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6,501
|
https://artofproblemsolving.com/wiki/index.php/2003_AIME_II_Problems/Problem_12
| 3
|
The members of a distinguished committee were choosing a president, and each member gave one vote to one of the 27 candidates. For each candidate, the exact percentage of votes the candidate got was smaller by at least 1 than the number of votes for that candidate. What was the smallest possible number of members of the committee?
|
Let $n$ be the total number of people in the committee, and $a_i$ be the number of votes candidate $i$ gets where $1 \leq i \leq 27$ . The problem tells us that \[\frac{100a_i}{n} \leq a_i - 1 \implies 100a_i \leq na_i - n \implies a_i \geq \frac{n}{n-100}.\] Therefore, \[\sum^{27}_{i=1} a_i = n \geq \sum^{27}_{i=1} \frac{n}{n-100} = \frac{27n}{n-100},\] and so $n(n-127) \geq 0 \implies n \geq 127$ . Trying $n = 127$ , we get that \[a_i \geq \frac{127}{27} \approx 4.7 \implies a_i \geq 5 \implies \sum^{27}_{a_i} a_i \geq 5 \cdot 27 = 135 \geq 127,\] a contradiction. Bashing out a few more, we find that $\boxed{134}$ works perfectly fine.
| 134
|
6,502
|
https://artofproblemsolving.com/wiki/index.php/2003_AIME_II_Problems/Problem_13
| 1
|
A bug starts at a vertex of an equilateral triangle. On each move, it randomly selects one of the two vertices where it is not currently located, and crawls along a side of the triangle to that vertex. Given that the probability that the bug moves to its starting vertex on its tenth move is $m/n,$ where $m$ and $n$ are relatively prime positive integers, find $m + n.$
|
Let $P_n$ represent the probability that the bug is at its starting vertex after $n$ moves. If the bug is on its starting vertex after $n$ moves, then it must be not on its starting vertex after $n-1$ moves. At this point it has $\frac{1}{2}$ chance of reaching the starting vertex in the next move. Thus $P_n=\frac{1}{2}(1-P_{n-1})$ $P_0=1$ , so now we can build it up:
$P_1=0$ $P_2=\frac{1}{2}$ $P_3=\frac{1}{4}$ $P_4=\frac{3}{8}$ $P_5=\frac{5}{16}$ $P_6=\frac{11}{32}$ $P_7=\frac{21}{64}$ $P_8=\frac{43}{128}$ $P_9=\frac{85}{256}$ $P_{10}=\frac{171}{512}$
Thus the answer is $171+512=$ $\boxed{683}$
| 683
|
6,503
|
https://artofproblemsolving.com/wiki/index.php/2003_AIME_II_Problems/Problem_13
| 2
|
A bug starts at a vertex of an equilateral triangle. On each move, it randomly selects one of the two vertices where it is not currently located, and crawls along a side of the triangle to that vertex. Given that the probability that the bug moves to its starting vertex on its tenth move is $m/n,$ where $m$ and $n$ are relatively prime positive integers, find $m + n.$
|
Consider there to be a clockwise and a counterclockwise direction around the triangle. Then, in order for the ant to return to the original vertex, the net number of clockwise steps must be a multiple of 3, i.e., $\#CW - \#CCW \equiv 0 \pmod{3}$ . Since $\#CW + \#CCW = 10$ , it is only possible that $(\#CW,\, \#CCW) = (5,5), (8,2), (2,8)$
In the first case, we pick $5$ out of the ant's $10$ steps to be clockwise, for a total of ${10 \choose 5}$ paths. In the second case, we choose $8$ of the steps to be clockwise, and in the third case we choose $2$ to be clockwise. Hence the total number of ways to return to the original vertex is ${10 \choose 5} + {10 \choose 8} + {10 \choose 2} = 252 + 45 + 45 = 342$ . Since the ant has two possible steps at each point, there are $2^{10}$ routes the ant can take, and the probability we seek is $\frac{342}{2^{10}} = \frac{171}{512}$ , and the answer is $\boxed{683}$
| 683
|
6,504
|
https://artofproblemsolving.com/wiki/index.php/2003_AIME_II_Problems/Problem_13
| 3
|
A bug starts at a vertex of an equilateral triangle. On each move, it randomly selects one of the two vertices where it is not currently located, and crawls along a side of the triangle to that vertex. Given that the probability that the bug moves to its starting vertex on its tenth move is $m/n,$ where $m$ and $n$ are relatively prime positive integers, find $m + n.$
|
Label the vertices of the triangle $A,B,C$ with the ant starting at $A$ . We will make a table of the number of ways to get to $A,B,C$ in $n$ moves $n\leq10$ . The values of the table are calculated from the fact that the number of ways from a vertex say $A$ in $n$ steps equals the number of ways to get to $B$ in $n-1$ steps plus the number of ways to get to $C$ in $n-1$ steps.
\[\begin{array}{|l|ccc|} \multicolumn{4}{c}{\text{Table}}\\\hline \text{Step}&A&B&C \\\hline 1 &0 &1 &1 \\ 2 &2 &1 &1 \\ 3 &2 &3 &3\\ \vdots &\vdots&\vdots&\vdots \\ 10 &342 &341 &341 \\\hline \end{array}\] Therefore, our answer is $512+171=\boxed{683}.$
| 683
|
6,505
|
https://artofproblemsolving.com/wiki/index.php/2003_AIME_II_Problems/Problem_13
| 4
|
A bug starts at a vertex of an equilateral triangle. On each move, it randomly selects one of the two vertices where it is not currently located, and crawls along a side of the triangle to that vertex. Given that the probability that the bug moves to its starting vertex on its tenth move is $m/n,$ where $m$ and $n$ are relatively prime positive integers, find $m + n.$
|
As a note, do NOT do this on the exam as it will eat up your time, but feel free to experiment around with this if you have a good enough understanding of linear algebra. This writeup will be extremely lengthy because I am assuming just very basic concepts of linear algebra. The concepts here extend to higher levels of mathematics, so feel free to explore more in depth so that you can end up solving almost any variation of this classic problem.
There are a possible of 3 states for this problem, so we can model the problem as a stochastic process. The resulting process has a transition matrix of:
\[\hat{T} = \begin{bmatrix} 0 & 0.5 & 0.5\\ 0.5 & 0 & 0.5\\ 0.5 & 0.5 & 0 \end{bmatrix}\]
We aim to diagonalize this transition matrix to make it easier to exponentiate by converting it into what's known as it's Jordan Canonical Form.
In order to do this, we must extract the eigenvalues and eigenvectors of the matrix. The eigenpolynomial for this matrix is obtained by calculating this matrix's determinant with $0-\lambda$ about it's main diagonal like so: \[\hat{T}_{\lambda} = \begin{bmatrix} 0-\lambda & 0.5 & 0.5\\ 0.5 & 0-\lambda & 0.5\\ 0.5 & 0.5 & 0-\lambda \end{bmatrix}\]
We have the matrix's eigenpolynomial to be $\lambda^3 - \frac{3\lambda}{4} + \frac{1}{4}$ , and extracting eigenvalues by setting the polynomial equal to $0$ , we have 2 eigenvalues: $\lambda_1 = 1$ of multiplicity 1, and $\lambda_2 = -\frac{1}{2}$ of multiplicity 2. To extract the eigenvectors, we must assess the kernel of this matrix (also known as the null space), or the linear subspace of the domain of $\hat{T}$ where everything gets mapped to the null vector.
We first do this for $\lambda_1$ . Taking $-\lambda_1$ across the diagonals to get $\hat{T}_{\lambda_1} = \begin{bmatrix} -1 & 0.5 & 0.5\\ 0.5 & -1 & 0.5\\ 0.5 & 0.5 & -1 \end{bmatrix}$ , we first reduce it to reduced row echelon form, which is $\begin{bmatrix} 1 & 0 & -1\\ 0 & 1 & -1\\ 0 & 0 & 0 \end{bmatrix}$ . From here, we compute the kernel by setting \[\begin{bmatrix} 1 & 0 & -1\\ 0 & 1 & -1\\ 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} x_1\\ x_2\\ x_3 \end{bmatrix} = 0\] . So if we take our free variable $x_3 = 0 = t$ , then that means that in the same fashion, $x_1 - x_1 = x_2 - x_2 = 0 = t$ , so hence, the kernel of $\hat{T}_{\lambda} = \begin{bmatrix} t\\ t\\ t \end{bmatrix}$ , or more simply, $t\begin{bmatrix} 1\\ 1\\ 1 \end{bmatrix}$ $\begin{bmatrix} 1\\ 1\\ 1 \end{bmatrix}$ is the eigenvector corresponding to $\lambda_1$ . We do the same computations for our second unique eigenvalue, but I will save the computation to you. There are actually 2 eigenvectors for $\lambda_2$ , because the reduced row echelon form for $\hat{T}_{\lambda_2}$ has 2 free variables instead of 1, so our eigenvectors for $\lambda_2$ are $\begin{bmatrix} -1\\ 1\\ 0 \end{bmatrix}, \begin{bmatrix} -1\\ 0\\ 1 \end{bmatrix}$ . We are now ready to begin finding the Jordan canonical form
In linear algebra, the JCF (which also goes by the name of Jordan Normal Form) is an upper triangular matrix representing a linear transformation over a finite-dimensional complex vector space. Any square matrix that has a Jordan Canonical Form has its field of coefficients extended into a field containing all it's eigenvalues. You can find more information about them on google, as well as exactly how to find them but for now let's get on with the problem. I will skip the computation in this step, largley because this writeup is already gargantuan for a simple AIME problem, and because there are countless resources explaining how to do so.
We aim to decompose $\hat{T}$ into the form $\hat{T} = SJS^{-1}$ , where $S$ is a matrix whose columns consist of the eigenvectors of $\hat{T}$ $J$ is the Jordan matrix, and $S^{-1}$ is, well, the inverse of $S$ . We have 1 eigenvalue of multiplicity 1, and 1 of multiplicity 2, so based on this info, we set our eigenvalues along the diagonals like so.
We have: \[J = \begin{bmatrix} -\frac{1}{2} & 0 & 0\\ 0 & -\frac{1}{2} & 0\\ 0 & 0 & 1 \end{bmatrix}, S = \begin{bmatrix} -1 & -1 & 1\\ 0 & 1 & 1\\ 1 & 0 & 1 \end{bmatrix}, S^{-1} = \frac{1}{3}\begin{bmatrix} -1 & -1 & 2\\ -1 & 2 & -1\\ 1 & 1 & 1 \end{bmatrix}\] and so: \[\hat{T} = \begin{bmatrix} 0 & 0.5 & 0.5\\ 0.5 & 0 & 0.5\\ 0.5 & 0.5 & 0 \end{bmatrix} = SJS^{-1} = \begin{bmatrix} -1 & -1 & 1\\ 0 & 1 & 1\\ 1 & 0 & 1 \end{bmatrix}\begin{bmatrix} -\frac{1}{2} & 0 & 0\\ 0 & -\frac{1}{2} & 0\\ 0 & 0 & 1 \end{bmatrix}\begin{bmatrix} -\frac{1}{3} & -\frac{1}{3} & \frac{2}{3}\\ -\frac{1}{3} & \frac{2}{3} & -\frac{1}{3}\\ \frac{1}{3} & \frac{1}{3} & \frac{1}{3} \end{bmatrix}\] Now that we have converted to Jordan Canonical Form, it is extremely easy to compute $\hat{T}^n$
It is an important fact that for any matrix $K$ with Jordan decomposition $SJ_kS^{-1}$ , we have that $K^n = S(J_k)^nS^{-1}$ Using this fact, we aim to find the general solution for the problem. $J^n = \begin{bmatrix} \left(-\frac{1}{2}\right)^n & 0 & 0\\ 0 & \left(-\frac{1}{2}\right)^n & 0\\ 0 & 0 & 1 \end{bmatrix}$ , and using the laws of matrix multiplication, \[SJ^n = \begin{bmatrix} (-1)^{n+1}\left(\frac{1}{2}\right)^n & (-1)^{n+1}\left(\frac{1}{2}\right)^n & 1\\ 0 & \left(-\frac{1}{2}\right)^n & 1\\ \left(-\frac{1}{2}\right)^n & 0 & 1 \end{bmatrix}\] So finally, our final, generalized transition matrix after any number of steps $n$ is: \[\hat{T}^n = \begin{bmatrix} \frac{1}{3} - \frac{1}{3}(-1)^{1+n}(2)^{1-n} & \frac{1}{3} + \frac{1}{3}(-1)^{n+1}(2)^{1-(1+n)} & \frac{1}{3} + \frac{1}{3}(-1)^{n+1}(2)^{1-(1+n)}\\ \frac{1}{3} + \frac{1}{3}(-1)^{n+1}(2)^{1-(1+n)} & \frac{1}{3} - \frac{1}{3}(-1)^{1+n}(2)^{1-n} & \frac{1}{3} + \frac{1}{3}(-1)^{n+1}(2)^{1-(1+n)}\\ \frac{1}{3} + \frac{1}{3}(-1)^{n+1}(2)^{1-(1+n)} & \frac{1}{3} + \frac{1}{3}(-1)^{n+1}(2)^{1-(1+n)} & \frac{1}{3} - \frac{1}{3}(-1)^{1+n}(2)^{1-n} \end{bmatrix}\]
For the sake of this problem, we seek the top left element, which is $\frac{1}{3} - \frac{1}{3}(-1)^{1+n}(2)^{1-n}$ . Substituting $n=10$ readily gives the probability of the bug reaching it's starting position within 10 moves to be $\frac{171}{512} \implies m+n = \boxed{683}$
| 683
|
6,506
|
https://artofproblemsolving.com/wiki/index.php/2003_AIME_II_Problems/Problem_13
| 5
|
A bug starts at a vertex of an equilateral triangle. On each move, it randomly selects one of the two vertices where it is not currently located, and crawls along a side of the triangle to that vertex. Given that the probability that the bug moves to its starting vertex on its tenth move is $m/n,$ where $m$ and $n$ are relatively prime positive integers, find $m + n.$
|
This method does not rigorously get the answer, but it works. As the bug makes more and more moves, the probability of it going back to the origin approaches closer and closer to 1/3. Therefore, after 10 moves, the probability gets close to $341.33/1024$ . We can either round up or down. If we round down, we see $341/1024$ cannot be reduced any further and because the only answers on the AIME are below 1000, this cannot be the right answer. However, if we round up, $342/1024$ can be reduced to $171/512$ where the sum 171+512= $\boxed{683}$ is an accepted answer.
| 683
|
6,507
|
https://artofproblemsolving.com/wiki/index.php/2003_AIME_II_Problems/Problem_14
| 2
|
Let $A = (0,0)$ and $B = (b,2)$ be points on the coordinate plane. Let $ABCDEF$ be a convex equilateral hexagon such that $\angle FAB = 120^\circ,$ $\overline{AB}\parallel \overline{DE},$ $\overline{BC}\parallel \overline{EF,}$ $\overline{CD}\parallel \overline{FA},$ and the y-coordinates of its vertices are distinct elements of the set $\{0,2,4,6,8,10\}.$ The area of the hexagon can be written in the form $m\sqrt {n},$ where $m$ and $n$ are positive integers and n is not divisible by the square of any prime. Find $m + n.$
|
[asy] size(200); draw((0,0)--(10/sqrt(3),2)--(18/sqrt(3),6)--(10/sqrt(3),10)--(0,8)--(-8/sqrt(3),4)--cycle); dot((0,0));dot((10/sqrt(3),2));dot((18/sqrt(3),6));dot((10/sqrt(3),10));dot((0,8));dot((-8/sqrt(3),4)); label("$A (0,0)$",(0,0),S);label("$B (b,2)$",(10/sqrt(3),2),SE);label("$C$",(18/sqrt(3),6),E);label("$D$",(10/sqrt(3),10),N);label("$E$",(0,8),NW);label("$F$",(-8/sqrt(3),4),W); [/asy] From this image, we can see that the y-coordinate of F is 4, and from this, we can gather that the coordinates of E, D, and C, respectively, are 8, 10, and 6.
[asy] size(200); draw((0,0)--(10/sqrt(3),2)--(18/sqrt(3),6)--(10/sqrt(3),10)--(0,8)--(-8/sqrt(3),4)--cycle); dot((0,0));dot((10/sqrt(3),2));dot((18/sqrt(3),6));dot((10/sqrt(3),10));dot((0,8));dot((-8/sqrt(3),4)); label("$A (0,0)$",(0,0),SE);label("$B (b,2)$",(10/sqrt(3),2),SE);label("$C$",(18/sqrt(3),6),E);label("$D$",(10/sqrt(3),10),N);label("$E$",(0,8),NW);label("$F$",(-8/sqrt(3),4),W); xaxis("$x$");yaxis("$y$"); pair b=foot((10/sqrt(3),2),(0,0),(10,0)); pair f=foot((-8/sqrt(3),4),(0,0),(-10,0)); draw(b--(10/sqrt(3),2),dotted); draw(f--(-8/sqrt(3),4),dotted); label("$\theta$",(0,0),7*dir((0,0)--(10/sqrt(3),2)+(4*sqrt(21)/3,0))); [/asy]
Let the angle between the $x$ -axis and segment $AB$ be $\theta$ , as shown above. Thus, as $\angle FAB=120^\circ$ , the angle between the $x$ -axis and segment $AF$ is $60-\theta$ , so $\sin{(60-\theta)}=2\sin{\theta}$ . Expanding, we have
Isolating $\sin{\theta}$ we see that $\frac{\sqrt{3}\cos{\theta}}{2}=\frac{5\sin{\theta}}{2}$ , or $\cos{\theta}=\frac{5}{\sqrt{3}}\sin{\theta}$ . Using the fact that $\sin^2{\theta}+\cos^2{\theta}=1$ , we have $\frac{28}{3}\sin^2{\theta}=1$ , or $\sin{\theta}=\sqrt{\frac{3}{28}}$ . Letting the side length of the hexagon be $y$ , we have $\frac{2}{y}=\sqrt{\frac{3}{28}}$ . After simplification we find that that $y=\frac{4\sqrt{21}}{3}$
In particular, note that by the Pythagorean theorem, $b^2+2^2=y^2$ , hence $b=10\sqrt{3}/3$ . Also, if $C=(c,6)$ , then $y^2=BC^2=4^2+(c-b)^2$ , hence $c-b=8\sqrt{3}/3,$ and thus $c=18\sqrt{3}/3$ . Using similar methods (or symmetry), we determine that $D=(10\sqrt{3}/3,10)$ $E=(0,8)$ , and $F=(-8\sqrt{3}/3,4)$ . By the Shoelace theorem, \[[ABCDEF]=\frac12\left|\begin{array}{cc} 0&0\\ 10\sqrt{3}/3&2\\ 18\sqrt{3}/3&6\\ 10\sqrt{3}/3&10\\ 0&8\\ -8\sqrt{3}/3&4\\ 0&0\\ \end{array}\right|=\frac12|60+180+80-36-60-(-64)|\sqrt{3}/3=48\sqrt{3}.\]
Hence the answer is $\boxed{51}$
| 51
|
6,508
|
https://artofproblemsolving.com/wiki/index.php/2002_AIME_I_Problems/Problem_2
| 1
|
The diagram shows twenty congruent circles arranged in three rows and enclosed in a rectangle. The circles are tangent to one another and to the sides of the rectangle as shown in the diagram. The ratio of the longer dimension of the rectangle to the shorter dimension can be written as $\dfrac{1}{2}(\sqrt{p}-q)$ where $p$ and $q$ are positive integers. Find $p+q$
|
Let the radius of the circles be $r$ . The longer dimension of the rectangle can be written as $14r$ , and by the Pythagorean Theorem , we find that the shorter dimension is $2r\left(\sqrt{3}+1\right)$
Therefore, $\frac{14r}{2r\left(\sqrt{3}+1\right)}= \frac{7}{\sqrt{3} + 1} \cdot \left[\frac{\sqrt{3}-1}{\sqrt{3}-1}\right] = \frac{1}{2}\left(7\sqrt{3} - 7\right) = \frac{1}{2}\left(\sqrt{p}-q\right)$ . Thus we have $p=147$ and $q=7$ , so $p+q=\boxed{154}$
| 154
|
6,509
|
https://artofproblemsolving.com/wiki/index.php/2002_AIME_I_Problems/Problem_2
| 2
|
The diagram shows twenty congruent circles arranged in three rows and enclosed in a rectangle. The circles are tangent to one another and to the sides of the rectangle as shown in the diagram. The ratio of the longer dimension of the rectangle to the shorter dimension can be written as $\dfrac{1}{2}(\sqrt{p}-q)$ where $p$ and $q$ are positive integers. Find $p+q$
|
Since we only care about the ratio between the longer side and shorter side, we can set the longer side to $14$ . So, this means that each of the radii is $1$ . Now, we connect the radii of three circles such that they form an equilateral triangle with side length 4. Obviously, the height this triangle is $2\sqrt{3}$ , and the shorter side if the triangle is therefore $2\sqrt{3}+2$ and we use simplification similar to as showed above, and we reach the result $\frac{1}{2} \cdot (\sqrt{147}-7)$ and the final answer is $147+7 = \boxed{154}$
| 154
|
6,510
|
https://artofproblemsolving.com/wiki/index.php/2002_AIME_I_Problems/Problem_5
| 1
|
Let $A_1,A_2,A_3,\cdots,A_{12}$ be the vertices of a regular dodecagon. How many distinct squares in the plane of the dodecagon have at least two vertices in the set $\{A_1,A_2,A_3,\cdots,A_{12}\} ?$
|
Proceed as above to initially get 198 squares (with overcounting). Then note that any square with all four vertices on the dodecagon has to have three sides "between" each vertex, giving us a total of three squares. However, we counted these squares with all four of their sides plus both of their diagonals, meaning we counted them 6 times. Therefore, our answer is $198-3(6-1)=198-15=\boxed{183}.$
| 183
|
6,511
|
https://artofproblemsolving.com/wiki/index.php/2002_AIME_I_Problems/Problem_7
| 1
|
The Binomial Expansion is valid for exponents that are not integers. That is, for all real numbers $x,y$ and $r$ with $|x|>|y|$
\[(x+y)^r=x^r+rx^{r-1}y+\dfrac{r(r-1)}{2}x^{r-2}y^2+\dfrac{r(r-1)(r-2)}{3!}x^{r-3}y^3 \cdots\]
What are the first three digits to the right of the decimal point in the decimal representation of $(10^{2002}+1)^{\frac{10}{7}}$
|
$1^n$ will always be 1, so we can ignore those terms, and using the definition ( $2002 / 7 = 286$ ):
\[(10^{2002} + 1)^{\frac {10}7} = 10^{2860}+\dfrac{10}{7}10^{858}+\dfrac{15}{49}10^{-1144}+\cdots\]
Since the exponent of the $10$ goes down extremely fast, it suffices to consider the first few terms. Also, the $10^{2860}$ term will not affect the digits after the decimal, so we need to find the first three digits after the decimal in
\[\dfrac{10}{7}10^{858}\]
(The remainder after this term is positive by the Remainder Estimation Theorem ). Since the repeating decimal of $\dfrac{10}{7}$ repeats every 6 digits, we can cut out a lot of 6's from $858$ to reduce the problem to finding the first three digits after the decimal of
$\dfrac{10}{7}$
That is the same as $1+\dfrac{3}{7}$ , and the first three digits after $\dfrac{3}{7}$ are $\boxed{428}$
| 428
|
6,512
|
https://artofproblemsolving.com/wiki/index.php/2002_AIME_I_Problems/Problem_8
| 1
|
Find the smallest integer $k$ for which the conditions
(1) $a_1,a_2,a_3\cdots$ is a nondecreasing sequence of positive integers
(2) $a_n=a_{n-1}+a_{n-2}$ for all $n>2$
(3) $a_9=k$
are satisfied by more than one sequence.
