id int64 1 7.14k | link stringlengths 75 84 | no int64 1 14 | problem stringlengths 14 5.33k | solution stringlengths 21 6.43k | answer int64 0 999 |
|---|---|---|---|---|---|
6,001 | https://artofproblemsolving.com/wiki/index.php/2013_AIME_I_Problems/Problem_13 | 2 | Triangle $AB_0C_0$ has side lengths $AB_0 = 12$ $B_0C_0 = 17$ , and $C_0A = 25$ . For each positive integer $n$ , points $B_n$ and $C_n$ are located on $\overline{AB_{n-1}}$ and $\overline{AC_{n-1}}$ , respectively, creating three similar triangles $\triangle AB_nC_n \sim \triangle B_{n-1}C_nC_{n-1} \sim \triangle AB_{... | Let $k$ be the coefficient of the similarity of triangles \[\triangle B_0 C_1 C_0 \sim \triangle AB_0 C_0 \implies k = \frac {B_0 C_0}{AC_0} = \frac {17}{25}.\] Then area $\frac {[B_0 C_1 C_0]}{[AB_0 C_0 ]} = k^2 \implies \frac {[AB_0 C_1]}{[AB_0 C_0]} = 1 – k^2.$
The height of triangles $\triangle B_0C_1A$ and $\trian... | 961 |
6,002 | https://artofproblemsolving.com/wiki/index.php/2013_AIME_I_Problems/Problem_15 | 1 | Let $N$ be the number of ordered triples $(A,B,C)$ of integers satisfying the conditions
(a) $0\le A<B<C\le99$ ,
(b) there exist integers $a$ $b$ , and $c$ , and prime $p$ where $0\le b<a<c<p$ ,
(c) $p$ divides $A-a$ $B-b$ , and $C-c$ , and
(d) each ordered triple $(A,B,C)$ and each ordered triple $(b,a,c)$ form ar... | From condition (d), we have $(A,B,C)=(B-D,B,B+D)$ and $(b,a,c)=(a-d,a,a+d)$ . Condition $\text{(c)}$ states that $p\mid B-D-a$ $p | B-a+d$ , and $p\mid B+D-a-d$ . We subtract the first two to get $p\mid-d-D$ , and we do the same for the last two to get $p\mid 2d-D$ . We subtract these two to get $p\mid 3d$ . So $p\mid ... | 272 |
6,003 | https://artofproblemsolving.com/wiki/index.php/2013_AIME_I_Problems/Problem_15 | 2 | Let $N$ be the number of ordered triples $(A,B,C)$ of integers satisfying the conditions
(a) $0\le A<B<C\le99$ ,
(b) there exist integers $a$ $b$ , and $c$ , and prime $p$ where $0\le b<a<c<p$ ,
(c) $p$ divides $A-a$ $B-b$ , and $C-c$ , and
(d) each ordered triple $(A,B,C)$ and each ordered triple $(b,a,c)$ form ar... | Let $(A, B, C)$ $(B-x, B, B+x)$ and $(b, a, c) = (a-y, a, a+y)$ . Now the 3 differences would be \begin{align} \label{1} &A-a = B-x-a \\ \label{2} &B - b = B-a+y \\ \label{3} &C - c = B+x-a-y \end{align}
Adding equations $(1)$ and $(3)$ would give $2B - 2a - y$ . Then doubling equation $(2)$ would give $2B - 2a + 2y$ ... | 272 |
6,004 | https://artofproblemsolving.com/wiki/index.php/2013_AIME_II_Problems/Problem_1 | 1 | Suppose that the measurement of time during the day is converted to the metric system so that each day has $10$ metric hours, and each metric hour has $100$ metric minutes. Digital clocks would then be produced that would read $\text{9:99}$ just before midnight, $\text{0:00}$ at midnight, $\text{1:25}$ at the former $\... | There are $24 \cdot 60=1440$ normal minutes in a day , and $10 \cdot 100=1000$ metric minutes in a day. The ratio of normal to metric minutes in a day is $\frac{1440}{1000}$ , which simplifies to $\frac{36}{25}$ . This means that every time 36 normal minutes pass, 25 metric minutes pass. From midnight to $\text{6:36}$ ... | 275 |
6,005 | https://artofproblemsolving.com/wiki/index.php/2013_AIME_II_Problems/Problem_1 | 2 | Suppose that the measurement of time during the day is converted to the metric system so that each day has $10$ metric hours, and each metric hour has $100$ metric minutes. Digital clocks would then be produced that would read $\text{9:99}$ just before midnight, $\text{0:00}$ at midnight, $\text{1:25}$ at the former $\... | First we want to find out what fraction of a day has passed at 6:36 AM. One hour is $\frac{1}{24}$ of a day, and 36 minutes is $\frac{36}{60}=\frac{3}{5}$ of an hour, so at 6:36 AM, $6 \cdot \frac{1}{24} + \frac{1}{24} \cdot \frac{3}{5}=\frac{1}{4}+\frac{1}{40}=\frac{11}{40}$ of a day has passed. Now the metric timing ... | 275 |
6,006 | https://artofproblemsolving.com/wiki/index.php/2013_AIME_II_Problems/Problem_2 | 1 | Positive integers $a$ and $b$ satisfy the condition \[\log_2(\log_{2^a}(\log_{2^b}(2^{1000}))) = 0.\] Find the sum of all possible values of $a+b$ | To simplify, we write this logarithmic expression as an exponential one. Just looking at the first log, it has a base of 2 and an argument of the expression in parenthesis. Therefore, we can make 2 the base, 0 the exponent, and the argument the result. That means $\log_{2^a}(\log_{2^b}(2^{1000}))=1$ (because $2^0=1$ )... | 881 |
6,007 | https://artofproblemsolving.com/wiki/index.php/2013_AIME_II_Problems/Problem_2 | 2 | Positive integers $a$ and $b$ satisfy the condition \[\log_2(\log_{2^a}(\log_{2^b}(2^{1000}))) = 0.\] Find the sum of all possible values of $a+b$ | We proceed as in Solution 1, raising $2$ to both sides to achieve $\log_{2^a}(\log_{2^b}(2^{1000})) = 1.$ We raise $2^a$ to both sides to get $\log_{2^b}(2^{1000})=2^a$ , then simplify to get $\dfrac{1000}b=2^a$
At this point, we want both $a$ and $b$ to be integers. Thus, $2^a$ can only be a power of $2$ . To help us ... | 881 |
