquad\textbf{(C)}\ -7007 \qquad\textbf{(D)}\ -6006 \qquad\textbf{(E)}\ -5005$ | We notice that the constant term of $f(x)=c$ and the constant term in $g(x)=10$ . Because $f(x)$ can be factored as $g(x) \cdot (x- r)$ (where $r$ is the unshared root of $f(x)$ , we see that using the constant term, $-10 \cdot r = c$ and therefore $r = -\frac{c}{10}$ .
Now we once again write $f(x)$ out in factored form:
\[f(x) = g(x)\cdot (x-r) = (x^3+ax^2+x+10)\left(x+\frac{c}{10}\right)\]
We can expand the expression on the right-hand side to get:
\[f(x) = x^4+\left(a+\frac{c}{10}\right)x^3+\left(1+\frac{ac}{10}\right)x^2+\left(10+\frac{c}{10}\right)x+c\]
Now we have $f(x) = x^4+\left(a+\frac{c}{10}\right)x^3+\left(1+\frac{ac}{10}\right)x^2+\left(10+\frac{c}{10} \right)x+c=x^4+x^3+bx^2+100x+c$
Simply looking at the coefficients for each corresponding term (knowing that they must be equal), we have the equations: \[10+\frac{c}{10}=100 \Rightarrow c=900\] \[a+\frac{c}{10} = 1, c=900 \Rightarrow a + 90 =1 \Rightarrow a= -89\]
and finally,
\[1+\frac{ac}{10} = b = 1+\frac{-89 \cdot 900}{10} = b = -8009\]
We know that $f(1)$ is the sum of its coefficients, hence $1+1+b+100+c$ . We substitute the values we obtained for $b$ and $c$ into this expression to get $f(1) = 1 + 1 + (-8009) + 100 + 900 = \boxed{7007}$ | C | 7007 |
a91a4d5f4ab36ee43189c02f60aa3ba2 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12A_Problems/Problem_23 | For certain real numbers $a$ $b$ , and $c$ , the polynomial \[g(x) = x^3 + ax^2 + x + 10\] has three distinct roots, and each root of $g(x)$ is also a root of the polynomial \[f(x) = x^4 + x^3 + bx^2 + 100x + c.\] What is $f(1)$
$\textbf{(A)}\ -9009 \qquad\textbf{(B)}\ -8008 \qquad\textbf{(C)}\ -7007 \qquad\textbf{(D)}\ -6006 \qquad\textbf{(E)}\ -5005$ | Let $r_1,r_2,$ and $r_3$ be the roots of $g(x)$ . Let $r_4$ be the additional root of $f(x)$ . Then from Vieta's formulas on the quadratic term of $g(x)$ and the cubic term of $f(x)$ , we obtain the following:
\begin{align*} r_1+r_2+r_3&=-a \\ r_1+r_2+r_3+r_4&=-1 \end{align*}
Thus $r_4=a-1$
Now applying Vieta's formulas on the constant term of $g(x)$ , the linear term of $g(x)$ , and the linear term of $f(x)$ , we obtain:
\begin{align*} r_1r_2r_3 & = -10\\ r_1r_2+r_2r_3+r_3r_1 &= 1\\ r_1r_2r_3+r_2r_3r_4+r_3r_4r_1+r_4r_1r_2 & = -100\\ \end{align*}
Substituting for $r_1r_2r_3$ in the bottom equation and factoring the remainder of the expression, we obtain:
\[-10+(r_1r_2+r_2r_3+r_3r_1)r_4=-10+r_4=-100\]
It follows that $r_4=-90$ . But $r_4=a-1$ so $a=-89$
Now we can factor $f(x)$ in terms of $g(x)$ as
\[f(x)=(x-r_4)g(x)=(x+90)g(x)\]
Then $f(1)=91g(1)$ and
\[g(1)=1^3-89\cdot 1^2+1+10=-77\]
Hence $f(1)=91\cdot(-77)=\boxed{7007}$ | C | 7007 |
a91a4d5f4ab36ee43189c02f60aa3ba2 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12A_Problems/Problem_23 | For certain real numbers $a$ $b$ , and $c$ , the polynomial \[g(x) = x^3 + ax^2 + x + 10\] has three distinct roots, and each root of $g(x)$ is also a root of the polynomial \[f(x) = x^4 + x^3 + bx^2 + 100x + c.\] What is $f(1)$
$\textbf{(A)}\ -9009 \qquad\textbf{(B)}\ -8008 \qquad\textbf{(C)}\ -7007 \qquad\textbf{(D)}\ -6006 \qquad\textbf{(E)}\ -5005$ | Let the roots of $g(x)$ be $r_1$ $r_2$ , and $r_3$ . Let the roots of $f(x)$ be $r_1$ $r_2$ $r_3$ , and $r_4$ . From Vieta's, we have: \begin{align*} r_1+r_2+r_3=-a \\ r_1+r_2+r_3+r_4=-1 \\ r_4=a-1 \end{align*} The fourth root is $a-1$ . Since $r_1$ $r_2$ , and $r_3$ are common roots, we have: \begin{align*} f(x)=g(x)(x-(a-1)) \\ f(1)=g(1)(1-(a-1)) \\ f(1)=(a+12)(2-a) \\ f(1)=-(a+12)(a-2) \\ \end{align*} Let $a-2=k$ \begin{align*} f(1)=-k(k+14) \end{align*} Note that $-7007=-1001\cdot(7)=-(7\cdot(11)\cdot(13))\cdot(7)=-91\cdot(77)$ This gives us a pretty good guess of $\boxed{7007}$ | C | 7007 |
a91a4d5f4ab36ee43189c02f60aa3ba2 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12A_Problems/Problem_23 | For certain real numbers $a$ $b$ , and $c$ , the polynomial \[g(x) = x^3 + ax^2 + x + 10\] has three distinct roots, and each root of $g(x)$ is also a root of the polynomial \[f(x) = x^4 + x^3 + bx^2 + 100x + c.\] What is $f(1)$
$\textbf{(A)}\ -9009 \qquad\textbf{(B)}\ -8008 \qquad\textbf{(C)}\ -7007 \qquad\textbf{(D)}\ -6006 \qquad\textbf{(E)}\ -5005$ | First off, let's get rid of the $x^4$ term by finding $h(x)=f(x)-xg(x)$ . This polynomial consists of the difference of two polynomials with $3$ common factors, so it must also have these factors. The polynomial is $h(x)=(1-a)x^3 + (b-1)x^2 + 90x + c$ , and must be equal to $(1-a)g(x)$ . Equating the coefficients, we get $3$ equations. We will tackle the situation one equation at a time, starting the $x$ terms. Looking at the coefficients, we get $\dfrac{90}{1-a} = 1$ \[\therefore 90=1-a.\] The solution to the previous is obviously $a=-89$ . We can now find $b$ and $c$ $\dfrac{b-1}{1-a} = a$ \[\therefore b-1=a(1-a)=-89\cdot90=-8010\] and $b=-8009$ . Finally $\dfrac{c}{1-a} = 10$ \[\therefore c=10(1-a)=10\cdot90=900\] Solving the original problem, $f(1)=1 + 1 + b + 100 + c = 102+b+c=102+900-8009=\boxed{7007}$ | null | 7007 |
a91a4d5f4ab36ee43189c02f60aa3ba2 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12A_Problems/Problem_23 | For certain real numbers $a$ $b$ , and $c$ , the polynomial \[g(x) = x^3 + ax^2 + x + 10\] has three distinct roots, and each root of $g(x)$ is also a root of the polynomial \[f(x) = x^4 + x^3 + bx^2 + 100x + c.\] What is $f(1)$
$\textbf{(A)}\ -9009 \qquad\textbf{(B)}\ -8008 \qquad\textbf{(C)}\ -7007 \qquad\textbf{(D)}\ -6006 \qquad\textbf{(E)}\ -5005$ | Simple polynomial division is a feasible method. Even though we have variables, we can equate terms at the end of the division so that we can cancel terms. Doing the division of $\frac{f(x)}{g(x)}$ eventually brings us the final step $(1-a)x^3 + (b-1)x^2 + 90x + c$ minus $(1-a)x^3 - (a-a^2)x^2 + (1-a)x + 10(1-a)$ after we multiply $f(x)$ by $(1-a)$ . Now we equate coefficients of same-degree $x$ terms. This gives us $10(1-a) = c, b-1 = a - a^2, 1-a = 90 \Rightarrow a = -89, c = 900, b = -8009$ . We are interested in finding $f(1)$ , which equals $1^4 + 1^3 -8009\cdot1^2 + 100\cdot1 + 900 = \boxed{7007}$ . ~skyscraper | C | 7007 |
f11e6458c1dcddfb1deca653594e4795 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12A_Problems/Problem_24 | Quadrilateral $ABCD$ is inscribed in circle $O$ and has side lengths $AB=3, BC=2, CD=6$ , and $DA=8$ . Let $X$ and $Y$ be points on $\overline{BD}$ such that $\frac{DX}{BD} = \frac{1}{4}$ and $\frac{BY}{BD} = \frac{11}{36}$ .
Let $E$ be the intersection of line $AX$ and the line through $Y$ parallel to $\overline{AD}$ . Let $F$ be the intersection of line $CX$ and the line through $E$ parallel to $\overline{AC}$ . Let $G$ be the point on circle $O$ other than $C$ that lies on line $CX$ . What is $XF\cdot XG$
$\textbf{(A) }17\qquad\textbf{(B) }\frac{59 - 5\sqrt{2}}{3}\qquad\textbf{(C) }\frac{91 - 12\sqrt{3}}{4}\qquad\textbf{(D) }\frac{67 - 10\sqrt{2}}{3}\qquad\textbf{(E) }18$ | Using the given ratios, note that $\frac{XY}{BD} = 1 - \frac{1}{4} - \frac{11}{36} = \frac{4}{9}.$
By AA Similarity, $\triangle AXD \sim \triangle EXY$ with a ratio of $\frac{DX}{XY} = \frac{9}{16}$ and $\triangle ACX \sim \triangle EFX$ with a ratio of $\frac{AX}{XE} = \frac{DX}{XY} = \frac{9}{16}$ , so $\frac{XF}{CX} = \frac{16}{9}$
Now we find the length of $BD$ . Because the quadrilateral is cyclic, we can simply use the Law of Cosines. \[BD^2=3^2+8^2-48\cos\angle BAD=2^2+6^2-24\cos (180-\angle BAD)=2^2+6^2+24\cos\angle BAD\] \[\rightarrow \cos\angle BAD = \frac{11}{24}\] \[\rightarrow BD=\sqrt{51}\] By Power of a Point, $CX\cdot XG = DX\cdot XB = \frac{\sqrt{51}}{4} \frac{3\sqrt{51}}{4}$ . Thus $XF\cdot XG = \frac{XF}{CX} CX\cdot XG = \frac{51}{3} = \boxed{17}.$ | A | 17 |
f11e6458c1dcddfb1deca653594e4795 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12A_Problems/Problem_24 | Quadrilateral $ABCD$ is inscribed in circle $O$ and has side lengths $AB=3, BC=2, CD=6$ , and $DA=8$ . Let $X$ and $Y$ be points on $\overline{BD}$ such that $\frac{DX}{BD} = \frac{1}{4}$ and $\frac{BY}{BD} = \frac{11}{36}$ .