|
From $(2)$ $a_9=$ $a_8+a_7=2a_7+a_6=3a_6+2a_5=5a_5+3a_4=8a_4+5a_3=13a_3+8a_2=21a_2+13a_1$ $=k$
Suppose that $a_1=x_0$ is the smallest possible value for $a_1$ that yields a good sequence, and $a_2=y_0$ in this sequence. So, $13x_0+21y_0=k$
Since $\gcd(13,21)=1$ , the next smallest possible value for $a_1$ that yields a good sequence is $a_1=x_0+21$ . Then, $a_2=y_0-13$
By $(1)$ $a_2 \ge a_1 \Rightarrow y_0-13 \ge x_0+21 \Rightarrow y_0 \ge x_0+34 \ge 35$ . So the smallest value of $k$ is attained when $(x_0,y_0)=(1,35)$ which yields $(a_1,a_2)=(1,35)$ or $(22,22)$
Thus, $k=13(1)+21(35)=\boxed{748}$ is the smallest possible value of $k$
| 748
|
6,513
|
https://artofproblemsolving.com/wiki/index.php/2002_AIME_I_Problems/Problem_9
| 1
|
Harold, Tanya, and Ulysses paint a very long picket fence.
Call the positive integer $100h+10t+u$ paintable when the triple $(h,t,u)$ of positive integers results in every picket being painted exactly once. Find the sum of all the paintable integers.
|
Note that it is impossible for any of $h,t,u$ to be $1$ , since then each picket will have been painted one time, and then some will be painted more than once.
$h$ cannot be $2$ , or that will result in painting the third picket twice. If $h=3$ , then $t$ may not equal anything not divisible by $3$ , and the same for $u$ . Now for fourth and fifth pickets to be painted, $t$ and $u$ must be $3$ as well. This configuration works, so $333$ is paintable.
If $h$ is $4$ , then $t$ must be even. The same for $u$ , except that it can't be $2 \mod 4$ . Thus $u$ is $0 \mod 4$ and $t$ is $2 \mod 4$ . Since this is all $\mod 4$ $t$ must be $2$ and $u$ must be $4$ , in order for $5,6$ to be paint-able. Thus $424$ is paintable.
$h$ cannot be greater than $4$ , since if that were the case then the answer would be greater than $999$ , which would be impossible for the AIME.
Thus the sum of all paintable numbers is $\boxed{757}$
| 757
|
6,514
|
https://artofproblemsolving.com/wiki/index.php/2002_AIME_I_Problems/Problem_9
| 2
|
Harold, Tanya, and Ulysses paint a very long picket fence.
Call the positive integer $100h+10t+u$ paintable when the triple $(h,t,u)$ of positive integers results in every picket being painted exactly once. Find the sum of all the paintable integers.
|
Again, note that $h,t,u \neq 1$ . The three conditions state that no picket number $n$ may satisfy any two of the conditions: $n \equiv 1 \pmod{h},\ n \equiv 2 \pmod{t},\ n \equiv 3 \pmod{u}$ . By the Chinese Remainder Theorem , the greatest common divisor of any pair of the three numbers $\{h,t,u\}$ cannot be $1$ (since otherwise without loss of generality consider $\text{gcd}\,(h,t) = 1$ ; then there will be a common solution $\pmod{h \times t}$ ).
Now for $4$ to be paint-able, we require either $h = 3$ or $t=2$ , but not both.
Thus the answer is $333+424 = \boxed{757}$
| 757
|
6,515
|
https://artofproblemsolving.com/wiki/index.php/2002_AIME_I_Problems/Problem_9
| 3
|
Harold, Tanya, and Ulysses paint a very long picket fence.
Call the positive integer $100h+10t+u$ paintable when the triple $(h,t,u)$ of positive integers results in every picket being painted exactly once. Find the sum of all the paintable integers.
|
The three conditions state that no picket number $n$ may satisfy any two of the conditions: $n \equiv 1 \pmod{h},\ n \equiv 2 \pmod{t},\ n \equiv 3 \pmod{u}$ . Note that the smallest number, $min \{ h,t,u \},$ divides the other $2$ , and the next smallest divide the largest number, otherwise there is a common solution by the Chinese Remainder Theorem . It is also a necessary condition so that it paints exactly once. Note that the smallest number can't be at least $5$ , otherwise not all picket will be painted. We are left with few cases (we could also exclude $1$ as the possibility) which could be done quickly. Thus the answer is $333+424 = \boxed{757}$
| 757
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6,516
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https://artofproblemsolving.com/wiki/index.php/2002_AIME_I_Problems/Problem_10
| 1
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In the diagram below, angle $ABC$ is a right angle. Point $D$ is on $\overline{BC}$ , and $\overline{AD}$ bisects angle $CAB$ . Points $E$ and $F$ are on $\overline{AB}$ and $\overline{AC}$ , respectively, so that $AE=3$ and $AF=10$ . Given that $EB=9$ and $FC=27$ , find the integer closest to the area of quadrilateral $DCFG$
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By the Pythagorean Theorem, $BC=35$ . Letting $BD=x$ we can use the Angle Bisector Theorem on triangle $ABC$ to get $x/12=(35-x)/37$ , and solving gives $BD=60/7$ and $DC=185/7$
The area of triangle $AGF$ is $10/3$ that of triangle $AEG$ , since they share a common side and angle, so the area of triangle $AGF$ is $10/13$ the area of triangle $AEF$
Since the area of a triangle is $\frac{ab\sin{C}}2$ , the area of $AEF$ is $525/37$ and the area of $AGF$ is $5250/481$
The area of triangle $ABD$ is $360/7$ , and the area of the entire triangle $ABC$ is $210$ . Subtracting the areas of $ABD$ and $AGF$ from $210$ and finding the closest integer gives $\boxed{148}$ as the answer.
| 148
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6,517
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https://artofproblemsolving.com/wiki/index.php/2002_AIME_I_Problems/Problem_10
| 2
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In the diagram below, angle $ABC$ is a right angle. Point $D$ is on $\overline{BC}$ , and $\overline{AD}$ bisects angle $CAB$ . Points $E$ and $F$ are on $\overline{AB}$ and $\overline{AC}$ , respectively, so that $AE=3$ and $AF=10$ . Given that $EB=9$ and $FC=27$ , find the integer closest to the area of quadrilateral $DCFG$
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By the Pythagorean Theorem, $BC=35$ . From now on, every number we get will be rounded to two decimal points. Using law of cosines on AEF, EF is around 9.46. Using the angle bisector theorem, GF and DC are 7.28 and (185/7), respectively. We also know the lengths of EG and BD, so a quick Stewart's Theorem for triangles ABC and AEF gives lengths 3.76 and 14.75 for AG and AD, respectively. Now we have all segments of triangles AGF and ADC. Joy! It's time for some Heron's Formula. This gives area 10.95 for triangle AGF and 158.68 for triangle ADC. Subtracting gives us 147.73. Round to the nearest whole number and we get $\boxed{148}$
| 148
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6,518
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https://artofproblemsolving.com/wiki/index.php/2002_AIME_I_Problems/Problem_10
| 3
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In the diagram below, angle $ABC$ is a right angle. Point $D$ is on $\overline{BC}$ , and $\overline{AD}$ bisects angle $CAB$ . Points $E$ and $F$ are on $\overline{AB}$ and $\overline{AC}$ , respectively, so that $AE=3$ and $AF=10$ . Given that $EB=9$ and $FC=27$ , find the integer closest to the area of quadrilateral $DCFG$
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By the Pythagorean Theorem, $BC=35$ . By the Angle Bisector Theorem $BD = 60/7$ and $DC = 185/7$ . We can find the coordinates of F, and use that the find the equation of line EF. Then, we can find the coordinates of G. Triangle $ADC = 1110/7$ and we can find the area triangle $AGF$ with the shoelace theorem, so subtracting that from $ADC$ gives us $\boxed{148}$ as the closest integer.
| 148
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6,519
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https://artofproblemsolving.com/wiki/index.php/2002_AIME_I_Problems/Problem_10
| 4
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In the diagram below, angle $ABC$ is a right angle. Point $D$ is on $\overline{BC}$ , and $\overline{AD}$ bisects angle $CAB$ . Points $E$ and $F$ are on $\overline{AB}$ and $\overline{AC}$ , respectively, so that $AE=3$ and $AF=10$ . Given that $EB=9$ and $FC=27$ , find the integer closest to the area of quadrilateral $DCFG$
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By the Pythagorean Theorem, $BC = 35$ , and by the Angle Bisector Theorem $BD = 60/7$ and $DC = 185/7$ . Draw a perpendicular from $F$ to $\overline{AE}$ . Let the intersection of $F$ and $\overline{AE}$ be $H$ . triangle $AHF$ is similar to $ABC$ by $AA$ similarity. thus, $AF/AC = HF/BC$ . We find that $HF = 350/37$ , so the area of $AEF = 525/37$ . The area of triangle $AGF$ is $10/3$ that of triangle $AEG$ , since they share a common side and angle, so the area of triangle $AGF$ is $10/13$ the area of triangle $AEF$ , so the area of $AGF = 5250/481$ . The area of triangle $ADC$ is $1110/7$ , since the base is $185/7$ and the height is $12$ . Thus, the area of $DCFG$ equals the area of $ADC - AGF$ , or rounded to the nearest integer, $\boxed{148}$
| 148
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6,520
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https://artofproblemsolving.com/wiki/index.php/2002_AIME_I_Problems/Problem_11
| 1
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Let $ABCD$ and $BCFG$ be two faces of a cube with $AB=12$ . A beam of light emanates from vertex $A$ and reflects off face $BCFG$ at point $P$ , which is 7 units from $\overline{BG}$ and 5 units from $\overline{BC}$ . The beam continues to be reflected off the faces of the cube. The length of the light path from the time it leaves point $A$ until it next reaches a vertex of the cube is given by $m\sqrt{n}$ , where $m$ and $n$ are integers and $n$ is not divisible by the square of any prime. Find $m+n$
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When a light beam reflects off a surface, the path is like that of a ball bouncing. Picture that, and also imagine X, Y, and Z coordinates for the cube vertices. The coordinates will all involve 0's and 12's only, so that means that the X, Y, and Z distance traveled by the light must all be divisible by 12. Since the light's Y changes by 5 and the X changes by 7 (the Z changes by 12, don't worry about that), and 5 and 7 are relatively prime to 12, the light must make 12 reflections onto the XY plane or the face parallel to the XY plane.
In each reflection, the distance traveled by the light is $\sqrt{ (12^2) + (5^2) + (7^2) }$ $\sqrt{218}$ . This happens 12 times, so the total distance is $12\sqrt{218}$ $m=12$ and $n=218$ , so therefore, the answer is $m+n=\boxed{230}$
| 230
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6,521
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https://artofproblemsolving.com/wiki/index.php/2002_AIME_I_Problems/Problem_11
| 2
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Let $ABCD$ and $BCFG$ be two faces of a cube with $AB=12$ . A beam of light emanates from vertex $A$ and reflects off face $BCFG$ at point $P$ , which is 7 units from $\overline{BG}$ and 5 units from $\overline{BC}$ . The beam continues to be reflected off the faces of the cube. The length of the light path from the time it leaves point $A$ until it next reaches a vertex of the cube is given by $m\sqrt{n}$ , where $m$ and $n$ are integers and $n$ is not divisible by the square of any prime. Find $m+n$
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We can use a similar trick as with reflections in 2D: Imagine that the entire space is divided into cubes identical to the one we have. Now let's follow two photons of light that start in $A$ at the same time: one of them will reflect as given in the problem statement, the second will simply fly straight through all cubes. It can easily be seen that at any moment in time the photons are in exactly the same position relative to the cubes they are inside at the moment. In other words, we can take the cube with the first photon, translate it and flip if necessary, to get the cube with the other photon.
It follows that both photons will hit a vertex at the same time, and at this moment they will have travelled the same distance.
Now, the path of the second photon is simply a half-line given by the vector $(12,7,5)$ . That is, the points visited by the photon are of the form $(12t,7t,5t)$ for $t\geq 0$ . We are looking for the smallest $t$ such that all three coordinates are integer multiples of $12$ (which is the length of the side of the cube).
Clearly $t$ must be an integer. As $7$ and $12$ are relatively prime, the smallest solution is $t=12$ . At this moment the second photon will be at the coordinates $(12\cdot 12,7\cdot 12,5\cdot 12)$
Then the distance it travelled is $\sqrt{ (12\cdot 12)^2 + (7\cdot 12)^2 + (5\cdot 12)^2 } = 12\sqrt{12^2 + 7^2 + 5^2}=12\sqrt{218}$ .
And as the factorization of $218$ is $218=2\cdot 109$ , we have $m=12$ and $n=218$ , hence $m+n=\boxed{230}$
| 230
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6,522
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https://artofproblemsolving.com/wiki/index.php/2002_AIME_I_Problems/Problem_12
| 1
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Let $F(z)=\dfrac{z+i}{z-i}$ for all complex numbers $z\neq i$ , and let $z_n=F(z_{n-1})$ for all positive integers $n$ . Given that $z_0=\dfrac{1}{137}+i$ and $z_{2002}=a+bi$ , where $a$ and $b$ are real numbers, find $a+b$
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Iterating $F$ we get:
\begin{align*} F(z) &= \frac{z+i}{z-i}\\ F(F(z)) &= \frac{\frac{z+i}{z-i}+i}{\frac{z+i}{z-i}-i} = \frac{(z+i)+i(z-i)}{(z+i)-i(z-i)}= \frac{z+i+zi+1}{z+i-zi-1}= \frac{(z+1)(i+1)}{(z-1)(1-i)}\\ &= \frac{(z+1)(i+1)^2}{(z-1)(1^2+1^2)}= \frac{(z+1)(2i)}{(z-1)(2)}= \frac{z+1}{z-1}i\\ F(F(F(z))) &= \frac{\frac{z+1}{z-1}i+i}{\frac{z+1}{z-1}i-i} = \frac{\frac{z+1}{z-1}+1}{\frac{z+1}{z-1}-1} = \frac{(z+1)+(z-1)}{(z+1)-(z-1)} = \frac{2z}{2} = z. \end{align*}
From this, it follows that $z_{k+3} = z_k$ , for all $k$ . Thus $z_{2002} = z_{3\cdot 667+1} = z_1 = \frac{z_0+i}{z_0-i} = \frac{(\frac{1}{137}+i)+i}{(\frac{1}{137}+i)-i}= \frac{\frac{1}{137}+2i}{\frac{1}{137}}= 1+274i.$
Thus $a+b = 1+274 = \boxed{275}$
| 275
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6,523
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https://artofproblemsolving.com/wiki/index.php/2002_AIME_I_Problems/Problem_13
| 5
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In triangle $ABC$ the medians $\overline{AD}$ and $\overline{CE}$ have lengths $18$ and $27$ , respectively, and $AB=24$ . Extend $\overline{CE}$ to intersect the circumcircle of $ABC$ at $F$ . The area of triangle $AFB$ is $m\sqrt{n}$ , where $m$ and $n$ are positive integers and $n$ is not divisible by the square of any prime. Find $m+n$
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Apply barycentric coordinates on $\triangle ABC$ . We know that $D=\left(0, \frac{1}{2}, \frac{1}{2}\right), E=\left(\frac{1}{2}, \frac{1}{2}, 0\right)$ . We can now get the displacement vectors $\overrightarrow{AD} = \left(1, -\frac{1}{2}, -\frac{1}{2}\right)$ and $\overrightarrow{CE}=\left(-\frac{1}{2}, -\frac{1}{2}, 1\right)$ . Now, applying the distance formula and simplifying gives us the two equations \begin{align*} 2b^2+2c^2-^2&=1296 \\ 2a^2+2b^2-c^2&=2916. \\ \end{align*} Substituting $c=24$ and solving with algebra now gives $a=6\sqrt{31}, b=3\sqrt{70}$ . Now we can find $F$ . Note that $CE$ can be parameterized as $(1:1:t)$ , so plugging into the circumcircle equation and solving for $t$ gives $t=\frac{-c^2}{a^2+b^2}$ so $F=(a^2+b^2:a^2+b^2:-c^2)$ . Plugging in for $a,b$ gives us $F=(1746:1746:-576)$ . Thus, by the area formula, we have \[\frac{[AFB]}{[ABC]}= \left|\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ \frac{97}{162} & \frac{97}{162} & -\frac{16}{81} \end{matrix}\right|=\frac{16}{81}.\] By Heron's Formula, we have $[ABC]=\frac{81\sqrt{55}}{2}$ which immediately gives $[AFB]=8\sqrt{55}$ from our ratio, extracting $\boxed{63}$
| 63
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6,524
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https://artofproblemsolving.com/wiki/index.php/2002_AIME_I_Problems/Problem_15
| 1
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Polyhedron $ABCDEFG$ has six faces. Face $ABCD$ is a square with $AB = 12;$ face $ABFG$ is a trapezoid with $\overline{AB}$ parallel to $\overline{GF},$ $BF = AG = 8,$ and $GF = 6;$ and face $CDE$ has $CE = DE = 14.$ The other three faces are $ADEG, BCEF,$ and $EFG.$ The distance from $E$ to face $ABCD$ is 12. Given that $EG^2 = p - q\sqrt {r},$ where $p, q,$ and $r$ are positive integers and $r$ is not divisible by the square of any prime, find $p + q + r.$
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We let $A$ be the origin, or $(0,0,0)$ $B = (0,0,12)$ , and $D = (12,0,0)$ . Draw the perpendiculars from F and G to AB, and let their intersections be X and Y, respectively. By symmetry, $FX = GY = \frac{12-6}2 = 3$ , so $G = (a,b,3)$ , where a and b are variables.
