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4ea49438d38831734f643090010d6dfc | https://artofproblemsolving.com/wiki/index.php/2024_AIME_II_Problems/Problem_14 | Let \(b\ge 2\) be an integer. Call a positive integer \(n\) \(b\text-\textit{eautiful}\) if it has exactly two digits when expressed in base \(b\) and these two digits sum to \(\sqrt n\). For example, \(81\) is \(13\text-\textit{eautiful}\) because \(81 = \underline{6} \ \underline{3}_{13} \) and \(6 + 3 = \sqrt{81}\). Find the least integer \(b\ge 2\) for which there are more than ten \(b\text-\textit{eautiful}\) integers. | We write the base- $b$ two-digit integer as $\left( xy \right)_b$ .
Thus, this number satisfies \[ \left( x + y \right)^2 = b x + y \] with $x \in \left\{ 1, 2, \cdots , b-1 \right\}$ and $y \in \left\{ 0, 1, \cdots , b - 1 \right\}$
The above conditions imply $\left( x + y \right)^2 < b^2$ . Thus, $x + y \leq b - 1$
The above equation can be reorganized as \[ \left( x + y \right) \left( x + y - 1 \right) = \left( b - 1 \right) x . \]
Denote $z = x + y$ and $b' = b - 1$ .
Thus, we have \[ z \left( z - 1 \right) = b' x , \hspace{1cm} (1) \] where $z \in \left\{ 2, 3, \cdots , b' \right\}$ and $x \in \left\{ 1, 2, \cdots , b' \right\}$
Next, for each $b'$ , we solve Equation (1).
We write $b'$ in the prime factorization form as $b' = \Pi_{i=1}^n p_i^{k_i}$ .
Let $\left(A, \bar A \right)$ be any ordered partition of $\left\{ 1, 2, \cdots , n \right\}$ (we allow one set to be empty).
Denote $P_A = \Pi_{i \in A} p_i^{k_i}$ and $P_{\bar A} = \Pi_{i \in \bar A} p_i^{k_i}$
Because ${\rm gcd} \left( z, z-1 \right) = 1$ , there must exist such an ordered partition, such that $P_A | z$ and $P_{\bar A} | z-1$
Next, we prove that for each ordered partition $\left( A, \bar A \right)$ , if a solution of $z$ exists, then it must be unique.
Suppose there are two solutions of $z$ under partition $\left( A, \bar A \right)$ $z_1 = c_1 P_A$ $z_1 - 1 = d_1 P_{\bar A}$ , and $z_2 = c_2 P_A$ $z_2 - 1 = d_2 P_{\bar A}$ .
W.L.O.G., assume $c_1 < c_2$ .
Hence, we have \[ \left( c_2 - c_1 \right) P_A = \left( d_2 - d_1 \right) P_{\bar A} . \]
Because ${\rm gcd} \left( P_A, P_{\bar A} \right) = 1$ and $c_1 < c_2$ , there exists a positive integer $m$ , such that $c_2 = c_1 + m P_{\bar A}$ and $d_2 = d_1 + m P_A$ .
Thus,
\begin{align*}
z_2 & = z_1 + m P_A P_{\bar A} \\
& = z_1 + m b' \\
& > b' .
\end{align*}
However, recall $z_2 \leq b'$ . We get a contradiction.
Therefore, under each ordered partition for $b'$ , the solution of $z$ is unique.
Note that if $b'$ has $n$ distinct prime factors, the number of ordered partitions is $2^n$ .
Therefore, to find a $b'$ such that the number of solutions of $z$ is more than 10, the smallest $n$ is 4.
With $n = 4$ , the smallest number is $2 \cdot 3 \cdot 5 \cdot 7 = 210$ .
Now, we set $b' = 210$ and check whether the number of solutions of $z$ under this $b'$ is more than 10.
We can easily see that all ordered partitions (except $A = \emptyset$ ) guarantee feasible solutions of $z$ .
Therefore, we have found a valid $b'$ .
Therefore, $b = b' + 1 = \boxed{211}$ | null | 211 |
dea2f8abb958fd795a6289c79f33e738 | https://artofproblemsolving.com/wiki/index.php/2024_AIME_II_Problems/Problem_15 | Find the number of rectangles that can be formed inside a fixed regular dodecagon ( $12$ -gon) where each side of the rectangle lies on either a side or a diagonal of the dodecagon. The diagram below shows three of those rectangles.
[asy] unitsize(0.6 inch); for(int i=0; i<360; i+=30) { dot(dir(i), 4+black); draw(dir(i)--dir(i+30)); } draw(dir(120)--dir(330)); filldraw(dir(210)--dir(240)--dir(30)--dir(60)--cycle, mediumgray, linewidth(1.5)); draw((0,0.366)--(0.366,0), linewidth(1.5)); [/asy] | By Furaken
There are two kinds of such rectangles: those whose sides are parallel to some edges of the regular 12-gon (Case 1, and those whose sides are not (Case 2).
For Case 1, WLOG assume that the rectangle's sides are horizontal and vertical (don't forget to multiply by 3 at the end of Case 1). Then the rectangle's sides coincide with these segments as shown in the diagram. [asy] real r = pi/6; pair A1 = (cos(r),sin(r)); pair A2 = (cos(2r),sin(2r)); pair A3 = (cos(3r),sin(3r)); pair A4 = (cos(4r),sin(4r)); pair A5 = (cos(5r),sin(5r)); pair A6 = (cos(6r),sin(6r)); pair A7 = (cos(7r),sin(7r)); pair A8 = (cos(8r),sin(8r)); pair A9 = (cos(9r),sin(9r)); pair A10 = (cos(10r),sin(10r)); pair A11 = (cos(11r),sin(11r)); pair A12 = (cos(12r),sin(12r)); dot(A1); dot(A2); dot(A3); dot(A4); dot(A5); dot(A6); dot(A7); dot(A8); dot(A9); dot(A10); dot(A11); dot(A12); pair B1 = (0.5,0.5); pair B2 = (-0.5,0.5); pair B3 = (-0.5,-0.5); pair B4 = (0.5,-0.5); dot(B1); dot(B2); dot(B3); dot(B4); draw(A1--A5--A7--A11--cycle); draw(A2--A4--A8--A10--cycle); draw(A3--A9); draw(A6--A12); label("$A_1$", A1, NE); label("$A_2$", A2, NE); label("$A_3$", A3, N); label("$A_4$", A4, NW); label("$A_5$", A5, NW); label("$A_6$", A6, W); label("$A_7$", A7, SW); label("$A_8$", A8, SW); label("$A_9$", A9, S); label("$A_{10}$", A10, SE); label("$A_{11}$", A11, SE); label("$A_{12}$", A12, E); label("$B_1$", B1, SW); label("$B_2$", B2, SE); label("$B_3$", B3, NE); label("$B_4$", B4, NW); [/asy] We use inclusion-exclusion for this. There are 30 valid rectangles contained in $A_1A_5A_7A_{11}$ , as well as 30 in $A_2A_4A_8A_{10}$ . However, the 9 rectangles contained in $B_1B_2B_3B_4$ have been counted twice, so we subtract 9 and we have 51 rectangles in the diagram. Multiplying by 3, we get 153 rectangles for Case 1.
For Case 2, we have this diagram. To be honest, you can count the rectangles here in whatever way you like. [asy] real r = pi/6; pair A1 = (cos(r),sin(r)); pair A2 = (cos(2r),sin(2r)); pair A3 = (cos(3r),sin(3r)); pair A4 = (cos(4r),sin(4r)); pair A5 = (cos(5r),sin(5r)); pair A6 = (cos(6r),sin(6r)); pair A7 = (cos(7r),sin(7r)); pair A8 = (cos(8r),sin(8r)); pair A9 = (cos(9r),sin(9r)); pair A10 = (cos(10r),sin(10r)); pair A11 = (cos(11r),sin(11r)); pair A12 = (cos(12r),sin(12r)); dot(A1); dot(A2); dot(A3); dot(A4); dot(A5); dot(A6); dot(A7); dot(A8); dot(A9); dot(A10); dot(A11); dot(A12); draw(A1--A6--A7--A12--cycle); draw(A3--A4--A9--A10--cycle); draw(A2--A5--A8--A11--cycle); label("$A_1$", A1, NE); label("$A_2$", A2, NE); label("$A_3$", A3, N); label("$A_4$", A4, NW); label("$A_5$", A5, NW); label("$A_6$", A6, W); label("$A_7$", A7, SW); label("$A_8$", A8, SW); label("$A_9$", A9, S); label("$A_{10}$", A10, SE); label("$A_{11}$", A11, SE); label("$A_{12}$", A12, E); [/asy] There are 36 rectangles contained within $A_2A_5A_8A_{11}$ , and 18 that use points outside $A_2A_5A_8A_{11}$ . So we get a total of $3(36+18)=162$ rectangles for Case 2.
Adding the two cases together, we get the answer $\boxed{315}$ | null | 315 |
dea2f8abb958fd795a6289c79f33e738 | https://artofproblemsolving.com/wiki/index.php/2024_AIME_II_Problems/Problem_15 | Find the number of rectangles that can be formed inside a fixed regular dodecagon ( $12$ -gon) where each side of the rectangle lies on either a side or a diagonal of the dodecagon. The diagram below shows three of those rectangles.
[asy] unitsize(0.6 inch); for(int i=0; i<360; i+=30) { dot(dir(i), 4+black); draw(dir(i)--dir(i+30)); } draw(dir(120)--dir(330)); filldraw(dir(210)--dir(240)--dir(30)--dir(60)--cycle, mediumgray, linewidth(1.5)); draw((0,0.366)--(0.366,0), linewidth(1.5)); [/asy] | Using the same diagram as Solution 1, we can get the number of rectangles from Case 1 by adding the number of rectangles of $A_2$ $A_8$ $A_8$ $A_{10}$ and $A_1$ $A_5$ $A_7$ $A_{11}$ and then subtracting the overlaps,
\[\binom{5}{2}\binom{3}{2} + \binom{5}{2}\binom{3}{2} - \binom{3}{2}\binom{3}{2}\] \[=51\]
We multiply this by 3 to get the total number of rectangles for Case 1, which is 153.
For Case 2, we can first get the total number of rectangles from $A_2A_3A_4A_5A_8A_9A_{10}A_{11}$ then add $A_1A_6A_7A_{12}$ and subtract by the overlaps, \[\binom{4}{2}\binom{4}{2} + \binom{6}{2} - \binom{4}{2} + \binom{6}{2} - \binom{4}{2}\] \[= 54\] Multiply that by 3 and add it to Case 1 to get $\boxed{315}$ | null | 315 |
dea2f8abb958fd795a6289c79f33e738 | https://artofproblemsolving.com/wiki/index.php/2024_AIME_II_Problems/Problem_15 | Find the number of rectangles that can be formed inside a fixed regular dodecagon ( $12$ -gon) where each side of the rectangle lies on either a side or a diagonal of the dodecagon. The diagram below shows three of those rectangles.
[asy] unitsize(0.6 inch); for(int i=0; i<360; i+=30) { dot(dir(i), 4+black); draw(dir(i)--dir(i+30)); } draw(dir(120)--dir(330)); filldraw(dir(210)--dir(240)--dir(30)--dir(60)--cycle, mediumgray, linewidth(1.5)); draw((0,0.366)--(0.366,0), linewidth(1.5)); [/asy] | We put the dodecagon in the right position that there exists a side whose slope is 0.
Note that finding a rectangle is equivalent to finding two pairs of lines, such that two lines in each pair are parallel and lines from different pairs are perpendicular.
Now, we use this property to count the number of rectangles.
Because two pairs of lines that form a rectangle are perpendicular, we only need to use the slope of one pair, denoted as $k$ , to determine the direction of the rectangle.
The slope of the other pair is thus $- \frac{1}{k}$ . To avoid overcounting, we do casework analysis by defining each case in term of $0 \leq k < \infty$ only (we make a convention that if $k = 0$ , then $- \frac{1}{k} = \infty$ ).
In our counting, we will frequently quantify the distance between two vertices of the regular dodecagon.
To characterize this in a straightforward way, we simply measure the number of vertices (on the minor arc) between our measured two vertices. For instance, two vertices on a side has distance 0. Distances between two vertices that are diagonals can be 1, 2, 3, 4, 5.
Case 1: $k = 0, \tan 30^\circ, \tan 60^\circ$
We only count for $k = 0$ . The number of solutions for $k = \tan 30^\circ$ and $\tan 60^\circ$ are the same.
Consider $k = 0$ .
We need to find a pair of horizontal segments and a pair of vertical segments to form a rectangle.
For $k = 0$ , the length of each horizontal segment can only be 0, 2, 4.
Denote by $2i$ the shorter length of two parallel horizontal segments.
Given $i$ , the number of pairs of two parallel horizontal segments is $1 + 2 \left( 4 - 2 i \right)$
Given $i$ , to form a rectangle, the number of pairs of vertical segments is $\binom{2i + 2}{2}$
Therefore, for $k = 0$ , the number of rectangles is
\begin{align*}
\sum_{i=0}^2 \left( 1 + 2 \left( 4 - 2 i \right) \right)
\binom{2i + 2}{2}
& = 54 .
\end{align*}
The number of rectangles for $k = \tan 30^\circ$ and $\tan 60^\circ$ are the same.
Therefore, the total number of rectangles in this case is $54 \cdot 3 = 162$
Case 2: $k = \tan 15^\circ$ $\tan 45^\circ$ $\tan 75^\circ$
The number of rectangles under all these $k$ s are the same.
So we only count for $k = \tan 15^\circ$
For $k = \tan 15^\circ$ , the length of each segment can only be 1, 3, 5.
However, there is only one segment with length 5.
So this cannot be the shorter length of two parallel segments with slope $\tan 15^\circ$
Denote by $2i + 1$ the shorter length of two parallel segments with slope $\tan 15^\circ$ .
Given $i$ , the number of pairs of two parallel segments is $1 + 2 \left( 3 - 2 i \right)$
Given $i$ , to form a rectangle, the number of pairs of vertical segments is $\binom{2i + 3}{2}$
Therefore, for $k = \tan 15^\circ$ , the number of rectangles is
\begin{align*}
\sum_{i=0}^1 \left( 1 + 2 \left( 3 - 2 i \right) \right)
\binom{2i + 3}{2}
& = 51 .
\end{align*}
The number of rectangles for $k = \tan 45^\circ$ and $\tan 75^\circ$ are the same.
Therefore, the total number of rectangles in this case is $51 \cdot 3 = 153$
Putting all cases together, the total number of rectangles is $162 + 153 = \boxed{315}$ | null | 315 |
d238278c57b88de1dd3978e48b14a21a | https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_1 | Five men and nine women stand equally spaced around a circle in random order. The probability that every man stands diametrically opposite a woman is $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$ | For simplicity purposes, we consider two arrangements different even if they only differ by rotations or reflections. In this way, there are $14!$ arrangements without restrictions.
First, there are $\binom{7}{5}$ ways to choose the man-woman diameters. Then, there are $10\cdot8\cdot6\cdot4\cdot2$ ways to place the five men each in a man-woman diameter. Finally, there are $9!$ ways to place the nine women without restrictions.
Together, the requested probability is \[\frac{\tbinom{7}{5}\cdot(10\cdot8\cdot6\cdot4\cdot2)\cdot9!}{14!} = \frac{21\cdot(10\cdot8\cdot6\cdot4\cdot2)}{14\cdot13\cdot12\cdot11\cdot10} = \frac{48}{143},\] from which the answer is $48+143 = \boxed{191}.$ | null | 191 |
d238278c57b88de1dd3978e48b14a21a | https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_1 | Five men and nine women stand equally spaced around a circle in random order. The probability that every man stands diametrically opposite a woman is $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$ | We can simply just loop through each of the men and find the probability that the person opposite from him is a woman.
Start by sitting down the $1$ st man. The probability that the person opposite to him is a woman is $\frac{9}{13}$ since out of the $13$ people who can sit opposite to him, $9$ can be a woman. With the $2$ nd man, we can use the same logic: there are $11$ people who can sit opposite to him, but only $8$ of them are a woman, so the probability is $\frac{8}{11}.$ We use the same logic for the $3$ rd, $4$ th and $5$ th men to get probabilities of $\frac{7}{9}$ $\frac{6}{7}$ and $\frac{5}{5},$ respectively.
Multiplying these probabilities, we get a final answer of \[\frac{9}{13}\cdot\frac{8}{11}\cdot\frac{7}{9}\cdot\frac{6}{7}\cdot\frac{5}{5}=\frac{48}{143}\longrightarrow\boxed{191}.\] | null | 191 |
d238278c57b88de1dd3978e48b14a21a | https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_1 | Five men and nine women stand equally spaced around a circle in random order. The probability that every man stands diametrically opposite a woman is $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$ | This problem is equivalent to solving for the probability that no man is standing diametrically opposite to another man. We can simply just construct this.
We first place the $1$ st man anywhere on the circle, now we have to place the $2$ nd man somewhere around the circle such that he is not diametrically opposite to the first man. This can happen with a probability of $\frac{12}{13}$ because there are $13$ available spots, and $12$ of them are not opposite to the first man.
We do the same thing for the $3$ rd man, finding a spot for him such that he is not opposite to the other $2$ men, which would happen with a probability of $\frac{10}{12}$ using similar logic. Doing this for the $4$ th and $5$ th men, we get probabilities of $\frac{8}{11}$ and $\frac{6}{10}$ respectively.
Multiplying these probabilities, we get, \[\frac{12}{13}\cdot\frac{10}{12}\cdot\frac{8}{11}\cdot\frac{6}{10}=\frac{48}{143}\longrightarrow\boxed{191}.\] | null | 191 |
d238278c57b88de1dd3978e48b14a21a | https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_1 | Five men and nine women stand equally spaced around a circle in random order. The probability that every man stands diametrically opposite a woman is $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$ | Assume that rotations and reflections are distinct arrangements, and replace men and women with identical M's and W's, respectively. (We can do that because the number of ways to arrange $5$ men in a circle and the number of ways to arrange $9$ women in a circle, are constants.) The total number of ways to arrange $5$ M's and $9$ W's is $\binom{14}{5} = 2002.$
To count the number of valid arrangements (i.e. arrangements where every M is diametrically opposite a W), we notice that exactly $2$ of the pairs of diametrically opposite positions must be occupied by $2$ W's. There are $\binom{7}{2} = 21$ ways to choose these $2$ pairs. For the remaining $5$ pairs, we have to choose which position is occupied by an M and which is occupied by a W. This can be done in $2^{5} = 32$ ways. Therefore, there are $21*32 = 672$ valid arrangements.
Therefore, the probability that an arrangement is valid is $\frac{672}{2002} = \frac{48}{143}$ for an answer of $\boxed{191}.$ | null | 191 |
d238278c57b88de1dd3978e48b14a21a | https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_1 | Five men and nine women stand equally spaced around a circle in random order. The probability that every man stands diametrically opposite a woman is $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$ | To start off, we calculate the total amount of ways to organize all $14$ people irrespective of any constraints. This is simply ${14\choose5} = 2002$ , because we just count how many ways we can place all $5$ men in any of the $14$ slots.
Since men cannot be diametrically opposite with each other, because of the constraints, placing down one man in any given spot will make another spot on the opposite side of the circle unable to hold any men. This means that placing down one man will effectively take away $2$ spots.
There are $14$ possible slots the first man can be placed. Once that man was placed, the next man only have $12$ possible slots because the slot that the first man is in is taken and the diametrically opposite spot to the first man can't have any men. Similar logic applies for the third man, who has $10$ possible slots. The fourth man has $8$ possible slots, and the fifth man has $6$ possible slots.
This means the number of ways you can place all $5$ men down is $14 \cdot 12 \cdot 10 \cdot 8 \cdot 6$ . However, since the men are all indistinct from each other, you also have to divide that value by $5! = 120$ , since there are $120$ ways to arrange the $5$ men in each possible positioning of the men on the circle. This means the total number of ways to arrange the men around the circle so that none of them are diametrically opposite of each other is: $\frac{14 \cdot 12 \cdot 10 \cdot 8 \cdot 6}{5!} = 672$ . The women simply fill in the rest of the available slots in each arrangement of men.
Thus, the final probability is $\frac{672}{2002} = \frac{48}{143}$ , meaning the answer is $48 + 143 = \boxed{191}$ | null | 191 |
d238278c57b88de1dd3978e48b14a21a | https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_1 | Five men and nine women stand equally spaced around a circle in random order. The probability that every man stands diametrically opposite a woman is $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$ | We will first assign seats to the men. The first man can be placed in any of the $14$ slots. The second man can be placed in any of the remaining $13$ seats, except for the one diametrically opposite to the first man. So, there are $13 - 1 = 12$ ways to seat him. With a similar argument, the third man can be seated in $10$ ways, the fourth man in $8$ ways and the last man in $6$ ways.
So, the total number of ways to arrange the men is $14 \cdot 12 \cdot 10 \cdot 8 \cdot 6$
The women go to the remaining $9$ spots. Note that since none of the seats diametrically opposite to the men is occupied, each man is opposite a woman. The number of ways to arrange the women is therefore, simply $9!$ , meaning that the total number of ways to arrange the people with restrictions is $14 \cdot 12 \cdot 10 \cdot 8 \cdot 6 \cdot 9!$ In general, there are $14!$ ways to arrange the people without restrictions. So, the probability is \[\frac{14 \cdot 12 \cdot 10 \cdot 8 \cdot 6 \cdot 9!}{14!} = \frac{8 \cdot 6}{13 \cdot 11} = \frac{48}{143}.\] The answer is $48 + 143 = \boxed{191}$ | null | 191 |
d238278c57b88de1dd3978e48b14a21a | https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_1 | Five men and nine women stand equally spaced around a circle in random order. The probability that every man stands diametrically opposite a woman is $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$ | First pin one man on one seat (to ensure no rotate situations). Then there are $13!$ arrangements.
Because $5$ men must have women at their opposite side, we consider the $2$ nd man and the woman opposite as one group and name it $P_2.$ There are $4$ groups, $P_1, P_2, P_3, P_4$ except the first man pinned on the same point. And for the rest $4$ women, name them $P_5$ and $P_6.$ First to order $P_1, P_2, P_3, P_4, P_5, P_6,$ there are $6!$ ways. For the $1$ st man, there are $9$ women to choose, $8$ for the $2$ nd, $\ldots,$ $5$ for the $5$ th, and then for the $2$ women pairs $3$ and $1.$ Because every $2$ person in the group have chance to change their position, there are $2^6$ possibilities.
So the possibility is \[P=\frac{6!\cdot 9\cdot 8\cdot 7\cdot 6\cdot 5\cdot 3\cdot 1 \cdot 2^6}{13!}=\frac{48}{143}.\]
The answer is $48+143=\boxed{191}.$ | null | 191 |
d238278c57b88de1dd3978e48b14a21a | https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_1 | Five men and nine women stand equally spaced around a circle in random order. The probability that every man stands diametrically opposite a woman is $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$ | We get around the condition that each man can't be opposite to another man by simply considering all $7$ diagonals, and choosing $5$ where there will be a single man. For each diagonal, the man can go on either side, and there are $\binom{14}{5}$ ways to arrange the men and the women in total. Thus our answer is $\frac{\binom{7}{5}\cdot 2^5}{\binom{14}{5}} = \frac{48}{143}.$ We get $48 + 143 = \boxed{191}$ | null | 191 |
d238278c57b88de1dd3978e48b14a21a | https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_1 | Five men and nine women stand equally spaced around a circle in random order. The probability that every man stands diametrically opposite a woman is $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$ | We can find the probability of one arrangement occurring, and multiply it by the total number of arrangements.
The probability of a man being in any specific position is $\frac{5}{14}.$ The probability of a woman being across from him is $\frac{9}{13}.$ The probability of a man being in any valid position is now $\frac{4}{12},$ and the probability of a woman being across from him is $\frac{8}{11},$ and so forth. We stop when there are no more men left. Multiplying these probabilities together,
\[P(\mathrm{One\ successful\ outcome})=\frac{5}{14}\cdot \frac{9}{13}\cdot \frac{4}{12}\cdot \frac{8}{11}\cdot \frac{3}{10}\cdot \frac{7}{9}\cdot \frac{2}{8}\cdot \frac{1}{6}\cdot \frac{6}{7}\cdot \frac{5}{5} = \frac{1}{2002}.\] To find the total number of successful outcomes, we consider the diagonals; the total number of diagonals to be made is $\binom75$ , since there are $7$ total diagonals, and we want to choose $5$ of them to connect a man to a woman. For each of these diagonals, the man can be on either side of the diagonal.
It follows that there are $2$ possibilities for each diagonal (man on one side, woman on the other, and vice versa). There are $5$ diagonals with a man and a woman, so there are $2^5$ different ways for these diagonals to appear.
There are $\binom75$ successful diagonals, and for each of these diagonals, there are $2^5$ ways to seat the men and the women, there are $\binom75$ $\cdot 2^5$ successful outcomes.
Recall that \[P(\mathrm{Any\ successful\ outcome})=P(\mathrm{One\ successful\ outcome})\cdot P(\mathrm{Total\ number\ of\ successful\ outcomes}).\] Therefore, \[P(\mathrm{Any\ successful\ outcome}) = \frac{1}{2002}\cdot 2^5\cdot \binom75 = \frac{1}{2002}\cdot 2^5\cdot 21 = \frac{2^5\cdot 21}{2002} = \frac{48}{143}.\] The requested sum is $48+143=\boxed{191}.$ | null | 191 |
923ba90c85c5bb7565d155c041eaecf1 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_3 | A plane contains $40$ lines, no $2$ of which are parallel. Suppose that there are $3$ points where exactly $3$ lines intersect, $4$ points where exactly $4$ lines intersect, $5$ points where exactly $5$ lines intersect, $6$ points where exactly $6$ lines intersect, and no points where more than $6$ lines intersect. Find the number of points where exactly $2$ lines intersect. | In this solution, let $\boldsymbol{n}$ -line points be the points where exactly $n$ lines intersect. We wish to find the number of $2$ -line points.