6,008 | https://artofproblemsolving.com/wiki/index.php/2013_AIME_II_Problems/Problem_3 | 1 | A large candle is $119$ centimeters tall. It is designed to burn down more quickly when it is first lit and more slowly as it approaches its bottom. Specifically, the candle takes $10$ seconds to burn down the first centimeter from the top, $20$ seconds to burn down the second centimeter, and $10k$ seconds to burn do... | We find that $T=10(1+2+\cdots +119)$ . From Gauss's formula, we find that the value of $T$ is $10(7140)=71400$ . The value of $\frac{T}{2}$ is therefore $35700$ . We find that $35700$ is $10(3570)=10\cdot \frac{k(k+1)}{2}$ , so $3570=\frac{k(k+1)}{2}$ . As a result, $7140=k(k+1)$ , which leads to $0=k^2+k-7140$ . We no... | 350 |
6,009 | https://artofproblemsolving.com/wiki/index.php/2013_AIME_II_Problems/Problem_4 | 4 | In the Cartesian plane let $A = (1,0)$ and $B = \left( 2, 2\sqrt{3} \right)$ . Equilateral triangle $ABC$ is constructed so that $C$ lies in the first quadrant. Let $P=(x,y)$ be the center of $\triangle ABC$ . Then $x \cdot y$ can be written as $\tfrac{p\sqrt{q}}{r}$ , where $p$ and $r$ are relatively prime positive... | Since $AC$ will be segment $AB$ rotated clockwise $60^{\circ}$ , we can use a rotation matrix to find $C$ . We first translate the triangle $1$ unit to the left, so $A'$ lies on the origin, and $B' = (1, 2\sqrt{3})$ . Rotating clockwise $60^{\circ}$ is the same as rotating $300^{\circ}$ counter-clockwise, so our rotati... | 40 |
6,010 | https://artofproblemsolving.com/wiki/index.php/2013_AIME_II_Problems/Problem_4 | 6 | In the Cartesian plane let $A = (1,0)$ and $B = \left( 2, 2\sqrt{3} \right)$ . Equilateral triangle $ABC$ is constructed so that $C$ lies in the first quadrant. Let $P=(x,y)$ be the center of $\triangle ABC$ . Then $x \cdot y$ can be written as $\tfrac{p\sqrt{q}}{r}$ , where $p$ and $r$ are relatively prime positive... | Labeling our points and sketching a graph we get that $C$ is to the right of $AB$ . Of course, we need to find $C$ . Note that the transformation from $A$ to $B$ is $[1,2\sqrt{3}]$ , and if we imagine a height dropped to $AB$ we see that a transformation from the midpoint $(\frac{3}{2},\sqrt {3})$ to $C$ is basically t... | 40 |
6,011 | https://artofproblemsolving.com/wiki/index.php/2013_AIME_II_Problems/Problem_6 | 1 | Find the least positive integer $N$ such that the set of $1000$ consecutive integers beginning with $1000\cdot N$ contains no square of an integer. | The difference between consecutive integral squares must be greater than 1000. $(x+1)^2-x^2\geq1000$ , so $x\geq\frac{999}{2}\implies x\geq500$ $x=500$ does not work, so $x>500$ . Let $n=x-500$ . By inspection, $n^2$ should end in a number close to but less than 1000 such that there exists $1000N$ within the difference... | 282 |
6,012 | https://artofproblemsolving.com/wiki/index.php/2013_AIME_II_Problems/Problem_6 | 2 | Find the least positive integer $N$ such that the set of $1000$ consecutive integers beginning with $1000\cdot N$ contains no square of an integer. | Let us first observe the difference between $x^2$ and $(x+1)^2$ , for any arbitrary $x\ge 0$ $(x+1)^2-x^2=2x+1$ . So that means for every $x\ge 0$ , the difference between that square and the next square have a difference of $2x+1$ . Now, we need to find an $x$ such that $2x+1\ge 1000$ . Solving gives $x\ge \frac{999}{... | 282 |
6,013 | https://artofproblemsolving.com/wiki/index.php/2013_AIME_II_Problems/Problem_6 | 3 | Find the least positive integer $N$ such that the set of $1000$ consecutive integers beginning with $1000\cdot N$ contains no square of an integer. | Let $x$ be the number being squared. Based on the reasoning above, we know that $N$ must be at least $250$ , so $x$ has to be at least $500$ . Let $k$ be $x-500$ . We can write $x^2$ as $(500+k)^2$ , or $250000+1000k+k^2$ . We can disregard $250000$ and $1000k$ , since they won't affect the last three digits, which det... | 282 |
6,014 | https://artofproblemsolving.com/wiki/index.php/2013_AIME_II_Problems/Problem_6 | 4 | Find the least positive integer $N$ such that the set of $1000$ consecutive integers beginning with $1000\cdot N$ contains no square of an integer. | The goal is to find the least $N \in \mathbb{N}$ such that $\exists m \in \mathbb{N}$ where $m^2 + 1 \leq 1000N, 1000N + 1000 \leq (m+1)^2$
Combining the two inequalities leads to $(m+1)^2 \geq m^2 + 1001, m \geq 500$
Let $m = k + 500$ , where $k \in \mathbb{W}$ , then the inequalities become,
$N \geq \frac{(k+500)^2 +... | 282 |
6,015 | https://artofproblemsolving.com/wiki/index.php/2013_AIME_II_Problems/Problem_7 | 1 | A group of clerks is assigned the task of sorting $1775$ files. Each clerk sorts at a constant rate of $30$ files per hour. At the end of the first hour, some of the clerks are reassigned to another task; at the end of the second hour, the same number of the remaining clerks are also reassigned to another task, and a s... | There are $x$ clerks at the beginning, and $t$ clerks are reassigned to another task at the end of each hour. So, $30x+30(x-t)+30(x-2t)+30\cdot\frac{10}{60} \cdot (x-3t)=1775$ , and simplify that we get $19x-21t=355$ .