We can now calculate the coordinates of E. Drawing the perpendicular from E to CD and letting the intersection be Z, we have $CZ = DZ = 6$ and $EZ = 4\sqrt{10}$ . Therefore, the x coordinate of $E$ is $12-\sqrt{(4\sqrt{10})^2-12^2}=12-\sqrt{16}=12-4=8$ , so $E = (8,12,6)$
We also know that $A,D,E,$ and $G$ are coplanar, so they all lie on the plane $z = Ax+By+C$ . Since $(0,0,0)$ is on it, then $C = 0$ . Also, since $(12,0,0)$ is contained, then $A = 0$ . Finally, since $(8,12,6)$ is on the plane, then $B = \frac 12$ . Therefore, $b = 6$ . Since $GA = 8$ , then $a^2+6^2+3^2=8^2$ , or $a = \pm \sqrt{19}$ . Therefore, the two permissible values of $EG^2$ are $(8 \pm \sqrt{19})^2+6^2+3^2 = 128 \pm 16\sqrt{19}$ . The only one that satisfies the conditions of the problem is $128 - 16\sqrt{19}$ , from which the answer is $128+16+19=\boxed{163}$
| 163
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6,525
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https://artofproblemsolving.com/wiki/index.php/2002_AIME_II_Problems/Problem_3
| 1
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It is given that $\log_{6}a + \log_{6}b + \log_{6}c = 6,$ where $a,$ $b,$ and $c$ are positive integers that form an increasing geometric sequence and $b - a$ is the square of an integer. Find $a + b + c.$
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$abc=6^6$ . Since they form an increasing geometric sequence, $b$ is the geometric mean of the product $abc$ $b=\sqrt[3]{abc}=6^2=36$
Since $b-a$ is the square of an integer, we can find a few values of $a$ that work: $11, 20, 27, 32,$ and $35$ . Out of these, the only value of $a$ that works is $a=27$ , from which we can deduce that $c=\dfrac{36}{27}\cdot 36=\dfrac{4}{3}\cdot 36=48$
Thus, $a+b+c=27+36+48=\boxed{111}$
| 111
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6,526
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https://artofproblemsolving.com/wiki/index.php/2002_AIME_II_Problems/Problem_3
| 2
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It is given that $\log_{6}a + \log_{6}b + \log_{6}c = 6,$ where $a,$ $b,$ and $c$ are positive integers that form an increasing geometric sequence and $b - a$ is the square of an integer. Find $a + b + c.$
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Let $r$ be the common ratio of the geometric sequence. Since it is increasing, that means that $b = ar$ , and $c = ar^2$ . Simplifying the logarithm, we get $\log_6(a^3*r^3) = 6$ . Therefore, $a^3*r^3 = 6^6$ . Taking the cube root of both sides, we see that $ar = 6^2 = 36$ . Now since $ar = b$ , that means $b = 36$ . Using the trial and error shown in solution 1, we get $a = 27$ , and $r = \frac{4}{3}$ . Now, $27*r^2= c = 48$ . Therefore, the answer is $27+36+48 = \boxed{111}$
| 111
|
6,527
|
https://artofproblemsolving.com/wiki/index.php/2002_AIME_II_Problems/Problem_4
| 1
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Patio blocks that are hexagons $1$ unit on a side are used to outline a garden by placing the blocks edge to edge with $n$ on each side. The diagram indicates the path of blocks around the garden when $n=5$
AIME 2002 II Problem 4.gif
If $n=202$ , then the area of the garden enclosed by the path, not including the path itself, is $m\left(\sqrt3/2\right)$ square units, where $m$ is a positive integer. Find the remainder when $m$ is divided by $1000$
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When $n>1$ , the path of blocks has $6(n-1)$ blocks total in it. When $n=1$ , there is just one lonely block. Thus, the area of the garden enclosed by the path when $n=202$ is
\[(1+6+12+18+\cdots +1200)A=(1+6(1+2+3...+200))A\]
where $A$ is the area of one block. Then, because $n(n+1)/2$ is equal to the sum of the first $n$ integers:
\[(1+6(1+2+3...+200))=(1+6((200)(201)/2))A=120601A\]
Since $A=\dfrac{3\sqrt{3}}{2}$ , the area of the garden is
\[120601\cdot \dfrac{3\sqrt{3}}{2}=\dfrac{361803\sqrt{3}}{2}\]
$m=361803$ $\dfrac{m}{1000}=361$ Remainder $\boxed{803}$
| 803
|
6,528
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https://artofproblemsolving.com/wiki/index.php/2002_AIME_II_Problems/Problem_7
| 1
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It is known that, for all positive integers $k$
Find the smallest positive integer $k$ such that $1^2+2^2+3^2+\ldots+k^2$ is a multiple of $200$
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$\frac{k(k+1)(2k+1)}{6}$ is a multiple of $200$ if $k(k+1)(2k+1)$ is a multiple of $1200 = 2^4 \cdot 3 \cdot 5^2$ .
So $16,3,25|k(k+1)(2k+1)$
Since $2k+1$ is always odd, and only one of $k$ and $k+1$ is even, either $k, k+1 \equiv 0 \pmod{16}$
Thus, $k \equiv 0, 15 \pmod{16}$
If $k \equiv 0 \pmod{3}$ , then $3|k$ . If $k \equiv 1 \pmod{3}$ , then $3|2k+1$ . If $k \equiv 2 \pmod{3}$ , then $3|k+1$
Thus, there are no restrictions on $k$ in $\pmod{3}$
It is easy to see that only one of $k$ $k+1$ , and $2k+1$ is divisible by $5$ . So either $k, k+1, 2k+1 \equiv 0 \pmod{25}$
Thus, $k \equiv 0, 24, 12 \pmod{25}$
From the Chinese Remainder Theorem $k \equiv 0, 112, 224, 175, 287, 399 \pmod{400}$ . Thus, the smallest positive integer $k$ is $\boxed{112}$
| 112
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6,529
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https://artofproblemsolving.com/wiki/index.php/2002_AIME_II_Problems/Problem_10
| 1
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While finding the sine of a certain angle, an absent-minded professor failed to notice that his calculator was not in the correct angular mode. He was lucky to get the right answer. The two least positive real values of $x$ for which the sine of $x$ degrees is the same as the sine of $x$ radians are $\frac{m\pi}{n-\pi}$ and $\frac{p\pi}{q+\pi}$ , where $m$ $n$ $p$ , and $q$ are positive integers. Find $m+n+p+q$
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Note that $x$ degrees is equal to $\frac{\pi x}{180}$ radians. Also, for $\alpha \in \left[0 , \frac{\pi}{2} \right]$ , the two least positive angles $\theta > \alpha$ such that $\sin{\theta} = \sin{\alpha}$ are $\theta = \pi-\alpha$ , and $\theta = 2\pi + \alpha$
Clearly $x > \frac{\pi x}{180}$ for positive real values of $x$
$\theta = \pi-\alpha$ yields: $x = \pi - \frac{\pi x}{180} \Rightarrow x = \frac{180\pi}{180+\pi} \Rightarrow (p,q) = (180,180)$
$\theta = 2\pi + \alpha$ yields: $x = 2\pi + \frac{\pi x}{180} \Rightarrow x = \frac{360\pi}{180-\pi} \Rightarrow (m,n) = (360,180)$
So, $m+n+p+q = \boxed{900}$
| 900
|
6,530
|
https://artofproblemsolving.com/wiki/index.php/2002_AIME_II_Problems/Problem_11
| 1
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Two distinct, real, infinite geometric series each have a sum of $1$ and have the same second term. The third term of one of the series is $1/8$ , and the second term of both series can be written in the form $\frac{\sqrt{m}-n}p$ , where $m$ $n$ , and $p$ are positive integers and $m$ is not divisible by the square of any prime. Find $100m+10n+p$
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Let the second term of each series be $x$ . Then, the common ratio is $\frac{1}{8x}$ , and the first term is $8x^2$
So, the sum is $\frac{8x^2}{1-\frac{1}{8x}}=1$ . Thus, $64x^3-8x+1 = (4x-1)(16x^2+4x-1) = 0 \Rightarrow x = \frac{1}{4}, \frac{-1 \pm \sqrt{5}}{8}$
The only solution in the appropriate form is $x = \frac{\sqrt{5}-1}{8}$ . Therefore, $100m+10n+p = \boxed{518}$
| 518
|
6,531
|
https://artofproblemsolving.com/wiki/index.php/2002_AIME_II_Problems/Problem_11
| 2
|
Two distinct, real, infinite geometric series each have a sum of $1$ and have the same second term. The third term of one of the series is $1/8$ , and the second term of both series can be written in the form $\frac{\sqrt{m}-n}p$ , where $m$ $n$ , and $p$ are positive integers and $m$ is not divisible by the square of any prime. Find $100m+10n+p$
|
Let's ignore the "two distinct, real, infinite geometric series" part for now and focus on what it means to be a geometric series.
Let the first term of the series with the third term equal to $\frac18$ be $a,$ and the common ratio be $r.$ Then, we get that $\frac{a}{1-r} = 1 \implies a = 1-r,$ and $ar^2 = \frac18 \implies (1-r)(r^2) = \frac18.$
We see that this cubic is equivalent to $r^3 - r^2 + \frac18 = 0.$ Through experimenting, we find that one of the solutions is $r = \frac12.$ Using synthetic division leads to the quadratic $4x^2 - 2x - 1 = 0.$ This has roots $\dfrac{2 \pm \sqrt{4 - 4(4)(-1)}}{8},$ or, when reduced, $\dfrac{1 \pm \sqrt{5}}{4}.$
It becomes clear that the two geometric series have common ratio $\frac{1 + \sqrt{5}}{4}$ and $\frac{1 - \sqrt{5}}{4}.$ Let $\frac{1 + \sqrt{5}}{4}$ be the ratio that we are inspecting. We see that in this case, $a = \dfrac{3 - \sqrt{5}}{4}.$
Since the second term in the series is $ar,$ we compute this and have that \[ar = \left(\dfrac{3 - \sqrt{5}}{4} \right)\left(\dfrac{1+\sqrt{5}}{4}\right) = \dfrac{\sqrt{5} - 1}{8},\] for our answer of $100 \cdot 5 + 1 \cdot 10 + 8 = \boxed{518}.$
| 518
|
6,532
|
https://artofproblemsolving.com/wiki/index.php/2002_AIME_II_Problems/Problem_12
| 1
|
A basketball player has a constant probability of $.4$ of making any given shot, independent of previous shots. Let $a_n$ be the ratio of shots made to shots attempted after $n$ shots. The probability that $a_{10} = .4$ and $a_n\le.4$ for all $n$ such that $1\le n\le9$ is given to be $p^aq^br/\left(s^c\right)$ where $p$ $q$ $r$ , and $s$ are primes, and $a$ $b$ , and $c$ are positive integers. Find $\left(p+q+r+s\right)\left(a+b+c\right)$
|
We graph the $10$ shots on a grid. Suppose that a made shot is represented by a step of $(0,1)$ , and a missed shot is represented by $(1,0)$ . Then the basketball player's shots can be represented by the number of paths from $(0,0)$ to $(6,4)$ that always stay below the line $y=\frac{2x}{3}$ . We can find the number of such paths using a Pascal's Triangle type method below, computing the number of paths to each point that only move right and up. [asy] size(150); for (int i=0;i<7;++i) {draw((i,0)--(i,4));} for (int i=0;i<5;++i) {draw((0,i)--(6,i));} draw((0,0)--(6,4),dashed); label("$1$",(0,0),SE,fontsize(8)); label("$1$",(1,0),SE,fontsize(8)); label("$1$",(2,0),SE,fontsize(8)); label("$1$",(2,1),SE,fontsize(8)); label("$1$",(3,0),SE,fontsize(8)); label("$2$",(3,1),SE,fontsize(8)); label("$2$",(3,2),SE,fontsize(8)); label("$1$",(4,0),SE,fontsize(8)); label("$3$",(4,1),SE,fontsize(8)); label("$5$",(4,2),SE,fontsize(8)); label("$1$",(5,0),SE,fontsize(8)); label("$4$",(5,1),SE,fontsize(8)); label("$9$",(5,2),SE,fontsize(8)); label("$9$",(5,3),SE,fontsize(8)); label("$1$",(6,0),SE,fontsize(8)); label("$5$",(6,1),SE,fontsize(8)); label("$14$",(6,2),SE,fontsize(8)); label("$23$",(6,3),SE,fontsize(8)); label("$23$",(6,4),SE,fontsize(8)); [/asy] Therefore, there are $23$ ways to shoot $4$ makes and $6$ misses under the given conditions. The probability of each possible sequence occurring is $(.4)^4(.6)^6$ . Hence the desired probability is \[\frac{23\cdot 2^4\cdot 3^6}{5^{10}},\] and the answer is $(23+2+3+5)(4+6+10)=\boxed{660}$
| 660
|
6,533
|
https://artofproblemsolving.com/wiki/index.php/2002_AIME_II_Problems/Problem_12
| 2
|
A basketball player has a constant probability of $.4$ of making any given shot, independent of previous shots. Let $a_n$ be the ratio of shots made to shots attempted after $n$ shots. The probability that $a_{10} = .4$ and $a_n\le.4$ for all $n$ such that $1\le n\le9$ is given to be $p^aq^br/\left(s^c\right)$ where $p$ $q$ $r$ , and $s$ are primes, and $a$ $b$ , and $c$ are positive integers. Find $\left(p+q+r+s\right)\left(a+b+c\right)$
|
The first restriction is that $a_{10} = .4$ , meaning that the player gets exactly 4 out of 10 baskets. The second restriction is $a_n\le.4$ . This means that the player may never have a shooting average over 40%. Thus, the first and second shots must fail, since $\frac{1}{1}$ and $\frac{1}{2}$ are both over $.4$ , but the player may make the third basket, since $\frac{1}{3} \le .4$ In other words, the earliest the first basket may be made is attempt 3. Using similar reasoning, the earliest the second basket may be made is attempt 5, the earliest the third basket may be made is attempt 8, and the earliest the fourth basket may be made is attempt 10.
Using X to represent a basket and O to represent a failure, this 'earliest' solution may be represented as:
OOXOXOOXOX
To simplify counting, note that the first, second, and tenth shots are predetermined. The first two shots must fail, and the last shot must succeed. Thus, only slots 3-9 need to be counted, and can be abbreviated as follows:
XOXOOXO
The problem may be separated into five cases, since the first shot may be made on attempt 3, 4, 5, 6, or 7. The easiest way to count the problem is to remember that each X may slide to the right, but NOT to the left.
First shot made on attempt 3:
XOXOOXO
XOXOOOX
XOOXOXO
XOOXOOX
XOOOXXO
XOOOXOX
XOOOOXX
Total - 7
First shot made on attempt 4:
Note that all that needs to be done is change each line in the prior case from starting with "XO....." to "OX.....".
Total - 7
First shot made on attempt 5:
OOXXOXO
OOXXOOX
OOXOXXO
OOXOXOX
OOXOOXX
Total - 5
First shot made on attempt 6:
OOOXXXO
OOOXXOX
OOOXOXX
Total - 3
First shot made on attempt 7:
OOOOXXX
Total - 1
The total number of ways the player may satisfy the requirements is $7+7+5+3+1=23$
The chance of hitting any individual combination (say, for example, OOOOOOXXXX) is $\left(\frac{3}{5}\right)^6\left(\frac{2}{5}\right)^4$
Thus, the chance of hitting any of these 23 combinations is $23\left(\frac{3}{5}\right)^6\left(\frac{2}{5}\right)^4 = \frac{23\cdot3^6\cdot2^4}{5^{10}}$
Thus, the final answer is $(23+3+2+5)(6+4+10)=\boxed{660}$
| 660
|
6,534
|
https://artofproblemsolving.com/wiki/index.php/2002_AIME_II_Problems/Problem_12
| 3
|
A basketball player has a constant probability of $.4$ of making any given shot, independent of previous shots. Let $a_n$ be the ratio of shots made to shots attempted after $n$ shots. The probability that $a_{10} = .4$ and $a_n\le.4$ for all $n$ such that $1\le n\le9$ is given to be $p^aq^br/\left(s^c\right)$ where $p$ $q$ $r$ , and $s$ are primes, and $a$ $b$ , and $c$ are positive integers. Find $\left(p+q+r+s\right)\left(a+b+c\right)$
|
Note $a_{10}=.4$ . Therefore the player made 4 shots out of 10. He must make the 10th shot, because if he doesn't, then $a_9=\frac{4}{9}>.4$ . Since $a_n\leq .4$ for all $n$ less than 11, we know that $a_1=a_2=0$ . Now we must look at the 3rd through 9th shot.
Now let's take a look at those un-determined shots. Let's put them into groups: the 3rd, 4th, and 5th shots in group A, and the 6th, 7th, 8th, and 9th shots in group B. The total number of shots made in groups A and B must be 3, since the player makes the 10th shot. We cannot have all three shots made in group A, since $a_5\leq .4$ . Therefore we can have two shots made, one shot made, or no shots made in group A.
Case 1: Group A contains no made shots.
The number of ways this can happen in group A is 1. Now we must arrange the shots in group B accordingly. There are four ways to arrange them total, and all of them work. There are $\textbf{4}$ possibilities here.
Case 2: Group A contains one made shot.
The number of ways this could happen in group A is 3. Now we must arrange the shots in group B accordingly. There are six ways to arrange them total, but the arrangement "hit hit miss miss" fails, because that would mean $a_7=\frac{3}{7}>.4$ . All the rest work. Therefore there are $3\cdot5=\textbf{15}$ possibilities here.
Case 3: Group A contains two made shots.
The number of ways this could happen in group A is 2 (hit hit miss doesn't work but the rest do). Now we must arrange the shots in group B accordingly. Note hit miss miss miss and miss hit miss miss fail. Therefore there are only 2 ways to do this, and there are $2\cdot 2=\textbf{4}$ total possibilities for this case.