Let $E$ be the intersection of line $AX$ and the line through $Y$ parallel to $\overline{AD}$ . Let $F$ be the intersection of line $CX$ and the line through $E$ parallel to $\overline{AC}$ . Let $G$ be the point on circle $O$ other than $C$ that lies on line $CX$ . What is $XF\cdot XG$
$\textbf{(A) }17\qquad\textbf{(B) }\frac{59 - 5\sqrt{2}}{3}\qquad\textbf{(C) }\frac{91 - 12\sqrt{3}}{4}\qquad\textbf{(D) }\frac{67 - 10\sqrt{2}}{3}\qquad\textbf{(E) }18$ | We shall make use of the pairs of similar triangles present in the problem, Ptolemy's Theorem, and Power of a Point.
Let $Z$ be the intersection of $AC$ and $BD$ . First, from $ABCD$ being a cyclic quadrilateral, we have that $\triangle BCZ \sim \triangle AZD$ $\triangle BZA \sim \triangle CDZ$ . Therefore, $\frac{2}{BZ} = \frac{8}{AZ}$ $\frac{6}{CZ} = \frac{3}{BZ}$ , and $\frac{2}{CZ} = \frac{8}{DZ}$ , so we have $BZ = \frac{1}{2}CZ$ $AZ = 2CZ$ , and $DZ = 4CZ$ . By Ptolemy's Theorem, \[(AB)(CD) + (BC)(DA) = (AC)(BD) = (AZ + ZC)(BZ + ZD)\] \[\rightarrow 3 \cdot 6 + 2 \cdot 8 = 34 = \left(2CZ + ZC\right)\left(\frac{1}{2}CZ + 4CZ\right) = \frac{27}{2}CZ^2.\] Thus, $CZ^2 = \frac{68}{27}$ . Then, by Power of a Point, $GX \cdot XC = BX \cdot XD = \frac{3}{4} \cdot \frac{1}{4}BD^2 = \frac{3}{16} \cdot \left(\frac{9}{2}CZ\right)^2 = \frac{9 \cdot 17}{16}$ . So, $XG = \frac{9 \cdot 17}{16XC}$ .
Next, observe that $\triangle ACX \sim \triangle EFX$ , so $\frac{XE}{XF} = \frac{AX}{XC}$ . Also, $\triangle{AXD} \sim \triangle{EXY}$ , so $\frac{8}{AX} = \frac{EY}{XE}$ . We can compute $EY = \frac{128}{9}$ after noticing that $XY = BD - BY - DX = BD - \frac{11}{36}BD - \frac{1}{4}BD = \frac{4}{9}BD$ and that $\frac{8}{DX} = \frac{32}{BD} = \frac{EY}{XY} = \frac{EY}{\frac{4}{9}BD}$ . So, $\frac{8}{AX} = \frac{128}{9XE}$ . Then, $\frac{XE}{AX} = \frac{XF}{XC} = \frac{16}{9} \rightarrow XF = \frac{16}{9}XC$
Multiplying our equations for $XF$ and $XG$ yields that $XF \cdot XG = \frac{9 \cdot 17}{16XC} \cdot \frac{16}{9}XC = \boxed{17}.$ | A | 17 |
f11e6458c1dcddfb1deca653594e4795 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12A_Problems/Problem_24 | Quadrilateral $ABCD$ is inscribed in circle $O$ and has side lengths $AB=3, BC=2, CD=6$ , and $DA=8$ . Let $X$ and $Y$ be points on $\overline{BD}$ such that $\frac{DX}{BD} = \frac{1}{4}$ and $\frac{BY}{BD} = \frac{11}{36}$ .
Let $E$ be the intersection of line $AX$ and the line through $Y$ parallel to $\overline{AD}$ . Let $F$ be the intersection of line $CX$ and the line through $E$ parallel to $\overline{AC}$ . Let $G$ be the point on circle $O$ other than $C$ that lies on line $CX$ . What is $XF\cdot XG$
$\textbf{(A) }17\qquad\textbf{(B) }\frac{59 - 5\sqrt{2}}{3}\qquad\textbf{(C) }\frac{91 - 12\sqrt{3}}{4}\qquad\textbf{(D) }\frac{67 - 10\sqrt{2}}{3}\qquad\textbf{(E) }18$ | Denote $P$ to be the intersection between line $AE$ and circle $O$ . Note that $\angle GFE = \angle ACG = \angle APG = 180 - \angle GPE$ , making $\angle GFE + \angle GDE = 180$ . Thus, $PEFG$ is a cyclic quadrilateral. Using Power of a Point on $X$ gives $XP \cdot XE = XG \cdot XF$
Since $\triangle ADX \sim \triangle EYX$ and $\triangle ACX \sim \triangle EFX$ $AX/XE = XD/YX = 9/16$ . Using Power of a Point on $X$ again, $(AX)(PX) = (BX)(DX)$ . Plugging in $AX=9/16 XE$ gives: \[\dfrac{9}{16}(XE)(PX) = (BX)(DX) = \dfrac{9}{16}(FX)(GX)\] By Law of Cosines , we can find $BD = \sqrt{51}$ , as in Solution 1. Now, $BX = 3/4 (\sqrt{51})$ and $DX = 1/4 (\sqrt{51})$ , making $\dfrac{9}{16}(FX)(GX) = \left( \dfrac{\sqrt{51}}{4}\right)\left( \dfrac{3\sqrt{51}}{4}\right) = \dfrac{3(51)}{16}$ . This gives us $FX \cdot GX = \boxed{17}$ as a result. | A | 17 |
f11e6458c1dcddfb1deca653594e4795 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12A_Problems/Problem_24 | Quadrilateral $ABCD$ is inscribed in circle $O$ and has side lengths $AB=3, BC=2, CD=6$ , and $DA=8$ . Let $X$ and $Y$ be points on $\overline{BD}$ such that $\frac{DX}{BD} = \frac{1}{4}$ and $\frac{BY}{BD} = \frac{11}{36}$ .
Let $E$ be the intersection of line $AX$ and the line through $Y$ parallel to $\overline{AD}$ . Let $F$ be the intersection of line $CX$ and the line through $E$ parallel to $\overline{AC}$ . Let $G$ be the point on circle $O$ other than $C$ that lies on line $CX$ . What is $XF\cdot XG$
$\textbf{(A) }17\qquad\textbf{(B) }\frac{59 - 5\sqrt{2}}{3}\qquad\textbf{(C) }\frac{91 - 12\sqrt{3}}{4}\qquad\textbf{(D) }\frac{67 - 10\sqrt{2}}{3}\qquad\textbf{(E) }18$ | $\because$ $AC \parallel EF$ $\quad \therefore$ $\triangle ACX \sim \triangle EFX$ $\quad \frac{XF}{XC} = \frac{XE}{XA}$
By Power of a Point, $XG \cdot XC = XD \cdot XB$
By multiplying the $2$ equations we get $XF \cdot XG = \frac{XE}{XA} \cdot XD \cdot XB$
$\because$ $YE \parallel AD$ $\quad \therefore$ $\triangle EYX \sim \triangle ADX$ $\quad \frac{XD}{XY} = \frac{XA}{XE}, \quad XD \cdot XE = XA \cdot XY, \quad XD = \frac{XA \cdot XY}{XE}$
By substitution, $XF \cdot XG = \frac{XE}{XA} \cdot \frac{XA \cdot XY}{XE} \cdot XB = XY \cdot XB = \frac{4}{9} BD \cdot \frac{3}{4} BD = \frac{BD^2}{3}$
Let $a = AB$ $b = BC$ $c = CD$ $d = AD$ $p = AC$ , and $q = BD$
By Ptolemy's theorem, $p \cdot q = a \cdot c + b \cdot d$
\[[ABD] = \frac12 \cdot ad \cdot \sin A, \quad [BCD] = \frac12 \cdot bc \cdot \sin C = \frac12 \cdot bc \cdot \sin A\]
\[[ABC] = \frac12 \cdot ab \cdot \sin B, \quad [ACD] = \frac12 \cdot cd \cdot \sin D = \frac12 \cdot cd \cdot \sin B\]
\[[ABCD] = [ABD] + [BCD] = \frac12 \cdot ad \cdot \sin A + \frac12 \cdot bc \cdot \sin A = \frac12 (ad + bc) \sin A\]
\[[ABCD] = [ABC] + [ACD] = \frac12 \cdot ab \cdot \sin B + \frac12 \cdot cd \cdot \sin B = \frac12 (ab + cd) \sin B\]
\[\frac{ab + cd}{ad + bc} = \frac{ \sin A }{ \sin B} = \frac{ \frac{q}{2R} }{ \frac{p}{2R} } = \frac{q}{p}, \quad p = \frac{q(ad + bc)}{ab + cd}\]
\[\frac{q(ad + bc)}{ab + cd} \cdot q = ac + bd\]
\[BD^2 = q^2 = \frac{ (ac + bd)(ab + cd) }{ad + bc} = \frac{(3 \cdot 6 + 2 \cdot 8)(3 \cdot 2 + 6 \cdot 8)}{3 \cdot 8 + 2 \cdot 6} = 51\]
\[XF \cdot XG = \frac{51}{3} = \boxed{17}\] | A | 17 |
a8953f0a328e2d3da982731cf159b7cc | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12B_Problems/Problem_1 | Kymbrea's comic book collection currently has $30$ comic books in it, and she is adding to her collection at the rate of $2$ comic books per month. LaShawn's collection currently has $10$ comic books in it, and he is adding to his collection at the rate of $6$ comic books per month. After how many months will LaShawn's collection have twice as many comic books as Kymbrea's?