Now the problem is to find a reasonable integer solution. Now we know $x= \frac{355+21t}{19}$ , so $1... | 945 |
6,016 | https://artofproblemsolving.com/wiki/index.php/2013_AIME_II_Problems/Problem_7 | 2 | A group of clerks is assigned the task of sorting $1775$ files. Each clerk sorts at a constant rate of $30$ files per hour. At the end of the first hour, some of the clerks are reassigned to another task; at the end of the second hour, the same number of the remaining clerks are also reassigned to another task, and a s... | We start with the same approach as solution 1 to get $19x-21t=355$ . Then notice that $21t + 355 \equiv 0 \pmod{19}$ , or $2t-6 \equiv 0 \pmod{19}$ , giving the smallest solution at $t=3$ . We find that $x=22$ . Then the number of files they sorted will be $30x+15(x-t)=660+285=\boxed{945}.$ | 945 |
6,017 | https://artofproblemsolving.com/wiki/index.php/2013_AIME_II_Problems/Problem_8 | 1 | A hexagon that is inscribed in a circle has side lengths $22$ $22$ $20$ $22$ $22$ , and $20$ in that order. The radius of the circle can be written as $p+\sqrt{q}$ , where $p$ and $q$ are positive integers. Find $p+q$ | Let us call the hexagon $ABCDEF$ , where $AB=CD=DE=AF=22$ , and $BC=EF=20$ .
We can just consider one half of the hexagon, $ABCD$ , to make matters simpler.
Draw a line from the center of the circle, $O$ , to the midpoint of $BC$ $X$ . Now, draw a line from $O$ to the midpoint of $AB$ $Y$ . Clearly, $\angle BXO=90^{\... | 272 |
6,018 | https://artofproblemsolving.com/wiki/index.php/2013_AIME_II_Problems/Problem_8 | 2 | A hexagon that is inscribed in a circle has side lengths $22$ $22$ $20$ $22$ $22$ , and $20$ in that order. The radius of the circle can be written as $p+\sqrt{q}$ , where $p$ and $q$ are positive integers. Find $p+q$ | Using the trapezoid $ABCD$ mentioned above, draw an altitude of the trapezoid passing through point $B$ onto $AD$ at point $J$ . Now, we can use the pythagorean theorem: $(22^2-(r-10)^2)+10^2=r^2$ . Expanding and combining like terms gives us the quadratic \[r^2-10r-242=0\] and solving for $r$ gives $r=5+\sqrt{267}$ . ... | 272 |
6,019 | https://artofproblemsolving.com/wiki/index.php/2013_AIME_II_Problems/Problem_8 | 3 | A hexagon that is inscribed in a circle has side lengths $22$ $22$ $20$ $22$ $22$ , and $20$ in that order. The radius of the circle can be written as $p+\sqrt{q}$ , where $p$ and $q$ are positive integers. Find $p+q$ | Join the diameter of the circle $AD$ and let the length be $d$ . By Ptolemy's Theorem on trapezoid $ADEF$ $(AD)(EF) + (AF)(DE) = (AE)(DF)$ . Since it is an isosceles trapezoid, both diagonals are equal. Let them be equal to $x$ each. Then
\[20d + 22^2 = x^2\]
Since $\angle AED$ is subtended by the diameter, it is right... | 272 |
6,020 | https://artofproblemsolving.com/wiki/index.php/2013_AIME_II_Problems/Problem_8 | 4 | A hexagon that is inscribed in a circle has side lengths $22$ $22$ $20$ $22$ $22$ , and $20$ in that order. The radius of the circle can be written as $p+\sqrt{q}$ , where $p$ and $q$ are positive integers. Find $p+q$ | As we can see this image, it is symmetrical hence the diameter divides the hexagon into two congruent quadrilateral. Now we can apply the Ptolemy's theorem. Denote the radius is r, we can get \[22*2x+440=\sqrt{4x^2-400}\sqrt{4x^2-484}\] , after simple factorization, we can get \[x^4-342x^2-2420x=0\] , it is easy to see... | 272 |
6,021 | https://artofproblemsolving.com/wiki/index.php/2013_AIME_II_Problems/Problem_8 | 5 | A hexagon that is inscribed in a circle has side lengths $22$ $22$ $20$ $22$ $22$ , and $20$ in that order. The radius of the circle can be written as $p+\sqrt{q}$ , where $p$ and $q$ are positive integers. Find $p+q$ | Using solution 1's diagram, extend line segments $AB$ and $CD$ upwards until they meet at point $G$ . Let point $O$ be the center of the hexagon. By the $AA$ postulate, $\Delta ADG \sim \Delta BCG \sim \Delta CDO$ . This means $\frac{DC}{AD} = \frac{22}{2r} = \frac{CO}{AG}$ , so $AG = r \times \frac{2r}{22} = \frac{r^2... | 272 |
6,022 | https://artofproblemsolving.com/wiki/index.php/2013_AIME_II_Problems/Problem_8 | 6 | A hexagon that is inscribed in a circle has side lengths $22$ $22$ $20$ $22$ $22$ , and $20$ in that order. The radius of the circle can be written as $p+\sqrt{q}$ , where $p$ and $q$ are positive integers. Find $p+q$ | Let $\angle{AOB} = \theta$ . So, we have $\sin \dfrac{\theta}{2} = \dfrac{11}{r}$ and $\cos \dfrac{\theta}{2} = \dfrac{\sqrt{r^{2} - 121}}{r}$ . So, $\sin \theta = 2 \sin \dfrac{\theta}{2} \cos \dfrac{\theta}{2} = \dfrac{22 \sqrt{r^{2} - 121}}{r^{2}}$ . Let $H$ be the foot of the perpendicular from $B$ to $\overline{AD... | 272 |
6,023 | https://artofproblemsolving.com/wiki/index.php/2013_AIME_II_Problems/Problem_8 | 7 | A hexagon that is inscribed in a circle has side lengths $22$ $22$ $20$ $22$ $22$ , and $20$ in that order. The radius of the circle can be written as $p+\sqrt{q}$ , where $p$ and $q$ are positive integers. Find $p+q$ | [asy] import olympiad; import math; real a; a=2*asin(11/(5+sqrt(267))); pair A,B,C,D,E,F; A=expi(pi); B=expi(pi-a); C=expi(a); D=expi(0); E=expi(-a); F=expi(pi+a); draw(A--B--C--D--E--F--A--D); draw(B--D); draw(A--C); dot(A); dot(B); dot(C); dot(D); dot(E); dot(F); draw(unitcircle); label("$A$",A,W);label("$B$",B,NW);... | 272 |
6,024 | https://artofproblemsolving.com/wiki/index.php/2013_AIME_II_Problems/Problem_9 | 1 | $7\times 1$ board is completely covered by $m\times 1$ tiles without overlap; each tile may cover any number of consecutive squares, and each tile lies completely on the board. Each tile is either red, blue, or green. Let $N$ be the number of tilings of the $7\times 1$ board in which all three colors are used at least ... | Firstly, we consider how many different ways possible to divide the $7\times 1$ board.