There are $\binom{40}{2}=780$ pairs of lines. Among them:
It follows that the $2$ -line points account for $780-9-24-50-90=\boxed{607}$ pairs of lines, where each pair intersect at a single point. | null | 607 |
5eb519ddb9d0c05ab135c7104c5b8b7e | https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_5 | Let $P$ be a point on the circle circumscribing square $ABCD$ that satisfies $PA \cdot PC = 56$ and $PB \cdot PD = 90.$ Find the area of $ABCD.$ | Ptolemy's theorem states that for cyclic quadrilateral $WXYZ$ $WX\cdot YZ + XY\cdot WZ = WY\cdot XZ$
We may assume that $P$ is between $B$ and $C$ . Let $PA = a$ $PB = b$ $PC = c$ $PD = d$ , and $AB = s$ . We have $a^2 + c^2 = AC^2 = 2s^2$ , because $AC$ is a diameter of the circle. Similarly, $b^2 + d^2 = 2s^2$ . Therefore, $(a+c)^2 = a^2 + c^2 + 2ac = 2s^2 + 2(56) = 2s^2 + 112$ . Similarly, $(b+d)^2 = 2s^2 + 180$
By Ptolemy's Theorem on $PCDA$ $as + cs = ds\sqrt{2}$ , and therefore $a + c = d\sqrt{2}$ . By Ptolemy's on $PBAD$ $bs + ds = as\sqrt{2}$ , and therefore $b + d = a\sqrt{2}$ . By squaring both equations, we obtain \begin{alignat*}{8} 2d^2 &= (a+c)^2 &&= 2s^2 + 112, \\ 2a^2 &= (b+d)^2 &&= 2s^2 + 180. \end{alignat*} Thus, $a^2 = s^2 + 90$ , and $d^2 = s^2 + 56$ . Plugging these values into $a^2 + c^2 = b^2 + d^2 = 2s^2$ , we obtain $c^2 = s^2 - 90$ , and $b^2 = s^2 - 56$ . Now, we can solve using $a$ and $c$ (though using $b$ and $d$ yields the same solution for $s$ ). \begin{align*} ac = (\sqrt{s^2 - 90})(\sqrt{s^2 + 90}) &= 56 \\ (s^2 + 90)(s^2 - 90) &= 56^2 \\ s^4 &= 90^2 + 56^2 = 106^2 \\ s^2 &= \boxed{106} ~mathboy100 | null | 106 |
5eb519ddb9d0c05ab135c7104c5b8b7e | https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_5 | Let $P$ be a point on the circle circumscribing square $ABCD$ that satisfies $PA \cdot PC = 56$ and $PB \cdot PD = 90.$ Find the area of $ABCD.$ | By the Inscribed Angle Theorem , we conclude that $\triangle PAC$ and $\triangle PBD$ are right triangles.
Let the brackets denote areas. We are given that \begin{alignat*}{8} 2[PAC] &= PA \cdot PC &&= 56, \\ 2[PBD] &= PB \cdot PD &&= 90. \end{alignat*} Let $O$ be the center of the circle, $X$ be the foot of the perpendicular from $P$ to $\overline{AC},$ and $Y$ be the foot of the perpendicular from $P$ to $\overline{BD},$ as shown below: [asy] /* Made by MRENTHUSIASM */ size(200); pair A, B, C, D, O, P, X, Y; A = (-sqrt(106)/2,sqrt(106)/2); B = (-sqrt(106)/2,-sqrt(106)/2); C = (sqrt(106)/2,-sqrt(106)/2); D = (sqrt(106)/2,sqrt(106)/2); O = origin; path p; p = Circle(O,sqrt(212)/2); draw(p); P = intersectionpoints(Circle(A,4),p)[1]; X = foot(P,A,C); Y = foot(P,B,D); draw(A--B--C--D--cycle); draw(P--A--C--cycle,red); draw(P--B--D--cycle,blue); draw(P--X,red+dashed); draw(P--Y,blue+dashed); markscalefactor=0.075; draw(rightanglemark(A,P,C),red); draw(rightanglemark(P,X,C),red); draw(rightanglemark(B,P,D),blue); draw(rightanglemark(P,Y,D),blue); dot("$A$", A, 1.5*NW, linewidth(4)); dot("$B$", B, 1.5*SW, linewidth(4)); dot("$C$", C, 1.5*SE, linewidth(4)); dot("$D$", D, 1.5*NE, linewidth(4)); dot("$P$", P, 1.5*dir(P), linewidth(4)); dot("$X$", X, 1.5*dir(20), linewidth(4)); dot("$Y$", Y, 1.5*dir(Y-P), linewidth(4)); dot("$O$", O, 1.5*E, linewidth(4)); [/asy] Let $d$ be the diameter of $\odot O.$ It follows that \begin{alignat*}{8} 2[PAC] &= d\cdot PX &&= 56, \\ 2[PBD] &= d\cdot PY &&= 90. \end{alignat*} Moreover, note that $OXPY$ is a rectangle. By the Pythagorean Theorem, we have \[PX^2+PY^2=PO^2.\] We rewrite this equation in terms of $d:$ \[\left(\frac{56}{d}\right)^2+\left(\frac{90}{d}\right)^2=\left(\frac d2\right)^2,\] from which $d^2=212.$ Therefore, we get \[[ABCD] = \frac{d^2}{2} = \boxed{106}.\] ~MRENTHUSIASM | null | 106 |
5eb519ddb9d0c05ab135c7104c5b8b7e | https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_5 | Let $P$ be a point on the circle circumscribing square $ABCD$ that satisfies $PA \cdot PC = 56$ and $PB \cdot PD = 90.$ Find the area of $ABCD.$ | [asy] /* Made by MRENTHUSIASM */ size(200); pair A, B, C, D, O, P, X, Y; A = (-sqrt(106)/2,sqrt(106)/2); B = (-sqrt(106)/2,-sqrt(106)/2); C = (sqrt(106)/2,-sqrt(106)/2); D = (sqrt(106)/2,sqrt(106)/2); O = origin; path p; p = Circle(O,sqrt(212)/2); draw(p); P = intersectionpoints(Circle(A,4),p)[1]; X = foot(P,A,C); Y = foot(P,B,D); draw(A--B--C--D--cycle); draw(P--A--C--cycle,red); draw(P--B--D--cycle,blue); draw(P--X,red+dashed); draw(P--Y,blue+dashed); markscalefactor=0.075; draw(rightanglemark(A,P,C),red); draw(rightanglemark(P,X,C),red); draw(rightanglemark(B,P,D),blue); draw(rightanglemark(P,Y,D),blue); dot("$A$", A, 1.5*NW, linewidth(4)); dot("$B$", B, 1.5*SW, linewidth(4)); dot("$C$", C, 1.5*SE, linewidth(4)); dot("$D$", D, 1.5*NE, linewidth(4)); dot("$P$", P, 1.5*dir(P), linewidth(4)); dot("$X$", X, 1.5*dir(20), linewidth(4)); dot("$Y$", Y, 1.5*dir(Y-P), linewidth(4)); dot("$O$", O, 1.5*E, linewidth(4)); [/asy] Let the center of the circle be $O$ , and the radius of the circle be $r$ . Since $ABCD$ is a rhombus with diagonals $2r$ and $2r$ , its area is $\dfrac{1}{2}(2r)(2r) = 2r^2$ . Since $AC$ and $BD$ are diameters of the circle, $\triangle APC$ and $\triangle BPD$ are right triangles. Let $X$ and $Y$ be the foot of the altitudes to $AC$ and $BD$ , respectively. We have \[[\triangle APC] = \frac{1}{2}(PA)(PC) = \frac{1}{2}(PX)(AC),\] so $PX = \dfrac{(PA)(PC)}{AC} = \dfrac{28}{r}$ . Similarly, \[[\triangle BPD] = \frac{1}{2}(PB)(PD) = \frac{1}{2}(PY)(PB),\] so $PY = \dfrac{(PB)(PD)}{BD} = \dfrac{45}{r}$ . Since $\triangle APX \sim \triangle PCX,$ \[\frac{AX}{PX} = \frac{PX}{CX}\] \[\frac{AO - XO}{PX} = \frac{PX}{OC + XO}.\] But $PXOY$ is a rectangle, so $PY = XO$ , and our equation becomes \[\frac{r - PY}{PX} = \frac{PX}{r + PY}.\] Cross multiplying and rearranging gives us $r^2 = PX^2 + PY^2 = \left(\dfrac{28}{r}\right)^2 + \left(\dfrac{45}{r}\right)^2$ , which rearranges to $r^4 = 2809$ . Therefore $[ABCD] = 2r^2 = \boxed{106}$ | null | 106 |
5eb519ddb9d0c05ab135c7104c5b8b7e | https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_5 | Let $P$ be a point on the circle circumscribing square $ABCD$ that satisfies $PA \cdot PC = 56$ and $PB \cdot PD = 90.$ Find the area of $ABCD.$ | Drop a height from point $P$ to line $\overline{AC}$ and line $\overline{BC}$ . Call these two points to be $X$ and $Y$ , respectively. Notice that the intersection of the diagonals of $\square ABCD$ meets at a right angle at the center of the circumcircle, call this intersection point $O$
Since $OXPY$ is a rectangle, $OX$ is the distance from $P$ to line $\overline{BD}$ . We know that $\tan{\angle{POX}} = \frac{PX}{XO} = \frac{28}{45}$ by triangle area and given information. Then, notice that the measure of $\angle{OCP}$ is half of $\angle{XOP}$
Using the half-angle formula for tangent,
\begin{align*} \frac{(2 \cdot \tan{\angle{OCP}})}{(1-\tan^2{\angle{OCP}})} = \tan{\angle{POX}} = \frac{28}{45} \\ 14\tan^2{\angle{OCP}} + 45\tan{\angle{OCP}} - 14 = 0 \end{align*}
Solving the equation above, we get that $\tan{\angle{OCP}} = -7/2$ or $2/7$ . Since this value must be positive, we pick $\frac{2}{7}$ . Then, $\frac{PA}{PC} = 2/7$ (since $\triangle CAP$ is a right triangle with line $\overline{AC}$ the diameter of the circumcircle) and $PA * PC = 56$ . Solving we get $PA = 4$ $PC = 14$ , giving us a diagonal of length $\sqrt{212}$ and area $\boxed{106}$ | null | 106 |
5eb519ddb9d0c05ab135c7104c5b8b7e | https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_5 | Let $P$ be a point on the circle circumscribing square $ABCD$ that satisfies $PA \cdot PC = 56$ and $PB \cdot PD = 90.$ Find the area of $ABCD.$ | Denote by $x$ the half length of each side of the square.
We put the square to the coordinate plane, with $A = \left( x, x \right)$ $B = \left( - x , x \right)$ $C = \left( - x , - x \right)$ $D = \left( x , - x \right)$
The radius of the circumcircle of $ABCD$ is $\sqrt{2} x$ .
Denote by $\theta$ the argument of point $P$ on the circle.
Thus, the coordinates of $P$ are $P = \left( \sqrt{2} x \cos \theta , \sqrt{2} x \sin \theta \right)$
Thus, the equations $PA \cdot PC = 56$ and $PB \cdot PD = 90$ can be written as \begin{align*} \sqrt{\left( \sqrt{2} x \cos \theta - x \right)^2 + \left( \sqrt{2} x \sin \theta - x \right)^2} \cdot \sqrt{\left( \sqrt{2} x \cos \theta + x \right)^2 + \left( \sqrt{2} x \sin \theta + x \right)^2} & = 56 \\ \sqrt{\left( \sqrt{2} x \cos \theta + x \right)^2 + \left( \sqrt{2} x \sin \theta - x \right)^2} \cdot \sqrt{\left( \sqrt{2} x \cos \theta - x \right)^2 + \left( \sqrt{2} x \sin \theta + x \right)^2} & = 90 \end{align*}
These equations can be reformulated as \begin{align*} x^4 \left( 4 - 2 \sqrt{2} \left( \cos \theta + \sin \theta \right) \right) \left( 4 + 2 \sqrt{2} \left( \cos \theta + \sin \theta \right) \right) & = 56^2 \\ x^4 \left( 4 + 2 \sqrt{2} \left( \cos \theta - \sin \theta \right) \right) \left( 4 - 2 \sqrt{2} \left( \cos \theta - \sin \theta \right) \right) & = 90^2 \end{align*}
These equations can be reformulated as \begin{align*} 2 x^4 \left( 1 - 2 \cos \theta \sin \theta \right) & = 28^2 \hspace{1cm} (1) \\ 2 x^4 \left( 1 + 2 \cos \theta \sin \theta \right) & = 45^2 \hspace{1cm} (2) \end{align*}
Taking $\frac{(1)}{(2)}$ , by solving the equation, we get \[ 2 \cos \theta \sin \theta = \frac{45^2 - 28^2}{45^2 + 28^2} . \hspace{1cm} (3) \]
Plugging (3) into (1), we get \begin{align*} {\rm Area} \ ABCD & = \left( 2 x \right)^2 \\ & = 4 \sqrt{\frac{28^2}{2 \left( 1 - 2 \cos \theta \sin \theta \right)}} \\ & = 2 \sqrt{45^2 + 28^2} \\ & = 2 \cdot 53 \\ & = \boxed{106} | null | 106 |
5eb519ddb9d0c05ab135c7104c5b8b7e | https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_5 | Let $P$ be a point on the circle circumscribing square $ABCD$ that satisfies $PA \cdot PC = 56$ and $PB \cdot PD = 90.$ Find the area of $ABCD.$ | WLOG, let $P$ be on minor arc $\overarc {AB}$ . Let $r$ and $O$ be the radius and center of the circumcircle respectively, and let $\theta = \angle AOP$
By the Pythagorean Theorem , the area of the square is $2r^2$ . We can use the Law of Cosines on isosceles triangles $\triangle AOP, \, \triangle COP, \, \triangle BOP, \, \triangle DOP$ to get
\begin{align*} PA^2 &= 2r^2(1 - \cos \theta), \\ PC^2 &= 2r^2(1 - \cos (180 - \theta)) = 2r^2(1 + \cos \theta), \\ PB^2 &= 2r^2(1 - \cos (90 - \theta)) = 2r^2(1 - \sin \theta), \\ PD^2 &= 2r^2(1 - \cos (90 + \theta)) = 2r^2(1 + \sin \theta). \end{align*}
Taking the products of the first two and last two equations, respectively, \[56^2 = (PA \cdot PC)^2 = 4r^4(1 - \cos \theta)(1 + \cos \theta) = 4r^4(1 - \cos^2 \theta) = 4r^4 \sin^2 \theta,\] and \[90^2 = (PB \cdot PD)^2 = 4r^4(1 - \sin \theta)(1 + \sin \theta) = 4r^4(1 - \sin^2 \theta) = 4r^4 \cos^2 \theta.\] Adding these equations, \[56^2 + 90^2 = 4r^4,\] so \[2r^2 = \sqrt{56^2+90^2} = 2\sqrt{28^2+45^2} = 2\sqrt{2809} = 2 \cdot 53 = \boxed{106}.\] ~OrangeQuail9 | null | 106 |
5eb519ddb9d0c05ab135c7104c5b8b7e | https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_5 | Let $P$ be a point on the circle circumscribing square $ABCD$ that satisfies $PA \cdot PC = 56$ and $PB \cdot PD = 90.$ Find the area of $ABCD.$ | First draw a diagram. [asy] pair A, B, C, D, O, P; A = (0,sqrt(106)); B = (0,0); C = (sqrt(106),0); D = (sqrt(106),sqrt(106)); O = (sqrt(106)/2, sqrt(106)/2); P = intersectionpoint(circle(A, sqrt(212)*sin(atan(28/45)/2)), circle(O, sqrt(212)/2)); draw(A--B--C--D--cycle); draw(circle(O, sqrt(212)/2)); label("$A$", A, NW); label("$B$", B, SW); label("$C$", C, SE); label("$D$", D, NE); label("$P$", P, NW); label("$O$", O, 1.5*S); label("$\theta$", O, dir(120)*5); draw(P--A--C--cycle, red); draw(P--B--D--cycle, blue); draw(P--O); draw(anglemark(P,O,A,30)); dot(P); dot(O); [/asy] Let's say that the radius is $r$ . Then the area of the $ABCD$ is $(\sqrt2r)^2 = 2r^2$ Using the formula for the length of a chord subtended by an angle, we get \[PA = 2r\sin\left(\dfrac{\theta}2\right)\] \[PC = 2r\sin\left(\dfrac{180-\theta}2\right) = 2r\sin\left(90 - \dfrac{\theta}2\right) = 2r\cos\left(\dfrac{\theta}2\right)\] Multiplying and simplifying these 2 equations gives \[PA \cdot PC = 4r^2 \sin \left(\dfrac{\theta}2 \right) \cos \left(\dfrac{\theta}2 \right) = 2r^2 \sin\left(\theta \right) = 56\] Similarly $PB = 2r\sin\left(\dfrac{90 +\theta}2\right)$ and $PD =2r\sin\left(\dfrac{90 -\theta}2\right)$ . Again, multiplying gives \[PB \cdot PD = 4r^2 \sin\left(\dfrac{90 +\theta}2\right) \sin\left(\dfrac{90 -\theta}2\right) = 4r^2 \sin\left(90 -\dfrac{90 -\theta}2\right) \sin\left(\dfrac{90 -\theta}2\right)\] \[=4r^2 \sin\left(\dfrac{90 -\theta}2\right) \cos\left(\dfrac{90 -\theta}2\right) = 2r^2 \sin\left(90 - \theta \right) = 2r^2 \cos\left(\theta \right) = 90\] Dividing $2r^2 \sin \left(\theta \right)$ by $2r^2 \cos \left( \theta \right)$ gives $\tan \left(\theta \right) = \dfrac{28}{45}$ , so $\theta = \tan^{-1} \left(\dfrac{28}{45} \right)$ .
Pluging this back into one of the equations, gives \[2r^2 = \dfrac{90}{\cos\left(\tan^{-1}\left(\dfrac{28}{45}\right)\right)}\] If we imagine a $28$ $45$ $53$ right triangle, we see that if $28$ is opposite and $45$ is adjacent, $\cos\left(\theta\right) = \dfrac{\text{adj}}{\text{hyp}} = \dfrac{45}{53}$ . Now we see that \[2r^2 = \dfrac{90}{\frac{45}{53}} = \boxed{106}.\] ~Voldemort101 | null | 106 |
5eb519ddb9d0c05ab135c7104c5b8b7e | https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_5 | Let $P$ be a point on the circle circumscribing square $ABCD$ that satisfies $PA \cdot PC = 56$ and $PB \cdot PD = 90.$ Find the area of $ABCD.$ | [asy] pair A,B,C,D,P; A=(-3,3); B=(3,3); C=(3,-3); D=(-3,-3); draw(A--B--C--D--cycle); label(A,"$A$",NW); label(B,"$B$",NE); label(C,"$C$",SE); label(D,"$D$",SW); draw(circle((0,0),4.24264068712)); P=(-1,4.12310562562); label(P,"$P$", NW); pen k=red+dashed; draw(P--A,k); draw(P--B,k); draw(P--C,k); draw(P--D,k); dot(P); [/asy] Let $P=(a,b)$ on the upper quarter of the circle, and let $k$ be the side length of the square. Hence, we want to find $k^2$ . Let the center of the circle be $(0,0)$ .
The two equations would thus become: \[\left(\left(a+\dfrac{k}2\right)^2+\left(b-\dfrac{k}2\right)^2\right)\left(\left(a-\dfrac{k}2\right)^2+\left(b+\dfrac{k}2\right)^2\right)=56^2\] \[\left(\left(a-\dfrac{k}2\right)^2+\left(b-\dfrac{k}2\right)^2\right)\left(\left(a+\dfrac{k}2\right)^2+\left(b+\dfrac{k}2\right)^2\right)=90^2\] Now, let $m=\left(a+\dfrac{k}2\right)^2$ $n=\left(a-\dfrac{k}2\right)^2$ $o=\left(b+\dfrac{k}2\right)^2$ , and $p=\left(b-\dfrac{k}2\right)^2$ . Our equations now change to $(m+p)(n+o)=56^2=mn+op+mo+pn$ and $(n+p)(m+o)=90^2=mn+op+no+pm$ . Subtracting the first from the second, we have $pm+no-mo-pn=p(m-n)-o(m-n)=(m-n)(p-o)=34\cdot146$ . Substituting back in and expanding, we have $2ak\cdot-2bk=34\cdot146$ , so $abk^2=-17\cdot73$ . We now have one of our terms we need ( $k^2$ ). Therefore, we only need to find $ab$ to find $k^2$ .
We now write the equation of the circle, which point $P$ satisfies: \[a^2+b^2=\left(\dfrac{k\sqrt{2}}{2}\right)^2=\dfrac{k^2}2\] We can expand the second equation, yielding \[\left(a^2+b^2+\dfrac{k^2}2+(ak+bk)\right)\left(a^2+b^2+\dfrac{k^2}2-(ak+bk)\right)=(k^2+k(a+b))(k^2-k(a+b))=8100.\] Now, with difference of squares, we get $k^4-k^2\cdot(a+b)^2=k^2(k^2-(a+b)^2)=8100$ . We can add $2abk^2=-17\cdot73\cdot2=-2482$ to this equation, which we can factor into $k^2(k^2-(a+b)^2+2ab)=k^2(k^2-(a^2+b^2))=8100-2482$ . We realize that $a^2+b^2$ is the same as the equation of the circle, so we plug its equation in: $k^2\left(k^2-\dfrac{k^2}2\right)=5618$ . We can combine like terms to get $k^2\cdot\dfrac{k^2}2=5618$ , so $(k^2)^2=11236$ .
Since the answer is an integer, we know $11236$ is a perfect square. Since it is even, it is divisible by $4$ , so we can factor $11236=2^2\cdot2809$ . With some testing with approximations and last-digit methods, we can find that $53^2=2809$ . Therefore, taking the square root, we find that $k^2$ , the area of square $ABCD$ , is $2\cdot53=\boxed{106}$ | null | 106 |
5eb519ddb9d0c05ab135c7104c5b8b7e | https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_5 | Let $P$ be a point on the circle circumscribing square $ABCD$ that satisfies $PA \cdot PC = 56$ and $PB \cdot PD = 90.$ Find the area of $ABCD.$ | WLOG, let $P$ be on minor arc $AD.$ Draw in $AP$ $BP$ $CP$ $DP$ and let $\angle ABP = x.$ We can see, by the inscribed angle theorem, that $\angle APB = \angle ACB = 45$ , and $\angle CPD = \angle CAD = 45.$ Then, $\angle PAB = 135-x$ $\angle PCD = \angle PAD = (135-x)-90 = 45-x$ , and $\angle PDC = 90+x.$ Letting $(PA, PB, PC, PD, AB) = (a,b,c,d,s)$ , we can use the law of sines on triangles $PAB$ and $PCD$ to get \[s\sqrt{2} = \frac{a}{\sin(x)} = \frac{b}{\sin(135-x)} = \frac{c}{\sin(90+x)} = \frac{d}{\sin(45-x)}.\] Making all the angles in the above equation acute gives \[s\sqrt{2} = \frac{a}{\sin(x)} = \frac{b}{\sin(45+x)} = \frac{c}{\sin(90-x)} = \frac{d}{\sin(45-x)}.\]
Note that we are looking for $s^{2}.$ We are given that $ac = 56$ and $bd = 90.$ This means that $s^{2}\sin(x)\sin(90-x) = 28$ and $s^{2}\sin(45+x)\sin(45-x) = 45.$ However, \[\sin(x)\sin(90-x) = \sin(x)\cos(x) = \frac{\sin(2x)}{2}\] and \[\sin(45+x)\sin(45-x) = \frac{(\cos(x) + \sin(x))(\cos(x) - \sin(x))}{2} = \frac{\cos^{2}(x) - \sin^{2}(x)}{2} = \frac{\cos(2x)}{2}.\] Therefore, $s^{2}\sin(2x) = 56$ and $s^{2}\cos(2x) = 90.$ Therefore, by the Pythagorean Identity, \[s^{2} = \sqrt{(s^{2}\sin(2x))^{2} + (s^{2}\cos(2x))^{2}} = \sqrt{56^{2} + 90^{2}} = \boxed{106}.\] | null | 106 |
5eb519ddb9d0c05ab135c7104c5b8b7e | https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_5 | Let $P$ be a point on the circle circumscribing square $ABCD$ that satisfies $PA \cdot PC = 56$ and $PB \cdot PD = 90.$ Find the area of $ABCD.$ | Similar to Solution 6, let $P$ be on minor arc $\overarc {AB}$ $r$ and $O$ be the radius and center of the circumcircle respectively, and $\theta = \angle AOP$ . Since $\triangle APC$ is a right triangle, $PA \cdot PC$ equals the hypotenuse, $2r$ , times its altitude, which can be represented as $r \sin \theta$ . Therefore, $2r^2 \sin \theta = 56$ . Applying similar logic to $\triangle BPD$ , we get $2r^2 \sin (90^\circ - \theta) = 2r^2 \cos \theta = 90$
Dividing the two equations, we have \begin{align*} \frac{\sin \theta}{\cos \theta} &= \frac{56}{90} \\ 56 \cos \theta &= 90 \sin \theta \\ (56 \cos \theta)^2 &= (90 \sin \theta)^2. \end{align*} Adding $(56 \sin \theta)^2$ to both sides allows us to get rid of $\cos \theta$ \begin{align*} (56 \cos \theta)^2 + (56 \sin \theta)^2 &= (90 \sin \theta)^2 + (56 \sin \theta)^2 \\ 56^2 &= (90^2 + 56^2)(\sin \theta)^2 \\ \frac{56^2}{90^2 + 56^2} &= (\sin \theta)^2 \\ \frac{28}{53} &= \sin \theta. \end{align*} Therefore, we have $2r^2\left(\frac{28}{53}\right) = 56$ , and since the area of the square can be represented as $2r^2$ , the answer is $56 \cdot \frac{53}{28} = \boxed{106}$ | null | 106 |
5e2937c4682db4c26e7015a33e9a241c | https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_6 | Alice knows that $3$ red cards and $3$ black cards will be revealed to her one at a time in random order. Before each card is revealed, Alice must guess its color. If Alice plays optimally, the expected number of cards she will guess correctly is $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$ | At any point in the game, Alice should guess whichever color has come up less frequently thus far (although if both colors have come up equally often, she may guess whichever she likes); using this strategy, her probability of guessing correctly is at least $\frac{1}{2}$ on any given card, as desired.
