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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 $\triangle AB_nC_n \sim \triangle B_{n-1}C_nC_{n-1} \sim \triangle AB_{n-1}C_{n-1}$ . The area of the union of all triangles $B_{n-1}C_nB_n$ for $n\geq1$ can be expressed as $\tfrac pq$ , where $p$ and $q$ are relatively prime positive integers. Find $q$
|
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 $\triangle AB_0C_0$ from $B_0$ is the same $\implies \frac {AC_1}{AC_0} = 1 – k^2.$
The coefficient of the similarity of triangles $\triangle AB_1C_1 \sim \triangle AB_0C_0$ is $\frac {AC_1}{AC_0} = 1 – k^2 \implies \frac {[B_1C_1C_2 ]}{[AB_0C_0 ]} = k^2 (1 – k^2)^2.$
Analogically the coefficient of the similarity of triangles $\triangle AB_2C_2 \sim \triangle AB_0C_0$ is $(1 – k^2)^2 \implies \frac {[B_2C_2C_3]}{[AB_0C_0 ]} = k^2 (1 – k^2)^4$ and so on.
The yellow area $[Y]$ is $\frac {[Y]}{[AB_0C_0 ]} = k^2 + k^2 (1 – k^2)^2 + k^2 (1 – k^2)^4 +.. = \frac {k^2}{1 – (1 – k^2)^2} = \frac{1}{2 – k^2}.$
The required area is $[AB_0C_0 ] – [Y] = [AB_0C_0 ] \cdot (1 – \frac{1}{2 – k^2}) = [AB_0C_0 ] \cdot \frac {1 – k^2}{2 – k^2} = [AB_0C_0 ] \cdot \frac {25^2 – 17^2} {2 \cdot 25^2 – 17^2} = [AB_0C_0 ] \cdot \frac {336}{961}.$
The number $961$ is prime, $[AB_0C_0]$ is integer but not $961,$ therefore the answer is $\boxed{961}$
| 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 arithmetic sequences. Find $N$
|
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 $p\mid 3$ or $p\mid d$ . The second case is clearly impossible, because that would make $c=a+d>p$ , violating condition $\text{(b)}$ . So we have $p\mid 3$ , meaning $p=3$ . Condition $\text{(b)}$ implies that $(b,a,c)=(0,1,2)$ or $(a,b,c)\in (1,0,2)\rightarrow (-2,0,2)\text{ }(D\equiv 2\text{ mod 3})$ . Now we return to condition $\text{(c)}$ , which now implies that $(A,B,C)\equiv(-2,0,2)\pmod{3}$ . Now, we set $B=3k$ for increasing positive integer values of $k$ $B=0$ yields no solutions. $B=3$ gives $(A,B,C)=(1,3,5)$ , giving us $1$ solution. If $B=6$ , we get $2$ solutions, $(4,6,8)$ and $(1,6,11)$ . Proceeding in the manner, we see that if $B=48$ , we get 16 solutions. However, $B=51$ still gives $16$ solutions because $C_\text{max}=2B-1=101>100$ . Likewise, $B=54$ gives $15$ solutions. This continues until $B=96$ gives one solution. $B=99$ gives no solution. Thus, $N=1+2+\cdots+16+16+15+\cdots+1=2\cdot\frac{16(17)}{2}=16\cdot 17=\boxed{272}$
| 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 arithmetic sequences. Find $N$
|
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 $2B - 2a + 2y$ . The difference between them would be $3y$ . Since $p|\{(1), (2), (3)\}$ , then $p|3y$ . Since $p$ is prime, $p|3$ or $p|y$ . However, since $p > y$ , we must have $p|3$ , which means $p=3$
If $p=3$ , the only possible values of $(b, a, c)$ are $(0, 1, 2)$ . Plugging this into our differences, we get \begin{align*} &A-a = B-x-1 \hspace{4cm}(4)\\ &B - b = B \hspace{5.35cm}(5)\\ &C - c = B+x-2 \hspace{4cm}(6) \end{align*} The difference between $(4)$ and $(5)$ is $x+1$ , which should be divisible by 3. So $x \equiv 2 \mod 3$ . Also note that since $3|(5)$ $3|B$ . Now we can try different values of $x$ and $B$
When $x=2$ $B=3, 6, ..., 96 \Rightarrow 17$ triples.
When $x=5$ $B=6, 9, ..., 93\Rightarrow 15$ triples..
... and so on until
When $x=44$ $B=45\Rightarrow 1$ triple.
So the answer is $17 + 15 + \cdots + 1 = \boxed{272}$
| 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 $\text{3:00}$ AM, and $\text{7:50}$ at the former $\text{6:00}$ PM. After the conversion, a person who wanted to wake up at the equivalent of the former $\text{6:36}$ AM would set his new digital alarm clock for $\text{A:BC}$ , where $\text{A}$ $\text{B}$ , and $\text{C}$ are digits. Find $100\text{A}+10\text{B}+\text{C}$
|
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 $\text{6:36}$ AM, $6 \cdot 60+36=396$ normal minutes pass. This can be viewed as $\frac{396}{36}=11$ cycles of 36 normal minutes, so 11 cycles of 25 metric minutes pass. Adding $25 \cdot 11=275$ to $\text{0:00}$ gives $\text{2:75}$ , so the answer is $\boxed{275}$
| 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 $\text{3:00}$ AM, and $\text{7:50}$ at the former $\text{6:00}$ PM. After the conversion, a person who wanted to wake up at the equivalent of the former $\text{6:36}$ AM would set his new digital alarm clock for $\text{A:BC}$ , where $\text{A}$ $\text{B}$ , and $\text{C}$ are digits. Find $100\text{A}+10\text{B}+\text{C}$
|
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 equivalent of $\frac{11}{40}$ of a day is $\frac{11}{40}\cdot 1000=275$ metric minutes, which is equivalent to 2 metric hours and 75 metric minutes, so our answer is $\boxed{275}$ - mathleticguyyy
| 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 $2^0=1$ ). Doing this again, we get $\log_{2^b}(2^{1000})=2^a$ . Doing the process one more time, we finally eliminate all of the logs, getting ${(2^{b})}^{(2^a)}=2^{1000}$ . Using the property that ${(a^x)^{y}}=a^{xy}$ , we simplify to $2^{b\cdot2^{a}}=2^{1000}$ . Eliminating equal bases leaves $b\cdot2^a=1000$ . The largest $a$ such that $2^a$ divides $1000$ is $3$ , so we only need to check $1$ $2$ , and $3$ . When $a=1$ $b=500$ ; when $a=2$ $b=250$ ; when $a=3$ $b=125$ . Summing all the $a$ 's and $b$ 's gives the answer of $\boxed{881}$
| 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 see the next step, we factorize $1000$ $\dfrac{2^35^3}b=2^a.$ It should be clear that $a$ must be from $1$ to $3$ ; when $a=1$ $b=500$ ; when $a=2$ $b=250$ ; and finally, when $a=3$ $b=125.$ We sum all the pairs to get $\boxed{881}.$
| 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 down the $k$ -th centimeter. Suppose it takes $T$ seconds for the candle to burn down completely. Then $\tfrac{T}{2}$ seconds after it is lit, the candle's height in centimeters will be $h$ . Find $10h$
|