Taking all these cases into account, we find that there are $4+15+4=23$ ways to have $a_{10} = .4$ and $a_n\leq .4$ . Each of these has a probability of $.4^4\cdot.6^6=\frac{2^4\cdot 3^6}{5^{10}}$ . Therefore the probability that we have $a_{10} = .4$ and $a_n\leq .4$ is $\frac{23\cdot2^4\cdot3^6}{5^{10}}$ . Now we are asked to find the product of the sum of the primes and the sum of the exponents, which is $(23+2+3+5)(4+6+10)=33\cdot20=\boxed{660}$
| 660
|
6,535
|
https://artofproblemsolving.com/wiki/index.php/2002_AIME_II_Problems/Problem_13
| 1
|
In triangle $ABC,$ point $D$ is on $\overline{BC}$ with $CD = 2$ and $DB = 5,$ point $E$ is on $\overline{AC}$ with $CE = 1$ and $EA = 3,$ $AB = 8,$ and $\overline{AD}$ and $\overline{BE}$ intersect at $P.$ Points $Q$ and $R$ lie on $\overline{AB}$ so that $\overline{PQ}$ is parallel to $\overline{CA}$ and $\overline{PR}$ is parallel to $\overline{CB}.$ It is given that the ratio of the area of triangle $PQR$ to the area of triangle $ABC$ is $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m + n$
|
First draw $\overline{CP}$ and extend it so that it meets with $\overline{AB}$ at point $X$
[asy] size(10cm); pair A,B,C,D,E,X,P,Q,R; A=(0,0); B=(8,0); C=(1.9375,3.4994); D=(3.6696,2.4996); E=(1.4531,2.6246); X=(4.3636,0); P=(2.9639,2.0189); Q=(1.8462,0); R=(6.4615,0); dot(A); dot(B); dot(C); dot(D); dot(E); dot(X); dot(P); dot(Q); dot(R); label("$A$",A,WSW); label("$B$",B,ESE); label("$C$",C,NNW); label("$D$",D,NE); label("$E$",E,WNW); label("$X$",X,SSE); label("$P$",P,NNE); label("$Q$",Q,SSW); label("$R$",R,SE); draw(A--B--C--cycle); draw(P--Q--R--cycle); draw(A--D); draw(B--E); draw(C--X); [/asy]
We have that $[ABC]=\frac{1}{2}\cdot AC \cdot BC\sin{C}=\frac{1}{2}\cdot 4\cdot {7}\sin{C}=14\sin{C}$
By Ceva's, \[3\cdot{\frac{2}{5}}\cdot{\frac{BX}{AX}}=1\implies BX=\frac{5\cdot AX}{6}\] That means that \[\frac{11\cdot {AX}}{6}=8\implies AX=\frac{48}{11} \ \text{and} \ BX=\frac{40}{11}\]
Now we apply mass points. Assume WLOG that $W_{A}=1$ . That means that
\[W_{C}=3, W_{B}=\frac{6}{5}, W_{X}=\frac{11}{5}, W_{D}=\frac{21}{5}, W_{E}=4, W_{P}=\frac{26}{5}\]
Notice now that $\triangle{PBQ}$ is similar to $\triangle{EBA}$ . Therefore,
\[\frac{PQ}{EA}=\frac{PB}{EB}\implies \frac{PQ}{3}=\frac{10}{13}\implies PQ=\frac{30}{13}\]
Also, $\triangle{PRA}$ is similar to $\triangle{DBA}$ . Therefore,
\[\frac{PA}{DA}=\frac{PR}{DB}\implies \frac{21}{26}=\frac{PR}{5}\implies PR=\frac{105}{26}\]
Because $\triangle{PQR}$ is similar to $\triangle{CAB}$ $\angle{C}=\angle{P}$
As a result, $[PQR]=\frac{1}{2}\cdot PQ \cdot PR \sin{C}=\frac{1}{2}\cdot \frac{30}{13}\cdot \frac{105}{26}\sin{P}=\frac{1575}{338}\sin{C}$
Therefore, \[\frac{[PQR]}{[ABC]}=\frac{\frac{1575}{338}\sin{C}}{14\sin{C}}=\frac{225}{676}\implies 225+676=\boxed{901}\]
| 901
|
6,536
|
https://artofproblemsolving.com/wiki/index.php/2002_AIME_II_Problems/Problem_13
| 2
|
In triangle $ABC,$ point $D$ is on $\overline{BC}$ with $CD = 2$ and $DB = 5,$ point $E$ is on $\overline{AC}$ with $CE = 1$ and $EA = 3,$ $AB = 8,$ and $\overline{AD}$ and $\overline{BE}$ intersect at $P.$ Points $Q$ and $R$ lie on $\overline{AB}$ so that $\overline{PQ}$ is parallel to $\overline{CA}$ and $\overline{PR}$ is parallel to $\overline{CB}.$ It is given that the ratio of the area of triangle $PQR$ to the area of triangle $ABC$ is $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m + n$
|
[asy] size(10cm); pair A,B,C,D,E,P,Q,R; A=(0,0); B=(8,0); C=(1.9375,3.4994); D=(3.6696,2.4996); E=(1.4531,2.6246); P=(2.9639,2.0189); Q=(1.8462,0); R=(6.4615,0); dot(A); dot(B); dot(C); dot(D); dot(E); dot(P); dot(Q); dot(R); label("$A$",A,WSW); label("$B$",B,ESE); label("$C$",C,NNW); label("$D$",D,NE); label("$E$",E,WNW); label("$P$",P,NNE); label("$Q$",Q,SSW); label("$R$",R,SE); draw(A--B--C--cycle); draw(P--Q--R--cycle); draw(A--D); draw(B--E); [/asy] Use the mass of point. Denoting the mass of $C=15,B=6,A=5,D=21,E=20$ , we can see that the mass of Q is $26$ , hence we know that $\frac{BP}{PE}=\frac{10}{3}$ , now we can find that $\frac{PQ}{AE}=\frac{10}{3}$ which implies $PQ=\frac{30}{13}$ , it is obvious that $\triangle{PQR}$ is similar to $\triangle{ACB}$ so we need to find the ration between PQ and AC, which is easy, it is $\frac{15}{26}$ , so our final answer is $\left( \frac{15}{26} \right)^2= \frac{225}{676}$ which is $\boxed{901}$ . ~bluesoul
| 901
|
6,537
|
https://artofproblemsolving.com/wiki/index.php/2002_AIME_II_Problems/Problem_15
| 1
|
Circles $\mathcal{C}_{1}$ and $\mathcal{C}_{2}$ intersect at two points, one of which is $(9,6)$ , and the product of the radii is $68$ . The x-axis and the line $y = mx$ , where $m > 0$ , are tangent to both circles. It is given that $m$ can be written in the form $a\sqrt {b}/c$ , where $a$ $b$ , and $c$ are positive integers, $b$ is not divisible by the square of any prime, and $a$ and $c$ are relatively prime. Find $a + b + c$
|
Let the smaller angle between the $x$ -axis and the line $y=mx$ be $\theta$ . Note that the centers of the two circles lie on the angle bisector of the angle between the $x$ -axis and the line $y=mx$ . Also note that if $(x,y)$ is on said angle bisector, we have that $\frac{y}{x}=\tan{\frac{\theta}{2}}$ . Let $\tan{\frac{\theta}{2}}=m_1$ , for convenience. Therefore if $(x,y)$ is on the angle bisector, then $x=\frac{y}{m_1}$ . Now let the centers of the two relevant circles be $(a/m_1 , a)$ and $(b/m_1 , b)$ for some positive reals $a$ and $b$ . These two circles are tangent to the $x$ -axis, so the radii of the circles are $a$ and $b$ respectively. We know that the point $(9,6)$ is a point on both circles, so we have that
\[(9-\frac{a}{m_1})^2+(6-a)^2=a^2\]
\[(9-\frac{b}{m_1})^2+(6-b)^2=b^2\]
Expanding these and manipulating terms gives
\[\frac{1}{m_1^2}a^2-[(18/m_1)+12]a+117=0\]
\[\frac{1}{m_1^2}b^2-[(18/m_1)+12]b+117=0\]
It follows that $a$ and $b$ are the roots of the quadratic
\[\frac{1}{m_1^2}x^2-[(18/m_1)+12]x+117=0\]
It follows from Vieta's Formulas that the product of the roots of this quadratic is $117m_1^2$ , but we were also given that the product of the radii was 68. Therefore $68=117m_1^2$ , or $m_1^2=\frac{68}{117}$ . Note that the half-angle formula for tangents is
\[\tan{\frac{\theta}{2}}=\sqrt{\frac{1-\cos{\theta}}{1+\cos{\theta}}}\]
Therefore
\[\frac{68}{117}=\frac{1-\cos{\theta}}{1+\cos{\theta}}\]
Solving for $\cos{\theta}$ gives that $\cos{\theta}=\frac{49}{185}$ . It then follows that $\sin{\theta}=\sqrt{1-\cos^2{\theta}}=\frac{12\sqrt{221}}{185}$
It then follows that $m=\tan{\theta}=\frac{12\sqrt{221}}{49}$ . Therefore $a=12$ $b=221$ , and $c=49$ . The desired answer is then $12+221+49=\boxed{282}$
| 282
|
6,538
|
https://artofproblemsolving.com/wiki/index.php/2002_AIME_II_Problems/Problem_15
| 2
|
Circles $\mathcal{C}_{1}$ and $\mathcal{C}_{2}$ intersect at two points, one of which is $(9,6)$ , and the product of the radii is $68$ . The x-axis and the line $y = mx$ , where $m > 0$ , are tangent to both circles. It is given that $m$ can be written in the form $a\sqrt {b}/c$ , where $a$ $b$ , and $c$ are positive integers, $b$ is not divisible by the square of any prime, and $a$ and $c$ are relatively prime. Find $a + b + c$
|
Let the centers of $C_1$ and $C_2$ be $A$ and $B$ , respectively, and let the point $(9, 6)$ be $P$
Because both $C_1$ and $C_2$ are tangent to the x-axis, and both of them pass through $P$ , both $A$ and $B$ must be equidistant from $P$ and the x-axis. Therefore, they must both be on the parabola with $P$ as the focus and the x-axis as the directrix. Since the coordinates of $P$ is $(9, 6)$ , we see that this parabola is the graph of the function \[y=\frac{1}{12}(x-9)^2+3=\frac{1}{12}x^2-\frac{3}{2}x+\frac{39}{4}.\]
Let $AB$ be $y=kx$ . Because $C_1$ and $C_2$ are both tangent to the x-axis, the y-coordinates of $A$ and $B$ are $r_1$ and $r_2$ , respectively, so the x-coordinates of $A$ and $B$ are $\frac{r_1}{k}$ and $\frac{r_2}{k}$ . But since $A$ and $B$ are also on the graph of the function $y=\frac{1}{12}x^2-\frac{3}{2}x+\frac{39}{4}$ , the x-coordinates of $A$ and $B$ are also the roots of the equation $kx=\frac{1}{12}x^2-\frac{3}{2}x+\frac{39}{4}$ , and by Vieta's Formulas, their product is $\frac{\frac{39}{4}}{\frac{1}{12}}=117$ . So we have $\frac{r_1}{k}\cdot \frac{r_2}{k}=117$
We are also given that $r_1r_2=68$ , so $k^2=\frac{r_1r_2}{117}=\frac{68}{117}$ , which means that $k=\sqrt{\frac{68}{117}}$ . Note that the line $AB$ is the angle bisector of the angle between the line $y=mx$ and the x-axis. Therefore, we apply the double-angle formula for tangents and get \[m=\frac{2k}{1-k^2}=\frac{2\sqrt{\frac{68}{117}}}{1-\frac{68}{117}}=\frac{12\sqrt{221}}{49}.\] Thus, the answer is $12+221+49=\boxed{282}$
| 282
|
6,539
|
https://artofproblemsolving.com/wiki/index.php/2001_AIME_I_Problems/Problem_1
| 1
|
Find the sum of all positive two-digit integers that are divisible by each of their digits.
|
Let our number be $10a + b$ $a,b \neq 0$ . Then we have two conditions: $10a + b \equiv 10a \equiv 0 \pmod{b}$ and $10a + b \equiv b \pmod{a}$ , or $a$ divides into $b$ and $b$ divides into $10a$ . Thus $b = a, 2a,$ or $5a$ (note that if $b = 10a$ , then $b$ would not be a digit).
If we ignore the case $b = 0$ as we have been doing so far, then the sum is $495 + 120 + 15 = \boxed{630}$
| 630
|
6,540
|
https://artofproblemsolving.com/wiki/index.php/2001_AIME_I_Problems/Problem_1
| 2
|
Find the sum of all positive two-digit integers that are divisible by each of their digits.
|
Using casework, we can list out all of these numbers: \[11+12+15+22+24+33+36+44+48+55+66+77+88+99=\boxed{630}.\]
| 630
|
6,541
|
https://artofproblemsolving.com/wiki/index.php/2001_AIME_I_Problems/Problem_2
| 1
|
A finite set $\mathcal{S}$ of distinct real numbers has the following properties: the mean of $\mathcal{S}\cup\{1\}$ is $13$ less than the mean of $\mathcal{S}$ , and the mean of $\mathcal{S}\cup\{2001\}$ is $27$ more than the mean of $\mathcal{S}$ . Find the mean of $\mathcal{S}$
|
Let $x$ be the mean of $\mathcal{S}$ . Let $a$ be the number of elements in $\mathcal{S}$ .
Then, the given tells us that $\frac{ax+1}{a+1}=x-13$ and $\frac{ax+2001}{a+1}=x+27$ . Subtracting, we have \begin{align*}\frac{ax+2001}{a+1}-40=\frac{ax+1}{a+1} \Longrightarrow \frac{2000}{a+1}=40 \Longrightarrow a=49\end{align*} We plug that into our very first formula, and get: \begin{align*}\frac{49x+1}{50}&=x-13 \\ 49x+1&=50x-650 \\ x&=\boxed{651}
| 651
|
6,542
|
https://artofproblemsolving.com/wiki/index.php/2001_AIME_I_Problems/Problem_2
| 2
|
A finite set $\mathcal{S}$ of distinct real numbers has the following properties: the mean of $\mathcal{S}\cup\{1\}$ is $13$ less than the mean of $\mathcal{S}$ , and the mean of $\mathcal{S}\cup\{2001\}$ is $27$ more than the mean of $\mathcal{S}$ . Find the mean of $\mathcal{S}$
|
Since this is a weighted average problem, the mean of $S$ is $\frac{13}{27}$ as far from $1$ as it is from $2001$ . Thus, the mean of $S$ is $1 + \frac{13}{13 + 27}(2001 - 1) = \boxed{651}$
| 651
|
6,543
|
https://artofproblemsolving.com/wiki/index.php/2001_AIME_I_Problems/Problem_3
| 1
|
Find the sum of the roots , real and non-real, of the equation $x^{2001}+\left(\frac 12-x\right)^{2001}=0$ , given that there are no multiple roots.
|
From Vieta's formulas , in a polynomial of the form $a_nx^n + a_{n-1}x^{n-1} + \cdots + a_0 = 0$ , then the sum of the roots is $\frac{-a_{n-1}}{a_n}$
From the Binomial Theorem , the first term of $\left(\frac 12-x\right)^{2001}$ is $-x^{2001}$ , but $x^{2001}+-x^{2001}=0$ , so the term with the largest degree is $x^{2000}$ . So we need the coefficient of that term, as well as the coefficient of $x^{1999}$
\begin{align*}\binom{2001}{1} \cdot (-x)^{2000} \cdot \left(\frac{1}{2}\right)^1&=\frac{2001x^{2000}}{2}\\ \binom{2001}{2} \cdot (-x)^{1999} \cdot \left(\frac{1}{2}\right)^2 &=\frac{-x^{1999}*2001*2000}{8}=-2001 \cdot 250x^{1999} \end{align*}
Applying Vieta's formulas, we find that the sum of the roots is $-\frac{-2001 \cdot 250}{\frac{2001}{2}}=250 \cdot 2=\boxed{500}$
| 500
|
6,544
|
https://artofproblemsolving.com/wiki/index.php/2001_AIME_I_Problems/Problem_3
| 2
|
Find the sum of the roots , real and non-real, of the equation $x^{2001}+\left(\frac 12-x\right)^{2001}=0$ , given that there are no multiple roots.
|
We find that the given equation has a $2000^{\text{th}}$ degree polynomial. Note that there are no multiple roots. Thus, if $\frac{1}{2} - x$ is a root, $x$ is also a root. Thus, we pair up $1000$ pairs of roots that sum to $\frac{1}{2}$ to get a sum of $\boxed{500}$
| 500
|
6,545
|
https://artofproblemsolving.com/wiki/index.php/2001_AIME_I_Problems/Problem_3
| 3
|
Find the sum of the roots , real and non-real, of the equation $x^{2001}+\left(\frac 12-x\right)^{2001}=0$ , given that there are no multiple roots.
|
Note that if $r$ is a root, then $\frac{1}{2}-r$ is a root and they sum up to $\frac{1}{2}.$ We make the substitution $y=x-\frac{1}{4}$ so \[(\frac{1}{4}+y)^{2001}+(\frac{1}{4}-y)^{2001}=0.\] Expanding gives \[2\cdot\frac{1}{4}\cdot\binom{2001}{1}y^{2000}-0y^{1999}+\cdots\] so by Vieta, the sum of the roots of $y$ is 0. Since $x$ has a degree of 2000, then $x$ has 2000 roots so the sum of the roots is \[2000(\sum_{n=1}^{2000} y+\frac{1}{4})=2000(0+\frac{1}{4})=\boxed{500}.\]
| 500
|
6,546
|
https://artofproblemsolving.com/wiki/index.php/2001_AIME_I_Problems/Problem_4
| 1
|
In triangle $ABC$ , angles $A$ and $B$ measure $60$ degrees and $45$ degrees, respectively. The bisector of angle $A$ intersects $\overline{BC}$ at $T$ , and $AT=24$ . The area of triangle $ABC$ can be written in the form $a+b\sqrt{c}$ , where $a$ $b$ , and $c$ are positive integers, and $c$ is not divisible by the square of any prime. Find $a+b+c$
|
After chasing angles, $\angle ATC=75^{\circ}$ and $\angle TCA=75^{\circ}$ , meaning $\triangle TAC$ is an isosceles triangle and $AC=24$
Using law of sines on $\triangle ABC$ , we can create the following equation:
$\frac{24}{\sin(\angle ABC)}$ $=$ $\frac{BC}{\sin(\angle BAC)}$
$\angle ABC=45^{\circ}$ and $\angle BAC=60^{\circ}$ , so $BC = 12\sqrt{6}$
We can then use the Law of Sines area formula $\frac{1}{2} \cdot BC \cdot AC \cdot \sin(\angle BCA)$ to find the area of the triangle.
$\sin(75)$ can be found through the sin addition formula.
$\sin(75)$ $=$ $\frac{\sqrt{6} + \sqrt{2}}{4}$
Therefore, the area of the triangle is $\frac{\sqrt{6} + \sqrt{2}}{4}$ $\cdot$ $24$ $\cdot$ $12\sqrt{6}$ $\cdot$ $\frac{1}{2}$
$72\sqrt{3} + 216$
$72 + 3 + 216 =$ $\boxed{291}$
| 291
|
6,547
|
https://artofproblemsolving.com/wiki/index.php/2001_AIME_I_Problems/Problem_4
| 2
|
In triangle $ABC$ , angles $A$ and $B$ measure $60$ degrees and $45$ degrees, respectively. The bisector of angle $A$ intersects $\overline{BC}$ at $T$ , and $AT=24$ . The area of triangle $ABC$ can be written in the form $a+b\sqrt{c}$ , where $a$ $b$ , and $c$ are positive integers, and $c$ is not divisible by the square of any prime. Find $a+b+c$
|
First, draw a good diagram.
We realize that $\angle C = 75^\circ$ , and $\angle CAT = 30^\circ$ . Therefore, $\angle CTA = 75^\circ$ as well, making $\triangle CAT$ an isosceles triangle. $AT$ and $AC$ are congruent, so $AC=24$ . We now drop an altitude from $C$ , and call the foot this altitude point $D$
By 30-60-90 triangles, $AD=12$ and $CD=12\sqrt{3}$
We also notice that $\triangle CDB$ is an isosceles right triangle. $CD$ is congruent to $BD$ , which makes $BD=12\sqrt{3}$ . The base $AB$ is $12+12\sqrt{3}$ , and the altitude $CD=12\sqrt{3}$ . We can easily find that the area of triangle $ABC$ is $216+72\sqrt{3}$ , so $a+b+c=\boxed{291}$
| 291
|
6,548
|
https://artofproblemsolving.com/wiki/index.php/2001_AIME_I_Problems/Problem_5
| 1
|
An equilateral triangle is inscribed in the ellipse whose equation is $x^2+4y^2=4$ . One vertex of the triangle is $(0,1)$ , one altitude is contained in the y-axis, and the square of the length of each side is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$
|
Denote the vertices of the triangle $A,B,$ and $C,$ where $B$ is in quadrant 4 and $C$ is in quadrant $3.$
Note that the slope of $\overline{AC}$ is $\tan 60^\circ = \sqrt {3}.$ Hence, the equation of the line containing $\overline{AC}$ is \[y = x\sqrt {3} + 1.\] This will intersect the ellipse when \begin{eqnarray*}4 = x^{2} + 4y^{2} & = & x^{2} + 4(x\sqrt {3} + 1)^{2} \\ & = & x^{2} + 4(3x^{2} + 2x\sqrt {3} + 1) \implies x(13x+8\sqrt 3)=0\implies x = \frac { - 8\sqrt {3}}{13}. \end{eqnarray*} We ignore the $x=0$ solution because it is not in quadrant 3.