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 20\qquad\textbf{(E)}\ 25$ | Kymbrea has $30$ comic books initially and every month, she adds two. This can be represented as $30 + 2x$ where x is the number of months elapsed. LaShawn's collection, similarly, is $10 + 6x$ . To find when LaShawn will have twice the number of comic books as Kymbrea, we solve for x with the equation $2(2x + 30) = 6x + 10$ and get $x = \boxed{25}$ | E | 25 |
8c02d125f106b5362bcd85e8fa82402f | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12B_Problems/Problem_3 | Supposed that $x$ and $y$ are nonzero real numbers such that $\frac{3x+y}{x-3y}=-2$ . What is the value of $\frac{x+3y}{3x-y}$
$\textbf{(A)}\ -3\qquad\textbf{(B)}\ -1\qquad\textbf{(C)}\ 1\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ 3$ | Rearranging, we find $3x+y=-2x+6y$ , or $5x=5y\implies x=y$ .
Substituting, we can convert the second equation into $\frac{x+3x}{3x-x}=\frac{4x}{2x}=\boxed{2}$ | D | 2 |
8c02d125f106b5362bcd85e8fa82402f | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12B_Problems/Problem_3 | Supposed that $x$ and $y$ are nonzero real numbers such that $\frac{3x+y}{x-3y}=-2$ . What is the value of $\frac{x+3y}{3x-y}$
$\textbf{(A)}\ -3\qquad\textbf{(B)}\ -1\qquad\textbf{(C)}\ 1\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ 3$ | Substituting each $x$ and $y$ with $1$ , we see that the given equation holds true, as $\frac{3(1)+1}{1-3(1)} = -2$ . Thus, $\frac{x+3y}{3x-y}=\boxed{2}$ | D | 2 |
8c02d125f106b5362bcd85e8fa82402f | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12B_Problems/Problem_3 | Supposed that $x$ and $y$ are nonzero real numbers such that $\frac{3x+y}{x-3y}=-2$ . What is the value of $\frac{x+3y}{3x-y}$
$\textbf{(A)}\ -3\qquad\textbf{(B)}\ -1\qquad\textbf{(C)}\ 1\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ 3$ | Let $y=ax$ . The first equation converts into $\frac{(3+a)x}{(1-3a)x}=-2$ , which simplifies to $3+a=-2(1-3a)$ . After a bit of algebra we found out $a=1$ , which means that $x=y$ . Substituting $y=x$ into the second equation it becomes $\frac{4x}{2x}=\boxed{2}$ - mathleticguyyy | D | 2 |
8c02d125f106b5362bcd85e8fa82402f | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12B_Problems/Problem_3 | Supposed that $x$ and $y$ are nonzero real numbers such that $\frac{3x+y}{x-3y}=-2$ . What is the value of $\frac{x+3y}{3x-y}$
$\textbf{(A)}\ -3\qquad\textbf{(B)}\ -1\qquad\textbf{(C)}\ 1\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ 3$ | Let $x=1$ . Then $y=1$ . So the desired result is $2$ . Select $\boxed{2}$ | D | 2 |
e788fbbd316a1a798ef1bfc26fd893e8 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12B_Problems/Problem_4 | Samia set off on her bicycle to visit her friend, traveling at an average speed of $17$ kilometers per hour. When she had gone half the distance to her friend's house, a tire went flat, and she walked the rest of the way at $5$ kilometers per hour. In all it took her $44$ minutes to reach her friend's house. In kilometers rounded to the nearest tenth, how far did Samia walk?
$\textbf{(A)}\ 2.0\qquad\textbf{(B)}\ 2.2\qquad\textbf{(C)}\ 2.8\qquad\textbf{(D)}\ 3.4\qquad\textbf{(E)}\ 4.4$ | Let's call the distance that Samia had to travel in total as $2x$ , so that we can avoid fractions. We know that the length of the bike ride and how far she walked are equal, so they are both $\frac{2x}{2}$ , or $x$ \[\] She bikes at a rate of $17$ kph, so she travels the distance she bikes in $\frac{x}{17}$ hours. She walks at a rate of $5$ kph, so she travels the distance she walks in $\frac{x}{5}$ hours. \[\] The total time is $\frac{x}{17}+\frac{x}{5} = \frac{22x}{85}$ . This is equal to $\frac{44}{60} = \frac{11}{15}$ of an hour. Solving for $x$ , we have: \[\] \[\frac{22x}{85} = \frac{11}{15}\] \[\frac{2x}{85} = \frac{1}{15}\] \[30x = 85\] \[6x = 17\] \[x = \frac{17}{6}\] \[\] Since $x$ is the distance of how far Samia traveled by both walking and biking, and we want to know how far Samia walked to the nearest tenth, we have that Samia walked about $\boxed{2.8}$ | C | 2.8 |
e788fbbd316a1a798ef1bfc26fd893e8 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12B_Problems/Problem_4 | Samia set off on her bicycle to visit her friend, traveling at an average speed of $17$ kilometers per hour. When she had gone half the distance to her friend's house, a tire went flat, and she walked the rest of the way at $5$ kilometers per hour. In all it took her $44$ minutes to reach her friend's house. In kilometers rounded to the nearest tenth, how far did Samia walk?
$\textbf{(A)}\ 2.0\qquad\textbf{(B)}\ 2.2\qquad\textbf{(C)}\ 2.8\qquad\textbf{(D)}\ 3.4\qquad\textbf{(E)}\ 4.4$ | Notice that Samia walks $\frac {17}{5}$ times slower than she bikes, and that she walks and bikes the same distance. Thus, the fraction of the total time that Samia will be walking is \[\frac{\frac{17}{5}}{\frac{17}{5}+\frac{5}{5}} = \frac{17}{22}\] Then, multiply this by the time \[\frac{17}{22} \cdot 44 \text{minutes} = 34 \text{minutes}\] 34 minutes is a little greater than $\frac {1}{2}$ of an hour so Samia traveled \[\sim \frac {1}{2} \cdot 5 = 2.5 \text{kilometers}\] The answer choice a little greater than 2.5 is $\boxed{2.8}$ by $5$ and gotten the exact answer as well) | C | 2.8 |
e788fbbd316a1a798ef1bfc26fd893e8 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12B_Problems/Problem_4 | Samia set off on her bicycle to visit her friend, traveling at an average speed of $17$ kilometers per hour. When she had gone half the distance to her friend's house, a tire went flat, and she walked the rest of the way at $5$ kilometers per hour. In all it took her $44$ minutes to reach her friend's house. In kilometers rounded to the nearest tenth, how far did Samia walk?
$\textbf{(A)}\ 2.0\qquad\textbf{(B)}\ 2.2\qquad\textbf{(C)}\ 2.8\qquad\textbf{(D)}\ 3.4\qquad\textbf{(E)}\ 4.4$ | Since the distance between biking and walking is equal, we can use the given rates' harmonic mean to find the average speed.
\[\frac{2ab}{a+b}\implies\frac{2 \cdot 17 \cdot 5}{17+5}=\frac{85}{11}\text{kph}\]
Then, multiplying by $\frac{11}{15}$ hours gives the overall distance $\frac{17}{3}$ kilometers. Samia only walks half of that, so $\frac{17}{6}\approx \boxed{2.8}$ kilometers. | C | 2.8 |
e788fbbd316a1a798ef1bfc26fd893e8 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12B_Problems/Problem_4 | Samia set off on her bicycle to visit her friend, traveling at an average speed of $17$ kilometers per hour. When she had gone half the distance to her friend's house, a tire went flat, and she walked the rest of the way at $5$ kilometers per hour. In all it took her $44$ minutes to reach her friend's house. In kilometers rounded to the nearest tenth, how far did Samia walk?
$\textbf{(A)}\ 2.0\qquad\textbf{(B)}\ 2.2\qquad\textbf{(C)}\ 2.8\qquad\textbf{(D)}\ 3.4\qquad\textbf{(E)}\ 4.4$ | Since the distance traveled by bicycle and foot are equal, we can substitute the time traveled by bike, or $b$ , as $\frac{5}{17}$ of the time traveled by foot, or $f$
Accordingly, we have $\frac{22}{17}f=44$ This comes out to $f=34$ This means that Samia traveled $34$ minutes on foot, and hence, $44-34=10$ minutes on bicycle
Because $10$ minutes on bike yields $\frac{17}{6}$ kilometers, and distance on bike = distance on foot, we have the final answer of $\frac{17}{6} \approx \boxed{2.8}$ | null | 2.8 |
258d68f46e74a6bc473986f937210798 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12B_Problems/Problem_5 | The data set $[6, 19, 33, 33, 39, 41, 41, 43, 51, 57]$ has median $Q_2 = 40$ , first quartile $Q_1 = 33$ , and third quartile $Q_3 = 43$ . An outlier in a data set is a value that is more than $1.5$ times the interquartile range below the first quartle ( $Q_1$ ) or more than $1.5$ times the interquartile range above the third quartile ( $Q_3$ ), where the interquartile range is defined as $Q_3 - Q_1$ . How many outliers does this data set have?