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 notice that $k=84$ , so the answer is $10(119-84)=\boxed{350}$
| 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 integers and $q$ is an integer that is not divisible by the square of any prime. Find $p+q+r$
|
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 rotation matrix is $\begin{bmatrix} \cos{300^{\circ}} & -\sin{300^{\circ}}\\ \sin{300^{\circ}} & \cos{300^{\circ}}\\ \end{bmatrix} = \begin{bmatrix} \frac{1}{2} & \frac{\sqrt{3}}{2}\\ -\frac{\sqrt{3}}{2} & \frac{1}{2}\\ \end{bmatrix}$ . Then $C' = \begin{bmatrix} \frac{1}{2} & \frac{\sqrt{3}}{2}\\ -\frac{\sqrt{3}}{2} & \frac{1}{2}\\ \end{bmatrix} \cdot \begin{bmatrix} 1\\ 2\sqrt{3}\\ \end{bmatrix} = \begin{bmatrix} \frac{7}{2}\\ \frac{\sqrt{3}}{2}\\ \end{bmatrix}$ . Thus, $C = (\frac{9}{2}, \frac{\sqrt{3}}{2})$ . Since the triangle is equilateral, the center of the triangle is the average of the coordinates of the vertices. Then $P = (\frac{1 + 2 + \frac{9}{2}}{3}, \frac{2\sqrt{3} + \frac{\sqrt{3}}{2}}{3}) = (\frac{5}{2}, \frac{5\sqrt{3}}{6})$ . Our answer is $\frac{5}{2} \cdot \frac{5\sqrt{3}}{6} = \frac{25\sqrt{3}}{12}$ $25 + 3 + 12 = \boxed{40}$
| 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 integers and $q$ is an integer that is not divisible by the square of any prime. Find $p+q+r$
|
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 the first transformation, with $\frac{\sqrt{3}}{2}$ the magnitude and the x and y switched– then multiply the new y by -1. Then, applying this transformation of $[3,\frac{-\sqrt{3}}{2}]$ we get that $C=(\frac{9}{2},\frac{\sqrt{3}}{2})$ which means that $P=(\frac{5}{2},\frac{5\sqrt{3}}{6})$ . Then our answer is $\boxed{40}$
| 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 of the two squares. Examine when $n^2=1000$ . Then, $n=10\sqrt{10}$ . One example way to estimate $\sqrt{10}$ follows.
Then, $n\approx 31.6$ $n^2<1000$ , so $n$ could be $31$ . Add 500 to get the first square and 501 to get the second. Then, the two integral squares are $531^2$ and $532^2$ . Checking, $531^2=281961$ and $532^2=283024$ $282,000$ straddles the two squares, which have a difference of 1063. The difference has been minimized, so $N$ is minimized $N=282000\implies\boxed{282}$
| 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 $x\ge \frac{999}{2}$ , so $x\ge 500$ . Now we need to find what range of numbers has to be square-free: $\overline{N000}\rightarrow \overline{N999}$ have to all be square-free.
Let us first plug in a few values of $x$ to see if we can figure anything out. $x=500$ $x^2=250000$ , and $(x+1)^2=251001$ . Notice that this does not fit the criteria, because $250000$ is a square, whereas $\overline{N000}$ cannot be a square. This means, we must find a square, such that the last $3$ digits are close to $1000$ , but not there, such as $961$ or $974$ . Now, the best we can do is to keep on listing squares until we hit one that fits. We do not need to solve for each square: remember that the difference between consecutive squares are $2x+1$ , so all we need to do is addition. After making a list, we find that $531^2=281961$ , while $532^2=283024$ . It skipped $282000$ , so our answer is $\boxed{282}$
| 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 determines if there are any squares between $\overline{N000}\rightarrow \overline{N999}$ . So we must find a square, $k^2$ , such that it is under $1000$ , but the next square is over $1000$ . We find that $k=31$ gives $k^2=961$ , and so $(k+1)^2=32^2=1024$ . We can be sure that this skips a thousand because the $1000k$ increments it up $1000$ each time. Now we can solve for $x$ $(500+31)^2=281961$ , while $(500+32)^2=283024$ . We skipped $282000$ , so the answer is $\boxed{282}$
| 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,
$N \geq \frac{(k+500)^2 + 1}{1000} = \frac{k^2 + 1}{1000} + k + 250$ , and
$N \leq \frac{(k+501)^2}{1000} - 1 = \frac{(k+1)^2}{1000} + k + 250.$
For $k=31$ , one can verify that $N = 282$ is the unique integer satisfying the inequalities.
For $k \leq 30$ $k + 250 < \frac{k^2 + 1}{1000} + k + 250 \leq N$ $\leq \frac{(k+1)^2}{1000} + k + 250 \leq \frac{(30+1)^2}{1000} + k + 250 < k + 251$
i.e., $k + 250 < N < k + 251$ , a contradiction.
Note $k \geq 32$ leads to larger $N$ (s).
Hence, the answer is $\boxed{282}$
| 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 similar assignment occurs at the end of the third hour. The group finishes the sorting in $3$ hours and $10$ minutes. Find the number of files sorted during the first one and a half hours of sorting.
|
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 $19$ divides $355+21t$ , AND as long as $t$ is a integer, $19$ must divide $2t+355$ . Now, we suppose that $19m=2t+355$ , similarly we get $t=\frac{19m-355}{2}$ , and so in order to get a minimum integer solution for $t$ , it is obvious that $m=19$ works. So we get $t=3$ and $x=22$ . One and a half hour's work should be $30x+15(x-t)$ , so the answer is $\boxed{945}$
| 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 similar assignment occurs at the end of the third hour. The group finishes the sorting in $3$ hours and $10$ minutes. Find the number of files sorted during the first one and a half hours of sorting.
|
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, $\angle BXO=90^{\circ}$ , because $BO=CO$ , and $\angle BYO=90^{\circ}$ , for similar reasons. Also notice that $\angle AOX=90^{\circ}$ .
Let us call $\angle BOY=\theta$ . Therefore, $\angle AOB=2\theta$ , and so $\angle BOX=90-2\theta$ . Let us label the radius of the circle $r$ . This means \[\sin{\theta}=\frac{BY}{r}=\frac{11}{r}\] \[\sin{(90-2\theta)}=\frac{BX}{r}=\frac{10}{r}\] Now we can use simple trigonometry to solve for $r$ .
Recall that $\sin{(90-\alpha)}=\cos(\alpha)$ : That means $\sin{(90-2\theta)}=\cos{2\theta}=\frac{10}{r}$ .
Recall that $\cos{2\alpha}=1-2\sin^2{\alpha}$ : That means $\cos{2\theta}=1-2\sin^2{\theta}=\frac{10}{r}$ .
