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7ddfc86c081a3e2bb0f1824477c7c839 | https://artofproblemsolving.com/wiki/index.php/2012_AIME_I_Problems/Problem_6 | The complex numbers $z$ and $w$ satisfy $z^{13} = w,$ $w^{11} = z,$ and the imaginary part of $z$ is $\sin{\frac{m\pi}{n}}$ , for relatively prime positive integers $m$ and $n$ with $m<n.$ Find $n.$ | Substituting the first equation into the second, we find that $(z^{13})^{11} = z$ and thus $z^{143} = z.$ We know that $z \neq 0,$ because we are given the imaginary part of $z,$ so we can divide by $z$ to get $z^{142} = 1.$ So, $z$ must be a $142$ nd root of unity, and thus, by De Moivre's theorem, the imaginary part of $z$ will be of the form $\sin{\frac{2k\pi}{142}} = \sin{\frac{k\pi}{71}},$ where $k \in \{1, 2, \ldots, 70\}.$ Note that $71$ is prime and $k<71$ by the conditions of the problem, so the denominator in the argument of this value will always be $71.$ Thus, $n = \boxed{071}.$ | null | 071 |
d1fca2a0a40ccb9308cc69150f61bf92 | https://artofproblemsolving.com/wiki/index.php/2012_AIME_I_Problems/Problem_7 | At each of the sixteen circles in the network below stands a student. A total of $3360$ coins are distributed among the sixteen students. All at once, all students give away all their coins by passing an equal number of coins to each of their neighbors in the network. After the trade, all students have the same number of coins as they started with. Find the number of coins the student standing at the center circle had originally. | Say the student in the center starts out with $a$ coins, the students neighboring the center student each start with $b$ coins, and all other students start out with $c$ coins. Then the $a$ -coin student has five neighbors, all the $b$ -coin students have three neighbors, and all the $c$ -coin students have four neighbors.
Now in order for each student's number of coins to remain equal after the trade, the number of coins given by each student must be equal to the number received, and thus
\begin{align*} a &= 5 \cdot \frac{b}{3}\\ b &= \frac{a}{5} + 2 \cdot \frac{c}{4}\\ c &= 2 \cdot \frac{c}{4} + 2 \cdot \frac{b}{3}. \end{align*}
Solving these equations, we see that $\frac{a}{5} = \frac{b}{3} = \frac{c}{4}.$ Also, the total number of coins is $a + 5b + 10c = 3360,$ so $a + 5 \cdot \frac{3a}{5} + 10 \cdot \frac{4a}{5} = 3360 \rightarrow a = \frac{3360}{12} = \boxed{280}.$ | null | 280 |
d1fca2a0a40ccb9308cc69150f61bf92 | https://artofproblemsolving.com/wiki/index.php/2012_AIME_I_Problems/Problem_7 | At each of the sixteen circles in the network below stands a student. A total of $3360$ coins are distributed among the sixteen students. All at once, all students give away all their coins by passing an equal number of coins to each of their neighbors in the network. After the trade, all students have the same number of coins as they started with. Find the number of coins the student standing at the center circle had originally. | Since the students give the same number of gifts of coins as they receive and still end up the same number of coins, we can assume that every gift of coins has the same number of coins. Let $x$ be the number of coins in each gift of coins. There $10$ people who give $4$ gifts of coins, $5$ people who give $3$ gifts of coins, and $1$ person who gives $5$ gifts of coins. Thus,
\begin{align*} 10(4x)+5(3x)+5x &= 3360\\ 40x+15x+5x &= 3360\\ 60x &= 3360\\ x &= 56 \end{align*} Therefore the answer is $5(56) = \boxed{280}.$ | null | 280 |
d1fca2a0a40ccb9308cc69150f61bf92 | https://artofproblemsolving.com/wiki/index.php/2012_AIME_I_Problems/Problem_7 | At each of the sixteen circles in the network below stands a student. A total of $3360$ coins are distributed among the sixteen students. All at once, all students give away all their coins by passing an equal number of coins to each of their neighbors in the network. After the trade, all students have the same number of coins as they started with. Find the number of coins the student standing at the center circle had originally. | Mark the number of coins from inside to outside as $a$ $b_1$ $b_2$ $b_3$ $b_4$ $b_5$ $c_1$ $c_2$ $c_3$ $c_4$ $c_5$ $d_1$ $d_2$ $d_3$ $d_4$ $d_5$ .
Then, we obtain \begin{align*} d_1 &= \frac{d_5}{4} + \frac{d_2}{4} + \frac{c_1}{4} + \frac{c_2}{4}\\ d_2 &= \frac{d_1}{4} + \frac{d_3}{4} + \frac{c_2}{4} + \frac{c_3}{4}\\ d_3 &= \frac{d_2}{4} + \frac{d_4}{4} + \frac{c_3}{4} + \frac{c_4}{4}\\ d_4 &= \frac{d_3}{4} + \frac{d_5}{4} + \frac{c_4}{4} + \frac{c_5}{4}\\ d_5 &= \frac{d_4}{4} + \frac{d_1}{4} + \frac{c_5}{4} + \frac{c_1}{4}\\ \end{align*} Letting $D = d_1 + d_2 + d_3 + d_4 + d_5$ $C = c_1 + c_2 + c_3 + c_4 + c_5$ gets us $D = \frac{D}{4} + \frac{D}{4} + \frac{C}{4} + \frac{C}{4}$ and $D = C$ .
In the same way, $C = \frac{D}{4} + \frac{D}{4} + \frac{B}{3} + \frac{B}{3}$ $B = \frac{3D}{4}$ $B = \frac{C}{4} + \frac{C}{4} + a$ $a = \frac{D}{4}$ .
Then, with $a + B + C + D = 3360$ $D = 1120$ $a = \boxed{280}$ | null | 280 |
d1fca2a0a40ccb9308cc69150f61bf92 | https://artofproblemsolving.com/wiki/index.php/2012_AIME_I_Problems/Problem_7 | At each of the sixteen circles in the network below stands a student. A total of $3360$ coins are distributed among the sixteen students. All at once, all students give away all their coins by passing an equal number of coins to each of their neighbors in the network. After the trade, all students have the same number of coins as they started with. Find the number of coins the student standing at the center circle had originally. | Define $x$ as the number of coins the student in the middle has. Since this student connects to $5$ other students, each of those students must have passed $\dfrac15 x$ coins to the center to maintain the same number of coins.
Each of these students connect to $3$ other students, passing $\dfrac15 x$ coins to each, so they must have $\dfrac35 x$ coins. These students must then recieve $\dfrac35 x$ coins, $\dfrac 15 x$ of which were given to by the center student. Thus, they must also have received $\dfrac25 x$ coins from the outer layer, and since this figure has symmetry, these must be the same.
Each of the next layer of students must have given $\dfrac15 x$ coins. Since they gave $4$ people coins, they must have started with $\dfrac45 x$ coins. They received $\dfrac25 x$ of them from the inner layer, making the two other connections have given them the same amount. By a similar argument, they recieved $\dfrac15 x$ from each of them.
By a similar argument, the outermost hexagon of students must have had $\dfrac45 x$ coins each. Summing this all up, we get the number of total coins passed out as a function of $x$ . This ends up to be \begin{align*} \dfrac45 x + \dfrac45 x + \dfrac45 x + \dfrac45 x + \dfrac45 x + \dfrac45 x + \dfrac45 x + \dfrac45 x + \dfrac45 x + \dfrac45 x + \dfrac35 x + \dfrac35 x + \dfrac35 x + \dfrac35 x + \dfrac35 x + x &= 10 \cdot \dfrac45 x + 5 \cdot \dfrac35 x + x \\ &= 8x + 3x + x \\ &= 12 x. \end{align*}
Since this all sums up to $3360$ , which is given, we find that \begin{align*} 12x &= 3360 \\ x &= 280 \end{align*}
We defined $x$ to be the number of coins that the center person has, so the answer is $x$ , which is $\boxed{280}$ | null | 280 |
d2509b04bb0dbe0d59e06a29f5b4aea7 | https://artofproblemsolving.com/wiki/index.php/2012_AIME_I_Problems/Problem_8 | Cube $ABCDEFGH,$ labeled as shown below, has edge length $1$ and is cut by a plane passing through vertex $D$ and the midpoints $M$ and $N$ of $\overline{AB}$ and $\overline{CG}$ respectively. The plane divides the cube into two solids. The volume of the larger of the two solids can be written in the form $\tfrac{p}{q},$ where $p$ and $q$ are relatively prime positive integers. Find $p+q.$ | Define a coordinate system with $D$ at the origin and $C,$ $A,$ and $H$ on the $x$ $y$ , and $z$ axes respectively. Then $D=(0,0,0),$ $M=(.5,1,0),$ and $N=(1,0,.5).$ It follows that the plane going through $D,$ $M,$ and $N$ has equation $2x-y-4z=0.$ Let $Q = (1,1,.25)$ be the intersection of this plane and edge $BF$ and let $P = (1,2,0).$ Now since $2(1) - 1(2) - 4(0) = 0,$ $P$ is on the plane. Also, $P$ lies on the extensions of segments $DM,$ $NQ,$ and $CB$ so the solid $DPCN = DMBCQN + MPBQ$ is a right triangular pyramid. Note also that pyramid $MPBQ$ is similar to $DPCN$ with scale factor $.5$ and thus the volume of solid $DMBCQN,$ which is one of the solids bounded by the cube and the plane, is $[DPCN] - [MPBQ] = [DPCN] - \left(\frac{1}{2}\right)^3[DPCN] = \frac{7}{8}[DPCN].$ But the volume of $DPCN$ is simply the volume of a pyramid with base $1$ and height $.5$ which is $\frac{1}{3} \cdot 1 \cdot .5 = \frac{1}{6}.$ So $[DMBCQN] = \frac{7}{8} \cdot \frac{1}{6} = \frac{7}{48}.$ Note, however, that this volume is less than $.5$ and thus this solid is the smaller of the two solids. The desired volume is then $[ABCDEFGH] - [DMBCQN] = 1 - \frac{7}{48} = \frac{41}{48} \rightarrow p+q = \boxed{089.}$ | null | 089. |
d2509b04bb0dbe0d59e06a29f5b4aea7 | https://artofproblemsolving.com/wiki/index.php/2012_AIME_I_Problems/Problem_8 | Cube $ABCDEFGH,$ labeled as shown below, has edge length $1$ and is cut by a plane passing through vertex $D$ and the midpoints $M$ and $N$ of $\overline{AB}$ and $\overline{CG}$ respectively. The plane divides the cube into two solids. The volume of the larger of the two solids can be written in the form $\tfrac{p}{q},$ where $p$ and $q$ are relatively prime positive integers. Find $p+q.$ | Define a coordinate system with $D = (0,0,0)$ $M = (1, \frac{1}{2}, 0)$ $N = (0,1,\frac{1}{2})$ . The plane formed by $D$ $M$ , and $N$ is $z = \frac{y}{2} - \frac{x}{4}$ . It intersects the base of the unit cube at $y = \frac{x}{2}$ . The z-coordinate of the plane never exceeds the height of the unit cube for $0 \leq x \leq 1, 0 \leq y \leq 1$ . Therefore, the volume of one of the two regions formed by the plane is \[\int_0^1 \int_{\frac{x}{2}}^1 \int_0^{\frac{y}{2}-\frac{x}{4}}dz\,dy\,dx = \frac{7}{48}\] Since $\frac{7}{48} < \frac{1}{2}$ , our answer is $1-\frac{7}{48} = \frac{41}{48} \rightarrow p+q = \boxed{089}$ | null | 089 |
d2509b04bb0dbe0d59e06a29f5b4aea7 | https://artofproblemsolving.com/wiki/index.php/2012_AIME_I_Problems/Problem_8 | Cube $ABCDEFGH,$ labeled as shown below, has edge length $1$ and is cut by a plane passing through vertex $D$ and the midpoints $M$ and $N$ of $\overline{AB}$ and $\overline{CG}$ respectively. The plane divides the cube into two solids. The volume of the larger of the two solids can be written in the form $\tfrac{p}{q},$ where $p$ and $q$ are relatively prime positive integers. Find $p+q.$ | Same coordinate system, but we will use domains instead of using triple integrals. By the way, the method to obtain the equation of the plane is the cross product ( $\vec{[1,.5,0]}\times\vec{[0,1,.5]}=\vec{[.25,-.5,1]}$ ). We can multiply this vector by $-4$ to make things look cleaner and get $\vec{[-1,2,-4]}$ . We then get the desired plane, $-x+2y-4z=0$ , or $z=\frac{2y-x}{4}$ . We use a double integral with a Type I domain. Observing the diagram, the domain is where $0 \leq x \leq 1, .5x \leq y \leq 1$ . The integral is then \[.25\int_0^1\int_{.5x}^1 2y - x\] which becomes \[.25\int_0^1 (y^2-xy)|_{.5x}^1\] which becomes \[.25\int_0^1 1-x+.25x^2\] which then becomes \[.25(1-\frac{1}{2}+\frac{1}{12})\] and finally \[\frac{7}{48}\] So our answer is $1^3-\frac{7}{48} = \frac{41}{48}, 41+48 = \boxed{089}$ | null | 089 |
d2509b04bb0dbe0d59e06a29f5b4aea7 | https://artofproblemsolving.com/wiki/index.php/2012_AIME_I_Problems/Problem_8 | Cube $ABCDEFGH,$ labeled as shown below, has edge length $1$ and is cut by a plane passing through vertex $D$ and the midpoints $M$ and $N$ of $\overline{AB}$ and $\overline{CG}$ respectively. The plane divides the cube into two solids. The volume of the larger of the two solids can be written in the form $\tfrac{p}{q},$ where $p$ and $q$ are relatively prime positive integers. Find $p+q.$ | Let $Q$ be the intersection of the plane with edge $FB,$ then $\triangle MQB$ is similar to $\triangle DNC$ and the volume $[DNCMQB]$ is a sum of areas of cross sections of similar triangles running parallel to face $ABFE.$ Let $x$ be the distance from face $ABFE,$ let $h$ be the height parallel to $AB$ of the cross-sectional triangle at that distance, and $a$ be the area of the cross-sectional triangle at that distance. $A(x)=\frac{h^2}{4},$ and $h=\frac{x+1}{2},$ then $A=\frac{(x+1)^2}{16}$ , and the volume $[DNCMQB]$ is $\int^1_0{A(x)}\,\mathrm{d}x=\int^1_0{\frac{(x+1)^2}{16}}\,\mathrm{d}x=\frac{7}{48}.$ Thus the volume of the larger solid is $1-\frac{7}{48}=\frac{41}{48} \rightarrow p+q = \boxed{089}$ | null | 089 |
d2509b04bb0dbe0d59e06a29f5b4aea7 | https://artofproblemsolving.com/wiki/index.php/2012_AIME_I_Problems/Problem_8 | Cube $ABCDEFGH,$ labeled as shown below, has edge length $1$ and is cut by a plane passing through vertex $D$ and the midpoints $M$ and $N$ of $\overline{AB}$ and $\overline{CG}$ respectively. The plane divides the cube into two solids. The volume of the larger of the two solids can be written in the form $\tfrac{p}{q},$ where $p$ and $q$ are relatively prime positive integers. Find $p+q.$ | The volume of a frustum is $\frac{h_2b_2 -h_1b_1}3$ where $b_i$ is the area of the base and $h_i$ is the height from the chopped off apex to the base.
We can easily see that from symmetry, the area of the smaller front base is $\frac{1}{16}$ and the area of the larger back base is $\frac{1}4$
Now to find the height of the apex.
Extend the $DM$ and (call the intersection of the plane with $FB$ G) $NG$ to meet at $x$ . Now from similar triangles $XMG$ and $XDN$ we can easily find the total height of the triangle $XDN$ to be $2$
Now straight from our formula, the volume is $\frac{7}{48}$ Thus the answer is:
$1-\text{Volume} \Longrightarrow \boxed{089}$ | null | 089 |
d2509b04bb0dbe0d59e06a29f5b4aea7 | https://artofproblemsolving.com/wiki/index.php/2012_AIME_I_Problems/Problem_8 | Cube $ABCDEFGH,$ labeled as shown below, has edge length $1$ and is cut by a plane passing through vertex $D$ and the midpoints $M$ and $N$ of $\overline{AB}$ and $\overline{CG}$ respectively. The plane divides the cube into two solids. The volume of the larger of the two solids can be written in the form $\tfrac{p}{q},$ where $p$ and $q$ are relatively prime positive integers. Find $p+q.$ | We will solve for the area of the smaller region, and then subtract it from 1.
Let $X$ be the point where plane $DMN$ intersects $FB$ . Then $DMBCNX$ can be split into triangular pyramid $DMBX$ and quadrilateral pyramid $BCNXD$
Pyramid $DMBX$ has base $DMB$ with area $\frac{1}2 \cdot 1 \div 2 = \frac{1}4$ . The height is $BX = \frac{1}4$ , so the volume of $DMBX$ is $\frac{1}4 \cdot \frac{1}4 \div 3 = \frac{1}{48}$
Similarly, pyramid $BCNXD$ has base $BCNX$ with area $(\frac{1}4 + \frac{1}2) \cdot 1 = \frac{3}8$ . The height is $CD = 1$ , so the volume of $BCNXD$ is $\frac{3}8 \cdot 1 \div 3 = \frac{1}8$
Adding up the volumes of $DMBX$ and $BCNXD$ , we find that the volume of $DMBCNX$ is $\frac{1}{48} + \frac{6}{48} = \frac{7}{48}$ . Therefore the volume of the larger solid is $1 - \frac{7}{48} = \frac{41}{48} \rightarrow p+q = \boxed{089}$ | null | 089 |
d2509b04bb0dbe0d59e06a29f5b4aea7 | https://artofproblemsolving.com/wiki/index.php/2012_AIME_I_Problems/Problem_8 | Cube $ABCDEFGH,$ labeled as shown below, has edge length $1$ and is cut by a plane passing through vertex $D$ and the midpoints $M$ and $N$ of $\overline{AB}$ and $\overline{CG}$ respectively. The plane divides the cube into two solids. The volume of the larger of the two solids can be written in the form $\tfrac{p}{q},$ where $p$ and $q$ are relatively prime positive integers. Find $p+q.$ | We use a coordinate system with $C = (0,0,0)$ $D = (1,0,0)$ $M = (\frac{1}{2},1,0)$ $N = (0,0,\frac{1}{2})$ . Then the plane going through $D$ $M$ , and $N$ has equation $z = \frac{1}{2} - \frac{1}{2}x - \frac{1}{4}y$ . We set up a double integral in this coordinate system. Consider the region to be integrated over in the $xy$ -plane. From $x=0$ to $x=\frac{1}{2}$ , the upper bound of the region is $y = 1$ . From $x= \frac{1}{2}$ to $x=1$ , the upper bound of the region is $y = 2 - 2x$ . In both cases, the lower bound of the region is $y = 0$ . Thus, we have the double integral $\int_{0}^{\frac{1}{2}} \int_{0}^{1} \frac{1}{2} - \frac{1}{2}x - \frac{1}{4}y dydx + \int_{\frac{1}{2}}^{1} \int_{0}^{2-2x} \frac{1}{2} - \frac{1}{2}x - \frac{1}{4}y dydx$ . We find that this sum evaluates to $\frac{7}{48}$ . However, this is the volume of the smaller region, so the larger region's volume is that of the cube minus that of the smaller region. Since the cube has side length $1$ , its volume is $1$ , so the volume of the larger region is $1 - \frac{7}{48} = \frac{41}{48}$ . Thus, our answer is $41 + 48 = \boxed{089}$ | null | 089 |
e849f52bf8ce85ea8a2956f9095c8fca | https://artofproblemsolving.com/wiki/index.php/2012_AIME_I_Problems/Problem_9 | Let $x,$ $y,$ and $z$ be positive real numbers that satisfy \[2\log_{x}(2y) = 2\log_{2x}(4z) = \log_{2x^4}(8yz) \ne 0.\] The value of $xy^5z$ can be expressed in the form $\frac{1}{2^{p/q}},$ where $p$ and $q$ are relatively prime positive integers. Find $p+q.$ | Since there are only two dependent equations given and three unknowns, the three expressions given can equate to any common value (that isn't 0, of course), so to simplify the problem let us assume without loss of generality that \[2\log_{x}(2y) = 2\log_{2x}(4z) = \log_{2x^4}(8yz) = 2.\] Then \begin{align*} 2\log_{x}(2y) = 2 &\implies x=2y\\ 2\log_{2x}(4z) = 2 &\implies 2x=4z\\ \log_{2x^4}(8yz) = 2 &\implies 4x^8 = 8yz \end{align*} Solving these equations, we quickly see that $4x^8 = (2y)(4z) = x(2x) \rightarrow x=2^{-1/6}$ and then $y=z=2^{-1/6 - 1} = 2^{-7/6}.$ Finally, our desired value is $2^{-1/6} \cdot (2^{-7/6})^5 \cdot 2^{-7/6} = 2^{-43/6}$ and thus $p+q = 43 + 6 = \boxed{049.}$ | null | 049. |
e849f52bf8ce85ea8a2956f9095c8fca | https://artofproblemsolving.com/wiki/index.php/2012_AIME_I_Problems/Problem_9 | Let $x,$ $y,$ and $z$ be positive real numbers that satisfy \[2\log_{x}(2y) = 2\log_{2x}(4z) = \log_{2x^4}(8yz) \ne 0.\] The value of $xy^5z$ can be expressed in the form $\frac{1}{2^{p/q}},$ where $p$ and $q$ are relatively prime positive integers. Find $p+q.$ | Notice that $2y\cdot 4z=8yz$ $2\log_2(2y)=\log_2\left(4y^2\right)$ and $2\log_2(4z)=\log_2\left(16z^2\right)$
From this, we see that $8yz$ is the geometric mean of $4y^2$ and $16z^2$ . So, for constant $C\ne 0$ \[\frac{\log 4y^2}{\log x}=\frac{\log 8yz}{\log 2x^4}=\frac{\log 16z^2}{\log 2x}=C\] Since $\log 4y^2,\log 8yz,\log 16z^2$ are in an arithmetic progression, so are $\log x,\log 2x^4,\log 2x$
Therefore, $2x^4$ is the geometric mean of $x$ and $2x$ \[2x^4=\sqrt{x\cdot 2x}\implies 4x^8=2x^2\implies 2x^6=1\implies x=2^{-1/6}\] We can plug $x$ in to any of the two equal fractions aforementioned. So, without loss of generality: \[\frac{\log 4y^2}{\log x}=\frac{\log 16z^2}{\log 2x}\implies \log\left(4y^2\right)\log\left(2x\right)=\log\left(16z^2\right)\log\left(x\right)\] \[\implies \log\left(4y^2\right)\cdot \frac{5}{6}\log 2=\log\left(16z^2\right)\cdot \frac{-1}{6}\log 2\] \[\implies 5\log\left(4y^2\right)=-\log\left(16z^2\right)\implies 5\log\left(4y^2\right)+\log\left(16z^2\right)=0\] \[\implies \left(4y^2\right)^5\cdot 16z^2=1\implies 16384y^{10}z^2=1\implies y^{10}z^2=\frac{1}{16384}\implies y^5z=\frac{1}{128}\]
Thus $xy^5z=2^{-\frac{1}{6}-7}=2^{-\frac{43}{6}}$ and $43+6=\boxed{049}$ | null | 049 |
e849f52bf8ce85ea8a2956f9095c8fca | https://artofproblemsolving.com/wiki/index.php/2012_AIME_I_Problems/Problem_9 | Let $x,$ $y,$ and $z$ be positive real numbers that satisfy \[2\log_{x}(2y) = 2\log_{2x}(4z) = \log_{2x^4}(8yz) \ne 0.\] The value of $xy^5z$ can be expressed in the form $\frac{1}{2^{p/q}},$ where $p$ and $q$ are relatively prime positive integers. Find $p+q.$ | Since we are given that $xy^5z = 2^{-p/q}$ , we may assume that $x, y$ , and $z$ are all powers of two. We shall thus let $x = 2^X$ $y = 2^Y$ , and $z = 2^Z$ . Let $a = \log_{2^X}(2^{Y+1})$ . From this we get the system of equations: \[\] $(1)$ \[a = \log_{2^X}(2^{Y+1}) \Rightarrow aX = Y + 1\] $(2)$ \[a = \log_{2^{X + 1}}(2^{Z + 2}) \Rightarrow aX + a = Z + 2\] $(3)$ \[2a = \log_{2^{4X + 1}}(2^{Y + Z + 3}) \Rightarrow 8aX + 2a = Y + Z + 3\]
Plugging equation $(1)$ into equation $(2)$ yields $Y + a = Z + 1$ . Plugging equation $(1)$ into equation $(3)$ and simplifying yields $7Y + 2a + 6 = Z + 1$ , and substituting $Y + a$ for $Z + 1$ and simplifying yields $Y + 1 = \frac{-a}{6}$ . But $Y + 1 = aX$ , so $aX = \frac{-a}{6}$ , so $X = \frac{-1}{6}$
Knowing this, we may substitute $\frac{-1}{6}$ for $X$ in equations $(1)$ and $(2)$ , yielding $\frac{-a}{6} = Y + 1$ and $\frac{5a}{6} = Z + 2$ . Thus, we have that $-5(Y + 1) = Z + 2 \rightarrow 5Y + Z = -7$ . We are looking for $xy^5z = 2^{X+ 5Y + Z}$ $X = \frac{-1}{6}$ and $5Y + Z = -7$ , so $xy^5z = 2^{-43/6} = \frac{1}{2^{43/6}}$ . The answer is $43+6=\boxed{049}$ | null | 049 |
e849f52bf8ce85ea8a2956f9095c8fca | https://artofproblemsolving.com/wiki/index.php/2012_AIME_I_Problems/Problem_9 | Let $x,$ $y,$ and $z$ be positive real numbers that satisfy \[2\log_{x}(2y) = 2\log_{2x}(4z) = \log_{2x^4}(8yz) \ne 0.\] The value of $xy^5z$ can be expressed in the form $\frac{1}{2^{p/q}},$ where $p$ and $q$ are relatively prime positive integers. Find $p+q.$ | We know that \[\frac{\log (4y^2)}{\log (x)} = \frac{\log (16z^2)}{\log (2x)} = \frac{2\log (8yz)}{\log (2x^2)}\] By the Mediant theorem
Substituting into the original equation yields us $\frac{2\log (8yz)}{\log (2x^2)} = \frac{\log (8yz)}{\log (2x^4)} \Rightarrow 2\log (2x^4) = \log (2x^2) \Rightarrow x=2^{-1/6}.$ For some constant $C\not= 0,$ Let $2\log_{x}(2y) = 2\log_{2x}(4z) = \log_{2x^4}(8yz) = C.$ Then, we obtain the system of equations \[y=2^{-13C/12}\] \[z=2^{-19C/12}\] \[8yz=2^{C/3}.\]
Adding the first two equations and subtracting the third, we find $C=1.$ Thus, \[xy^5z=2^{-1/6} \cdot 2^{-65/12} \cdot 2^{-19/12}=2^{-43/6} \Rightarrow p+q=\boxed{049}.\] | null | 049 |
a2c2dc84a4b9e0f1d766dba57e13afb7 | https://artofproblemsolving.com/wiki/index.php/2012_AIME_I_Problems/Problem_10 | Let $\mathcal{S}$ be the set of all perfect squares whose rightmost three digits in base $10$ are $256$ . Let $\mathcal{T}$ be the set of all numbers of the form $\frac{x-256}{1000}$ , where $x$ is in $\mathcal{S}$ . In other words, $\mathcal{T}$ is the set of numbers that result when the last three digits of each number in $\mathcal{S}$ are truncated. Find the remainder when the tenth smallest element of $\mathcal{T}$ is divided by $1000$ | It is apparent that for a perfect square $s^2$ to satisfy the constraints, we must have $s^2 - 256 = 1000n$ or $(s+16)(s-16) = 1000n.$ Now in order for $(s+16)(s-16)$ to be a multiple of $1000,$ at least one of $s+16$ and $s-16$ must be a multiple of $5,$ and since $s+16 \not\equiv s-16 \pmod{5},$ one term must have all the factors of $5$ and thus must be a multiple of $125.$ Furthermore, each of $s+16$ and $s-16$ must have at least two factors of $2,$ since otherwise $(s+16)(s-16)$ could not possibly be divisible by $8.$ So therefore the conditions are satisfied if either $s+16$ or $s-16$ is divisible by $500,$ or equivalently if $s = 500n \pm 16.$ Counting up from $n=0$ to $n=5,$ we see that the tenth value of $s$ is $500 \cdot 5 - 16 = 2484$ and thus the corresponding element in $\mathcal{T}$ is $\frac{2484^2 - 256}{1000} = 6170 \rightarrow \boxed{170.}$ | null | 170. |
a2c2dc84a4b9e0f1d766dba57e13afb7 | https://artofproblemsolving.com/wiki/index.php/2012_AIME_I_Problems/Problem_10 | Let $\mathcal{S}$ be the set of all perfect squares whose rightmost three digits in base $10$ are $256$ . Let $\mathcal{T}$ be the set of all numbers of the form $\frac{x-256}{1000}$ , where $x$ is in $\mathcal{S}$ . In other words, $\mathcal{T}$ is the set of numbers that result when the last three digits of each number in $\mathcal{S}$ are truncated. Find the remainder when the tenth smallest element of $\mathcal{T}$ is divided by $1000$ | Notice that is $16^2=256$ $1016^2$ ends in $256$ . In general, if $x^2$ ends in $256$ $(x+1000)^2=x^2+2000x+1000000$ ends in 256 because $1000|2000x$ and $1000|2000000$ . It is clear that we want all numbers whose squares end in $256$ that are less than $1000$
Firstly, we know the number has to end in a $4$ or a $6$ . Let's look at the ones ending in $6$
Assume that the second digit of the three digit number is $0$ . We find that the last $3$ digits of $\overline{a06}^2$ is in the form $12a \cdot 100 + 3 \cdot 10 + 6$ . However, the last two digits need to be a $56$ . Thus, similarly trying all numbers from $0$ to $10$ , we see that only 1 for the middle digit works. Carrying out the multiplication, we see that the last $3$ digits of $\overline{a06}^2$ is in the form $(12a + 2) \cdot 100 + 5 \cdot 10 + 6$ . We want $(12a + 2)\pmod{10}$ to be equal to $2$ . Thus, we see that a is $0$ or $5$ . Thus, the numbers that work in this case are $016$ and $516$
Next, let's look at the ones ending in $4$ . Carrying out a similar technique as above, we see that the last $3$ digits of $\overline{a84}^2$ is in the form $((8a+10) \cdot 100+ 5 \cdot 10 + 6$ . We want $(8a + 10)\pmod{10}$ to be equal to $2$ . We see that only $4$ and $9$ work. Thus, we see that only $484$ and $984$ work.