$\textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 4$ | The interquartile range is defined as $Q3 - Q1$ , which is $43 - 33 = 10$ $1.5$ times this value is $15$ , so all values more than $15$ below $Q1$ $33 - 15 = 18$ is an outlier. The only one that fits this is $6$ . All values more than $15$ above $Q3 = 43 + 15 = 58$ are also outliers, of which there are none so there is only $\boxed{1}$ outlier in total. | B | 1 |
8a0f796722de43b3be8cd441e89f2c10 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12B_Problems/Problem_9 | A circle has center $(-10, -4)$ and has radius $13$ . Another circle has center $(3, 9)$ and radius $\sqrt{65}$ . The line passing through the two points of intersection of the two circles has equation $x+y=c$ . What is $c$
$\textbf{(A)}\ 3\qquad\textbf{(B)}\ 3\sqrt{3}\qquad\textbf{(C)}\ 4\sqrt{2}\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ \frac{13}{2}$ | The equations of the two circles are $(x+10)^2+(y+4)^2=169$ and $(x-3)^2+(y-9)^2=65$ . Rearrange them to $(x+10)^2+(y+4)^2-169=0$ and $(x-3)^2+(y-9)^2-65=0$ , respectively. Their intersection points are where these two equations gain equality. The two points lie on the line with the equation $(x+10)^2+(y+4)^2-169=(x-3)^2+(y-9)^2-65$ . We can simplify this like the following. $(x+10)^2+(y+4)^2-169=(x-3)^2+(y-9)^2-65 \rightarrow (x^2+20x+100)+(y^2+8y+16)-(x^2-6x+9)-(y^2-18y+81)=104 \rightarrow 26x+26y+26=104 \rightarrow 26x+26y=78 \rightarrow x+y=3$ . Thus, $c = \boxed{3}$ | A | 3 |
8a0f796722de43b3be8cd441e89f2c10 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12B_Problems/Problem_9 | A circle has center $(-10, -4)$ and has radius $13$ . Another circle has center $(3, 9)$ and radius $\sqrt{65}$ . The line passing through the two points of intersection of the two circles has equation $x+y=c$ . What is $c$
$\textbf{(A)}\ 3\qquad\textbf{(B)}\ 3\sqrt{3}\qquad\textbf{(C)}\ 4\sqrt{2}\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ \frac{13}{2}$ | Note the specificity of the radii, $13$ and $\sqrt{65}$ , and that specificity is often deliberately added to simplify the solution to a problem.
One may recognize $13$ as the hypotenuse of the $\text{5-12-13}$ right triangle and $\sqrt{65}$ as the hypotenuse of the right triangle with legs $1$ and $8$ . We can suppose that the legs of these triangles connect the circles' centers to their intersection along the gridlines of the plane.
If we suspect that one of the intersections lies $12$ units to the right of and $5$ units above the center of the first circle, we find the point $(-10 + 12,-4 + 5) = (2,1)$ , which is in fact $1$ unit to the left of and $8$ units below the center of the second circle at $(3,9)$
Plugging $(2,1)$ into $x + y$ gives us $c = 2 + 1 = \boxed{3}$ | A | 3 |
416eb7b201a7abbfdc566bc6be4fdc5d | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12B_Problems/Problem_11 | Call a positive integer monotonous if it is a one-digit number or its digits, when read from left to right, form either a strictly increasing or a strictly decreasing sequence. For example, $3$ $23578$ , and $987620$ are monotonous, but $88$ $7434$ , and $23557$ are not. How many monotonous positive integers are there?
$\textbf{(A)}\ 1024\qquad\textbf{(B)}\ 1524\qquad\textbf{(C)}\ 1533\qquad\textbf{(D)}\ 1536\qquad\textbf{(E)}\ 2048$ | Case 1: monotonous numbers with digits in ascending order
There are $\sum_{n=1}^{9} \binom{9}{n}$ ways to choose n digits from the digits 1 to 9. For each of these ways, we can generate exactly one monotonous number by ordering the chosen digits in ascending order. Note that 0 is not included since it will always be a leading digit and that is not allowed. Also, $\emptyset$ (the empty set) isn't included because it doesn't generate a number. The sum is equivalent to $\sum_{n=0}^{9} \binom{9}{n} -\binom{9}{0} = 2^9 - 1 = 511.$
Case 2: monotonous numbers with digits in descending order
There are $\sum_{n=1}^{10} \binom{10}{n}$ ways to choose n digits from the digits 0 to 9. For each of these ways, we can generate exactly one monotonous number by ordering the chosen digits in descending order. Note that 0 is included since we are allowed to end numbers with zeros. However, $\emptyset$ (the empty set) still isn't included because it doesn't generate a number. The sum is equivalent to $\sum_{n=0}^{10} \binom{10}{n} -\binom{10}{0} = 2^{10} - 1 = 1023.$ We discard the number 0 since it is not positive. Thus there are $1022$ here.
Since the 1-digit numbers 1 to 9 satisfy both case 1 and case 2, we have overcounted by 9. Thus there are $511+1022-9=\boxed{1524}$ monotonous numbers. | B | 1524 |
416eb7b201a7abbfdc566bc6be4fdc5d | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12B_Problems/Problem_11 | Call a positive integer monotonous if it is a one-digit number or its digits, when read from left to right, form either a strictly increasing or a strictly decreasing sequence. For example, $3$ $23578$ , and $987620$ are monotonous, but $88$ $7434$ , and $23557$ are not. How many monotonous positive integers are there?
$\textbf{(A)}\ 1024\qquad\textbf{(B)}\ 1524\qquad\textbf{(C)}\ 1533\qquad\textbf{(D)}\ 1536\qquad\textbf{(E)}\ 2048$ | Like Solution 1, divide the problem into an increasing and decreasing case:
$\bullet$ Case 1: Monotonous numbers with digits in ascending order.
Arrange the digits 1 through 9 in increasing order, and exclude 0 because a positive integer cannot begin with 0.
To get a monotonous number, we can either include or exclude each of the remaining 9 digits, and there are $2^9 = 512$ ways to do this. However, we cannot exclude every digit at once, so we subtract 1 to get $512-1=511$ monotonous numbers for this case.
$\bullet$ Case 2: Monotonous numbers with digits in descending order.
This time, we arrange all 10 digits in decreasing order and repeat the process to find $2^{10} = 1024$ ways to include or exclude each digit. We cannot exclude every digit at once, and we cannot include only 0, so we subtract 2 to get $1024-2=1022$ monotonous numbers for this case.
At this point, we have counted all of the single-digit monotonous numbers twice, so we must subtract 9 from our total.
Thus our final answer is $511+1022-9 = \boxed{1524}$ | B | 1524 |
416eb7b201a7abbfdc566bc6be4fdc5d | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12B_Problems/Problem_11 | Call a positive integer monotonous if it is a one-digit number or its digits, when read from left to right, form either a strictly increasing or a strictly decreasing sequence. For example, $3$ $23578$ , and $987620$ are monotonous, but $88$ $7434$ , and $23557$ are not. How many monotonous positive integers are there?
$\textbf{(A)}\ 1024\qquad\textbf{(B)}\ 1524\qquad\textbf{(C)}\ 1533\qquad\textbf{(D)}\ 1536\qquad\textbf{(E)}\ 2048$ | Unlike the first two solutions, we can do our casework based off of whether or not the number contains a $0$
If it does, then we know the $0$ must be the last digit in the number (and hence, the number has to be decreasing). Because $0$ is not positive, $0$ is not monotonous. So, we need to pick at least $1$ number in the set $[1, 9].$ After choosing our numbers, there will be just $1$ way to arrange them so that the overall number is monotonous.
In total, each of the $9$ digits can either be in the monotonous number or not, yielding $2^9 = 512$ total solutions. However, we said earlier that $0$ cannot be by itself so we have to subtract out the case in which we picked none of the numbers $1-9$ . So, this case gives us $511$
Onto the second case, if there are no $0$ s, then the number can either be arranged in ascending order or in descending order. So, for each selection of the digits $1- 9$ , there are $2$ solutions. This gives \[2 \cdot (2^9 - 1) = 2 \cdot 511 = 1022\] possibilities. Note that we subtracted out the $1$ because we cannot choose none of the numbers.
However, realize that if we pick just $1$ digit, then there is only $1$ arrangement. We cannot put a single digit in both ascending and descending order. So, we must subtract out $9$ from there (because there are $9$ possible ways to select one number and for each case, we overcounted by $1$ ).
All in all, that gives $511 + 1022 - 9 = \boxed{1524}$ monotonous numbers. | B | 1524 |
416eb7b201a7abbfdc566bc6be4fdc5d | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12B_Problems/Problem_11 | Call a positive integer monotonous if it is a one-digit number or its digits, when read from left to right, form either a strictly increasing or a strictly decreasing sequence. For example, $3$ $23578$ , and $987620$ are monotonous, but $88$ $7434$ , and $23557$ are not. How many monotonous positive integers are there?
$\textbf{(A)}\ 1024\qquad\textbf{(B)}\ 1524\qquad\textbf{(C)}\ 1533\qquad\textbf{(D)}\ 1536\qquad\textbf{(E)}\ 2048$ | Let $n$ be the number of digits of a monotonous number. Notice for an increasing monotonous number with $n \ge 2$ , we can obtain 2 more monotonous numbers that are decreasing by reversing its digits and adding a $0$ to the end of the reversed digits. Whenever $n$ digits are chosen, the order is fixed. There are $\binom{9}{n}$ ways to obtain an increasing monotonous number with $n$ digits. So, there are $3\cdot \sum_{n=2}^{9} \binom{9}{n}$ monotonous numbers when $n \ge 2$ . When $n=1$ , there is no reverse but we could add $0$ to the end, so there are $2 \cdot \binom{9}{1}$ monotonous numbers.
The answer is:
$3\cdot \sum_{n=2}^{9} \binom{9}{n} + 2 \cdot \binom{9}{1}$ $=3\cdot \sum_{n=1}^{9} \binom{9}{n} - \binom{9}{1}$ $=3\cdot \left( \sum_{n=0}^{9} \binom{9}{n} - \binom{9}{0} \right) - \binom{9}{1}$ $= 3 \cdot (2^9-1) - 9$ $=\boxed{1524}$ | B | 1524 |
416eb7b201a7abbfdc566bc6be4fdc5d | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12B_Problems/Problem_11 | Call a positive integer monotonous if it is a one-digit number or its digits, when read from left to right, form either a strictly increasing or a strictly decreasing sequence. For example, $3$ $23578$ , and $987620$ are monotonous, but $88$ $7434$ , and $23557$ are not. How many monotonous positive integers are there?
$\textbf{(A)}\ 1024\qquad\textbf{(B)}\ 1524\qquad\textbf{(C)}\ 1533\qquad\textbf{(D)}\ 1536\qquad\textbf{(E)}\ 2048$ | We have 2 cases.