Let $\sin{\theta}=x$ .
Substitute to get $x=\frac{11}{r}$ and $1-2x^2=\frac{10}{r}$ Now substitute the first equation into the second equation: $1-2\left(\frac{11}{r}\right)^2=\frac{10}{r}$ Multiplying both sides by $r^2$ and reordering gives us the quadratic \[r^2-10r-242=0\] Using the quadratic equation to solve, we get that $r=5+\sqrt{267}$ (because $5-\sqrt{267}$ gives a negative value), so the answer is $5+267=\boxed{272}$
| 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 $r=5+\sqrt{267}$ . So the solution is $5+267=\boxed{272}$
| 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. Hence by the Pythagorean Theorem with right $\triangle AED$
\[(AE)^2 + (ED)^2 = (AD)^2\] \[x^2 + 22^2 = d^2\]
From the above equations, we have: \[x^2 = d^2 - 22^2 = 20d + 22^2\] \[d^2 - 20d = 2\times22^2\] \[d^2 - 20d + 100 = 968+100 = 1068\] \[(d-10) = \sqrt{1068}\] \[d = \sqrt{1068} + 10 = 2\times(\sqrt{267}+5)\]
Since the radius is half the diameter, it is $\sqrt{267}+5$ , so the answer is $5+267 \Rightarrow \boxed{272}$
| 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 that $x=-10, x=0$ are two solutions for the equation, so we can factorize that into \[x(x+10)(x^2-10x-242)\] so we only need to find the solution for \[x^2-10x-242=0\] and we can get $x=(\sqrt{267}+5)$ is the desired answer for the problem, and our answer is $5+267 \Rightarrow \boxed{272}$ .～bluesoul
| 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 $AG = r \times \frac{2r}{22} = \frac{r^2}{11}$ . We then solve for $AB$ \[\frac{AD-BC}{AD} = \frac{2r-20}{2r} = \frac{AB}{AG}\] \[AB = \frac{r^2}{11} \times \frac{2r-20}{2r} = \frac{r^2-10r}{11}\] Remember that $AB=22$ as well, so $22 = \frac{r^2-10r}{11} \Rightarrow r^2-10r-242=0$ . Solving for $r$ gives $r=5+\sqrt{267}$ . So the solution is $5+267=\boxed{272}$
| 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 $\overline{AD}$ . We have $BF = 2 BH = 2 r \sin \theta = \dfrac{44 \sqrt{r^{2} - 121}}{r}$ . Using Pythagorean theorem on $\triangle BCF$ , to get $(\dfrac{44 \sqrt{r^{2} - 121}}{r})^{2} + 20^{2} = (2r)^{2}$ , or $\dfrac{44^{2}r^{2} - 44^{2} \cdot 121}{r^{2}} + 20^{2} = 4r^{4}$ . Multiplying by $r^{2}$ , we get $44^{2} r^{2} - 44^{2} \cdot 121 + 20^{2} r^{2} = 4r^{4}$ . Rearranging and simplifying, we get a quadratic in $r^{2}$ \[r^{4} - 584r^{2} + 242^{2} = 0 \text\] which gives us $r^{2} = 292 \pm 10\sqrt{267}$ . Because $r$ is in the form $p + \sqrt{q}$ , we know to choose the larger option, meaning $r^2 = 292 + 10\sqrt{267}$ , so $p\sqrt{q} = 5\sqrt{267}$ and $p^2 + q = 292$ . By inspection, we get $(p, q) = (5, 267)$ , so our answer is $5 + 267 = \boxed{272}$
| 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);label("$C$",C,NE);label("$D$",D,dir(0));label("$E$",E,SE);label("$F$",F,SW); [/asy]
We know that $AD=x$ is a diameter, hence $ABD$ and $ACD$ are right triangles. Let $AB=BC=22$ , and $CD=20.$ Hence, $ABD$ is a right triangle with legs $22,\sqrt{x^2-484},$ and hypotenuse, $x,$ and $ACD$ is a right triangle with legs $20, \sqrt{x^2-400},$ with hypotenuse $x$ . By Ptolemy's we have \[22(x+20)=\sqrt{x^2-400}\sqrt{x^2-484}\] .
We square both sides to get \[484(x+20)^2=(x^2-400)(x^2-484) \implies 484(x+20)=(x-20)(x^2-484) \implies 484x=x^3-20x^2-484x \implies x(x^2-20x-968)=0\]
We solve for $x$ via the Quadratic Formula and receive $x=10+2\sqrt{267}$ , but we must divide by $2$ since we want the radius, and hence $267+5=\boxed{272}.$ ~SirAppel
| 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 once. For example, a $1\times 1$ red tile followed by a $2\times 1$ green tile, a $1\times 1$ green tile, a $2\times 1$ blue tile, and a $1\times 1$ green tile is a valid tiling. Note that if the $2\times 1$ blue tile is replaced by two $1\times 1$ blue tiles, this results in a different tiling. Find the remainder when $N$ is divided by $1000$
|
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: $15\times 6+20\times 36+15\times 150+6\times 540+1\times 1806= 8106$
So the answer is $\boxed{106}$
| 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 once. For example, a $1\times 1$ red tile followed by a $2\times 1$ green tile, a $1\times 1$ green tile, a $2\times 1$ blue tile, and a $1\times 1$ green tile is a valid tiling. Note that if the $2\times 1$ blue tile is replaced by two $1\times 1$ blue tiles, this results in a different tiling. Find the remainder when $N$ is divided by $1000$
|
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 found using PIE giving $3^i-3\cdot2^i+3$ . Plus, we don't have to worry about the cases where $i=1$ or $I=2$ since they both give no solutions. So our equation becomes:
$\sum_{i=3}^{7} \left(\dbinom{6}{I}\right)\cdot\left(3^i-3\cdot2^i+3\right)$
Writing it all out and keeping the numbers small with mod 1000, we will eventually arrive at the answer of $\boxed{106}$
| 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 once. For example, a $1\times 1$ red tile followed by a $2\times 1$ green tile, a $1\times 1$ green tile, a $2\times 1$ blue tile, and a $1\times 1$ green tile is a valid tiling. Note that if the $2\times 1$ blue tile is replaced by two $1\times 1$ blue tiles, this results in a different tiling. Find the remainder when $N$ is divided by $1000$
|
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 an mx1 as $f_1(m)$
Now let's look at two colors. Let's see if we can get a two-tiling of an $(n+1) \times 1$ based off a $n \times 1$ . There are 2 cases we should consider:
1) The $n \times 1$ was a one-coloring and the block we are going to add consists of the second color. The number of ways we can do this is $2f_1(n)$
2) The $n \times 1$ was a two-color tiling so now we've got 3 cases to form the $(n+1) \times 1$ : We can either add a block of the first color, the second color, or we can adjoin a block to the last block in the $n \times 1$
This gives us $f_2(n+1)=2f_1(n)+3f_2(n)$
Time to tackle the 3-color tiling. Again, we split into 2 cases:
1) The $n \times 1$ was a two-color tiling, and the block we are adding is of the 3rd color. This gives $f_2(n)$ ways but we have to multiply by $3C2 = 3$ because we have to pick 2 different colors for the two-color tiling.