We ignore the cases of 1 or 2 pieces since we need at least one tile of each color.
Secondly, we use Principle of Inclusion-Exclusion to consider how many ways to color them:
Finally, we combine them together: $15\times 6+20\times 36... | 106 |
6,025 | https://artofproblemsolving.com/wiki/index.php/2013_AIME_II_Problems/Problem_9 | 2 | $7\times 1$ board is completely covered by $m\times 1$ tiles without overlap; each tile may cover any number of consecutive squares, and each tile lies completely on the board. Each tile is either red, blue, or green. Let $N$ be the number of tilings of the $7\times 1$ board in which all three colors are used at least ... | This solution is basically solution 1 with more things done at once. The game plan:
$\sum_{i=0}^{7} ($ the amount of ways to divide the board into $i$ pieces $) \cdot ($ the amount of ways to color the respective divisions)
The amount of ways to divide the board is determined using stars and bars. The colorings are fou... | 106 |
6,026 | https://artofproblemsolving.com/wiki/index.php/2013_AIME_II_Problems/Problem_9 | 3 | $7\times 1$ board is completely covered by $m\times 1$ tiles without overlap; each tile may cover any number of consecutive squares, and each tile lies completely on the board. Each tile is either red, blue, or green. Let $N$ be the number of tilings of the $7\times 1$ board in which all three colors are used at least ... | 3 colors is a lot. How many ways can we tile an $n \times 1$ board with one color? It's going to be $2^{n-1}$ because if you draw out the $n$ spaces, you can decide whether each of the borders between the tiles are either there or not there. There are $n-1$ borders so there are $2^{n-1}$ tilings. Define a one-tiling of... | 106 |
6,027 | https://artofproblemsolving.com/wiki/index.php/2013_AIME_II_Problems/Problem_9 | 4 | $7\times 1$ board is completely covered by $m\times 1$ tiles without overlap; each tile may cover any number of consecutive squares, and each tile lies completely on the board. Each tile is either red, blue, or green. Let $N$ be the number of tilings of the $7\times 1$ board in which all three colors are used at least ... | Let $n$ be the length of the board and $x$ be the number of colors. We will find the number of ways to tile the $n \times 1$ board with no color restrictions (some colors may be unused) and then use PIE.
By stars and bars, the number of ways to divide the board into $k$ pieces is ${n-1 \choose k-1}$ . There are $x^k$ w... | 106 |
6,028 | https://artofproblemsolving.com/wiki/index.php/2013_AIME_II_Problems/Problem_10 | 1 | Given a circle of radius $\sqrt{13}$ , let $A$ be a point at a distance $4 + \sqrt{13}$ from the center $O$ of the circle. Let $B$ be the point on the circle nearest to point $A$ . A line passing through the point $A$ intersects the circle at points $K$ and $L$ . The maximum possible area for $\triangle BKL$ can be wri... | [asy] import math; import olympiad; import graph; pair A, B, K, L; B = (sqrt(13), 0); A=(4+sqrt(13), 0); dot(B); dot(A); draw(Circle((0,0), sqrt(13))); label("$O$", (0,0), S);label("$B$", B, SW);label("$A$", A, S); dot((0,0)); [/asy]
Now we put the figure in the Cartesian plane, let the center of the circle $O (0,0... | 146 |
6,029 | https://artofproblemsolving.com/wiki/index.php/2013_AIME_II_Problems/Problem_10 | 2 | Given a circle of radius $\sqrt{13}$ , let $A$ be a point at a distance $4 + \sqrt{13}$ from the center $O$ of the circle. Let $B$ be the point on the circle nearest to point $A$ . A line passing through the point $A$ intersects the circle at points $K$ and $L$ . The maximum possible area for $\triangle BKL$ can be wri... | [asy] import math; import olympiad; import graph; pair A, B, K, L; B = (sqrt(13), 0); A=(4+sqrt(13), 0); dot(B); dot(A); draw(Circle((0,0), sqrt(13))); label("$O$", (0,0), S);label("$B$", B, SW);label("$A$", A, S); dot((0,0)); [/asy]
Draw $OC$ perpendicular to $KL$ at $C$ . Draw $BD$ perpendicular to $KL$ at $D$
\[... | 146 |
6,030 | https://artofproblemsolving.com/wiki/index.php/2013_AIME_II_Problems/Problem_10 | 3 | Given a circle of radius $\sqrt{13}$ , let $A$ be a point at a distance $4 + \sqrt{13}$ from the center $O$ of the circle. Let $B$ be the point on the circle nearest to point $A$ . A line passing through the point $A$ intersects the circle at points $K$ and $L$ . The maximum possible area for $\triangle BKL$ can be wri... | Let $N,M$ les on $AL$ such that $BM\bot AL, ON\bot AL$ , call $BM=h, ON=k,LN=KN=d$ We call $\angle{LON}=\alpha$ By similar triangle, we have $\frac{h}{k}=\frac{4}{4+\sqrt{13}}, h=\frac{4k}{4+\sqrt{13}}$ . Then, we realize the area is just $dh=d\cdot \frac{4K}{4+\sqrt{13}}$ As $\sin \alpha=\frac{d}{\sqrt{13}}, \cos \alp... | 146 |
6,031 | https://artofproblemsolving.com/wiki/index.php/2013_AIME_II_Problems/Problem_10 | 4 | Given a circle of radius $\sqrt{13}$ , let $A$ be a point at a distance $4 + \sqrt{13}$ from the center $O$ of the circle. Let $B$ be the point on the circle nearest to point $A$ . A line passing through the point $A$ intersects the circle at points $K$ and $L$ . The maximum possible area for $\triangle BKL$ can be wri... | Let C and D be the base of perpendiculars dropped from points O and B to AK. Denote BD = h, OC = H. \[\triangle ABD \sim \triangle AOC \implies \frac {h}{H} = \frac {4}{4 + \sqrt{13}}.\] $KL$ is the base of triangles $\triangle OKL$ and $\triangle BKL \implies \frac {[BKL]}{[OKL]} = \frac{h}{H} =$ const $\implies$ Th... | 146 |
6,032 | https://artofproblemsolving.com/wiki/index.php/2013_AIME_II_Problems/Problem_11 | 1 | Let $A = \{1, 2, 3, 4, 5, 6, 7\}$ , and let $N$ be the number of functions $f$ from set $A$ to set $A$ such that $f(f(x))$ is a constant function. Find the remainder when $N$ is divided by $1000$ | Any such function can be constructed by distributing the elements of $A$ on three tiers.