Case 1: Ascending order
We can set up a 1-1 Correspondence. For any subset of all the digits $1$ to $9$ $0$ cannot be a digit for ascending order), we can always rearrange them into an ascending monotonous number. Therefore, the number of subsets of the integers $1$ to $9$ is equivalent to the number of ascending integers. So, $2^9=512$ . However, the empty set ( $\emptyset$ ) is not an integer, so we must subtract 1. Thus, $512-1=511$
Case 2: Descending order
Similarly, any subset of the digits $0$ to $9$ can be rearranged into a descending monotonous number. So, $2^{10}=1024$ . However, $\emptyset$ and $0$ are not positive integers, so we must subtract 2. Thus, $1024-2=1022$
We have covered all the cases. We add $511$ to $1022$ , giving us $1533$ . So now we just innocently go ahead and choose $\textbf{(C) } 1533$ as our answer, right? No! We overcounted the $9$ single-digit integers . The answer is actually $1533-9=\boxed{1524}$ | B | 1524 |
99c5ba211cb755dde9eda00c6ce9e509 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12B_Problems/Problem_13 | In the figure below, $3$ of the $6$ disks are to be painted blue, $2$ are to be painted red, and $1$ is to be painted green. Two paintings that can be obtained from one another by a rotation or a reflection of the entire figure are considered the same. How many different paintings are possible?
[asy] size(100); pair A, B, C, D, E, F; A = (0,0); B = (1,0); C = (2,0); D = rotate(60, A)*B; E = B + D; F = rotate(60, A)*C; draw(Circle(A, 0.5)); draw(Circle(B, 0.5)); draw(Circle(C, 0.5)); draw(Circle(D, 0.5)); draw(Circle(E, 0.5)); draw(Circle(F, 0.5)); [/asy]
$\textbf{(A) } 6 \qquad \textbf{(B) } 8 \qquad \textbf{(C) } 9 \qquad \textbf{(D) } 12 \qquad \textbf{(E) } 15$ | Looking at the answer choices, we see that the possibilities are indeed countable. Thus, we will utilize that approach in the form of two separate cases, as rotation and reflection take care of numerous possibilities. First, consider the case that the green disk is in a corner. This yields $6$ possible arrangements for the $3$ blue disks and $2$ red disks in the remaining available slots. Now, consider the case that the green disk is on an edge. This yields $6$ more possible arrangements for the $3$ blue disks and $2$ red disks in the remaining available slots. Thus, our answer is $6 + 6 = \boxed{12}$ | D | 12 |
99c5ba211cb755dde9eda00c6ce9e509 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12B_Problems/Problem_13 | In the figure below, $3$ of the $6$ disks are to be painted blue, $2$ are to be painted red, and $1$ is to be painted green. Two paintings that can be obtained from one another by a rotation or a reflection of the entire figure are considered the same. How many different paintings are possible?
[asy] size(100); pair A, B, C, D, E, F; A = (0,0); B = (1,0); C = (2,0); D = rotate(60, A)*B; E = B + D; F = rotate(60, A)*C; draw(Circle(A, 0.5)); draw(Circle(B, 0.5)); draw(Circle(C, 0.5)); draw(Circle(D, 0.5)); draw(Circle(E, 0.5)); draw(Circle(F, 0.5)); [/asy]
$\textbf{(A) } 6 \qquad \textbf{(B) } 8 \qquad \textbf{(C) } 9 \qquad \textbf{(D) } 12 \qquad \textbf{(E) } 15$ | Burnside's Lemma can be applied to the problem. Number the disks 1-6 from top to bottom and left to right. First, we count the number of group actions. There are $6$ in total: $3$ rotations, $0$ °, $120$ °, and $240$ °. Additionally, there are $3$ different reflections, which are over lines passing through the middle of disks $1$ $4$ , and $6$ . The $0$ ° rotation has the total number of diagrams fixed, which is $6!/3!\cdot2!=60$ . The $120$ ° and $240$ ° rotations each have $0$ fixed diagrams, as one would need the same color for disks $1, 4, 6$ and the same color for disks $2, 3, 5,$ which isn't possible. Consider the reflection over the line passing through disk $1$ .
The green can be placed in $2$ places, disk $1$ or $5$ , and then, the blue must go in either disks $2$ and $3$ or disks $4$ and $6$ , so the number of fixed diagrams is $2\cdot2=4$ . The other two reflections also have $4$ unique fixed diagrams. Therefore, applying Burnside's Lemma, we get $(60+4+4+4)/6 = \boxed{12}$ | D | 12 |
bd666a589736ea3ad22ec6ec2cd8af9b | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12B_Problems/Problem_15 | Let $ABC$ be an equilateral triangle. Extend side $\overline{AB}$ beyond $B$ to a point $B'$ so that $BB'=3 \cdot AB$ . Similarly, extend side $\overline{BC}$ beyond $C$ to a point $C'$ so that $CC'=3 \cdot BC$ , and extend side $\overline{CA}$ beyond $A$ to a point $A'$ so that $AA'=3 \cdot CA$ . What is the ratio of the area of $\triangle A'B'C'$ to the area of $\triangle ABC$
$\textbf{(A)}\ 9\qquad\textbf{(B)}\ 16\qquad\textbf{(C)}\ 25\qquad\textbf{(D)}\ 36\qquad\textbf{(E)}\ 37$ | Note that by symmetry, $\triangle A'B'C'$ is also equilateral. Therefore, we only need to find one of the sides of $A'B'C'$ to determine the area ratio. WLOG, let $AB = BC = CA = 1$ . Therefore, $BB' = 3$ and $BC' = 4$ . Also, $\angle B'BC' = 120^{\circ}$ , so by the Law of Cosines, $B'C' = \sqrt{37}$ . Therefore, the answer is $(\sqrt{37})^2 : 1^2 = \boxed{37}$ | E | 37 |
bd666a589736ea3ad22ec6ec2cd8af9b | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12B_Problems/Problem_15 | Let $ABC$ be an equilateral triangle. Extend side $\overline{AB}$ beyond $B$ to a point $B'$ so that $BB'=3 \cdot AB$ . Similarly, extend side $\overline{BC}$ beyond $C$ to a point $C'$ so that $CC'=3 \cdot BC$ , and extend side $\overline{CA}$ beyond $A$ to a point $A'$ so that $AA'=3 \cdot CA$ . What is the ratio of the area of $\triangle A'B'C'$ to the area of $\triangle ABC$
$\textbf{(A)}\ 9\qquad\textbf{(B)}\ 16\qquad\textbf{(C)}\ 25\qquad\textbf{(D)}\ 36\qquad\textbf{(E)}\ 37$ | As mentioned in the first solution, $\triangle A'B'C'$ is equilateral. WLOG, let $AB=2$ . Let $D$ be on the line passing through $AB$ such that $A'D$ is perpendicular to $AB$ . Note that $\triangle A'DA$ is a 30-60-90 with right angle at $D$ . Since $AA'=6$ $AD=3$ and $A'D=3\sqrt{3}$ . So we know that $DB'=11$ . Note that $\triangle A'DB'$ is a right triangle with right angle at $D$ . So by the Pythagorean theorem, we find $A'B'= \sqrt{(3\sqrt{3})^2 + 11^2} = 2\sqrt{37}.$ Therefore, the answer is $(2\sqrt{37})^2 : 2^2 = \boxed{37}$ | E | 37 |
bd666a589736ea3ad22ec6ec2cd8af9b | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12B_Problems/Problem_15 | Let $ABC$ be an equilateral triangle. Extend side $\overline{AB}$ beyond $B$ to a point $B'$ so that $BB'=3 \cdot AB$ . Similarly, extend side $\overline{BC}$ beyond $C$ to a point $C'$ so that $CC'=3 \cdot BC$ , and extend side $\overline{CA}$ beyond $A$ to a point $A'$ so that $AA'=3 \cdot CA$ . What is the ratio of the area of $\triangle A'B'C'$ to the area of $\triangle ABC$
$\textbf{(A)}\ 9\qquad\textbf{(B)}\ 16\qquad\textbf{(C)}\ 25\qquad\textbf{(D)}\ 36\qquad\textbf{(E)}\ 37$ | Let $AB=BC=CA=x$ . We start by noting that we can just write $AB'$ as just $AB+BB'=4AB$ .
Similarly $BC'=4BC$ , and $CA'=4CA$ . We can evaluate the area of triangle $ABC$ by simply using Heron's formula, $[ABC]=\sqrt{\frac{3x}{2}\cdot {\left(\frac{3x}{2}-x\right)}^3}=\frac{x^2\sqrt{3}}{4}$ .
Next in order to evaluate $A'B'C'$ we need to evaluate the area of the larger triangles $AA'B',BB'C', \text{ and } CC'A'$ .
In this solution we shall just compute $1$ of these as the others are trivially equivalent.
In order to compute the area of $\Delta{AA'B'}$ we can use the formula $[XYZ]=\frac{1}{2}xy\cdot\sin{z}$ .
Since $ABC$ is equilateral and $A$ $B$ $B'$ are collinear, we already know $\angle{A'AB'}=180-60=120$ Similarly from above we know $AB'$ and $A'A$ to be $4x$ , and $3x$ respectively. Thus the area of $\Delta{AA'B'}$ is $\frac{1}{2}\cdot 4x\cdot 3x \cdot \sin{120}=3x^2\cdot\sqrt{3}$ . Likewise we can find $BB'C', \text{ and } CC'A'$ to also be $3x^2\cdot\sqrt{3}$ $[A'B'C']=[AA'B']+[BB'C']+[CC'A']+[ABC]=3\cdot3x^2\cdot\sqrt{3}+\frac{x^2\sqrt{3}}{4}=\sqrt{3}\cdot\left(9x^2+\frac{x^2}{4}\right)$ .