2) The $n \times 1$ was a 3-color tiling, and we have to consider what we can do with the block that we are adding. It can be any of the 3 colors, or we can adjoin it to whatever was the last block in the $n \times 1$ . This gives $4f_3(n)$
So in total we have $f_3(n+1)=3f_2(n)+4f_3(n)$
Finally, we just sorta bash through the computation to get $f_3(7)=8\boxed{106}$
| 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 once. For example, a $1\times 1$ red tile followed by a $2\times 1$ green tile, a $1\times 1$ green tile, a $2\times 1$ blue tile, and a $1\times 1$ green tile is a valid tiling. Note that if the $2\times 1$ blue tile is replaced by two $1\times 1$ blue tiles, this results in a different tiling. Find the remainder when $N$ is divided by $1000$
|
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 $x^k$ ways to color each of these divisions. Therefore, the total number of ways to divide and color the board is \begin{align*} \chi(n, x) &:= \sum_{k=1}^n {n-1 \choose k-1} x^k \\ &= x\sum_{k=0}^{n-1} {n-1 \choose k} x^k \\ &= x(x+1)^{n-1}. \end{align*}
In the given problem, we have $n=7$ . By PIE, we have \begin{align*} &\quad {3 \choose 3} \chi(7, 3) - {3 \choose 2} \, \chi(7, 2) + {3 \choose 1} \, \chi(7, 1) \\ &= 3 \cdot 4^6 - 3(2 \cdot 3^6) + 3(1 \cdot 2^6) \\ &= 8106 \rightarrow \boxed{106}
| 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 written in the form $\frac{a - b\sqrt{c}}{d}$ , where $a$ $b$ $c$ , and $d$ are positive integers, $a$ and $d$ are relatively prime, and $c$ is not divisible by the square of any prime. Find $a+b+c+d$
|
[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 $O (0,0)$ , then $B (\sqrt{13},0)$ , and $A(4+\sqrt{13},0)$
The equation for Circle O is $x^2+y^2=13$ , and let the slope of the line $AKL$ be $k$ , then the equation for line $AKL$ is $y=k(x-4-\sqrt{13})$
Then we get $(k^2+1)x^2-2k^2(4+\sqrt{13})x+k^2\cdot (4+\sqrt{13})^2-13=0$ . According to Vieta's Formulas , we get
$x_1+x_2=\frac{2k^2(4+\sqrt{13})}{k^2+1}$ , and $x_1x_2=\frac{(4+\sqrt{13})^2\cdot k^2-13}{k^2+1}$
So, $LK=\sqrt{1+k^2}\cdot \sqrt{(x_1+x_2)^2-4x_1x_2}$
Also, the distance between $B$ and $LK$ is $\frac{k\times \sqrt{13}-(4+\sqrt{13})\cdot k}{\sqrt{1+k^2}}=\frac{-4k}{\sqrt{1+k^2}}$
So the area $S=0.5ah=\frac{-4k\sqrt{(16+8\sqrt{13})k^2-13}}{k^2+1}$
Then the maximum value of $S$ is $\frac{104-26\sqrt{13}}{3}$
So the answer is $104+26+13+3=\boxed{146}$
| 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 written in the form $\frac{a - b\sqrt{c}}{d}$ , where $a$ $b$ $c$ , and $d$ are positive integers, $a$ and $d$ are relatively prime, and $c$ is not divisible by the square of any prime. Find $a+b+c+d$
|
[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 $D$
\[\frac{\triangle OKL}{\triangle BKL} = \frac{OC}{BD} = \frac{AO}{AB} = \frac{4+\sqrt{13}}{4}\]
Therefore, to maximize area of $\triangle BKL$ , we need to maximize area of $\triangle OKL$
\[\triangle OKL = \frac12 r^2 \sin{\angle KOL}\]
So when area of $\triangle OKL$ is maximized, $\angle KOL = \frac{\pi}{2}$
Eventually, we get \[\triangle BKL= \frac12 \cdot (\sqrt{13})^2\cdot(\frac{4}{4+\sqrt{13}})=\frac{104-26\sqrt{13}}{3}\]
So the answer is $104+26+13+3=\boxed{146}$
| 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 written in the form $\frac{a - b\sqrt{c}}{d}$ , where $a$ $b$ $c$ , and $d$ are positive integers, $a$ and $d$ are relatively prime, and $c$ is not divisible by the square of any prime. Find $a+b+c+d$
|
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 $\sin \alpha=\frac{d}{\sqrt{13}}, \cos \alpha=\frac{k}{\sqrt{13}}$ . Now, we have to maximize $\frac{52\sin \alpha \cos \alpha}{4+\sqrt{13}}=\frac{26\sin 2\alpha}{4+\sqrt{13}}$ , which is obviously reached when $\alpha=45^{\circ}$ , the answer is $\frac{104-26\sqrt{13}}{3}$ leads to $\boxed{146}$
| 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 written in the form $\frac{a - b\sqrt{c}}{d}$ , where $a$ $b$ $c$ , and $d$ are positive integers, $a$ and $d$ are relatively prime, and $c$ is not divisible by the square of any prime. Find $a+b+c+d$
|
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 $\implies$ The maximum possible area for $\triangle BKL$ and $\triangle OKL$ are at the same position of point $K$
$\triangle OKL$ has sides $OK = OL = \sqrt{13}\implies \max[\triangle OKL] = \frac {OK^2}{2} = \frac {13}{2}$
in the case $\angle KOL = 90^\circ.$ It is possible – if we rotate such triangle, we can find position when $A$ lies on $KL.$ \[\max[\triangle BKL] = \max[\triangle OKL] \cdot \frac {4}{4+\sqrt{13}} = \frac {26}{4+\sqrt{13}} \implies \boxed{146}\] vladimir.shelomovskii@gmail.com, vvsss
| 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 $f(x)$ equals an element on the middle tier.
There are $7$ choices for $c$ . Then for a given $k$ , there are $\tbinom6k$ ways to choose the elements on the middle tier, and then $k^{6-k}$ ways to draw arrows down from elements on the top tier to elements on the middle tier.