There are ${6 \choose 3} = 20$ possible orderings of cards, all equally likely (since any of the $6! = 720$ permutations of the cards is equally likely, and each ordering covers $3!^2 = 6^2 = 36$ permutations).
Each of the $10$ orderings that start with red cards corresponds with one that starts with a black card; the problem is symmetrical with respect to red and black cards, so we can, without loss of generality, consider only the orderings that start with red cards.
We then generate a tally table showing whether Alice's guesses are correct for each ordering; for a given card, she guesses correctly if fewer than half the previously shown cards were the same color, guesses incorrectly if more than half were the same color, and guesses correctly with probability $\frac{1}{2}$ if exactly half were the same color.
In this table, $\mid$ denotes a correct guess, $\--$ denotes an incorrect guess, and $/$ denotes a guess with $\frac{1}{2}$ probability of being correct.
Now we sum the tallies across orderings, obtaining $41$ , and finally divide by the number of orderings ( $10$ ) to obtain the expected number of correct guesses, $\frac{41}{10}$ , which yields an answer of $41 + 10 = \boxed{051}.$ | null | 051 |
5e2937c4682db4c26e7015a33e9a241c | https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_6 | Alice knows that $3$ red cards and $3$ black cards will be revealed to her one at a time in random order. Before each card is revealed, Alice must guess its color. If Alice plays optimally, the expected number of cards she will guess correctly is $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$ | Denote by $N \left( a, b \right)$ the optimal expected number of cards that Alice guesses correctly, where the number of red and black cards are $a$ and $b$ , respectively.
Thus, for $a, b \geq 1$ , we have \begin{align*} N \left( a, b \right) & = \max \left\{ \frac{a}{a+b} \left( 1 + N \left( a - 1 , b \right) \right) + \frac{b}{a+b} N \left( a , b - 1 \right) , \right. \\ & \hspace{1cm} \left. \frac{a}{a+b} N \left( a - 1 , b \right) + \frac{b}{a+b} \left( 1 + N \left( a , b - 1 \right) \right) \right\} . \end{align*}
For $a = 0$ , Alice always guesses black. So $N \left( 0 , b \right) = b$
For $b = 0$ , Alice always guesses red. So $N \left( a , 0 \right) = a$
To solve this dynamic program, we can also exploit its symmetry that $N \left( a , b \right) = N \left( b , a \right)$
By solving this dynamic program, we get $N \left( 1, 1 \right) = \frac{3}{2}$ $N \left( 1, 2 \right) = \frac{7}{3}$ $N \left( 1 , 3 \right) = \frac{13}{4}$ $N \left( 2 , 2 \right) = \frac{17}{6}$ $N \left( 2 , 3 \right) = \frac{18}{5}$ $N \left( 3, 3 \right) = \frac{41}{10}$
Therefore, the answer is $41 + 10 = \boxed{051}$ | null | 051 |
5e2937c4682db4c26e7015a33e9a241c | https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_6 | Alice knows that $3$ red cards and $3$ black cards will be revealed to her one at a time in random order. Before each card is revealed, Alice must guess its color. If Alice plays optimally, the expected number of cards she will guess correctly is $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$ | Denote by $N_{i,j}$ the optimal expected number of cards that Alice guesses correctly, where the number of cards are $i$ and $j \ge i.$
If $i = 0$ then Alice guesses correctly all cards, so $N_{0,j} = j.$
If $j = i$ then Alice guesses next card with probability $\frac {1}{2} \implies N_{i,i} = \frac {1}{2} + N_{i-1,i}.$
If $j = i+1$ then Alice guesses next card with probability $\frac {i+1}{2i+1} \implies N_{i,i+1} = \frac {i+1}{2i+1} (1+ N_{i,i}) + \frac{i}{2i+1} N_{i-1,i+1}.$
If $j = i+2$ then Alice guesses next card with probability $\frac {i+2}{2i+2} \implies N_{i,i+2} = \frac {i+2}{2i+2} (1+ N_{i,i+1}) + \frac{i}{2i+2} N_{i-1,i+2}.$
One can find consistently: $N_{1,1} = \frac {1}{2} + N_{0,1} = \frac {3}{2},$ \[N_{1,2} = \frac {2}{3} (1 + N_{1,1}) + \frac {1}{3} N_{0,2} = \frac {7}{3}.\] \[N_{2,2} = \frac {1}{2} + N_{1,2} = \frac {17}{6}.\] \[N_{1,3} = \frac {3}{4} (1 + N_{1,2}) + \frac {1}{4} N_{0,3} = \frac {13}{4}.\] \[N_{2,3} = \frac {3}{5} (1 + N_{2,2}) + \frac {2}{5} N_{1,3} = \frac {18}{5}.\] \[N_{3,3} = \frac {1}{2} + N_{2,3} = \frac {41}{10}.\] Therefore, the answer is $41 + 10 = \boxed{051}$ | null | 051 |
5e2937c4682db4c26e7015a33e9a241c | https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_6 | Alice knows that $3$ red cards and $3$ black cards will be revealed to her one at a time in random order. Before each card is revealed, Alice must guess its color. If Alice plays optimally, the expected number of cards she will guess correctly is $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$ | Denote $E_n$ the expected number of cards Alice guesses correctly given $n$ red cards and $n$ black cards. We want to find $E_3$
Alice has a $\frac{1}{2}$ chance of guessing the first card. WLOG assume the first card color is red. For the next card, Alice has a $\frac{3}{5}$ chance of guessing the card if she chooses black; if they guess right, there's one less red and black card, so the expected number of cards Alice guesses from here is $E_2$ . If Alice does not guess correctly (which occurs with probability $\frac{2}{5}$ ), this means that there's 3 black cards and 1 red card left, so Alice should guess black next with a $\frac{3}{4}$ chance of being right. If Alice is wrong (with probability $\frac{1}{4}$ ), there are only 3 black cards left, so Alice can guess these with certainty; if Alice is right, there are 2 blacks and 1 red left, so Alice should again guess black. If Alice is right (with probability $\frac{2}{3}$ ), there is now 1 black and red card each, so the expected number of cards guessed is $E_1$ ; if she is wrong (with probability $\frac{1}{3}$ ), there are 2 black cards left, so Alice can guess these with certainty.
Summing this up into a formula: \[E_3 = \frac{1}{2} + \frac{3}{5} \left(1 + E_2 \right) + \frac{2}{5} \left( \frac{1}{4}(3) + \frac{3}{4}\left(1 + \frac{2}{3}\left(1 + E_1 \right) + \frac{1}{3}(2)\right) \right)\]
We can apply similar logic to compute $E_2$ and get \[E_2 = \frac{1}{2} + \frac{2}{3}(1 + E_1) + \frac{1}{3}(2)\]
To compute $E_1$ , we know that Alice can guess the last card with certainty, and there's a $\frac{1}{2}$ chance they get the first card as well, so $E_1 = \frac{3}{2}$
Thus, $E_2 = \frac{17}{6}$ , and after long computation, we get $E_3 = \frac{41}{10}$ . The requested answer is $41 + 10 = \boxed{51}$ | null | 51 |
0b190233a003f0dd285e62b4ab2fb671 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_7 | Call a positive integer $n$ extra-distinct if the remainders when $n$ is divided by $2, 3, 4, 5,$ and $6$ are distinct. Find the number of extra-distinct positive integers less than $1000$ | $n$ can either be $0$ or $1$ (mod $2$ ).
Case 1: $n \equiv 0 \pmod{2}$
Then, $n \equiv 2 \pmod{4}$ , which implies $n \equiv 1 \pmod{3}$ and $n \equiv 4 \pmod{6}$ , and therefore $n \equiv 3 \pmod{5}$ . Using CRT , we obtain $n \equiv 58 \pmod{60}$ , which gives $16$ values for $n$
Case 2: $n \equiv 1 \pmod{2}$
$n$ is then $3 \pmod{4}$ . If $n \equiv 0 \pmod{3}$ $n \equiv 3 \pmod{6}$ , a contradiction. Thus, $n \equiv 2 \pmod{3}$ , which implies $n \equiv 5 \pmod{6}$ $n$ can either be $0 \pmod{5}$ , which implies that $n \equiv 35 \pmod{60}$ by CRT, giving $17$ cases; or $4 \pmod{5}$ , which implies that $n \equiv 59 \pmod{60}$ by CRT, giving $16$ cases.
The total number of extra-distinct numbers is thus $16 + 16 + 17 = \boxed{049}$ | null | 049 |
0b190233a003f0dd285e62b4ab2fb671 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_7 | Call a positive integer $n$ extra-distinct if the remainders when $n$ is divided by $2, 3, 4, 5,$ and $6$ are distinct. Find the number of extra-distinct positive integers less than $1000$ | Because the LCM of all of the numbers we are dividing by is $60$ , we know that all of the remainders are $0$ again at $60$ , meaning that we have a cycle that repeats itself every $60$ numbers.
After listing all of the remainders up to $60$ , we find that $35$ $58$ , and $59$ are extra-distinct. So, we have $3$ numbers every $60$ which are extra-distinct. $60\cdot16 = 960$ and $3\cdot16 = 48$ , so we have $48$ extra-distinct numbers in the first $960$ numbers. Because of our pattern, we know that the numbers from $961$ thru $1000$ will have the same remainders as $1$ thru $40$ , so we have $1$ other extra-distinct number ( $35$ ).
$48 + 1 = \boxed{049}$ | null | 049 |
0b190233a003f0dd285e62b4ab2fb671 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_7 | Call a positive integer $n$ extra-distinct if the remainders when $n$ is divided by $2, 3, 4, 5,$ and $6$ are distinct. Find the number of extra-distinct positive integers less than $1000$ | $\textbf{Case 0: } {\rm Rem} \ \left( n, 6 \right) = 0$
We have ${\rm Rem} \ \left( n, 2 \right) = 0$ . This violates the condition that $n$ is extra-distinct.
Therefore, this case has no solution.
$\textbf{Case 1: } {\rm Rem} \ \left( n, 6 \right) = 1$
We have ${\rm Rem} \ \left( n, 2 \right) = 1$ . This violates the condition that $n$ is extra-distinct.
Therefore, this case has no solution.
$\textbf{Case 2: } {\rm Rem} \ \left( n, 6 \right) = 2$
We have ${\rm Rem} \ \left( n, 3 \right) = 2$ . This violates the condition that $n$ is extra-distinct.
Therefore, this case has no solution.
$\textbf{Case 3: } {\rm Rem} \ \left( n, 6 \right) = 3$
The condition ${\rm Rem} \ \left( n, 6 \right) = 3$ implies ${\rm Rem} \ \left( n, 2 \right) = 1$ ${\rm Rem} \ \left( n, 3 \right) = 0$
Because $n$ is extra-distinct, ${\rm Rem} \ \left( n, l \right)$ for $l \in \left\{ 2, 3, 4 \right\}$ is a permutation of $\left\{ 0, 1 ,2 \right\}$ .
Thus, ${\rm Rem} \ \left( n, 4 \right) = 2$
However, ${\rm Rem} \ \left( n, 4 \right) = 2$ conflicts ${\rm Rem} \ \left( n, 2 \right) = 1$ .
Therefore, this case has no solution.
$\textbf{Case 4: } {\rm Rem} \ \left( n, 6 \right) = 4$
The condition ${\rm Rem} \ \left( n, 6 \right) = 4$ implies ${\rm Rem} \ \left( n, 2 \right) = 0$ and ${\rm Rem} \ \left( n, 3 \right) = 1$
Because $n$ is extra-distinct, ${\rm Rem} \ \left( n, l \right)$ for $l \in \left\{ 2, 3, 4 , 5 \right\}$ is a permutation of $\left\{ 0, 1 ,2 , 3 \right\}$
Because ${\rm Rem} \ \left( n, 2 \right) = 0$ , we must have ${\rm Rem} \ \left( n, 4 \right) = 2$ . Hence, ${\rm Rem} \ \left( n, 5 \right) = 3$
Hence, $n \equiv -2 \pmod{{\rm lcm} \left( 4, 5 , 6 \right)}$ .
Hence, $n \equiv - 2 \pmod{60}$
We have $1000 = 60 \cdot 16 + 40$ .
Therefore, the number extra-distinct $n$ in this case is 16.
$\textbf{Case 5: } {\rm Rem} \ \left( n, 6 \right) = 5$
The condition ${\rm Rem} \ \left( n, 6 \right) = 5$ implies ${\rm Rem} \ \left( n, 2 \right) = 1$ and ${\rm Rem} \ \left( n, 3 \right) = 2$
Because $n$ is extra-distinct, ${\rm Rem} \ \left( n, 4 \right)$ and ${\rm Rem} \ \left( n, 5 \right)$ are two distinct numbers in $\left\{ 0, 3, 4 \right\}$ .
Because ${\rm Rem} \ \left( n, 4 \right) \leq 3$ and $n$ is odd, we have ${\rm Rem} \ \left( n, 4 \right) = 3$ .
Hence, ${\rm Rem} \ \left( n, 5 \right) = 0$ or 4.
$\textbf{Case 5.1: } {\rm Rem} \ \left( n, 6 \right) = 5$ ${\rm Rem} \ \left( n, 4 \right) = 3$ ${\rm Rem} \ \left( n, 5 \right) = 0$
We have $n \equiv 35 \pmod{60}$
We have $1000 = 60 \cdot 16 + 40$ .
Therefore, the number extra-distinct $n$ in this subcase is 17.
$\textbf{Case 5.2: } {\rm Rem} \ \left( n, 6 \right) = 5$ ${\rm Rem} \ \left( n, 4 \right) = 3$ ${\rm Rem} \ \left( n, 5 \right) = 4$
$n \equiv - 1 \pmod{60}$
We have $1000 = 60 \cdot 16 + 40$ .
Therefore, the number extra-distinct $n$ in this subcase is 16.
Putting all cases together, the total number of extra-distinct $n$ is $16 + 17 + 16 = \boxed{049}$ | null | 049 |
1fdc3db56a887c8dd07bd2990be10766 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_8 | Rhombus $ABCD$ has $\angle BAD < 90^\circ.$ There is a point $P$ on the incircle of the rhombus such that the distances from $P$ to the lines $DA,AB,$ and $BC$ are $9,$ $5,$ and $16,$ respectively. Find the perimeter of $ABCD.$ | This solution refers to the Diagram section.
Let $O$ be the incenter of $ABCD$ for which $\odot O$ is tangent to $\overline{DA},\overline{AB},$ and $\overline{BC}$ at $X,Y,$ and $Z,$ respectively. Moreover, suppose that $R,S,$ and $T$ are the feet of the perpendiculars from $P$ to $\overleftrightarrow{DA},\overleftrightarrow{AB},$ and $\overleftrightarrow{BC},$ respectively, such that $\overline{RT}$ intersects $\odot O$ at $P$ and $Q.$
We obtain the following diagram: [asy] /* Made by MRENTHUSIASM; inspired by Math Jams. */ size(300); pair A, B, C, D, O, P, R, S, T, X, Y, Z, Q; A = origin; B = (125/4,0); C = B + 125/4 * dir((3,4)); D = A + 125/4 * dir((3,4)); O = (25,25/2); P = (15,5); R = foot(P,A,D); S = foot(P,A,B); T = foot(P,B,C); X = (15,20); Y = (25,0); Z = (35,5); Q = intersectionpoints(Circle(O,25/2),R--T)[1]; fill(R--T--Z--X--cycle,cyan); markscalefactor=0.15; draw(rightanglemark(P,R,D)^^rightanglemark(P,S,B)^^rightanglemark(P,T,C),red); draw(Circle(O,25/2)^^A--B--C--D--cycle^^B--T); draw(P--R^^P--S^^P--T,red+dashed); draw(O--X^^O--Y^^O--Z); dot("$A$",A,1.5*dir(225),linewidth(4.5)); dot("$B$",B,1.5*dir(-45),linewidth(4.5)); dot("$C$",C,1.5*dir(45),linewidth(4.5)); dot("$D$",D,1.5*dir(135),linewidth(4.5)); dot("$P$",P,1.5*dir(60),linewidth(4.5)); dot("$R$",R,1.5*dir(135),linewidth(4.5)); dot("$S$",S,1.5*dir(-90),linewidth(4.5)); dot("$T$",T,1.5*dir(-45),linewidth(4.5)); dot("$O$",O,1.5*dir(45),linewidth(4.5)); dot("$X$",X,1.5*dir(135),linewidth(4.5)); dot("$Y$",Y,1.5*dir(-90),linewidth(4.5)); dot("$Z$",Z,1.5*dir(-45),linewidth(4.5)); dot("$Q$",Q,1.5*dir(60),linewidth(4.5)); label("$9$",midpoint(P--R),dir(A-D),red); label("$5$",midpoint(P--S),dir(180),red); label("$16$",midpoint(P--T),dir(A-D),red); [/asy] Note that $\angle RXZ = \angle TZX = 90^\circ$ by the properties of tangents, so $RTZX$ is a rectangle. It follows that the diameter of $\odot O$ is $XZ = RT = 25.$
Let $x=PQ$ and $y=RX=TZ.$ We apply the Power of a Point Theorem to $R$ and $T:$ \begin{align*} y^2 &= 9(9+x), \\ y^2 &= 16(16-x). \end{align*} We solve this system of equations to get $x=7$ and $y=12.$ Alternatively, we can find these results by the symmetry on rectangle $RTZX$ and semicircle $\widehat{XPZ}.$
We extend $\overline{SP}$ beyond $P$ to intersect $\odot O$ and $\overleftrightarrow{CD}$ at $E$ and $F,$ respectively, where $E\neq P.$ So, we have $EF=SP=5$ and $PE=25-SP-EF=15.$ On the other hand, we have $PX=15$ by the Pythagorean Theorem on right $\triangle PRX.$ Together, we conclude that $E=X.$ Therefore, points $S,P,$ and $X$ must be collinear.
Let $G$ be the foot of the perpendicular from $D$ to $\overline{AB}.$ Note that $\overline{DG}\parallel\overline{XP},$ as shown below: [asy] /* Made by MRENTHUSIASM; inspired by Math Jams. */ size(300); pair A, B, C, D, O, P, R, S, T, X, Y, Z, Q, G; A = origin; B = (125/4,0); C = B + 125/4 * dir((3,4)); D = A + 125/4 * dir((3,4)); O = (25,25/2); P = (15,5); R = foot(P,A,D); S = foot(P,A,B); T = foot(P,B,C); X = (15,20); Y = (25,0); Z = (35,5); Q = intersectionpoints(Circle(O,25/2),R--T)[1]; G = foot(D,A,B); fill(D--A--G--cycle,green); fill(P--R--X--cycle,yellow); markscalefactor=0.15; draw(rightanglemark(P,R,D)^^rightanglemark(D,G,A),red); draw(Circle(O,25/2)^^A--B--C--D--cycle^^X--P^^D--G); draw(P--R,red+dashed); dot("$A$",A,1.5*dir(225),linewidth(4.5)); dot("$B$",B,1.5*dir(-45),linewidth(4.5)); dot("$C$",C,1.5*dir(45),linewidth(4.5)); dot("$D$",D,1.5*dir(135),linewidth(4.5)); dot("$P$",P,1.5*dir(60),linewidth(4.5)); dot("$R$",R,1.5*dir(135),linewidth(4.5)); dot("$O$",O,1.5*dir(45),linewidth(4.5)); dot("$X$",X,1.5*dir(135),linewidth(4.5)); dot("$G$",G,1.5*dir(-90),linewidth(4.5)); draw(P--X,MidArrow(0.3cm,Fill(red))); draw(G--D,MidArrow(0.3cm,Fill(red))); label("$9$",midpoint(P--R),dir(A-D),red); label("$12$",midpoint(R--X),dir(135),red); label("$15$",midpoint(X--P),dir(0),red); label("$25$",midpoint(G--D),dir(0),red); [/asy] As $\angle PRX = \angle AGD = 90^\circ$ and $\angle PXR = \angle ADG$ by the AA Similarity, we conclude that $\triangle PRX \sim \triangle AGD.$ The ratio of similitude is \[\frac{PX}{AD} = \frac{RX}{GD}.\] We get $\frac{15}{AD} = \frac{12}{25},$ from which $AD = \frac{125}{4}.$
Finally, the perimeter of $ABCD$ is $4AD = \boxed{125}.$ | null | 125 |
1fdc3db56a887c8dd07bd2990be10766 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_8 | Rhombus $ABCD$ has $\angle BAD < 90^\circ.$ There is a point $P$ on the incircle of the rhombus such that the distances from $P$ to the lines $DA,AB,$ and $BC$ are $9,$ $5,$ and $16,$ respectively. Find the perimeter of $ABCD.$ | This solution refers to the Diagram section.
Define points $O,R,S,$ and $T$ as Solution 1 does. Moreover, let $H$ be the foot of the perpendicular from $P$ to $\overleftrightarrow{CD},$ $M$ be the foot of the perpendicular from $O$ to $\overleftrightarrow{HS},$ and $N$ be the foot of the perpendicular from $O$ to $\overleftrightarrow{RT}.$
We obtain the following diagram: [asy] /* Made by MRENTHUSIASM; inspired by Math Jams. */ size(300); pair A, B, C, D, O, P, R, S, T, H, M, N; A = origin; B = (125/4,0); C = B + 125/4 * dir((3,4)); D = A + 125/4 * dir((3,4)); O = (25,25/2); P = (15,5); R = foot(P,A,D); S = foot(P,A,B); T = foot(P,B,C); H = foot(S,C,D); M = foot(O,S,H); N = foot(O,R,T); fill(O--M--P--cycle,yellow); fill(O--N--P--cycle,green); markscalefactor=0.15; draw(rightanglemark(P,R,D)^^rightanglemark(P,S,B)^^rightanglemark(P,T,C)^^rightanglemark(O,M,P)^^rightanglemark(O,N,P)^^rightanglemark(S,H,D),red); draw(Circle(O,25/2)^^A--B--C--D--cycle^^B--T^^D--H^^O--M^^O--N^^O--P); draw(P--R^^P--S^^P--T^^P--H,red+dashed); dot("$A$",A,1.5*dir(225),linewidth(4.5)); dot("$B$",B,1.5*dir(-45),linewidth(4.5)); dot("$C$",C,1.5*dir(45),linewidth(4.5)); dot("$D$",D,1.5*dir(90),linewidth(4.5)); dot("$P$",P,1.5*dir(60),linewidth(4.5)); dot("$R$",R,1.5*dir(135),linewidth(4.5)); dot("$S$",S,1.5*dir(-90),linewidth(4.5)); dot("$T$",T,1.5*dir(-45),linewidth(4.5)); dot("$O$",O,1.5*dir(45),linewidth(4.5)); dot("$H$",H,1.5*dir(90),linewidth(4.5)); dot("$M$",M,1.5*dir(180),linewidth(4.5)); dot("$N$",N,1.5*dir(15),linewidth(4.5)); label("$9$",midpoint(P--R),dir(A-D),red); label("$5$",midpoint(P--S),dir(180),red); label("$16$",midpoint(P--T),dir(A-D),red); [/asy] Note that the diameter of $\odot O$ is $HS=RT=25,$ so $OP=\frac{25}{2}.$ It follows that:
Since $\overline{MO}\parallel\overline{AB}$ and $\overline{ON}\parallel\overline{DA},$ we conclude that $\angle A = \angle MON.$ We apply the Sine of a Sum Formula: \begin{align*} \sin\angle A &= \sin\angle MON \\ &= \sin(\angle MOP + \angle PON) \\ &= \sin\angle MOP \cos\angle PON + \cos\angle MOP \sin\angle PON \\ &= \frac{3}{5}\cdot\frac{24}{25} + \frac{4}{5}\cdot\frac{7}{25} \\ &= \frac{4}{5}. \end{align*} Note that \[\sin\angle A = \frac{HS}{DA},\] from which $\frac{4}{5} = \frac{25}{DA}.$ We solve this equation to get $DA=\frac{125}{4}.$
Finally, the perimeter of $ABCD$ is $4DA = \boxed{125}.$ | null | 125 |
1fdc3db56a887c8dd07bd2990be10766 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_8 | Rhombus $ABCD$ has $\angle BAD < 90^\circ.$ There is a point $P$ on the incircle of the rhombus such that the distances from $P$ to the lines $DA,AB,$ and $BC$ are $9,$ $5,$ and $16,$ respectively. Find the perimeter of $ABCD.$ | Label the points of the rhombus to be $X$ $Y$ $Z$ , and $W$ and the center of the incircle to be $O$ so that $9$ $5$ , and $16$ are the distances from point $P$ to side $ZW$ , side $WX$ , and $XY$ respectively. Through this, we know that the distance from the two pairs of opposite lines of rhombus $XYZW$ is $25$ and circle $O$ has radius $\frac{25}{2}$
Call the feet of the altitudes from $P$ to side $ZW$ , side $WX$ , and side $XY$ to be $A$ $B$ , and $C$ respectively. Additionally, call the feet of the altitudes from $O$ to side $ZW$ , side $WX$ , and side $XY$ to be $D$ $E$ , and $F$ respectively.
Draw a line segment from $P$ to $\overline{OD}$ so that it is perpendicular to $\overline{OD}$ . Notice that this segment length is equal to $AD$ and is $\sqrt{\left(\frac{25}{2}\right)^2-\left(\frac{7}{2}\right)^2}=12$ by Pythagorean Theorem.