We order these numbers to get $16$ $484$ $516$ $984$ ... We want the $10th$ number in order which is $2484^2 = 6\boxed{170}256$ | null | 170 |
a2c2dc84a4b9e0f1d766dba57e13afb7 | https://artofproblemsolving.com/wiki/index.php/2012_AIME_I_Problems/Problem_10 | Let $\mathcal{S}$ be the set of all perfect squares whose rightmost three digits in base $10$ are $256$ . Let $\mathcal{T}$ be the set of all numbers of the form $\frac{x-256}{1000}$ , where $x$ is in $\mathcal{S}$ . In other words, $\mathcal{T}$ is the set of numbers that result when the last three digits of each number in $\mathcal{S}$ are truncated. Find the remainder when the tenth smallest element of $\mathcal{T}$ is divided by $1000$ | The condition implies $x^2\equiv 256 \pmod{1000}$ . Rearranging and factoring, \[(x-16)(x+16)\equiv 0\pmod {1000}.\] This can be expressed with the system of congruences \[\begin{cases} (x-16)(x+16)\equiv 0\pmod{125} \\ (x-16)(x+16)\equiv 0\pmod{8} \end{cases}\] Observe that $x\equiv {109} \pmod {125}$ or $x\equiv{16}\pmod {125}$ . Similarly, it can be seen that $x\equiv{0}\pmod{8}$ or $x\equiv{4}\pmod{8}$ . By CRT, there are four solutions to this modulo $1000$ (one for each case e.g. $x\equiv{109}$ and $x\equiv{0}$ or $x\equiv{125}$ and $x\equiv{4}$ . These solutions are (working modulo $1000$ \[\begin{cases} x=16 \\ x=484 \\ x= 516 \\ x=984 \end{cases}\] The tenth solution is $x=2484,$ which gives an answer of $\boxed{170}$ | null | 170 |
a2c2dc84a4b9e0f1d766dba57e13afb7 | https://artofproblemsolving.com/wiki/index.php/2012_AIME_I_Problems/Problem_10 | Let $\mathcal{S}$ be the set of all perfect squares whose rightmost three digits in base $10$ are $256$ . Let $\mathcal{T}$ be the set of all numbers of the form $\frac{x-256}{1000}$ , where $x$ is in $\mathcal{S}$ . In other words, $\mathcal{T}$ is the set of numbers that result when the last three digits of each number in $\mathcal{S}$ are truncated. Find the remainder when the tenth smallest element of $\mathcal{T}$ is divided by $1000$ | An element in S can be represented by $y^2 = 1000a + 256$ , where $y^2$ is the element in S. Since the right hand side must be even, we let $y = 2y_1$ and substitute to get $y_1^2 = 250a + 64$ . However, the right hand side is still even, so we make the substitution $y_1 = 2y_2$ to get $y_2^2 = 125a/2 + 16$ . Because both sides must be an integer, we know that $a = 2a_1$ for some integer $a_1$ . Our equation then becomes $y_2^2 = 125a_1 + 16$ , and we can simplify no further.
Rearranging terms, we get $y_2^2 - 16 = 125a_1$ , whence difference of squares gives $(y_2 + 4)(y_2 - 4) = 125a_1$ . Note that this equation tells us that one of $y_2 + 4$ and $y_2 - 4$ contains a nonnegative multiple of $125$ . Hence, listing out the smallest possible values of $y_2$ , we have $y_2 = 4, 121, 129, \cdots, 621$ . The tenth term is $621$ , whence $y = 4y_2 = 2484$ . The desired result can then be calculated to be $\boxed{170}$ . - Spacesam | null | 170 |
a2c2dc84a4b9e0f1d766dba57e13afb7 | https://artofproblemsolving.com/wiki/index.php/2012_AIME_I_Problems/Problem_10 | Let $\mathcal{S}$ be the set of all perfect squares whose rightmost three digits in base $10$ are $256$ . Let $\mathcal{T}$ be the set of all numbers of the form $\frac{x-256}{1000}$ , where $x$ is in $\mathcal{S}$ . In other words, $\mathcal{T}$ is the set of numbers that result when the last three digits of each number in $\mathcal{S}$ are truncated. Find the remainder when the tenth smallest element of $\mathcal{T}$ is divided by $1000$ | From the conditions, we can let every element in $\mathcal{S}$ be written as $y^2=1000x+256$ , where $x$ and $y$ are integers. Since there are no restrictions on $y$ , we let $y_1$ be equal to $y+16$ $y-16$ works as well). Then the $256$ cancels out and we're left with \[y_1^2+32y_1=1000x\] which can be factored as \[y_1(y_1+32)=1000x\] Since the RHS is even, $y_1$ must be even, so we let $y_1=2y_2$ , to get \[y_2(y_2+16)=250x\] Again, because the RHS is even, the LHS must be even too, so substituting $y_3=\frac{1}{2}y_2$ we have \[y_3(y_3+8)=125\cdot\frac{1}{2}x\] Since the LHS is an integer, the RHS must thus be an integer, so substituting $x=2x_1$ we get \[y_3(y_3+8)=125x_1\] Then we can do casework on the values of $y_3$ , as only one of $y_3$ and $y_3+8$ can be multiples of $125$
Case 1 $125|y_3$
Since we're trying to find the values of $x_1$ , we can let $y_4=\frac{1}{125}y_3$ , to get \[x_1=y_4(125y_4+8)\] or \[x=2y_4(125y_4+8)\]
Case 2 $125|y_3+8$
Similar to Case 1, only the equation is \[x=2y_4(125y_4-8)\]
In whole, the values of $x$ (i.e. the elements in $\mathcal{T}$ ) are of the form \[x=2k(125k\pm8)\] where $k$ is any integer. It can easily be seen that if $k<0$ , then $x$ is negative, thus $k\geq0$ . Also, note that when $k=0$ , there is only one value, because one of the factors is $0$ $k$ ). Thus the $10^{th}$ smallest number in the set $\mathcal{T}$ is when the $\pm$ sign takes the minus side and $k=5$ , giving $6170$ , so the answer is $\boxed{170}$ | null | 170 |
fa78e9d0c17c2aa0b6e3dfe9b21af9ef | https://artofproblemsolving.com/wiki/index.php/2012_AIME_I_Problems/Problem_11 | A frog begins at $P_0 = (0,0)$ and makes a sequence of jumps according to the following rule: from $P_n = (x_n, y_n),$ the frog jumps to $P_{n+1},$ which may be any of the points $(x_n + 7, y_n + 2),$ $(x_n + 2, y_n + 7),$ $(x_n - 5, y_n - 10),$ or $(x_n - 10, y_n - 5).$ There are $M$ points $(x, y)$ with $|x| + |y| \le 100$ that can be reached by a sequence of such jumps. Find the remainder when $M$ is divided by $1000.$ | First of all, it is easy to see by induction that for any $P(x,y)$ in the frog's jump sequence, $x+y$ will be a multiple of $3$ and $x-y$ will be a multiple of $5.$ The base case $(x,y) = (0,0)$ obviously satisfies the constraints and if $x+y = 3n$ and $x-y = 5m,$ any of the four transformations will sustain this property:
\begin{align*} (x+7)+(y+2) = x+y+9 \rightarrow 3(n+3) &\text{ and } (x+7)-(y+2) = x-y+5 \rightarrow 5(m+1)\\ (x+2)+(y+7) = x+y+9 \rightarrow 3(n+3) &\text{ and } (x+2)-(y+7) = x-y-5 \rightarrow 5(m-1)\\ (x-5)+(y-10) = x+y-15 \rightarrow 3(n-5) &\text{ and } (x-5)-(y-10) = x-y+5 \rightarrow 5(m+1)\\ (x-10)+(y-5) = x+y-15 \rightarrow 3(n-5) &\text{ and } (x-10)-(y-5) = x-y-5 \rightarrow 5(m-1).\\ \end{align*} So we know that any point the frog can reach will satisfy $x+y = 3n$ and $x-y = 5m.$
$\textbf{Lemma:}$ Any point $(x,y)$ such that there exists 2 integers $m$ and $n$ that satisfy $x+y = 3n$ and $x-y = 5m$ is reachable.
$\textbf{Proof:}$ Denote the total amounts of each specific transformation in the frog's jump sequence to be $a,$ $b,$ $c,$ and $d$ respectively. Then
$x=7a+2b-5c-10d$
$y=2a+7b-10c-5d$
$x+y = 9(a+b)-15(c+d) = 3n$ , and
$x-y = 5(a-b)+5(c-d) = 5m$
together must have integral solutions. But
$3(a+b)-5(c+d) = n$ implies
$(c+d) \equiv n \mod 3$ and thus
$(a+b) = \lfloor{n/3}\rfloor + 2(c+d).$
Similarly, $(a-b)+(c-d) = m$ implies that $(a-b)$ and $(c-d)$ have the same parity. Now in order for an integral solution to exist, there must always be a way to ensure that the pairs $(a+b)$ and $(a-b)$ and $(c+d)$ and $(c-d)$ have identical parities. The parity of $(a+b)$ is completely dependent on $n,$ so the parities of $(a-b)$ and $(c-d)$ must be chosen to match this value. But the parity of $(c+d)$ can then be adjusted by adding or subtracting $3$ until it is identical to the parity of $(c-d)$ as chosen before, so we conclude that it is always possible to find an integer solution for $(a,b,c,d)$ and thus any point that satisfies $x+y = 3n$ and $x-y = 5m$ can be reached by the frog.
To count the number of such points in the region $|x| + |y| \le 100,$ we first note that any such point will lie on the intersection of one line of the form $y=x-5m$ and another line of the form $y=-x+3n.$ The intersection of two such lines will yield the point $\left(\frac{3n+5m}{2},\frac{3n-5m}{2}\right),$ which will be integral if and only if $m$ and $n$ have the same parity. Now since $|x| + |y| = |x \pm y|,$ we find that
\begin{align*} |x + y| = |3n| \le 100 &\rightarrow -33 \le n \le 33\\ |x - y| = |5m| \le 100 &\rightarrow -20 \le m \le 20. \end{align*}
So there are $34$ possible odd values and $33$ possible even values for $n,$ and $20$ possible odd values and $21$ possible even values for $m.$ Every pair of lines described above will yield a valid accessible point for all pairs of $m$ and $n$ with the same parity, and the number of points $M$ is thus $34 \cdot 20 + 33 \cdot 21 = 1373 \rightarrow \boxed{373}$ | null | 373 |
d1c7877bdaa453b5247a4caea7d87758 | https://artofproblemsolving.com/wiki/index.php/2012_AIME_I_Problems/Problem_12 | Let $\triangle ABC$ be a right triangle with right angle at $C.$ Let $D$ and $E$ be points on $\overline{AB}$ with $D$ between $A$ and $E$ such that $\overline{CD}$ and $\overline{CE}$ trisect $\angle C.$ If $\frac{DE}{BE} = \frac{8}{15},$ then $\tan B$ can be written as $\frac{m \sqrt{p}}{n},$ where $m$ and $n$ are relatively prime positive integers, and $p$ is a positive integer not divisible by the square of any prime. Find $m+n+p.$ | We have $\angle BCE = \angle ECD = \angle ECA = \tfrac 13 \cdot 90^\circ = 30^\circ$ . Drop the altitude from $D$ to $CB$ and call the foot $F$
Let $CD = 8a$ . Using angle bisector theorem on $\triangle CDB$ , we get $CB = 15a$ . Now $CDF$ is a $30$ $60$ $90$ triangle, so $CF = 4a$ $FD = 4a\sqrt{3}$ , and $FB = 11a$ . Finally, $\tan{B} = \tfrac{DF}{FB}=\tfrac{4\sqrt{3}a}{11a} = \tfrac{4\sqrt{3}}{11}$ . Our final answer is $4 + 3 + 11 = \boxed{018}$ | null | 018 |
d1c7877bdaa453b5247a4caea7d87758 | https://artofproblemsolving.com/wiki/index.php/2012_AIME_I_Problems/Problem_12 | Let $\triangle ABC$ be a right triangle with right angle at $C.$ Let $D$ and $E$ be points on $\overline{AB}$ with $D$ between $A$ and $E$ such that $\overline{CD}$ and $\overline{CE}$ trisect $\angle C.$ If $\frac{DE}{BE} = \frac{8}{15},$ then $\tan B$ can be written as $\frac{m \sqrt{p}}{n},$ where $m$ and $n$ are relatively prime positive integers, and $p$ is a positive integer not divisible by the square of any prime. Find $m+n+p.$ | Without loss of generality, set $CB = 1$ . Then, by the Angle Bisector Theorem on triangle $DCB$ , we have $CD = \frac{8}{15}$ . We apply the Law of Cosines to triangle $DCB$ to get $1 + \frac{64}{225} - \frac{8}{15} = BD^{2}$ , which we can simplify to get $BD = \frac{13}{15}$
Now, we have $\cos \angle B = \frac{1 + \frac{169}{225} - \frac{64}{225}}{\frac{26}{15}}$ by another application of the Law of Cosines to triangle $DCB$ , so $\cos \angle B = \frac{11}{13}$ . In addition, $\sin \angle B = \sqrt{1 - \frac{121}{169}} = \frac{4\sqrt{3}}{13}$ , so $\tan \angle B = \frac{4\sqrt{3}}{11}$
Our final answer is $4+3+11 = \boxed{018}$ | null | 018 |
d1c7877bdaa453b5247a4caea7d87758 | https://artofproblemsolving.com/wiki/index.php/2012_AIME_I_Problems/Problem_12 | Let $\triangle ABC$ be a right triangle with right angle at $C.$ Let $D$ and $E$ be points on $\overline{AB}$ with $D$ between $A$ and $E$ such that $\overline{CD}$ and $\overline{CE}$ trisect $\angle C.$ If $\frac{DE}{BE} = \frac{8}{15},$ then $\tan B$ can be written as $\frac{m \sqrt{p}}{n},$ where $m$ and $n$ are relatively prime positive integers, and $p$ is a positive integer not divisible by the square of any prime. Find $m+n+p.$ | (This solution does not use the Angle Bisector Theorem or the Law of Cosines, but it uses the Law of Sines and more trig)
Find values for all angles in terms of $\angle B$ $\angle CEB = 150-B$ $\angle CED = 30+B$ $\angle CDE = 120-B$ $\angle CDA = 60+B$ , and $\angle A = 90-B$
Use the law of sines on $\triangle CED$ and $\triangle CEB$
In $\triangle CED$ $\frac{8}{\sin 30} = \frac{CE}{\sin (120-B)}$ . This simplifies to $16 = \frac{CE}{\sin (120-B)}$
In $\triangle CEB$ $\frac{15}{\sin 30} = \frac{CE}{\sin B}$ . This simplifies to $30 = \frac{CE}{\sin B}$
Solve for $CE$ and equate them so that you get $16\sin (120-B) = 30\sin B$
From this, $\frac{8}{15} = \frac{\sin B}{\sin (120-B)}$
Use a trig identity on the denominator on the right to obtain: $\frac{8}{15} = \frac{\sin B}{\sin 120 \cos B - \cos 120 \sin B}$
This simplifies to $\frac{8}{15} = \frac{\sin B}{\frac{\sqrt{3}\cos B}{2} + \frac{\sin B}{2}} = \frac{\sin B}{\frac{\sqrt{3} \cos B + \sin B}{2}} = \frac{2\sin B}{\sqrt{3}\cos B + \sin B}$
This gives $8\sqrt{3}\cos B+8\sin B=30\sin B$ Dividing by $\cos B$ , we have ${8\sqrt{3}}= 22\tan B$
$\tan{B} = \frac{8\sqrt{3}}{22} = \frac{4\sqrt{3}}{11}$ . Our final answer is $4 + 3 + 11 = \boxed{018}$ | null | 018 |
d1c7877bdaa453b5247a4caea7d87758 | https://artofproblemsolving.com/wiki/index.php/2012_AIME_I_Problems/Problem_12 | Let $\triangle ABC$ be a right triangle with right angle at $C.$ Let $D$ and $E$ be points on $\overline{AB}$ with $D$ between $A$ and $E$ such that $\overline{CD}$ and $\overline{CE}$ trisect $\angle C.$ If $\frac{DE}{BE} = \frac{8}{15},$ then $\tan B$ can be written as $\frac{m \sqrt{p}}{n},$ where $m$ and $n$ are relatively prime positive integers, and $p$ is a positive integer not divisible by the square of any prime. Find $m+n+p.$ | (This solution avoids advanced trigonometry)
Let $X$ be the foot of the perpendicular from $D$ to $\overline{BC}$ , and let $Y$ be the foot of the perpendicular from $E$ to $\overline{BC}$
Now let $EY=x$ . Clearly, triangles $EYB$ and $DXB$ are similar with $\frac{BE}{BD}=\frac{15}{15+8}=\frac{15}{23}=\frac{EY}{DX}$ , so $DX=\frac{23}{15}x$
Since triangles $CDX$ and $CEY$ are 30-60-90 right triangles, we can easily find other lengths in terms of $x$ . For example, we see that $CY=x\sqrt{3}$ and $CX=\frac{\frac{23}{15}x}{\sqrt{3}}=\frac{23\sqrt{3}}{45}x$ . Therefore $XY=CY-CX=x\sqrt{3}-\frac{23\sqrt{3}}{45}x=\frac{22\sqrt{3}}{45}x$
Again using the fact that triangles $EYB$ and $DXB$ are similar, we see that $\frac{BX}{BY}=\frac{XY+BY}{BY}=\frac{XY}{BY}+1=\frac{23}{15}$ , so $BY=\frac{15}{8}XY=\frac{15}{8}*\frac{22\sqrt{3}}{45}=\frac{11\sqrt{3}}{2}$
Thus $\tan \angle B = \frac{x}{\frac{11\sqrt{3}}{12}x}=\frac{4\sqrt{3}}{11}$ , and our answer is $4+3+11=\boxed{018}$ | null | 018 |
d1c7877bdaa453b5247a4caea7d87758 | https://artofproblemsolving.com/wiki/index.php/2012_AIME_I_Problems/Problem_12 | Let $\triangle ABC$ be a right triangle with right angle at $C.$ Let $D$ and $E$ be points on $\overline{AB}$ with $D$ between $A$ and $E$ such that $\overline{CD}$ and $\overline{CE}$ trisect $\angle C.$ If $\frac{DE}{BE} = \frac{8}{15},$ then $\tan B$ can be written as $\frac{m \sqrt{p}}{n},$ where $m$ and $n$ are relatively prime positive integers, and $p$ is a positive integer not divisible by the square of any prime. Find $m+n+p.$ | (Another solution without trigonometry)
Extend $CD$ to point $F$ such that $\overline{AF} \parallel \overline{CB}$ . It is then clear that $\triangle AFD$ is similar to $\triangle BCD$
Let $AC=p$ $BC=q$ . Then $\tan \angle B = p/q$
With the Angle Bisector Theorem, we get that $CD=\frac{8}{15}q$ . From 30-60-90 $\triangle CAF$ , we get that $AF=\frac{1}{\sqrt{3}}p$ and $FD=FC-CD=\frac{2}{\sqrt{3}}p-\frac{8}{15}q$
From $\triangle AFD \sim \triangle BCD$ , we have that $\frac{FD}{CD}=\frac{FA}{CB}=\frac{\frac{2}{\sqrt{3}}p-\frac{8}{15}q}{\frac{8}{15}q}=\frac{\frac{1}{\sqrt{3}}p}{q}$ . Simplifying yields $\left(\frac{p}{q}\right)\left(\frac{2\sqrt{3}}{3}*\frac{15}{8}-\frac{\sqrt{3}}{3}\right)=1$ , and $\tan \angle B=\frac{p}{q}=\frac{4\sqrt{3}}{11}$ , so our answer is $4+3+11=\boxed{018}$ | null | 018 |