Thus $N=7\cdot\sum_{k=1}^6\tbinom6k\cdot k^{6-k}=7399$ , giving the answer $\boxed{399}$
| 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 can be $f(x)$ for all those values should satisfy $f(f(x))=c$ . Let's $S$ be that set of values. First things first, we must have $c$ to be part of $S$ , for the $S$ is part of the domain of $x$ . Since the values in $i\in S$ all satisfy $f(i) = c$ , we have $c$ to be a value that $f(x)$ can be. Now, for the elements other than $5$
If we have $k$ elements other than $5$ that can be part of $S$ , we will have $\binom{6}{k}$ ways to choose those values. There will also be $k$ ways for each of the elements in $A$ other than $c$ and those in set $S$ (for when function $f$ is applied on those values, we already know it would be $c$ ). There are $6-k$ elements in $A$ other than $c$ and those in set $S$ . Thus, there should be $6^{6-k}$ ways to match the domain $x$ to the values of $f(x)$ . Summing up all possible values of $k$ $[1,6]$ ), we have
\[\sum_{k=1}^6 k^{6-k} = 6\cdot 1 + 15\cdot 16 + 20\cdot 27 + 15\cdot 16 + 6\cdot 5 + 1) = 1057\]
Multiplying that by the original $7$ for the choice of $c$ , we have $7 \cdot 1057 = 7\boxed{399}.$
| 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 $c=-r|\omega|^2$ . But $\omega+\omega^*=2\Re{(\omega)}$ (i.e., adding the conjugates cancels the imaginary part). Therefore, to make $a$ an integer, $2\Re{(\omega)}$ must be an integer. Conversely, if $\omega+\omega^*=2\Re{(\omega)}$ is an integer, then $a,b,$ and $c$ are clearly integers. Therefore $2\Re{(\omega)}\in \mathbb{Z}$ is equivalent to the desired property. Let $\omega=\alpha+i\beta$
In this case, $\omega$ lies on a circle of radius $20$ in the complex plane. As $\omega$ is nonreal, we see that $\beta\ne 0$ . Hence $-20<\Re{(\omega)}< 20$ , or rather $-40<2\Re{(\omega)}< 40$ . We count $79$ integers in this interval, each of which corresponds to a unique complex number on the circle of radius $20$ with positive imaginary part.
In this case, $\omega$ lies on a circle of radius $13$ in the complex plane. As $\omega$ is nonreal, we see that $\beta\ne 0$ . Hence $-13<\Re{(\omega)}< 13$ , or rather $-26<2\Re{(\omega)}< 26$ . We count $51$ integers in this interval, each of which corresponds to a unique complex number on the circle of radius $13$ with positive imaginary part.
Therefore, there are $79+51=130$ choices for $\omega$ . We also have $4$ choices for $r$ , hence there are $4\cdot 130=520$ total polynomials in this case.
In this case, there are four possible real roots, namely $\pm 13, \pm20$ . Let $p$ be the number of times that $13$ appears among $r_1,r_2,r_3$ , and define $q,r,s$ similarly for $-13,20$ , and $-20$ , respectively. Then $p+q+r+s=3$ because there are three roots. We wish to find the number of ways to choose nonnegative integers $p,q,r,s$ that satisfy that equation. By balls and urns, these can be chosen in $\binom{6}{3}=20$ ways.
Therefore, there are a total of $520+20=\boxed{540}$ polynomials with the desired property.
| 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 two are the same.
This is obviously $\dbinom{4}{3}=4$
Sub-case 1.2: Exactly two are the same.
There are four ways to choose the root that will repeat twice, and three ways to choose the remaining root. For this sub-case, $4\cdot 3=12$
Sub-case 1.3: All three are the same.
This is obviously $4$
Thus for case one, we have $4+12+4=20$ polynomials in $S$ . We now have case two, which we state below.
Case 2: Two roots are imaginary and one is real.
Let these roots be $p-qi$ $p+qi$ , and $r$ . Then by Vieta's formulas
Since $a$ $b$ $c$ , and $r$ are integers, we have that $p=\frac{1}{2}k$ for some integer $k$ . Case two splits into two sub-cases now:
Sub-case 2.1: $|p-qi|=|p+qi|=13$ .
Obviously, $|p|<13$ . The $51$ cases in which $p$ is either $0,\pm\frac{1}{2},\pm\frac{2}{2},\pm\frac{3}{2},\ldots,\pm\frac{25}{2}$ are acceptable. Each can pair with one value of $q$ and four values of $r$ , adding $51\cdot 4=204$ polynomials to $S$
Sub-case 2.2: $|p-qi|=|p+qi|=20$ .
Obviously, $|p|<20$ . Here, the $79$ cases in which $p$ is either $0,\pm\frac{1}{2},\pm\frac{2}{2},\pm\frac{3}{2},\ldots,\pm\frac{39}{2}$ are acceptable. Again, each can pair with a single value of $q$ as well as four values of $r$ , adding $79\cdot 4=316$ polynomials to $S$
Thus for case two, $204+316=520$ polynomials are part of $S$
All in all, $20+204+316=\boxed{540}$ polynomials can call $S$ home.
| 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 $n$ is divided by $k_M$ $(n+1)/2$ for odd $n$ and $k_M$ $(n+2)/2$ for even $n$
Using the lemma above:
$\sum\limits_{n=20}^{100} F(n) = \sum\limits_{n=20}^{100} [(n - 6)/3 + (2/3)*n \pmod{3}]$ $= [(120*81/2)/3 - 2*81 + (2/3)*81] = 1512$
So the answer is $\boxed{512}$
| 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 \[\cos^2 C + \cos^2 A + 2 \sin C \sin A \cos B = \frac{p-q\sqrt{r}}{s},\] where $p+q$ and $s$ are relatively prime and $r$ is not divisible by the square of any prime. Find $p+q+r+s$
|
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-\sin^2B + 2\sin A\sin B\cos C \\ &= 2-(\sin^2A + \sin^2B - 2\sin A\sin B\cos C) \end{align*}
Now we can use the Law of Cosines to simplify this:
\[= 2-\sin^2C\]
Therefore: \[\sin C = \sqrt{\dfrac{1}{8}},\cos C = \sqrt{\dfrac{7}{8}}.\] Similarly, \[\sin A = \sqrt{\dfrac{4}{9}},\cos A = \sqrt{\dfrac{5}{9}}.\] Note that the desired value is equivalent to $2-\sin^2B$ , which is $2-\sin^2(A+C)$ . All that remains is to use the sine addition formula and, after a few minor computations, we obtain a result of $\dfrac{111-4\sqrt{35}}{72}$ . Thus, the answer is $111+4+35+72 = \boxed{222}$
| 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 \[\cos^2 C + \cos^2 A + 2 \sin C \sin A \cos B = \frac{p-q\sqrt{r}}{s},\] where $p+q$ and $s$ are relatively prime and $r$ is not divisible by the square of any prime. Find $p+q+r+s$
|
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 get: \[2 \cos^2 A + \cos^2 B + \cos^2 C + 2 \sin A( \sin B \cos C + \sin C \cos B) = \frac{15}{8} + x\] or \[2 \cos^2 A + \cos^2 B + \cos^2 C + 2 \sin A \sin (B+C) = \frac{15}{8} + x\] or \[2 \cos^2 A + \cos^2 B + \cos^2 C + 2 \sin ^2 A = \frac{15}{8} + x\] or \[\cos^2 B + \cos^2 C = x - \frac{1}{8} \text{ ------ (4)}\]
Similarly adding (2) and (3) we get: \[\cos^2 A + \cos^2 B = x - \frac{4}{9} \text{ ------ (5)}\] Similarly adding (1) and (2) we get: \[\cos^2 A + \cos^2 C = \frac{14}{9} - \frac{1}{8} \text{ ------ (6)}\]
And (4) - (5) gives: \[\cos^2 C - \cos^2 A = \frac{4}{9} - \frac{1}{8} \text{ ------ (7)}\]
Now (6) - (7) gives: $\cos^2 A = \frac{5}{9}$ or $\cos A = \sqrt{\dfrac{5}{9}}$ and $\sin A = \frac{2}{3}$ so $\cos C$ is $\sqrt{\dfrac{7}{8}}$ and therefore $\sin C$ is $\sqrt{\dfrac{1}{8}}$
Now $\sin B = \sin(A+C)$ can be computed first and then $\cos^2 B$ is easily found.