Since the triangle is symmetric with respect to the y-axis, the coordinates of $B$ and $C$ are now $\left(\frac {8\sqrt {3}}{13},y_{0}\right)$ and $\left(\frac { - 8\sqrt {3}}{13},y_{0}\right),$ respectively, for some value of $y_{0}.$
It is clear that the value of $y_{0}$ is irrelevant to the length of $BC$ . Our answer is \[BC = 2*\frac {8\sqrt {3}}{13}=\sqrt {4\left(\frac {8\sqrt {3}}{13}\right)^{2}} = \sqrt {\frac {768}{169}}\implies m + n = \boxed{937}.\]
| 937
|
6,549
|
https://artofproblemsolving.com/wiki/index.php/2001_AIME_I_Problems/Problem_5
| 2
|
An equilateral triangle is inscribed in the ellipse whose equation is $x^2+4y^2=4$ . One vertex of the triangle is $(0,1)$ , one altitude is contained in the y-axis, and the square of the length of each side is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$
|
Solving for $y$ in terms of $x$ gives $y=\sqrt{4-x^2}/2$ , so the two other points of the triangle are $(x,\sqrt{4-x^2}/2)$ and $(-x,\sqrt{4-x^2}/2)$ , which are a distance of $2x$ apart. Thus $2x$ equals the distance between $(x,\sqrt{4-x^2}/2)$ and $(0,1)$ , so by the distance formula we have
\[2x=\sqrt{x^2+(1-\sqrt{4-x^2}/2)^2}.\]
Squaring both sides and simplifying through algebra yields $x^2=192/169$ , so $2x=\sqrt{768/169}$ and the answer is $\boxed{937}$
| 937
|
6,550
|
https://artofproblemsolving.com/wiki/index.php/2001_AIME_I_Problems/Problem_5
| 3
|
An equilateral triangle is inscribed in the ellipse whose equation is $x^2+4y^2=4$ . One vertex of the triangle is $(0,1)$ , one altitude is contained in the y-axis, and the square of the length of each side is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$
|
Since the altitude goes along the $y$ axis, this means that the base is a horizontal line, which means that the endpoints of the base are $(x,y)$ and $(-x,y)$ , and WLOG, we can say that $x$ is positive.
Now, since all sides of an equilateral triangle are the same, we can do this (distance from one of the endpoints of the base to the vertex and the length of the base):
$\sqrt{x^2 + (y-1)^2} = 2x$
Square both sides,
$x^2 + (y-1)^2 = 4x^2\implies (y-1)^2 = 3x^2$
Now, with the equation of the ellipse: $x^2 + 4y^2 = 4$
$x^2 = 4-4y^2$
$3x^2 = 12-12y^2$
Substituting,
$12-12y^2 = y^2 - 2y +1$
Moving stuff around and solving:
$y = \frac{-11}{13}, 1$
The second is found to be extraneous, so, when we go back and figure out $x$ and then $2x$ (which is the side length), we find it to be:
$\sqrt{\frac{768}{169}}$
and so we get the desired answer of $\boxed{937}$
| 937
|
6,551
|
https://artofproblemsolving.com/wiki/index.php/2001_AIME_I_Problems/Problem_5
| 4
|
An equilateral triangle is inscribed in the ellipse whose equation is $x^2+4y^2=4$ . One vertex of the triangle is $(0,1)$ , one altitude is contained in the y-axis, and the square of the length of each side is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$
|
Denote $(0,1)$ as vertex $A,$ $B$ as the vertex to the left of the $y$ -axis and $C$ as the vertex to the right of the $y$ -axis. Let $D$ be the intersection of $BC$ and the $y$ -axis.
Let $x_0$ be the $x$ -coordinate of $C.$ This implies \[C=\left(x_0 , \sqrt{\frac{4-x_0^2}{4}}\right)\] and \[B=\left(-x_0 , \sqrt{\frac{4-x_0^2}{4}}\right).\] Note that $BC=2x_0$ and \[\frac{BC}{\sqrt3}=AD=1-\sqrt{\frac{4-x_0^2}{4}}.\] This yields \[\frac{2x_0}{\sqrt3}=1-\sqrt{\frac{4-x_0^2}{4}}.\] Re-arranging and squaring, we have \[\frac{4-x_0^2}{4}=\frac{4x_0^2}{3}-\frac{4x_0}{\sqrt3} +1.\] Simplifying and solving for $x_0$ , we have \[x_0=\frac{48}{13\sqrt 3}.\] As the length of each side is $2x_0,$ our desired length is \[4x_0^2=\frac{768}{169}\] which means our desired answer is \[768+169=\boxed{937}\]
| 937
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6,552
|
https://artofproblemsolving.com/wiki/index.php/2001_AIME_I_Problems/Problem_5
| 6
|
An equilateral triangle is inscribed in the ellipse whose equation is $x^2+4y^2=4$ . One vertex of the triangle is $(0,1)$ , one altitude is contained in the y-axis, and the square of the length of each side is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$
|
Consider the transformation $(x,y)$ to $(x/2, y).$ This sends the ellipse to the unit circle. If we let $n$ be one-fourth of the side length of the triangle, the equilateral triangle is sent to an isosceles triangle with side lengths $2n, n\sqrt{13}, n\sqrt{13}.$ Let the triangle be $ABC$ such that $AB=AC.$ Let the foot of the altitude from A be $X.$ Then $BX=n,$ and $AX=2n\sqrt{3}.$ Let $C$ be a point such that $AC$ is a diameter of the unit circle. Then $XC=2-2n\sqrt{3}.$ Using power of a point on X, \[n^2=2n\sqrt{3}(2-2n\sqrt{3})\] Simplifying gets us to $13n^2=4n\sqrt{3}.$ Then, $n=\dfrac{4\sqrt{3}}{13}$ which means the side length is $\dfrac{16\sqrt{3}}{13}=\sqrt{\dfrac{768}{169}}.$ Thus, the answer is $768+169=\boxed{937}.$
| 937
|
6,553
|
https://artofproblemsolving.com/wiki/index.php/2001_AIME_I_Problems/Problem_6
| 4
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A fair die is rolled four times. The probability that each of the final three rolls is at least as large as the roll preceding it may be expressed in the form $\frac{m}{n}$ where $m$ and $n$ are relatively prime positive integers . Find $m + n$
|
Call the dice rolls $a, b, c, d$ . The difference between the $a$ and $d$ distinguishes the number of possible rolls there are.
Continuing, we see that the sum is equal to $\sum_{i = 0}^{5}{i+2\choose2}(6 - i) = 6 + 15 + 24 + 30 + 30 + 21 = 126$ . The requested probability is $\frac{126}{6^4} = \frac{7}{72}$ and our answer is $\boxed{79}$
| 79
|
6,554
|
https://artofproblemsolving.com/wiki/index.php/2001_AIME_I_Problems/Problem_6
| 9
|
A fair die is rolled four times. The probability that each of the final three rolls is at least as large as the roll preceding it may be expressed in the form $\frac{m}{n}$ where $m$ and $n$ are relatively prime positive integers . Find $m + n$
|
Let the rolls be $a,b,c$ and $d\newline$ let $z=a-1, e=b-a, f=c-b, g=d-c, h=6-d\newline$ $z+e+f+g+h=5\newline$ This equation has $C(5+5-1, 5-1)=126$ integer solutions $\newline$ $126/1296=7/72\newline$ $7+72=\boxed{79}$ ~ryanbear
| 79
|
6,555
|
https://artofproblemsolving.com/wiki/index.php/2001_AIME_I_Problems/Problem_7
| 1
|
Triangle $ABC$ has $AB=21$ $AC=22$ and $BC=20$ . Points $D$ and $E$ are located on $\overline{AB}$ and $\overline{AC}$ , respectively, such that $\overline{DE}$ is parallel to $\overline{BC}$ and contains the center of the inscribed circle of triangle $ABC$ . Then $DE=m/n$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$
|
Let $I$ be the incenter of $\triangle ABC$ , so that $BI$ and $CI$ are angle bisectors of $\angle ABC$ and $\angle ACB$ respectively. Then, $\angle BID = \angle CBI = \angle DBI,$ so $\triangle BDI$ is isosceles , and similarly $\triangle CEI$ is isosceles. It follows that $DE = DB + EC$ , so the perimeter of $\triangle ADE$ is $AD + AE + DE = AB + AC = 43$ . Hence, the ratio of the perimeters of $\triangle ADE$ and $\triangle ABC$ is $\frac{43}{63}$ , which is the scale factor between the two similar triangles, and thus $DE = \frac{43}{63} \times 20 = \frac{860}{63}$ . Thus, $m + n = \boxed{923}$
| 923
|
6,556
|
https://artofproblemsolving.com/wiki/index.php/2001_AIME_I_Problems/Problem_7
| 2
|
Triangle $ABC$ has $AB=21$ $AC=22$ and $BC=20$ . Points $D$ and $E$ are located on $\overline{AB}$ and $\overline{AC}$ , respectively, such that $\overline{DE}$ is parallel to $\overline{BC}$ and contains the center of the inscribed circle of triangle $ABC$ . Then $DE=m/n$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$
|
The semiperimeter of $ABC$ is $s = \frac{20 + 21 + 22}{2} = \frac{63}{2}$ . By Heron's formula , the area of the whole triangle is $A = \sqrt{s(s-a)(s-b)(s-c)} = \frac{21\sqrt{1311}}{4}$ . Using the formula $A = rs$ , we find that the inradius is $r = \frac{A}{s} = \frac{\sqrt{1311}}6$ . Since $\triangle ADE \sim \triangle ABC$ , the ratio of the heights of triangles $ADE$ and $ABC$ is equal to the ratio between sides $DE$ and $BC$ . From $A=\frac{1}{2}bh$ , we find $h_{ABC} = \frac{21\sqrt{1311}}{40}$ . Thus, we have
Solving for $DE$ gives $DE=\frac{860}{63},$ so the answer is $m+n=\boxed{923}$
| 923
|
6,557
|
https://artofproblemsolving.com/wiki/index.php/2001_AIME_I_Problems/Problem_7
| 3
|
Triangle $ABC$ has $AB=21$ $AC=22$ and $BC=20$ . Points $D$ and $E$ are located on $\overline{AB}$ and $\overline{AC}$ , respectively, such that $\overline{DE}$ is parallel to $\overline{BC}$ and contains the center of the inscribed circle of triangle $ABC$ . Then $DE=m/n$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$
|
Let $P$ be the incenter ; then it is be the intersection of all three angle bisectors . Draw the bisector $AP$ to where it intersects $BC$ , and name the intersection $F$
Using the angle bisector theorem , we know the ratio $BF:CF$ is $21:22$ , thus we shall assign a weight of $22$ to point $B$ and a weight of $21$ to point $C$ , giving $F$ a weight of $43$ . In the same manner, using another bisector, we find that $A$ has a weight of $20$ . So, now we know $P$ has a weight of $63$ , and the ratio of $FP:PA$ is $20:43$ . Therefore, the smaller similar triangle $ADE$ is $43/63$ the height of the original triangle $ABC$ . So, $DE$ is $43/63$ the size of $BC$ . Multiplying this ratio by the length of $BC$ , we find $DE$ is $860/63 = m/n$ . Therefore, $m+n=\boxed{923}$
| 923
|
6,558
|
https://artofproblemsolving.com/wiki/index.php/2001_AIME_I_Problems/Problem_7
| 4
|
Triangle $ABC$ has $AB=21$ $AC=22$ and $BC=20$ . Points $D$ and $E$ are located on $\overline{AB}$ and $\overline{AC}$ , respectively, such that $\overline{DE}$ is parallel to $\overline{BC}$ and contains the center of the inscribed circle of triangle $ABC$ . Then $DE=m/n$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$
|
More directly than Solution 2, we have \[DE=BC\left(\frac{h_a-r}{h_a}\right)=20\left(1-\frac{r}{\frac{[ABC]}{\frac{BC}{2}}}\right)=20\left(1-\frac{10r}{sr}\right)=20\left(1-\frac{10}{\frac{63}{2}}\right)=\frac{860}{63}\implies \boxed{923}.\]
| 923
|
6,559
|
https://artofproblemsolving.com/wiki/index.php/2001_AIME_I_Problems/Problem_7
| 5
|
Triangle $ABC$ has $AB=21$ $AC=22$ and $BC=20$ . Points $D$ and $E$ are located on $\overline{AB}$ and $\overline{AC}$ , respectively, such that $\overline{DE}$ is parallel to $\overline{BC}$ and contains the center of the inscribed circle of triangle $ABC$ . Then $DE=m/n$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$
|
Diagram borrowed from Solution 3.
Let the angle bisector of $\angle{A}$ intersects $BC$ at $F$
Applying the Angle Bisector Theorem on $\angle{A}$ we have \[\frac{AB}{BF}=\frac{AC}{CF}\] \[BF=BC\cdot(\frac{AB}{AB+AC})\] \[BF=\frac{420}{43}\] Since $BP$ is the angle bisector of $\angle{B}$ , we can once again apply the Angle Bisector Theorem on $\angle{B}$ which gives \[\frac{BA}{AP}=\frac{BF}{FP}\] \[\frac{AP}{PF}=\frac{AB}{BF}=\frac{41}{20}\] Since $\bigtriangleup ADE\sim\bigtriangleup ABC$ we have \[\frac{DE}{BC}=\frac{AP}{AF}\] \[DE=BC\cdot(\frac{AP}{(\frac{61}{41})\cdot AP})\] Solving gets $DE=\frac{860}{63}$ . Thus $m+n=860+63=\boxed{923}$
| 923
|
6,560
|
https://artofproblemsolving.com/wiki/index.php/2001_AIME_I_Problems/Problem_7
| 6
|
Triangle $ABC$ has $AB=21$ $AC=22$ and $BC=20$ . Points $D$ and $E$ are located on $\overline{AB}$ and $\overline{AC}$ , respectively, such that $\overline{DE}$ is parallel to $\overline{BC}$ and contains the center of the inscribed circle of triangle $ABC$ . Then $DE=m/n$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$
|
Label $P$ the point the angle bisector of $A$ intersects ${BC}$ . First we find ${BP}$ and ${PC}$ . By the Angle Bisector Theorem, $\frac{BP}{PC} = \frac{21}{22}$ and solving for each using the fact that ${BC} = 20$ , we see that ${BP} = \frac{420}{43}$ and $PC = \frac{440}{43}$
\[{AP} = 21*22 - \frac{440}{43}\cdot\frac{420}{43}\] \[{AP} = 21*22 - \frac{440\cdot420}{43^2}\]
Now we can calculate what ${AO}$ is. Using the formula to find the distance from a vertex to the incenter, ${AO} = \frac{43}{63} \cdot[21\cdot22 - \frac{420*440}{43^2}] = \frac{43^2\cdot22 - 20\cdot440}{43\cdot3}$
Now because $\triangle{APE} ~ \triangle{ABC}$ , we can find ${DE}$ by $\frac{AO}{AP} \cdot 20$ . Dividing and simplifying, we see that $\frac{1}{21}\cdot\frac{43}{3}\cdot20 = \frac{860}{63}$ . So the answer is $\boxed{923}$
| 923
|
6,561
|
https://artofproblemsolving.com/wiki/index.php/2001_AIME_I_Problems/Problem_8
| 1
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Call a positive integer $N$ 7-10 double if the digits of the base- $7$ representation of $N$ form a base- $10$ number that is twice $N$ . For example, $51$ is a 7-10 double because its base- $7$ representation is $102$ . What is the largest 7-10 double?
|
Let $A$ be the base $10$ representation of our number, and let $B$ be its base $7$ representation.
Given this is an AIME problem, $A<1000$ . If we look at $B$ in base $10$ , it must be equal to $2A$ , so $B<2000$ when $B$ is looked at in base $10.$
If $B$ in base $10$ is less than $2000$ , then $B$ as a number in base $7$ must be less than $2*7^3=686$
$686$ is non-existent in base $7$ , so we're gonna have to bump that down to $666_7$
This suggests that $A$ is less than $\frac{666}{2}=333$
Guess and check shows that $A<320$ , and checking values in that range produces $\boxed{315}$
| 315
|
6,562
|
https://artofproblemsolving.com/wiki/index.php/2001_AIME_I_Problems/Problem_10
| 1
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Let $S$ be the set of points whose coordinates $x,$ $y,$ and $z$ are integers that satisfy $0\le x\le2,$ $0\le y\le3,$ and $0\le z\le4.$ Two distinct points are randomly chosen from $S.$ The probability that the midpoint of the segment they determine also belongs to $S$ is $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$
|
The distance between the $x$ $y$ , and $z$ coordinates must be even so that the midpoint can have integer coordinates. Therefore,
However, we have $3\cdot 4\cdot 5 = 60$ cases where we have simply taken the same point twice, so we subtract those. Therefore, our answer is $\frac {5\cdot 8\cdot 13 - 60}{60\cdot 59} = \frac {23}{177}\Longrightarrow m+n = \boxed{200}$
| 200
|
6,563
|
https://artofproblemsolving.com/wiki/index.php/2001_AIME_I_Problems/Problem_10
| 2
|
Let $S$ be the set of points whose coordinates $x,$ $y,$ and $z$ are integers that satisfy $0\le x\le2,$ $0\le y\le3,$ and $0\le z\le4.$ Two distinct points are randomly chosen from $S.$ The probability that the midpoint of the segment they determine also belongs to $S$ is $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$
|
There are $(2 + 1)(3 + 1)(4 + 1) = 60$ points in total. We group the points by parity of each individual coordinate -- that is, if $x$ is even or odd, $y$ is even or odd, and $z$ is even or odd. Note that to have something that works, the two points must have this same type of classification (otherwise, if one doesn't match, the resulting sum for the coordinates will be odd at that particular spot).
There are $12$ EEEs (the first position denotes the parity of $x,$ the second $y,$ and the third $z.$ ), $8$ EEOs, $12$ EOEs, $6$ OEEs, $8$ EOOs, $4$ OEOs, $6$ OOEs, and $4$ OOOs. Doing a sanity check, $12 + 8 + 12 + 6 + 8 + 4 + 6 + 4 = 60,$ which is the total number of points.
Now, we can see that there are $12 \cdot 11$ ways to choose two EEEs (respective to order), $8 \cdot 7$ ways to choose two EEOs, and so on. Therefore, we get \[12\cdot11 + 8\cdot7 + 12\cdot11 + 6\cdot5 + 8\cdot7 + 4\cdot3 + 6\cdot5 + 4\cdot3 = 460\] ways to choose two points where order matters. There are $60 \cdot 59$ total ways to do this, so we get a final answer of \[\dfrac{460}{60 \cdot 59} = \dfrac{23}{3 \cdot 59} = \dfrac{23}{177},\] for our answer of $23 + 177 = \boxed{200}.$
| 200
|
6,564
|
https://artofproblemsolving.com/wiki/index.php/2001_AIME_I_Problems/Problem_10
| 3
|
Let $S$ be the set of points whose coordinates $x,$ $y,$ and $z$ are integers that satisfy $0\le x\le2,$ $0\le y\le3,$ and $0\le z\le4.$ Two distinct points are randomly chosen from $S.$ The probability that the midpoint of the segment they determine also belongs to $S$ is $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$
|
Similarly to Solution 2, we note that there are $60$ points and that the parities of the two points' coordinates must be the same in order for the midpoint to be in $S$
Ignore the distinct points condition. The probability that the midpoint is in $S$ is then \[\left(\left(\frac 23\right)^2+\left(\frac 13\right)^2\right)\left(\left(\frac 24\right)^2+\left(\frac 24\right)^2\right)\left(\left(\frac 35\right)^2+\left(\frac 25\right)^2\right)=\frac{13}{90}.\]
Note that $\frac{13}{90}=\frac{520}{3600}$ . Since there are $3600$ total ways to choose $2$ points from $S$ , there must be $520$ pairs of points that have their midpoint in $S$ . Of these pairs, $60$ of them contain identical points (not distinct).
Subtracting these cases, our answer is $\frac{520-60}{3600-60}=\frac{23}{177}\implies\boxed{200}$
| 200
|
6,565
|
https://artofproblemsolving.com/wiki/index.php/2001_AIME_I_Problems/Problem_10
| 4
|
Let $S$ be the set of points whose coordinates $x,$ $y,$ and $z$ are integers that satisfy $0\le x\le2,$ $0\le y\le3,$ and $0\le z\le4.$ Two distinct points are randomly chosen from $S.$ The probability that the midpoint of the segment they determine also belongs to $S$ is $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$
|
There are $(2 + 1)(3 + 1)(4 + 1) = 60$ points in total. Note that in order for the midpoint of the line segment to be a lattice point, the lengths on the x, y, and z axis must be even numbers. We will define all segments by denoting the amount they extend in each dimension: $(x, y, z)$ . For example, the longest diagonal possible will be $(2,3,4)$ , the space diagonal of the box. Thus, any line segment must have dimensions that are even. For $x$ the segment may have a value of $0$ for $x$ , (in which case the segment would be two dimensional) or a value of $2$ . The same applies for $y$ , because although it is three units long the longest even integer is two. For $z$ the value may be $0$ $2$ , or $4$ . Notice that if a value is zero, then the segment will pertain to only two dimensions. If two values are zero then the line segment becomes one dimensional.