Similarly, perform the same operations with perpendicular from $P$ to $\overline{OE}$ to get $BE=10$
By equal tangents, $WD=WE$ . Now, label the length of segment $WA=n$ and $WB=n+2$
Using Pythagorean Theorem again, we get
\begin{align*} WA^2+PA^2&=WB^2+PB^2 \\ n^2+9^2&=(n+2)^2+5^2 \\ n&=13. \end{align*}
Which also gives us $\tan{\angle{OWX}}=\frac{1}{2}$ and $OW=\frac{25\sqrt{5}}{2}$
Since the diagonals of the rhombus intersect at $O$ and are angle bisectors and are also perpendicular to each other, we can get that
\begin{align*} \frac{OX}{OW}&=\tan{\angle{OWX}} \\ OX&=\frac{25\sqrt{5}}{4} \\ WX^2&=OW^2+OX^2 \\ WX&=\frac{125}{4} \\ 4WX&=\boxed{125} | null | 125 |
1fdc3db56a887c8dd07bd2990be10766 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_8 | Rhombus $ABCD$ has $\angle BAD < 90^\circ.$ There is a point $P$ on the incircle of the rhombus such that the distances from $P$ to the lines $DA,AB,$ and $BC$ are $9,$ $5,$ and $16,$ respectively. Find the perimeter of $ABCD.$ | Denote by $O$ the center of $ABCD$ .
We drop an altitude from $O$ to $AB$ that meets $AB$ at point $H$ .
We drop altitudes from $P$ to $AB$ and $AD$ that meet $AB$ and $AD$ at $E$ and $F$ , respectively.
We denote $\theta = \angle BAC$ .
We denote the side length of $ABCD$ as $d$
Because the distances from $P$ to $BC$ and $AD$ are $16$ and $9$ , respectively, and $BC \parallel AD$ , the distance between each pair of two parallel sides of $ABCD$ is $16 + 9 = 25$ .
Thus, $OH = \frac{25}{2}$ and $d \sin \theta = 25$
We have \begin{align*} \angle BOH & = 90^\circ - \angle HBO \\ & = 90^\circ - \angle HBD \\ & = 90^\circ - \frac{180^\circ - \angle C}{2} \\ & = 90^\circ - \frac{180^\circ - \theta}{2} \\ & = \frac{\theta}{2} . \end{align*}
Thus, $BH = OH \tan \angle BOH = \frac{25}{2} \tan \frac{\theta}{2}$
In $FAEP$ , we have $\overrightarrow{FA} + \overrightarrow{AE} + \overrightarrow{EP} + \overrightarrow{PF} = 0$ .
Thus, \[ AF + AE e^{i \left( \pi - \theta \right)} + EP e^{i \left( \frac{3 \pi}{2} - \theta \right)} - PF i . \]
Taking the imaginary part of this equation and plugging $EP = 5$ and $PF = 9$ into this equation, we get \[ AE = \frac{9 + 5 \cos \theta}{\sin \theta} . \]
We have \begin{align*} OP^2 & = \left( OH - EP \right)^2 + \left( AH - AE \right)^2 \\ & = \left( \frac{25}{2} - 5 \right)^2 + \left( d - \frac{25}{2} \tan \frac{\theta}{2} - \frac{9 + 5 \cos \theta}{\sin \theta} \right) \\ & = \left( \frac{15}{2} \right)^2 + \left( \frac{25}{\sin \theta} - \frac{25}{2} \tan \frac{\theta}{2} - \frac{9 + 5 \cos \theta}{\sin \theta} \right) . \hspace{1cm} (\bigstar) \end{align*}
Because $P$ is on the incircle of $ABCD$ $OP = \frac{25}{2}$ . Plugging this into $(\bigstar)$ , we get the following equation \[ 20 \sin \theta - 15 \cos \theta = 7 . \]
By solving this equation, we get $\sin \theta = \frac{4}{5}$ and $\cos \theta = \frac{3}{5}$ .
Therefore, $d = \frac{25}{\sin \theta} = \frac{125}{4}$
Therefore, the perimeter of $ABCD$ is $4d = \boxed{125}$ | null | 125 |
1fdc3db56a887c8dd07bd2990be10766 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_8 | Rhombus $ABCD$ has $\angle BAD < 90^\circ.$ There is a point $P$ on the incircle of the rhombus such that the distances from $P$ to the lines $DA,AB,$ and $BC$ are $9,$ $5,$ and $16,$ respectively. Find the perimeter of $ABCD.$ | The center of the incircle is $O.$ Denote the points in which the incircle meets $\overline{AB},$ $\overline{BC},$ $\overline{CD},$ and $\overline{DA}$ as $W,$ $X,$ $Y,$ and $Z,$ respectively. Next, also denote the base of the perpendicular from $P$ to $\overline{AB},$ $\overline{AD},$ $\overline{OW},$ and $\overline{OZ}$ as $M,$ $N,$ $S,$ and $T,$ respectively.
We can easily see that the radius of the circle is $\frac{25}{2}.$ Using this and Pythagorus on right $\triangle OSP$ and $\triangle OTP,$ we find that $MW = PS = 10$ and $NZ = PT = 12.$
Since $AW = AZ$ by properties of circle tangents, we can deduce by the above information that $AM = AN+2.$ Doing Pythagorus on right $\triangle AMP$ and $\triangle ANP$ we find that $a^2 = b^2 + 56$ (because $a^2+25=b^2+81.$ ) From solving the $2$ just derived equations, we find that $AM=15$ and $AN=13.$
Next, we use Pythagorus on right $\triangle AOB$ (we can see it's right because of properties of rhombuses.) We get \[AB^2 = AO^2 + BO^2.\] We know $AB = AW + WB = 25 + WB.$ By Pythagorus on $\triangle AWO$ and $\triangle BWO,$ we also know $AO^2 = 25^2+\left(\frac{25}{2}\right)^2$ and $BO^2=WB^2+\left(\frac{25}{2}\right)^2.$ Substituting these in, we have \[25^2 + 50WB + WB^2 = 25^2+\left(\frac{25}{2}\right)^2+\left(\frac{25}{2}\right)^2+WB^2.\] Solving for $WB,$ we get $WB = \frac{25}{4}.$ Now we find that each side of the rhombus $=AB=25+\frac{25}{4}=\frac{125}{4}.$ The perimeter of the rhombus would be that times $4.$ Our final answer is \[\frac{125}{4}\cdot4=\boxed{125}.\] | null | 125 |
1fdc3db56a887c8dd07bd2990be10766 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_8 | Rhombus $ABCD$ has $\angle BAD < 90^\circ.$ There is a point $P$ on the incircle of the rhombus such that the distances from $P$ to the lines $DA,AB,$ and $BC$ are $9,$ $5,$ and $16,$ respectively. Find the perimeter of $ABCD.$ | Notation is shown on diagram, $RT \perp AD, FG \perp AB, E = AD \cap \omega, E' = FG \cap AD.$ $RT = 9 + 16 = 25 = FG$ as hights of rhombus. \[RP = QT = 9, PQ = 16 - 9 = 7, GE' = PF = 5,\] \[PE' = 25 - 5 - 5 = 15, RE = \sqrt{RP \cdot RQ} = \sqrt{9 \cdot 16} = 12.\] \[PE = \sqrt{RP^2 + RE^2} = 15 \implies E = E'.\] \[\sin \alpha = \frac {RE}{PE} = \frac {GF}{AD} \implies AD = \frac {15 \cdot 25}{12} = \frac {125}{4}.\] The perimeter of $ABCD$ is $\frac{125}{4}\cdot4=\boxed{125}.$ | null | 125 |
301c10fb692392bf6eabe45fc075d09d | https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_9 | Find the number of cubic polynomials $p(x) = x^3 + ax^2 + bx + c,$ where $a, b,$ and $c$ are integers in $\{-20,-19,-18,\ldots,18,19,20\},$ such that there is a unique integer $m \not= 2$ with $p(m) = p(2).$ | Plugging $2$ and $m$ into $P(x)$ and equating them, we get $8+4a+2b+c = m^3+am^2+bm+c$ . Rearranging, we have \[(m^3-8) + (m^2 - 4)a + (m-2)b = 0.\] Note that the value of $c$ won't matter as it can be anything in the provided range, giving a total of $41$ possible choices for $c.$ So what we just need to do is to just find the number of ordered pairs $(a, b)$ that work, and multiply it by $41.$ We can start by first dividing both sides by $m-2.$ (Note that this is valid since $m\neq2:$ \[m^2 + 2m + 4 + (m+2)a + b = 0.\] We can rearrange this so it is a quadratic in $m$ \[m^2 + (a+2)m + (4 + 2a + b) = 0.\] Remember that $m$ has to be unique and not equal to $2.$ We can split this into two cases: case $1$ being that $m$ has exactly one solution, and it isn't equal to $2$ ; case $2$ being that $m$ has two solutions, one being equal to $2,$ but the other is a unique solution not equal to $2.$
$\textbf{Case 1:}$
There is exactly one solution for $m,$ and that solution is not $2.$ This means that the discriminant of the quadratic equation is $0,$ using that, we have $(a+2)^2 = 4(4 + 2a + b),$ rearranging in a neat way, we have \[(a-2)^2 = 4(4 + b)\Longrightarrow a = 2\pm2\sqrt{4+b}.\] Using the fact that $4+b$ must be a perfect square, we can easily see that the values for $b$ can be $-4, -3, 0, 5,$ and $12.$ Also since it's a " $\pm$ " there will usually be $2$ solutions for $a$ for each value of $b.$ The two exceptions for this would be if $b = -4$ and $b = 12.$ For $b=-4$ because it would be a $\pm0,$ which only gives one solution, instead of two. And for $b=12$ because then $a = -6$ and the solution for $m$ would equal to $2,$ and we don't want this. (We can know this by putting the solutions back into the quadratic formula).
So we have $5$ solutions for $b,$ each of which give $2$ values for $a,$ except for $2,$ which only give one. So in total, there are $5*2 - 2 = 8$ ordered pairs of $(a,b)$ in this case.
$\textbf{Case 2:}$
$m$ has two solutions, but exactly one of them isn't equal to $2.$ This ensures that $1$ of the solutions is equal to $2.$
Let $r$ be the other value of $m$ that isn't $2.$ By Vieta: \begin{align*} r+2 &= -a-2\\ 2r &= 4+2a+b. \end{align*} From the first equation, we subtract both sides by $2$ and double both sides to get $2r = -2a - 8$ which also equals to $4+2a+b$ from the second equation. Equating both, we have $4a + b + 12 = 0.$ We can easily count that there would be $11$ ordered pairs $(a,b)$ that satisfy that.
However, there's an outlier case in which $r$ happens to also equal to $2,$ and we don't want that. We can reverse engineer and find out that $r=2$ when $(a,b) = (-6, 12),$ which we overcounted. So we subtract by one and we conclude that there are $10$ ordered pairs of $(a,b)$ that satisfy this case.
This all shows that there are a total of $8+10 = 18$ amount of ordered pairs $(a,b).$ Multiplying this by $41$ (the amount of values for $c$ ) we get $18\cdot41=\boxed{738}$ as our final answer. | null | 738 |
301c10fb692392bf6eabe45fc075d09d | https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_9 | Find the number of cubic polynomials $p(x) = x^3 + ax^2 + bx + c,$ where $a, b,$ and $c$ are integers in $\{-20,-19,-18,\ldots,18,19,20\},$ such that there is a unique integer $m \not= 2$ with $p(m) = p(2).$ | $p(x)-p(2)$ is a cubic with at least two integral real roots, therefore it has three real roots, which are all integers.
There are exactly two distinct roots, so either $p(x)=p(2)+(x-2)^2(x-m)$ or $p(x)=p(2)+(x-2)(x-m)^2$ , with $m\neq 2$
In the first case $p(x)=x^3-(4+m)x^2+(4+4m)x-4m+p(2)$ , with $|4+4m|\leq 20$ (which entails $|4+m|\leq 20$ ), so $m$ can be $-6,-5,-4,-3,-2,-1,0,1, (\textbf{not 2}!), 3,4$ and $-4m+p(2)$ can be any integer from $-20$ to $20$ , giving $410$ polynomials (since the coefficients are given by linear functions of $m$ and thus are distinct).
In the second case $p(x)=x^3-(2+2m)x^2+(4m+m^2)x-2m^2+p(2)$ , and $m$ can be $-6,-5,-4,-3,-2,-1,0,1$ and $-4m+p(2)$ can be any integer from $-20$ to $20$ , giving $328$ polynomials.
The total is $\boxed{738}$ | null | 738 |
a3b12607844df814117f255d2482e53a | https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_10 | There exists a unique positive integer $a$ for which the sum \[U=\sum_{n=1}^{2023}\left\lfloor\dfrac{n^{2}-na}{5}\right\rfloor\] is an integer strictly between $-1000$ and $1000$ . For that unique $a$ , find $a+U$
(Note that $\lfloor x\rfloor$ denotes the greatest integer that is less than or equal to $x$ .) | Define $\left\{ x \right\} = x - \left\lfloor x \right\rfloor$
First, we bound $U$
We establish an upper bound of $U$ . We have \begin{align*} U & \leq \sum_{n=1}^{2023} \frac{n^2 - na}{5} \\ & = \frac{1}{5} \sum_{n=1}^{2023} n^2 - \frac{a}{5} \sum_{n=1}^{2023} n \\ & = \frac{1012 \cdot 2023}{5} \left( 1349 - a \right) \\ & \triangleq UB . \end{align*}
We establish a lower bound of $U$ . We have \begin{align*} U & = \sum_{n=1}^{2023} \left( \frac{n^2 - na}{5} - \left\{ \frac{n^2 - na}{5} \right\} \right) \\ & = \sum_{n=1}^{2023} \frac{n^2 - na}{5} - \sum_{n=1}^{2023} \left\{ \frac{n^2 - na}{5} \right\} \\ & = UB - \sum_{n=1}^{2023} \left\{ \frac{n^2 - na}{5} \right\} \\ & \geq UB - \sum_{n=1}^{2023} \mathbf 1 \left\{ \frac{n^2 - na}{5} \notin \Bbb Z \right\} . \end{align*}
We notice that if $5 | n$ , then $\frac{n^2 - na}{5} \in \Bbb Z$ .
Thus, \begin{align*} U & \geq UB - \sum_{n=1}^{2023} \mathbf 1 \left\{ \frac{n^2 - na}{5} \notin \Bbb Z \right\} \\ & \geq UB - \sum_{n=1}^{2023} \mathbf 1 \left\{ 5 \nmid n \right\} \\ & = UB - \left( 2023 - \left\lfloor \frac{2023}{5} \right\rfloor \right) \\ & = UB - 1619 \\ & \triangleq LB . \end{align*}
Because $U \in \left[ - 1000, 1000 \right]$ and $UB - LB = 1619 < \left( 1000 - \left( - 1000 \right) \right)$ , we must have either $UB \in \left[ - 1000, 1000 \right]$ or $LB \in \left[ - 1000, 1000 \right]$
For $UB \in \left[ - 1000, 1000 \right]$ , we get a unique $a = 1349$ .
For $LB \in \left[ - 1000, 1000 \right]$ , there is no feasible $a$
Therefore, $a = 1349$ . Thus $UB = 0$
Next, we compute $U$
Let $n = 5 q + r$ , where $r = {\rm Rem} \ \left( n, 5 \right)$
We have \begin{align*} \left\{ \frac{n^2 - na}{5} \right\} & = \left\{ \frac{\left( 5 q + r \right)^2 - \left( 5 q + r \right)\left( 1350 - 1 \right)}{5} \right\} \\ & = \left\{ 5 q^2 + 2 q r - \left( 5 q + r \right) 270 + q + \frac{r^2 + r}{5} \right\} \\ & = \left\{\frac{r^2 + r}{5} \right\} \\ & = \left\{ \begin{array}{ll} 0 & \mbox{ if } r = 0, 4 \\ \frac{2}{5} & \mbox{ if } r = 1, 3 \\ \frac{1}{5} & \mbox{ if } r = 2 \end{array} \right. . \end{align*}
Therefore, \begin{align*} U & = \sum_{n=1}^{2023} \left( \frac{n^2 - na}{5} - \left\{ \frac{n^2 - na}{5} \right\} \right) \\ & = UB - \sum_{n=1}^{2023} \left\{ \frac{n^2 - na}{5} \right\} \\ & = - \sum_{n=1}^{2023} \left\{ \frac{n^2 - na}{5} \right\} \\ & = - \sum_{q=0}^{404} \sum_{r=0}^4 \left\{\frac{r^2 + r}{5} \right\} + \left\{ \frac{0^2 - 0 \cdot a}{5} \right\} + \left\{ \frac{2024^2 - 2024a}{5} \right\} \\ & = - \sum_{q=0}^{404} \left( 0 + 0 + \frac{2}{5} + \frac{2}{5} + \frac{1}{5} \right) + 0 + 0 \\ & = - 405 . \end{align*}
Therefore, $a + U = 1349 - 405 = \boxed{944}$ | null | 944 |
a3b12607844df814117f255d2482e53a | https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_10 | There exists a unique positive integer $a$ for which the sum \[U=\sum_{n=1}^{2023}\left\lfloor\dfrac{n^{2}-na}{5}\right\rfloor\] is an integer strictly between $-1000$ and $1000$ . For that unique $a$ , find $a+U$
(Note that $\lfloor x\rfloor$ denotes the greatest integer that is less than or equal to $x$ .) | We define $U' = \sum^{2023}_{n=1} {\frac{n^2-na}{5}}$ . Since for any real number $x$ $\lfloor x \rfloor \le x \le \lfloor x \rfloor + 1$ , we have $U \le U' \le U + 2023$ . Now, since $-1000 \le U \le 1000$ , we have $-1000 \le U' \le 3023$
Now, we can solve for $U'$ in terms of $a$ . We have: \begin{align*} U' &= \sum^{2023}_{n=1} {\frac{n^2-na}{5}} \\ &= \sum^{2023}_{n=1} {\frac{n^2}{5} - \frac{na}{5}} \\ &= \sum^{2023}_{n=1} {\frac{n^2}{5}} - \sum^{2023}_{n=1} {\frac{na}{5}} \\ &= \frac{\sum^{2023}_{n=1} {{n^2}} - \sum^{2023}_{n=1} {na}}{5} \\ &= \frac{\frac{2023(2023+1)(2023 \cdot 2 + 1)}{6} - \frac{a \cdot 2023(2023+1)}{2} }{5} \\ &= \frac{2023(2024)(4047-3a)}{30} \\ \end{align*} So, we have $U' = \frac{2023(2024)(4047-3a)}{30}$ , and $-1000 \le U' \le 3023$ , so we have $-1000 \le \frac{2023(2024)(4047-3a)}{30} \le 3023$ , or $-30000 \le 2023(2024)(4047-3a) \le 90690$ . Now, $2023 \cdot 2024$ is much bigger than $90690$ or $30000$ , and since $4047-3a$ is an integer, to satsify the inequalities, we must have $4047 - 3a = 0$ , or $a = 1349$ , and $U' = 0$
Now, we can find $U - U'$ . We have: \begin{align*} U - U' &= \sum^{2023}_{n=1} {\lfloor \frac{n^2-1349n}{5} \rfloor} - \sum^{2023}_{n=1} {\frac{n^2-1349n}{5}} \\ &= \sum^{2023}_{n=1} {\lfloor \frac{n^2-1349n}{5} \rfloor - \frac{n^2-1349n}{5}} \end{align*}. Now, if $n^2-1349n \equiv 0 \text{ (mod 5)}$ , then $\lfloor \frac{n^2-1349n}{5} \rfloor - \frac{n^2-1349n}{5} = 0$ , and if $n^2-1349n \equiv 1 \text{ (mod 5)}$ , then $\lfloor \frac{n^2-1349n}{5} \rfloor - \frac{n^2-1349n}{5} = -\frac{1}{5}$ , and so on. Testing with $n \equiv 0,1,2,3,4, \text{ (mod 5)}$ , we get $n^2-1349n \equiv 0,2,1,2,0 \text{ (mod 5)}$ respectively. From 1 to 2023, there are 405 numbers congruent to 1 mod 5, 405 numbers congruent to 2 mod 5, 405 numbers congruent to 3 mod 5, 404 numbers congruent to 4 mod 5, and 404 numbers congruent to 0 mod 5. So, solving for $U - U'$ , we get: \begin{align*} U - U' &= \sum^{2023}_{n=1} {\lfloor \frac{n^2-1349n}{5} \rfloor - \frac{n^2-1349n}{5}} \\ &= 404 \cdot 0 - 405 \cdot \frac{2}{5} - 405 \cdot \frac{1}{5} - 405 \cdot \frac{2}{5} - 404 \cdot t0 \\ &= -405(\frac{2}{5}+\frac{1}{5}+\frac{2}{5}) \\ &= -405 \end{align*} Since $U' = 0$ , this gives $U = -405$ , and we have $a + U = 1349-405 = \boxed{944}$ | null | 944 |
a3b12607844df814117f255d2482e53a | https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_10 | There exists a unique positive integer $a$ for which the sum \[U=\sum_{n=1}^{2023}\left\lfloor\dfrac{n^{2}-na}{5}\right\rfloor\] is an integer strictly between $-1000$ and $1000$ . For that unique $a$ , find $a+U$
(Note that $\lfloor x\rfloor$ denotes the greatest integer that is less than or equal to $x$ .) | We can view the floor function in this problem as simply subtracting the remainder of $n^2 - na$ (mod $5$ ) from the numerator of $\frac{n^2-na}{5}$ . For example, $\left\lfloor \frac{7}{5} \right\rfloor = \frac{7-2}{5} = 1$
Note that the congruence of $n^2 - na$ (mod $5$ ) loops every time $n$ increases by 5. Also, note that the congruence of $a$ (mod $5$ ) determines the set of congruences of $n^2 - na$ for each congruence of $n$ (mod $5$ ).
For example, if $a \equiv 1$ (mod $5$ ), the set of remainders is $(0, 2, 1, 2, 0)$ for $n \equiv 1,2,3,4,0$ (mod $5$ ). Let the sum of these elements be $s$ . Note that for each “loop” of the numerator (mod $5$ ), each element of the set will be subtracted exactly once, meaning $s$ is subtracted once for each loop. The value of the numerator will loop $404$ times (mod $5$ ) throughout the sum, as $5 \cdot 404=2020$ . Then
$U \approx \frac {\left( \frac {n(n+1)(2n+1)}{6} - \frac{(a)(n)(n+1)}{2} -404s \right)}{5}$
Where $n=2023$ . Note that since $5 \cdot 404=2020$ , this is an approximation for $U$ because the equation disregards the remainder (mod $5$ ) when $n=2021, 2022$ , and $2023$ so we must subtract the first 3 terms of our set of congruences one extra time to get the exact value of $U$ (*). However, we will find that this is a negligible error when it comes to the inequality $-1000<U<1000$ , so we can proceed with this approximation to solve for $a$
Factoring our approximation gives $U \approx \frac {\frac{(n)(n+1)(2n+1 - 3a)}{6}-404s}{5}$
We set $a= \frac{(2n+1)}{3} = 1349$ to make $\frac{(n)(n+1)(2n+1 - 3a)}{6}=0$ , accordingly minimizing $|U|$ , yielding $U \approx \frac{-404s}{5}$
If $a$ increases or decreases by $1$ , then $U$ changes by $\frac {(n)(n+1)}{2 \cdot 5} = \frac {2023 \cdot 2024}{10}$ which clearly breaks the inequality on $U$ . Therefore $a=1349 \equiv 4$ (mod $5$ ) giving the set of remainders $(2,1,2,0,0)$ , so $s=5$ and our approximation yields $U \approx -404$ . However, we must subtract 2, 1, and 2 (*) giving us $U = - 404 - \frac{(2+1+2)}{5} = - 405$ , giving an answer of $1349-405= \boxed{944}$ | null | 944 |
a3b12607844df814117f255d2482e53a | https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_10 | There exists a unique positive integer $a$ for which the sum \[U=\sum_{n=1}^{2023}\left\lfloor\dfrac{n^{2}-na}{5}\right\rfloor\] is an integer strictly between $-1000$ and $1000$ . For that unique $a$ , find $a+U$
(Note that $\lfloor x\rfloor$ denotes the greatest integer that is less than or equal to $x$ .) | Consider the integral \[\int_{0}^{2023} \dfrac{n^2-na}{5} \, dn.\] We hope this will give a good enough appoximation of $U$ to find $a.$ However, this integral can be easily evaluated(if you know calculus) to be \[\dfrac{1}{15}2023^3-\dfrac{a}{10}2023^2=2023^2\left(\dfrac{2023}{15}-\dfrac{a}{10}\right).\] Because we want this to be as close to $0$ as possible, we find that $a$ should equal $1349.$ Then, evaluating the sum becomes trivial. Set \[U'=\sum_{n=1}^{2023}\dfrac{n^2-1349n}{5}\] and \[U''=\sum_{n=1}^{2023}\{\dfrac{n^2-1349n}{5}\}.\] Then $U=U'-U''.$ We can evaluate $U'$ to be $0$ and $U''$ to be $-405$ (using some basic number theory). Thus, $U=-405$ and the answer is \[1349+(-405)=\boxed{944}.\] BS2012 | null | 944 |
a7c138b9986e3075b952c56cdd0239a4 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_11 | Find the number of subsets of $\{1,2,3,\ldots,10\}$ that contain exactly one pair of consecutive integers. Examples of such subsets are $\{\mathbf{1},\mathbf{2},5\}$ and $\{1,3,\mathbf{6},\mathbf{7},10\}.$ | Define $f(x)$ to be the number of subsets of $\{1, 2, 3, 4, \ldots x\}$ that have $0$ consecutive element pairs, and $f'(x)$ to be the number of subsets that have $1$ consecutive pair.