Thus $\cos^2 B$ and $\cos^2 C$ can be plugged into (4) above to give x = $\dfrac{111-4\sqrt{35}}{72}$
Hence the answer is = $111+4+35+72 = \boxed{222}$
| 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 \[\cos^2 C + \cos^2 A + 2 \sin C \sin A \cos B = \frac{p-q\sqrt{r}}{s},\] where $p+q$ and $s$ are relatively prime and $r$ is not divisible by the square of any prime. Find $p+q+r+s$
|
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 $\cos^2 A + \cos^2 B + 2 \sin^2 A \sin^2 B - 2 \sin A \sin B \cos A \cos B = \frac{15}{8}$
Using the double angle formula $\cos^2 k = \frac{\cos (2k) + 1}{2}$ , we can substitute for each of the squares $\cos^2 A$ and $\cos^2 B$ . Next we can use the Pythagorean identity on the $\sin^2 A$ and $\sin^2 B$ terms. Lastly we can use the sine double angle to simplify.
$\cos^2 A + \cos^2 B + 2(1 - \cos^2 A)(1 - \cos^2 B) - \frac{1}{2} \cdot \sin 2A \sin 2B = \frac{15}{8}$
Expanding and canceling yields, and again using double angle substitution,
$1 + 2 \cdot \frac{\cos (2A) + 1}{2} \cdot \frac{\cos (2B) + 1}{2} - \frac{1}{2} \cdot \sin 2A \sin 2B = \frac{15}{8}$
Further simplifying yields:
$\frac{3}{2} + \frac{\cos 2A \cos 2B - \sin 2A \sin 2B}{2} = \frac{15}{8}$
Using cosine angle addition formula and simplifying further yields, and applying the same logic to Equation $2$ yields:
$\cos (2A + 2B) = \frac{3}{4}$ and $\cos (2B + 2C) = \frac{1}{9}$
Substituting the identity $\cos (2A + 2B) = \cos(2C)$ , we get:
$\cos (2C) = \frac{3}{4}$ and $\cos (2A) = \frac{1}{9}$
Since the third expression simplifies to the expression $\frac{3}{2} + \frac{\cos (2A + 2C)}{2}$ , taking inverse cosine and using the angles in angle addition formula yields the answer, $\frac{111 - 4\sqrt{35}}{72}$ , giving us the answer $\boxed{222}$
| 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 \[\cos^2 C + \cos^2 A + 2 \sin C \sin A \cos B = \frac{p-q\sqrt{r}}{s},\] where $p+q$ and $s$ are relatively prime and $r$ is not divisible by the square of any prime. Find $p+q+r+s$
|
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 simplify to $2=sin^2C+\frac{15}{8}$ . Similarly, we can generalize $2=sin^2A+\frac{14}{9}$ . After solving, we can get that $sinA=\frac{2}{3}; cosA=\frac{\sqrt{5}}{3}; sinC=\frac{\sqrt{2}}{4}; cosC=\frac{\sqrt{14}}{4}$ Assume the value we are looking for is $x$ , we get $sin^2B+x=2$ , while $sinB=sin(180^{\circ}-A-C)=sin(A+C)$ which is $\frac{2\sqrt{14}+\sqrt{10}}{12}$ , so $x=\frac{111 - 4\sqrt{35}}{72}$ , giving us the answer $\boxed{222}$ .~bluesoul
| 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 the first, last, and middle terms of the original sequence.
|
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 $\frac{715}{11} = 65,$ and the first and last terms average to this middle term, so the desired sum is simply three times the middle term, or $\boxed{195}.$
| 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 the first, last, and middle terms of the original sequence.
|
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 around the mean, our final answer is $65 \cdot 3 = \boxed{195}$
| 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 the first, last, and middle terms of the original sequence.
|
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 problem. Assume WLOG that $a_1 = 60$ and $d = 1$ . Then the first term of the corresponding arithmetic sequence will be $60$ , the sixth (middle) term will be $65$ , and the eleventh (largest) term will be $70$ . Thus, our final answer is $60 + 65 + 70 = \boxed{195}$
| 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 exactly one person receives the type of meal ordered by that person.
|
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 are $2$ ways to serve their meals. Thus, there are $(3\cdot2)^3=\boxed{216}$ ways to serve their meals.