The bottom tier contains the constant value, $c=f(f(x))$ for any $x$ . (Obviously $f(c)=c$ .)
The middle tier contains $k$ elements $x\ne c$ such that $f(x)=c$ , where $1\le k\le 6$
The top tier contains $6-k$ elements such that $f... | 399 |
6,033 | https://artofproblemsolving.com/wiki/index.php/2013_AIME_II_Problems/Problem_11 | 2 | Let $A = \{1, 2, 3, 4, 5, 6, 7\}$ , and let $N$ be the number of functions $f$ from set $A$ to set $A$ such that $f(f(x))$ is a constant function. Find the remainder when $N$ is divided by $1000$ | Define the three layers as domain $x$ codomain $f(x)$ , and codomain $f(f(x))$ . Each one of them is contained in the set $A$ . We know that $f(f(x))$ is a constant function , or in other words, can only take on one value. So, we can start off by choosing that value $c$ in $7$ ways. So now, we choose the values that ca... | 399 |
6,034 | https://artofproblemsolving.com/wiki/index.php/2013_AIME_II_Problems/Problem_12 | 1 | Let $S$ be the set of all polynomials of the form $z^3 + az^2 + bz + c$ , where $a$ $b$ , and $c$ are integers. Find the number of polynomials in $S$ such that each of its roots $z$ satisfies either $|z| = 20$ or $|z| = 13$ | Every cubic with real coefficients has to have either three real roots or one real and two nonreal roots which are conjugates. This follows from Vieta's formulas
The real root $r$ must be one of $-20$ $20$ $-13$ , or $13$ . By Viète's formulas, $a=-(r+\omega+\omega^*)$ $b=|\omega|^2+r(\omega+\omega^*)$ , and $c=-r|\ome... | 540 |
6,035 | https://artofproblemsolving.com/wiki/index.php/2013_AIME_II_Problems/Problem_12 | 2 | Let $S$ be the set of all polynomials of the form $z^3 + az^2 + bz + c$ , where $a$ $b$ , and $c$ are integers. Find the number of polynomials in $S$ such that each of its roots $z$ satisfies either $|z| = 20$ or $|z| = 13$ | There are two cases: either all the roots are real, or one is real and two are imaginary.
Case 1: All roots are real.
Then each of the roots is a member of the set $\{-20, 20, -13, 13\}$ . It splits into three sub-cases: either no two are the same, exactly two are the same, or all three are the same.
Sub-case 1.1: No t... | 540 |
6,036 | https://artofproblemsolving.com/wiki/index.php/2013_AIME_II_Problems/Problem_14 | 2 | For positive integers $n$ and $k$ , let $f(n, k)$ be the remainder when $n$ is divided by $k$ , and for $n > 1$ let $F(n) = \max_{\substack{1\le k\le \frac{n}{2}}} f(n, k)$ . Find the remainder when $\sum\limits_{n=20}^{100} F(n)$ is divided by $1000$ | $Lemma:$ Highest remainder when $n$ is divided by $1\leq k\leq n/2$ is obtained for $k_0 = (n + (3 - n \pmod{3}))/3$ and the remainder thus obtained is $(n - k_0*2) = [(n - 6)/3 + (2/3)*n \pmod{3}]$
$Note:$ This is the second highest remainder when $n$ is divided by $1\leq k\leq n$ and the highest remainder occurs when... | 512 |
6,037 | https://artofproblemsolving.com/wiki/index.php/2013_AIME_II_Problems/Problem_15 | 1 | Let $A,B,C$ be angles of a triangle with \begin{align*} \cos^2 A + \cos^2 B + 2 \sin A \sin B \cos C &= \frac{15}{8} \text{ and} \\ \cos^2 B + \cos^2 C + 2 \sin B \sin C \cos A &= \frac{14}{9} \end{align*} There are positive integers $p$ $q$ $r$ , and $s$ for which \[\cos^2 C + \cos^2 A + 2 \sin C \sin A \cos B = \frac... | Let's draw the triangle. Since the problem only deals with angles, we can go ahead and set one of the sides to a convenient value. Let $BC = \sin{A}$
By the Law of Sines, we must have $CA = \sin{B}$ and $AB = \sin{C}$
Now let us analyze the given:
\begin{align*} \cos^2A + \cos^2B + 2\sin A\sin B\cos C &= 1-\sin^2A + 1-... | 222 |
6,038 | https://artofproblemsolving.com/wiki/index.php/2013_AIME_II_Problems/Problem_15 | 2 | Let $A,B,C$ be angles of a triangle with \begin{align*} \cos^2 A + \cos^2 B + 2 \sin A \sin B \cos C &= \frac{15}{8} \text{ and} \\ \cos^2 B + \cos^2 C + 2 \sin B \sin C \cos A &= \frac{14}{9} \end{align*} There are positive integers $p$ $q$ $r$ , and $s$ for which \[\cos^2 C + \cos^2 A + 2 \sin C \sin A \cos B = \frac... | Let \begin{align*} \cos^2 A + \cos^2 B + 2 \sin A \sin B \cos C &= \frac{15}{8} \text{ ------ (1)}\\ \cos^2 B + \cos^2 C + 2 \sin B \sin C \cos A &= \frac{14}{9} \text{ ------ (2)}\\ \cos^2 C + \cos^2 A + 2 \sin C \sin A \cos B &= x \text{ ------ (3)}\\ \end{align*}
Adding (1) and (3) we ... | 222 |
6,039 | https://artofproblemsolving.com/wiki/index.php/2013_AIME_II_Problems/Problem_15 | 3 | Let $A,B,C$ be angles of a triangle with \begin{align*} \cos^2 A + \cos^2 B + 2 \sin A \sin B \cos C &= \frac{15}{8} \text{ and} \\ \cos^2 B + \cos^2 C + 2 \sin B \sin C \cos A &= \frac{14}{9} \end{align*} There are positive integers $p$ $q$ $r$ , and $s$ for which \[\cos^2 C + \cos^2 A + 2 \sin C \sin A \cos B = \frac... | Let's take the first equation $\cos^2 A + \cos^2 B + 2 \sin A \sin B \cos C = \frac{15}{8}$ . Substituting $180 - A - B$ for C, given A, B, and C form a triangle, and that $\cos C = \cos(A + B)$ , gives us:
$\cos^2 A + \cos^2 B + 2 \sin A \sin B \cos (A+B) = \frac{15}{8}$
Expanding out gives us $\cos^2 A + \cos^2 B + 2... | 222 |
6,040 | https://artofproblemsolving.com/wiki/index.php/2013_AIME_II_Problems/Problem_15 | 5 | Let $A,B,C$ be angles of a triangle with \begin{align*} \cos^2 A + \cos^2 B + 2 \sin A \sin B \cos C &= \frac{15}{8} \text{ and} \\ \cos^2 B + \cos^2 C + 2 \sin B \sin C \cos A &= \frac{14}{9} \end{align*} There are positive integers $p$ $q$ $r$ , and $s$ for which \[\cos^2 C + \cos^2 A + 2 \sin C \sin A \cos B = \frac... | According to LOC $a^2+b^2-2ab\cos{\angle{c}}=c^2$ , we can write it into $\sin^2{\angle{A}}+\sin^2{\angle{B}}-2\sin{\angle{A}}\sin{\angle{B}}\cos{\angle{C}}=\sin^2{\angle{C}}$ $\sin^2{\angle{A}}+\sin^2{\angle{B}}-2\sin{\angle{A}}\sin{\angle{B}}\cos{\angle{C}}+cos^2A+cos^2B+2sinAsinBcosC=\frac{15}{8}+sin^2C$ We can simp... | 222 |