d1c7877bdaa453b5247a4caea7d87758 | https://artofproblemsolving.com/wiki/index.php/2012_AIME_I_Problems/Problem_12 | Let $\triangle ABC$ be a right triangle with right angle at $C.$ Let $D$ and $E$ be points on $\overline{AB}$ with $D$ between $A$ and $E$ such that $\overline{CD}$ and $\overline{CE}$ trisect $\angle C.$ If $\frac{DE}{BE} = \frac{8}{15},$ then $\tan B$ can be written as $\frac{m \sqrt{p}}{n},$ where $m$ and $n$ are relatively prime positive integers, and $p$ is a positive integer not divisible by the square of any prime. Find $m+n+p.$ | Let $CB = 1$ , and let the feet of the altitudes from $D$ and $E$ to $\overline{CB}$ be $D'$ and $E'$ , respectively. Also, let $DE = 8k$ and $EB = 15k$ . We see that $BD' = 15k\cos B$ and $BE' = 23k\cos B$ by right triangles $\triangle{BDD'}$ and $\triangle{BEE'}$ . From this we have that $D'E' = 8k\cos B$ . With the same triangles we have $DD' = 23k\sin B$ and $EE' = 15k\sin B$ . From 30-60-90 triangles $\triangle{CDD'}$ and $\triangle{CEE'}$ , we see that $CD' = \frac{23k\sqrt{3}\sin B}{3}$ and $CE' = 15k\sqrt{3}\sin B$ , so $D'E' = \frac{22k\sqrt{3}\sin B}{3}$ . From our two values of $D'E'$ we get: $8k\cos B = \frac{22k\sqrt{3}\sin B}{3}$
$\frac{\sin B}{\cos B} = \frac{8k}{\frac{22k\sqrt{3}}{3}} = \tan B$
$\tan B = \frac{8}{\frac{22\sqrt{3}}{3}} = \frac{24}{22\sqrt{3}} = \frac{8\sqrt{3}}{22} = \frac{4\sqrt{3}}{11}$ Our answer is then $4+3+11 = \boxed{018}$ | null | 018 |
8e07c0c14f2c7aca5fe492a692f8c262 | https://artofproblemsolving.com/wiki/index.php/2012_AIME_I_Problems/Problem_13 | Three concentric circles have radii $3,$ $4,$ and $5.$ An equilateral triangle with one vertex on each circle has side length $s.$ The largest possible area of the triangle can be written as $a + \tfrac{b}{c} \sqrt{d},$ where $a,$ $b,$ $c,$ and $d$ are positive integers, $b$ and $c$ are relatively prime, and $d$ is not divisible by the square of any prime. Find $a+b+c+d.$ | Reinterpret the problem in the following manner. Equilateral triangle $ABC$ has a point $X$ on the interior such that $AX = 5,$ $BX = 4,$ and $CX = 3.$ $60^\circ$ counter-clockwise rotation about vertex $A$ maps $X$ to $X'$ and $B$ to $C.$
Note that angle $XAX'$ is $60$ and $XA = X'A = 5$ which tells us that triangle $XAX'$ is equilateral and that $XX' = 5.$ We now notice that $XC = 3$ and $X'C = 4$ which tells us that angle $XCX'$ is $90$ because there is a $3$ $4$ $5$ Pythagorean triple. Now note that $\angle ABC + \angle ACB = 120^\circ$ and $\angle XCA + \angle XBA = 90^\circ,$ so $\angle XCB+\angle XBC = 30^\circ$ and $\angle BXC = 150^\circ.$ Applying the law of cosines on triangle $BXC$ yields
\[BC^2 = BX^2+CX^2 - 2 \cdot BX \cdot CX \cdot \cos150^\circ = 4^2+3^2-24 \cdot \frac{-\sqrt{3}}{2} = 25+12\sqrt{3}\]
and thus the area of $ABC$ equals \[\frac{\sqrt{3}}{4}\cdot BC^2 = 25\frac{\sqrt{3}}{4}+9.\]
so our final answer is $3+4+25+9 = \boxed{041}.$ | null | 041 |
8e07c0c14f2c7aca5fe492a692f8c262 | https://artofproblemsolving.com/wiki/index.php/2012_AIME_I_Problems/Problem_13 | Three concentric circles have radii $3,$ $4,$ and $5.$ An equilateral triangle with one vertex on each circle has side length $s.$ The largest possible area of the triangle can be written as $a + \tfrac{b}{c} \sqrt{d},$ where $a,$ $b,$ $c,$ and $d$ are positive integers, $b$ and $c$ are relatively prime, and $d$ is not divisible by the square of any prime. Find $a+b+c+d.$ | Let's call the circle center $X$ . It has a distance of 3, 4, 5 to an equilateral triangle $LMN$ . Consider $X$ ’s pedal triangle $ABC$ . Since $X$ ’s antipedal triangle is equilateral, $X$ must be the one of the isogonic centers of $\triangle{ABC}$ . We’ll take the one inside $ABC$ , i.e., the Fermat point, because it leads to larger $\triangle LMN$ . Now we construct the three equilateral triangles $ABD$ $ACE$ , and $BCF$ , the same way the Fermat point is constructed. Then we have $\angle DXE = \angle EXF = \angle FXE = 120$ . Since $AEMCX$ is concyclic with $XM$ =4 as diameter, we have $AC=4\sin(60)$ . Similarly, $AB=3\sin(60)$ , and $BC=5\sin(60)$ . So $\triangle ABC$ is a 3-4-5 right triangle with $\angle BAC=90$ . With some more angle chasing we get \[\angle MXC+\angle LXB = \angle MAC + \angle LAB = 180 – \angle BAC = 90\] \[\angle LXM = 360 – (\angle MXC + \angle LXB + \angle BXC) = 360 –(90+120)=150\] By Law of Cosines, we have \[LM^2 = 3^2+4^2-2*3*4\cos(150)=25+12\sqrt 3\] And the area follows. \[[LMN] = \frac{25}{4}\sqrt{3} + 9; \boxed{041}.\] By Mathdummy | null | 041 |
8e07c0c14f2c7aca5fe492a692f8c262 | https://artofproblemsolving.com/wiki/index.php/2012_AIME_I_Problems/Problem_13 | Three concentric circles have radii $3,$ $4,$ and $5.$ An equilateral triangle with one vertex on each circle has side length $s.$ The largest possible area of the triangle can be written as $a + \tfrac{b}{c} \sqrt{d},$ where $a,$ $b,$ $c,$ and $d$ are positive integers, $b$ and $c$ are relatively prime, and $d$ is not divisible by the square of any prime. Find $a+b+c+d.$ | [asy] import cse5; size(350); defaultpen(linewidth(0.6)+fontsize(12)); pair O=origin; pair Op,Bp,Cp; path c3 = CR(O,3), c4 = CR(O,4), c5 = CR(O,5); var theta=55.75; pair A = 3*dir(theta), Ap = rotate(150,O)*A, F=IP(c4,O--2*Ap), C=rotate(60,A)*F, E=rotate(60,A)*C, B=IP(c5,O--E), D=foot(C,O,E); filldraw(A--O--B--cycle, rgb(255,255,200)); draw(c3, cyan); draw(c4, green); draw(c5, purple); draw(A--F--C--A--E--C, red+0.8); draw(D--B--C--D--O--C, purple+0.6); draw(A--O^^E--B, cyan+0.6); draw(A--B^^O--F, heavygreen+0.6); draw(A--D); dot("$A$",A,N); dot("$O$",O,dir(A-B)); dot("$F$",F,dir(F-E)); dot("$C$",C,SE); dot("$B$",B,dir(-45)); dot("$D$",D,dir(B-C)); dot("$E$",E,dir(E-F)); label("$s$",A--F,dir(A-E),red); label("$s$",F--C,dir(C-B),red); label("$s$",A--C,dir(C-E),red); label("$s$",A--E,dir(E-C),red); label("$s$",C--E,dir(C),red); label("$x$",B--E,down,royalblue); label("$x$",O--A,dir(B-A),royalblue); label("$y$",A--B,dir(F-A),heavygreen); label("$y$",O--F,dir(C),heavygreen); label("$z$",C--B,dir(E-A),purple); label("$z$",O--C,dir(F),purple); label("$m$",A--D,dir(0),blue+fontsize(9)); markscalefactor=0.03; draw(rightanglemark(C,D,B), gray); MA("\varphi",A,D,O,0.25,blue); [/asy] We have $x=3$ $y=4$ , and $z=5$ . Because $AD=m$ is the median of $\triangle AOB$ , by Stewart's Theorem we have \[m^2=\frac 12 (x^2+y^2)-\frac 14 \cdot z^2\quad \Rightarrow \quad m = \frac 52.\] Because $CD$ is the altitude of equilateral triangle $OBC$ , we have $CD=\frac{\sqrt{3}}2\cdot z$ . Then in $\triangle ADC$ , we have $\angle ADC=\varphi+90^\circ$ , so $\cos(\angle ADC)=-\sin\varphi$ , and the Law of Cosines gives \[s^2=m^2+\frac 34\cdot z^2 + mz\sqrt{3}\sin\varphi = 25 \left(1+\frac {\sqrt{3}}{2}\cdot\sin\varphi \right)\] To calculate $\sin\phi$ we apply the Law of Cosines to $\triangle ADO$ to get \[mz\cos\varphi = m^2+\frac 14\cdot z^2 - x^2 \quad \Rightarrow \quad \cos\varphi = \frac 7{25}\quad \Rightarrow \quad \sin\varphi = \frac {24}{25}.\] Finally, we get $s^2=25+12\sqrt{3}$ and and thus the area of $\triangle ABC$ equals \[\frac{\sqrt{3}}{4}\cdot s^2 = 25\frac{\sqrt{3}}{4}+9.\] so our final answer is $3+4+25+9 = \boxed{041}.$ | null | 041 |
a12b5a7f7cf92f325ca8a0df883f858e | https://artofproblemsolving.com/wiki/index.php/2012_AIME_I_Problems/Problem_14 | Complex numbers $a,$ $b,$ and $c$ are zeros of a polynomial $P(z) = z^3 + qz + r,$ and $|a|^2 + |b|^2 + |c|^2 = 250.$ The points corresponding to $a,$ $b,$ and $c$ in the complex plane are the vertices of a right triangle with hypotenuse $h.$ Find $h^2.$ | By Vieta's formula, the sum of the roots is equal to 0, or $a+b+c=0$ . Therefore, $\frac{(a+b+c)}{3}=0$ . Because the centroid of any triangle is the average of its vertices, the centroid of this triangle is the origin. Let one leg of the right triangle be $x$ and the other leg be $y$ . Without the loss of generality, let $\overline{ac}$ be the hypotenuse. The magnitudes of $a$ $b$ , and $c$ are just $\frac{2}{3}$ of the medians because the origin, or the centroid in this case, cuts the median in a ratio of $2:1$ . So, $|a|^2=\frac{4}{9}\cdot((\frac{x}{2})^2+y^2)=\frac{x^2}{9}+\frac{4y^2}{9}$ because $|a|$ is two thirds of the median from $a$ . Similarly, $|c|^2=\frac{4}{9}\cdot(x^2+(\frac{y}{2})^2)=\frac{4x^2}{9}+\frac{y^2}{9}$ . The median from $b$ is just half the hypotenuse because the median of any right triangle is just half the hypotenuse. So, $|b|^2=\frac{4}{9}\cdot\frac{x^2+y^2}{4}=\frac{x^2}{9}+\frac{y^2}{9}$ . Hence, $|a|^2+|b|^2+|c|^2=\frac{6x^2+6y^2}{9}=\frac{2x^2+2y^2}{3}=250$ . Therefore, $h^2=x^2+y^2=\frac{3}{2}\cdot250=\boxed{375}$ | null | 375 |
a12b5a7f7cf92f325ca8a0df883f858e | https://artofproblemsolving.com/wiki/index.php/2012_AIME_I_Problems/Problem_14 | Complex numbers $a,$ $b,$ and $c$ are zeros of a polynomial $P(z) = z^3 + qz + r,$ and $|a|^2 + |b|^2 + |c|^2 = 250.$ The points corresponding to $a,$ $b,$ and $c$ in the complex plane are the vertices of a right triangle with hypotenuse $h.$ Find $h^2.$ | Assume $q$ and $r$ are real, so at least one of $a,$ $b,$ and $c$ must be real, with the remaining roots being pairs of complex conjugates. Without loss of generality, we assume $a$ is real and $b$ and $c$ are $x + yi$ and $x - yi$ respectively. By symmetry, the triangle described by $a,$ $b,$ and $c$ must be isosceles and is thus an isosceles right triangle with hypotenuse $\overline{bc}.$ Now since $P(z)$ has no $z^2$ term, we must have $a+b+c = a + (x + yi) + (x - yi) = 0$ and thus $a = -2x.$ Also, since the length of the altitude from the right angle of an isosceles triangle is half the length of the hypotenuse, $a-x=y$ and thus $y=-3x.$ We can then solve for $x$
\begin{align*} |a|^2 + |b|^2 + |c|^2 &= 250\\ |-2x|^2 + |x-3xi|^2 + |x+3xi|^2 &= 250\\ 4x^2 + (x^2 + 9x^2) + (x^2 + 9x^2) &= 250\\ x^2 &= \frac{250}{24} \end{align*}
Now $h$ is the distance between $b$ and $c,$ so $h = 2y = -6x$ and thus $h^2 = 36x^2 = 36 \cdot \frac{250}{24} = \boxed{375.}$ | null | 375. |
a12b5a7f7cf92f325ca8a0df883f858e | https://artofproblemsolving.com/wiki/index.php/2012_AIME_I_Problems/Problem_14 | Complex numbers $a,$ $b,$ and $c$ are zeros of a polynomial $P(z) = z^3 + qz + r,$ and $|a|^2 + |b|^2 + |c|^2 = 250.$ The points corresponding to $a,$ $b,$ and $c$ in the complex plane are the vertices of a right triangle with hypotenuse $h.$ Find $h^2.$ | Let the roots $a$ $b$ , and $c$ each be represented by complex numbers $m + ni$ $p + qi$ , and $r + ti$ .
By Vieta's formulas, their sum is 0. Breaking into real and imaginary components, we get:
$m + p + r = 0$
$n + q + t = 0$
And, we know that the sum of the squares of the magnitudes of each is 250, so
$m^2 + n^2 + p^2 + q^2 + r^2 + t^2 = 250$
Given the complex plane, we set each of these complex numbers to points: $(m, n)$ $(p, q)$ $(r, t)$ .
WLOG let $(r, t)$ be the vertex opposite the hypotenuse.
If the three points form a right triangle, the vectors from $(r, t)$ to $(m, m)$ and $(p, q)$ 's dot product is 0. $mp + r^2 - r(m + p) + nq + t^2 - t(n + q) = 0$
Substituting $m + p + r = 0$ and likewise, simplifying: $mp + 2r^2 + nq + 2t^2 = 0$
Rearranging we get: $r^2 + t^2 = -\frac{mp + nq}{2}$
The answer is the distance from $(m, n)$ to $(p, q)$ $m^2 + n^2 + p^2 + q^2 - 2(mp + nq)$ .
Substituting the equation equal to 250,
$= 250 - r^2 - t^2 - 2(mp + nq)$ $= 250 + \frac{mp + nq}{2} - 2(mp + nq)$ $= 250 - \frac{3}{2} \cdot (mp + nq)$
Taking our original equations summing to 0, and squaring each we get:
$n + q = -t$ $m + p = -r$
$n^2 + 2nq + q^2 = t^2$ $m^2 + 2mp + p^2 = r^2$
Adding, we get:
$m^2 + n^2 + p^2 + q^2 + 2(mp + nq) = r^2 + t^2$
Substituting again we obtain:
$250 - r^2 - t^2 + 2(mp + nq) = r^2 + t^2$ $2(r^2 + t^2) = 250 + 2(mp + nq)$ $r^2 + t^2 = 125 + (mp + nq)$
Substituting the equivalence of $r^2 + t^2$
$-\frac{mp + nq}{2} = 125 + (mp + nq)$
Solving for $mp + nq$ , we find it equal to $-\frac{250}{3}$
Substituting this value into our answer expression, we get:
$250 - \frac{3}{2} \cdot (-\frac{250}{3})$ ,
Answer = $\boxed{375}$ | null | 375 |
a12b5a7f7cf92f325ca8a0df883f858e | https://artofproblemsolving.com/wiki/index.php/2012_AIME_I_Problems/Problem_14 | Complex numbers $a,$ $b,$ and $c$ are zeros of a polynomial $P(z) = z^3 + qz + r,$ and $|a|^2 + |b|^2 + |c|^2 = 250.$ The points corresponding to $a,$ $b,$ and $c$ in the complex plane are the vertices of a right triangle with hypotenuse $h.$ Find $h^2.$ | As noted in the previous solutions, $a+b+c = 0$ .
Let $a = a_1+a_2 i$ $b = b_1+b_2 i$ $c = c_1+c_2 i$ and we have $a_1 + b_1 + c_1 = a_2 + b_2 + c_2 = 0$ .
Then the given $|a|^2 + |b|^2 + |c|^2 = 250$ translates to $\sum_{} ( {a_1}^2 + {a_2}^2 ) = 250.$ Note that in a right triangle, the sum of the squares of the three sides is equal to two times the square of the hypotenuse, by the pythagorean theorem.
Thus, we have \[2h^2 = (a_1 - b_1)^2 + (a_2 - b_2)^2 + (b_1 - c_1)^2 + (b_2 - c_2)^2 + (a_1 - c_1)^2 + (a_2 - c_2)^2\] \[= 2 \left( \sum_{} ( {a_1}^2 + {a_2}^2 ) \right) - 2 \left( \sum_{cyc} a_1 b_1 + \sum_{cyc} a_2 b_2 \right)\] \[= 500 - \left( (a_1 + b_1 + c_1)^2 + (a_2 + b_2 + c_2)^2 - \sum_{cyc} ( {a_1}^2 + {a_2}^2 ) \right)\] \[= 500 - (0^2 + 0^2 - 250)\] so $h^2 = \boxed{375}$ and we may conclude.
~ rzlng | null | 375 |
a12b5a7f7cf92f325ca8a0df883f858e | https://artofproblemsolving.com/wiki/index.php/2012_AIME_I_Problems/Problem_14 | Complex numbers $a,$ $b,$ and $c$ are zeros of a polynomial $P(z) = z^3 + qz + r,$ and $|a|^2 + |b|^2 + |c|^2 = 250.$ The points corresponding to $a,$ $b,$ and $c$ in the complex plane are the vertices of a right triangle with hypotenuse $h.$ Find $h^2.$ | First, note that the roots of this cubic will be $a, b$ and $-(a+b)$ due to Vieta's, which means that the sum of the roots are 0.
Now, we use some god level wishful thinking. It would really be nice if one of these roots was on the real axis, and it was a right isosceles triangle, because that would be very convenient and easy to work with. The neat part is that it actually works
Set one of the roots as $r$ , where $r$ is any real number. WLOG, assume that this is the right angle. With symmetry to respect of the x axis (because symmetry makes the imaginary parts of the other 2 roots cancel out, besides the fact that complex conjugate root theorem forces it). This way, we can set the other 2 roots as $\frac{r}{2}+ni$ and $\frac{r}{2}-ni$ , making the roots add up to 0.
Now, as we want the roots to satisfy the original condition (right triangle) we are going to have to set an equation to find $n$ out. We use the fact that it is an isosceles right triangle to find that $\frac{3r}{2}=n$ , which means that the 2 other roots are now $\frac{r}{2}+\frac{3r}{2}i$ and $\frac{r}{2}-\frac{3r}{2}i$
Now we use the fact that $|a|^2+|b|^2+|c|^2=250$ . Clearly one of these is $r$ away from the origin, so that gets $r^2$ , and then we get $2*\frac{r}{2}^2+\frac{3r}{2}^2$ which gets us $5r^2$ , getting $r^2+5r^2=250$ , so $r=\sqrt{\frac{250}{6}}$ . So the final answer comes out to \[(\frac{250}{6}*\frac{9}{4}*2)^2=\boxed{375}\] | null | 375 |
a12b5a7f7cf92f325ca8a0df883f858e | https://artofproblemsolving.com/wiki/index.php/2012_AIME_I_Problems/Problem_14 | Complex numbers $a,$ $b,$ and $c$ are zeros of a polynomial $P(z) = z^3 + qz + r,$ and $|a|^2 + |b|^2 + |c|^2 = 250.$ The points corresponding to $a,$ $b,$ and $c$ in the complex plane are the vertices of a right triangle with hypotenuse $h.$ Find $h^2.$ | As shown in the other solutions, $a+b+c = 0$
Without loss of generality, let $b$ be the complex number opposite the hypotenuse.
Note that there is an isomorphism between $\mathbb{C}$ under $+$ and $\mathbb{R}^2$ under $+$
Let $\Vec{a}$ $\Vec{b}$ , and $\Vec{c}$ be the corresponding vectors to $a$ $b$ , and $c$
Thus $\Vec{a} + \Vec{b} + \Vec{c} = \Vec{0}$
$\Rightarrow 0 = \Vec{0}\cdot \Vec{0} = (\Vec{a} + \Vec{b} + \Vec{c})\cdot (\Vec{a} + \Vec{b} + \Vec{c}) = \Vec{a}\cdot \Vec{a} + \Vec{b}\cdot \Vec{b} + \Vec{c}\cdot \Vec{c} + 2(\Vec{a}\cdot \Vec{b} + \Vec{a}\cdot \Vec{c} + \Vec{b}\cdot \Vec{c})$
Now $|a|^2 + |b|^2 + |c|^2 = 250$ implies that $\lVert \Vec{a}\rVert^2 + \lVert \Vec{b}\rVert^2 + \lVert \Vec{c}\rVert^2 = 250$
$\Rightarrow \Vec{a}\cdot \Vec{a} + \Vec{b}\cdot \Vec{b} + \Vec{c}\cdot \Vec{c} = \lVert \Vec{a}\rVert^2 + \lVert \Vec{b}\rVert^2 + \lVert \Vec{c}\rVert^2 = 250$
Also note that because there is a right angle at $b$ $\Vec{a} - \Vec{b}$ and $\Vec{c} - \Vec{b}$ are perpendicular.