Then the total number of possibilities will be $2 \cdot 2 \cdot 3$
Listing them out appears as follows:
$2,2,4$
$2,2,2$
$2,2,0$
$2,0,4$
$2,0,2$
$2,0,0$
$0,2,4$
$0,2,2$
$0,2,0$
$0,0,4$
$0,0,2$
$0,0,0$ * this value is a single point
Now, picture every line segment to be the space diagonal of a box. Allow this box to define the space the segment occupies. The question now transforms into "how many ways can we arrange this smaller box in the two by three by four?".
Notice that the amount an edge can shift inside the larger box is the length of an edge of the larger box (2, 3, or 4) minus the edge of the smaller box (also known as the edge), plus one. For example, (0, 2, 2) would be $3 \cdot 2 \cdot 3$ . Repeat this process.
$2,2,4$
$2,2,2$
$2,2,0$ 10
$2,0,4$
$2,0,2$ 12
$2,0,0$ 20
$0,2,4$
$0,2,2$ 18
$0,2,0$ 30
$0,0,4$ 12
$0,0,2$ 36
$0,0,0$ 60 * this won't be included, but notice that sixty the number of lattice points
Finally, we remember that there are four distinct space diagonals in a box, so we should multiply every value by four, right? Unfortunately we forgot to consider that some values have only one or two dimensions. They should be multiplied by one or two, respectively. This is because segments with two dimensions are the diagonals of a rectangle and thus have two orientations. Then any value on our list without any zeroes will be multiplied by four, and any value on our list with only one zero will be multiplied by two, and finally any value on our list with two zeroes will be multiplied by one:
$2,2,4$ 2 8
$2,2,2$ 6 24
$2,2,0$ 10 20
$2,0,4$ 4 8
$2,0,2$ 12 24
$2,0,0$ 20 20
$0,2,4$ 6 12
$0,2,2$ 18 36
$0,2,0$ 30 30
$0,0,4$ 12 12
$0,0,2$ 36 36
$0,0,0$ 60 * it's nice to point out that this value will be multiplied by zero
add every value on the rightmost side of each term and we will receive $230$ . Multiply by two because each segment can be flipped, to receive $460$ . There are $60 \cdot 59$ ways to choose two distinct points, so we get \[\dfrac{460}{60 \cdot 59} = \dfrac{23}{3 \cdot 59} = \dfrac{23}{177},\] for our answer of $23 + 177 = \boxed{200}$
| 200
|
6,566
|
https://artofproblemsolving.com/wiki/index.php/2001_AIME_I_Problems/Problem_11
| 1
|
In a rectangular array of points, with 5 rows and $N$ columns, the points are numbered consecutively from left to right beginning with the top row. Thus the top row is numbered 1 through $N,$ the second row is numbered $N + 1$ through $2N,$ and so forth. Five points, $P_1, P_2, P_3, P_4,$ and $P_5,$ are selected so that each $P_i$ is in row $i.$ Let $x_i$ be the number associated with $P_i.$ Now renumber the array consecutively from top to bottom, beginning with the first column. Let $y_i$ be the number associated with $P_i$ after the renumbering. It is found that $x_1 = y_2,$ $x_2 = y_1,$ $x_3 = y_4,$ $x_4 = y_5,$ and $x_5 = y_3.$ Find the smallest possible value of $N.$
|
Let each point $P_i$ be in column $c_i$ . The numberings for $P_i$ can now be defined as follows. \begin{align*}x_i &= (i - 1)N + c_i\\ y_i &= (c_i - 1)5 + i \end{align*}
We can now convert the five given equalities. \begin{align}x_1&=y_2 & \Longrightarrow & & c_1 &= 5 c_2-3\\ x_2&=y_1 & \Longrightarrow & & N+c_2 &= 5 c_1-4\\ x_3&=y_4 & \Longrightarrow & & 2 N+c_3 &= 5 c_4-1\\ x_4&=y_5 & \Longrightarrow & & 3 N+c_4 &= 5 c_5\\ x_5&=y_3 & \Longrightarrow & & 4 N+c_5 &= 5 c_3-2 \end{align} Equations $(1)$ and $(2)$ combine to form \[N = 24c_2 - 19\] Similarly equations $(3)$ $(4)$ , and $(5)$ combine to form \[117N +51 = 124c_3\] Take this equation modulo 31 \[24N+20\equiv 0 \pmod{31}\] And substitute for N \[24 \cdot 24 c_2 - 24 \cdot 19 +20\equiv 0 \pmod{31}\] \[18 c_2 \equiv 2 \pmod{31}\]
Thus the smallest $c_2$ might be is $7$ and by substitution $N = 24 \cdot 7 - 19 = 149$
The column values can also easily be found by substitution \begin{align*}c_1&=32\\ c_2&=7\\ c_3&=141\\ c_4&=88\\ c_5&=107 \end{align*} As these are all positive and less than $N$ $\boxed{149}$ is the solution.
| 149
|
6,567
|
https://artofproblemsolving.com/wiki/index.php/2001_AIME_I_Problems/Problem_11
| 2
|
In a rectangular array of points, with 5 rows and $N$ columns, the points are numbered consecutively from left to right beginning with the top row. Thus the top row is numbered 1 through $N,$ the second row is numbered $N + 1$ through $2N,$ and so forth. Five points, $P_1, P_2, P_3, P_4,$ and $P_5,$ are selected so that each $P_i$ is in row $i.$ Let $x_i$ be the number associated with $P_i.$ Now renumber the array consecutively from top to bottom, beginning with the first column. Let $y_i$ be the number associated with $P_i$ after the renumbering. It is found that $x_1 = y_2,$ $x_2 = y_1,$ $x_3 = y_4,$ $x_4 = y_5,$ and $x_5 = y_3.$ Find the smallest possible value of $N.$
|
If we express all the $c_i$ in terms of $N$ , we have \[24c_1=5N+23\] \[24c_2=N+19\] \[124c_3=117N+51\] \[124c_4=73N+35\] \[124c_5=89N+7\]
It turns out that there exists such an array satisfying the problem conditions if and only if \[N\equiv 149 \pmod{744}\]
In addition, the first two equation can be written $n = 5mod24$ , and chasing variables in the last three equation gives us $89n + 7 = 124e$ . With these two equations you may skip a lot of rewriting and testing. $\boxed{149}$ still appears as our answer.
| 149
|
6,568
|
https://artofproblemsolving.com/wiki/index.php/2001_AIME_I_Problems/Problem_13
| 1
|
In a certain circle , the chord of a $d$ -degree arc is $22$ centimeters long, and the chord of a $2d$ -degree arc is $20$ centimeters longer than the chord of a $3d$ -degree arc, where $d < 120.$ The length of the chord of a $3d$ -degree arc is $- m + \sqrt {n}$ centimeters, where $m$ and $n$ are positive integers. Find $m + n.$
|
Note that a cyclic quadrilateral in the form of an isosceles trapezoid can be formed from three chords of three $d$ -degree arcs and one chord of one $3d$ -degree arc. The diagonals of this trapezoid turn out to be two chords of two $2d$ -degree arcs. Let $AB$ $AC$ , and $BD$ be the chords of the $d$ -degree arcs, and let $CD$ be the chord of the $3d$ -degree arc. Also let $x$ be equal to the chord length of the $3d$ -degree arc. Hence, the length of the chords, $AD$ and $BC$ , of the $2d$ -degree arcs can be represented as $x + 20$ , as given in the problem.
Using Ptolemy's theorem,
\[AB(CD) + AC(BD) = AD(BC)\] \[22x + 22(22) = (x + 20)^2\] \[22x + 484 = x^2 + 40x + 400\] \[0 = x^2 + 18x - 84\]
We can then apply the quadratic formula to find the positive root to this equation since polygons obviously cannot have sides of negative length. \[x = \frac{-18 + \sqrt{18^2 + 4(84)}}{2}\] \[x = \frac{-18 + \sqrt{660}}{2}\]
$x$ simplifies to $\frac{-18 + 2\sqrt{165}}{2},$ which equals $-9 + \sqrt{165}.$ Thus, the answer is $9 + 165 = \boxed{174}$
| 174
|
6,569
|
https://artofproblemsolving.com/wiki/index.php/2001_AIME_I_Problems/Problem_13
| 2
|
In a certain circle , the chord of a $d$ -degree arc is $22$ centimeters long, and the chord of a $2d$ -degree arc is $20$ centimeters longer than the chord of a $3d$ -degree arc, where $d < 120.$ The length of the chord of a $3d$ -degree arc is $- m + \sqrt {n}$ centimeters, where $m$ and $n$ are positive integers. Find $m + n.$
|
Let $z=\frac{d}{2},$ and $R$ be the circumradius. From the given information, \[2R\sin z=22\] \[2R(\sin 2z-\sin 3z)=20\] Dividing the latter by the former, \[\frac{2\sin z\cos z-(3\cos^2z\sin z-\sin^3 z)}{\sin z}=2\cos z-(3\cos^2z-\sin^2z)=1+2\cos z-4\cos^2z=\frac{10}{11}\] \[4\cos^2z-2\cos z-\frac{1}{11}=0 (1)\] We want to find \[\frac{22\sin (3z)}{\sin z}=22(3-4\sin^2z)=22(4\cos^2z-1).\] From $(1),$ this is equivalent to $44\cos z-20.$ Using the quadratic formula, we find that the desired length is equal to $\sqrt{165}-9,$ so our answer is $\boxed{174}$
| 174
|
6,570
|
https://artofproblemsolving.com/wiki/index.php/2001_AIME_I_Problems/Problem_13
| 3
|
In a certain circle , the chord of a $d$ -degree arc is $22$ centimeters long, and the chord of a $2d$ -degree arc is $20$ centimeters longer than the chord of a $3d$ -degree arc, where $d < 120.$ The length of the chord of a $3d$ -degree arc is $- m + \sqrt {n}$ centimeters, where $m$ and $n$ are positive integers. Find $m + n.$
|
Let $z=\frac{d}{2}$ $R$ be the circumradius, and $a$ be the length of 3d degree chord. Using the extended sine law, we obtain: \[22=2R\sin(z)\] \[20+a=2R\sin(2z)\] \[a=2R\sin(3z)\] Dividing the second from the first we get $\cos(z)=\frac{20+a}{44}$ By the triple angle formula we can manipulate the third equation as follows: $a=2R\times \sin(3z)=\frac{22}{\sin(z)} \times (3\sin(z)-4\sin^3(z)) = 22(3-4\sin^2(z))=22(4\cos^2(z)-1)=\frac{(20+a)^2}{22}-22$ Solving the quadratic equation gives the answer to be $\boxed{174}$
| 174
|
6,571
|
https://artofproblemsolving.com/wiki/index.php/2001_AIME_I_Problems/Problem_14
| 1
|
A mail carrier delivers mail to the nineteen houses on the east side of Elm Street. The carrier notices that no two adjacent houses ever get mail on the same day, but that there are never more than two houses in a row that get no mail on the same day. How many different patterns of mail delivery are possible?
|
Let $0$ represent a house that does not receive mail and $1$ represent a house that does receive mail. This problem is now asking for the number of $19$ -digit strings of $0$ 's and $1$ 's such that there are no two consecutive $1$ 's and no three consecutive $0$ 's.
The last two digits of any $n$ -digit string can't be $11$ , so the only possibilities are $00$ $01$ , and $10$
Let $a_n$ be the number of $n$ -digit strings ending in $00$ $b_n$ be the number of $n$ -digit strings ending in $01$ , and $c_n$ be the number of $n$ -digit strings ending in $10$
If an $n$ -digit string ends in $00$ , then the previous digit must be a $1$ , and the last two digits of the $n-1$ digits substring will be $10$ . So \[a_{n} = c_{n-1}.\]
If an $n$ -digit string ends in $01$ , then the previous digit can be either a $0$ or a $1$ , and the last two digits of the $n-1$ digits substring can be either $00$ or $10$ . So \[b_{n} = a_{n-1} + c_{n-1}.\]
If an $n$ -digit string ends in $10$ , then the previous digit must be a $0$ , and the last two digits of the $n-1$ digits substring will be $01$ . So \[c_{n} = b_{n-1}.\]
Clearly, $a_2=b_2=c_2=1$ . Using the recursive equations and initial values: \[\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|} \multicolumn{19}{c}{}\\\hline n&2&3&4&5&6&7&8&9&10&11&12&13&14&15&16&17&18&19\\\hline a_n&1&1&1&2&2&3&4&5&7&9&12&16&21&28&37&49&65&86\\\hline b_n&1&2&2&3&4&5&7&9&12&16&21&28&37&49&65&86&114&151\\\hline c_n&1&1&2&2&3&4&5&7&9&12&16&21&28&37&49&65&86&114\\\hline \end{array}\]
As a result $a_{19}+b_{19}+c_{19}=\boxed{351}$
| 351
|
6,572
|
https://artofproblemsolving.com/wiki/index.php/2001_AIME_I_Problems/Problem_14
| 2
|
A mail carrier delivers mail to the nineteen houses on the east side of Elm Street. The carrier notices that no two adjacent houses ever get mail on the same day, but that there are never more than two houses in a row that get no mail on the same day. How many different patterns of mail delivery are possible?
|
We split the problem into cases using the number of houses that get mail. Let "|" represent a house that gets mail, and "o" represent a house that doesn't. With a fixed number of |, an o can be inserted between 2 |'s or on the very left or right. There cannot be more than one o that is free to arrange to be placed between two |'s because no three o's can be adjacent, but there can be a maximum of two o's placed on the very left or right. Note that according to the Pigeonhole Principle , no more than 10 houses can get mail on the same day.
Case 1: 10 houses get mail. No 2 adjacent houses can get mail on the same day, so there must be an o between every two |. $10-1=9$ o's are fixed so we count the number of ways to insert $19 - 10 - 9 = 0$ o's to $10+1 = 11$ spots, or $\binom{11}{0} = 1$
Case 2: 9 houses get mail. In this case, $9-1 = 8$ o's are fixed so we count the number of ways to insert $19 - 9 - 8 = 2$ o's to $9+1=10$ spots. However, there is also the case where two o's are both on the very left / right. When both o's that are free to arrange are put on a side, there are $10-1=9$ spots left to insert $2-2=0$ o's. Hence the total number of ways in this case is $\binom{10}{2} + 2\binom{9}{0} = 47$
Case 3: 8 houses get mail. In this case, $8-1=7$ o's are fixed so we count the number of ways to insert $19-8-7=4$ o's to $8+1=9$ spots. When two o's are put to the very left / right, there are $9-1=8$ spots left to insert $4-2=2$ o's. We also need to take care of the case where two o's are on the very left and two o's are on the very right: we have $9-1-1=7$ spots to insert $4-2-2=0$ o's. Hence the total number of ways in this case is $\binom{9}{4} + 2\binom{8}{2} + \binom{7}{0} = 183$
Case 4: 7 houses get mail. In this case, $7-1=6$ o's are fixed so we count the number of ways to insert $19-7-6=6$ o's to $7+1=8$ spots. When two o's are put to the very left / right, there are $8-1=7$ spots left to insert $6-2=4$ o's. When two o's are on the very left and two o's are on the very right, we have $8-1-1=6$ spots to insert $6-2-2=2$ o's. Hence the total number of ways in this case is $\binom{8}{6} + 2\binom{7}{4} + \binom{6}{2} = 113$
Case 5: 6 houses get mail. We have to be careful in this case: $6-1=5$ o's are fixed so we are inserting $19-6-5=8$ o's to $6+1=7$ spots, which means that at least 1 of the 2 sides must have two o's. When 1 of the 2 sides have two o's, there are $7-1=6$ spots to insert $8-2=6$ o's. When both sides have two o's, there are $7-1-1=5$ spots to insert $8-2-2=4$ o's. Hence the total number of ways in this case is $2\binom{6}{6} + \binom{5}{4} = 7$
When less than 6 houses get(s) mail, it's again not possible since at least three o's must be together (again, according to the Pigeonhole Principle). Therefore, the desired answer is $1+47+183+113+7=\boxed{351}$
| 351
|
6,573
|
https://artofproblemsolving.com/wiki/index.php/2001_AIME_I_Problems/Problem_14
| 3
|
A mail carrier delivers mail to the nineteen houses on the east side of Elm Street. The carrier notices that no two adjacent houses ever get mail on the same day, but that there are never more than two houses in a row that get no mail on the same day. How many different patterns of mail delivery are possible?
|
There doesn't seem to be anything especially noticeable about the number nineteen in this problem, meaning that we can replace the number nineteen with any number without a big effect on the logic that we use to solve the problem. This pits the problem as a likely candidate for recursion.
At first, it's not immediately clear how to relate the state of $n$ houses in general to that of $n - 1, n - 2,$ or $n - 3.$ We thus break it up into cases, based on whether the first house gets mail or not.
Let $p_n$ be the number of ways to distribute the mail to $n$ houses. Assume that the first house gets mail. Therefore, since no two adjacent houses get mail on the same day, the second house must not get mail. Starting from the third house, however, things start to look messy, and it looks like we have to break our recurrence down into even smaller cases, which is something that we don't like -- we want to keep our relations as simple as possible. Therefore, seeing that we can't work forwards anymore, we try to work backwards.
Once the mail carrier delivers the mail to the first and (lack of mail) to the second houses, have him deliver mail to the remaining $n - 3$ houses at the end of the row, skipping the third house. There are $p_{n - 3}$ ways to do this. Now, we see that the availability of mail at the third house is fixed -- if the fourth house doesn't receive mail, the third one must, and if the fourth house receives mail, the third one can't. Therefore, there are simply $p_{n-3}$ ways to deliver the mail if the first house gets mail.
If the first house doesn't get mail, then we use the same logic -- have the mail carrier skip the second house and deliver the remaining mail to the $n - 2$ houses in $p_{n-2}$ ways. Then, the availability of mail for the second house is fixed, so there are $p_{n - 2}$ ways to deliver the mail in this case.
We thus have established a recurrence relation -- since the first house either gets mail or it doesn't, and cannot achieve both at the same time, we are confident about the validity of our relation: \[p_n = p_{n-2} + p_{n-3}.\] Now, we simply calculate $p_1, p_2,$ and $p_3.$ Then, it's off to the races for computation!
$p_1 = 2,$ because the first house can either gets mail or it doesn't -- there are no restrictions.
$p_2 = 3,$ because all of the possible deliveries are valid (of which there are $2 \cdot 2 = 4$ ) except the one where both houses receive mail.
$p_3 = 4,$ as there are $4$ possible ways (here, M represents that that house gets mail and N represents no mail): MNM, MNN, NNM, NMN.
Using our recurrence relation, we eventually get that $p_{19} = \boxed{351},$ and we're done.
| 351
|
6,574
|
https://artofproblemsolving.com/wiki/index.php/2001_AIME_I_Problems/Problem_14
| 4
|
A mail carrier delivers mail to the nineteen houses on the east side of Elm Street. The carrier notices that no two adjacent houses ever get mail on the same day, but that there are never more than two houses in a row that get no mail on the same day. How many different patterns of mail delivery are possible?
|
Let $w_n$ be the number of possible ways if the last house has mail, and $b_n$ be the number of possible ways if the last house does not have mail.
If the last house has mail, then, the next house can't have mail, meaning that $b_n = w_{n - 1}$
If the last house doesn't have mail, then the next house can either have mail or not have mail. If the next house has mail, then we simply count the number of ways that the row ends in a house with mail, so that means so far, our recursive rule is $w_n = b_{n - 1} + \text{something}$ . If the next house does not have mail, then the next house after that must have mail, meaning that $w_n = b_{n - 1} + b_{n - 2}$
Recursing all the way up to $b_{19}$ and $w_{19}$ , we get $100 + 251 = \boxed{351}$
| 351
|
6,575
|
https://artofproblemsolving.com/wiki/index.php/2001_AIME_I_Problems/Problem_14
| 5
|
A mail carrier delivers mail to the nineteen houses on the east side of Elm Street. The carrier notices that no two adjacent houses ever get mail on the same day, but that there are never more than two houses in a row that get no mail on the same day. How many different patterns of mail delivery are possible?
|
Let $a_n$ be the number of ways if the first house has mail, and let $b_n$ be the number of ways if the first house does not get mail.