6,041 | https://artofproblemsolving.com/wiki/index.php/2012_AIME_I_Problems/Problem_2 | 1 | The terms of an arithmetic sequence add to $715$ . The first term of the sequence is increased by $1$ , the second term is increased by $3$ , the third term is increased by $5$ , and in general, the $k$ th term is increased by the $k$ th odd positive integer. The terms of the new sequence add to $836$ . Find the sum of... | If the sum of the original sequence is $\sum_{i=1}^{n} a_i$ then the sum of the new sequence can be expressed as $\sum_{i=1}^{n} a_i + (2i - 1) = n^2 + \sum_{i=1}^{n} a_i.$ Therefore, $836 = n^2 + 715 \rightarrow n=11.$ Now the middle term of the original sequence is simply the average of all the terms, or $\frac{715}{... | 195 |
6,042 | https://artofproblemsolving.com/wiki/index.php/2012_AIME_I_Problems/Problem_2 | 2 | The terms of an arithmetic sequence add to $715$ . The first term of the sequence is increased by $1$ , the second term is increased by $3$ , the third term is increased by $5$ , and in general, the $k$ th term is increased by the $k$ th odd positive integer. The terms of the new sequence add to $836$ . Find the sum of... | After the adding of the odd numbers, the total of the sequence increases by $836 - 715 = 121 = 11^2$ . Since the sum of the first $n$ positive odd numbers is $n^2$ , there must be $11$ terms in the sequence, so the mean of the sequence is $\dfrac{715}{11} = 65$ . Since the first, last, and middle terms are centered aro... | 195 |
6,043 | https://artofproblemsolving.com/wiki/index.php/2012_AIME_I_Problems/Problem_2 | 3 | The terms of an arithmetic sequence add to $715$ . The first term of the sequence is increased by $1$ , the second term is increased by $3$ , the third term is increased by $5$ , and in general, the $k$ th term is increased by the $k$ th odd positive integer. The terms of the new sequence add to $836$ . Find the sum of... | Proceed as in Solution 2 until it is noted that there are 11 terms in the sequence. Since the sum of the terms of the original arithmetic sequence is 715, we note that $\frac{2a_1 + 10d}{2} \cdot 11 = 715$ or $2a_1 + 10d = 130$ for all sets of first terms and common differences that fit the conditions given in the prob... | 195 |
6,044 | https://artofproblemsolving.com/wiki/index.php/2012_AIME_I_Problems/Problem_3 | 2 | Nine people sit down for dinner where there are three choices of meals. Three people order the beef meal, three order the chicken meal, and three order the fish meal. The waiter serves the nine meals in random order. Find the number of ways in which the waiter could serve the meal types to the nine people so that exact... | We only need to figure out the number of ways to order the string $BBBCCCFFF$ , where exactly one $B$ is in the first three positions, one $C$ is in the $4^{th}$ to $6^{th}$ positions, and one $F$ is in the last three positions. There are $3^3=27$ ways to place the first $3$ meals. Then for the other two people, there ... | 216 |
6,045 | https://artofproblemsolving.com/wiki/index.php/2012_AIME_I_Problems/Problem_4 | 1 | Butch and Sundance need to get out of Dodge. To travel as quickly as possible, each alternates walking and riding their only horse, Sparky, as follows. Butch begins by walking while Sundance rides. When Sundance reaches the first of the hitching posts that are conveniently located at one-mile intervals along their rout... | When they meet at the milepost, Sparky has been ridden for $n$ miles total. Assume Butch rides Sparky for $a$ miles, and Sundance rides for $n-a$ miles. Thus, we can set up an equation, given that Sparky takes $\frac{1}{6}$ hours per mile, Butch takes $\frac{1}{4}$ hours per mile, and Sundance takes $\frac{2}{5}$ hours... | 279 |
6,046 | https://artofproblemsolving.com/wiki/index.php/2012_AIME_I_Problems/Problem_5 | 1 | Let $B$ be the set of all binary integers that can be written using exactly $5$ zeros and $8$ ones where leading zeros are allowed. If all possible subtractions are performed in which one element of $B$ is subtracted from another, find the number of times the answer $1$ is obtained. | When $1$ is subtracted from a binary number, the number of digits will remain constant if and only if the original number ended in $10.$ Therefore, every subtraction involving two numbers from $B$ will necessarily involve exactly one number ending in $10.$ To solve the problem, then, we can simply count the instances o... | 330 |
6,047 | https://artofproblemsolving.com/wiki/index.php/2012_AIME_I_Problems/Problem_7 | 1 | 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 ... | 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 neighb... | 280 |
6,048 | https://artofproblemsolving.com/wiki/index.php/2012_AIME_I_Problems/Problem_7 | 2 | 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 ... | 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 ... | 280 |
6,049 | https://artofproblemsolving.com/wiki/index.php/2012_AIME_I_Problems/Problem_7 | 3 | 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 ... | 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}\\ ... | 280 |
6,050 | https://artofproblemsolving.com/wiki/index.php/2012_AIME_I_Problems/Problem_7 | 4 | 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 ... | 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 ... | 280 |
6,051 | https://artofproblemsolving.com/wiki/index.php/2012_AIME_I_Problems/Problem_10 | 3 | 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 numb... | 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}... | 170 |
6,052 | https://artofproblemsolving.com/wiki/index.php/2012_AIME_I_Problems/Problem_10 | 4 | 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 numb... | 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 b... | 170 |
6,053 | https://artofproblemsolving.com/wiki/index.php/2012_AIME_I_Problems/Problem_10 | 5 | 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 numb... | 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... | 170 |
6,054 | https://artofproblemsolving.com/wiki/index.php/2012_AIME_I_Problems/Problem_11 | 1 | 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| \l... | 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 prope... | 373 |
6,055 | https://artofproblemsolving.com/wiki/index.php/2012_AIME_I_Problems/Problem_14 | 1 | 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, ... | 375 |