$\Rightarrow (\Vec{a} - \Vec{b})\cdot (\Vec{c} - \Vec{b}) = 0$
$\Rightarrow 0 = (\Vec{a} - \Vec{b})\cdot (\Vec{c} - \Vec{b}) = \Vec{a}\cdot \Vec{c} + \Vec{b} \cdot \Vec{b} - \Vec{a} \cdot \Vec{b} - \Vec{b} \cdot \Vec{c}$
Note that $h^2 = |a-c|^2$
$\Rightarrow h^2 = \lVert \Vec{a} - \Vec{c} \rVert^2 = (\Vec{a} - \Vec{c})\cdot (\Vec{a} - \Vec{c}) = \Vec{a} \cdot \Vec{a} + \Vec{c}\cdot \Vec{c} - \Vec{a}\cdot \Vec{c} - \Vec{a}\cdot \Vec{c} = \Vec{a} \cdot \Vec{a} + \Vec{c}\cdot \Vec{c} - 2\Vec{a}\cdot \Vec{c}$
$\Rightarrow 0 = \Vec{a}\cdot \Vec{a} + \Vec{b}\cdot \Vec{b} + \Vec{c}\cdot \Vec{c} + 2(\Vec{a}\cdot \Vec{b} + \Vec{a}\cdot \Vec{c} + \Vec{b}\cdot \Vec{c}) = 250 + 2(\Vec{a}\cdot \Vec{b} + \Vec{a}\cdot \Vec{c} + \Vec{b}\cdot \Vec{c})$
$\Rightarrow -250 = 2(\Vec{a}\cdot \Vec{b} + \Vec{a}\cdot \Vec{c} + \Vec{b}\cdot \Vec{c})$
$\Rightarrow -125 = \Vec{a}\cdot \Vec{b} + \Vec{a}\cdot \Vec{c} + \Vec{b}\cdot \Vec{c}$
$\Rightarrow -125 = -125 + 0 = (\Vec{a}\cdot \Vec{b} + \Vec{a}\cdot \Vec{c} + \Vec{b}\cdot \Vec{c}) + (\Vec{a}\cdot \Vec{c} + \Vec{b} \cdot \Vec{b} - \Vec{a} \cdot \Vec{b} - \Vec{b} \cdot \Vec{c}) = 2\Vec{a}\cdot \Vec{c} + \Vec{b} \cdot \Vec{b}$
$\Rightarrow 125 = - \Vec{b} \cdot \Vec{b} - 2\Vec{a}\cdot \Vec{c}$
$\Rightarrow 375 = 250 + 125 = \Vec{a}\cdot \Vec{a} + \Vec{b}\cdot \Vec{b} + \Vec{c}\cdot \Vec{c} - \Vec{b} \cdot \Vec{b} - 2\Vec{a}\cdot \Vec{c} = \Vec{a}\cdot \Vec{a} + \Vec{c}\cdot \Vec{c} - 2\Vec{a}\cdot \Vec{c} = h^2$
$\Rightarrow h^2 = \boxed{375}$ | null | 375 |
a12b5a7f7cf92f325ca8a0df883f858e | https://artofproblemsolving.com/wiki/index.php/2012_AIME_I_Problems/Problem_14 | Complex numbers $a,$ $b,$ and $c$ are zeros of a polynomial $P(z) = z^3 + qz + r,$ and $|a|^2 + |b|^2 + |c|^2 = 250.$ The points corresponding to $a,$ $b,$ and $c$ in the complex plane are the vertices of a right triangle with hypotenuse $h.$ Find $h^2.$ | Note that the roots of the polynomial $(a,b,c)$ must sum to $0$ due to the $z^2$ coefficient equaling $0$ because of Vieta's Formulas. This tells us that $a+b+c = 0$ and $\overline{a} + \overline{b} + \overline{c} = 0$ so $(a+b+c)(\overline{a+b+c}) = |a|^2 + |b|^2 + |c|^2 + a\overline{b} + \overline{a}b + b\overline{c} + \overline{b}c + c\overline{a} + \overline{c}a = 250 + a\overline{b} + \overline{a}b + b\overline{c} + \overline{b}c + c\overline{a} + \overline{c}a = 0$ so $a\overline{b} + \overline{a}b + b\overline{c} + \overline{b}c + c\overline{a} + \overline{c}a = -250.$
Note that terms similar to $\overline{a}b + \overline{b}a$ appear in $|a-b|^2$ so we decide to sum $|a-b|^2 + |b-c|^2 + |c-a|^2$ out of intuition. Note that this corresponds to the sum of the squares of the sidelengths of the right triangle and if WLOG the $|c-a|^2$ side is the hypotenuse then our sum is equal to $2|c-a|^2 = 2h^2.$
\[2|a|^2 + 2|b|^2 + 2|c|^2 -a\overline{b} - \overline{a}b - b\overline{c} - \overline{b}c - c\overline{a} - \overline{c}a = 500 + 250 = 2h^2\]
As a result we know that $h^2 = \boxed{375}$ ~SailS | null | 375 |
f6a6c4c763e967d26c4fbd6bdfd5128b | https://artofproblemsolving.com/wiki/index.php/2012_AIME_I_Problems/Problem_15 | There are $n$ mathematicians seated around a circular table with $n$ seats numbered $1,$ $2,$ $3,$ $...,$ $n$ in clockwise order. After a break they again sit around the table. The mathematicians note that there is a positive integer $a$ such that
Find the number of possible values of $n$ with $1 < n < 1000.$ | It is a well-known fact that the set $0, a, 2a, ... (n-1)a$ forms a complete set of residues if and only if $a$ is relatively prime to $n$
Thus, we have $a$ is relatively prime to $n$ . In addition, for any seats $p$ and $q$ , we must have $ap - aq$ not be equivalent to either $p - q$ or $q - p$ modulo $n$ to satisfy our conditions. These simplify to $(a-1)p \not\equiv (a-1)q$ and $(a+1)p \not\equiv (a+1)q$ modulo $n$ , so multiplication by both $a-1$ and $a+1$ must form a complete set of residues mod $n$ as well.
Thus, we have $a-1$ $a$ , and $a+1$ are relatively prime to $n$ . We must find all $n$ for which such an $a$ exists. $n$ obviously cannot be a multiple of $2$ or $3$ , but for any other $n$ , we can set $a = n-2$ , and then $a-1 = n-3$ and $a+1 = n-1$ . All three of these will be relatively prime to $n$ , since two numbers $x$ and $y$ are relatively prime if and only if $x-y$ is relatively prime to $x$ . In this case, $1$ $2$ , and $3$ are all relatively prime to $n$ , so $a = n-2$ works.
Now we simply count all $n$ that are not multiples of $2$ or $3$ , which is easy using inclusion-exclusion. We get a final answer of $998 - (499 + 333 - 166) = \boxed{332}$ | null | 332 |
481bc8c74cf92c369f9ccc883d4611f8 | https://artofproblemsolving.com/wiki/index.php/2012_AIME_II_Problems/Problem_1 | Find the number of ordered pairs of positive integer solutions $(m, n)$ to the equation $20m + 12n = 2012$ | Solving for $m$ gives us $m = \frac{503-3n}{5},$ so in order for $m$ to be an integer, we must have $3n \equiv 503 \mod 5 \longrightarrow n \equiv 1 \mod 5.$ The smallest possible value of $n$ is obviously $1,$ and the greatest is $\frac{503 - 5}{3} = 166,$ so the total number of solutions is $\frac{166-1}{5}+1 = \boxed{34}$ | null | 34 |
481bc8c74cf92c369f9ccc883d4611f8 | https://artofproblemsolving.com/wiki/index.php/2012_AIME_II_Problems/Problem_1 | Find the number of ordered pairs of positive integer solutions $(m, n)$ to the equation $20m + 12n = 2012$ | Dividing by $4$ gives us $5m + 3n = 503$ . Thus, we have $503-5m\equiv 0 \pmod 3$ since $n$ is an integer. Rearranging then gives $m\equiv 1\pmod 3.$ Since $503-5m>0,$ we know that $m<503/5.$ Because $m$ is an integer, we can rewrite this as $m\le 100.$ Therefore, $m$ ranges from \[0\cdot 3+1\quad \text{to} \quad 33\cdot 3+1,\] giving $\boxed{034}$ values.
~vaporwave | null | 034 |
481bc8c74cf92c369f9ccc883d4611f8 | https://artofproblemsolving.com/wiki/index.php/2012_AIME_II_Problems/Problem_1 | Find the number of ordered pairs of positive integer solutions $(m, n)$ to the equation $20m + 12n = 2012$ | Because the x-intercept of the equation is $\frac{2012}{20}$ , and the y-intercept is $\frac{2012}{12}$ , the slope is $\frac{\frac{-2012}{12}}{\frac{2012}{20}} = \frac{-5}{3}$ . Now, notice the first obvious solution: (100,1). From it, we derive all the other solutions by applying the slope in reverse, i.e: $(100,1), (97,6), (94,11)...$ Because the solutions are only positive, we can generate only 33 more solutions, so in total we have $33+1=\boxed{34}$ solutions. | null | 34 |
5fecb8bdf2c500e0fa6404789d3af42f | https://artofproblemsolving.com/wiki/index.php/2012_AIME_II_Problems/Problem_2 | Two geometric sequences $a_1, a_2, a_3, \ldots$ and $b_1, b_2, b_3, \ldots$ have the same common ratio, with $a_1 = 27$ $b_1=99$ , and $a_{15}=b_{11}$ . Find $a_9$ | Call the common ratio $r.$ Now since the $n$ th term of a geometric sequence with first term $x$ and common ratio $y$ is $xy^{n-1},$ we see that $a_1 \cdot r^{14} = b_1 \cdot r^{10} \implies r^4 = \frac{99}{27} = \frac{11}{3}.$ But $a_9$ equals $a_1 \cdot r^8 = a_1 \cdot (r^4)^2=27\cdot {\left(\frac{11}{3}\right)}^2=27\cdot \frac{121} 9=\boxed{363}$ | null | 363 |
51fb28612caf4ddc33d6c5383a59eb0d | https://artofproblemsolving.com/wiki/index.php/2012_AIME_II_Problems/Problem_3 | At a certain university, the division of mathematical sciences consists of the departments of mathematics, statistics, and computer science. There are two male and two female professors in each department. A committee of six professors is to contain three men and three women and must also contain two professors from each of the three departments. Find the number of possible committees that can be formed subject to these requirements. | There are two cases:
Case 1: One man and one woman is chosen from each department.
Case 2: Two men are chosen from one department, two women are chosen from another department, and one man and one woman are chosen from the third department.
For the first case, in each department there are ${{2}\choose{1}} \times {{2}\choose{1}} = 4$ ways to choose one man and one woman. Thus there are $4^3 = 64$ total possibilities conforming to case 1.
For the second case, there is only ${{2}\choose{2}} = 1$ way to choose two professors of the same gender from a department, and again there are $4$ ways to choose one man and one woman. Thus there are $1 \cdot 1 \cdot 4 = 4$ ways to choose two men from one department, two women from another department, and one man and one woman from the third department. However, there are $3! = 6$ different department orders, so the total number of possibilities conforming to case 2 is $4 \cdot 6 = 24$
Summing these two values yields the final answer: $64 + 24 = \boxed{088}$ | null | 088 |
51fb28612caf4ddc33d6c5383a59eb0d | https://artofproblemsolving.com/wiki/index.php/2012_AIME_II_Problems/Problem_3 | At a certain university, the division of mathematical sciences consists of the departments of mathematics, statistics, and computer science. There are two male and two female professors in each department. A committee of six professors is to contain three men and three women and must also contain two professors from each of the three departments. Find the number of possible committees that can be formed subject to these requirements. | Use generating functions . For each department, there is 1 way to pick 2 males, 4 ways to pick one of each, and 1 way to pick 2 females. Since there are three departments in total, and we wish for three males and three females, the answer will be equal to the coefficient of $x^3y^3$ in the expansion of $(x^2+4xy+y^2)^3$ . The requested coefficient is $\boxed{088}$ | null | 088 |
c580f9e892dc2a8347d3c059024874a2 | https://artofproblemsolving.com/wiki/index.php/2012_AIME_II_Problems/Problem_4 | Ana, Bob, and CAO bike at constant rates of $8.6$ meters per second, $6.2$ meters per second, and $5$ meters per second, respectively. They all begin biking at the same time from the northeast corner of a rectangular field whose longer side runs due west. Ana starts biking along the edge of the field, initially heading west, Bob starts biking along the edge of the field, initially heading south, and Cao bikes in a straight line across the field to a point $D$ on the south edge of the field. Cao arrives at point $D$ at the same time that Ana and Bob arrive at $D$ for the first time. The ratio of the field's length to the field's width to the distance from point $D$ to the southeast corner of the field can be represented as $p : q : r$ , where $p$ $q$ , and $r$ are positive integers with $p$ and $q$ relatively prime. Find $p+q+r$ | [asy]draw((1.2,0)--(0,0)--(0,1.4)--(6,1.4)--(6,0)--(1.2,0)--(6,1.4)); label("$D$", (1.2,0),dir(-90)); dot((6,1.4)); dot((1.2,0)); label("$a$", (0.6,0),dir(-90)); label("$b$", (3.6,0),dir(-90)); label("$c$", (6,0.7),dir(0));[/asy]
Let $a,b,c$ be the labeled lengths as shown in the diagram. Also, assume WLOG the time taken is $1$ second.
Observe that $\dfrac{2a+b+c}{8.6}=1$ or $2a+b+c=8.6$ , and $\dfrac{b+c}{6.2}=1$ or $b+c=6.2$ . Subtracting the second equation from the first gives $2a=2.4$ , or $a=1.2$
Now, let us solve $b$ and $c$ . Note that $\dfrac{\sqrt{b^2+c^2}}{5}=1$ , or $b^2+c^2=25$ . We also have $b+c=6.2$
We have a system of equations: \[\left\{\begin{array}{l}b+c=6.2\\ b^2+c^2=25\end{array}\right.\]
Squaring the first equation gives $b^2+2bc+c^2=38.44$ , and subtracting the second from this gives $2bc=13.44$ . Now subtracting this from $b^2+c^2=25$ gives $b^2-2bc+c^2=(b-c)^2=11.56$ , or $b-c=3.4$ . Now we have the following two equations:
\[\left\{\begin{array}{l}b+c=6.2\\ b-c=3.4\end{array}\right.\]
Adding the equations and dividing by two gives $b=4.8$ , and it follows that $c=1.4$
The ratios we desire are therefore $1.4:6:4.8=7:30:24$ , and our answer is $7+30+24=\boxed{061}$ | null | 061 |
716abb5b3f2244871b23393c63040a94 | https://artofproblemsolving.com/wiki/index.php/2012_AIME_II_Problems/Problem_5 | In the accompanying figure, the outer square $S$ has side length $40$ . A second square $S'$ of side length $15$ is constructed inside $S$ with the same center as $S$ and with sides parallel to those of $S$ . From each midpoint of a side of $S$ , segments are drawn to the two closest vertices of $S'$ . The result is a four-pointed starlike figure inscribed in $S$ . The star figure is cut out and then folded to form a pyramid with base $S'$ . Find the volume of this pyramid. | The volume of this pyramid can be found by the equation $V=\frac{1}{3}bh$ , where $b$ is the base and $h$ is the height. The base is easy, since it is a square and has area $15^2=225$
To find the height of the pyramid, the height of the four triangles is needed, which will be called $h^\prime$ . By drawing a line through the middle of the larger square, we see that its length is equal to the length of the smaller rectangle and two of the triangle's heights. Then $40=2h^\prime +15$ , which means that $h^\prime=12.5$
When the pyramid is made, you see that the height is the one of the legs of a right triangle, with the hypotenuse equal to $h^\prime$ and the other leg having length equal to half of the side length of the smaller square, or $7.5$ . So, the Pythagorean Theorem can be used to find the height.
$h=\sqrt{12.5^2-7.5^2}=\sqrt{100}=10$
Finally, $V=\frac{1}{3}bh=\frac{1}{3}(225)(10)=\boxed{750}$ | null | 750 |
7e9ae42eef8f3eb96a24244341444514 | https://artofproblemsolving.com/wiki/index.php/2012_AIME_II_Problems/Problem_6 | Let $z=a+bi$ be the complex number with $\vert z \vert = 5$ and $b > 0$ such that the distance between $(1+2i)z^3$ and $z^5$ is maximized, and let $z^4 = c+di$ . Find $c+d$ | Let's consider the maximization constraint first: we want to maximize the value of $|z^5 - (1+2i)z^3|$ Simplifying, we have
$|z^3| * |z^2 - (1+2i)|$
$=|z|^3 * |z^2 - (1+2i)|$
$=125|z^2 - (1+2i)|$
Thus we only need to maximize the value of $|z^2 - (1+2i)|$
To maximize this value, we must have that $z^2$ is in the opposite direction of $1+2i$ . The unit vector in the complex plane in the desired direction is $\frac{-1}{\sqrt{5}} + \frac{-2}{\sqrt{5}} i$ . Furthermore, we know that the magnitude of $z^2$ is $25$ , because the magnitude of $z$ is $5$ . From this information, we can find that $z^2 = \sqrt{5} (-5 - 10i)$
Squaring, we get $z^4 = 5 (25 - 100 + 100i) = -375 + 500i$ . Finally, $c+d = -375 + 500 = \boxed{125}$ | null | 125 |
7e9ae42eef8f3eb96a24244341444514 | https://artofproblemsolving.com/wiki/index.php/2012_AIME_II_Problems/Problem_6 | Let $z=a+bi$ be the complex number with $\vert z \vert = 5$ and $b > 0$ such that the distance between $(1+2i)z^3$ and $z^5$ is maximized, and let $z^4 = c+di$ . Find $c+d$ | WLOG, let $z_{1}=5(\cos{\theta_{1}}+i\sin{\theta_{1}})$ and
$z_{2}=1+2i=\sqrt{5}(\cos{\theta_{2}+i\sin{\theta_{2}}})$
This means that
$z_{1}^3=125(\cos{3\theta_{1}}+i\sin{3\theta_{1}})$
$z_{1}^4=625(\cos{4\theta_{1}}+i\sin{4\theta_{1}})$
Hence, this means that
$z_{2}z_{1}^3=125\sqrt{5}(\cos({\theta_{2}+3\theta_{1}})+i\sin({\theta_{2}+3\theta_{1}}))$
And
$z_{1}^5=3125(\cos{5\theta_{1}}+i\sin{5\theta_{1}})$
Now, common sense tells us that the distance between these two complex numbers is maxed when they both are points satisfying the equation of the line $yi=mx$ , or when they are each a $180^{\circ}$ rotation away from each other.
Hence, we must have that $5\theta_{1}=3\theta_{1}+\theta_{2}+180^{\circ}\implies\theta_{1}=\frac{\theta_{2}+180^{\circ}}{2}$
Now, plug this back into $z_{1}^4$ (if you want to know why, reread what we want in the problem!)
So now, we have that $z_{1}^4=625(\cos{2\theta_{2}}+i\sin{2\theta_{2}})$
Notice that $\cos\theta_{2}=\frac{1}{\sqrt{5}}$ and $\sin\theta_{2}=\frac{2}{\sqrt{5}}$
Then, we have that $\cos{2\theta_{2}}=\cos^2{\theta_{2}}-\sin^2{\theta_{2}}=-\frac{3}{5}$ and $\sin{2\theta_{2}}=2\sin{\theta_{2}}\cos{\theta_{2}}=\frac{4}{5}$
Finally, plugging back in, we find that $z_{1}^4=625(-\frac{3}{5}+\frac{4i}{5})=-375+500i$
$-375+500=\boxed{125}$ | null | 125 |
4c79e4018d20234226594e596a676064 | https://artofproblemsolving.com/wiki/index.php/2012_AIME_II_Problems/Problem_7 | Let $S$ be the increasing sequence of positive integers whose binary representation has exactly $8$ ones. Let $N$ be the 1000th number in $S$ . Find the remainder when $N$ is divided by $1000$ | Okay, an exercise in counting (lots of binomials to calculate!). In base 2, the first number is $11111111$ , which is the only way to choose 8 1's out of 8 spaces, or $\binom{8}{8}$ . What about 9 spaces? Well, all told, there are $\binom{9}{8}=9$ , which includes the first 1. Similarly, for 10 spaces, there are $\binom{10}{8}=45,$ which includes the first 9. For 11 spaces, there are $\binom{11}{8}=165$ , which includes the first 45. You're getting the handle. For 12 spaces, there are $\binom{12}{8}=495$ , which includes the first 165; for 13 spaces, there are $\binom{13}{8}=13 \cdot 99 > 1000$ , so we now know that $N$ has exactly 13 spaces, so the $2^{12}$ digit is 1.
Now we just proceed with the other 12 spaces with 7 1's, and we're looking for the $1000-495=505th$ number. Well, $\binom{11}{7}=330$ , so we know that the $2^{11}$ digit also is 1, and we're left with finding the $505-330=175th$ number with 11 spaces and 6 1's. Now $\binom{10}{6}=210,$ which is too big, but $\binom{9}{6}=84.$ Thus, the $2^9$ digit is 1, and we're now looking for the $175-84=91st$ number with 9 spaces and 5 1's. Continuing the same process, $\binom{8}{5}=56$ , so the $2^8$ digit is 1, and we're left to look for the $91-56=35th$ number with 8 spaces and 4 1's. But here $\binom{7}{4}=35$ , so N must be the last or largest 7-digit number with 4 1's. Thus the last 8 digits of $N$ must be $01111000$ , and to summarize, $N=1101101111000$ in base $2$ . Therefore, $N = 8+16+32+64+256+512+2048+4096 \equiv 32 \pmod{1000}$ , and the answer is $\boxed{032}$ | null | 032 |
40836e7f6ea315e4d08e831b9204a387 | https://artofproblemsolving.com/wiki/index.php/2012_AIME_II_Problems/Problem_8 | The complex numbers $z$ and $w$ satisfy the system \[z + \frac{20i}w = 5+i\] \[w+\frac{12i}z = -4+10i\] Find the smallest possible value of $\vert zw\vert^2$ | Multiplying the two equations together gives us \[zw + 32i - \frac{240}{zw} = -30 + 46i\] and multiplying by $zw$ then gives us a quadratic in $zw$ \[(zw)^2 + (30-14i)zw - 240 =0.\] Using the quadratic formula, we find the two possible values of $zw$ to be $7i-15 \pm \sqrt{(15-7i)^2 + 240}$ $6+2i,$ $12i - 36.$ The smallest possible value of $\vert zw\vert^2$ is then obviously $6^2 + 2^2 = \boxed{040}$ | null | 040 |
9ede860bb47d00e9b3aacb1b9a2560c1 | https://artofproblemsolving.com/wiki/index.php/2012_AIME_II_Problems/Problem_9 | Let $x$ and $y$ be real numbers such that $\frac{\sin x}{\sin y} = 3$ and $\frac{\cos x}{\cos y} = \frac12$ . The value of $\frac{\sin 2x}{\sin 2y} + \frac{\cos 2x}{\cos 2y}$ can be expressed in the form $\frac pq$ , where $p$ and $q$ are relatively prime positive integers. Find $p+q$ | Let the equation $\frac{\sin x}{\sin y} = 3$ be equation 1, and let the equation $\frac{\cos x}{\cos y} = \frac12$ be equation 2.