Therefore the ratio of $[A'B'C']$ to $[ABC]$ is $\frac{\sqrt{3}\cdot\left(9x^2+\frac{x^2}{4}\right)}{\frac{x^2\sqrt{3}}{4}}=\boxed{37}$ | E | 37 |
bd666a589736ea3ad22ec6ec2cd8af9b | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12B_Problems/Problem_15 | Let $ABC$ be an equilateral triangle. Extend side $\overline{AB}$ beyond $B$ to a point $B'$ so that $BB'=3 \cdot AB$ . Similarly, extend side $\overline{BC}$ beyond $C$ to a point $C'$ so that $CC'=3 \cdot BC$ , and extend side $\overline{CA}$ beyond $A$ to a point $A'$ so that $AA'=3 \cdot CA$ . What is the ratio of the area of $\triangle A'B'C'$ to the area of $\triangle ABC$
$\textbf{(A)}\ 9\qquad\textbf{(B)}\ 16\qquad\textbf{(C)}\ 25\qquad\textbf{(D)}\ 36\qquad\textbf{(E)}\ 37$ | Looking at the answer choices, we see that all but ${\textbf{(E)}}$ has a perfect square in the ratio. With some intuition, we can guess that the sidelength of the new triangle formed is not an integer, thus we pick $\boxed{37}$ | E | 37 |
bd666a589736ea3ad22ec6ec2cd8af9b | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12B_Problems/Problem_15 | Let $ABC$ be an equilateral triangle. Extend side $\overline{AB}$ beyond $B$ to a point $B'$ so that $BB'=3 \cdot AB$ . Similarly, extend side $\overline{BC}$ beyond $C$ to a point $C'$ so that $CC'=3 \cdot BC$ , and extend side $\overline{CA}$ beyond $A$ to a point $A'$ so that $AA'=3 \cdot CA$ . What is the ratio of the area of $\triangle A'B'C'$ to the area of $\triangle ABC$
$\textbf{(A)}\ 9\qquad\textbf{(B)}\ 16\qquad\textbf{(C)}\ 25\qquad\textbf{(D)}\ 36\qquad\textbf{(E)}\ 37$ | We use barycentric coordinates wrt $\triangle ABC$ , to which we can easily obtain that $A'=(4,0,-3)$ $B'=(-3,4,0)$ , and $C'=(0,-3,4)$ . Now, since the coordinates are homogenized ( $-3+4=1$ ), we can directly apply the area formula to obtain that \[[A'B'C']=[ABC]\cdot\left| \begin{array}{ccc} 4 & 0 & -3 \\ -3 & 4 & 0 \\ 0 & -3 & 4 \end{array} \right| = (64-27)[ABC]=37[ABC],\] so the answer is $\boxed{37}$ | E | 37 |
bd666a589736ea3ad22ec6ec2cd8af9b | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12B_Problems/Problem_15 | Let $ABC$ be an equilateral triangle. Extend side $\overline{AB}$ beyond $B$ to a point $B'$ so that $BB'=3 \cdot AB$ . Similarly, extend side $\overline{BC}$ beyond $C$ to a point $C'$ so that $CC'=3 \cdot BC$ , and extend side $\overline{CA}$ beyond $A$ to a point $A'$ so that $AA'=3 \cdot CA$ . What is the ratio of the area of $\triangle A'B'C'$ to the area of $\triangle ABC$
$\textbf{(A)}\ 9\qquad\textbf{(B)}\ 16\qquad\textbf{(C)}\ 25\qquad\textbf{(D)}\ 36\qquad\textbf{(E)}\ 37$ | First, comparing bases yields that $[BA'B']=3[AA'B]=9[ABC]\implies [AA'B']=12$ . By congruent triangles, \[[AA'B']=[BB'C']=[CC'A']\implies [A'B'C']=(12+12+12+1)[ABC],\] so $[A'B'C']:[ABC]=\boxed{37}$ | E | 37 |
bd666a589736ea3ad22ec6ec2cd8af9b | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12B_Problems/Problem_15 | Let $ABC$ be an equilateral triangle. Extend side $\overline{AB}$ beyond $B$ to a point $B'$ so that $BB'=3 \cdot AB$ . Similarly, extend side $\overline{BC}$ beyond $C$ to a point $C'$ so that $CC'=3 \cdot BC$ , and extend side $\overline{CA}$ beyond $A$ to a point $A'$ so that $AA'=3 \cdot CA$ . What is the ratio of the area of $\triangle A'B'C'$ to the area of $\triangle ABC$
$\textbf{(A)}\ 9\qquad\textbf{(B)}\ 16\qquad\textbf{(C)}\ 25\qquad\textbf{(D)}\ 36\qquad\textbf{(E)}\ 37$ | Scale down the figure so that the area formulas for the $120^\circ$ and equilateral triangles become proportional with proportionality constant equivalent to the product of the corresponding sides. By the proportionality, it becomes clear that the answer is $3*4*3+1*1=37, \boxed{37}$ .
~ Solution by mathchampion1 | E | 37 |
bd666a589736ea3ad22ec6ec2cd8af9b | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12B_Problems/Problem_15 | Let $ABC$ be an equilateral triangle. Extend side $\overline{AB}$ beyond $B$ to a point $B'$ so that $BB'=3 \cdot AB$ . Similarly, extend side $\overline{BC}$ beyond $C$ to a point $C'$ so that $CC'=3 \cdot BC$ , and extend side $\overline{CA}$ beyond $A$ to a point $A'$ so that $AA'=3 \cdot CA$ . What is the ratio of the area of $\triangle A'B'C'$ to the area of $\triangle ABC$
$\textbf{(A)}\ 9\qquad\textbf{(B)}\ 16\qquad\textbf{(C)}\ 25\qquad\textbf{(D)}\ 36\qquad\textbf{(E)}\ 37$ | Drawing the diagram, we see that the large triangle, $A'B'C'$ , is composed of three congruent triangles with the triangle $ABC$ at the center. Let each of the sides of triangle $ABC$ be $x$ . Therefore, using the equilateral triangle area formula, the $[ABC] = \frac{x^2\sqrt{3}}{4}$ . We also know now that the sides of the triangles are $3x$ and $3x + x$ , or $4x$ . We also know that since $BB'$ are collinear, as are the others, angle $C'BB'$ is $180 - 60$ , which is $120$ degrees. Because that angle is an included angle, we get the area of all three congruent triangle's are $\frac{12x^2\sin120}{2} \cdot 3$ . Simplifying that yields $\frac{36x^2\sqrt{3}}{4}$ . Adding that to the $[ABC]$ yields $\frac{37x^2\sqrt{3}}{4}$ . From this, we can compare the ratios by canceling everything out except for the $37$ , so the answer is $\boxed{37}$ | E | 37 |
bd666a589736ea3ad22ec6ec2cd8af9b | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12B_Problems/Problem_15 | Let $ABC$ be an equilateral triangle. Extend side $\overline{AB}$ beyond $B$ to a point $B'$ so that $BB'=3 \cdot AB$ . Similarly, extend side $\overline{BC}$ beyond $C$ to a point $C'$ so that $CC'=3 \cdot BC$ , and extend side $\overline{CA}$ beyond $A$ to a point $A'$ so that $AA'=3 \cdot CA$ . What is the ratio of the area of $\triangle A'B'C'$ to the area of $\triangle ABC$
$\textbf{(A)}\ 9\qquad\textbf{(B)}\ 16\qquad\textbf{(C)}\ 25\qquad\textbf{(D)}\ 36\qquad\textbf{(E)}\ 37$ | Solution by HydroQuantum
Let $AB=BC=CA=x$
Recall The Law of Cosines. Letting $A'B'=B'C'=C'A'=y$ \[y^2=(3x)^2+(x+3x)^2-2(3x)(x+3x)(\cos 120) =\] \[(3x)^2+(4x)^2-2(3x)(4x)(\cos 120)=9x^2+16x^2-24x(\cos 120)=25x^2+12x^2=37x^2.\] Since both $\triangle ABC$ and $\triangle A'B'C'$ are both equilateral triangles, they must be similar due to $AA$ similarity. This means that $\frac{A'B'}{AB}$ $=$ $\frac{B'C'}{BC}$ $=$ $\frac{C'A'}{CA}$ $=$ $\frac{[\triangle A'B'C']}{[\triangle ABC]}$ $=$ $\frac{37}{1}$
Therefore, our answer is $\boxed{37}$ | E | 37 |
bd666a589736ea3ad22ec6ec2cd8af9b | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12B_Problems/Problem_15 | Let $ABC$ be an equilateral triangle. Extend side $\overline{AB}$ beyond $B$ to a point $B'$ so that $BB'=3 \cdot AB$ . Similarly, extend side $\overline{BC}$ beyond $C$ to a point $C'$ so that $CC'=3 \cdot BC$ , and extend side $\overline{CA}$ beyond $A$ to a point $A'$ so that $AA'=3 \cdot CA$ . What is the ratio of the area of $\triangle A'B'C'$ to the area of $\triangle ABC$
$\textbf{(A)}\ 9\qquad\textbf{(B)}\ 16\qquad\textbf{(C)}\ 25\qquad\textbf{(D)}\ 36\qquad\textbf{(E)}\ 37$ | Note that the height and base of $\triangle A'CC'$ are respectively 4 times and 3 times that of $\triangle ABC$ . Therefore the area of $\triangle A'CC'$ is 12 times that of $\triangle ABC$
By symmetry, $\triangle A'CC' \cong \triangle B'AA' \cong \triangle C'BB'$ . Adding the areas of these three triangles and $\triangle ABC$ for the total area of $\triangle A'B'C'$ gives a ratio of $(12 + 12 + 12 + 1) : 1$ , or $\boxed{37}$ | E | 37 |
bd666a589736ea3ad22ec6ec2cd8af9b | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12B_Problems/Problem_15 | Let $ABC$ be an equilateral triangle. Extend side $\overline{AB}$ beyond $B$ to a point $B'$ so that $BB'=3 \cdot AB$ . Similarly, extend side $\overline{BC}$ beyond $C$ to a point $C'$ so that $CC'=3 \cdot BC$ , and extend side $\overline{CA}$ beyond $A$ to a point $A'$ so that $AA'=3 \cdot CA$ . What is the ratio of the area of $\triangle A'B'C'$ to the area of $\triangle ABC$
$\textbf{(A)}\ 9\qquad\textbf{(B)}\ 16\qquad\textbf{(C)}\ 25\qquad\textbf{(D)}\ 36\qquad\textbf{(E)}\ 37$ | First we note that $A'B'C'\sim ABC$ due to symmetry. WLOG, let $B = (0, 0)$ and $AB = 1$ Therefore, $C = (1, 0), A = \frac{1}{2}, \left(\frac{\sqrt{3}}{2}\right)$ . Using the condition that $CC' = 3$ , we get $C' = (4, 0)$ and $B' = \left(\frac{-3}{2}, \frac{-3\sqrt{3}}{2}\right)$ . It is easy to check that $B'C' = \sqrt{37}$ . Since the area ratios of two similar figures is the square of the ratio of their lengths, the ratio is $\boxed{37}$ | E | 37 |
bd666a589736ea3ad22ec6ec2cd8af9b | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12B_Problems/Problem_15 | Let $ABC$ be an equilateral triangle. Extend side $\overline{AB}$ beyond $B$ to a point $B'$ so that $BB'=3 \cdot AB$ . Similarly, extend side $\overline{BC}$ beyond $C$ to a point $C'$ so that $CC'=3 \cdot BC$ , and extend side $\overline{CA}$ beyond $A$ to a point $A'$ so that $AA'=3 \cdot CA$ . What is the ratio of the area of $\triangle A'B'C'$ to the area of $\triangle ABC$