6,056 | https://artofproblemsolving.com/wiki/index.php/2012_AIME_I_Problems/Problem_14 | 3 | 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 + ... | 375 |
6,057 | https://artofproblemsolving.com/wiki/index.php/2012_AIME_I_Problems/Problem_14 | 5 | 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 ... | 375 |
6,058 | https://artofproblemsolving.com/wiki/index.php/2012_AIME_I_Problems/Problem_14 | 6 | 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 $\V... | 375 |
6,059 | https://artofproblemsolving.com/wiki/index.php/2012_AIME_I_Problems/Problem_14 | 7 | 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}... | 375 |
6,060 | https://artofproblemsolving.com/wiki/index.php/2012_AIME_I_Problems/Problem_15 | 1 | 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 o... | 332 |
6,061 | https://artofproblemsolving.com/wiki/index.php/2012_AIME_II_Problems/Problem_1 | 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 = \boxe... | 34 |
6,062 | https://artofproblemsolving.com/wiki/index.php/2012_AIME_II_Problems/Problem_1 | 3 | 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), ... | 34 |
6,063 | https://artofproblemsolving.com/wiki/index.php/2012_AIME_II_Problems/Problem_2 | 1 | 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... | 363 |
6,064 | https://artofproblemsolving.com/wiki/index.php/2012_AIME_II_Problems/Problem_5 | 1 | 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 ... | 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 throu... | 750 |
6,065 | https://artofproblemsolving.com/wiki/index.php/2012_AIME_II_Problems/Problem_6 | 1 | 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 opposi... | 125 |
6,066 | https://artofproblemsolving.com/wiki/index.php/2012_AIME_II_Problems/Problem_6 | 2 | 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\... | 125 |
6,067 | https://artofproblemsolving.com/wiki/index.php/2012_AIME_II_Problems/Problem_9 | 1 | 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 ... | 107 |
6,068 | https://artofproblemsolving.com/wiki/index.php/2012_AIME_II_Problems/Problem_9 | 2 | 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{\c... | 107 |
6,069 | https://artofproblemsolving.com/wiki/index.php/2012_AIME_II_Problems/Problem_9 | 3 | 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{... | 107 |
6,070 | https://artofproblemsolving.com/wiki/index.php/2012_AIME_II_Problems/Problem_9 | 4 | 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}}... | 107 |
6,071 | https://artofproblemsolving.com/wiki/index.php/2012_AIME_II_Problems/Problem_10 | 1 | 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... | 496 |
6,072 | https://artofproblemsolving.com/wiki/index.php/2012_AIME_II_Problems/Problem_10 | 2 | 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 <... | 496 |
6,073 | https://artofproblemsolving.com/wiki/index.php/2012_AIME_II_Problems/Problem_10 | 3 | 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 $n$ | 496 |
6,074 | https://artofproblemsolving.com/wiki/index.php/2012_AIME_II_Problems/Problem_10 | 4 | 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 exi... | 496 |
6,075 | https://artofproblemsolving.com/wiki/index.php/2012_AIME_II_Problems/Problem_13 | 1 | 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 l... | 677 |
6,076 | https://artofproblemsolving.com/wiki/index.php/2012_AIME_II_Problems/Problem_15 | 1 | 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^... | 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 e... | 919 |
6,077 | https://artofproblemsolving.com/wiki/index.php/2012_AIME_II_Problems/Problem_15 | 2 | 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^... | 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 $\triangl... | 919 |
6,078 | https://artofproblemsolving.com/wiki/index.php/2012_AIME_II_Problems/Problem_15 | 3 | 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^... | 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^\ci... | 919 |
6,079 | https://artofproblemsolving.com/wiki/index.php/2012_AIME_II_Problems/Problem_15 | 4 | 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^... | 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}}... | 919 |
6,080 | https://artofproblemsolving.com/wiki/index.php/2012_AIME_II_Problems/Problem_15 | 5 | 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^... | 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 \... | 919 |
6,081 | https://artofproblemsolving.com/wiki/index.php/2011_AIME_I_Problems/Problem_1 | 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 en... | 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... | 85 |
6,082 | https://artofproblemsolving.com/wiki/index.php/2011_AIME_I_Problems/Problem_2 | 1 | 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 $m \sqrt{n} - p$ , wh... | 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... | 36 |
6,083 | https://artofproblemsolving.com/wiki/index.php/2011_AIME_I_Problems/Problem_2 | 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 $m \sqrt{n} - p$ , wh... | 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: $3+21+12=\boxed{36}$ | 36 |
6,084 | https://artofproblemsolving.com/wiki/index.php/2011_AIME_I_Problems/Problem_5 | 1 | 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 no... | 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... | 144 |
6,085 | https://artofproblemsolving.com/wiki/index.php/2011_AIME_I_Problems/Problem_5 | 2 | 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 no... | 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... | 144 |
6,086 | https://artofproblemsolving.com/wiki/index.php/2011_AIME_I_Problems/Problem_8 | 1 | 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 th... | 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 $... | 318 |
6,087 | https://artofproblemsolving.com/wiki/index.php/2011_AIME_I_Problems/Problem_8 | 2 | 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 th... | 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}... | 318 |
6,088 | https://artofproblemsolving.com/wiki/index.php/2011_AIME_I_Problems/Problem_8 | 3 | 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 th... | 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} =... | 318 |