Hungry for the widely-used identity $\sin^2\theta + \cos^2\theta = 1$ , we cross multiply equation 1 by $\sin y$ and multiply equation 2 by $\cos y$
Equation 1 then becomes:
$\sin x = 3\sin y$
Equation 2 then becomes:
$\cos x = \frac{1}{2} \cos y$
Aha! We can square both of the resulting equations and match up the resulting LHS with the resulting RHS:
$1 = 9\sin^2 y + \frac{1}{4} \cos^2 y$
Applying the identity $\cos^2 y = 1 - \sin^2 y$ (which is similar to $\sin^2\theta + \cos^2\theta = 1$ but a bit different), we can change $1 = 9\sin^2 y + \frac{1}{4} \cos^2 y$ into:
$1 = 9\sin^2 y + \frac{1}{4} - \frac{1}{4} \sin^2 y$
Rearranging, we get $\frac{3}{4} = \frac{35}{4} \sin^2 y$
So, $\sin^2 y = \frac{3}{35}$
Squaring Equation 1 (leading to $\sin^2 x = 9\sin^2 y$ ), we can solve for $\sin^2 x$
$\sin^2 x = 9\left(\frac{3}{35}\right) = \frac{27}{35}$
Using the identity $\cos 2\theta = 1 - 2\sin^2\theta$ , we can solve for $\frac{\cos 2x}{\cos 2y}$
$\cos 2x = 1 - 2\sin^2 x = 1 - 2\cdot\frac{27}{35} = 1 - \frac{54}{35} = -\frac{19}{35}$
$\cos 2y = 1 - 2\sin^2 y = 1 - 2\cdot\frac{3}{35} = 1 - \frac{6}{35} = \frac{29}{35}$
Thus, $\frac{\cos 2x}{\cos 2y} = \frac{-19/35}{29/35} = -\frac{19}{29}$
Plugging in the numbers we got back into the original equation :
We get $\frac{\sin 2x}{\sin 2y} + \frac{\cos 2x}{\cos 2y} = \frac32 + \left(-\frac{19}{29} \right) = \frac{49}{58}$
So, the answer is $49+58=\boxed{107}$ | null | 107 |
9ede860bb47d00e9b3aacb1b9a2560c1 | https://artofproblemsolving.com/wiki/index.php/2012_AIME_II_Problems/Problem_9 | Let $x$ and $y$ be real numbers such that $\frac{\sin x}{\sin y} = 3$ and $\frac{\cos x}{\cos y} = \frac12$ . The value of $\frac{\sin 2x}{\sin 2y} + \frac{\cos 2x}{\cos 2y}$ can be expressed in the form $\frac pq$ , where $p$ and $q$ are relatively prime positive integers. Find $p+q$ | As mentioned above, the first term is clearly $\frac{3}{2}.$ For the second term, we first wish to find $\frac{\cos 2x}{\cos 2y} =\frac{2\cos^2 x - 1}{2 \cos^2y -1}.$ Now we first square the first equation getting $\frac{\sin^2x}{\sin^2y} =\frac{1-\cos^2x}{1 - \cos^2y} =9.$ Squaring the second equation yields $\frac{\cos^2x}{\cos^2y} =\frac{1}{4}.$ Let $\cos^2x = a$ and $\cos^2y = b.$ We have the system of equations \begin{align*} 1-a &= 9-9b \\ 4a &= b \\ \end{align*} Multiplying the first equation by $4$ yields $4-4a = 36 - 36b$ and so $4-b =36 - 36b \implies b =\frac{32}{35}.$ We then find $a =\frac{8}{35}.$ Therefore the second fraction ends up being $\frac{\frac{64}{35}-1}{\frac{16}{35}-1} = -\frac{19}{29}$ so that means our desired sum is $\frac{49}{58}$ so the desired sum is $\boxed{107}.$ | null | 107 |
9ede860bb47d00e9b3aacb1b9a2560c1 | https://artofproblemsolving.com/wiki/index.php/2012_AIME_II_Problems/Problem_9 | Let $x$ and $y$ be real numbers such that $\frac{\sin x}{\sin y} = 3$ and $\frac{\cos x}{\cos y} = \frac12$ . The value of $\frac{\sin 2x}{\sin 2y} + \frac{\cos 2x}{\cos 2y}$ can be expressed in the form $\frac pq$ , where $p$ and $q$ are relatively prime positive integers. Find $p+q$ | Dgxje.png
We draw 2 right triangles with angles x and y that have the same hypotenuse.
We get $b^2 + 9a^2 = 4b^2 + a^2$ . Then, we find $8a^2 = 3b^2$
Now, we can scale the triangle such that $a = \sqrt{3}$ $b = \sqrt{8}$ . We find all the side lengths, and we find the hypotenuse of both these triangles to equal $\sqrt{35}$ This allows us to find sin and cos easily.
The first term is $\frac{3}{2}$ , refer to solution 1 for how to find it.
The second term is $\frac{\cos^2(x) - \sin^2(x)}{\cos^2(y) - \sin^2(y)}$ . Using the diagram, we can easily compute this as $\frac{\frac{8}{35} - \frac{27}{35}}{\frac{32}{35} - \frac{3}{35}} = \frac{-19}{29}$
Summing these you get $\frac{3}{2} + \frac{-19}{29} = \frac{49}{58} \implies \boxed{107}$ | null | 107 |
9ede860bb47d00e9b3aacb1b9a2560c1 | https://artofproblemsolving.com/wiki/index.php/2012_AIME_II_Problems/Problem_9 | Let $x$ and $y$ be real numbers such that $\frac{\sin x}{\sin y} = 3$ and $\frac{\cos x}{\cos y} = \frac12$ . The value of $\frac{\sin 2x}{\sin 2y} + \frac{\cos 2x}{\cos 2y}$ can be expressed in the form $\frac pq$ , where $p$ and $q$ are relatively prime positive integers. Find $p+q$ | Let $a = \sin(x), b = \sin(y)$ The first equation yields $\frac{a}{b} = 3.$ Using $sin^2(x) + cos^2(x) = 1$ the second equation yields \[\frac{\sqrt{1-a^2}}{\sqrt{1-b^2}} = \frac{1}{2} \rightarrow \frac{1-a^2}{1-b^2} = \frac{1}{4}\]
Solving this yields $\left(a, b\right) = \left(3\sqrt{\frac{3}{35}},\sqrt{\frac{3}{35}}\right).$ Finding the first via double angle for sin yields \[\frac{\sin(2x)}{\sin(2y)} = \frac{2\sin{x}\cos{x}}{2\sin{y}\cos{y}} = 3 \cdot \frac{1}{2} = \frac{3}{2}\] Double angle for cosine is \[\cos(2x) = 1-2\sin^2{x}\] so $\frac{\cos(2x)}{\sin(2x)} = \frac{1-2a^2}{1-2b^2} = -\frac{19}{29}.$ Adding yields $\frac{49}{58} \rightarrow 49 + 58 = \boxed{107}$ | null | 107 |
e58ab0d3fd8cdb863bfbbbecb6e8c5fa | https://artofproblemsolving.com/wiki/index.php/2012_AIME_II_Problems/Problem_10 | Find the number of positive integers $n$ less than $1000$ for which there exists a positive real number $x$ such that $n=x\lfloor x \rfloor$
Note: $\lfloor x \rfloor$ is the greatest integer less than or equal to $x$ | We know that $x$ cannot be irrational because the product of a rational number and an irrational number is irrational (but $n$ is an integer). Therefore $x$ is rational.
Let $x = a + \frac{b}{c}$ where $a,b,c$ are nonnegative integers and $0 \le b < c$ (essentially, $x$ is a mixed number). Then, \[n = \left(a + \frac{b}{c}\right) \left\lfloor a +\frac{b}{c} \right\rfloor \Rightarrow n = \left(a + \frac{b}{c}\right)a = a^2 + \frac{ab}{c}\]
Here it is sufficient for $\frac{ab}{c}$ to be an integer. We can use casework to find values of $n$ based on the value of $a$
$a = 0 \implies$ nothing because n is positive
$a = 1 \implies \frac{b}{c} = \frac{0}{1}$
$a = 2 \implies \frac{b}{c} = \frac{0}{2},\frac{1}{2}$
$a = 3 \implies\frac{b}{c} =\frac{0}{3},\frac{1}{3},\frac{2}{3}$
The pattern continues up to $a = 31$ . Note that if $a = 32$ , then $n > 1000$ . However if $a = 31$ , the largest possible $x$ is $31 + \frac{30}{31}$ , in which $n$ is still less than $1000$ . Therefore the number of positive integers for $n$ is equal to $1+2+3+...+31 = \frac{31 \cdot 32}{2} = \boxed{496}.$ | null | 496 |
e58ab0d3fd8cdb863bfbbbecb6e8c5fa | https://artofproblemsolving.com/wiki/index.php/2012_AIME_II_Problems/Problem_10 | Find the number of positive integers $n$ less than $1000$ for which there exists a positive real number $x$ such that $n=x\lfloor x \rfloor$
Note: $\lfloor x \rfloor$ is the greatest integer less than or equal to $x$ | Notice that $x\lfloor x\rfloor$ is continuous over the region $x \in [k, k+1)$ for any integer $k$ . Therefore, it takes all values in the range $[k\lfloor k\rfloor, (k+1)\lfloor k+1\rfloor) = [k^2, (k+1)k)$ over that interval. Note that if $k>32$ then $k^2 > 1000$ and if $k=31$ , the maximum value attained is $31*32 < 1000$ . It follows that the answer is $\sum_{k=1}^{31} (k+1)k-k^2 = \sum_{k=1}^{31} k = \frac{31\cdot 32}{2} = \boxed{496}.$ | null | 496 |
e58ab0d3fd8cdb863bfbbbecb6e8c5fa | https://artofproblemsolving.com/wiki/index.php/2012_AIME_II_Problems/Problem_10 | Find the number of positive integers $n$ less than $1000$ for which there exists a positive real number $x$ such that $n=x\lfloor x \rfloor$
Note: $\lfloor x \rfloor$ is the greatest integer less than or equal to $x$ | Bounding gives $x^2\le n<x^2+x$ . Thus there are a total of $x$ possible values for $n$ , for each value of $x^2$ . Checking, we see $31^2+31=992<1000$ , so there are \[1+2+3+...+31= \boxed{496}\] such values for $n$ | null | 496 |
e58ab0d3fd8cdb863bfbbbecb6e8c5fa | https://artofproblemsolving.com/wiki/index.php/2012_AIME_II_Problems/Problem_10 | Find the number of positive integers $n$ less than $1000$ for which there exists a positive real number $x$ such that $n=x\lfloor x \rfloor$
Note: $\lfloor x \rfloor$ is the greatest integer less than or equal to $x$ | After a bit of experimenting, we let $n=l^2+s, s < 2n+1$ . We claim that I (the integer part of $x$ ) = $l$ . (Prove it yourself using contradiction !) so now we get that $x=l+\frac{s}{l}$ . This implies that solutions exist iff $s<l$ , or for all natural numbers of the form $l^2+s$ where $s<l$ .
Hence, 1 solution exists for $l=1$ ! 2 for $l=2$ and so on. Therefore our final answer is $31+30+\dots+1= \boxed{496}$ | null | 496 |
67389835cbcaba2302975b4b6d3b7ab2 | https://artofproblemsolving.com/wiki/index.php/2012_AIME_II_Problems/Problem_11 | Let $f_1(x) = \frac23 - \frac3{3x+1}$ , and for $n \ge 2$ , define $f_n(x) = f_1(f_{n-1}(x))$ . The value of $x$ that satisfies $f_{1001}(x) = x-3$ can be expressed in the form $\frac mn$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | After evaluating the first few values of $f_k (x)$ , we obtain $f_4(x) = f_1(x) = \frac{2}{3} - \frac{3}{3x+1} = \frac{6x-7}{9x+3}$ . Since $1001 \equiv 2 \mod 3$ $f_{1001}(x) = f_2(x) = \frac{3x+7}{6-9x}$ . We set this equal to $x-3$ , i.e.
$\frac{3x+7}{6-9x} = x-3 \Rightarrow x = \frac{5}{3}$ . The answer is thus $5+3 = \boxed{008}$ | null | 008 |
e5eed0001ca2cb267426f0647f9582a0 | https://artofproblemsolving.com/wiki/index.php/2012_AIME_II_Problems/Problem_13 | Equilateral $\triangle ABC$ has side length $\sqrt{111}$ . There are four distinct triangles $AD_1E_1$ $AD_1E_2$ $AD_2E_3$ , and $AD_2E_4$ , each congruent to $\triangle ABC$ ,
with $BD_1 = BD_2 = \sqrt{11}$ . Find $\sum_{k=1}^4(CE_k)^2$ | Screen Shot 2020-02-17 at 2.24.50 PM.png
We create a diagram. Each of the red quadrilaterals are actually 60 degree rotations about point A. We easily get that $\overline{C E_1} = \sqrt{11}$ , and $\overline{C E_3} = \sqrt{11}$ . If we set $\angle B A D_2 = \theta$ , we can start angle chasing. In particular, we will like to find $\angle D E_4 C$ , and $\angle D E_2 C$ , since then we will be able to set up some Law Of Cosines. $\angle D E_4 C = \angle D E_4 A + \angle A E_4 C = 90 - \frac{\theta}{2} + 30 + \frac{\theta}{2} = 120^{\circ}$ That was convenient! We can do it with the other angle as well. $\angle D E_2 C = \angle D E_2 A - \angle C E_2 A = 90 - \frac{\theta}{2} - (30 - \frac{\theta}{2}) = 60^{\circ}$ . That means we are able to set up Law of Cosines, on triangles $\triangle D E_4 C$ and $\triangle D E_2 C$ , with some really convenient angles.
Let $CE_2 = x$ , and $CE_4 = y$ \[333 = 11 + x^2 - \sqrt{11} x\] \[333 = 11 + y^2 + \sqrt{11} y\] We subtract and get: \[0 = (x+y)(x-y-\sqrt{11})\] $x+y$ obviously can't be 0, so $x-y = \sqrt{11}$ We add and get: \[666 = 22 + x^2 + y^2 + \sqrt{11} (y-x)\] $y-x = -\sqrt{11}$ . Thus, we can fill in and solve. \[666 = 22 + x^2 + y^2 - 11\] \[655 = x^2 + y^2\] Thus our answer is $C E_1^2 + C E_2^2 + C E_2^2 + C E_4^2 = 11 + 11 + C E_2^2 + C E_4^2 = 11 + 11 + x^2 + y^2 = 11 + 11 + 655 = \boxed{677}$ | null | 677 |
61a1f7b3bca880023835a9d0fb596cbf | https://artofproblemsolving.com/wiki/index.php/2012_AIME_II_Problems/Problem_14 | In a group of nine people each person shakes hands with exactly two of the other people from the group. Let $N$ be the number of ways this handshaking can occur. Consider two handshaking arrangements different if and only if at least two people who shake hands under one arrangement do not shake hands under the other arrangement. Find the remainder when $N$ is divided by $1000$ | Given that each person shakes hands with two people, we can view all of these through graph theory as 'rings'. This will split it into four cases: Three rings of three, one ring of three and one ring of six, one ring of four and one ring of five, and one ring of nine. (All other cases that sum to nine won't work, since they have at least one 'ring' of two or fewer points, which doesn't satisfy the handshaking conditions of the problem.)
Case 1: To create our groups of three, there are $\dfrac{\dbinom{9}{3}\dbinom{6}{3}\dbinom{3}{3}}{3!}$ . In general, the number of ways we can arrange people within the rings to count properly is $\dfrac{(n-1)!}{2}$ , since there are $(n-1)!$ ways to arrange items in the circle, and then we don't want to want to consider reflections as separate entities. Thus, each of the three cases has $\dfrac{(3-1)!}{2}=1$ arrangements. Therefore, for this case, there are $\left(\dfrac{\dbinom{9}{3}\dbinom{6}{3}\dbinom{3}{3}}{3!}\right)(1)^3=280$
Case 2: For three and six, there are $\dbinom{9}{6}=84$ sets for the rings. For organization within the ring, as before, there is only one way to arrange the ring of three. For six, there is $\dfrac{(6-1)!}{2}=60$ . This means there are $(84)(1)(60)=5040$ arrangements.
Case 3: For four and five, there are $\dbinom{9}{5}=126$ sets for the rings. Within the five, there are $\dfrac{4!}{2}=12$ , and within the four there are $\dfrac{3!}{2}=3$ arrangements. This means the total is $(126)(12)(3)=4536$
Case 4: For the nine case, there is $\dbinom{9}{9}=1$ arrangement for the ring. Within it, there are $\dfrac{8!}{2}=20160$ arrangements.
Summing the cases, we have $280+5040+4536+20160=30016 \to \boxed{016}$ | null | 016 |
61a1f7b3bca880023835a9d0fb596cbf | https://artofproblemsolving.com/wiki/index.php/2012_AIME_II_Problems/Problem_14 | In a group of nine people each person shakes hands with exactly two of the other people from the group. Let $N$ be the number of ways this handshaking can occur. Consider two handshaking arrangements different if and only if at least two people who shake hands under one arrangement do not shake hands under the other arrangement. Find the remainder when $N$ is divided by $1000$ | Let $f_N$ be the number of ways a group of $N$ people can shake hands with exactly two of the other people from the group, where $N \ge 3$
We can easily find that $f_3=1$
Continuing on, we will label the people as $A$ $B$ $C$ ,... corresponding with $N$
$f_4$ : There are $\dbinom{3}{2}=3$ possible ways for person $A$ to shake hands with two others (WLOG assume $A$ shakes hands with $B$ and $C$ ), and there is only one possible outcome thereon ( $B$ and $C$ both shakes hands with $D$ ).
Therefore, we can conclude that $f_4=3$
$f_5$ : There are $\dbinom{4}{2}=6$ possible ways for person $A$ to shake hands with two others (WLOG assume they are $B$ and $C$ ). Then, $B$ and $C$ must also shake hands with the remaining two people ( $2$ ways), and those last $2$ people must shake hands with each other ( $1$ way).
Therefore, $f_5=\dbinom{4}{2}\ast(2)\ast(1)=12$
$f_6$ : There are $\dbinom{5}{2}=10$ possible ways for person $A$ to shake hands with two others (WLOG assume they are $B$ and $C$ ). However, $B$ and $C$ shake hands with each other, then there are $f_3$ ways for the rest of the people to shake hands. If $B$ and $C$ shake hands with two others out of the three remaining people (3 $\ast$ 2 ways), those $2$ people must shake hands with the last person ( $1$ way).
Therefore, $f_5=\dbinom{5}{2}\ast(f_3)+3\ast(2)\ast(1)=70$
Now we have enough information to find $f_9$
$f_9$ : There are $\dbinom{8}{2}=28$ possible ways for person $A$ to shake hands with two others (WLOG assume they are $B$ and $C$ ).
Case 1: $B$ and $C$ shake hands with each other
There are $f_6$ ways for $D$ $E$ $F$ $G$ $H$ , and $I$ to shake hands
Case 2: $B$ and $C$ shake hands with one of the others out of the $6$ remaining people ( $6$ ways)
There are $f_5$ ways for $E$ $F$ $G$ $H$ , and $I$ to shake hands
Case 3: If $B$ and $C$ shake hands with two different people (there are 6 $\ast$ 5 ways to choose the people but WLOG assume they are $D$ and $E$ ). Calculations of Case 3 are in its subcases.
Subcase 3.1: $D$ and $E$ shake hands with each other
There are $f_4$ ways for $F$ $G$ $H$ , and $I$ to shake hands
Subcase 3.2: $D$ and $E$ each shake hands with one other person (there are $4$ ways to choose the person, WLOG let that be $F$ )
There are $f_3$ ways for $G$ $H$ , and $I$ to shake hands
Subcase 3.3: $D$ and $E$ each shake hands with two different people (there are 4 $\ast$ 3 ways, WLOG let that be $F$ and $G$ )
There are $2\ast1$ ways for $F$ and $G$ to then shake hands with $H$ and $I$ , and $H$ and $I$ will shake hands with each other.
We have
\begin{align*} 28(f_6+6\ast(f_5)+6\ast(5)\ast(f_4)+4\ast(f_3)+4\ast(3)\ast(2)\ast(1) &= 28(70+6\ast(12)+6\ast(5)\ast(3+4\ast(1)+4\ast(3)\ast(2)\ast(1) \\ &= 30016 \end{align*}
which gives us the answer of $\boxed{016}$ | null | 016 |
9f38d901faba7c2a093387d757c47f50 | https://artofproblemsolving.com/wiki/index.php/2012_AIME_II_Problems/Problem_15 | Triangle $ABC$ is inscribed in circle $\omega$ with $AB=5$ $BC=7$ , and $AC=3$ . The bisector of angle $A$ meets side $\overline{BC}$ at $D$ and circle $\omega$ at a second point $E$ . Let $\gamma$ be the circle with diameter $\overline{DE}$ . Circles $\omega$ and $\gamma$ meet at $E$ and a second point $F$ . Then $AF^2 = \frac mn$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | Take a force-overlaid inversion about $A$ and note $D$ and $E$ map to each other. As $DE$ was originally the diameter of $\gamma$ $DE$ is still the diameter of $\gamma$ . Thus $\gamma$ is preserved. Note that the midpoint $M$ of $BC$ lies on $\gamma$ , and $BC$ and $\omega$ are swapped. Thus points $F$ and $M$ map to each other, and are isogonal. It follows that $AF$ is a symmedian of $\triangle{ABC}$ , or that $ABFC$ is harmonic. Then $(AB)(FC)=(BF)(CA)$ , and thus we can let $BF=5x, CF=3x$ for some $x$ . By the LoC, it is easy to see $\angle{BAC}=120^\circ$ so $(5x)^2+(3x)^2-2\cos{60^\circ}(5x)(3x)=49$ . Solving gives $x^2=\frac{49}{19}$ , from which by Ptolemy's we see $AF=\frac{30}{\sqrt{19}}$ . We conclude the answer is $900+19=\boxed{919}$ | null | 919 |
9f38d901faba7c2a093387d757c47f50 | https://artofproblemsolving.com/wiki/index.php/2012_AIME_II_Problems/Problem_15 | Triangle $ABC$ is inscribed in circle $\omega$ with $AB=5$ $BC=7$ , and $AC=3$ . The bisector of angle $A$ meets side $\overline{BC}$ at $D$ and circle $\omega$ at a second point $E$ . Let $\gamma$ be the circle with diameter $\overline{DE}$ . Circles $\omega$ and $\gamma$ meet at $E$ and a second point $F$ . Then $AF^2 = \frac mn$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | Use the angle bisector theorem to find $CD=\tfrac{21}{8}$ $BD=\tfrac{35}{8}$ , and use Stewart's Theorem to find $AD=\tfrac{15}{8}$ . Use Power of Point $D$ to find $DE=\tfrac{49}{8}$ , and so $AE=8$ . Use law of cosines to find $\angle CAD = \tfrac{\pi} {3}$ , hence $\angle BAD = \tfrac{\pi}{3}$ as well, and $\triangle BCE$ is equilateral, so $BC=CE=BE=7$ [asy] size(150); defaultpen(fontsize(9pt)); picture pic; pair A,B,C,D,E,F,W; B=MP("B",origin,dir(180)); C=MP("C",(7,0),dir(0)); A=MP("A",IP(CR(B,5),CR(C,3)),N); D=MP("D",extension(B,C,A,bisectorpoint(C,A,B)),dir(220)); path omega=circumcircle(A,B,C); E=MP("E",OP(omega,A--(A+20*(D-A))),S); path gamma=CR(midpoint(D--E),length(D-E)/2); F=MP("F",OP(omega,gamma),SE); draw(omega^^A--B--C--cycle^^gamma); draw(pic, A--E--F--cycle, gray); add(pic); dot("$W$",circumcenter(A,B,C),dir(180)); label("$\gamma$",gamma,dir(180)); [/asy] In triangle $AEF$ , let $X$ be the foot of the altitude from $A$ ; then $EF=EX+XF$ , where we use signed lengths. Writing $EX=AE \cdot \cos \angle AEF$ and $XF=AF \cdot \cos \angle AFE$ , we get \begin{align}\tag{1} EF = AE \cdot \cos \angle AEF + AF \cdot \cos \angle AFE. \end{align} Note $\angle AFE = \angle ACE$ , and the Law of Cosines in $\triangle ACE$ gives $\cos \angle ACE = -\tfrac 17$ .