| 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 route, he ties Sparky to the post and begins walking. When Butch reaches Sparky, he rides until he passes Sundance, then leaves Sparky at the next hitching post and resumes walking, and they continue in this manner. Sparky, Butch, and Sundance walk at $6,$ $4,$ and $2.5$ miles per hour, respectively. The first time Butch and Sundance meet at a milepost, they are $n$ miles from Dodge, and they have been traveling for $t$ minutes. Find $n + t$
|
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 per mile:
\[\frac{a}{6} + \frac{n-a}{4} = \frac{n-a}{6} + \frac{2a}{5} \rightarrow a = \frac{5}{19}n.\]
The smallest possible integral value of $n$ is $19$ , so we plug in $n = 19$ and $a = 5$ and get $t = \frac{13}{3}$ hours, or $260$ minutes. So our answer is $19 + 260 = \boxed{279}$
| 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 of such numbers. With the $10$ in place, the seven remaining $1$ 's can be distributed in any of the remaining $11$ spaces, so the answer is ${11 \choose 7} = \boxed{330}$
| 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 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 $a + 5 \cdot \frac{3a}{5} + 10 \cdot \frac{4a}{5} = 3360 \rightarrow a = \frac{3360}{12} = \boxed{280}.$
| 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 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 $5(56) = \boxed{280}.$
| 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 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 $a + B + C + D = 3360$ $D = 1120$ $a = \boxed{280}$
| 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 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 $\boxed{280}$
| 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 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 $\boxed{170}$
| 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 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
| 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 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 $\boxed{170}$
| 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 $|x| + |y| \le 100$ 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 $34 \cdot 20 + 33 \cdot 21 = 1373 \rightarrow \boxed{373}$
| 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, 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, $h^2=x^2+y^2=\frac{3}{2}\cdot250=\boxed{375}$
| 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
$m^2 + n^2 + p^2 + q^2 + r^2 + t^2 = 250$
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 = $\boxed{375}$
| 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 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 \[(\frac{250}{6}*\frac{9}{4}*2)^2=\boxed{375}\]
| 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 $\Vec{a} + \Vec{b} + \Vec{c} = \Vec{0}$
$\Rightarrow 0 = \Vec{0}\cdot \Vec{0} = (\Vec{a} + \Vec{b} + \Vec{c})\cdot (\Vec{a} + \Vec{b} + \Vec{c}) = \Vec{a}\cdot \Vec{a} + \Vec{b}\cdot \Vec{b} + \Vec{c}\cdot \Vec{c} + 2(\Vec{a}\cdot \Vec{b} + \Vec{a}\cdot \Vec{c} + \Vec{b}\cdot \Vec{c})$
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 $h^2 = |a-c|^2$
$\Rightarrow h^2 = \lVert \Vec{a} - \Vec{c} \rVert^2 = (\Vec{a} - \Vec{c})\cdot (\Vec{a} - \Vec{c}) = \Vec{a} \cdot \Vec{a} + \Vec{c}\cdot \Vec{c} - \Vec{a}\cdot \Vec{c} - \Vec{a}\cdot \Vec{c} = \Vec{a} \cdot \Vec{a} + \Vec{c}\cdot \Vec{c} - 2\Vec{a}\cdot \Vec{c}$
$\Rightarrow 0 = \Vec{a}\cdot \Vec{a} + \Vec{b}\cdot \Vec{b} + \Vec{c}\cdot \Vec{c} + 2(\Vec{a}\cdot \Vec{b} + \Vec{a}\cdot \Vec{c} + \Vec{b}\cdot \Vec{c}) = 250 + 2(\Vec{a}\cdot \Vec{b} + \Vec{a}\cdot \Vec{c} + \Vec{b}\cdot \Vec{c})$
$\Rightarrow -250 = 2(\Vec{a}\cdot \Vec{b} + \Vec{a}\cdot \Vec{c} + \Vec{b}\cdot \Vec{c})$
$\Rightarrow -125 = \Vec{a}\cdot \Vec{b} + \Vec{a}\cdot \Vec{c} + \Vec{b}\cdot \Vec{c}$
$\Rightarrow -125 = -125 + 0 = (\Vec{a}\cdot \Vec{b} + \Vec{a}\cdot \Vec{c} + \Vec{b}\cdot \Vec{c}) + (\Vec{a}\cdot \Vec{c} + \Vec{b} \cdot \Vec{b} - \Vec{a} \cdot \Vec{b} - \Vec{b} \cdot \Vec{c}) = 2\Vec{a}\cdot \Vec{c} + \Vec{b} \cdot \Vec{b}$
$\Rightarrow 125 = - \Vec{b} \cdot \Vec{b} - 2\Vec{a}\cdot \Vec{c}$
$\Rightarrow 375 = 250 + 125 = \Vec{a}\cdot \Vec{a} + \Vec{b}\cdot \Vec{b} + \Vec{c}\cdot \Vec{c} - \Vec{b} \cdot \Vec{b} - 2\Vec{a}\cdot \Vec{c} = \Vec{a}\cdot \Vec{a} + \Vec{c}\cdot \Vec{c} - 2\Vec{a}\cdot \Vec{c} = h^2$
$\Rightarrow h^2 = \boxed{375}$
| 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 $(a+b+c)(\overline{a+b+c}) = |a|^2 + |b|^2 + |c|^2 + a\overline{b} + \overline{a}b + b\overline{c} + \overline{b}c + c\overline{a} + \overline{c}a = 250 + a\overline{b} + \overline{a}b + b\overline{c} + \overline{b}c + c\overline{a} + \overline{c}a = 0$ 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
| 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 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 $998 - (499 + 333 - 166) = \boxed{332}$
| 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 $\frac{166-1}{5}+1 = \boxed{34}$
| 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: $(100,1), (97,6), (94,11)...$ Because the solutions are only positive, we can generate only 33 more solutions, so in total we have $33+1=\boxed{34}$ solutions.
| 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 $a_1 \cdot r^8 = a_1 \cdot (r^4)^2=27\cdot {\left(\frac{11}{3}\right)}^2=27\cdot \frac{121} 9=\boxed{363}$
| 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 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, $V=\frac{1}{3}bh=\frac{1}{3}(225)(10)=\boxed{750}$
| 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 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, $c+d = -375 + 500 = \boxed{125}$
| 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
$z_{2}z_{1}^3=125\sqrt{5}(\cos({\theta_{2}+3\theta_{1}})+i\sin({\theta_{2}+3\theta_{1}}))$
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 $z_{1}^4=625(-\frac{3}{5}+\frac{4i}{5})=-375+500i$
$-375+500=\boxed{125}$
| 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:
$\sin x = 3\sin y$
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 $49+58=\boxed{107}$
| 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 $\frac{\cos^2x}{\cos^2y} =\frac{1}{4}.$ 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 $\boxed{107}.$
| 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 $\sqrt{35}$ 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 $\frac{3}{2} + \frac{-19}{29} = \frac{49}{58} \implies \boxed{107}$
| 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 $\left(a, b\right) = \left(3\sqrt{\frac{3}{35}},\sqrt{\frac{3}{35}}\right).$ 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 $\frac{49}{58} \rightarrow 49 + 58 = \boxed{107}$
| 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, \[n = \left(a + \frac{b}{c}\right) \left\lfloor a +\frac{b}{c} \right\rfloor \Rightarrow n = \left(a + \frac{b}{c}\right)a = a^2 + \frac{ab}{c}\]
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 $1+2+3+...+31 = \frac{31 \cdot 32}{2} = \boxed{496}.$
| 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 $31*32 < 1000$ . It follows that the answer is $\sum_{k=1}^{31} (k+1)k-k^2 = \sum_{k=1}^{31} k = \frac{31\cdot 32}{2} = \boxed{496}.$
| 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 exists for $l=1$ ! 2 for $l=2$ and so on. Therefore our final answer is $31+30+\dots+1= \boxed{496}$
| 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 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 $C E_1^2 + C E_2^2 + C E_2^2 + C E_4^2 = 11 + 11 + C E_2^2 + C E_4^2 = 11 + 11 + x^2 + y^2 = 11 + 11 + 655 = \boxed{677}$
| 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 $AF^2 = \frac mn$ , 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 $900+19=\boxed{919}$
| 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 $AF^2 = \frac mn$ , 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 $\boxed{919}$
| 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 $AF^2 = \frac mn$ , 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 $900+19=\boxed{919}$
| 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 $AF^2 = \frac mn$ , 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 $900+19=\boxed{919}$
| 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 $AF^2 = \frac mn$ , 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 \[AF = AM' = \frac {AD \cdot AE}{AM} = \frac {4b c}{\sqrt{2 b^2 + 2 c^2 – a^2}} = \frac {4 \cdot 3 \cdot 5}{\sqrt{ 50 + 18 – 49}} = \frac {30}{\sqrt{19}} \implies \boxed{919}.\]
| 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 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 $80 + 2 + 3 = \boxed{85}$
| 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 $m \sqrt{n} - p$ , 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:
$3+21+12=\boxed{36}$
| 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 $m \sqrt{n} - p$ , 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: $3+21+12=\boxed{36}$
| 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 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 $2(2 \times 6 \times 6)=\boxed{144}$
| 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 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.