$\textbf{(A)}\ 9\qquad\textbf{(B)}\ 16\qquad\textbf{(C)}\ 25\qquad\textbf{(D)}\ 36\qquad\textbf{(E)}\ 37$ | Note that angle $C'BB'$ is $120$ °, as it is supplementary to the equilateral triangle. Then, using area $= \frac{1}{2}ab\sin\theta$ and letting side $AB = 1$ for ease, we get: $4\cdot3\cdot\frac{\sin120}{2} = 3\sqrt{3}$ as the area of $C'BB'$ . Then, the area of $ABC$ is $\frac{\sqrt{3}}{4}$ , so the ratio is $\frac{3(3\sqrt{3})+\frac{\sqrt{3}}{4}}{\frac{\sqrt{3}}{4}} = \boxed{37}$ | E | 37 |
dc2a5c2a18b684866411520a74a7cefe | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12B_Problems/Problem_19 | Let $N=123456789101112\dots4344$ be the $79$ -digit number that is formed by writing the integers from $1$ to $44$ in order, one after the other. What is the remainder when $N$ is divided by $45$
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 18\qquad\textbf{(E)}\ 44$ | We will consider this number $\bmod\ 5$ and $\bmod\ 9$ . By looking at the last digit, it is obvious that the number is $\equiv 4\bmod\ 5$ . To calculate the number $\bmod\ 9$ , note that
\[123456\cdots 4344 \equiv 1+2+3+4+5+6+7+8+9+(1+0)+(1+1)+\cdots+(4+3)+(4+4) \equiv 1+2+\cdots+44 \bmod\ 9,\]
so it is equivalent to
\[\frac{44\cdot 45}{2} = 22\cdot 45 \equiv 0\bmod\ 9.\]
Let $x$ be the remainder when this number is divided by $45$ . We know that $x\equiv 0 \pmod {9}$ and $x\equiv 4 \pmod {5}$ , so by the Chinese remainder theorem, since $9(-1)\equiv 1 \pmod{5}$ $x\equiv 5(0)+9(-1)(4) \pmod {5\cdot 9}$ , or $x\equiv -36 \equiv 9 \pmod {45}$ . So the answer is $\boxed{9}$ | C | 9 |
dc2a5c2a18b684866411520a74a7cefe | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12B_Problems/Problem_19 | Let $N=123456789101112\dots4344$ be the $79$ -digit number that is formed by writing the integers from $1$ to $44$ in order, one after the other. What is the remainder when $N$ is divided by $45$
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 18\qquad\textbf{(E)}\ 44$ | We know that this number is divisible by $9$ because the sum of the digits is $270$ , which is divisible by $9$ . If we subtracted $9$ from the integer we would get $1234 \cdots 4335$ , which is also divisible by $5$ and by $45$ . Thus the remainder is $9$ , or $\boxed{9}$ | C | 9 |
dc2a5c2a18b684866411520a74a7cefe | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12B_Problems/Problem_19 | Let $N=123456789101112\dots4344$ be the $79$ -digit number that is formed by writing the integers from $1$ to $44$ in order, one after the other. What is the remainder when $N$ is divided by $45$
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 18\qquad\textbf{(E)}\ 44$ | To find the sum of digits of our number, we break it up into $5$ cases, starting with $0$ $1$ $2$ $3$ , or $4$
Case 1: $1+2+3+\cdots+9 = 45$
Case 2: $1+0+1+1+1+2+\cdots+1+8+1+9 = 55$ (We add 10 to the previous cases, as we are in the next ten's place)
Case 3: $2+0+2+1+\cdots+2+9 = 65$
Case 4: $3+0+3+1+\cdots+3+9 = 75$
Case 5: $4+0+4+1+\cdots+4+4 = 30$
Thus the sum of the digits is $45+55+65+75+30 = 270$ , so the number is divisible by $9$ . We notice that the number ends in " $4$ ", which is $9$ more than a multiple of $5$ . Thus if we subtracted $9$ from our number it would be divisible by $5$ , and $5\cdot 9 = 45$ . (Multiple of n - n = Multiple of n)
So our remainder is $\boxed{9}$ , the value we need to add to the multiple of $45$ to get to our number. | C | 9 |
dc2a5c2a18b684866411520a74a7cefe | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12B_Problems/Problem_19 | Let $N=123456789101112\dots4344$ be the $79$ -digit number that is formed by writing the integers from $1$ to $44$ in order, one after the other. What is the remainder when $N$ is divided by $45$
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 18\qquad\textbf{(E)}\ 44$ | We notice that $10^{k}\equiv 10 \pmod {45}$
Hence $N = 44+43\cdot10^{2}+42\cdot10^{4}+\cdots+10^{78} \equiv 44+10\cdot(1+2+3+\cdots+43)\equiv 9 \pmod {45}.$
Choose $\boxed{9}$ | C | 9 |
cfa9ee62bc46a4c03cdcf982674f888c | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12B_Problems/Problem_21 | Last year, Isabella took 7 math tests and received 7 different scores, each an integer between 91 and 100, inclusive. After each test she noticed that the average of her test scores was an integer. Her score on the seventh test was 95. What was her score on the sixth test?
$\textbf{(A)}\ 92\qquad\textbf{(B)}\ 94\qquad\textbf{(C)}\ 96\qquad\textbf{(D)}\ 98\qquad\textbf{(E)}\ 100$ | First, remove all the 90s, since they make no impact. So, we have numbers from $1$ to $10$ . Then, $5$ is the 7th number. Let the sum of the first $6$ numbers be $k$ . Then, $k\equiv 0 \mod 6$ and $k\equiv 3 \mod 7$ . We easily solve this as $k \equiv 24 \mod 42$ . Clearly, the sum of the first $6$ numbers must be $570$ . In order for the first $5$ numbers to sum to a multiple of $5$ , the sixth number must also be a multiple of $5$ , since $30$ is a multiple of $5$ . Thus, the only option is the sixth number is $10$ , which gives $10+90= \boxed{100}$ | E | 100 |
cfa9ee62bc46a4c03cdcf982674f888c | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12B_Problems/Problem_21 | Last year, Isabella took 7 math tests and received 7 different scores, each an integer between 91 and 100, inclusive. After each test she noticed that the average of her test scores was an integer. Her score on the seventh test was 95. What was her score on the sixth test?
$\textbf{(A)}\ 92\qquad\textbf{(B)}\ 94\qquad\textbf{(C)}\ 96\qquad\textbf{(D)}\ 98\qquad\textbf{(E)}\ 100$ | Let us simplify the problem. Since all of Isabella's test scores can be expressed as the sum of $90$ and an integer between $1$ and $10$ , we rewrite the problem into receiving scores between $1$ and $10$ . Later, we can add $90$ to her score to obtain the real answer.
From this point of view, the problem states that Isabella's score on the seventh test was $5$ . We note that Isabella received $7$ integer scores out of $1$ to $10$ . Since $5$ is already given as the seventh test score, the possible scores for Isabella on the other six tests are $S={1,2,3,4,6,7,8,9,10}$
The average score for the seven tests must be an integer. In other words, six distinct integers must be picked from set $S$ above, and their sum with $5$ must be a multiple of $7$ . The interval containing the possible sums of the six numbers in S are from $1 +2+3+4+6+7=23$ to $4+6+7+8+9+10=44$ . We must now find multiples of $7$ from the interval $23+5 = 28$ to $44+5=49$ . There are four possibilities: $28$ $35$ $42$ $49$ .
However, we also note that the sum of the six numbers (besides $5$ ) must be a multiple of $6$ as well. Thus, $35$ is the only valid choice.(The six numbers sum to $30$ .)
Thus the sum of the six numbers equals to $30$ . We apply the logic above in a similar way for the sum of the scores from the first test to the fifth test. The sum must be a multiple of $5$ . The possible interval is from $1+2+3+4+6=16$ to $6+7+8+9+10=40$ . Since the sum of the five scores must be less than $30$ , the only possibilities are $20$ and $25$ . However, we notice that $25$ does not work because the seventh score turns out to be $5$ from the calculation. Therefore, the sum of Isabella's scores from test $1$ to $5$ is $20$ . Therefore, her score on the sixth test is $10$ .
Our final answer is $10+90= \boxed{100}$ | E | 100 |
cfa9ee62bc46a4c03cdcf982674f888c | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12B_Problems/Problem_21 | Last year, Isabella took 7 math tests and received 7 different scores, each an integer between 91 and 100, inclusive. After each test she noticed that the average of her test scores was an integer. Her score on the seventh test was 95. What was her score on the sixth test?
$\textbf{(A)}\ 92\qquad\textbf{(B)}\ 94\qquad\textbf{(C)}\ 96\qquad\textbf{(D)}\ 98\qquad\textbf{(E)}\ 100$ | Let $n$ be Isabella's average after $6$ tests. $6n+95 \equiv 0 \pmod{7}$ , so $n \equiv 4 \pmod{7}$ . The only integer between $90$ and $100$ that satisfies this condition is $95$ . Let $m$ be Isabella's average after $5$ tests, and let $a$ be her sixth test score. $5m+a= 6\cdot95 \equiv 0 \pmod{5}$ , so $a$ is a multiple of $5$ . Since $100$ is the only choice that is a multiple of $5$ , the answer is $\boxed{100}$ | E | 100 |
cfa9ee62bc46a4c03cdcf982674f888c | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12B_Problems/Problem_21 | Last year, Isabella took 7 math tests and received 7 different scores, each an integer between 91 and 100, inclusive. After each test she noticed that the average of her test scores was an integer. Her score on the seventh test was 95. What was her score on the sixth test?