Also, $\angle AEF = \angle DEF$ , and $\angle DFE = \tfrac{\pi}{2}$ $DE$ is a diameter), so $\cos \angle AEF = \tfrac{EF}{DE} = \tfrac{8}{49}\cdot EF$
Plugging in all our values into equation $(1)$ , we get: \[EF = \tfrac{64}{49} EF -\tfrac{1}{7} AF \quad \Longrightarrow \quad EF = \tfrac{7}{15} AF.\] The Law of Cosines in $\triangle AEF$ , with $EF=\tfrac 7{15}AF$ and $\cos\angle AFE = -\tfrac 17$ gives \[8^2 = AF^2 + \tfrac{49}{225} AF^2 + \tfrac 2{15} AF^2 = \tfrac{225+49+30}{225}\cdot AF^2\] Thus $AF^2 = \frac{900}{19}$ . The answer is $\boxed{919}$ | null | 919 |
9f38d901faba7c2a093387d757c47f50 | https://artofproblemsolving.com/wiki/index.php/2012_AIME_II_Problems/Problem_15 | Triangle $ABC$ is inscribed in circle $\omega$ with $AB=5$ $BC=7$ , and $AC=3$ . The bisector of angle $A$ meets side $\overline{BC}$ at $D$ and circle $\omega$ at a second point $E$ . Let $\gamma$ be the circle with diameter $\overline{DE}$ . Circles $\omega$ and $\gamma$ meet at $E$ and a second point $F$ . Then $AF^2 = \frac mn$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | Let $a = BC$ $b = CA$ $c = AB$ for convenience. Let $M$ be the midpoint of segment $BC$ . We claim that $\angle MAD=\angle DAF$
$\textit{Proof}$ . Since $AE$ is the angle bisector, it follows that $EB = EC$ and consequently $EM\perp BC$ . Therefore, $M\in \gamma$ . Now let $X = FD\cap \omega$ . Since $\angle EFX=90^\circ$ $EX$ is a diameter, so $X$ lies on the perpendicular bisector of $BC$ ; hence $E$ $M$ $X$ are collinear. From $\angle DAG = \angle DMX = 90^\circ$ , quadrilateral $ADMX$ is cyclic. Therefore, $\angle MAD = \angle MXD$ . But $\angle MXD$ and $\angle EAF$ are both subtended by arc $EF$ in $\omega$ , so they are equal. Thus $\angle MAD=\angle DAF$ , as claimed. [asy] size(175); defaultpen(fontsize(10pt)); picture pic; pair A,B,C,D,E,F,W; B=MP("B",origin,dir(180)); C=MP("C",(7,0),dir(0)); A=MP("A",IP(CR(B,5),CR(C,3)),N); D=MP("D",extension(B,C,A,bisectorpoint(C,A,B)),dir(220)); path omega=circumcircle(A,B,C); E=MP("E",OP(omega,A--(A+20*(D-A))),S); path gamma=CR(midpoint(D--E),length(D-E)/2); F=MP("F",OP(omega,gamma),SE); pair X=MP("X",IP(omega,F--(F+2*(D-F))),N); pair M=MP("M",midpoint(B--C),dir(220)); draw(omega^^A--B--C--cycle^^gamma); draw(A--E--F--cycle, gray); draw(E--X--F, gray); draw(pic, A--M--C--cycle^^A--B--F--cycle); draw(A--M, royalblue); dot("$W$",circumcenter(A,B,C),dir(180)); dot(circumcenter(D,E,F)); label("$\gamma$",gamma,dir(180)); draw(A--B--F--cycle, black+1); [/asy] As a result, $\angle CAM = \angle FAB$ . Combined with $\angle BFA=\angle MCA$ , we get $\triangle ABF\sim\triangle AMC$ and therefore \[\frac c{AM}=\frac {AF}b\qquad \Longrightarrow \qquad AF^2=\frac{b^2c^2}{AM^2} = \frac{15^2}{AM^2}\] By Stewart's Theorem on $\triangle ABC$ (with cevian $AM$ ), we get \[AM^2 = \tfrac 12 (b^2+c^2)-\tfrac 14 a^2 = \tfrac{19}{4},\] so $AF^2 = \tfrac{900}{19}$ , so the answer is $900+19=\boxed{919}$ | null | 919 |
9f38d901faba7c2a093387d757c47f50 | https://artofproblemsolving.com/wiki/index.php/2012_AIME_II_Problems/Problem_15 | Triangle $ABC$ is inscribed in circle $\omega$ with $AB=5$ $BC=7$ , and $AC=3$ . The bisector of angle $A$ meets side $\overline{BC}$ at $D$ and circle $\omega$ at a second point $E$ . Let $\gamma$ be the circle with diameter $\overline{DE}$ . Circles $\omega$ and $\gamma$ meet at $E$ and a second point $F$ . Then $AF^2 = \frac mn$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | Use the angle bisector theorem to find $CD=\tfrac{21}{8}$ $BD=\tfrac{35}{8}$ , and use Stewart's Theorem to find $AD=\tfrac{15}{8}$ . Use Power of Point $D$ to find $DE=\tfrac{49}{8}$ , and so $AE=8$ . Then use the Extended Law of Sine to find that the length of the circumradius of $\triangle ABC$ is $\tfrac{7\sqrt{3}}{3}$ [asy] size(175); defaultpen(fontsize(9pt)); pair A,B,C,D,E,F,W; B=MP("B",origin,dir(180)); C=MP("C",(7,0),dir(0)); A=MP("A",IP(CR(B,5),CR(C,3)),N); D=MP("D",extension(B,C,A,bisectorpoint(C,A,B)),dir(220)); path omega=circumcircle(A,B,C); E=MP("E",OP(omega,A--(A+20*(D-A))),S); path gamma=CR(midpoint(D--E),length(D-E)/2); F=MP("F",OP(omega,gamma),SE); pair X=MP("X",IP(omega,F--(F+2*(D-F))),N); draw(omega^^A--B--C--cycle^^gamma); draw(A--E--F--cycle, gray); draw(E--X--F, royalblue); dot("$W$",circumcenter(A,B,C),dir(180)); dot(circumcenter(D,E,F)); label("$\gamma$",gamma,dir(180)); label("$u$",X--D,dir(60)); label("$v$",D--F,dir(70)); [/asy] Since $DE$ is the diameter of circle $\gamma$ $\angle DFE$ is $90^\circ$ . Extending $DF$ to intersect circle $\omega$ at $X$ , we find that $XE$ is the diameter of $\omega$ (since $\angle DFE$ is $90^\circ$ ). Therefore, $XE=\tfrac{14\sqrt{3}}{3}$
Let $EF=x$ $XD=u$ , and $DF=v$ . Then $XE^2-XF^2=EF^2=DE^2-DF^2$ , so we get \[(u+v)^2-v^2=\frac{196}{3}-\frac{2401}{64}\] which simplifies to \[u^2+2uv = \frac{5341}{192}.\] By Power of Point $D$ $uv=BD \cdot DC=735/64$ . Combining with above, we get \[XD^2=u^2=\frac{931}{192}.\] Note that $\triangle XDE\sim \triangle ADF$ and the ratio of similarity is $\rho = AD : XD = \tfrac{15}{8}:u$ . Then $AF=\rho\cdot XE = \tfrac{15}{8u}\cdot R$ and \[AF^2 = \frac{225}{64}\cdot \frac{R^2}{u^2} = \frac{900}{19}.\] The answer is $900+19=\boxed{919}$ | null | 919 |
9f38d901faba7c2a093387d757c47f50 | https://artofproblemsolving.com/wiki/index.php/2012_AIME_II_Problems/Problem_15 | Triangle $ABC$ is inscribed in circle $\omega$ with $AB=5$ $BC=7$ , and $AC=3$ . The bisector of angle $A$ meets side $\overline{BC}$ at $D$ and circle $\omega$ at a second point $E$ . Let $\gamma$ be the circle with diameter $\overline{DE}$ . Circles $\omega$ and $\gamma$ meet at $E$ and a second point $F$ . Then $AF^2 = \frac mn$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | Denote $AB = c, BC = a, AC = b, \angle A = 2 \alpha.$ Let M be midpoint BC. Let $\theta$ be the circle centered at $A$ with radius $\sqrt{AB \cdot AC} =\sqrt{bc}.$
We calculate the length of some segments.
The median $AM = \sqrt{\frac {b^2}{2} + \frac {c^2}{2} - \frac {a^2}{4}}.$ The bisector $AD = \frac {2 b c \cos \alpha}{b+c}.$ One can use Stewart's Theorem in both cases.
$AD$ is bisector of $\angle A \implies BD = \frac {a c}{b + c}, CD = \frac {a b}{b + c} \implies$ \[BD \cdot CD = \frac {a^2 bc }{(b+c)^2}.\] We use Power of Point $D$ and get $AD \cdot DE = BD \cdot CD.$ \[AE = AD + DE = AD + \frac {BD \cdot CD}{AD},\] \[AE =\frac {2 b c \cos \alpha}{b+c} + \frac {a^2 bc \cdot (b+c) }{(b+c)^2 \cdot 2 b c \cos \alpha} =\] \[= \frac {b c \cos^2 \alpha + a^2}{2(b+c)\cos \alpha} =\frac {4bc \cos^2 \alpha + b^2 +c^2 -2 b c \cos 2\alpha}{2(b+c) \cos \alpha} = \frac {b+c}{2} \implies AD \cdot AE = 2 bc \cos \alpha.\] We consider the inversion with respect $\theta.$
$B$ swap $B' \implies AB' = AC, B' \in AB \implies B'$ is symmetric to $C$ with respect to $AE.$
$C$ swap $C' \implies AC' = AB, C'$ lies on line $AC \implies C'$ is symmetric to $B$ with respect to $AE.$
$BC^2 = AB^2 + AC^2 + AB \cdot BC \implies \alpha = 60^\circ \implies AD \cdot AE = bc \implies D$ swap $E.$
Points $D$ and $E$ lies on $\Gamma \implies \Gamma$ swap $\Gamma.$
$DE$ is diameter $\Gamma, \angle DME = 90^\circ \implies M \in \Gamma.$ Therefore $M$ is crosspoint of $BC$ and $\Gamma.$
Let $\Omega$ be circumcircle $AB'C'. \Omega$ is image of line $BC.$ Point $M$ maps into $M' \implies M' = \Gamma \cap \Omega.$
Points $A, B',$ and $C'$ are symmetric to $A, C,$ and $B,$ respectively.
Point $M'$ lies on $\Gamma$ which is symmetric with respect to $AE$ and on $\Omega$ which is symmetric to $\omega$ with respect to $AE \implies$
$M'$ is symmetric $F$ with respect to $AE \implies AM' = AF.$
We use Power of Point $A$ and get \[AF = AM' = \frac {AD \cdot AE}{AM} = \frac {4b c}{\sqrt{2 b^2 + 2 c^2 – a^2}} = \frac {4 \cdot 3 \cdot 5}{\sqrt{ 50 + 18 – 49}} = \frac {30}{\sqrt{19}} \implies \boxed{919}.\] | null | 919 |
2699b5d8056e122b8a660a7bb23622ba | https://artofproblemsolving.com/wiki/index.php/2011_AIME_I_Problems/Problem_1 | Jar A contains four liters of a solution that is 45% acid. Jar B contains five liters of a solution that is 48% acid. Jar C contains one liter of a solution that is $k\%$ acid. From jar C, $\frac{m}{n}$ liters of the solution is added to jar A, and the remainder of the solution in jar C is added to jar B. At the end both jar A and jar B contain solutions that are 50% acid. Given that $m$ and $n$ are relatively prime positive integers, find $k + m + n$ | Jar A contains $\frac{11}{5}$ liters of water, and $\frac{9}{5}$ liters of acid; jar B contains $\frac{13}{5}$ liters of water and $\frac{12}{5}$ liters of acid.
The gap between the amount of water and acid in the first jar, $\frac{2}{5}$ , is double that of the gap in the second jar, $\frac{1}{5}$ . Therefore, we must add twice as much of jar C into the jar $A$ over jar $B$ . So, we must add $\frac{2}{3}$ of jar C into jar $A$ , so $m = 2, n=3$
Since jar C contains $1$ liter of solution, we are adding $\frac{2}{3}$ of a liter of solution to jar $A$ . In order to close the gap between water and acid, there must be $\frac{2}{5}$ more liters of acid than liters of water in these $\frac{2}{3}$ liters of solution. So, in the $\frac{2}{3}$ liters of solution, there are $\frac{2}{15}$ liters of water, and $\frac{8}{15}$ liters of acid. So, 80% of the $\frac{2}{3}$ sample is acid, so overall, in jar C, 80% of the sample is acid.
Therefore, our answer is $80 + 2 + 3 = \boxed{85}$ | null | 85 |
2699b5d8056e122b8a660a7bb23622ba | https://artofproblemsolving.com/wiki/index.php/2011_AIME_I_Problems/Problem_1 | Jar A contains four liters of a solution that is 45% acid. Jar B contains five liters of a solution that is 48% acid. Jar C contains one liter of a solution that is $k\%$ acid. From jar C, $\frac{m}{n}$ liters of the solution is added to jar A, and the remainder of the solution in jar C is added to jar B. At the end both jar A and jar B contain solutions that are 50% acid. Given that $m$ and $n$ are relatively prime positive integers, find $k + m + n$ | There are $\frac{45}{100}(4)=\frac{9}{5}$ L of acid in Jar A. There are $\frac{48}{100}(5)=\frac{12}{5}$ L of acid in Jar B. And there are $\frac{k}{100}$ L of acid in Jar C. After transferring the solutions from jar C, there will be $4+\frac{m}{n}$ L of solution in Jar A and $\frac{9}{5}+\frac{k}{100}\cdot\frac{m}{n}$ L of acid in Jar A. $6-\frac{m}{n}$ L of solution in Jar B and $\frac{12}{5}+\frac{k}{100}\cdot \left(1-\frac{m}{n}\right)=\frac{12}{5}+\frac{k}{100}-\frac{mk}{100n}$ of acid in Jar B. Since the solutions are 50% acid, we can multiply the amount of acid for each jar by 2, then equate them to the amount of solution. \[\frac{18}{5}+\frac{km}{50n}=4+\frac{m}{n}\] \[\frac{24}{5}-\frac{km}{50n}+\frac{k}{50}=6-\frac{m}{n}\] Add the equations to get \[\frac{42}{5}+\frac{k}{50}=10\] Solving gives $k=80$ If we substitute back in the original equation we get $\frac{m}{n}=\frac{2}{3}$ so $3m=2n$ . Since $m$ and $n$ are relatively prime, $m=2$ and $n=3$ . Thus $k+m+n=80+2+3=\boxed{085}$ | null | 085 |
2699b5d8056e122b8a660a7bb23622ba | https://artofproblemsolving.com/wiki/index.php/2011_AIME_I_Problems/Problem_1 | Jar A contains four liters of a solution that is 45% acid. Jar B contains five liters of a solution that is 48% acid. Jar C contains one liter of a solution that is $k\%$ acid. From jar C, $\frac{m}{n}$ liters of the solution is added to jar A, and the remainder of the solution in jar C is added to jar B. At the end both jar A and jar B contain solutions that are 50% acid. Given that $m$ and $n$ are relatively prime positive integers, find $k + m + n$ | One might cleverly change the content of both Jars.
Since the end result of both Jars are $50\%$ acid, we can turn Jar A into a 1 gallon liquid with $50\%-4(5\%) = 30\%$ acid
and Jar B into 1 gallon liquid with $50\%-5(2\%) =40\%$ acid.
Now, since Jar A and Jar B contain the same amount of liquid, twice as much liquid will be pour into Jar A than Jar B, so $\dfrac{2}{3}$ of Jar C will be pour into Jar A.
Thus, $m=2$ and $n=3$
$\dfrac{30\% + \frac{2}{3} \cdot k\%}{\frac{5}{3}} = 50\%$
Solving for $k$ yields $k=80$
So the answer is $80+2+3 = \boxed{085}$ | null | 085 |
2699b5d8056e122b8a660a7bb23622ba | https://artofproblemsolving.com/wiki/index.php/2011_AIME_I_Problems/Problem_1 | Jar A contains four liters of a solution that is 45% acid. Jar B contains five liters of a solution that is 48% acid. Jar C contains one liter of a solution that is $k\%$ acid. From jar C, $\frac{m}{n}$ liters of the solution is added to jar A, and the remainder of the solution in jar C is added to jar B. At the end both jar A and jar B contain solutions that are 50% acid. Given that $m$ and $n$ are relatively prime positive integers, find $k + m + n$ | One may first combine all three jars in to a single container. That container will have $10$ liters of liquid, and it should be $50\%$ acidic. Thus there must be $5$ liters of acid.
Jar A contained $45\% \cdot 4L$ , or $1.8L$ of acid, and jar B $48\% \cdot 5L$ or $2.4L$ . Solving for the amount of acid in jar C, $k = (5 - 2.4 - 1.8) = .8$ , or $80\%$
Once one knowss that the jar C is $80\%$ acid, use solution 1 to figure out m and n for $k+m+n=80+2+3=\boxed{085}$ | null | 085 |
f222bdda0e8a20f7591ebe54efa0c756 | https://artofproblemsolving.com/wiki/index.php/2011_AIME_I_Problems/Problem_2 | In rectangle $ABCD$ $AB = 12$ and $BC = 10$ . Points $E$ and $F$ lie inside rectangle $ABCD$ so that $BE = 9$ $DF = 8$ $\overline{BE} \parallel \overline{DF}$ $\overline{EF} \parallel \overline{AB}$ , and line $BE$ intersects segment $\overline{AD}$ . The length $EF$ can be expressed in the form $m \sqrt{n} - p$ , where $m$ $n$ , and $p$ are positive integers and $n$ is not divisible by the square of any prime. Find $m + n + p$ | Let us call the point where $\overline{EF}$ intersects $\overline{AD}$ point $G$ , and the point where $\overline{EF}$ intersects $\overline{BC}$ point $H$ . Since angles $FHB$ and $EGA$ are both right angles, and angles $BEF$ and $DFE$ are congruent due to parallelism, right triangles $BHE$ and $DGF$ are similar. This implies that $\frac{BH}{GD} = \frac{9}{8}$ . Since $BC=10$ $BH+GD=BH+HC=BC=10$ . ( $HC$ is the same as $GD$ because they are opposite sides of a rectangle.) Now, we have a system:
$\frac{BH}{GD}=\frac{9}8$
$BH+GD=10$
Solving this system (easiest by substitution), we get that:
$BH=\frac{90}{17}$
$GD=\frac{80}{17}$
Using the Pythagorean Theorem, we can solve for the remaining sides of the two right triangles:
$\sqrt{9^2-\left(\frac{90}{17}\right)^2}$ and $\sqrt{8^2-\left(\frac{80}{17}\right)^2}$
Notice that adding these two sides would give us twelve plus the overlap $EF$ . This means that:
$EF= \sqrt{9^2-\left(\frac{90}{17}\right)^2}+\sqrt{8^2-\left(\frac{80}{17}\right)^2}-12=3\sqrt{21}-12$
Since $21$ isn't divisible by any perfect square, our answer is:
$3+21+12=\boxed{36}$ | null | 36 |
f222bdda0e8a20f7591ebe54efa0c756 | https://artofproblemsolving.com/wiki/index.php/2011_AIME_I_Problems/Problem_2 | In rectangle $ABCD$ $AB = 12$ and $BC = 10$ . Points $E$ and $F$ lie inside rectangle $ABCD$ so that $BE = 9$ $DF = 8$ $\overline{BE} \parallel \overline{DF}$ $\overline{EF} \parallel \overline{AB}$ , and line $BE$ intersects segment $\overline{AD}$ . The length $EF$ can be expressed in the form $m \sqrt{n} - p$ , where $m$ $n$ , and $p$ are positive integers and $n$ is not divisible by the square of any prime. Find $m + n + p$ | Extend lines $BE$ and $CD$ to meet at point $G$ .
Draw the altitude $GH$ from point $G$ to line $BA$ extended.
$GE=DF=8,$ $GB=17$
In right $\bigtriangleup GHB$ $GH=10$ $GB=17$ , thus by Pythagoras Theorem we have: $HB=\sqrt{17^2-10^2}=3\sqrt{21}$
$HA=EF=3\sqrt{21}-12$
Thus our answer is: $3+21+12=\boxed{36}$ | null | 36 |
f222bdda0e8a20f7591ebe54efa0c756 | https://artofproblemsolving.com/wiki/index.php/2011_AIME_I_Problems/Problem_2 | In rectangle $ABCD$ $AB = 12$ and $BC = 10$ . Points $E$ and $F$ lie inside rectangle $ABCD$ so that $BE = 9$ $DF = 8$ $\overline{BE} \parallel \overline{DF}$ $\overline{EF} \parallel \overline{AB}$ , and line $BE$ intersects segment $\overline{AD}$ . The length $EF$ can be expressed in the form $m \sqrt{n} - p$ , where $m$ $n$ , and $p$ are positive integers and $n$ is not divisible by the square of any prime. Find $m + n + p$ | We notice that since $\overline{BE}||\overline{DF}$ $\overline{EF}$ is the diagonal of a rectangle. Now, let us extend the width lines to intersect with $\overline{BE}$ and $\overline{DF}$ , respectively, to form this rectangle. Let us call the length of $\overline{EF}$ $d$ , the perpendicular distance between $\overline{EF}$ and $\overline{AD}$ $k$ , and the perpendicular distance between $\overline{EF}$ and $\overline{CD}$ $x$ . We now can begin the similar triangles. When drawing the diagram (where $E$ is closer to $\overline{AD}$ than $F$ ), we put the similar right triangles in the same position so that we can begin solving for our variables and finding ratios. Since all of these triangles are similar, we find that $\frac{dx}{8}=\frac{10d-dx}{9}$ , which solves for $x=\frac{80}{17}$ . Completing the same thing for $k$ , we see $\frac{12d-dk}{9}=\frac{dk+d^2}{8}$ , which solves for $k=\frac{96-9d}{17}$ . We are now ready to find both the length and the width of the rectangle. We find these to be $\frac{10d}{17}$ and $\frac{12d+d^2}{17}$ . Now let us use the Pythagorean to solve for $d$ . We square the length and the width and multiply both sides by 289 (the denominator of the LHS) to reach the equation $(12+d^2)^2+100d^2=289d^2$ . Expanding and simplifying, we find that $d^4+24d^3=45d^2$ . Dividing both sides by $d^2$ and moving everything to the LHS, we see that $d^2+24d-45=0$ . Applying the quadratic formula, $d=\frac{-24\pm \sqrt{756}}{2}=-12\pm \sqrt{189}$ . Since $d>0$ $d=\sqrt{189}-12=3\sqrt{21}-12$ . Finally, $m+n+p=3+21+12=\boxed{036}$ . -Gideontz | null | 036 |
9a5d4fd73db0dd2007fd4001c823b972 | https://artofproblemsolving.com/wiki/index.php/2011_AIME_I_Problems/Problem_3 | Let $L$ be the line with slope $\frac{5}{12}$ that contains the point $A=(24,-1)$ , and let $M$ be the line perpendicular to line $L$ that contains the point $B=(5,6)$ . The original coordinate axes are erased, and line $L$ is made the $x$ -axis and line $M$ the $y$ -axis. In the new coordinate system, point $A$ is on the positive $x$ -axis, and point $B$ is on the positive $y$ -axis. The point $P$ with coordinates $(-14,27)$ in the original system has coordinates $(\alpha,\beta)$ in the new coordinate system. Find $\alpha+\beta$ | Given that $L$ has slope $\frac{5}{12}$ and contains the point $A=(24,-1)$ , we may write the point-slope equation for $L$ as $y+1=\frac{5}{12}(x-24)$ .
Since $M$ is perpendicular to $L$ and contains the point $B=(5,6)$ , we have that the slope of $M$ is $-\frac{12}{5}$ , and consequently that the point-slope equation for $M$ is $y-6=-\frac{12}{5}(x-5)$
Converting both equations to the form $0=Ax+By+C$ , we have that $L$ has the equation $0=5x-12y-132$ and that $M$ has the equation $0=12x+5y-90$ .
Applying the point-to-line distance formula, $\frac{|Ax+By+C|}{\sqrt{A^2+B^2}}$ , to point $P$ and lines $L$ and $M$ , we find that the distance from $P$ to $L$ and $M$ are $\frac{526}{13}$ and $\frac{123}{13}$ , respectively.