Using casework on where the consecutive element pair is, there is a unique consecutive element pair that satisfies the conditions. It is easy to see that \[f'(10) = 2f(7) + 2f(6) + 2f(1)f(5) + 2f(2)f(4) + f(3)^2.\]
We see that $f(1) = 2$ $f(2) = 3$ , and $f(n) = f(n-1) + f(n-2)$ . This is because if the element $n$ is included in our subset, then there are $f(n-2)$ possibilities for the rest of the elements (because $n-1$ cannot be used), and otherwise there are $f(n-1)$ possibilities. Thus, by induction, $f(n)$ is the $n+1$ th Fibonacci number.
This means that $f'(10) = 2(34) + 2(21) + 2(2)(13) + 2(3)(8) + 5^2 = \boxed{235}$ | null | 235 |
a7c138b9986e3075b952c56cdd0239a4 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_11 | Find the number of subsets of $\{1,2,3,\ldots,10\}$ that contain exactly one pair of consecutive integers. Examples of such subsets are $\{\mathbf{1},\mathbf{2},5\}$ and $\{1,3,\mathbf{6},\mathbf{7},10\}.$ | We can solve this problem using casework, with one case for each possible pair of consecutive numbers.
$\textbf{Case 1: (1,2)}$
If we have (1,2) as our pair, we are left with the numbers from 3-10 as elements that can be added to our subset. So, we must compute how many ways we can pick these numbers so that the set has no consecutive numbers other than (1,2). Our first option is to pick no more numbers, giving us $8 \choose {0}$ . We can also pick one number, giving us $7 \choose {1}$ because 3 cannot be picked. Another choice is to pick two numbers and in order to make sure they are not consecutive we must fix one number in between them, giving us $6 \choose {2}$ . This pattern continues for each amount of numbers, yielding $5 \choose {3}$ for 3 numbers and $4 \choose {4}$ for four numbers. Adding these up, we have $8 \choose {0}$ $7 \choose {1}$ $6 \choose {2}$ $5 \choose {3}$ $4 \choose {4}$ $\textbf{34}$
$\textbf{Case 2: (2,3)}$
If we have (2,3) as our pair, everything works the same as with (1,2), because 1 is still unusable as it is consecutive with 2. The only difference is we now have only 4-10 to work with. Using the same pattern as before, we have $7 \choose {0}$ $6 \choose {1}$ $5 \choose {2}$ $4 \choose {3}$ $\textbf{21}$
$\textbf{Case 3: (3,4)}$
This case remains pretty much the same except we now have an option of whether or not to include 1. If we want to represent this like we have with our other choices, we would say $2 \choose {0}$ for choosing no numbers and $1 \choose {1}$ for choosing 1, leaving us with $2 \choose {0}$ $1 \choose {1}$ = 2 choices (either including the number 1 in our subset or not including it). As far as the numbers from 5-10, our pattern from previous cases still holds. We have $6 \choose {0}$ $5 \choose {1}$ $4 \choose {2}$ $3 \choose {3}$ = 13. With 2 choices on one side and 13 choices on the other side, we have $2\cdot13$ $\textbf{26}$ combinations in all.
$\textbf{Case 4: (4,5)}$
Following the patterns we have already created in our previous cases, for the numbers 1-3 we have $3 \choose {0}$ $2 \choose {1}$ = 3 choices (1, 2, or neither) and for the numbers 6-10 we have $5 \choose {0}$ $4 \choose {1}$ $3 \choose {2}$ = 8 choices. With 3 choices on one side and 8 choices on the other side, we have $3\cdot8$ $\textbf{24}$ combinations in all.
$\textbf{Case 5: (5,6)}$
Again following the patterns we have already created in our previous cases, for the numbers 1-4 we have $4 \choose {0}$ $3 \choose {1}$ $2 \choose {2}$ = 5 choices and for the numbers 5-10 we have the same $4 \choose {0}$ $3 \choose {1}$ $2 \choose {2}$ = 5 choices. $5\cdot5$ $\textbf{25}$ combinations in all.
$\textbf{Rest of the cases}$
By symmetry, the case with (6,7) will act the same as case 4 with (4,5). This goes the same for (7,8) and case 3, (8.9) and case 2, and (9,10) and case 1.
Now, we simply add up all of the possibilities for each case to get our final answer. 34 + 21 + 26 + 24 + 25 + 24 + 26 + 21 + 34 = $\boxed{235}$ | null | 235 |
a7c138b9986e3075b952c56cdd0239a4 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_11 | Find the number of subsets of $\{1,2,3,\ldots,10\}$ that contain exactly one pair of consecutive integers. Examples of such subsets are $\{\mathbf{1},\mathbf{2},5\}$ and $\{1,3,\mathbf{6},\mathbf{7},10\}.$ | Denote by $N_1 \left( m \right)$ the number of subsets of a set $S$ that consists of $m$ consecutive integers, such that each subset contains exactly one pair of consecutive integers.
Denote by $N_0 \left( m \right)$ the number of subsets of a set $S$ that consists of $m$ consecutive integers, such that each subset does not contain any consecutive integers.
Denote by $a$ the smallest number in set $S$
First, we compute $N_1 \left( m \right)$
Consider $m \geq 3$ .
We do casework analysis.
Case 1: A subset does not contain $a$
The number of subsets that has exactly one pair of consecutive integers is $N_1 \left( m - 1 \right)$
Case 2: A subset contains $a$ but does not contain $a + 1$
The number of subsets that has exactly one pair of consecutive integers is $N_1 \left( m - 2 \right)$
Case 3: A subset contains $a$ and $a + 1$
To have exactly one pair of consecutive integers, this subset cannot have $a + 2$ , and cannot have consecutive integers in $\left\{ a+3, a+4, \cdots , a + m - 1 \right\}$
Thus, the number of subsets that has exactly one pair of consecutive integers is $N_0 \left( m - 3 \right)$
Therefore, for $m \geq 3$ \[N_1 \left( m \right) = N_1 \left( m - 1 \right) + N_1 \left( m - 2 \right) + N_0 \left( m - 3 \right) .\]
For $m = 1$ , we have $N_1 \left( 1 \right) = 0$ .
For $m = 2$ , we have $N_1 \left( 2 \right) = 1$
Second, we compute $N_0 \left( m \right)$
Consider $m \geq 2$ .
We do casework analysis.
Case 1: A subset does not contain $a$
The number of subsets that has no consecutive integers is $N_0 \left( m - 1 \right)$
Case 2: A subset contains $a$
To avoid having consecutive integers, the subset cannot have $a + 1$
Thus, the number of subsets that has no consecutive integers is $N_0 \left( m - 2 \right)$
Therefore, for $m \geq 2$ \[N_0 \left( m \right) = N_0 \left( m - 1 \right) + N_0 \left( m - 2 \right) .\]
For $m = 0$ , we have $N_0 \left( 0 \right) = 1$ .
For $m = 1$ , we have $N_0 \left( 1 \right) = 2$
By solving the recursive equations above, we get $N_1 \left( 10 \right) = \boxed{235}$ | null | 235 |
a7c138b9986e3075b952c56cdd0239a4 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_11 | Find the number of subsets of $\{1,2,3,\ldots,10\}$ that contain exactly one pair of consecutive integers. Examples of such subsets are $\{\mathbf{1},\mathbf{2},5\}$ and $\{1,3,\mathbf{6},\mathbf{7},10\}.$ | Let $a_n$ be the number of subsets of the set $\{1,2,3,\ldots,n\}$ such that there exists exactly 1 pair of consecutive elements.
Let $b_n$ be the number of subsets of the set $\{1, 2, 3\ldots, n\}$ such that there doesn't exist any pair of consecutive elements.
First, lets see how we can construct $a_n.$ For each subset $S$ counted in $a_n,$ either:
1. $\{n-1, n\}\subseteq S,$ 2. $n\not\in S$ , or
3. $n-1 \not\in S$ and $n\in S.$ The first case counts $b_{n-3}$ subsets (as $n-1$ cannot be included and the rest cannot have any consecutive elements), The second counts $a_{n-1},$ and the third counts $a_{n-2}.$ Thus, \[a_n = a_{n-1} + a_{n-2} + b_{n-3}.\] Next, Lets try to construct $b_n.$ For each subset $T$ counted in $b_n,$ either:
1. $n \not\in T,$ or
2. $n \in T.$ The first case counts $b_{n-1}$ subsets and the second counts $b_{n-2}.$ Thus, \[b_n = b_{n-1} + b_{n-2}.\] Since $b_1 = 2$ and $b_2 = 3,$ we have that $b_n = F_{n+1},$ so $a_n = a_{n-1} + a_{n-2} + F_{n-2}.$ (The $F_i$ is the $i$ th Fibonacci number). From here, we can construct a table of the values of $a_n$ until $n = 10.$ By listing out possibilities, we can solve for our first 3 values.
\[\begin{array}{r|l} n & a_n \\ \hline 1 & 0 \\ 2 & 1\\ 3 & 2\\ 4 & 2 + 1 + F_2 = 5\\5 & 5 + 2 + F_3 = 10\\6 & 10 + 5 + F_4 = 20 \\ 7 & 20 + 10 + F_5 = 38\\ 8 & 38 + 20 + F_6 = 71 \\ 9& 71 + 38 + F_7 = 130\\ 10 & 130 + 71 + F_8 = 235 \end{array}\] Our answer is $a_{10} = \boxed{235}.$ ~AtharvNaphade | null | 235 |
a7c138b9986e3075b952c56cdd0239a4 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_11 | Find the number of subsets of $\{1,2,3,\ldots,10\}$ that contain exactly one pair of consecutive integers. Examples of such subsets are $\{\mathbf{1},\mathbf{2},5\}$ and $\{1,3,\mathbf{6},\mathbf{7},10\}.$ | Note: This is a very common stars and bars application.
Casework on number of terms, let the number of terms be $n$ . We can come up with a generalized formula for the number of subsets with n terms.
Let $d_1, d_2, ..., d_{n-1}$ be the differences between the n terms. For example, in the set {2, 3, 6}, $d_1 = 1; d_2 = 3; d_3 = 5$
Let the range of the set be k for now, $d_1 + ... + d_{n-1} = k$ . We select one pair of terms to be consecutive by selecting one of the (n-1) terms to be 1. WLOG, let $d_1 = 1$ $1 + d_2 + ... + d_{n-1} = k$
To ensure the other $d_2, d_3, ..., d_{n-1}$ are greater than 1 such that no two other terms are consecutive or the same, let $D_2 = d_2 + 1; D_3 = d_3 + 1; ...$ $(n-2) + 1 + D_2 + ... + D_{n-1} = k$ where $D_2, D_3, ..., D_{n-1}$ are positive integers.
Finally, we add in $D_0$ , the distance between 0 and the first term of the set, and $D_n$ , the distance between the last term and 11. This way, the "distance" from 0 to 11 is "bridged" by $D_0$ , k, and $D_N$ \[D_0 + k + D_n = D_0 + ((n-1) + D_2 + ... + D_{n-1}) + D_N = 11\] \[D_0 + D_2 + D_3 + ... + D_n = 12 - n\] There are n positive terms , by Balls and Urns , there are ${(12-n)-1 \choose (n)-1} = {11-n \choose n-1}$ ways of doing this. However, recall that there were $(n-1)$ ways, and we had used a WLOG to choose which two digits are consecutive. The final formula for the number of valid n-element subsets is hence $(n-1){11-n \choose n-1}$ for $n > 2$
Case 1: Two terms $n = 2$ , so $(2-1){11-2 \choose 2-1} = 1{9 \choose 1} = 9$
Case 2: Three terms $n = 3$ , so $(3-1){11-3 \choose 3-1} = 2{8 \choose 2} = 56$
Case 3: Four terms $n = 4$ , so $(4-1){11-4 \choose 4-1} = 3{7 \choose 3} = 105$
Case 4: Five terms $n = 5$ , so $(5-1){11-5 \choose 5-1} = 4{6 \choose 4} = 60$
Case 5: Six terms $n = 6$ , so $(6-1){11-6 \choose 6-1} = 5{5 \choose 5} = 5$
We can check by the Pigeonhole principle that there cannot be more than six terms, so the answer is $9+56+105+60+5=\boxed{235}$ | null | 235 |
a7c138b9986e3075b952c56cdd0239a4 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_11 | Find the number of subsets of $\{1,2,3,\ldots,10\}$ that contain exactly one pair of consecutive integers. Examples of such subsets are $\{\mathbf{1},\mathbf{2},5\}$ and $\{1,3,\mathbf{6},\mathbf{7},10\}.$ | Note that there are $F_{n+2}$ subsets of a set of $n$ consecutive integers that contains no two consecutive integers. (This can be proven by induction.)
Now, notice that if we take $i$ and $i+1$ as the consecutive integers in our subset, we need to make a subset of the remaining integers such that it doesn't contain any two consecutive integers. Clearly, $i-1$ and $i+2$ cannot be chosen, and since $i-2$ and $i+3$ are sufficiently far apart, it is obvious we do not need to be concerned that an element of the set $\{1, 2, ... , i-2 \}$ is consecutive with any element of the set $\{ i+3, i+4, ... , 10 \}$
Thus, we can count the number of ways to choose a subset from the first set without any two elements being consecutive and multiply this by the number of ways to choose a subset from the second set without any two elements being consecutive. From above, and noting that the first set has $i-2$ consecutive integer elements and the second set has $8-i$ consecutive integer elements, we know that this is $F_i F_{10-i}.$
Summing this over for all $1 \leq i \leq 9$ yields \[\sum_{i=1}^9 F_i F_{10-i} = F_1F_9 + F_2F_8 + F_3F_7 + F_4F_6 + F_5 F_5 + F_6 F_4 + F_7 F_3 + F_8 F_2 + F_9 F_1 = 2(F_1F_9 + F_2F_8 + F_3F_7+F_4F_6) + F_5^2 = 2(1 \cdot 34 + 1 \cdot 21 + 2 \cdot 13 + 3 \cdot 8) + 5^2 = 2 \cdot 105+25 = \boxed{235}.\] | null | 235 |
a7c138b9986e3075b952c56cdd0239a4 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_11 | Find the number of subsets of $\{1,2,3,\ldots,10\}$ that contain exactly one pair of consecutive integers. Examples of such subsets are $\{\mathbf{1},\mathbf{2},5\}$ and $\{1,3,\mathbf{6},\mathbf{7},10\}.$ | The problem is the same as laying out a line of polynomoes to cover spots $0,1,...10$ : 1 triomino ( $RGG$ ), $n$ dominoes ( $RG$ ), and $8-2n$ monominoes ( $R$ ). The $G$ spots cover the members of the subset.
The total number spots is 11, because one $R$ spot always covers the 0, and the other spots cover 1 through 10.
There are 5 ways to choose polyomino sets, and many ways to order each set:
$R + RG + RGG =$ Polyominoes $\rightarrow$ Orderings
$0 + 4 + 1 = 5 \rightarrow 5! / 0!4!1! = ~~~5$
$2 + 3 + 1 = 6 \rightarrow 6! / 2!3!1! = ~60$
$4 + 2 + 1 = 7 \rightarrow 7! / 4!2!1! = 105$
$6 + 1 + 1 = 8 \rightarrow 8! / 6!1!1! = ~56$
$8 + 0 + 1 = 9 \rightarrow 9! / 8!0!1! = ~~~9$
The sum is $\boxed{235}$ | null | 235 |
a7c138b9986e3075b952c56cdd0239a4 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_11 | Find the number of subsets of $\{1,2,3,\ldots,10\}$ that contain exactly one pair of consecutive integers. Examples of such subsets are $\{\mathbf{1},\mathbf{2},5\}$ and $\{1,3,\mathbf{6},\mathbf{7},10\}.$ | Let $dp(i)$ be the number of subsets of a set $i$ consecutive integers such that the maximum value in the set is $i$ and there exists exactly one pair of consecutive integers. Define $dp2(i)$ similarly, but without any pair of consecutive integers. The base cases are $dp2(1)=dp2(2)=1$ $dp(1)=0$ , and $dp(2)=1$
The transitions are:
\[dp2(i)=\sum_{j=1}^{i-2}(dp(j))+1\]
\[dp(i)=dp2(i-1)+\sum_{j=1}^{i-2}dp(j)\]
Note that $dp2$ is the Fibonacci numbers.
\[\begin{array}{rcl} i & dp & dp2\\ \hline 1 & 0 & 1 \\ 2 & 1 & 1 \\ 3 & 1 & 2 \\ 4 & 3 & 3 \\ 5 & 5 & 5 \\ 6 & 10 & 8 \\ 7 & 18 & 13 \\ 8 & 33 & 21 \\ 9 & 59 & 34 \\ 10 & 105 & 55 \\ \end{array}\]
Summing over $dp$ yields $1+1+3+5+10+18+33+59+105=\boxed{235}$ | null | 235 |
eb5188fef8fa8f43660daeda2d0bc7e1 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_12 | Let $\triangle ABC$ be an equilateral triangle with side length $55.$ Points $D,$ $E,$ and $F$ lie on $\overline{BC},$ $\overline{CA},$ and $\overline{AB},$ respectively, with $BD = 7,$ $CE=30,$ and $AF=40.$ Point $P$ inside $\triangle ABC$ has the property that \[\angle AEP = \angle BFP = \angle CDP.\] Find $\tan^2(\angle AEP).$ | By Miquel's theorem, $P=(AEF)\cap(BFD)\cap(CDE)$ (intersection of circles). The law of cosines can be used to compute $DE=42$ $EF=35$ , and $FD=13$ . Toss the points on the coordinate plane; let $B=(-7, 0)$ $D=(0, 0)$ , and $C=(48, 0)$ , where we will find $\tan^{2}\left(\measuredangle CDP\right)$ with $P=(BFD)\cap(CDE)$
By the extended law of sines, the radius of circle $(BFD)$ is $\frac{13}{2\sin 60^{\circ}}=\frac{13}{3}\sqrt{3}$ . Its center lies on the line $x=-\frac{7}{2}$ , and the origin is a point on it, so $y=\frac{23}{6}\sqrt{3}$
The radius of circle $(CDE)$ is $\frac{42}{2\sin 60^{\circ}}=14\sqrt{3}$ . The origin is also a point on it, and its center is on the line $x=24$ , so $y=2\sqrt{3}$
The equations of the two circles are \begin{align*}(BFD)&:\left(x+\tfrac{7}{2}\right)^{2}+\left(y-\tfrac{23}{6}\sqrt{3}\right)^{2}=\tfrac{169}{3} \\ (CDE)&:\left(x-24\right)^{2}+\left(y-2\sqrt{3}\right)^{2}=588\end{align*} These equations simplify to \begin{align*}(BFD)&:x^{2}+7x+y^{2}-\tfrac{23}{3}\sqrt{3}y=0 \\ (CDE)&: x^{2}-48x+y^{2}-4\sqrt{3}y=0\end{align*} Subtracting these two equations gives that both their points of intersection, $D$ and $P$ , lie on the line $55x-\tfrac{11}{3}\sqrt{3}y=0$ . Hence, $\tan^{2}\left(\measuredangle AEP\right)=\tan^{2}\left(\measuredangle CDP\right)=\left(\frac{55}{\tfrac{11}{3}\sqrt{3}}\right)^{2}=3\left(\tfrac{55}{11}\right)^{2}=\boxed{075}$ . To scale, the configuration looks like the figure below: [asy] /* Made by MRENTHUSIASM */ size(400); pair A, B, C, D, E, F, P; A = 55*sqrt(3)/3 * dir(90); B = 55*sqrt(3)/3 * dir(210); C = 55*sqrt(3)/3 * dir(330); D = B + 7*dir(0); E = A + 25*dir(C-A); F = A + 40*dir(B-A); P = intersectionpoints(Circle(D,54*sqrt(19)/19),Circle(F,5*sqrt(19)/19))[0]; filldraw(D--E--F--cycle,yellow); draw(A--B--C--cycle); draw(circumcircle(A,E,F)^^circumcircle(B,D,F)^^circumcircle(C,D,E),blue); dot("$A$",A,1.5*dir(A),linewidth(4)); dot("$B$",B,1.5*dir(B),linewidth(4)); dot("$C$",C,1.5*dir(C),linewidth(4)); dot("$D$",D,1.5*S,linewidth(4)); dot("$E$",E,1.5*dir(30),linewidth(4)); dot("$F$",F,1.5*dir(150),linewidth(4)); dot("$P$",P,1.5*dir(-30),linewidth(4)); label("$7$",midpoint(B--D),1.5*S,red); label("$30$",midpoint(C--E),1.5*dir(30),red); label("$40$",midpoint(A--F),1.5*dir(150),red); [/asy] | null | 075 |
eb5188fef8fa8f43660daeda2d0bc7e1 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_12 | Let $\triangle ABC$ be an equilateral triangle with side length $55.$ Points $D,$ $E,$ and $F$ lie on $\overline{BC},$ $\overline{CA},$ and $\overline{AB},$ respectively, with $BD = 7,$ $CE=30,$ and $AF=40.$ Point $P$ inside $\triangle ABC$ has the property that \[\angle AEP = \angle BFP = \angle CDP.\] Find $\tan^2(\angle AEP).$ | Denote $\theta = \angle AEP$
In $AFPE$ , we have $\overrightarrow{AF} + \overrightarrow{FP} + \overrightarrow{PE} + \overrightarrow{EA} = 0$ .