$a_n=a_{n-2}+a_{n-3}$ because if the first house gets mail, the next house that gets mail must either be the third or fourth house.
$b_n=a_{n-1}+a_{n-2}$ because if the first house does not get mail, the next house that gets mail must either be the second or third house.
Note that we only need list out values of $a_n$ as $b$ depends on $a$ $a_1=1, a_2=1, a_3=2, a_4=2, \ldots$
$a_{19}+b_{19}=a_{17}+a_{16}+a_{18}+a_{17}=a_{16}+2\cdot a_{17}+a_{18}=65+2\cdot 86+114=\boxed{351}$
| 351
|
6,576
|
https://artofproblemsolving.com/wiki/index.php/2001_AIME_II_Problems/Problem_1
| 1
|
Let $N$ be the largest positive integer with the following property: reading from left to right, each pair of consecutive digits of $N$ forms a perfect square. What are the leftmost three digits of $N$
|
The two-digit perfect squares are $16, 25, 36, 49, 64, 81$ . We try making a sequence starting with each one:
The largest is $81649$ , so our answer is $\boxed{816}$
| 816
|
6,577
|
https://artofproblemsolving.com/wiki/index.php/2001_AIME_II_Problems/Problem_2
| 1
|
Each of the $2001$ students at a high school studies either Spanish or French, and some study both. The number who study Spanish is between $80$ percent and $85$ percent of the school population, and the number who study French is between $30$ percent and $40$ percent. Let $m$ be the smallest number of students who could study both languages, and let $M$ be the largest number of students who could study both languages. Find $M-m$
|
Let $S$ be the percent of people who study Spanish, $F$ be the number of people who study French, and let $S \cap F$ be the number of students who study both. Then $\left\lceil 80\% \cdot 2001 \right\rceil = 1601 \le S \le \left\lfloor 85\% \cdot 2001 \right\rfloor = 1700$ , and $\left\lceil 30\% \cdot 2001 \right\rceil = 601 \le F \le \left\lfloor 40\% \cdot 2001 \right\rfloor = 800$ . By the Principle of Inclusion-Exclusion
\[S+F- S \cap F = S \cup F = 2001\]
For $m = S \cap F$ to be smallest, $S$ and $F$ must be minimized.
\[1601 + 601 - m = 2001 \Longrightarrow m = 201\]
For $M = S \cap F$ to be largest, $S$ and $F$ must be maximized.
\[1700 + 800 - M = 2001 \Longrightarrow M = 499\]
Therefore, the answer is $M - m = 499 - 201 = \boxed{298}$
| 298
|
6,578
|
https://artofproblemsolving.com/wiki/index.php/2001_AIME_II_Problems/Problem_3
| 1
|
Given that
\begin{align*}x_{1}&=211,\\ x_{2}&=375,\\ x_{3}&=420,\\ x_{4}&=523,\ \text{and}\\ x_{n}&=x_{n-1}-x_{n-2}+x_{n-3}-x_{n-4}\ \text{when}\ n\geq5, \end{align*}
find the value of $x_{531}+x_{753}+x_{975}$
|
We find that $x_5 = 267$ by the recursive formula. Summing the recursions
\begin{align*} x_{n}&=x_{n-1}-x_{n-2}+x_{n-3}-x_{n-4} \\ x_{n-1}&=x_{n-2}-x_{n-3}+x_{n-4}-x_{n-5} \end{align*}
yields $x_{n} = -x_{n-5}$ . Thus $x_n = (-1)^k x_{n-5k}$ . Since $531 = 106 \cdot 5 + 1,\ 753 = 150 \cdot 5 + 3,\ 975 = 194 \cdot 5 + 5$ , it follows that
\[x_{531} + x_{753} + x_{975} = (-1)^{106} x_1 + (-1)^{150} x_3 + (-1)^{194} x_5 = 211 + 420 + 267 = \boxed{898}.\]
| 898
|
6,579
|
https://artofproblemsolving.com/wiki/index.php/2001_AIME_II_Problems/Problem_3
| 2
|
Given that
\begin{align*}x_{1}&=211,\\ x_{2}&=375,\\ x_{3}&=420,\\ x_{4}&=523,\ \text{and}\\ x_{n}&=x_{n-1}-x_{n-2}+x_{n-3}-x_{n-4}\ \text{when}\ n\geq5, \end{align*}
find the value of $x_{531}+x_{753}+x_{975}$
|
The recursive formula suggests telescoping. Indeed, if we add $x_n$ and $x_{n-1}$ , we have $x_n + x_{n-1} = (x_{n-1} - x_{n-2} + x_{n-3} - x_{n-4}) + (x_{n-2} - x_{n-3} + x_{n-4} - x_{n-5}) = x_{n-1} - x_{n-5}$
Subtracting $x_{n-1}$ yields $x_n = -x_{n-5} \implies x_n = -(-(x_{n-10})) = x_{n-10}$
Thus,
\[x_{531} + x_{753} + x_{975} = x_1 + x_3 + x_5 = x_1 + x_3 + (x_4 - x_3 + x_2 - x_1) = x_2 + x_4 = 375 + 523 = \boxed{898}.\]
| 898
|
6,580
|
https://artofproblemsolving.com/wiki/index.php/2001_AIME_II_Problems/Problem_3
| 3
|
Given that
\begin{align*}x_{1}&=211,\\ x_{2}&=375,\\ x_{3}&=420,\\ x_{4}&=523,\ \text{and}\\ x_{n}&=x_{n-1}-x_{n-2}+x_{n-3}-x_{n-4}\ \text{when}\ n\geq5, \end{align*}
find the value of $x_{531}+x_{753}+x_{975}$
|
Calculate the first few terms:
\[211,375,420,523,267,-211,-375,-420,-523,\dots\]
At this point it is pretty clear that the sequence is periodic with period 10 (one may prove it quite easily like in solution 1) so our answer is obviously $211+420+267=\boxed{898}$
| 898
|
6,581
|
https://artofproblemsolving.com/wiki/index.php/2001_AIME_II_Problems/Problem_6
| 1
|
Square $ABCD$ is inscribed in a circle . Square $EFGH$ has vertices $E$ and $F$ on $\overline{CD}$ and vertices $G$ and $H$ on the circle. If the area of square $ABCD$ is $1$ , then the area of square $EFGH$ can be expressed as $\frac {m}{n}$ where $m$ and $n$ are relatively prime positive integers and $m < n$ . Find $10n + m$
|
Let $O$ be the center of the circle, and $2a$ be the side length of $ABCD$ $2b$ be the side length of $EFGH$ . By the Pythagorean Theorem , the radius of $\odot O = OC = a\sqrt{2}$
Now consider right triangle $OGI$ , where $I$ is the midpoint of $\overline{GH}$ . Then, by the Pythagorean Theorem,
\begin{align*} OG^2 = 2a^2 &= OI^2 + GI^2 = (a+2b)^2 + b^2 \\ 0 &= a^2 - 4ab - 5b^2 = (a - 5b)(a + b) \end{align*}
Thus $a = 5b$ (since lengths are positive, we discard the other root). The ratio of the areas of two similar figures is the square of the ratio of their corresponding side lengths, so $\frac{[EFGH]}{[ABCD]} = \left(\frac 15\right)^2 = \frac{1}{25}$ , and the answer is $10n + m = \boxed{251}$
| 251
|
6,582
|
https://artofproblemsolving.com/wiki/index.php/2001_AIME_II_Problems/Problem_6
| 2
|
Square $ABCD$ is inscribed in a circle . Square $EFGH$ has vertices $E$ and $F$ on $\overline{CD}$ and vertices $G$ and $H$ on the circle. If the area of square $ABCD$ is $1$ , then the area of square $EFGH$ can be expressed as $\frac {m}{n}$ where $m$ and $n$ are relatively prime positive integers and $m < n$ . Find $10n + m$
|
Let point $A$ be the top-left corner of square $ABCD$ and the rest of the vertices be arranged, in alphabetical order, in a clockwise arrangement from there. Let $D$ have coordinates $(0,0)$ and the side length of square $ABCD$ be $a$ . Let $DF$ $b$ and diameter $HI$ go through $J$ the midpoint of $EF$ . Since a diameter always bisects a chord perpendicular to it, $DJ$ $JC$ and since $F$ and $E$ must be symmetric around the diameter, $FJ = JE$ and it follows that $DF = EC = b.$ Hence $FE$ the side of square $EFGH$ has length $a - 2b$ $F$ has coordinates $(b,0)$ and $G$ has coordinates $(b, 2b - a).$ We know that point $G$ must be on the circle $O$ - hence it must satisfy the circle equation. Since the center of the circle is at the center of the square $(a/2, a/2)$ and has radius $a *$ $\sqrt{2} / 2$ , half the diagonal of the square, $(x - a/2)^2 + (y - a/2)^2 = 1/2a^2$ follows as the circle equation. Then substituting coordinates of $G$ into the equation, $(b - a/2)^2 + (2b - a - a/2)^2 = a^2/2$ . Simplifying and factoring, we get $2a^2-7ab+5b^2 = (2a-5b)(a-b) = 0.$ Since $a = b$ would imply $m = n$ , and $m < n$ in the problem, we must use the other factor. We get $b = 2/5a$ , meaning the ratio of areas $((a-2b)/a)^2$ $(1/5)^2$ $1/25$ $m/n.$ Then $10n + m = 25 * 10 + 1 = \boxed{251}$
| 251
|
6,583
|
https://artofproblemsolving.com/wiki/index.php/2001_AIME_II_Problems/Problem_7
| 1
|
Let $\triangle{PQR}$ be a right triangle with $PQ = 90$ $PR = 120$ , and $QR = 150$ . Let $C_{1}$ be the inscribed circle . Construct $\overline{ST}$ with $S$ on $\overline{PR}$ and $T$ on $\overline{QR}$ , such that $\overline{ST}$ is perpendicular to $\overline{PR}$ and tangent to $C_{1}$ . Construct $\overline{UV}$ with $U$ on $\overline{PQ}$ and $V$ on $\overline{QR}$ such that $\overline{UV}$ is perpendicular to $\overline{PQ}$ and tangent to $C_{1}$ . Let $C_{2}$ be the inscribed circle of $\triangle{RST}$ and $C_{3}$ the inscribed circle of $\triangle{QUV}$ . The distance between the centers of $C_{2}$ and $C_{3}$ can be written as $\sqrt {10n}$ . What is $n$
|
Let $P = (0,0)$ be at the origin. Using the formula $A = rs$ on $\triangle PQR$ , where $r_{1}$ is the inradius (similarly define $r_2, r_3$ to be the radii of $C_2, C_3$ ), $s = \frac{PQ + QR + RP}{2} = 180$ is the semiperimeter , and $A = \frac 12 bh = 5400$ is the area, we find $r_{1} = \frac As = 30$ . Or, the inradius could be directly by using the formula $\frac{a+b-c}{2}$ , where $a$ and $b$ are the legs of the right triangle and $c$ is the hypotenuse. (This formula should be used only for right triangles .) Thus $ST, UV$ lie respectively on the lines $y = 60, x = 60$ , and so $RS = 60, UQ = 30$
Note that $\triangle PQR \sim \triangle STR \sim \triangle UQV$ . Since the ratio of corresponding lengths of similar figures are the same, we have
\[\frac{r_{1}}{PR} = \frac{r_{2}}{RS} \Longrightarrow r_{2} = 15\ \text{and} \ \frac{r_{1}}{PQ} = \frac{r_{3}}{UQ} \Longrightarrow r_{3} = 10.\]
Let the centers of $\odot C_2, C_3$ be $O_2 = (0 + r_{2}, 60 + r_{2}) = (15, 75), O_3 = (60 + r_{3}, 0 + r_{3}) = (70,10)$ , respectively; then by the distance formula we have $O_2O_3 = \sqrt{55^2 + 65^2} = \sqrt{10 \cdot 725}$ . Therefore, the answer is $n = \boxed{725}$
| 725
|
6,584
|
https://artofproblemsolving.com/wiki/index.php/2001_AIME_II_Problems/Problem_7
| 2
|
Let $\triangle{PQR}$ be a right triangle with $PQ = 90$ $PR = 120$ , and $QR = 150$ . Let $C_{1}$ be the inscribed circle . Construct $\overline{ST}$ with $S$ on $\overline{PR}$ and $T$ on $\overline{QR}$ , such that $\overline{ST}$ is perpendicular to $\overline{PR}$ and tangent to $C_{1}$ . Construct $\overline{UV}$ with $U$ on $\overline{PQ}$ and $V$ on $\overline{QR}$ such that $\overline{UV}$ is perpendicular to $\overline{PQ}$ and tangent to $C_{1}$ . Let $C_{2}$ be the inscribed circle of $\triangle{RST}$ and $C_{3}$ the inscribed circle of $\triangle{QUV}$ . The distance between the centers of $C_{2}$ and $C_{3}$ can be written as $\sqrt {10n}$ . What is $n$
|
We compute $r_1 = 30, r_2 = 15, r_3 = 10$ as above. Let $A_1, A_2, A_3$ respectively the points of tangency of $C_1, C_2, C_3$ with $QR$
By the Two Tangent Theorem , we find that $A_{1}Q = 60$ $A_{1}R = 90$ . Using the similar triangles, $RA_{2} = 45$ $QA_{3} = 20$ , so $A_{2}A_{3} = QR - RA_2 - QA_3 = 85$ . Thus $(O_{2}O_{3})^{2} = (15 - 10)^{2} + (85)^{2} = 7250\implies n=\boxed{725}$
| 725
|
6,585
|
https://artofproblemsolving.com/wiki/index.php/2001_AIME_II_Problems/Problem_7
| 3
|
Let $\triangle{PQR}$ be a right triangle with $PQ = 90$ $PR = 120$ , and $QR = 150$ . Let $C_{1}$ be the inscribed circle . Construct $\overline{ST}$ with $S$ on $\overline{PR}$ and $T$ on $\overline{QR}$ , such that $\overline{ST}$ is perpendicular to $\overline{PR}$ and tangent to $C_{1}$ . Construct $\overline{UV}$ with $U$ on $\overline{PQ}$ and $V$ on $\overline{QR}$ such that $\overline{UV}$ is perpendicular to $\overline{PQ}$ and tangent to $C_{1}$ . Let $C_{2}$ be the inscribed circle of $\triangle{RST}$ and $C_{3}$ the inscribed circle of $\triangle{QUV}$ . The distance between the centers of $C_{2}$ and $C_{3}$ can be written as $\sqrt {10n}$ . What is $n$
|
The radius of an incircle is $r=A_t/\text{semiperimeter}$ . The area of the triangle is equal to $\frac{90\times120}{2} = 5400$ and the semiperimeter is equal to $\frac{90+120+150}{2} = 180$ . The radius, therefore, is equal to $\frac{5400}{180} = 30$ . Thus using similar triangles the dimensions of the triangle circumscribing the circle with center $C_2$ are equal to $120-2(30) = 60$ $\frac{1}{2}(90) = 45$ , and $\frac{1}{2}\times150 = 75$ . The radius of the circle inscribed in this triangle with dimensions $45\times60\times75$ is found using the formula mentioned at the very beginning. The radius of the incircle is equal to $15$
Defining $P$ as $(0,0)$ $C_2$ is equal to $(60+15,15)$ or $(75,15)$ . Also using similar triangles, the dimensions of the triangle circumscribing the circle with center $C_3$ are equal to $90-2(30)$ $\frac{1}{3}\times120$ $\frac{1}{3}\times150$ or $30,40,50$ . The radius of $C_3$ by using the formula mentioned at the beginning is $10$ . Using $P$ as $(0,0)$ $C_3$ is equal to $(10, 60+10)$ or $(10,70)$ . Using the distance formula, the distance between $C_2$ and $C_3$ $\sqrt{(75-10)^2 +(15-70)^2}$ this equals $\sqrt{7250}$ or $\sqrt{725\times10}$ , thus $n$ is $\boxed{725}$
| 725
|
6,586
|
https://artofproblemsolving.com/wiki/index.php/2001_AIME_II_Problems/Problem_7
| 4
|
Let $\triangle{PQR}$ be a right triangle with $PQ = 90$ $PR = 120$ , and $QR = 150$ . Let $C_{1}$ be the inscribed circle . Construct $\overline{ST}$ with $S$ on $\overline{PR}$ and $T$ on $\overline{QR}$ , such that $\overline{ST}$ is perpendicular to $\overline{PR}$ and tangent to $C_{1}$ . Construct $\overline{UV}$ with $U$ on $\overline{PQ}$ and $V$ on $\overline{QR}$ such that $\overline{UV}$ is perpendicular to $\overline{PQ}$ and tangent to $C_{1}$ . Let $C_{2}$ be the inscribed circle of $\triangle{RST}$ and $C_{3}$ the inscribed circle of $\triangle{QUV}$ . The distance between the centers of $C_{2}$ and $C_{3}$ can be written as $\sqrt {10n}$ . What is $n$
|
We can calculate the inradius of each triangle as with the previous solutions. Now, notice that the hexagon $PSTVU$ is a square with its corner cut off. We literally complete the square and mark the last corner as point X. Now, construct right triangle with $C_3C_2$ as its hypotenuse. The right angle will be at point $Y$ . We will now find the length of each leg and use the Pythagorean Theorem to compute the desired length. We see that the length of $C_3Y$ is $50 + 15 = 65$ , as seen by the inradius of $C_2$ and $10$ less than the square's side length. $C_2Y$ is $45 + 10 = 55$ , which is $15$ less than the square plus the inradius of $C_3$ . Our final answer is $\sqrt{65^2 + 55^2} = \sqrt{7250} = \sqrt{(10)(725)} \rightarrow \boxed{725}$
| 725
|
6,587
|
https://artofproblemsolving.com/wiki/index.php/2001_AIME_II_Problems/Problem_8
| 1
|
A certain function $f$ has the properties that $f(3x) = 3f(x)$ for all positive real values of $x$ , and that $f(x) = 1-|x-2|$ for $1\le x \le 3$ . Find the smallest $x$ for which $f(x) = f(2001)$
|
Iterating the condition $f(3x) = 3f(x)$ , we find that $f(x) = 3^kf\left(\frac{x}{3^k}\right)$ for positive integers $k$ . We know the definition of $f(x)$ from $1 \le x \le 3$ , so we would like to express $f(2001) = 3^kf\left(\frac{2001}{3^k}\right),\ 1 \le \frac{2001}{3^k} \le 3 \Longrightarrow k = 6$ . Indeed,
\[f(2001) = 729\left[1 - \left| \frac{2001}{729} - 2\right|\right] = 186.\]
We now need the smallest $x$ such that $f(x) = 3^kf\left(\frac{x}{3^k}\right) = 186$ . The range of $f(x),\ 1 \le x \le 3$ , is $0 \le f(x) \le 1$ . So when $1 \le \frac{x}{3^k} \le 3$ , we have $0 \le f\left(\frac{x}{3^k}\right) = \frac{186}{3^k} \le 1$ . Multiplying by $3^k$ $0 \le 186 \le 3^k$ , so the smallest value of $k$ is $k = 5$ . Then,
\[186 = {3^5}f\left(\frac{x}{3^5}\right).\]
Because we forced $1 \le \frac{x}{3^5} \le 3$ , so
\[186 = {3^5}f\left(\frac{x}{3^5}\right) = 243\left[1 - \left| \frac{x}{243} - 2\right|\right] \Longrightarrow x = \pm 57 + 2 \cdot 243.\]
We want the smaller value of $x = \boxed{429}$
| 429
|
6,588
|
https://artofproblemsolving.com/wiki/index.php/2001_AIME_II_Problems/Problem_8
| 2
|
A certain function $f$ has the properties that $f(3x) = 3f(x)$ for all positive real values of $x$ , and that $f(x) = 1-|x-2|$ for $1\le x \le 3$ . Find the smallest $x$ for which $f(x) = f(2001)$
|
First, we start by graphing the function when $1\leq{x}\leq3$ , which consists of the lines $y=x-1$ and $y=3-x$ that intersect at $(2,1)$ . Similarly, using $f(3x)=3f(x)$ , we get a dilation of our initial figure by a factor of 3 for the next interval and so on.