Since $A$ and $B$ lie on the positive axes of the shifted coordinate plane, we may show by graphing the given system that point P will lie in the second quadrant in the new coordinate system. Therefore, the $x$ -coordinate of $P$ is negative, and is therefore $-\frac{123}{13}$ ; similarly, the $y$ -coordinate of $P$ is positive and is therefore $\frac{526}{13}$
Thus, we have that $\alpha=-\frac{123}{13}$ and that $\beta=\frac{526}{13}$ . It follows that $\alpha+\beta=-\frac{123}{13}+\frac{526}{13}=\frac{403}{13}=\boxed{031}$ | null | 031 |
9a5d4fd73db0dd2007fd4001c823b972 | https://artofproblemsolving.com/wiki/index.php/2011_AIME_I_Problems/Problem_3 | Let $L$ be the line with slope $\frac{5}{12}$ that contains the point $A=(24,-1)$ , and let $M$ be the line perpendicular to line $L$ that contains the point $B=(5,6)$ . The original coordinate axes are erased, and line $L$ is made the $x$ -axis and line $M$ the $y$ -axis. In the new coordinate system, point $A$ is on the positive $x$ -axis, and point $B$ is on the positive $y$ -axis. The point $P$ with coordinates $(-14,27)$ in the original system has coordinates $(\alpha,\beta)$ in the new coordinate system. Find $\alpha+\beta$ | The equations for the axes are $\frac{5}{12} (x-24) = y+1$ and $-\frac{12}{5}(x-5) = y - 6$ . We can solve the system to find that they intersect at the point $\left( \frac{1740}{169},\frac{-1134}{169} \right)$
The unit basis vectors of our new axes are $\begin{pmatrix} 12/13 \\ 5/13 \end{pmatrix}$ and $\begin{pmatrix} -5/13 \\ 12/13 \end{pmatrix}$ for the $x$ and $y$ axes respectively (taking into account which direction is positive).
Then, we solve the following system for $\alpha$ and $\beta$
\[\alpha \begin{pmatrix} 12/13 \\ 5/13 \end{pmatrix} + \beta \begin{pmatrix} -5/13 \\ 12/13 \end{pmatrix} + \begin{pmatrix} 1740/169 \\ -1134/169 \end{pmatrix} = \begin{pmatrix} -146 \\ 27 \end{pmatrix}\]
Painful bashing gives $\alpha = -\frac{123}{13}$ and $\beta = \frac{526}{13}$ . Adding gives $\alpha + \beta = \frac{403}{13} = \boxed{031}$ | null | 031 |
b5ad0eb7721991a36b3999f77127f13b | https://artofproblemsolving.com/wiki/index.php/2011_AIME_I_Problems/Problem_4 | In triangle $ABC$ $AB=125$ $AC=117$ and $BC=120$ . The angle bisector of angle $A$ intersects $\overline{BC}$ at point $L$ , and the angle bisector of angle $B$ intersects $\overline{AC}$ at point $K$ . Let $M$ and $N$ be the feet of the perpendiculars from $C$ to $\overline{BK}$ and $\overline{AL}$ , respectively. Find $MN$ | Extend ${CM}$ and ${CN}$ such that they intersect line ${AB}$ at points $P$ and $Q$ , respectively. [asy] defaultpen(fontsize(10)+0.8); size(200); pen p=fontsize(9)+linewidth(3); pair A,B,C,D,K,L,M,N,P,Q; A=origin; B=(125,0); C=IP(CR(A,117),CR(B,120)); L=extension(B,C,A,bisectorpoint(B,A,C)); K=extension(A,C,B,bisectorpoint(C,B,A)); M=foot(C,B,K); N=foot(C,A,L); draw(A--B--C--A); draw(A--L^^B--K, gray+dashed+0.5); draw(M--C--N^^N--extension(A,B,C,N)^^M--extension(A,B,C,M), gray+0.5); dot("$A$",A,dir(200),p); dot("$B$",B,right,p); dot("$C$",C,up,p); dot("$L$",L,2*dir(70),p); dot("$N$",N,2*dir(-90),p); dot("$M$",M,2*dir(-90),p); dot("$P$",extension(A,B,C,M),2*down,p); dot("$Q$",extension(A,B,C,N),2*down,p); label("$125$",A--B,down,fontsize(10)); label("$117$",A--C,2*dir(130),fontsize(10)); label("$120$",B--C,1.5*dir(30),fontsize(10)); [/asy] Since ${BM}$ is the angle bisector of angle $B$ and ${CM}$ is perpendicular to ${BM}$ $\triangle BCP$ must be an isoceles triangle, so $BP=BC=120$ , and $M$ is the midpoint of ${CP}$ . For the same reason, $AQ=AC=117$ , and $N$ is the midpoint of ${CQ}$ .
Hence $MN=\tfrac 12 PQ$ . Since \[PQ=BP+AQ-AB=120+117-125=112,\] so $MN=\boxed{056}$ | null | 056 |
b5ad0eb7721991a36b3999f77127f13b | https://artofproblemsolving.com/wiki/index.php/2011_AIME_I_Problems/Problem_4 | In triangle $ABC$ $AB=125$ $AC=117$ and $BC=120$ . The angle bisector of angle $A$ intersects $\overline{BC}$ at point $L$ , and the angle bisector of angle $B$ intersects $\overline{AC}$ at point $K$ . Let $M$ and $N$ be the feet of the perpendiculars from $C$ to $\overline{BK}$ and $\overline{AL}$ , respectively. Find $MN$ | Let $I$ be the incenter of $ABC$ . Since $I$ lies on $BM$ and $AN$ $IM \perp MC$ and $IN \perp NC$ , so $\angle IMC + \angle INC = 180^\circ$ . This means that $CMIN$ is a cyclic quadrilateral. By the Law of Sines, $\frac{MN}{\sin \angle MIN} = \frac{2R}{\sin \angle CMI} = 2R = CI$ , where $R$ is the radius of the circumcircle of $CMIN$ . Since $\sin \angle MIN = \sin \angle BIA = \sin (90^\circ + \tfrac 12 \angle BCA) = \cos \tfrac 12 \angle BCA = \cos \angle BCI$ , we have that $MN = CI \cdot \sin \angle MIN = CI \cdot \cos \angle BCI$ . Letting $H$ be the point of contact of the incircle of $ABC$ with side $BC$ , we have $MN = CI \cdot \cos \angle BCI = CI \cdot \frac{CH}{CI} = CH$ . Thus, $MN = s - AB = \frac{117+120-125}{2}=\boxed{056}$ | null | 056 |
b5ad0eb7721991a36b3999f77127f13b | https://artofproblemsolving.com/wiki/index.php/2011_AIME_I_Problems/Problem_4 | In triangle $ABC$ $AB=125$ $AC=117$ and $BC=120$ . The angle bisector of angle $A$ intersects $\overline{BC}$ at point $L$ , and the angle bisector of angle $B$ intersects $\overline{AC}$ at point $K$ . Let $M$ and $N$ be the feet of the perpendiculars from $C$ to $\overline{BK}$ and $\overline{AL}$ , respectively. Find $MN$ | Project $I$ onto $AC$ and $BC$ as $D$ and $E$ $ID$ and $IE$ are both in-radii of $\triangle ABC$ so we get right triangles with legs $r$ (the in-radius length) and $s - c = 56$ . Since $IC$ is the hypotenuse for the 4 triangles ( $\triangle INC, \triangle IMC, \triangle IDC,$ and $\triangle IEC$ ), $C, D, M, I, N, E$ are con-cyclic on a circle we shall denote as $\omega$ which is also the circumcircle of $\triangle CMN$ and $\triangle CDE$ . To find $MN$ , we can use the Law of Cosines on $\angle MON \implies MN^2 = 2R^2(1 - \cos{2\angle MCN})$ where $O$ is the center of $\omega$ . Now, the circumradius $R$ can be found with Pythagorean Theorem with $\triangle CDI$ or $\triangle CEI$ $r^2 + 56^2 = (2R)^2$ . To find $r$ , we can use the formula $rs = [ABC]$ and by Heron's, $[ABC] = \sqrt{181 \cdot 61 \cdot 56 \cdot 64} \implies r = \sqrt{\frac{61 \cdot 56 \cdot 64}{181}} \implies 2R^2 = \frac{393120}{181}$ . To find $\angle MCN$ , we can find $\angle MIN$ since $\angle MCN = 180 - \angle MIN$ $\angle MIN = \angle MIC + \angle NIC = 180 - \angle BIC + 180 - \angle AIC = 180 - (180 - \frac{\angle A + \angle C}{2}) + 180 - (180 - \frac{\angle B + \angle C}{2}) = \frac{\angle A + \angle B + \angle C}{2} + \frac{\angle C}{2}$ . Thus, $\angle MCN = 180 - \frac{\angle A + \angle B + \angle C}{2} - \frac{\angle C}{2}$ and since $\angle A + \angle B + \angle C = 180$ , we have $\angle A + \angle B + \angle C - \frac{\angle A + \angle B + \angle C}{2} - \frac{\angle C}{2} = \frac{\angle A + \angle B}{2}$ . Plugging this into our Law of Cosines (LoC) formula gives $MN^2 = 2R^2(1 - \cos{\angle A + \angle B}) = 2R^2(1 + \cos{\angle C})$ . To find $\cos{\angle C}$ , we use LoC on $\triangle ABC \implies \cos{\angle C} = \frac{120^2 + 117^2 - 125^2}{2 \cdot 117 \cdot 120} = \frac{41 \cdot 19}{117 \cdot 15}$ . Our formula now becomes $MN^2 = \frac{393120}{181} + \frac{2534}{15 \cdot 117}$ . After simplifying, we get $MN^2 = 3136 \implies MN = \boxed{056}$ | null | 056 |
b5ad0eb7721991a36b3999f77127f13b | https://artofproblemsolving.com/wiki/index.php/2011_AIME_I_Problems/Problem_4 | In triangle $ABC$ $AB=125$ $AC=117$ and $BC=120$ . The angle bisector of angle $A$ intersects $\overline{BC}$ at point $L$ , and the angle bisector of angle $B$ intersects $\overline{AC}$ at point $K$ . Let $M$ and $N$ be the feet of the perpendiculars from $C$ to $\overline{BK}$ and $\overline{AL}$ , respectively. Find $MN$ | Because $\angle CMI = \angle CNI = 90$ $CMIN$ is cyclic.
Applying Ptolemy's theorem on CMIN:
$CN \cdot MI+CM \cdot IN=CI \cdot MN$
$CI^2(\cos \angle ICN \sin \angle ICM + \cos \angle ICM \sin \angle ICN) = CI \cdot MN$
$MN = CI \sin \angle MCN$ by sine angle addition formula.
$\angle MCN = 180 - \angle MIN = 90 - \angle BCI$
Let $H$ be where the incircle touches $BC$ , then $CI \cos \angle BCI = CH = \frac{a+b-c}{2}$ $a=120, b=117, c=125$ , for a final answer of $\boxed{056}$ | null | 056 |
e4cfd87f9de8885d61446f81467087e1 | https://artofproblemsolving.com/wiki/index.php/2011_AIME_I_Problems/Problem_5 | The vertices of a regular nonagon (9-sided polygon) are to be labeled with the digits 1 through 9 in such a way that the sum of the numbers on every three consecutive vertices is a multiple of 3. Two acceptable arrangements are considered to be indistinguishable if one can be obtained from the other by rotating the nonagon in the plane. Find the number of distinguishable acceptable arrangements. | First, we determine which possible combinations of digits $1$ through $9$ will yield sums that are multiples of $3$ . It is simplest to do this by looking at each of the digits $\bmod{3}$
We see that the numbers $1, 4,$ and $7$ are congruent to $1 \pmod{3}$ , that the numbers $2, 5,$ and $8$ are congruent to $2 \pmod{3}$ , and that the numbers $3, 6,$ and $9$ are congruent to $0 \pmod{3}$ . In order for a sum of three of these numbers to be a multiple of three, the mod $3$ sum must be congruent to $0$ . Quick inspection reveals that the only possible combinations are $0+0+0, 1+1+1, 2+2+2,$ and $0+1+2$ . However, every set of three consecutive vertices must sum to a multiple of three, so using any of $0+0+0, 1+1+1$ , or $2+2+2$ would cause an adjacent sum to include exactly $2$ digits with the same $\bmod{3}$ value, and this is an unacceptable arrangement. Thus the only possible groupings are composed of three digits congruent to three different $\bmod{3}$ values.
We see also that there are two possible arrangements for these trios on the nonagon: a digit congruent to $1 \pmod{3}$ can be located counterclockwise of a digit congruent to $0$ and clockwise of a digit congruent to $2 \pmod{3}$ , or the reverse can be true.
We set the first digit as $3$ avoid overcounting rotations, so we have one option as a choice for the first digit. The other two $0 \pmod{3}$ numbers can be arranged in $2!=2$ ways. The three $1 \pmod{3}$ and three $2 \pmod{3}$ can both be arranged in $3!=6$ ways. Therefore, the desired result is $2(2 \times 6 \times 6)=\boxed{144}$ | null | 144 |
e4cfd87f9de8885d61446f81467087e1 | https://artofproblemsolving.com/wiki/index.php/2011_AIME_I_Problems/Problem_5 | The vertices of a regular nonagon (9-sided polygon) are to be labeled with the digits 1 through 9 in such a way that the sum of the numbers on every three consecutive vertices is a multiple of 3. Two acceptable arrangements are considered to be indistinguishable if one can be obtained from the other by rotating the nonagon in the plane. Find the number of distinguishable acceptable arrangements. | Notice that there are three triplets of congruent integers $\mod 3$ $(1,4,7),$ $(2,5,8),$ and $(3,6,9).$ There are $3!$ ways to order each of the triplets individually and $3!$ ways to order the triplets as a group (see solution 1). Rotations are indistinguishable, so there are $(3!)^4/9=\boxed{144}$ total arrangements. | null | 144 |
3ecc91f30bda2a491e521863ba9e9481 | https://artofproblemsolving.com/wiki/index.php/2011_AIME_I_Problems/Problem_6 | Suppose that a parabola has vertex $\left(\frac{1}{4},-\frac{9}{8}\right)$ and equation $y = ax^2 + bx + c$ , where $a > 0$ and $a + b + c$ is an integer. The minimum possible value of $a$ can be written in the form $\frac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p + q$ | If the vertex is at $\left(\frac{1}{4}, -\frac{9}{8}\right)$ , the equation of the parabola can be expressed in the form \[y=a\left(x-\frac{1}{4}\right)^2-\frac{9}{8}.\] Expanding, we find that \[y=a\left(x^2-\frac{x}{2}+\frac{1}{16}\right)-\frac{9}{8},\] and \[y=ax^2-\frac{ax}{2}+\frac{a}{16}-\frac{9}{8}.\] From the problem, we know that the parabola can be expressed in the form $y=ax^2+bx+c$ , where $a+b+c$ is an integer. From the above equation, we can conclude that $a=a$ $-\frac{a}{2}=b$ , and $\frac{a}{16}-\frac{9}{8}=c$ . Adding up all of these gives us \[\frac{9a-18}{16}=a+b+c.\] We know that $a+b+c$ is an integer, so $9a-18$ must be divisible by $16$ . Let $9a=z$ . If ${z-18}\equiv {0} \pmod{16}$ , then ${z}\equiv {2} \pmod{16}$ . Therefore, if $9a=2$ $a=\frac{2}{9}$ . Adding up gives us $2+9=\boxed{011}$ | null | 011 |
3ecc91f30bda2a491e521863ba9e9481 | https://artofproblemsolving.com/wiki/index.php/2011_AIME_I_Problems/Problem_6 | Suppose that a parabola has vertex $\left(\frac{1}{4},-\frac{9}{8}\right)$ and equation $y = ax^2 + bx + c$ , where $a > 0$ and $a + b + c$ is an integer. The minimum possible value of $a$ can be written in the form $\frac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p + q$ | Complete the square. Since $a>0$ , the parabola must be facing upwards. $a+b+c=\text{integer}$ means that $f(1)$ must be an integer. The function can be recasted into $a\left(x-\frac{1}{4}\right)^2-\frac{9}{8}$ because the vertex determines the axis of symmetry and the critical value of the parabola. The least integer greater than $-\frac{9}{8}$ is $-1$ . So the $y$ -coordinate must change by $\frac{1}{8}$ and the $x$ -coordinate must change by $1-\frac{1}{4}=\frac{3}{4}$ . Thus, $a\left(\frac{3}{4}\right)^2=\frac{1}{8}\implies \frac{9a}{16}=\frac{1}{8}\implies a=\frac{2}{9}$ . So $2+9=\boxed{011}$ | null | 011 |
3ecc91f30bda2a491e521863ba9e9481 | https://artofproblemsolving.com/wiki/index.php/2011_AIME_I_Problems/Problem_6 | Suppose that a parabola has vertex $\left(\frac{1}{4},-\frac{9}{8}\right)$ and equation $y = ax^2 + bx + c$ , where $a > 0$ and $a + b + c$ is an integer. The minimum possible value of $a$ can be written in the form $\frac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p + q$ | To do this, we can use the formula for the minimum (or maximum) value of the $x$ coordinate at a vertex of a parabola, $-\frac{b}{2a}$ and equate this to $\frac{1}{4}$ . Solving, we get $-\frac{a}{2}=b$ . Enter $x=\frac{1}{4}$ to get $-\frac{9}{8}=\frac{a}{16}+\frac{b}{4}+c=-\frac{a}{16}+c$ so $c=\frac{a-18}{16}$ . This means that $\frac{9a-18}{16}\in \mathbb{Z}$ so the minimum of $a>0$ is when the fraction equals -1, so $a=\frac{2}{9}$ . Therefore, $p+q=2+9=\boxed{011}$ .
-Gideontz | null | 011 |
e71a98d1bf9ac91da3e823b676cfb1cf | https://artofproblemsolving.com/wiki/index.php/2011_AIME_I_Problems/Problem_7 | Find the number of positive integers $m$ for which there exist nonnegative integers $x_0$ $x_1$ $\dots$ $x_{2011}$ such that \[m^{x_0} = \sum_{k = 1}^{2011} m^{x_k}.\] | $m^{x_0}= m^{x_1} +m^{x_2} + .... + m^{x_{2011}}$ . Now, divide by $m^{x_0}$ to get $1= m^{x_1-x_0} +m^{x_2-x_0} + .... + m^{x_{2011}-x_0}$ . Notice that since we can choose all nonnegative $x_0,...,x_{2011}$ , we can make $x_n-x_0$ whatever we desire. WLOG, let $x_0\geq...\geq x_{2011}$ and let $a_n=x_n-x_0$ . Notice that, also, $m^{a_{2011}}$ doesn't matter if we are able to make $m^{a_1} +m^{a_2} + .... + m^{a_{2010}}$ equal to $1-\left(\frac{1}{m}\right)^x$ for any power of $x$ . Consider $m=2$ . We can achieve a sum of $1-\left(\frac{1}{2}\right)^x$ by doing $\frac{1}{2}+\frac{1}{4}+...$ (the "simplest" sequence). If we don't have $\frac{1}{2}$ , to compensate, we need $2\cdot 1\frac{1}{4}$ 's. Now, let's try to generalize. The "simplest" sequence is having $\frac{1}{m}$ $m-1$ times, $\frac{1}{m^2}$ $m-1$ times, $\ldots$ . To make other sequences, we can split $m-1$ $\frac{1}{m^i}$ s into $m(m-1)$ $\text{ }\frac{1}{m^{i+1}}$ s since $(m-1)\cdot\frac{1}{m^{i}} = m(m-1)\cdot\frac{1}{m^{i+1}}$ . Since we want $2010$ terms, we have $\sum$ $(m-1)\cdot m^x=2010$ . However, since we can set $x$ to be anything we want (including 0), all we care about is that $(m-1)\text{ }|\text{ } 2010$ which happens $\boxed{016}$ times. | null | 016 |
e71a98d1bf9ac91da3e823b676cfb1cf | https://artofproblemsolving.com/wiki/index.php/2011_AIME_I_Problems/Problem_7 | Find the number of positive integers $m$ for which there exist nonnegative integers $x_0$ $x_1$ $\dots$ $x_{2011}$ such that \[m^{x_0} = \sum_{k = 1}^{2011} m^{x_k}.\] | Let \[P(m) = m^{x_0} - m^{x_1} -m^{x_2} - .... - m^{x_{2011}}\] . The problem then becomes finding the number of positive integer roots $m$ for which $P(m) = 0$ and $x_0, x_1, ..., x_{2011}$ are nonnegative integers. We plug in $m = 1$ and see that \[P(1) = 1 - 1 - 1... -1 = 1-2011 = -2010\] Now, we can say that \[P(m) = (m-1)Q(m) - 2010\] for some polynomial $Q(m)$ with integer coefficients. Then if $P(m) = 0$ $(m-1)Q(m) = 2010$ . Thus, if $P(m) = 0$ , then $m-1 | 2010$ .
Now, we need to show that \[m^{x_{0}}=\sum_{k = 1}^{2011}m^{x_{k}}\forall m-1|2011\] We try with the first few $m$ that satisfy this.