Thus, \[ AF + FP e^{i \theta} + PE e^{i \left( \theta + 60^\circ \right)} + EA e^{- i 120^\circ} = 0. \]
Taking the real and imaginary parts, we get \begin{align*} AF + FP \cos \theta + PE \cos \left( \theta + 60^\circ \right) + EA \cos \left( - 120^\circ \right) & = 0 \hspace{1cm} (1) \\ FP \sin \theta + PE \sin \left( \theta + 60^\circ \right) + EA \sin \left( - 120^\circ \right) & = 0 \hspace{1cm} (2) \end{align*}
In $BDPF$ , analogous to the analysis of $AFPE$ above, we get \begin{align*} BD + DP \cos \theta + PF \cos \left( \theta + 60^\circ \right) + FB \cos \left( - 120^\circ \right) & = 0 \hspace{1cm} (3) \\ DP \sin \theta + PF \sin \left( \theta + 60^\circ \right) + FB \sin \left( - 120^\circ \right) & = 0 \hspace{1cm} (4) \end{align*}
Taking $(1) \cdot \sin \left( \theta + 60^\circ \right) - (2) \cdot \cos \left( \theta + 60^\circ \right)$ , we get \[ AF \sin \left( \theta + 60^\circ \right) + \frac{\sqrt{3}}{2} FP - EA \sin \theta = 0 . \hspace{1cm} (5) \]
Taking $(3) \cdot \sin \theta - (4) \cdot \cos \theta$ , we get \[ BD \sin \theta - \frac{\sqrt{3}}{2} FP + FB \sin \left( \theta + 120^\circ \right) . \hspace{1cm} (6) \]
Taking $(5) + (6)$ , we get \[ AF \sin \left( \theta + 60^\circ \right) - EA \sin \theta + BD \sin \theta + FB \sin \left( \theta + 120^\circ \right) . \]
Therefore, \begin{align*} \tan \theta & = \frac{\frac{\sqrt{3}}{2} \left( AF + FB \right)} {\frac{FB}{2} + EA - \frac{AF}{2} - BD} \\ & = 5 \sqrt{3} . \end{align*}
Therefore, $\tan^2 \theta = \boxed{075}$ | null | 075 |
eb5188fef8fa8f43660daeda2d0bc7e1 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_12 | Let $\triangle ABC$ be an equilateral triangle with side length $55.$ Points $D,$ $E,$ and $F$ lie on $\overline{BC},$ $\overline{CA},$ and $\overline{AB},$ respectively, with $BD = 7,$ $CE=30,$ and $AF=40.$ Point $P$ inside $\triangle ABC$ has the property that \[\angle AEP = \angle BFP = \angle CDP.\] Find $\tan^2(\angle AEP).$ | Drop the perpendiculars from $P$ to $\overline{AB}$ $\overline{AC}$ $\overline{BC}$ , and call them $Q,R,$ and $S$ respectively. This gives us three similar right triangles $FQP$ $ERP$ , and $DSP.$
The sum of the perpendiculars to a point $P$ within an equilateral triangle is always constant, so we have that $PQ+PR+PS=\dfrac{55 \sqrt{3}}{2}.$
The sum of the lengths of the alternating segments split by the perpendiculars from a point $P$ within an equilateral triangle is always equal to half the perimeter, so $QA+RC+SB = \dfrac{165}{2},$ which means that $FQ+ER+DS = QA+RC+SB - CE - AF - BD = \dfrac{165}{2} - 30 - 40 - 7 = \dfrac{11}{2}.$
Finally, $\tan AEP = \dfrac{PQ}{FQ} = \dfrac{PR}{ER} = \dfrac{PS}{DS} = \dfrac{PQ+PR+PS}{FQ+ER+DS} = 5 \sqrt{3}.$
Thus, $\tan^2 AEP = \boxed{075}.$ | null | 075 |
eb5188fef8fa8f43660daeda2d0bc7e1 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_12 | Let $\triangle ABC$ be an equilateral triangle with side length $55.$ Points $D,$ $E,$ and $F$ lie on $\overline{BC},$ $\overline{CA},$ and $\overline{AB},$ respectively, with $BD = 7,$ $CE=30,$ and $AF=40.$ Point $P$ inside $\triangle ABC$ has the property that \[\angle AEP = \angle BFP = \angle CDP.\] Find $\tan^2(\angle AEP).$ | This solution is inspired by AIME 1999 Problem 14 Solution 2 (a similar question)
Draw line segments from $P$ to points $A$ $B$ , and $C$ . And label the angle measure of $\angle{BFP}$ $\angle{CDP}$ , and $\angle{AEP}$ to be $\alpha$
Using Law of Cosines (note that $\cos{\angle{AFP}}=\cos{\angle{BDP}}=\cos{\angle{CEP}}=\cos{180^\circ-\alpha}=-\cos{\alpha}$
\begin{align*} (1) \ BP^2 &= FP^2+15^2-2\cdot FP\cdot15\cdot\cos(\alpha)\\ (2) \ BP^2 &= DP^2+7^2+2\cdot DP\cdot7\cdot\cos(\alpha)\\ (3) \ CP^2 &= DP^2+48^2-2\cdot DP\cdot48\cdot\cos(\alpha)\\ (4) \ CP^2 &= EP^2+30^2+2\cdot EP\cdot30\cdot\cos(\alpha)\\ (5) \ AP^2 &= EP^2+25^2-2\cdot EP\cdot25\cdot\cos(\alpha)\\ (6) \ AP^2 &= FP^2+40^2+2\cdot FP\cdot40\cdot\cos(\alpha)\\ \end{align*}
We can perform this operation $(1) - (2) + (3) - (4) + (5) - (6)$
Leaving us with (after combining and simplifying) \[\cos{\alpha}=\frac{-11}{2\cdot(DP+EP+FP)}\]
Therefore, we want to solve for $DP+EP+FP$
Notice that $\angle{DPE}=\angle{EPF}=\angle{FPD}=120^\circ$
We can use Law of Cosines again to solve for the sides of $\triangle{DEF}$ , which have side lengths of $13$ $42$ , and $35$ , and area $120\sqrt{3}$
Label the lengths of $PD$ $PE$ , and $PF$ to be $x$ $y$ , and $z$
Therefore, using the $\sin$ area formula,
\begin{align*} [\triangle{DEF}] &= \frac{1}{2}\cdot\sin{120°}\cdot(xy+yz+zx) = 120\sqrt{3} \\ xy+yz+zx &= 2^5\cdot3\cdot5 \end{align*}
In addition, we know that
\begin{align*} x^2+y^2+xy&=42^2\\ y^2+z^2+yz&=35^2\\ z^2+x^2+zx&=13^2\\ \end{align*}
By using Law of Cosines for $\triangle{DPE}$ $\triangle{EPF}$ , and $\triangle{FPD}$ respectively
Because we want $DP+EP+FP$ , which is $x+y+z$ , we see that
\begin{align*} (x+y+z)^2 &= \frac{x^2+y^2+x+y^2+z^2+yz+z^2+x^2+3(zx+xy+yz+zx)}{2} \\ (x+y+z)^2 &= \frac{42^2+35^2+13^2+3\cdot2^5\cdot3\cdot5}{2} \\ (x+y+z)^2 &= 2299 \\ x+y+z &= 11\sqrt{19} \end{align*}
So plugging the results back into the equation before, we get
\begin{align*} \cos{\alpha} &= \frac{-1}{2\sqrt{19}}\\ \sin{\alpha} &= \frac{5\sqrt{3}}{2\sqrt{19}} \end{align*}
Giving us \[\tan^2{\alpha}=\boxed{075}\] | null | 075 |
eb5188fef8fa8f43660daeda2d0bc7e1 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_12 | Let $\triangle ABC$ be an equilateral triangle with side length $55.$ Points $D,$ $E,$ and $F$ lie on $\overline{BC},$ $\overline{CA},$ and $\overline{AB},$ respectively, with $BD = 7,$ $CE=30,$ and $AF=40.$ Point $P$ inside $\triangle ABC$ has the property that \[\angle AEP = \angle BFP = \angle CDP.\] Find $\tan^2(\angle AEP).$ | We begin by using the fact stated in Solution 3 that, for any point in an equilateral triangle, the lengths of the three perpendicular lines dropped to the sides of the triangle add up to the altitude of that triangle. To make things simple, let's assign $\angle AEP = \angle BFP = \angle CDP = \alpha$ . We can label these three perpendiculars as: \[PD\cdot\sin{\alpha} + PE\cdot\sin{\alpha} + PF\cdot\sin{\alpha} = \dfrac{55 \sqrt{3}}{2}\] Simplifying, we get \[PD + PE + PF = \dfrac{55 \sqrt{3}}{2\cdot\sin{\alpha}}\] Now, as stated and quoting Solution 4,
"Draw line segments from $P$ to points $A$ $B$ , and $C$ . [We know that] the angle measure of $\angle{AEP}$ $\angle{BFP}$ , and $\angle{CDP}$ is $\alpha$
Using Law of Cosines (note that $\cos{\angle{AFP}}=\cos{\angle{BDP}}=\cos{\angle{CEP}}=\cos{180^\circ-\alpha}=-\cos{\alpha}$
\begin{align*} (1) \ BP^2 &= FP^2+15^2-2\cdot FP\cdot15\cdot\cos(\alpha)\\ (2) \ BP^2 &= DP^2+7^2+2\cdot DP\cdot7\cdot\cos(\alpha)\\ (3) \ CP^2 &= DP^2+48^2-2\cdot DP\cdot48\cdot\cos(\alpha)\\ (4) \ CP^2 &= EP^2+30^2+2\cdot EP\cdot30\cdot\cos(\alpha)\\ (5) \ AP^2 &= EP^2+25^2-2\cdot EP\cdot25\cdot\cos(\alpha)\\ (6) \ AP^2 &= FP^2+40^2+2\cdot FP\cdot40\cdot\cos(\alpha)\\ \end{align*}
We can perform this operation $(1) - (2) + (3) - (4) + (5) - (6)$
Leaving us with (after combining and simplifying) \[PD + PE + PF=\frac{11}{2\cdot\cos{\alpha}}\] ".
Now, we can use our previous equation along with this one to get: \[\frac{11}{2\cdot\cos{\alpha}} = \dfrac{55 \sqrt{3}}{2\cdot\sin{\alpha}}\]
This equation becomes: \[\tan{\alpha} = 5\sqrt{3}\] As so, our answer is \[\left(5\sqrt3\right)^2=\boxed{075}.\] ~Solution by armang32324 (Mathemagics Club) | null | 075 |
eb5188fef8fa8f43660daeda2d0bc7e1 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_12 | Let $\triangle ABC$ be an equilateral triangle with side length $55.$ Points $D,$ $E,$ and $F$ lie on $\overline{BC},$ $\overline{CA},$ and $\overline{AB},$ respectively, with $BD = 7,$ $CE=30,$ and $AF=40.$ Point $P$ inside $\triangle ABC$ has the property that \[\angle AEP = \angle BFP = \angle CDP.\] Find $\tan^2(\angle AEP).$ | By the law of cosines, \[FE=\sqrt{AF^2+AE^2-2AF\cdot AE\cos\angle FAE}=35.\] Similarly we get $FD=13$ and $DE=42$ $\angle AEP=\angle BFP=\angle CDP\overset\triangle=\theta$ implies that $AFPE$ $BDPF$ , and $CDPE$ are three cyclic quadrilaterals, as shown below: [asy] /* Made by MRENTHUSIASM */ size(400); pair A, B, C, D, E, F, P; A = 55*sqrt(3)/3 * dir(90); B = 55*sqrt(3)/3 * dir(210); C = 55*sqrt(3)/3 * dir(330); D = B + 7*dir(0); E = A + 25*dir(C-A); F = A + 40*dir(B-A); P = intersectionpoints(Circle(D,54*sqrt(19)/19),Circle(F,5*sqrt(19)/19))[0]; draw(anglemark(A,E,P,20),red); draw(anglemark(B,F,P,20),red); draw(anglemark(C,D,P,20),red); add(pathticks(anglemark(A,E,P,20), n = 1, r = 0.2, s = 12, red)); add(pathticks(anglemark(B,F,P,20), n = 1, r = 0.2, s = 12, red)); add(pathticks(anglemark(C,D,P,20), n = 1, r = 0.2, s = 12, red)); draw(A--B--C--cycle^^P--E^^P--F^^P--D); draw(P--A^^P--B^^P--C,dashed); draw(circumcircle(A,E,F)^^circumcircle(B,D,F)^^circumcircle(C,D,E),blue); dot("$A$",A,1.5*dir(A),linewidth(4)); dot("$B$",B,1.5*dir(B),linewidth(4)); dot("$C$",C,1.5*dir(C),linewidth(4)); dot("$D$",D,1.5*S,linewidth(4)); dot("$E$",E,1.5*dir(30),linewidth(4)); dot("$F$",F,1.5*dir(150),linewidth(4)); dot("$P$",P,1.5*dir(-30),linewidth(4)); label("$7$",midpoint(B--D),1.5*S,red); label("$30$",midpoint(C--E),1.5*dir(30),red); label("$40$",midpoint(A--F),1.5*dir(150),red); [/asy] Using the law of sines in each, \[\frac{AP}{35}=\frac{AP}{FE}=\frac{CP}{42}=\frac{CP}{ED}=\frac{BP}{13}=\frac{BP}{DF}=\frac{\sin\theta}{\sin\frac\pi3}.\] So we can set $AP=35k$ $BP=13k$ , and $CP=42k$ . Let $PD=d$ $PE=e$ , and $PF=f$ . Applying Ptolemy theorem in the cyclic quadrilaterals, \[\begin{cases}AP\cdot FE=AF\cdot PE+AE\cdot PF,\\CP\cdot ED=CE\cdot PD+CD\cdot PE,\\BP\cdot DF=BD\cdot PF+BF\cdot PD.\end{cases} \implies \begin{cases} 1225k=40e+25f,\\1764k=30d+48e,\\169k=15d+7f, \end{cases}\] We can solve out $d=\frac{54k}5$ $e=30k$ $f=k$ . By the law of cosines in $\triangle PEF$ $FE=\sqrt{900k^2+k^2-60k\cdot\left(\frac{-1}2\right)}=\sqrt{931}k$ . The law of sines yield $\frac{\sin\angle AEP}{\sin\angle FAE}=\frac{AP}{FE}=\frac{35k}{\sqrt{931}k}=\frac{35}{\sqrt{931}}$ .
Lastly, $\sin\angle AEP=\frac{5\sqrt{57}}{38}$ , then $\tan\angle AEP=5\sqrt3$ . The answer is \[\left(5\sqrt3\right)^2=\boxed{075}.\] | null | 075 |
355b1e9348c8f71fad096a83edff94ea | https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_13 | Each face of two noncongruent parallelepipeds is a rhombus whose diagonals have lengths $\sqrt{21}$ and $\sqrt{31}$ .
The ratio of the volume of the larger of the two polyhedra to the volume of the smaller is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$ . A parallelepiped is a solid with six parallelogram faces
such as the one shown below.
[asy] unitsize(2cm); pair o = (0, 0), u = (1, 0), v = 0.8*dir(40), w = dir(70); draw(o--u--(u+v)); draw(o--v--(u+v), dotted); draw(shift(w)*(o--u--(u+v)--v--cycle)); draw(o--w); draw(u--(u+w)); draw(v--(v+w), dotted); draw((u+v)--(u+v+w)); [/asy] | Denote $\alpha = \tan^{-1} \frac{\sqrt{21}}{\sqrt{31}}$ .
Denote by $d$ the length of each side of a rhombus.
Now, we put the solid to the 3-d coordinate space.
We put the bottom face on the $x-O-y$ plane.
For this bottom face, we put a vertex with an acute angle $2 \alpha$ at the origin, denoted as $O$ .
For two edges that are on the bottom face and meet at $O$ , we put one edge on the positive side of the $x$ -axis. The endpoint is denoted as $A$ . Hence, $A = \left( d , 0 , 0 \right)$ .
We put the other edge in the first quadrant of the $x-O-y$ plane. The endpoint is denoted as $B$ . Hence, $B = \left( d \cos 2 \alpha , d \sin 2 \alpha , 0 \right)$
For the third edge that has one endpoint $O$ , we denote by $C$ its second endpoint.
We denote $C = \left( u , v , w \right)$ .
Without loss of generality, we set $w > 0$ .
Hence, \[ u^2 + v^2 + w^2 = d^2 . \hspace{1cm} (1) \]
We have \begin{align*} \cos \angle AOC & = \frac{\overrightarrow{OA} \cdot \overrightarrow{OC}}{|OA| \cdot |OC|} \\ & = \frac{u}{d} , \hspace{1cm} (2) \end{align*} and \begin{align*} \cos \angle BOC & = \frac{\overrightarrow{OB} \cdot \overrightarrow{OC}}{|OB| \cdot |OC|} \\ & = \frac{u \cos 2 \alpha + v \sin 2 \alpha}{d} . \hspace{1cm} (3) \end{align*}
Case 1: $\angle AOC = \angle BOC = 2 \alpha$ or $2 \left( 90^\circ - \alpha \right)$
By solving (2) and (3), we get \begin{align*} u & = \pm d \cos 2 \alpha , \\ v & = \pm d \cos 2 \alpha \frac{1 - \cos 2 \alpha}{\sin 2 \alpha} \\ & = \pm d \cos 2 \alpha \tan \alpha . \end{align*}
Plugging these into (1), we get \begin{align*} w & = d \sqrt{1 - \cos^2 2 \alpha - \cos^2 2 \alpha \tan^2 \alpha} \\ & = d \sqrt{\sin^2 2 \alpha - \cos^2 2 \alpha \tan^2 \alpha} . \hspace{1cm} (4) \end{align*}
Case 2: $\angle AOC = 2 \alpha$ and $\angle BOC = 2 \left( 90^\circ - \alpha \right)$ , or $\angle BOC = 2 \alpha$ and $\angle AOC = 2 \left( 90^\circ - \alpha \right)$
By solving (2) and (3), we get \begin{align*} u & = \pm d \cos 2 \alpha , \\ v & = \mp d \cos 2 \alpha \frac{1 + \cos 2 \alpha}{\sin 2 \alpha} \\ & = \mp d \cos 2 \alpha \cot \alpha . \end{align*}
Plugging these into (1), we get \begin{align*} w & = d \sqrt{1 - \cos^2 2 \alpha - \cos^2 2 \alpha \cot^2 \alpha} \\ & = d \sqrt{\sin^2 2 \alpha - \cos^2 2 \alpha \cot^2 \alpha} . \hspace{1cm} (5) \end{align*}
We notice that $(4) > (5)$ . Thus, (4) (resp. (5)) is the parallelepiped with a larger (resp. smaller) height.
Therefore, the ratio of the volume of the larger parallelepiped to the smaller one is \begin{align*} \frac{(4)}{(5)} & = \frac{\sqrt{\sin^2 2 \alpha - \cos^2 2 \alpha \tan^2 \alpha}} {\sqrt{\sin^2 2 \alpha - \cos^2 2 \alpha \cot^2 \alpha}} \\ & = \sqrt{\frac{\tan^2 2 \alpha - \tan^2 \alpha}{\tan^2 2 \alpha - \cot^2 \alpha}} . \end{align*}
Recall that $\tan \alpha = \frac{\sqrt{21}}{\sqrt{31}}$ .
Thus, $\tan 2 \alpha = \frac{2 \tan \alpha}{1 - \tan^2 \alpha} = \frac{\sqrt{21 \cdot 31}}{5}$ .
Plugging this into the equation above, we get \begin{align*} \frac{(4)}{(5)} & = \frac{63}{62}. \end{align*}
Therefore, the answer is $63 + 62 = \boxed{125}$ | null | 125 |
355b1e9348c8f71fad096a83edff94ea | https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_13 | Each face of two noncongruent parallelepipeds is a rhombus whose diagonals have lengths $\sqrt{21}$ and $\sqrt{31}$ .
The ratio of the volume of the larger of the two polyhedra to the volume of the smaller is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$ . A parallelepiped is a solid with six parallelogram faces
such as the one shown below.
[asy] unitsize(2cm); pair o = (0, 0), u = (1, 0), v = 0.8*dir(40), w = dir(70); draw(o--u--(u+v)); draw(o--v--(u+v), dotted); draw(shift(w)*(o--u--(u+v)--v--cycle)); draw(o--w); draw(u--(u+w)); draw(v--(v+w), dotted); draw((u+v)--(u+v+w)); [/asy] | Let one of the vertices be at the origin and the three adjacent vertices be $u$ $v$ , and $w$ . For one of the parallelepipeds, the three diagonals involving the origin have length $\sqrt {21}$ . Hence, $(u+v)\cdot (u+v)=u\cdot u+v\cdot v+2u\cdot v=21$ and $(u-v)\cdot (u-v)=u\cdot u+v\cdot v-2u\cdot v=31$ . Since all of $u$ $v$ , and $w$ have equal length, $u\cdot u=13$ $v\cdot v=13$ , and $u\cdot v=-2.5$ . Symmetrically, $w\cdot w=13$ $u\cdot w=-2.5$ , and $v\cdot w=-2.5$ . Hence the volume of the parallelepiped is given by $\sqrt{\operatorname{det}\begin{pmatrix}13&-2.5&-2.5\\-2.5&13&-2.5\\-2.5&-2.5&13\end{pmatrix}}=\sqrt{\operatorname{det}\begin{pmatrix}15.5&-15.5&0\\-2.5&13&-2.5\\0&-15.5&15.5\end{pmatrix}}=\sqrt{15.5^2\operatorname\det\begin{pmatrix}1&-1&0\\-2.5&13&-2.5\\0&-1&1\end{pmatrix}}=\sqrt{15.5^2\cdot 8}$
For the other parallelepiped, the three diagonals involving the origin are of length $\sqrt{31}$ and the volume is $\sqrt{\operatorname{det}\begin{pmatrix}13&2.5&2.5\\2.5&13&2.5\\2.5&2.5&13\end{pmatrix}}=\sqrt{\operatorname{det}\begin{pmatrix}10.5&-10.5&0\\2.5&13&2.5\\0&-10.5&10.5\end{pmatrix}}=\sqrt{10.5^2\operatorname\det\begin{pmatrix}1&-1&0\\2.5&13&2.5\\0&-1&1\end{pmatrix}}=\sqrt{10.5^2\cdot 18}$
Consequently, the answer is $\sqrt\frac{10.5^2\cdot 18}{15.5^2\cdot 8}=\frac{63}{62}$ , giving $\boxed{125}$ | null | 125 |
355b1e9348c8f71fad096a83edff94ea | https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_13 | Each face of two noncongruent parallelepipeds is a rhombus whose diagonals have lengths $\sqrt{21}$ and $\sqrt{31}$ .
The ratio of the volume of the larger of the two polyhedra to the volume of the smaller is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$ . A parallelepiped is a solid with six parallelogram faces
such as the one shown below.
[asy] unitsize(2cm); pair o = (0, 0), u = (1, 0), v = 0.8*dir(40), w = dir(70); draw(o--u--(u+v)); draw(o--v--(u+v), dotted); draw(shift(w)*(o--u--(u+v)--v--cycle)); draw(o--w); draw(u--(u+w)); draw(v--(v+w), dotted); draw((u+v)--(u+v+w)); [/asy] | Observe that both parallelepipeds have two vertices (one on each base) that have three congruent angles meeting at them. Denote the parallelepiped with three acute angles meeting $P$ , and the one with three obtuse angles meeting $P'$
The area of a parallelepiped is simply the base area times the height, but because both parallelepipeds have the same base, what we want is just the ratio of the heights.
Denote the point with three acute angles meeting at it in $P$ as $A$ , and its neighbors $B$ $C$ , and $D$ . Similarly, denote the point with three obtuse angles meeting at it in $P'$ as $A'$ , and its neighbors $B'$ $C'$ , and $D'$
We have the following equations:
\[\textrm{Height of }P\textrm{ from }ACD = \frac{\textrm{Vol}(ABCD) \cdot 3}{[ACD]},\] \[\textrm{Height of }P'\textrm{ from }A'C'D' = \frac{\textrm{Vol}(A'B'C'D') \cdot 3}{[A'C'D']}.\]
However, $ACD$ and $A'C'D'$ are both half the area of a rhombus with diagonals $\sqrt{31}$ and $\sqrt{21}$ , so our ratio is really
\[\frac{P}{P'} = \frac{\textrm{Vol}(ABCD)}{\textrm{Vol}(A'B'C'D')}.\]
Because the diagonals of all of the faces are $\sqrt{31}$ and $\sqrt{21}$ , each edge of the parallelepipeds is $\sqrt{13}$ by the Pythagorean theorem.
We have $AB = AC = AD = \sqrt{13}$ , and $BC = CD = BD = \sqrt{21}$ . When we drop a perpendicular to the centroid of $BCD$ from $A$ (let's call this point $O$ ), we have $BO = \frac{\sqrt{21}}{\sqrt{3}} = \sqrt{7}$ . Thus,
\[AB^2 - BO^2 = AO^2\] \[13 - 7 = AO^2 = 6\] \[AO = \sqrt{6}.\]
The area of base $BCD$ is $\frac{21\sqrt{3}}{4}$ . Hence,
\[\textrm{Vol}(ABCD) = \frac{\sqrt{6}\cdot\frac{21\sqrt{3}}{4}}{3}\] \[= \frac{63\sqrt{2}}{12}.\]
We can apply a similar approach to $A'B'C'D'$
$A'B' = A'C' = A'D' = \sqrt{13}$ , and $B'C' = C'D' = B'D' = \sqrt{31}$ . When we drop a perpendicular to the centroid of $B'C'D'$ from $A'$ (let's call this point $O'$ ), we have $B'O' = \frac{\sqrt{31}}{\sqrt{3}} = \sqrt{\frac{31}{3}}$ . Thus,
\[A'B'^2 - B'O'^2 = A'O'^2\] \[13 - \frac{31}{3} = A'O'^2\] \[A'O' = \sqrt{8}{3} = \frac{2\sqrt{6}}{3}.\]
The area of base $B'C'D'$ is $\frac{31\sqrt{3}}{4}$ . Hence,
\[\textrm{Vol}(A'B'C'D') = \frac{\frac{2\sqrt{6}}{3}\cdot\frac{31\sqrt{3}}{4}}{3}\] \[= \frac{186\sqrt{2}}{36}\] \[= \frac{62\sqrt{2}}{12}.\]
Finally,
\[\frac{P}{P'} = \frac{\textrm{Vol}(ABCD)}{\textrm{Vol}(A'B'C'D')} = \frac{\frac{63\sqrt{2}}{12}}{\frac{62\sqrt{2}}{12}} = \frac{63}{62}.\]
Our answer is $63 + 62 = \boxed{125}$ | null | 125 |
355b1e9348c8f71fad096a83edff94ea | https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_13 | Each face of two noncongruent parallelepipeds is a rhombus whose diagonals have lengths $\sqrt{21}$ and $\sqrt{31}$ .
The ratio of the volume of the larger of the two polyhedra to the volume of the smaller is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$ . A parallelepiped is a solid with six parallelogram faces
such as the one shown below.
[asy] unitsize(2cm); pair o = (0, 0), u = (1, 0), v = 0.8*dir(40), w = dir(70); draw(o--u--(u+v)); draw(o--v--(u+v), dotted); draw(shift(w)*(o--u--(u+v)--v--cycle)); draw(o--w); draw(u--(u+w)); draw(v--(v+w), dotted); draw((u+v)--(u+v+w)); [/asy] | Since the two parallelepipeds have the same base, all we need to do is to find their respective heights.
[asy] unitsize(2cm); pair a = (0, 0), b = (1, 0), c = 0.8*dir(40), d = dir(70), p = 0.33*dir(20), o = (b+c)/2; label("A",a,S); label("B",b,S); label("C",c,S); label("D",d,N); label("P",p,S); label("O",o,E); draw(a--b--(b+c)); draw(a--c--(b+c), dotted); draw(shift(d)*(a--b--(b+c)--c--cycle)); draw(a--d); draw(b--(b+d)); draw(c--(c+d), dotted); draw((b+c)--(b+c+d)); draw(d--p, dotted); draw(c--b, dotted); draw(a--(b+c), dotted); draw(p--c, dotted); draw(d--c, dotted); [/asy]
As illustrated in the above diagram, drop a perpendicular from $D$ onto the base at $P$ . Denote the center of the base by $O$ . By symmetry, $P$ must be on $AO$ . Now we need to find $DP$
Apply Pythagorean theorem to $\triangle DPA$ we have \[DP^2 = AD^2 - AP^2.\]
Apply Pythagorean theorem to $\triangle DPC$ and then $\triangle CPO$ we have \[DP^2 = DC^2 - CP^2 = DC^2 - (CO^2 + OP^2) = DC^2 - (CO^2 + (AO-AP)^2) = DC^2 - CO^2 - (AO-AP)^2.\]
Combining the above two, we have \[AD^2 - AP^2 = DC^2 - CO^2 - (AO-AP)^2.\]
Since $AD=\sqrt{13}$ $DC=\sqrt{21}$ $CO=\frac{\sqrt{21}}{2}$ $AO=\frac{\sqrt{31}}{2}$ , plug them into the above equation and solve for the only unknown variable $AP$ , we get $AP=\frac{5}{\sqrt{31}}.$
Thus the height \[DP = \sqrt{AD^2 - AP^2} = \sqrt{13 - \frac{25}{31}} = \sqrt{\frac{378}{31}}.\]
[asy] unitsize(2cm); pair a = (0, 0), b = (1, 0), c = 0.8*dir(40), d = 0.7*dir(80), e = c+d, p = 0.9*dir(10), o = (b+c)/2; label("A'",a,S); label("B'",b,S); label("C'",c,W); label("E'",e,N); label("P'",p,S); label("O'",o,W); draw(a--b--(b+c)); draw(a--c--(b+c), dotted); draw(shift(d)*(a--b--(b+c)--c--cycle)); draw(a--d); draw(b--(b+d)); draw(c--(c+d), dotted); draw((b+c)--(b+c+d)); draw(e--p, dotted); draw(c--b, dotted); draw(a--(b+c), dotted); draw(p--a, dotted); draw(e--a, dotted); [/asy]
For the other parallelepiped, using the same approach and drop a perpendicular from $E'$ onto the base at $P'$ . Similarly applying Pythagorean theorem to $\triangle E'P'C'$ $\triangle E'P'A'$ and $\triangle A'P'O'$ we have \[C'E'^2 - C'P'^2 = A'E'^2 - A'O'^2 - (C'P'-C'O')^2.\]
Plugging known values into the above equation and solve for the only unknown variable $C'P'$ , we get $C'P'=\frac{5}{\sqrt{21}}.$
Thus the height \[E'P' = \sqrt{C'E'^2 - C'P'^2} = \sqrt{13 - \frac{25}{21}} = \sqrt{\frac{248}{21}}.\]
The ratio between the two is therefore \[\frac{DP}{E'P'} = \frac{\sqrt{\frac{378}{31}}}{\sqrt{\frac{248}{21}}} = \sqrt{\frac{2\cdot3^3\cdot7}{31}\cdot\frac{3\cdot7}{2^3\cdot31}} = \frac{3^2\cdot7}{2\cdot31} = \frac{63}{62}\] , giving $\boxed{125}$ | null | 125 |
564234f8c645a8cec1238954374fbbea | https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_14 | The following analog clock has two hands that can move independently of each other. [asy] unitsize(2cm); draw(unitcircle,black+linewidth(2)); for (int i = 0; i < 12; ++i) { draw(0.9*dir(30*i)--dir(30*i)); } for (int i = 0; i < 4; ++i) { draw(0.85*dir(90*i)--dir(90*i),black+linewidth(2)); } for (int i = 1; i < 13; ++i) { label("\small" + (string) i, dir(90 - i * 30) * 0.75); } draw((0,0)--0.6*dir(90),black+linewidth(2),Arrow(TeXHead,2bp)); draw((0,0)--0.4*dir(90),black+linewidth(2),Arrow(TeXHead,2bp)); [/asy] Initially, both hands point to the number $12$ . The clock performs a sequence of hand movements so that on each movement, one of the two hands moves clockwise to the next number on the clock face while the other hand does not move.