Observe that the intersection of two lines always has coordinates $(2y,y)$ where $y=3^a$ for some $a$ . First, we compute $f(2001)$ . The nearest intersection point is $(1458,729)$ when $a=7$ . Therefore, we can safely assume that $f(2001)$ is somewhere on the line with a slope of $-1$ that intersects at that nearest point. Using the fact that the slope of the line is $-1$ , we compute $f(2001)=729-543=186$ . However, we want the minimum value such that $f(x)=186$ and we see that there is another intersection point on the left which has a $y>186$ , namely $(486,243)$ . Therefore, we want the point that lies on the line with slope $1$ that intersects this point. Once again, since the slope of the line is $1$ , we get $x=486-57=\boxed{429}$
| 429
|
6,589
|
https://artofproblemsolving.com/wiki/index.php/2001_AIME_II_Problems/Problem_9
| 1
|
Each unit square of a 3-by-3 unit-square grid is to be colored either blue or red. For each square, either color is equally likely to be used. The probability of obtaining a grid that does not have a 2-by-2 red square is $\frac {m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$
|
We can use complementary counting , counting all of the colorings that have at least one red $2\times 2$ square.
By the Principle of Inclusion-Exclusion , there are (alternatively subtracting and adding) $128-40+8-1=95$ ways to have at least one red $2 \times 2$ square.
There are $2^9=512$ ways to paint the $3 \times 3$ square with no restrictions, so there are $512-95=417$ ways to paint the square with the restriction. Therefore, the probability of obtaining a grid that does not have a $2 \times 2$ red square is $\frac{417}{512}$ , and $417+512=\boxed{929}$
| 929
|
6,590
|
https://artofproblemsolving.com/wiki/index.php/2001_AIME_II_Problems/Problem_9
| 2
|
Each unit square of a 3-by-3 unit-square grid is to be colored either blue or red. For each square, either color is equally likely to be used. The probability of obtaining a grid that does not have a 2-by-2 red square is $\frac {m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$
|
We consider how many ways we can have 2*2 grid
$(1)$ : All the grids are red-- $1$ case
$(2)$ : One unit square is blue--The blue lies on the center of the bigger square, makes no 2*2 grid $9-1=8$ cases
$(3)$ : Two unit squares are blue--one of the squares lies in the center of the bigger square, makes no 2*2 grid, $8$ cases.
Or, two squares lie on second column, first row, second column third row; second row first column, second row third column, 2 extra cases. $\binom 9 2-8-2=26$ cases
$(4)$ Three unit squares are blue. We find that if a 2*2 square is formed, there are 5 extra unit squares can be painted. But cases that three squares in the same column or same row is overcomunted. So in this case, there are $4\cdot (\binom 5 3)-4=36$
$(5)$ Four unit squares are blue, no overcomunted case will be considered. there are $4\cdot \binom 5 4=20$
$(6)$ Five unit squares are blue, $4$ cases in all
Sum up those cases, there are $1+8+26+36+20+4=95$ cases that a 2*2 grid can be formed.
In all, there are $2^9=512$ possible ways to paint the big square, so the answer is $1-\frac{95}{512}=\frac{417}{512}$ leads to $\boxed{929}$
| 929
|
6,591
|
https://artofproblemsolving.com/wiki/index.php/2001_AIME_II_Problems/Problem_9
| 3
|
Each unit square of a 3-by-3 unit-square grid is to be colored either blue or red. For each square, either color is equally likely to be used. The probability of obtaining a grid that does not have a 2-by-2 red square is $\frac {m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$
|
\[\begin{array}{|c|c|c|} \hline C_{11} & C_{12} & C_{13}\\ \hline C_{21} & C_{22} & C_{23}\\ \hline C_{31} & C_{32} & C_{33}\\ \hline \end{array}\]
Case 1: The 3-by-3 unit-square grid has exactly $1$ 2-by-2 red square
Assume the 2-by-2 red square is at $C_{11}, C_{12}, C_{21}, C_{22}$ . To make sure there are no more 2-by-2 red squares, $C_{31} \text{and} C_{32}$ can't both be red and $C_{13} \text{and} C_{23}$ can't both be red. Meaning that there are $2^2-1=3$ coloring methods for $C_{31} \text{and} C_{32}$ and $C_{13} \text{and} C_{23}$ $C_{33}$ can be colored with either colors. However, the coloring method where $C_{23}, C_{32}, C_{33}$ are all red needs to be removed. For exactly one 2-by-2 red square at $C_{11}, C_{12}, C_{21}, C_{22}$ , there are $3 \cdot 3 \cdot 2 -1=17$ coloring methods. As there are $4$ locations for the 2-by-2 red square on the 3-by-3 unit-square grid, there are $17 \cdot 4 = 68$ coloring methods.
Case 2: The 3-by-3 unit-square grid has exactly $2$ 2-by-2 red squares
Case 2.1: $2$ 2-by-2 red squares take up $6$ unit grids
Assume the $2$ 2-by-2 red squares are at $C_{11}, C_{12}, C_{21}, C_{22}, C_{31}, C_{32}$ . To make sure there are no more 2-by-2 red squares, $C_{13} \text{and} C_{23}$ can't both be red. Meaning that there are $2^2-1=3$ coloring methods for $C_{13} \text{and} C_{23}$ $C_{33}$ can be colored with either colors. However, the coloring method where $C_{13}, C_{23}, C_{33}$ are all red needs to be removed. For exactly $2$ 2-by-2 red squares at $C_{11}, C_{12}, C_{21}, C_{22}, C_{31}, C_{32}$ , there are $3 \cdot 2 -1=5$ coloring methods. As there are $4$ locations for the $2$ 2-by-2 red squares on the 3-by-3 unit-square grid, there are $5 \cdot 4 = 20$ coloring methods.
Case 2.2: $2$ 2-by-2 red squares take up $7$ unit grids
Assume the $2$ 2-by-2 red squares are at $C_{11}, C_{12}, C_{21}, C_{22}, C_{23}, C_{32}, C_{33}$ . To make sure there are no more 2-by-2 red squares, $C_{13}$ and $C_{31}$ can't be red. Meaning that there is $1$ coloring method for $C_{13}$ and $C_{31}$ . For exactly $2$ 2-by-2 red squares at $C_{11}, C_{12}, C_{21}, C_{22}, C_{23}, C_{32}, C_{33}$ , there is $1$ coloring method. As there are $2$ locations for the $2$ 2-by-2 red squares on the 3-by-3 unit-square grid, there are $1 \cdot 2 = 2$ coloring methods.
Hence, if the 3-by-3 unit-square grid has exactly $2$ 2-by-2 red squares, there are $20+2 = 22$ coloring methods.
Case 3: The 3-by-3 unit-square grid has exactly $3$ 2-by-2 red squares
Assume the $3$ 2-by-2 red squares are at $C_{11}, C_{12}, C_{13}, C_{21}, C_{22}, C_{23}, C_{32}, C_{33}$ . To make sure there are no more 2-by-2 red squares, $C_{33}$ can't be red. Meaning that there is $1$ coloring method for $C_{33}$ . For exactly $3$ 2-by-2 red squares at $C_{11}, C_{12}, C_{13}, C_{21}, C_{22}, C_{23}, C_{32}, C_{33}$ , there is $1$ coloring method. As there are $4$ locations for the $3$ 2-by-2 red squares on the 3-by-3 unit-square grid, there are $1 \cdot 4 = 4$ coloring methods.
Case 4: The 3-by-3 unit-square grid has exactly $4$ 2-by-2 red squares
If the 3-by-3 unit-square grid has exactly $4$ 2-by-2 red squares, all $9$ unit grids are red and there is $1$ coloring method.
In total, there are $68+22+4+1=95$ coloring methods with 2-by-2 red squares. \[\frac{m}{n}=1-\frac{95}{2^9}=\frac{417}{512}\]
\[m+n=417+512=\boxed{929}\]
| 929
|
6,592
|
https://artofproblemsolving.com/wiki/index.php/2001_AIME_II_Problems/Problem_10
| 1
|
How many positive integer multiples of $1001$ can be expressed in the form $10^{j} - 10^{i}$ , where $i$ and $j$ are integers and $0\leq i < j \leq 99$
|
Observation: We see that there is a pattern with $10^k \pmod{1001}$ \[10^0 \equiv 1 \pmod{1001}\] \[10^1 \equiv 10 \pmod{1001}\] \[10^2 \equiv 100 \pmod{1001}\] \[10^3 \equiv -1 \pmod{1001}\] \[10^4 \equiv -10 \pmod{1001}\] \[10^5 \equiv -100 \pmod{1001}\] \[10^6 \equiv 1 \pmod{1001}\] \[10^7 \equiv 10 \pmod{1001}\] \[10^8 \equiv 100 \pmod{1001}\]
So, this pattern repeats every 6.
Also, $10^j-10^i \equiv 0 \pmod{1001}$ , so $10^j \equiv 10^i \pmod{1001}$ , and thus, \[j \equiv i \pmod{6}\] . Continue with the 2nd paragraph of solution 1, and we get the answer of $\boxed{784}$
| 784
|
6,593
|
https://artofproblemsolving.com/wiki/index.php/2001_AIME_II_Problems/Problem_10
| 2
|
How many positive integer multiples of $1001$ can be expressed in the form $10^{j} - 10^{i}$ , where $i$ and $j$ are integers and $0\leq i < j \leq 99$
|
Note that $1001=7\cdot 11\cdot 13,$ and note that $10^3 \equiv \pmod{p}$ for prime $p | 1001$ ; therefore, the order of 10 modulo $7,11$ , and $13$ must divide 6. A quick check on 7 reveals that it is indeed 6. Therefore we note that $i-j=6k$ for some natural number k. From here, we note that for $j=0,1,2,3,$ we have 16 options and we have 15,14,...,1 option(s) for the next 90 numbers (6 each), so our total is $4\cdot 16 + 6 \cdot \frac{15 \cdot 16}{2} = \boxed{784}$
| 784
|
6,594
|
https://artofproblemsolving.com/wiki/index.php/2001_AIME_II_Problems/Problem_11
| 1
|
Club Truncator is in a soccer league with six other teams, each of which it plays once. In any of its 6 matches, the probabilities that Club Truncator will win, lose, or tie are each $\frac {1}{3}$ . The probability that Club Truncator will finish the season with more wins than losses is $\frac {m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$
|
Note that the probability that Club Truncator will have more wins than losses is equal to the probability that it will have more losses than wins; the only other possibility is that they have the same number of wins and losses. Thus, by the complement principle , the desired probability is half the probability that Club Truncator does not have the same number of wins and losses.
The possible ways to achieve the same number of wins and losses are $0$ ties, $3$ wins and $3$ losses; $2$ ties, $2$ wins, and $2$ losses; $4$ ties, $1$ win, and $1$ loss; or $6$ ties. Since there are $6$ games, there are $\frac{6!}{3!3!}$ ways for the first, and $\frac{6!}{2!2!2!}$ $\frac{6!}{4!}$ , and $1$ ways for the rest, respectively, out of a total of $3^6$ . This gives a probability of $141/729$ . Then the desired answer is $\frac{1 - \frac{141}{729}}{2} = \frac{98}{243}$ , so the answer is $m+n = \boxed{341}$
| 341
|
6,595
|
https://artofproblemsolving.com/wiki/index.php/2001_AIME_II_Problems/Problem_11
| 2
|
Club Truncator is in a soccer league with six other teams, each of which it plays once. In any of its 6 matches, the probabilities that Club Truncator will win, lose, or tie are each $\frac {1}{3}$ . The probability that Club Truncator will finish the season with more wins than losses is $\frac {m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$
|
At first, it wins $6$ games, only one way
Secondly, it wins $5$ games, the other game can be either win or loss, there are $\binom{6}{5}\cdot 2=12$ ways
Thirdly, it wins $4$ games, still the other two games can be either win or loss, there are $\binom{6}{4}\cdot 2^2=60$ ways
Fourthly, it wins $3$ games, this time, it can't lose $3$ games but other arrangements of the three non-winning games are fine, there are $\binom{6}{3}\cdot (2^3-1)=140$ ways
Fifth case, it wins $2$ games, only $0/1$ lose and $4/3$ draw is ok, so there are $\binom{6}{2}(1+\binom{4}{1})=75$ cases
Last case, it only wins $1$ game so the rest games must be all draw, $1$ game
The answer is $\frac{1+12+60+140+75+6}{3^6}=\frac{98}{243}$ leads to $\boxed{341}$
| 341
|
6,596
|
https://artofproblemsolving.com/wiki/index.php/2001_AIME_II_Problems/Problem_12
| 1
|
Given a triangle , its midpoint triangle is obtained by joining the midpoints of its sides. A sequence of polyhedra $P_{i}$ is defined recursively as follows: $P_{0}$ is a regular tetrahedron whose volume is 1. To obtain $P_{i + 1}$ , replace the midpoint triangle of every face of $P_{i}$ by an outward-pointing regular tetrahedron that has the midpoint triangle as a face. The volume of $P_{3}$ is $\frac {m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$
|
On the first construction, $P_1$ , four new tetrahedra will be constructed with side lengths $\frac 12$ of the original one. Since the ratio of the volume of similar polygons is the cube of the ratio of their corresponding lengths, it follows that each of these new tetrahedra will have volume $\left(\frac 12\right)^3 = \frac 18$ . The total volume added here is then $\Delta P_1 = 4 \cdot \frac 18 = \frac 12$
We now note that for each midpoint triangle we construct in step $P_{i}$ , there are now $6$ places to construct new midpoint triangles for step $P_{i+1}$ . The outward tetrahedron for the midpoint triangle provides $3$ of the faces, while the three equilateral triangles surrounding the midpoint triangle provide the other $3$ . This is because if you read this question carefully, it asks to add new tetrahedra to each face of $P_{i}$ which also includes the ones that were left over when we did the previous addition of tetrahedra. However, the volume of the tetrahedra being constructed decrease by a factor of $\frac 18$ . Thus we have the recursion $\Delta P_{i+1} = \frac{6}{8} \Delta P_i$ , and so $\Delta P_i = \frac 12 \cdot \left(\frac{3}{4}\right)^{i-1} P_1$
The volume of $P_3 = P_0 + \Delta P_1 + \Delta P_2 + \Delta P_3 = 1 + \frac 12 + \frac 38 + \frac 9{32} = \frac{69}{32}$ , and $m+n=\boxed{101}$ . Note that the summation was in fact a geometric series
| 101
|
6,597
|
https://artofproblemsolving.com/wiki/index.php/2001_AIME_II_Problems/Problem_14
| 1
|
There are $2n$ complex numbers that satisfy both $z^{28} - z^{8} - 1 = 0$ and $\mid z \mid = 1$ . These numbers have the form $z_{m} = \cos\theta_{m} + i\sin\theta_{m}$ , where $0\leq\theta_{1} < \theta_{2} < \ldots < \theta_{2n} < 360$ and angles are measured in degrees. Find the value of $\theta_{2} + \theta_{4} + \ldots + \theta_{2n}$
|
$z$ can be written in the form $\text{cis\,}\theta$ . Rearranging, we find that $\text{cis\,}{28}\theta = \text{cis\,}{8}\theta+1$
Since the real part of $\text{cis\,}{28}\theta$ is one more than the real part of $\text{cis\,} {8}\theta$ and their imaginary parts are equal, it is clear that either $\text{cis\,}{28}\theta = \frac{1}{2}+\frac {\sqrt{3}}{2}i$ and $\text{cis\,} {8}\theta = -\frac{1}{2}+\frac {\sqrt{3}}{2}i$ , or $\text{cis\,}{28}\theta = \frac{1}{2} - \frac{\sqrt{3}}{2}i$ and $\text{cis\,} {8}\theta = -\frac{1}{2}- \frac{\sqrt{3}}{2}i$
Setting up and solving equations, $Z^{28}= \text{cis\,}{60^\circ}$ and $Z^8= \text{cis\,}{120^\circ}$ , we see that the solutions common to both equations have arguments $15^\circ , 105^\circ, 195^\circ,$ and $\ 285^\circ$ . We can figure this out by adding 360 repeatedly to the number 60 to try and see if it will satisfy what we need. We realize that it does not work in the integer values.
Again setting up equations ( $Z^{28}= \text{cis\,}{300^\circ}$ and $Z^{8} = \text{cis\,}{240^\circ}$ ) we see that the common solutions have arguments of $75^\circ, 165^\circ, 255^\circ,$ and $345^\circ$
Listing all of these values, we find that $\theta_{2} + \theta_{4} + \ldots + \theta_{2n}$ is equal to $(75 + 165 + 255 + 345) ^\circ$ which is equal to $\boxed{840}$ degrees. We only want the sum of a certain number of theta, not all of it.
| 840
|
6,598
|
https://artofproblemsolving.com/wiki/index.php/2001_AIME_II_Problems/Problem_15
| 1
|
Let $EFGH$ $EFDC$ , and $EHBC$ be three adjacent square faces of a cube , for which $EC = 8$ , and let $A$ be the eighth vertex of the cube. Let $I$ $J$ , and $K$ , be the points on $\overline{EF}$ $\overline{EH}$ , and $\overline{EC}$ , respectively, so that $EI = EJ = EK = 2$ . A solid $S$ is obtained by drilling a tunnel through the cube. The sides of the tunnel are planes parallel to $\overline{AE}$ , and containing the edges, $\overline{IJ}$ $\overline{JK}$ , and $\overline{KI}$ . The surface area of $S$ , including the walls of the tunnel, is $m + n\sqrt {p}$ , where $m$ $n$ , and $p$ are positive integers and $p$ is not divisible by the square of any prime. Find $m + n + p$
|
Set the coordinate system so that vertex $E$ , where the drilling starts, is at $(8,8,8)$ . Using a little visualization (involving some similar triangles , because we have parallel lines) shows that the tunnel meets the bottom face (the xy plane one) in the line segments joining $(1,0,0)$ to $(2,2,0)$ , and $(0,1,0)$ to $(2,2,0)$ , and similarly for the other three faces meeting at the origin (by symmetry). So one face of the tunnel is the polygon with vertices (in that order), $S(1,0,0), T(2,0,2), U(8,6,8), V(8,8,6), W(2,2,0)$ , and the other two faces of the tunnel are congruent to this shape.
Observe that this shape is made up of two congruent trapezoids each with height $\sqrt {2}$ and bases $7\sqrt {3}$ and $6\sqrt {3}$ . Together they make up an area of $\sqrt {2}(7\sqrt {3} + 6\sqrt {3}) = 13\sqrt {6}$ . The total area of the tunnel is then $3\cdot13\sqrt {6} = 39\sqrt {6}$ . Around the corner $E$ we're missing an area of $6$ , the same goes for the corner opposite $E$ . So the outside area is $6\cdot 64 - 2\cdot 6 = 372$ . Thus the the total surface area is $372 + 39\sqrt {6}$ , and the answer is $372 + 39 + 6 = \boxed{417}$
| 417
|
6,599
|
https://artofproblemsolving.com/wiki/index.php/2000_AIME_I_Problems/Problem_1
| 1
|
Find the least positive integer $n$ such that no matter how $10^{n}$ is expressed as the product of any two positive integers, at least one of these two integers contains the digit $0$
|
If a factor of $10^{n}$ has a $2$ and a $5$ in its prime factorization , then that factor will end in a $0$ . Therefore, we have left to consider the case when the two factors have the $2$ s and the $5$ s separated, so we need to find the first power of 2 or 5 that contains a 0.
For $n = 1:$ \[2^1 = 2 , 5^1 = 5\] $n = 2:$ \[2^2 = 4 , 5 ^ 2 =25\] $n = 3:$ \[2^3 = 8 , 5 ^3 = 125\]
and so on, until,
$n = 8:$ $2^8 = 256$ $5^8 = 390625$
We see that $5^8$ contains the first zero, so $n = \boxed{8}$
| 8
|
6,600
|
https://artofproblemsolving.com/wiki/index.php/2000_AIME_I_Problems/Problem_3
| 1
|
In the expansion of $(ax + b)^{2000},$ where $a$ and $b$ are relatively prime positive integers, the coefficients of $x^{2}$ and $x^{3}$ are equal. Find $a + b$
|
Using the binomial theorem $\binom{2000}{1998} b^{1998}a^2 = \binom{2000}{1997}b^{1997}a^3 \Longrightarrow b=666a$
Since $a$ and $b$ are positive relatively prime integers, $a=1$ and $b=666$ , and $a+b=\boxed{667}$
| 667
|
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