For $m = 2$ , we see we can satisfy this if $x_0 = 2010$ $x_1 = 2009$ $x_2 = 2008$ $\cdots$ $x_{2008} = 2$ $x_{2009} = 1$ $x_{2010} = 0$ $x_{2011} = 0$ , because \[2^{2009} + 2^{2008} + \cdots + 2^1 + 2^0 +2^ 0 = 2^{2009} + 2^{2008} + \cdots + 2^1 + 2^1 = \cdots\] (based on the idea $2^n + 2^n = 2^{n+1}$ , leading to a chain of substitutions of this kind) \[= 2^{2009} + 2^{2008} + 2^{2008} = 2^{2009} + 2^{2009} = 2^{2010}\] . Thus $2$ is a possible value of $m$ . For other values, for example $m = 3$ , we can use the same strategy, with \[x_{2011} = x_{2010} = x_{2009} = 0\] \[x_{2008} = x_{2007} = 1\] \[x_{2006} = x_{2005} = 2\] \[\cdots\] \[x_2 = x_1 = 1004\] and \[x_0 = 1005\] , because \[3^0 + 3^0 + 3^0 +3^1+3^1+3^2+3^2+\cdots+3^{1004} +3^{1004} = 3^1+3^1+3^1+3^2+3^2+\cdots+3^{1004} +3^{1004} = 3^2+3^2+3^2+\cdots+3^{1004} +3^{1004}\] \[=\cdots = 3^{1004} +3^{1004}+3^{1004} = 3^{1005}\] It's clearly seen we can use the same strategy for all $m-1 |2010$ . We count all positive $m$ satisfying $m-1 |2010$ , and see there are $\boxed{016}$ | null | 016 |
e71a98d1bf9ac91da3e823b676cfb1cf | https://artofproblemsolving.com/wiki/index.php/2011_AIME_I_Problems/Problem_7 | Find the number of positive integers $m$ for which there exist nonnegative integers $x_0$ $x_1$ $\dots$ $x_{2011}$ such that \[m^{x_0} = \sum_{k = 1}^{2011} m^{x_k}.\] | One notices that $m-1 \mid 2010$ if and only if there exist non-negative integers $x_0,x_1,\ldots,x_{2011}$ such that $m^{x_0} = \sum_{k=1}^{2011}m^{x_k}$
To prove the forward case, we proceed by directly finding $x_0,x_1,\ldots,x_{2011}$ . Suppose $m$ is an integer such that $m^{x_0} = \sum_{k=1}^{2011}m^{x_k}$ . We will count how many $x_k = 0$ , how many $x_k = 1$ , etc. Suppose the number of $x_k = 0$ is non-zero. Then, there must be at least $m$ such $x_k$ since $m$ divides all the remaining terms, so $m$ must also divide the sum of all the $m^0$ terms. Thus, if we let $x_k = 0$ for $k = 1,2,\ldots,m$ , we have, \[m^{x_0} = m + \sum_{k=m+1}^{2011}m^{x_k}.\] Well clearly, $m^{x_0}$ is greater than $m$ , so $m^2 \mid m^{x_0}$ $m^2$ will also divide every term, $m^{x_k}$ , where $x_k \geq 2$ . So, all the terms, $m^{x_k}$ , where $x_k < 2$ must sum to a multiple of $m^2$ . If there are exactly $m$ terms where $x_k = 0$ , then we must have at least $m-1$ terms where $x_k = 1$ . Suppose there are exactly $m-1$ such terms and $x_k = 1$ for $k = m+1,m+2,2m-1$ . Now, we have, \[m^{x_0} = m^2 + \sum_{k=2m}^{2011}m^{x_k}.\] One can repeat this process for successive powers of $m$ until the number of terms reaches 2011. Since there are $m + j(m-1)$ terms after the $j$ th power, we will only hit exactly 2011 terms if $m-1$ is a factor of 2010. To see this, \[m+j(m-1) = 2011 \Rightarrow m-1+j(m-1) = 2010 \Rightarrow (m-1)(j+1) = 2010.\] Thus, when $j = 2010/(m-1) - 1$ (which is an integer since $m-1 \mid 2010$ by assumption, there are exactly 2011 terms. To see that these terms sum to a power of $m$ , we realize that the sum is a geometric series: \[1 + (m-1) + (m-1)m+(m-1)m^2 + \cdots + (m-1)m^j = 1+(m-1)\frac{m^{j+1}-1}{m-1} = m^{j+1}.\] Thus, we have found a solution for the case $m-1 \mid 2010$
Now, for the reverse case, we use the formula \[x^k-1 = (x-1)(x^{k-1}+x^{k-2}+\cdots+1).\] Suppose $m^{x_0} = \sum_{k=1}^{2011}m^{x^k}$ has a solution. Subtract 2011 from both sides to get \[m^{x_0}-1-2010 = \sum_{k=1}^{2011}(m^{x^k}-1).\] Now apply the formula to get \[(m-1)a_0-2010 = \sum_{k=1}^{2011}[(m-1)a_k],\] where $a_k$ are some integers. Rearranging this equation, we find \[(m-1)A = 2010,\] where $A = a_0 - \sum_{k=1}^{2011}a_k$ . Thus, if $m$ is a solution, then $m-1 \mid 2010$
So, there is one positive integer solution corresponding to each factor of 2010. Since $2010 = 2\cdot 3\cdot 5\cdot 67$ , the number of solutions is $2^4 = \boxed{016}$ | null | 016 |
e71a98d1bf9ac91da3e823b676cfb1cf | https://artofproblemsolving.com/wiki/index.php/2011_AIME_I_Problems/Problem_7 | Find the number of positive integers $m$ for which there exist nonnegative integers $x_0$ $x_1$ $\dots$ $x_{2011}$ such that \[m^{x_0} = \sum_{k = 1}^{2011} m^{x_k}.\] | The problem is basically asking how many integers $m$ have a power that can be expressed as the sum of 2011 other powers of $m$ (not necessarily distinct). Notice that $2+2+4+8+16=32$ $3+3+3+9+9+27+27+81+81=243$ , and $4+4+4+4+16+16+16+64+64+64+256+256+256=1024$ . Thus, we can safely assume that the equation $2011 = (m-1)x + m$ must have an integer solution $x$ . To find the number of $m$ -values that allow the aforementioned equation to have an integer solution, we can subtract 1 from the constant $m$ to make the equation equal a friendlier number, $2010$ , instead of the ugly prime number $2011$ $2010 = (m-1)x+(m-1)$ . Factor the equation and we get $2010 = (m-1)(x+1)$ . The number of values of $m-1$ that allow $x+1$ to be an integer is quite obviously the number of factors of $2010$ . Factoring $2010$ , we obtain $2010 = 2 \times 3 \times 5 \times 67$ , so the number of positive integers $m$ that satisfy the required condition is $2^4 = \boxed{016}$ | null | 016 |
e71a98d1bf9ac91da3e823b676cfb1cf | https://artofproblemsolving.com/wiki/index.php/2011_AIME_I_Problems/Problem_7 | Find the number of positive integers $m$ for which there exist nonnegative integers $x_0$ $x_1$ $\dots$ $x_{2011}$ such that \[m^{x_0} = \sum_{k = 1}^{2011} m^{x_k}.\] | First of all, note that the nonnegative integer condition really does not matter, since even if we have a nonnegative power, there is always a power of $m$ we can multiply to get to non-negative powers.
Now we see that our problem is just a matter of m-chopping blocks.
What is meant by $m$ -chopping is taking an existing block of say $m^{k}$ and turning it into $m$ blocks of $m^{k-1}$ . This process increases the total number of blocks by $m-1$ per chop.
The problem wants us to find the number of positive integers $m$ where some number of chops will turn $1$ block into $2011$ such blocks, thus increasing the total amount by $2010= 2 \cdot 3 \cdot 5 \cdot 67$ . Thus $m-1 | 2010$ , and a cursory check on extreme cases will confirm that there are indeed $\boxed{016}$ possible $m$ s. | null | 016 |
e71a98d1bf9ac91da3e823b676cfb1cf | https://artofproblemsolving.com/wiki/index.php/2011_AIME_I_Problems/Problem_7 | Find the number of positive integers $m$ for which there exist nonnegative integers $x_0$ $x_1$ $\dots$ $x_{2011}$ such that \[m^{x_0} = \sum_{k = 1}^{2011} m^{x_k}.\] | The problem is basically saying that we want to find 2011 powers of m that add together to equal another power of m. If we had a power of m, m^n, then to get to m^(n+1) or m*(m^n), we have to add m^n, m-1 times. Then when we are at m^(n+1), to get to m^(n+2), it is similar. So we have to have m^(some number) = m^n + (m-1)(m^n) + (m-1)(m^(n+1))...<=This expression has 1 + (m-1) + (m-1) + ... = 1 + k(m-1) terms, and the number of terms must be equal to 2011 (the problem said). Then 1+k(m-1) = 2011, and we get the equation k*(m-1) = 2010 = 2*3*5*67. 2010 has 16 factors, and setting m-1 = each of these factors makes $\boxed{016}$ values of m. | null | 016 |
b07d0db9abcf7085bd2afa0639cad635 | https://artofproblemsolving.com/wiki/index.php/2011_AIME_I_Problems/Problem_8 | In triangle $ABC$ $BC = 23$ $CA = 27$ , and $AB = 30$ . Points $V$ and $W$ are on $\overline{AC}$ with $V$ on $\overline{AW}$ , points $X$ and $Y$ are on $\overline{BC}$ with $X$ on $\overline{CY}$ , and points $Z$ and $U$ are on $\overline{AB}$ with $Z$ on $\overline{BU}$ . In addition, the points are positioned so that $\overline{UV}\parallel\overline{BC}$ $\overline{WX}\parallel\overline{AB}$ , and $\overline{YZ}\parallel\overline{CA}$ . Right angle folds are then made along $\overline{UV}$ $\overline{WX}$ , and $\overline{YZ}$ . The resulting figure is placed on a leveled floor to make a table with triangular legs. Let $h$ be the maximum possible height of a table constructed from triangle $ABC$ whose top is parallel to the floor. Then $h$ can be written in the form $\frac{k\sqrt{m}}{n}$ , where $k$ and $n$ are relatively prime positive integers and $m$ is a positive integer that is not divisible by the square of any prime. Find $k+m+n$ | Note that triangles $\triangle AUV, \triangle BYZ$ and $\triangle CWX$ all have the same height because when they are folded up to create the legs of the table, the top needs to be parallel to the floor. We want to find the maximum possible value of this height, given that no two of $\overline{UV}, \overline{WX}$ and $\overline{YZ}$ intersect inside $\triangle ABC$ . Let $h_{A}$ denote the length of the altitude dropped from vertex $A,$ and define $h_{B}$ and $h_{C}$ similarly. Also let $\{u, v, w, x, y, z\} = \{AU, AV, CW, CX, BY, BZ\}$ . Then by similar triangles \begin{align} \frac{u}{AB}=\frac{v}{AC}=\frac{h}{h_{A}}, \\ \frac{w}{CA}=\frac{x}{CB}=\frac{h}{h_{C}}, \\ \frac{y}{BC}=\frac{z}{BA}=\frac{h}{h_{B}}. \end{align} Since $h_{A}=\frac{2K}{23}$ and similarly for $27$ and $30,$ where $K$ is the area of $\triangle ABC,$ we can write \begin{align} \frac{u}{30}=\frac{v}{27}=\frac{h}{\tfrac{2K}{23}}, \\ \frac{w}{27}=\frac{x}{23}=\frac{h}{\tfrac{2K}{30}}, \\ \frac{y}{23}=\frac{z}{30}=\frac{h}{\tfrac{2K}{27}}. \end{align} and simplifying gives $u=x=\frac{690h}{2K}, v=y=\frac{621h}{2K}, w=z=\frac{810h}{2K}$ . Because no two segments can intersect inside the triangle, we can form the inequalities $v+w\leq 27, x+y\leq 23,$ and $z+u\leq 30$ . That is, all three of the inequalities \begin{align} \frac{621h+810h}{2K}\leq 27, \\ \frac{690h+621h}{2K}\leq 23, \\ \frac{810h+690h}{2K}\leq 30. \end{align} must hold. Dividing both sides of each equation by the RHS, we have \begin{align} \frac{53h}{2K}\leq 1\, \text{since}\, \frac{1431}{27}=53, \\ \frac{57h}{2K}\leq 1\, \text{since}\, \frac{1311}{23}=57, \\ \frac{50h}{2K}\leq 1\, \text{since}\, \frac{1500}{30}=50. \end{align} It is relatively easy to see that $\frac{57h}{2K}\leq 1$ restricts us the most since it cannot hold if the other two do not hold. The largest possible value of $h$ is thus $\frac{2K}{57},$ and note that by Heron's formula the area of $\triangle ABC$ is $20\sqrt{221}$ . Then $\frac{2K}{57}=\frac{40\sqrt{221}}{57},$ and the answer is $40+221+57=261+57=\boxed{318}$ | null | 318 |
b07d0db9abcf7085bd2afa0639cad635 | https://artofproblemsolving.com/wiki/index.php/2011_AIME_I_Problems/Problem_8 | In triangle $ABC$ $BC = 23$ $CA = 27$ , and $AB = 30$ . Points $V$ and $W$ are on $\overline{AC}$ with $V$ on $\overline{AW}$ , points $X$ and $Y$ are on $\overline{BC}$ with $X$ on $\overline{CY}$ , and points $Z$ and $U$ are on $\overline{AB}$ with $Z$ on $\overline{BU}$ . In addition, the points are positioned so that $\overline{UV}\parallel\overline{BC}$ $\overline{WX}\parallel\overline{AB}$ , and $\overline{YZ}\parallel\overline{CA}$ . Right angle folds are then made along $\overline{UV}$ $\overline{WX}$ , and $\overline{YZ}$ . The resulting figure is placed on a leveled floor to make a table with triangular legs. Let $h$ be the maximum possible height of a table constructed from triangle $ABC$ whose top is parallel to the floor. Then $h$ can be written in the form $\frac{k\sqrt{m}}{n}$ , where $k$ and $n$ are relatively prime positive integers and $m$ is a positive integer that is not divisible by the square of any prime. Find $k+m+n$ | Note that the area is given by Heron's formula and it is $20\sqrt{221}$ . Let $h_i$ denote the length of the altitude dropped from vertex i. It follows that $h_b = \frac{40\sqrt{221}}{27}, h_c = \frac{40\sqrt{221}}{30}, h_a = \frac{40\sqrt{221}}{23}$ . From similar triangles we can see that $\frac{27h}{h_a}+\frac{27h}{h_c} \le 27 \rightarrow h \le \frac{h_ah_c}{h_a+h_c}$ . We can see this is true for any combination of a,b,c and thus the minimum of the upper bounds for h yields $h = \frac{40\sqrt{221}}{57} \rightarrow \boxed{318}$ | null | 318 |
b07d0db9abcf7085bd2afa0639cad635 | https://artofproblemsolving.com/wiki/index.php/2011_AIME_I_Problems/Problem_8 | In triangle $ABC$ $BC = 23$ $CA = 27$ , and $AB = 30$ . Points $V$ and $W$ are on $\overline{AC}$ with $V$ on $\overline{AW}$ , points $X$ and $Y$ are on $\overline{BC}$ with $X$ on $\overline{CY}$ , and points $Z$ and $U$ are on $\overline{AB}$ with $Z$ on $\overline{BU}$ . In addition, the points are positioned so that $\overline{UV}\parallel\overline{BC}$ $\overline{WX}\parallel\overline{AB}$ , and $\overline{YZ}\parallel\overline{CA}$ . Right angle folds are then made along $\overline{UV}$ $\overline{WX}$ , and $\overline{YZ}$ . The resulting figure is placed on a leveled floor to make a table with triangular legs. Let $h$ be the maximum possible height of a table constructed from triangle $ABC$ whose top is parallel to the floor. Then $h$ can be written in the form $\frac{k\sqrt{m}}{n}$ , where $k$ and $n$ are relatively prime positive integers and $m$ is a positive integer that is not divisible by the square of any prime. Find $k+m+n$ | As from above, we can see that the length of the altitude from A is the longest. Thus the highest table is formed when X and Y meet up. Let the distance of this point from B be $x$ , making the distance from C $23 - x$ . Let $h$ be the height of the table. From similar triangles, we have $\frac{x}{23} = \frac{h}{h_b} = \frac{27h}{2A}$ where A is the area of ABC. Similarly, $\frac{23-x}{23}=\frac{h}{h_c}=\frac{30h}{2A}$ . Therefore, $1-\frac{x}{23}=\frac{30h}{2A} \rightarrow1-\frac{27h}{2A}=\frac{30h}{2A}$ and hence $h = \frac{2A}{57} = \frac{40\sqrt{221}}{57}\rightarrow \boxed{318}$ | null | 318 |
3155d93e70a2442cf02440cf0adf1a5d | https://artofproblemsolving.com/wiki/index.php/2011_AIME_I_Problems/Problem_9 | Suppose $x$ is in the interval $[0, \pi/2]$ and $\log_{24\sin x} (24\cos x)=\frac{3}{2}$ . Find $24\cot^2 x$ | We can rewrite the given expression as \[\sqrt{24^3\sin^3 x}=24\cos x\] Square both sides and divide by $24^2$ to get \[24\sin ^3 x=\cos ^2 x\] Rewrite $\cos ^2 x$ as $1-\sin ^2 x$ \[24\sin ^3 x=1-\sin ^2 x\] \[24\sin ^3 x+\sin ^2 x - 1=0\] Testing values using the rational root theorem gives $\sin x=\frac{1}{3}$ as a root, $\sin^{-1} \frac{1}{3}$ does fall in the first quadrant so it satisfies the interval.
There are now two ways to finish this problem.
First way: Since $\sin x=\frac{1}{3}$ , we have \[\sin ^2 x=\frac{1}{9}\] Using the Pythagorean Identity gives us $\cos ^2 x=\frac{8}{9}$ . Then we use the definition of $\cot ^2 x$ to compute our final answer. $24\cot ^2 x=24\frac{\cos ^2 x}{\sin ^2 x}=24\left(\frac{\frac{8}{9}}{\frac{1}{9}}\right)=24(8)=\boxed{192}$ | null | 192 |
3155d93e70a2442cf02440cf0adf1a5d | https://artofproblemsolving.com/wiki/index.php/2011_AIME_I_Problems/Problem_9 | Suppose $x$ is in the interval $[0, \pi/2]$ and $\log_{24\sin x} (24\cos x)=\frac{3}{2}$ . Find $24\cot^2 x$ | Like Solution 1, we can rewrite the given expression as \[24\sin^3x=\cos^2x\] Divide both sides by $\sin^3x$ \[24 = \cot^2x\csc x\] Square both sides. \[576 = \cot^4x\csc^2x\] Substitute the identity $\csc^2x = \cot^2x + 1$ \[576 = \cot^4x(\cot^2x + 1)\] Let $a = \cot^2x$ . Then \[576 = a^3 + a^2\] .
Since $\sqrt[3]{576} \approx 8$ , we can easily see that $a = 8$ is a solution. Thus, the answer is $24\cot^2x = 24a = 24 \cdot 8 = \boxed{192}$ | null | 192 |
161f64706304e5e8af78ddeb1518f435 | https://artofproblemsolving.com/wiki/index.php/2011_AIME_I_Problems/Problem_10 | The probability that a set of three distinct vertices chosen at random from among the vertices of a regular n-gon determine an obtuse triangle is $\frac{93}{125}$ . Find the sum of all possible values of $n$ | Inscribe the regular polygon inside a circle. A triangle inside this circle will be obtuse if and only if its three vertices lie on one side of a diameter of the circle. (This is because if an inscribed angle on a circle is obtuse, the arc it spans must be 180 degrees or greater).
Break up the problem into two cases: an even number of sides $2n$ , or an odd number of sides $2n-1$ . For polygons with $2n$ sides, the circumdiameter has endpoints on $2$ vertices. There are $n-1$ points on one side of a diameter, plus $1$ of the endpoints of the diameter for a total of $n$ points. For polygons with $2n - 1$ points, the circumdiameter has $1$ endpoint on a vertex and $1$ endpoint on the midpoint of the opposite side. There are also $n - 1$ points on one side of the diameter, plus the vertex for a total of $n$ points on one side of the diameter.
Case 1: $2n$ -sided polygon. There are clearly $\binom{2n}{3}$ different triangles total. To find triangles that meet the criteria, choose the left-most point. There are obviously $2n$ choices for this point. From there, the other two points must be within the $n-1$ points remaining on the same side of the diameter. So our desired probability is $\frac{2n\binom{n-1}{2}}{\binom{2n}{3}}$ $=\frac{n(n-1)(n-2)}{\frac{2n(2n-1)(2n-2)}{6}}$ $=\frac{6n(n-1)(n-2)}{2n(2n-1)(2n-2)}$ $=\frac{3(n-2)}{2(2n-1)}$
so $\frac{93}{125}=\frac{3(n-2)}{2(2n-1)}$
$186(2n-1)=375(n-2)$
$372n-186=375n-750$
$3n=564$
$n=188$ and so the polygon has $376$ sides.
Case 2: $2n-1$ -sided polygon. Similarly, $\binom{2n-1}{3}$ total triangles. Again choose the leftmost point, with $2n-1$ choices. For the other two points, there are again $\binom{n-1}{2}$ possibilities.
The probability is $\frac{(2n-1)\binom{n-1}{2}}{\binom{2n-1}{3}}$
$=\frac{3(2n-1)(n-1)(n-2)}{(2n-1)(2n-2)(2n-3)}$
$=\frac{3(n-2)}{2(2n-3)}$
so $\frac{93}{125}=\frac{3(n-2)}{2(2n-3)}$
$186(2n-3)=375(n-2)$
$375n-750=372n-558$
$3n=192$
$n=64$ and our polygon has $127$ sides.
Adding, $127+376=\boxed{503}$ | null | 503 |
161f64706304e5e8af78ddeb1518f435 | https://artofproblemsolving.com/wiki/index.php/2011_AIME_I_Problems/Problem_10 | The probability that a set of three distinct vertices chosen at random from among the vertices of a regular n-gon determine an obtuse triangle is $\frac{93}{125}$ . Find the sum of all possible values of $n$ | We use casework on the locations of the vertices, if we choose the locations of vertices $v_a, v_b, v_c$ on the n-gon (where the vertices of the n-gon are $v_0, v_1, v_2, ... v_{n-1},$ in clockwise order) to be the vertices of triangle ABC, in order, with the restriction that $a<b<c$
By symmetry, we can assume W/O LOG that the location of vertex A is vertex $v_0$
Now, vertex B can be any of $v_1, v_2, ... v_{n-2}$ . We start in on casework.
Case 1: vertex B is at one of the locations $v_{n-2}, v_{n-3}, ... v_{\lfloor n/2 \rfloor +1}$ . (The ceiling function is necessary for the cases in which n is odd.)
Now, since the clockwise arc from A to B measures more than 180 degrees; for every location of vertex C we can choose in the above restrictions, angle C will be an obtuse angle.
There are $\lceil n/2 \rceil - 2$ choices for vertex B now (again, the ceiling function is necessary to satisfy both odd and even cases of n). If vertex B is placed at $v_m$ , there are $n - m - 1$ possible places for vertex C.
Summing over all these possibilities, we obtain that the number of obtuse triangles obtainable from this case is $\frac{(n- \lceil n/2 \rceil - 2)(n - \lceil n/2 \rceil) - 1}{2}$
Case 2: vertex B is at one of the locations not covered in the first case.
Note that this will result in the same number of obtuse triangles as case 1, but multiplied by 2. This is because fixing vertex B in $v_0$ , then counting up the cases for vertices C, and again for vertices C and A, respectively, is combinatorially equivalent to fixing vertex A at $v_0$ , then counting cases for vertex B, as every triangle obtained in this way can be rotated in the n-gon to place vertex A at $v_0$ , and will not be congruent to any obtuse triangle obtained in case 1, as there will be a different side opposite the obtuse angle in this case.
Therefore, there are $\frac{3(n- \lceil n/2 \rceil - 2)(n - \lceil n/2 \rceil - 1)}{2}$ total obtuse triangles obtainable.
The total number of triangles obtainable is $1+2+3+...+(n-2) = \frac{(n-2)(n-1)}{2}$
The ratio of obtuse triangles obtainable to all triangles obtainable is therefore
$\frac{\frac{3(n- \lceil n/2 \rceil - 2)(n - \lceil n/2 \rceil - 1)}{2}}{\frac{(n-2)(n-1)}{2}} = \frac{3(n- \lceil n/2 \rceil - 2)(n - \lceil n/2 \rceil - 1)}{(n-2)(n-1)} = \frac{93}{125}$
So, $\frac{(n- \lceil n/2 \rceil - 2)(n - \lceil n/2 \rceil - 1)}{(n-2)(n-1)} = \frac{31}{125}$
Now, we have that $(n-2)(n-1)$ is divisible by $125 = 5^3$ . It is now much easier to perform trial-and-error on possible values of n, because we see that $n \equiv 1,2 \pmod{125}$
We find that $n = 127$ and $n = 376$ both work, so the final answer is $127 + 376 = \boxed{503}$ | null | 503 |
1fece03f90c5f4895e9a75d8a686cfd4 | https://artofproblemsolving.com/wiki/index.php/2011_AIME_I_Problems/Problem_11 | Let $R$ be the set of all possible remainders when a number of the form $2^n$ $n$ a nonnegative integer, is divided by 1000. Let $S$ be the sum of the elements in $R$ . Find the remainder when $S$ is divided by 1000. | Obviously, $2^i$ will have to repeat at some point, and our goal is just to find when it repeats. Suppose $2^a$ is the first time the powers of 2 repeat mod 1000, and that it is the same as $2^b$ where $b < a.$ We have \[2^a \equiv 2^b \mod 1000 \rightarrow 2^a - 2^b \equiv 0 \mod 1000\] We can factor out a $2^b$ to get \[2^b\left(2^{a-b} - 1\right) \equiv 0 \mod 1000\] Now, let's apply CRT. Obviously, if it is 0 mod 1000, this means directly that it is 0 mod 8 and 0 mod 125. Since $2^{a-b} - 1$ has to be odd, this necessarily means $b \ge 3$ for $8 \div 2^b.$ This means that 125 has to divide $2^{a-b} - 1.$ We wish to find the minimum $a - b$ such that this is true, or similarly, we just wish to find $\text{ord}_{125}(2).$ Note that by Euler's Totient Theorem, that $2^{100} \equiv 1 \mod 125.$ After checking, and seeing that $2^{50} \equiv -1 \mod 125$ , this directly means that $\text{ord}_{125}(2) = 100$ or $a - b = 100.$ Since $b \ge 3$ , the smallest $(a, b)$ pair is $(103, 3).$ What this really means is that the sequence of remainders will start repeating at $2^{103}$ or namely that $\{2^0, 2^1, ..., 2^{102}\}$ are all distinct residues mod 1000. Adding these yields \[2^{103} - 1 \equiv 8 - 1 \equiv \boxed{7} \mod 1000\] | null | 7 |