$\textbf{(A)}\ 92\qquad\textbf{(B)}\ 94\qquad\textbf{(C)}\ 96\qquad\textbf{(D)}\ 98\qquad\textbf{(E)}\ 100$ | Let $T$ be the total sum of Isabella's first five test scores, and let $S$ be her score on the sixth test. It follows that $T\equiv 0\pmod {5}$ $T+S\equiv 0\pmod {6}$ , and $T+S+95\equiv 0\pmod {7}$ , since at each step, her average score was an integer. Using the last equivalence, $T+S\equiv -95\equiv 3\pmod{7}$ , so we have a system of equivalences for $T+S$ . Solving this using the Chinese Remainder Theorem, we get $T+S\equiv (0)(7)(1)+(3)(6)(-1) \equiv -18 \equiv 24 \pmod{42}$
Now let's put a bound on $T$ . Using the given information that each test score was a distinct integer from $91$ to $100$ inclusive and that the seventh score was 95, we get $91+92+93+94+96\leq T\leq 100+99+98+97+96$ . Since $T\equiv 0 \pmod{5}$ , we get $T=470,475,480,485,490$ . Therefore, $T\equiv 8,13,18,23,28\pmod{42}$
The last preparation step will involve calculating all the possible test scores $\pmod {42}$ . Here they are: $91\equiv 7\pmod{42},92\equiv 8\pmod{42},93\equiv 9\pmod{42},94\equiv 10\pmod{42},95\equiv 11\pmod{42},96\equiv 12\pmod{42},97\equiv 13\pmod{42},98\equiv 14\pmod{42},99\equiv 15\pmod{42},100\equiv 16\pmod{42}$ . This means that $S\equiv 7,8,9,10,12,13,14,15,16\pmod{42}$ . Note that $11$ is not in the previous list because it corresponds to a score of $95$ , which we cannot have.
We must have $T+S\equiv 24\pmod{42}$ , and using the possible values we found for $T$ and $S$ , the only two that sum to $24$ are $T\equiv 8\pmod{42}$ and $S\equiv 16\pmod{42}$ . This corresponds to an $S$ value of $100$ , so the answer is $\boxed{100}$ | E | 100 |
cfa9ee62bc46a4c03cdcf982674f888c | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12B_Problems/Problem_21 | Last year, Isabella took 7 math tests and received 7 different scores, each an integer between 91 and 100, inclusive. After each test she noticed that the average of her test scores was an integer. Her score on the seventh test was 95. What was her score on the sixth test?
$\textbf{(A)}\ 92\qquad\textbf{(B)}\ 94\qquad\textbf{(C)}\ 96\qquad\textbf{(D)}\ 98\qquad\textbf{(E)}\ 100$ | We know that Isabella's score on the 7th test is 95, and we work backwards. Let $a_n$ denote the average of the first $n$ scores. We are given that all $a_n$ are integers. If the $a_6$ is different from the $a_7$ , then the last score must be a multiple of 7 away from $a_6$ . However, $95-7=88$ and $95+7=102$ clearly cannot be $a_6$ since they are outside the range of 91 to 100. Thus, $a_6$ must be 95. Additionally, $a_5$ cannot be 95 since that would mean the sixth score is 95, which isn't possible since scores can't repeat (also, it's not an answer choice). Intuitively, $a_5$ should be close to 95 due to the constraint on the range of scores. We can try 94 and 96 for $a_5$ . Again, the sixth score must be a multiple of 6 (preferably only one multiple) away from $a_5$ . The only one that works is $94+6=100$ , and we can check that the rest don't work since they exceed the range. The answer is $\boxed{100}$ | E | 100 |
cfa9ee62bc46a4c03cdcf982674f888c | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12B_Problems/Problem_21 | Last year, Isabella took 7 math tests and received 7 different scores, each an integer between 91 and 100, inclusive. After each test she noticed that the average of her test scores was an integer. Her score on the seventh test was 95. What was her score on the sixth test?
$\textbf{(A)}\ 92\qquad\textbf{(B)}\ 94\qquad\textbf{(C)}\ 96\qquad\textbf{(D)}\ 98\qquad\textbf{(E)}\ 100$ | Say the sum of the first 6 scores is $n$ . Then, we know that $n + 95 \equiv 0 \pmod{7}$ . Thus, $n \equiv 3 \pmod{7}$ . Additionally, since the average of these 6 scores was an integer, we know that $n \equiv 0 \pmod{6}$ . We find that the smallest $n$ to satisfy both of these inequalities is 24 (series below). \[3, 10, 17, 24, ...\] To get to the next one, we have to add $6 \cdot 7 = 42$ so as not to ruin the moduli.
Next, we know that $n$ is between $91 \cdot 6 = 546$ and $100 \cdot 6 = 600$ . It can't take on either value as all the scores have to be different. Now, we calculate the first $n$ that satisfies both moduli equations:
\[546 < 42 + 24x\] \[91 < 7 + 4x\] \[84 < 4x\] \[21 < x\]
$x = 22$ works while $x = 23$ does not (as it hits the upper limit of 600), and we find that $n = 42 + 24 \cdot 22 = 570$ . So, the sum of the first 6 scores is 570. We also know that the sum of the first 5 scores is a multiple of 5. Since $5~\mid~570$ , the 6th score must be a multiple of 5. 95 is taken, so the only possibility $\boxed{100}$ | E | 100 |
eccaf701517e2aa284d3c7684ff0bdc5 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12B_Problems/Problem_25 | A set of $n$ people participate in an online video basketball tournament. Each person may be a member of any number of $5$ -player teams, but no two teams may have exactly the same $5$ members. The site statistics show a curious fact: The average, over all subsets of size $9$ of the set of $n$ participants, of the number of complete teams whose members are among those $9$ people is equal to the reciprocal of the average, over all subsets of size $8$ of the set of $n$ participants, of the number of complete teams whose members are among those $8$ people. How many values $n$ $9\leq n\leq 2017$ , can be the number of participants?
$\textbf{(A) } 477 \qquad \textbf{(B) } 482 \qquad \textbf{(C) } 487 \qquad \textbf{(D) } 557 \qquad \textbf{(E) } 562$ | Let there be $T$ teams. For each team, there are ${n-5\choose 4}$ different subsets of $9$ players that includes a given full team, so the total number of team-(group of 9) pairs is
\[T{n-5\choose 4}.\]
Thus, the expected value of the number of full teams in a random set of $9$ players is
\[\frac{T{n-5\choose 4}}{{n\choose 9}}.\]
Similarly, the expected value of the number of full teams in a random set of $8$ players is
\[\frac{T{n-5\choose 3}}{{n\choose 8}}.\]
The condition is thus equivalent to the existence of a positive integer $T$ such that
\[\frac{T{n-5\choose 4}}{{n\choose 9}}\frac{T{n-5\choose 3}}{{n\choose 8}} = 1.\]
\[T^2\frac{(n-5)!(n-5)!8!9!(n-8)!(n-9)!}{n!n!(n-8)!(n-9)!3!4!} = 1\]
\[T^2 = \big((n)(n-1)(n-2)(n-3)(n-4)\big)^2 \frac{3!4!}{8!9!}\]
\[T^2 = \big((n)(n-1)(n-2)(n-3)(n-4)\big)^2 \frac{144}{7!7!8\cdot8\cdot9}\]
\[T^2 = \big((n)(n-1)(n-2)(n-3)(n-4)\big)^2 \frac{1}{4\cdot7!7!}\]
\[T = \frac{(n)(n-1)(n-2)(n-3)(n-4)}{2^5\cdot3^2\cdot5\cdot7}\]
Note that this is always less than ${n\choose 5}$ , so as long as $T$ is integral, $n$ is a possibility. Thus, we have that this is equivalent to
\[2^5\cdot3^2\cdot5\cdot7\big|(n)(n-1)(n-2)(n-3)(n-4).\]
It is obvious that $5$ divides the RHS, and that $7$ does iff $n\equiv 0,1,2,3,4\mod 7$ . Also, $3^2$ divides it iff $n\not\equiv 5,8\mod 9$ . One can also bash out that $2^5$ divides it in $16$ out of the $32$ possible residues $\mod 32$
Note that $2016 = 7*9*32$ so by using all numbers from $2$ to $2017$ , inclusive, it is clear that each possible residue $\mod 7,9,32$ is reached an equal number of times, so the total number of working $n$ in that range is $5\cdot 7\cdot 16 = 560$ . However, we must subtract the number of "working" $2\leq n\leq 8$ , which is $3$ . Thus, the answer is $\boxed{557}$ | D | 557 |
fabf50d3c9036707a682a56bd4792fb1 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_1 | What is the value of $\dfrac{11!-10!}{9!}$
$\textbf{(A)}\ 99\qquad\textbf{(B)}\ 100\qquad\textbf{(C)}\ 110\qquad\textbf{(D)}\ 121\qquad\textbf{(E)}\ 132$ | We can use subtraction of fractions to get \[\frac{11!-10!}{9!} = \frac{11!}{9!} - \frac{10!}{9!} = 110 -10 = \boxed{100}.\] | B | 100 |
fabf50d3c9036707a682a56bd4792fb1 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_1 | What is the value of $\dfrac{11!-10!}{9!}$
$\textbf{(A)}\ 99\qquad\textbf{(B)}\ 100\qquad\textbf{(C)}\ 110\qquad\textbf{(D)}\ 121\qquad\textbf{(E)}\ 132$ | Factoring out $9!$ gives $\frac{11!-10!}{9!} = \frac{9!(11 \cdot 10 - 10)}{9!} = 110-10=\boxed{100}$ | B | 100 |
fabf50d3c9036707a682a56bd4792fb1 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_1 | What is the value of $\dfrac{11!-10!}{9!}$
$\textbf{(A)}\ 99\qquad\textbf{(B)}\ 100\qquad\textbf{(C)}\ 110\qquad\textbf{(D)}\ 121\qquad\textbf{(E)}\ 132$ | We are given the equation $\frac{11!-10!}{9!}$
This is equivalent to $\frac{11(10!) - 1(10!)}{9!}$ Simplifying, we get $\frac{(11-1)(10!)}{9!}$ , which equals $10 \cdot 10$
Therefore, the answer is $10^2$ $\boxed{100}$ | B | 100 |
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