Let $N$ be the number of sequences of $144$ hand movements such that during the sequence, every possible positioning of the hands appears exactly once, and at the end of the $144$ movements, the hands have returned to their initial position. Find the remainder when $N$ is divided by $1000$ | This problem is, in essence, the following: A $12\times12$ coordinate grid is placed on a (flat) torus; how many loops are there that pass through each point while only moving up or right? In other words, Felix the frog starts his journey at $(0,0)$ in the coordinate plane. Every second, he jumps either to the right or up, until he reaches an $x$ - or $y$ -coordinate of $12$ . At this point, if he tries to jump to a coordinate outside the square from $(0,0)$ to $(11,11)$ , he "wraps around" and ends up at an $x$ - or $y$ - coordinate of $0$ . How many ways are there for Felix to jump on every grid point in this square, so that he ends at $(0,0)$ ? This is consistent with the construction of the flat torus as $\mathbb Z^2/12\mathbb Z^2$ (2-dimensional modular arithmetic. $(\mathbb{Z}_{12})^2$
Moving on, define a $\textit{path}$ from point $A$ to point $B$ to be a sequence of "up"s and "right"s that takes Felix from $A$ to $B$ . The $\textit{distance}$ from $A$ to $B$ is the length of the shortest path from $A$ to $B$ . At the crux of this problem is the following consideration: The points $A_i=(i,12-i), i\in{0,...,11}$ are pairwise equidistant, each pair having distance of $12$ in both directions.
[asy] size(7cm); for (int x=0; x<12; ++x){ for (int y=0; y<12; ++y){ fill(circle((x,y),0.05));}} for (int i=0; i<12; ++i){ fill(circle((i,11-i),0.1),red);} pen p=green+dashed; path u=(3,8)--(4,8)--(4,9)--(4,10)--(4,11)--(5,11)--(5,11.5); path v=(5,-0.5)--(5,0)--(5,1)--(6,1)--(6,2)--(6,3)--(6,4)--(7,4); draw(u,p); draw(v,p); pen p=blue+dashed; path u=(4,7)--(5,7)--(5,8)--(5,9)--(5,10)--(6,10)--(6,11)--(6,11.5); path v=(6,-0.5)--(6,0)--(7,0)--(7,1)--(7,2)--(7,3)--(8,3); draw(u,p); draw(v,p); [/asy]
A valid complete path then joins two $A_i$ 's, say $A_i$ and $A_j$ . In fact, a link between some $A_i$ and $A_j$ fully determines the rest of the cycle, as the path from $A_{i+1}$ must "hug" the path from $A_i$ , to ensure that there are no gaps. We therefore see that if $A_0$ leads to $A_k$ , then $A_i$ leads to $A_{i+k}$ . Only the values of $k$ relatively prime to $12$ result in solutions, though, because otherwise $A_0$ would only lead to $\{A_i:\exists n\in \mathbb Z:i\equiv kn\quad\text{mod 12}\}$ . The number of paths from $A_0$ to $A_k$ is ${12\choose k}$ , and so the answer is
\[{12\choose1}+{12\choose5}+{12\choose7}+{12\choose11}=1\boxed{608}.\] | null | 608 |
564234f8c645a8cec1238954374fbbea | https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_14 | The following analog clock has two hands that can move independently of each other. [asy] unitsize(2cm); draw(unitcircle,black+linewidth(2)); for (int i = 0; i < 12; ++i) { draw(0.9*dir(30*i)--dir(30*i)); } for (int i = 0; i < 4; ++i) { draw(0.85*dir(90*i)--dir(90*i),black+linewidth(2)); } for (int i = 1; i < 13; ++i) { label("\small" + (string) i, dir(90 - i * 30) * 0.75); } draw((0,0)--0.6*dir(90),black+linewidth(2),Arrow(TeXHead,2bp)); draw((0,0)--0.4*dir(90),black+linewidth(2),Arrow(TeXHead,2bp)); [/asy] Initially, both hands point to the number $12$ . The clock performs a sequence of hand movements so that on each movement, one of the two hands moves clockwise to the next number on the clock face while the other hand does not move.
Let $N$ be the number of sequences of $144$ hand movements such that during the sequence, every possible positioning of the hands appears exactly once, and at the end of the $144$ movements, the hands have returned to their initial position. Find the remainder when $N$ is divided by $1000$ | This is more of a solution sketch and lacks rigorous proof for interim steps, but illustrates some key observations that lead to a simple solution.
Note that one can visualize this problem as walking on a $N \times N$ grid where the edges warp. Your goal is to have a single path across all nodes on the grid leading back to $(0,\ 0)$ . For convenience, any grid position are presumed to be in $\mod N$
Note that there are exactly two ways to reach node $(i,\ j)$ , namely $(i - 1,\ j)$ and $(i,\ j - 1)$
As a result, if a path includes a step from $(i,\ j)$ to $(i + 1,\ j)$ , there cannot be a step from $(i,\ j)$ to $(i,\ j + 1)$ . However, a valid solution must reach $(i,\ j + 1)$ , and the only valid step is from $(i - 1,\ j + 1)$
So a solution that includes a step from $(i,\ j)$ to $(i + 1,\ j)$ dictates a step from $(i - 1,\ j + 1)$ to $(i,\ j + 1)$ and by extension steps from $(i - a,\ j + a)$ to $(i - a + 1,\ j + a)$ . We observe the equivalent result for steps in the orthogonal direction.
This means that in constructing a valid solution, taking one step in fact dictates N steps, thus it's sufficient to count valid solutions with $N = a + b$ moves of going right $a$ times and $b$ times up the grid. The number of distinct solutions can be computed by permuting 2 kinds of indistinguishable objects $\binom{N}{a}$
Here we observe, without proof, that if $\gcd(a, b) \neq 1$ , then we will return to the origin prematurely. For $N = 12$ , we only want to count the number of solutions associated with $12 = 1 + 11 = 5 + 7 = 7 + 5 = 11 + 1$
(For those attempting a rigorous proof, note that $\gcd(a, b) = \gcd(a + b, b) = \gcd(N, b) = \gcd(N, a)$ ).
The total number of solutions, noting symmetry, is thus
\[2\cdot\left(\binom{12}{1} + \binom{12}{5}\right) = 1608\]
This yields $\boxed{608}$ as our desired answer. | null | 608 |
564234f8c645a8cec1238954374fbbea | https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_14 | The following analog clock has two hands that can move independently of each other. [asy] unitsize(2cm); draw(unitcircle,black+linewidth(2)); for (int i = 0; i < 12; ++i) { draw(0.9*dir(30*i)--dir(30*i)); } for (int i = 0; i < 4; ++i) { draw(0.85*dir(90*i)--dir(90*i),black+linewidth(2)); } for (int i = 1; i < 13; ++i) { label("\small" + (string) i, dir(90 - i * 30) * 0.75); } draw((0,0)--0.6*dir(90),black+linewidth(2),Arrow(TeXHead,2bp)); draw((0,0)--0.4*dir(90),black+linewidth(2),Arrow(TeXHead,2bp)); [/asy] Initially, both hands point to the number $12$ . The clock performs a sequence of hand movements so that on each movement, one of the two hands moves clockwise to the next number on the clock face while the other hand does not move.
Let $N$ be the number of sequences of $144$ hand movements such that during the sequence, every possible positioning of the hands appears exactly once, and at the end of the $144$ movements, the hands have returned to their initial position. Find the remainder when $N$ is divided by $1000$ | Define a $12 \times 12$ matrix $X$ .
Each entry $x_{i, j}$ denotes the number of movements the longer hand moves, given that two hands jointly make $12 \left( i - 1 \right) + \left( j - 1 \right)$ movements.
Thus, the number of movements the shorter hand moves is $12 \left( i - 1 \right) + \left( j - 1 \right) - x_{i, j}$
Denote by $r_{i, j}$ the remainder of $x_{i, j}$ divided by 12.
Denote by $R$ this remainder matrix.
If two hands can return to their initial positions after 144 movements, then $r_{12, 12} = 0$ or 11.
Denote by $S_0$ (resp. $S_{11}$ ) the collection of feasible sequences of movements, such that $r_{12, 12} = 0$ (resp. $r_{12, 12} = 11$ ).
Define a function $f : S_0 \rightarrow S_{11}$ , such that for any $\left\{ x_{i,j} , \ \forall \ i, j \in \left\{ 1, 2, \cdots , 12 \right\} \right\} \in S_0$ , the functional value of the entry indexed as $\left( i, j \right)$ is $12 \left( i - 1 \right) + \left( j - 1 \right) - x_{i, j}$ .
Thus, function $f$ is bijective. This implies $| S_0 | = | S_{11} |$
In the rest of analysis, we count $| S_0 |$
We make the following observations:
\begin{enumerate}
\item $x_{1, 1} = 0$ and $12 | x_{12, 12}$
These follow from the definition of $S_0$
\item Each column of $R$ is a permutation of $\left\{ 0, 1, \cdots , 11 \right\}$
The reasoning is as follows. Suppose there exist $i < i'$ $j$ , such that $r_{i, j} = r_{i', j}$ . Then this entails that the positions of two hands after the $\left( 12 \left( i' - 1 \right) + \left( j - 1 \right) \right)$ th movement coincide with their positions after the $\left( 12 \left( i - 1 \right) + \left( j - 1 \right) \right)$ th movement.
\item For any $j \in \left\{ 1, 2 ,\cdots , 11 \right\}$ $x_{i, j+1} - x_{i, j}$ is equal to either 0 for all $i$ or 1 for all $i$
The reasoning is as follows. If this does not hold and the $j$ th column in $R$ is a permutation of $\left\{ 0, 1, \cdots , 12 \right\}$ , then the $j+1$ th column is no longer a permutation of $\left\{ 0, 1, \cdots , 12 \right\}$ . This leads to the infeasibility of the movements.
\item $x_{i+1, 1} = x_{i, 12}$ for any $i \in \left\{ 1, 2, \cdots , 11 \right\}$
This follows from the conditions that the $12$ th column in $R$ excluding $r_{12, 12}$ and the first column in $R$ excluding $x_{1, 1}$ are both permutations of $\left\{ 1, 2, \cdots , 11 \right\}$
\end{enumerate}
All observations jointly imply that $x_{i, 12} = i \cdot x_{1, 12}$ .
Thus, $\left\{ r_{1, 12}, r_{2, 12} , \cdots , r_{11, 12} \right\}$ is a permutation of $\left\{ 1, 2, \cdots , 11 \right\}$ .
Thus, $x_{1, 12}$ is relatively prime to 12.
Because $x_{1, 1} = 0$ and $x_{1, 12} - x_{1, 1} \leq 11$ , we have $x_{1, 12} = 1$ , 5, 7, or 11.
Recall that when we move from $x_{1, 1}$ to $x_{1, 12}$ , there are 11 steps of movements. Each movement has $x_{1, j+1} - x_{i, j} = 0$ or 1.
Thus, for each given $x_{1, 12}$ , the number of feasible movements from $x_{1, 1}$ to $x_{1, 12}$ is $\binom{11}{x_{1, 12}}$
Therefore, the total number of feasible movement sequences in this problem is \begin{align*} | S_0 | + | S_{11} | & = 2 | S_0 | \\ & = 2 \cdot \sum_{x_{1, 12} = 1, 5, 7, 11} \binom{11}{x_{1, 12}} \\ & = 2 \left( 11 + 462 + 330 + 1 \right) \\ & = 1608 . \end{align*}
Therefore, the answer is $\boxed{608}$ | null | 608 |
10995028a4036dbe2836069ac9f62ae0 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_15 | Find the largest prime number $p<1000$ for which there exists a complex number $z$ satisfying | Assume that $z=a+bi$ . Then, \[z^3=(a^3-3ab^2)+(3a^2b-b^3)i\] Note that by the Triangle Inequality, \[|(a^3-3ab^2)-(3a^2b-b^3)|<p\implies |a^3+b^3-3ab^2-3a^2b|<a^2+b^2\] Thus, we know \[|a+b||a^2+b^2-4ab|<a^2+b^2\] Without loss of generality, assume $a>b$ (as otherwise, consider $i^3\overline z=b+ai$ ). If $|a/b|\geq 4$ , then \[17b^2\geq a^2+b^2>|a+b||a^2+b^2-4ab|\geq |b-4b||16b^2-16b^2+b^2|=3b^3\] `Thus, this means $b\leq\frac{17}3$ or $b\leq 5$ . Also note that the roots of $x^2-4x+1$ are $2\pm\sqrt 3$ , so thus if $b\geq 6$ \[2\sqrt 3b=(2(2-\sqrt 3)-4)b<a<4b\] Note that \[1000>p=a^2+b^2\geq 12b^2+b^2=13b^2\] so $b^2<81$ , and $b<9$ . If $b=8$ , then $16\sqrt 3\leq a\leq 32$ . Note that $\gcd(a,b)=1$ , and $a\not\equiv b\pmod 2$ , so $a=29$ or $31$ . However, then $5\mid a^2+b^2$ , absurd.
If $b=7$ , by similar logic, we have that $14\sqrt 3 <a< 28$ , so $b=26$ . However, once again, $5\mid a^2+b^2$ . If $b=6$ , by the same logic, $12\sqrt3<a<24$ , so $a=23$ , where we run into the same problem. Thus $b\leq 5$ indeed.
If $b=5$ , note that \[(a+5)(a^2+25-20a)<a^2+25\implies a<20\] We note that $p=5^2+18^2=349$ works. Thus, we just need to make sure that if $b\leq 4$ $a\leq 18$ . But this is easy, as \[p>(a+b)(a^2+b^2-4ab)\geq (4+18)(4^2+18^2-4\cdot 4\cdot 18)>1000\] absurd. Thus, the answer is $\boxed{349}$ | null | 349 |
10995028a4036dbe2836069ac9f62ae0 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_15 | Find the largest prime number $p<1000$ for which there exists a complex number $z$ satisfying | Denote $z = a + i b$ . Thus, $a^2 + b^2 = p$
Thus, \[z^3 = a \left( a^2 - 3 b^2 \right) + i b \left( - b^2 + 3 a^2 \right) .\]
Because $p$ ${\rm Re} \left( z^3 \right)$ ${\rm Im} \left( z^3 \right)$ are three sides of a triangle, we have ${\rm Re} \left( z^3 \right) > 0$ and ${\rm Im} \left( z^3 \right) > 0$ .
Thus, \begin{align*} a \left( a^2 - 3 b^2 \right) & > 0 , \hspace{1cm} (1) \\ b \left( - b^2 + 3 a^2 \right) & > 0. \hspace{1cm} (2) \end{align*}
Because $p$ ${\rm Re} \left( z^3 \right)$ ${\rm Im} \left( z^3 \right)$ are three sides of a triangle, we have the following triangle inequalities: \begin{align*} {\rm Re} \left( z^3 \right) + {\rm Im} \left( z^3 \right) & > p \hspace{1cm} (3) \\ p + {\rm Re} \left( z^3 \right) & > {\rm Im} \left( z^3 \right) \hspace{1cm} (4) \\ p + {\rm Im} \left( z^3 \right) & > {\rm Re} \left( z^3 \right) \hspace{1cm} (5) \end{align*}
We notice that $| z^3 | = p^{3/2}$ , and ${\rm Re} \left( z^3 \right)$ ${\rm Im} \left( z^3 \right)$ , and $| z^3 |$ form a right triangle. Thus, ${\rm Re} z^3 + {\rm Im} z^3 > p^{3/2}$ .
Because $p > 1$ $p^{3/2} > p$ .
Therefore, (3) holds.
Conditions (4) and (5) can be written in the joint form as \[\left| {\rm Re} \left( z^3 \right) - {\rm Im} \left( z^3 \right) \right| < p . \hspace{1cm} (4)\]
We have \begin{align*} {\rm Re} \left( z^3 \right) - {\rm Im} \left( z^3 \right) & = \left( a^3 - 3 a b^2 \right) - \left( - b^3 + 3 a^2 b \right) \\ & = \left( a + b \right) \left( a^2 - 4 ab + b^2 \right) \end{align*} and $p = a^2 + b^2$
Thus, (5) can be written as \[\left| \left( a + b \right) \left( a^2 - 4 ab + b^2 \right) \right| < a^2 + b^2 . \hspace{1cm} (6)\]
Therefore, we need to jointly solve (1), (2), (6).
From (1) and (2), we have either $a, b >0$ , or $a, b < 0$ .
In (6), by symmetry, without loss of generality, we assume $a, b > 0$
Thus, (1) and (2) are reduced to \[a > \sqrt{3} b . \hspace{1cm} (7)\]
Let $a = \lambda b$ . Plugging this into (6), we get \begin{align*} \left| \left( \left( \lambda - 2 \right)^2 - 3 \right) \right| < \frac{1}{b} \frac{\lambda^2 + 1}{\lambda + 1} . \hspace{1cm} (8) \end{align*}
Because $p= a^2 + b^2$ is a prime, $a$ and $b$ are relatively prime.
Therefore, we can use (7), (8), $a^2 + b^2 <1000$ , and $a$ and $b$ are relatively prime to solve the problem.
To facilitate efficient search, we apply the following criteria:
\begin{enumerate*}
\item To satisfy (7) and $a^2 + b^2 < 1000$ , we have $1 \leq b \leq 15$ .
In the outer layer, we search for $b$ in a decreasing order.
In the inner layer, for each given $b$ , we search for $a$ .
\item Given $b$ , we search for $a$ in the range $\sqrt{3} b < a < \sqrt{1000 - b^2}$ .
\item We can prove that for $b \geq 9$ , there is no feasible $a$ .
The proof is as follows.
For $b \geq 9$ , to satisfy $a^2 + b^2 < 1000$ , we have $a \leq 30$ .
Thus, $\sqrt{3} < \lambda \leq \frac{30}{9}$ .
Thus, the R.H.S. of (8) has the following upper bound \begin{align*} \frac{1}{b} \frac{\lambda^2 + 1}{\lambda + 1} & < \frac{1}{b} \frac{\lambda^2 + \lambda}{\lambda + 1} \\ & = \frac{\lambda}{b} \\ & \leq \frac{\frac{30}{9}}{9} \\ & < \frac{10}{27} . \end{align*}
Hence, to satisfy (8), a necessary condition is \begin{align*} \left| \left( \left( \lambda - 2 \right)^2 - 3 \right) \right| < \frac{10}{27} . \end{align*}
However, this cannot be satisfied for $\sqrt{3} < \lambda \leq \frac{30}{9}$ .
Therefore, there is no feasible solution for $b \geq 9$ .
Therefore, we only need to consider $b \leq 8$
\item We eliminate $a$ that are not relatively prime to $b$
\item We use the following criteria to quickly eliminate $a$ that make $a^2 + b^2$ a composite number.
\item For the remaining $\left( b, a \right)$ , check whether (8) and the condition that $a^2 + b^2$ is prime are both satisfied.
The first feasible solution is $b = 5$ and $a = 18$ .
Thus, $a^2 + b^2 = 349$
\item For the remaining search, given $b$ , we only search for $a \geq \sqrt{349 - b^2}$
Following the above search criteria, we find the final answer as $b = 5$ and $a = 18$ .
Thus, the largest prime $p$ is $p = a^2 + b^2 = \boxed{349}$ | null | 349 |
588106cf2f1937bc0a0b960ac3895513 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_II_Problems/Problem_1 | The numbers of apples growing on each of six apple trees form an arithmetic sequence where the greatest number of apples growing on any of the six trees is double the least number of apples growing on any of the six trees. The total number of apples growing on all six trees is $990.$ Find the greatest number of apples growing on any of the six trees. | In the arithmetic sequence, let $a$ be the first term and $d$ be the common difference, where $d>0.$ The sum of the first six terms is \[a+(a+d)+(a+2d)+(a+3d)+(a+4d)+(a+5d) = 6a+15d.\] We are given that \begin{align*} 6a+15d &= 990, \\ 2a &= a+5d. \end{align*} The second equation implies that $a=5d.$ Substituting this into the first equation, we get \begin{align*} 6(5d)+15d &=990, \\ 45d &= 990 \\ d &= 22. \end{align*} It follows that $a=110.$ Therefore, the greatest number of apples growing on any of the six trees is $a+5d=\boxed{220}.$ | null | 220 |
588106cf2f1937bc0a0b960ac3895513 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_II_Problems/Problem_1 | The numbers of apples growing on each of six apple trees form an arithmetic sequence where the greatest number of apples growing on any of the six trees is double the least number of apples growing on any of the six trees. The total number of apples growing on all six trees is $990.$ Find the greatest number of apples growing on any of the six trees. | Let the terms in the sequence be defined as \[a_1, a_2, ..., a_6.\]
Since this is an arithmetic sequence, we have $a_1+a_6=a_2+a_5=a_3+a_4.$ So, \[\sum_{i=1}^6 a_i=3(a_1+a_6)=990.\] Hence, $(a_1+a_6)=330.$ And, since we are given that $a_6=2a_1,$ we get $3a_1=330\implies a_1=110$ and $a_6=\boxed{220}.$ | null | 220 |
fe9ec693a1b5155f9771e0364c9600c0 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_II_Problems/Problem_3 | Let $\triangle ABC$ be an isosceles triangle with $\angle A = 90^\circ.$ There exists a point $P$ inside $\triangle ABC$ such that $\angle PAB = \angle PBC = \angle PCA$ and $AP = 10.$ Find the area of $\triangle ABC.$ | This solution refers to the Diagram section.