6,089 | https://artofproblemsolving.com/wiki/index.php/2011_AIME_I_Problems/Problem_9 | 1 | 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 ... | 192 |
6,090 | https://artofproblemsolving.com/wiki/index.php/2011_AIME_I_Problems/Problem_9 | 2 | 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]{57... | 192 |
6,091 | https://artofproblemsolving.com/wiki/index.php/2011_AIME_I_Problems/Problem_10 | 1 | 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: a... | 503 |
6,092 | https://artofproblemsolving.com/wiki/index.php/2011_AIME_I_Problems/Problem_10 | 2 | 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 ... | 503 |
6,093 | https://artofproblemsolving.com/wiki/index.php/2011_AIME_I_Problems/Problem_12 | 1 | Six men and some number of women stand in a line in random order. Let $p$ be the probability that a group of at least four men stand together in the line, given that every man stands next to at least one other man. Find the least number of women in the line such that $p$ does not exceed 1 percent. | Let $n$ be the number of women present, and let _ be some positive number of women between groups of men. Since the problem states that every man stands next to another man, there cannot be isolated men. Thus, there are five cases to consider, where $(k)$ refers to a consecutive group of $k$ men:
For the first case, ... | 594 |
6,094 | https://artofproblemsolving.com/wiki/index.php/2011_AIME_I_Problems/Problem_13 | 1 | A cube with side length 10 is suspended above a plane. The vertex closest to the plane is labeled $A$ . The three vertices adjacent to vertex $A$ are at heights 10, 11, and 12 above the plane. The distance from vertex $A$ to the plane can be expressed as $\frac{r-\sqrt{s}}{t}$ , where $r$ $s$ , and $t$ are positive int... | Set the cube at the origin with the three vertices along the axes and the plane equal to $ax+by+cz+d=0$ , where $a^2+b^2+c^2=1$ . The distance from a point $(X,Y,Z)$ to a plane with equation $Ax+By+Cz+D=0$ is \[\frac{AX+BY+CZ+D}{\sqrt{A^2+B^2+C^2}},\] so the (directed) distance from any point $(x,y,z)$ to the plane is... | 330 |
6,095 | https://artofproblemsolving.com/wiki/index.php/2011_AIME_I_Problems/Problem_13 | 2 | A cube with side length 10 is suspended above a plane. The vertex closest to the plane is labeled $A$ . The three vertices adjacent to vertex $A$ are at heights 10, 11, and 12 above the plane. The distance from vertex $A$ to the plane can be expressed as $\frac{r-\sqrt{s}}{t}$ , where $r$ $s$ , and $t$ are positive int... | Let the vertices with distance $10,11,12$ be $B,C,D$ , respectively. An equilateral triangle $\triangle BCD$ is formed with side length $10\sqrt{2}$ . We care only about the $z$ coordinate: $B=10,C=11,D=12$ . It is well known that the centroid of a triangle is the average of the coordinates of its three vertices, so $\... | 330 |
6,096 | https://artofproblemsolving.com/wiki/index.php/2011_AIME_II_Problems/Problem_3 | 1 | The degree measures of the angles in a convex 18-sided polygon form an increasing arithmetic sequence with integer values. Find the degree measure of the smallest angle | Another way to solve this problem would be to use exterior angles. Exterior angles of any polygon add up to $360^{\circ}$ . Since there are $18$ exterior angles in an 18-gon, the average measure of an exterior angles is $\frac{360}{18}=20^\circ$ . We know from the problem that since the exterior angles must be in an ar... | 143 |
6,097 | https://artofproblemsolving.com/wiki/index.php/2011_AIME_II_Problems/Problem_3 | 2 | The degree measures of the angles in a convex 18-sided polygon form an increasing arithmetic sequence with integer values. Find the degree measure of the smallest angle | The sum of the angles in a 18-gon is $(18-2) \cdot 180^\circ = 2880 ^\circ.$ Because the angles are in an arithmetic sequence, we can also write the sum of the angles as $a+(a+d)+(a+2d)+\dots+(a+17d)=18a+153d,$ where $a$ is the smallest angle and $d$ is the common difference. Since these two are equal, we know that $18... | 143 |
6,098 | https://artofproblemsolving.com/wiki/index.php/2011_AIME_II_Problems/Problem_3 | 3 | The degree measures of the angles in a convex 18-sided polygon form an increasing arithmetic sequence with integer values. Find the degree measure of the smallest angle | Each individual angle in a $18$ -gon is $\frac {(18-2) \cdot 180^\circ}{18} = 160^\circ$ . Since no angle in a convex polygon can be larger than $180^\circ$ , the smallest angle possible is in the set $159, 161, 157, 163, 155, 165, 153, 167, 151, 169, 149, 171, 147, 173, 145, 175, 143, 177$
Our smallest possible angle ... | 143 |
6,099 | https://artofproblemsolving.com/wiki/index.php/2011_AIME_II_Problems/Problem_4 | 1 | In triangle $ABC$ $AB=20$ and $AC=11$ . The angle bisector of $\angle A$ intersects $BC$ at point $D$ , and point $M$ is the midpoint of $AD$ . Let $P$ be the point of the intersection of $AC$ and $BM$ . The ratio of $CP$ to $PA$ can be expressed in the form $\dfrac{m}{n}$ , where $m$ and $n$ are relatively prime posit... | [asy] pointpen = black; pathpen = linewidth(0.7); pair A = (0,0), C= (11,0), B=IP(CR(A,20),CR(C,18)), D = IP(B--C,CR(B,20/31*abs(B-C))), M = (A+D)/2, P = IP(M--2*M-B, A--C), D2 = IP(D--D+P-B, A--C); D(MP("A",D(A))--MP("B",D(B),N)--MP("C",D(C))--cycle); D(A--MP("D",D(D),NE)--MP("D'",D(D2))); D(B--MP("P",D(P))); D(MP("... | 51 |
6,100 | https://artofproblemsolving.com/wiki/index.php/2011_AIME_II_Problems/Problem_4 | 2 | In triangle $ABC$ $AB=20$ and $AC=11$ . The angle bisector of $\angle A$ intersects $BC$ at point $D$ , and point $M$ is the midpoint of $AD$ . Let $P$ be the point of the intersection of $AC$ and $BM$ . The ratio of $CP$ to $PA$ can be expressed in the form $\dfrac{m}{n}$ , where $m$ and $n$ are relatively prime posit... | Assign mass points as follows: by Angle-Bisector Theorem, $BD / DC = 20/11$ , so we assign $m(B) = 11, m(C) = 20, m(D) = 31$ . Since $AM = MD$ , then $m(A) = 31$ , and $\frac{CP}{PA} = \frac{m(A) }{ m(C)} = \frac{31}{20}$ , so $m+n = \boxed{51}$ | 51 |
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