| 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 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 $40+221+57=261+57=\boxed{318}$
| 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 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 $h = \frac{40\sqrt{221}}{57} \rightarrow \boxed{318}$
| 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 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 $h = \frac{2A}{57} = \frac{40\sqrt{221}}{57}\rightarrow \boxed{318}$
| 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 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. $24\cot ^2 x=24\frac{\cos ^2 x}{\sin ^2 x}=24\left(\frac{\frac{8}{9}}{\frac{1}{9}}\right)=24(8)=\boxed{192}$
| 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 $\sqrt[3]{576} \approx 8$ , we can easily see that $a = 8$ is a solution. Thus, the answer is $24\cot^2x = 24a = 24 \cdot 8 = \boxed{192}$
| 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: 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, $127+376=\boxed{503}$
| 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 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 $127 + 376 = \boxed{503}$
| 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, we can place the three groups of men in between women. We can think of the groups of men as dividers splitting up the $n$ women. Since there are $n+1$ possible places to insert the dividers, and we need to choose any three of these locations, we have $\dbinom{n+1}{3}$ ways.
The second, third, and fourth cases are like the first, only that we need to insert two dividers among the $n+1$ possible locations. Each gives us $\dbinom{n+1}{2}$ ways, for a total of $3\dbinom{n+1}{2}$ ways.
The last case gives us $\dbinom{n+1}{1}=n+1$ ways.
Therefore, the total number of possible ways where there are no isolated men is
\[\dbinom{n+1}{3}+3\dbinom{n+1}{2}+(n+1).\]
The total number of ways where there is a group of at least four men together is the sum of the third, fourth, and fifth case, or
\[2\dbinom{n+1}{2}+(n+1).\]
Thus, we want to find the minimum possible value of $n$ where $n$ is a positive integer such that
\[\dfrac{2\dbinom{n+1}{2}+(n+1)}{\dbinom{n+1}{3}+3\dbinom{n+1}{2}+(n+1)}\le\dfrac{1}{100}.\]
After simplification, we arrive at \[\dfrac{6(n+1)}{n^2+8n+6}\le\dfrac{1}{100}.\]
Simplifying again, we see that we seek the smallest positive integer value of $n$ such that $n(n-592)\ge594$ . Clearly $n>592$ , or the left side will not even be positive; we quickly see that $n=593$ is too small but $n=\boxed{594}$ satisfies the inequality.
| 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 integers. Find $r+s+t$
|
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 $ax+by+cz+d$ . So, by looking at the three vertices, we have $10a+d=10, 10b+d=11, 10c+d=12$ , and by rearranging and summing, \[(10-d)^2+(11-d)^2+(12-d)^2= 100\cdot(a^2+b^2+c^2)=100.\]
Solving the equation is easier if we substitute $11-d=y$ , to get $3y^2+2=100$ , or $y=\sqrt {98/3}$ . The distance from the origin to the plane is simply $d$ , which is equal to $11-\sqrt{98/3} =(33-\sqrt{294})/3$ , so $33+294+3=\boxed{330}$
| 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 integers. Find $r+s+t$
|
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 $\text{centroid}=(10+11+12)/3=11$ . Designate the midpoint of $BD$ as $M$ . Notice that median $CM$ is parallel to the plane because the $\text{centroid}$ and vertex $C$ have the same $z$ coordinate, $11$ , and the median contains $C$ and the $\text{centroid}$ . We seek the angle $\theta$ of the line: $(1)$ through the centroid $(2)$ perpendicular to the plane formed by $\triangle BCD$ $(3)$ with the plane under the cube. Since the median is parallel to the plane, this orthogonal line is also perpendicular $\textit{in slope}$ to $BD$ . Since $BD$ makes a $2-14-10\sqrt{2}$ right triangle, the orthogonal line makes the same right triangle rotated $90^\circ$ . Therefore, $\sin\theta=\frac{14}{10\sqrt{2}}=\frac{7\sqrt{2}}{10}$
It is also known that the centroid of $\triangle BCD$ is a third of the way between vertex $A$ and $H$ , the vertex farthest from the plane. Since $AH$ is a diagonal of the cube, $AH=10\sqrt{3}$ . So the distance from the $\text{centroid}$ to $A$ is $10/\sqrt{3}$ . So, the $\Delta z$ from $A$ to the centroid is $\frac{10}{\sqrt{3}}\sin\theta=\frac{10}{\sqrt{3}}\left(\frac{7\sqrt{2}}{10}\right)=\frac{7\sqrt{6}}{3}$
Thus the distance from $A$ to the plane is $11-\frac{7\sqrt{6}}{3}=\frac{33-7\sqrt{6}}{3}=\frac{33-\sqrt{294}}{3}$ , and $33+294+3=\boxed{330}$
| 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 arithmetic sequence, the median and average of them is $20$ . Since there are even number of exterior angles, the middle two must be $19^\circ$ and $21^\circ$ , and the difference between terms must be $2$ . Check to make sure the smallest exterior angle is greater than $0$ $19-2(8)=19-16=3^\circ$ . It is, so the greatest exterior angle is $21+2(8)=21+16=37^\circ$ and the smallest interior angle is $180-37=\boxed{143}$
| 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 $18a+153d = 2880 ^\circ,$ or $2a+17d = 320^\circ.$ The smallest value of $d$ that satisfies this is $d=2,$ so $a=143.$ Other values of $d$ and $a$ satisfy that equation, but if we tried any of them the last angle would be greater than $180,$ so the only value of $a$ that works is $a=\boxed{143}$
| 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 is $\boxed{143}$
| 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 positive integers. Find $m+n$
|
[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("M",M,NW)); MP("20",(B+D)/2,ENE); MP("11",(C+D)/2,ENE); [/asy] Let $D'$ be on $\overline{AC}$ such that $BP \parallel DD'$ . It follows that $\triangle BPC \sim \triangle DD'C$ , so \[\frac{PC}{D'C} = 1 + \frac{BD}{DC} = 1 + \frac{AB}{AC} = \frac{31}{11}\] by the Angle Bisector Theorem . Similarly, we see by the Midline Theorem that $AP = PD'$ . Thus, \[\frac{CP}{PA} = \frac{1}{\frac{PD'}{PC}} = \frac{1}{1 - \frac{D'C}{PC}} = \frac{31}{20},\] and $m+n = \boxed{51}$
| 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 positive integers. Find $m+n$
|
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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