Let $\angle PAB = \angle PBC = \angle PCA = \theta,$ from which $\angle PAC = 90^\circ-\theta,$ and $\angle APC = 90^\circ.$
Moreover, we have $\angle PBA = \angle PCB = 45^\circ-\theta,$ as shown below: [asy] /* Made by MRENTHUSIASM */ size(250); pair A, B, C, P; A = origin; B = (0,10*sqrt(5)); C = (10*sqrt(5),0); P = intersectionpoints(Circle(A,10),Circle(C,20))[0]; dot("$A$",A,1.5*SW,linewidth(4)); dot("$B$",B,1.5*NW,linewidth(4)); dot("$C$",C,1.5*SE,linewidth(4)); dot("$P$",P,1.5*NE,linewidth(4)); markscalefactor=0.125; draw(rightanglemark(B,A,C,10),red); draw(rightanglemark(A,P,C,10),red); draw(anglemark(P,A,B,25),red); draw(anglemark(P,B,C,25),red); draw(anglemark(P,C,A,25),red); draw(anglemark(A,B,P,25),green); draw(anglemark(B,C,P,25),green); draw(anglemark(C,A,P,25),green); add(pathticks(anglemark(P,A,B,25), n = 1, r = 0.1, s = 10, red)); add(pathticks(anglemark(P,B,C,25), n = 1, r = 0.1, s = 10, red)); add(pathticks(anglemark(P,C,A,25), n = 1, r = 0.1, s = 10, red)); draw(A--B--C--cycle^^P--A^^P--B^^P--C); label("$10$",midpoint(A--P),dir(-30),blue); label("$\theta$",A,9.5*dir(76),red); label("$\theta$",C,9.5*dir(168),red); label("$\theta$",B,9*dir(305),red); label("$45^\circ-\theta$",B,6*dir(235),green); label("$45^\circ-\theta$",C,6*dir(85),green); label("$90^\circ-\theta$",A,2*dir(-40),green); [/asy] Note that $\triangle PAB \sim \triangle PBC$ by AA Similarity. The ratio of similitude is $\frac{PA}{PB} = \frac{PB}{PC} = \frac{AB}{BC},$ so $\frac{10}{PB} = \frac{1}{\sqrt2}$ and thus $PB=10\sqrt2.$ Similarly, we can figure out that $PC=20$
Finally, $AC=\sqrt{10^2+20^2}=10\sqrt{5}$ , so the area of $\triangle ABC$ is \[\frac12\cdot AB\cdot AC = \frac12\cdot (10\sqrt{5})^2 = \boxed{250}.\] | null | 250 |
fe9ec693a1b5155f9771e0364c9600c0 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_II_Problems/Problem_3 | Let $\triangle ABC$ be an isosceles triangle with $\angle A = 90^\circ.$ There exists a point $P$ inside $\triangle ABC$ such that $\angle PAB = \angle PBC = \angle PCA$ and $AP = 10.$ Find the area of $\triangle ABC.$ | Since the triangle is a right isosceles triangle, $\angle B = \angle C = 45^\circ$
Let the common angle be $\theta$ . Note that $\angle PAC = 90^\circ-\theta$ , thus $\angle APC = 90^\circ$ . From there, we know that $AC = \frac{10}{\sin\theta}$
Note that $\angle ABP = 45^\circ-\theta$ , so from law of sines we have \[\frac{10}{\sin\theta \cdot \frac{\sqrt{2}}{2}}=\frac{10}{\sin(45^\circ-\theta)}.\] Dividing by $10$ and multiplying across yields \[\sqrt{2}\sin(45^\circ-\theta)=\sin\theta.\] From here use the sine subtraction formula, and solve for $\sin\theta$ \begin{align*} \cos\theta-\sin\theta&=\sin\theta \\ 2\sin\theta&=\cos\theta \\ 4\sin^2\theta&=\cos^2\theta \\ 4\sin^2\theta&=1-\sin^2\theta \\ 5\sin^2\theta&=1 \\ \sin\theta&=\frac{1}{\sqrt{5}}. \end{align*} Substitute this to find that $AC=10\sqrt{5}$ , thus the area is $\frac{(10\sqrt{5})^2}{2}=\boxed{250}$ | null | 250 |
fe9ec693a1b5155f9771e0364c9600c0 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_II_Problems/Problem_3 | Let $\triangle ABC$ be an isosceles triangle with $\angle A = 90^\circ.$ There exists a point $P$ inside $\triangle ABC$ such that $\angle PAB = \angle PBC = \angle PCA$ and $AP = 10.$ Find the area of $\triangle ABC.$ | Since the triangle is a right isosceles triangle, $\angle B = \angle C = 45^\circ$
Do some angle chasing yielding:
We have $AC=\frac{10}{\sin\theta}$ since $\triangle APC$ is a right triangle. Since $\triangle ABC$ is a $45^\circ$ $45^\circ$ $90^\circ$ triangle, $AB=\frac{10}{\sin\theta}$ , and $BC=\frac{10\sqrt{2}}{\sin\theta}$
Note that $\triangle APB \sim \triangle BPC$ by a factor of $\sqrt{2}$ . Thus, $BP = 10\sqrt{2}$ , and $PC = 20$
From Pythagorean theorem, $AC=10\sqrt{5}$ so the area of $\triangle ABC$ is $\frac{(10\sqrt{5})^2}{2}=\boxed{250}$ | null | 250 |
fe9ec693a1b5155f9771e0364c9600c0 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_II_Problems/Problem_3 | Let $\triangle ABC$ be an isosceles triangle with $\angle A = 90^\circ.$ There exists a point $P$ inside $\triangle ABC$ such that $\angle PAB = \angle PBC = \angle PCA$ and $AP = 10.$ Find the area of $\triangle ABC.$ | Since the triangle is a right isosceles triangle, $\angle B = \angle C = 45^\circ$
Notice that in triangle $PBC$ $\angle PBC + 45-\angle PCA = 45^\circ$ , so $\angle BPC = 135^\circ$ . Similar logic shows $\angle APC = 135^\circ$
Now, we see that $\triangle APB \sim \triangle BPC$ with ratio $1:\sqrt{2}$ (as $\triangle ABC$ is a $45^\circ$ $45^\circ$ $90^\circ$ triangle). Hence, $\overline{PB}=10\sqrt{2}$ . We use the Law of Cosines to find $AB$ \begin{align*} AB^2&=BP^2+AP^2-2ab\cos(APB) \\ &=100+200-2(10)(10\sqrt{2}\cos(135^\circ)) \\ &=300+200\cdot\sqrt{2}\cdot\frac{1}{\sqrt{2}} \\ &=500. \end{align*} Since $\triangle ABC$ is a right triangle, the area is $\frac{AB^2}{2}=\frac{500}{2}=\boxed{250}$ | null | 250 |
fe9ec693a1b5155f9771e0364c9600c0 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_II_Problems/Problem_3 | Let $\triangle ABC$ be an isosceles triangle with $\angle A = 90^\circ.$ There exists a point $P$ inside $\triangle ABC$ such that $\angle PAB = \angle PBC = \angle PCA$ and $AP = 10.$ Find the area of $\triangle ABC.$ | Denote the area of $X$ by $[X].$ As in previous solutions, we see that $\angle APC = 90 ^\circ, \triangle BPC \sim \triangle APB$ with ratio $k = \sqrt{2}\implies$ \[\frac {PC}{PB} = \frac {PB}{PA} = k \implies PC = k^2 \cdot AP = 20 \implies [APC] = \frac {AP \cdot PC}{2} = 100.\] \[[BPC] = k^2 [APB] = 2 [APB].\] \[AB = BC, \angle PCA = \angle PAB \implies \frac {[APC]}{[APB]} = \frac {PC}{PA} = 2 \implies\] \[[ABC] = [APB] + [APC] + [BPC] = [APC] \cdot (\frac {1}{2} + 1 + 2 \cdot \frac {1}{2}) = \frac {5}{2} \cdot [APC] = \boxed{250}.\] [email protected], vvsss | null | 250 |
fe9ec693a1b5155f9771e0364c9600c0 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_II_Problems/Problem_3 | Let $\triangle ABC$ be an isosceles triangle with $\angle A = 90^\circ.$ There exists a point $P$ inside $\triangle ABC$ such that $\angle PAB = \angle PBC = \angle PCA$ and $AP = 10.$ Find the area of $\triangle ABC.$ | Denote $\angle PCA = \theta$ . Then, by trig Ceva's: \begin{align*} \frac{\sin^3(\theta)}{\sin(90-\theta) \cdot \left(\sin(45-\theta)\right)^2} &= 1 \\ \sin^3(\theta) &= \cos(\theta) \cdot \left(\sin(45) \cos(\theta) - \cos(45) \sin(\theta)\right)^2 \\ 2\sin^3(\theta) &= \cos(\theta) \cdot \left(\cos(\theta) - \sin(\theta)\right)^2 \\ 2\sin^2(\theta) &= \cot(\theta) \cdot \left(1 - 2\sin(\theta)\cos(\theta)\right) \\ 2\sin^2(\theta) &= \cot(\theta) - 2\cos^2(\theta) \\ \cot(\theta) &= 2 \\ \sin(\theta) &= \frac{\sqrt{5}}{5}. \end{align*} Note that $\angle APC$ is a right angle. Therefore: \begin{align*} \sin(\theta) &= \frac{AP}{AC} \\ AC &= \frac{10}{\frac{\sqrt{5}}{5}} \\ &= 10\sqrt{5} \\ |ABC| &= \frac{AC^2}{2} \\ &= \boxed{250} ~ConcaveTriangle | null | 250 |
fe9ec693a1b5155f9771e0364c9600c0 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_II_Problems/Problem_3 | Let $\triangle ABC$ be an isosceles triangle with $\angle A = 90^\circ.$ There exists a point $P$ inside $\triangle ABC$ such that $\angle PAB = \angle PBC = \angle PCA$ and $AP = 10.$ Find the area of $\triangle ABC.$ | Notice that point $P$ is one of the two Brocard Points of $\triangle ABC$ . (The angle equalities given in the problem are equivalent to the definition of a Brocard point.) By the Brocard point formula, \begin{align*} \cot(\phi) = \cot(A)+\cot(B)+\cot(C) \end{align*} , where $\phi$ is equal to $\angle PAB$ .(This is also called the Brocard angle of triangle ABC). Because the triangle is an isosceles right triangle, the cotangents are easy to compute: \begin{align*}\cot(\phi) = 0 + 1 + 1 \\ \cot(\phi) = 2\end{align*} By definition, $\cot(\phi) = \frac{\cos(\phi)}{\sin(\phi)}$ . By the Pythagorean identity, $\cos(\phi)=\frac{2\sqrt{5}}{5}$ and $\sin(\phi) = \cos(\phi)=\frac{\sqrt{5}}{5}$ . Consider triangle $APB$ . By the problem condition, $\angle PBA = 45-\phi$ , so $\angle BPA = 135^{\circ}$ \begin{align*}\sin{45-\theta} = \sin{45}\cos{\phi}-\cos{45}\sin{\phi} = \frac{\sqrt{10}}{10}\end{align*} Now, we can use the Law of Sines. \begin{align*} \frac{AP}{\sin{45-\theta}}&=\frac{AB}{\sin{135}} \\ 10 \sqrt{10} &= \sqrt{2} AB \\ AB &= 10 \sqrt{5} \end{align*} Therefore, the answer is \[[ABC] = \frac{1}{2} (AB)^2 = \boxed{250}.\] ~ewei12 | null | 250 |
5bcc35f7680198395166038bc199705b | https://artofproblemsolving.com/wiki/index.php/2023_AIME_II_Problems/Problem_4 | Let $x,y,$ and $z$ be real numbers satisfying the system of equations \begin{align*} xy + 4z &= 60 \\ yz + 4x &= 60 \\ zx + 4y &= 60. \end{align*} Let $S$ be the set of possible values of $x.$ Find the sum of the squares of the elements of $S.$ | We first subtract the second equation from the first, noting that they both equal $60$ \begin{align*} xy+4z-yz-4x&=0 \\ 4(z-x)-y(z-x)&=0 \\ (z-x)(4-y)&=0 \end{align*}
Case 1: Let $y=4$
The first and third equations simplify to: \begin{align*} x+z&=15 \\ xz&=44 \end{align*} from which it is apparent that $x=4$ and $x=11$ are solutions.
Case 2: Let $x=z$
The first and third equations simplify to: \begin{align*} xy+4x&=60 \\ x^2+4y&=60 \end{align*}
We subtract the following equations, yielding: \begin{align*} x^2+4y-xy-4x&=0 \\ x(x-4)-y(x-4)&=0 \\ (x-4)(x-y)&=0 \end{align*}
We thus have $x=4$ and $x=y$ , substituting in $x=y=z$ and solving yields $x=6$ and $x=-10$
Then, we just add the squares of the solutions (make sure not to double count the $4$ ), and get \[4^2+11^2+(-6)^2+10^2=16+121+36+100=\boxed{273}.\] ~SAHANWIJETUNGA | null | 273 |
5bcc35f7680198395166038bc199705b | https://artofproblemsolving.com/wiki/index.php/2023_AIME_II_Problems/Problem_4 | Let $x,y,$ and $z$ be real numbers satisfying the system of equations \begin{align*} xy + 4z &= 60 \\ yz + 4x &= 60 \\ zx + 4y &= 60. \end{align*} Let $S$ be the set of possible values of $x.$ Find the sum of the squares of the elements of $S.$ | We index these equations as (1), (2), and (3), respectively.
Taking $(1)-(2)$ , we get \[ \left( x - z \right) \left( y - 4 \right) = 0 . \]
Denote $x' = x - 4$ $y' = y - 4$ $z' = z - 4$ .
Thus, the above equation can be equivalently written as \[ \left( x' - z' \right) y' = 0 . \hspace{1cm} (1') \]
Similarly, by taking $(2)-(3)$ , we get \[ \left( y' - x' \right) z' = 0 . \hspace{1cm} (2') \]
By taking $(3) - (1)$ , we get \[ \left( z' - y' \right) x' = 0 . \hspace{1cm} (3') \]
From $(3')$ , we have the following two cases.
Case 1: $x' = 0$
Plugging this into $(1')$ and $(2')$ , we get $y'z' = 0$ .
Thus, $y' = 0$ or $z' = 0$
Because we only need to compute all possible values of $x$ , without loss of generality, we only need to analyze one case that $y' = 0$
Plugging $x' = 0$ and $y' = 0$ into (1), we get a feasible solution $x = 4$ $y = 4$ $z = 11$
Case 2: $x' \neq 0$ and $z' - y' = 0$
Plugging this into $(1')$ and $(2')$ , we get $\left( x' - y' \right) y' = 0$
Case 2.1: $y' = 0$
Thus, $z' = 0$ . Plugging $y' = 0$ and $z' = 0$ into (1), we get a feasible solution $x = 11$ $y = 4$ $z = 4$
Case 2.2: $y' \neq 0$ and $x' = y'$
Thus, $x' = y' = z'$ . Plugging these into (1), we get $\left( x, y, z \right) = \left( -10, -10, -10 \right)$ or $\left( 6, 6, 6 \right)$
Putting all cases together, $S = \left\{ 4, 11, -10, 6 \right\}$ .
Therefore, the sum of the squares of the elements of $S$ is \begin{align*} 4^2 + 11^2 + \left( -10 \right)^2 + 6^2 = \boxed{273} | null | 273 |
5bcc35f7680198395166038bc199705b | https://artofproblemsolving.com/wiki/index.php/2023_AIME_II_Problems/Problem_4 | Let $x,y,$ and $z$ be real numbers satisfying the system of equations \begin{align*} xy + 4z &= 60 \\ yz + 4x &= 60 \\ zx + 4y &= 60. \end{align*} Let $S$ be the set of possible values of $x.$ Find the sum of the squares of the elements of $S.$ | We index these equations as (1), (2), and (3), respectively.
Using equation (1), we get $z = \frac{60 - xy}{4} = 15 - \frac{xy}{4}$ We need to solve for x, so we plug this value of z into equation (3) to get:
\[15x - \frac{x^2y}{4} - 4y = 60\] \[\frac{y}{4} * x^2 - 15x + (60 + 4y) = 0\]
We use the quadratic formula to get possible values of x:
\[x = \frac{15 \pm \sqrt{15^2 - 4(\frac{y}{4})(60 + 4y)}}{\frac{y}{2}}\] \[x = \frac{30 \pm 2\sqrt{225 - 60y + 4y^2}}{y}\] \[x = \frac{30 \pm 2\sqrt{(2y-15)^2}}{y}\] \[x = \frac{30 \pm 2(2y - 15)}{y}\] \[x = \frac{30 \pm (4y - 30)}{y}\]
Here, we have two cases, (plus) and (minus)
In the plus case, we have: \[x = \frac{(30 + 4y - 30)}{y}\] \[x = 4\]
So, our first case gives us one value of x, which is 4.
In the minus case, we have: \[x = \frac{30 - (4y - 30)}{y}\] \[x = \frac{60 - 4y}{y}\] \[x = \frac{60}{y} - 4\]
For this case, we now have values of x in terms of y.
Plugging this expression for x in equation (1), we get \[60 - 4y + 4z = 60\] \[4z = 4y\] \[z = y\]
So we know that for this case, z = y.
Using this information in equation (2), we get \[y^2 + \frac{240}{y} - 16 = 60\] \[y^2 + \frac{240}{y} - 76 = 0\] Multiplying both sides by y, we get a cubic expression: \[y^3 + 0y^2 - 76y + 240 = 0\] Here we just have to figure out the values of y that make this equation true. I used Vieta's Formulas to get a possible list, but you could also use the rational root theorem and synthetic division to find these. We call the three values of y that solve this equation: $y_{1},y_{2},y_{3}$ Using Vieta's Formulas, you get these three expressions: \[y_{1} + y_{2} + y_{3} = 0\] \[y_{1} * y_{2} * y_{3} = -240\] \[(y_{1} * y_{2}) + (y_{2} * y_{3}) + (y_{1} * y_{3}) = -76\]
In addition, we know that $y \vert 60$ , because of our expression for x.
Since the three values of y multiply to a negative number but also add to 0, we know that one value is negative and the other two are positive, and that the absolute value of the negative value is greater than both of the positive values.
List of possible values for y are $\{-60,-30,-20,-15,-12,-10,-6,-5,-4,-3,-2,-1,1,2,3,4,5,6,10,12,15,20,30,60\}$ From a list of these values, the only values that work are $y_1 = -10, y_2 = 6, y_3 = 4$ because \[-10 + 6 + 4 = 0\] \[-10 * 6 * 4 = -240\] \[(-10 * 6) + (-10 * 4) + (6 * 4) = -60 - 40 + 24 = -76\]
Plugging in these values for y into our expression for x, we get: \[x = \frac{60}{-10} - 4 = -10\] \[x = \frac{60}{6} - 4 = 6\] \[x = \frac{60}{4} - 4 = 11\]
So, now we have accounted for both cases, and we have 4 values of x = $\{-10,4,6,11\}$ Squaring all these terms we get: 100 + 16 + 36 + 121 = 273, so our answer is $\boxed{273}$ | null | 273 |
5bcc35f7680198395166038bc199705b | https://artofproblemsolving.com/wiki/index.php/2023_AIME_II_Problems/Problem_4 | Let $x,y,$ and $z$ be real numbers satisfying the system of equations \begin{align*} xy + 4z &= 60 \\ yz + 4x &= 60 \\ zx + 4y &= 60. \end{align*} Let $S$ be the set of possible values of $x.$ Find the sum of the squares of the elements of $S.$ | Since all three equations are in the form $\frac{K}{a} + 4a = 60$ where $K = xyz$ , we can rearrange this to see that $x$ $y$ , and $z$ all satisfy
\[ 4a^2 - 60a + K = 0. \]
Let this quadratic have roots $a_1$ and $a_2$ . Then, there are two cases to consider: two of $x$ $y$ $z$ are equal to $a_1$ and the third is equal to $a_2$ , or all of $x$ $y$ $z$ are equal to $a_1$
Case 1: WLOG let $x = y = a_1$ and $z = a_2$
Then by Vieta's,
\[x + z = 15 \hspace{1cm} (1)\] \[xz = \frac{x^2z}{4} \hspace{1cm} (2)\]
which gives $x = y = 4$ and $z = 11$ . But we can swap $x$ $y$ , and $z$ however we like, so this also gives $x = 11$ as a solution. In total, this case yields $x = 4$ and $x = 11$ as possible values of $x$
Case 2: WLOG let $x = y = z = a_1$
Again, by Vieta's,
\[x + a_2 = 15 \hspace{1cm} (1)\] \[xa_2 = \frac{x^3}{4} \hspace{1cm} (2)\]
We can use $(2)$ to isolate $a_2$ in terms of $x$ , then plug that into $(1)$ to get that $x^2 + 4x - 60 = 0$ . This yields $x = -10$ and $x = 6$ as additional possible values of $x$
In all, $x$ can be any of $\{ 4, 11, -10, 6 \}$ , so the requested answer is $4^2 + 11^2 + (-10)^2 + 6^2 = \boxed{273}.$ | null | 273 |
5bcc35f7680198395166038bc199705b | https://artofproblemsolving.com/wiki/index.php/2023_AIME_II_Problems/Problem_4 | Let $x,y,$ and $z$ be real numbers satisfying the system of equations \begin{align*} xy + 4z &= 60 \\ yz + 4x &= 60 \\ zx + 4y &= 60. \end{align*} Let $S$ be the set of possible values of $x.$ Find the sum of the squares of the elements of $S.$ | We index these equations as (1), (2), and (3) respectively (same as solution 2). There are two possible cases:
Case 1: $x = \pm 4$
In this case, we simply plug in $x = 4$ and $x = -4$ . We note that $x=4$ is a valid case.
Case 2: $x \neq \pm 4$
In this case, using equation (3), we get $y = 15 - \frac{xz}{4}$ . Plugging that into equation (1), we get $z = \frac{240-60x}{16-x^2}$ . Plugging that expression back into the original expression for $y$ we obtain $y = 15 - \frac{60x-15x^2}{16-x^2}$ .
Now we plug these two expressions into equation (2): \[\left(15 - \frac{60x-15x^2}{16-x^2}\right)\left(\frac{240-60x}{16-x^2}\right) = 60-4x\] multiplying both sides by $(16-x^2)$ and factoring, we get: \[\left(15(4-x)(4+x)-60x+15x^2\right)\left(\frac{60(4-x)}{(4+x)(4-x)}\right) = 4(15 - x)(4 + x)(4 - x)\] which simplifies to: \[x^3 - 7x^2 - 104x + 660 = 0\] we note that $x = 6$ is a root. Factoring, we get the other roots, -10 and 11.
Our desired answer is the sum of the square of all these roots: \[4^2 + 6^2 + (-10)^2 + 11^2 = \boxed{273}\] | null | 273 |
ade59ef785ad8645236caab21029ffed | https://artofproblemsolving.com/wiki/index.php/2023_AIME_II_Problems/Problem_5 | Let $S$ be the set of all positive rational numbers $r$ such that when the two numbers $r$ and $55r$ are written as fractions in lowest terms, the sum of the numerator and denominator of one fraction is the same as the sum of the numerator and denominator of the other fraction. The sum of all the elements of $S$ can be expressed in the form $\frac{p}{q},$ where $p$ and $q$ are relatively prime positive integers. Find $p+q.$ | Denote $r = \frac{a}{b}$ , where $\left( a, b \right) = 1$ .
We have $55 r = \frac{55a}{b}$ .
Suppose $\left( 55, b \right) = 1$ , then the sum of the numerator and the denominator of $55r$ is $55a + b$ .
This cannot be equal to the sum of the numerator and the denominator of $r$ $a + b$ .
Therefore, $\left( 55, b \right) \neq 1$
Case 1: $b$ can be written as $5c$ with $\left( c, 11 \right) = 1$
Thus, $55r = \frac{11a}{c}$
Because the sum of the numerator and the denominator of $r$ and $55r$ are the same, \[ a + 5c = 11a + c . \]
Hence, $2c = 5 a$
Because $\left( a, b \right) = 1$ $\left( a, c \right) = 1$ .
Thus, $a = 2$ and $c = 5$ .
Therefore, $r = \frac{a}{5c} = \frac{2}{25}$
Case 2: $b$ can be written as $11d$ with $\left( d, 5 \right) = 1$
Thus, $55r = \frac{5a}{c}$
Because the sum of the numerator and the denominator of $r$ and $55r$ are the same, \[ a + 11c = 5a + c . \]
Hence, $2a = 5 c$
Because $\left( a, b \right) = 1$ $\left( a, c \right) = 1$ .
Thus, $a = 5$ and $c = 2$ .
Therefore, $r = \frac{a}{11c} = \frac{5}{22}$
Case 3: $b$ can be written as $55 e$
Thus, $55r = \frac{a}{c}$
Because the sum of the numerator and the denominator of $r$ and $55r$ are the same, \[ a + 55c = a + c . \]
Hence, $c = 0$ . This is infeasible.
Thus, there is no solution in this case.
Putting all cases together, $S = \left\{ \frac{2}{25}, \frac{5}{22} \right\}$ .
Therefore, the sum of all numbers in $S$ is \[ \frac{2}{25} + \frac{5}{22} = \frac{169}{550} . \]
Therefore, the answer is $169 + 550 = \boxed{719}$ | null | 719 |
d59714d615a1262fb813437141ed2c62 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_II_Problems/Problem_7 | Each vertex of a regular dodecagon ( $12$ -gon) is to be colored either red or blue, and thus there are $2^{12}$ possible colorings. Find the number of these colorings with the property that no four vertices colored the same color are the four vertices of a rectangle. | Note that the condition is equivalent to stating that there are no 2 pairs of oppositely spaced vertices with the same color.
Case 1: There are no pairs. This yields $2$ options for each vertices 1-6, and the remaining vertices 7-12 are set, yielding $2^6=64$ cases.
Case 2: There is one pair. Again start with 2 options for each vertex in 1-6, but now multiply by 6 since there are 6 possibilities for which pair can have the same color assigned instead of the opposite. Thus, the cases are: $2^6*6=384$
case 3: There are two pairs, but oppositely colored. Start with $2^6$ for assigning 1-6, then multiply by 6C2=15 for assigning which have repeated colors. Divide by 2 due to half the cases having the same colored opposites. $2^6*15/2=480$
It is apparent that no other cases exist, as more pairs would force there to be 2 pairs of same colored oppositely spaced vertices with the same color. Thus, the answer is: $64+384+480=\boxed{928}$ | null | 928 |
d59714d615a1262fb813437141ed2c62 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_II_Problems/Problem_7 | Each vertex of a regular dodecagon ( $12$ -gon) is to be colored either red or blue, and thus there are $2^{12}$ possible colorings. Find the number of these colorings with the property that no four vertices colored the same color are the four vertices of a rectangle. | First, we identify the geometric condition for the sentence ``four vertices colored the
same color are the four vertices of a rectangle Consider any four vertices on the dodecagon, $A$ $B$ $C$ $D$ .
Denote by $O$ the center of the dodecagon.
Because $OA = OB = OC$ $\angle OAB = \angle OBA$ and $\angle OBC = \angle OCB$
Thus, \begin{align*} \angle ABC & = \angle OBA + \angle OBC \\ & = \frac{\angle OBA + \angle OAB}{2} + \frac{\angle OBC + \angle OCB}{2} \\ & = \frac{180^\circ - \angle AOB}{2} + \frac{180^\circ - \angle COB}{2} \\ & = 180^\circ - \frac{\angle AOB + \angle COB}{2} \\ & = 180^\circ - \frac{\angle AOC}{2} . \end{align*}
Hence, $\angle ABC = 90^\circ$ if and only if $\angle AOC = 180^\circ$ .
Similarly, $\angle ADC = 90^\circ$ if and only if $\angle AOC = 180^\circ$ , and $\angle BCD = 90^\circ$ (or $\angle DAB = 90^\circ$ ) if and only if $\angle BOD = 180^\circ$
Therefore, $ABCD$ is a rectangle if and only if two diagonals both pass through $O$
Now, we categorize 12 vertices into 6 groups. Each group contains 2 diagonal vertices.
Next, we compute the number of coloring configurations such that the above same-color rectangles do not exist.
Case 1: Two vertices in each group has distinct colors.
For each group, we only need to determine which vertex is red. The other one must be blue.
Therefore, the number of configurations in this case is $2^6$
Case 2: There is one group who vertices have the same color. All other groups are with vertices that have distinct colors.
We construct such an instance in the following steps.
Step 1: We determine which group has two vertices that have the same color.
The number of ways is 6.
Step 2: For the selected group, we choose a color for its two vertices.
The number of ways is 2.
Step 3: For each unselected group, we determine which vertex is red.
The number of ways is $2^5$
Following from the rule of product, the total number of configurations in this case is $6 \cdot 2 \cdot 2^5 = 6 \cdot 2^6$
Case 3: One group has two red vertices, one group has two blue vertices, and each of the other four groups has vertices with distinct colors.
We construct such an instance in the following steps.
Step 1: We determine which group has two vertices that have both red color.
The number of ways is 6.
Step 2: We determine which group has two vertices that have both blue color.
The number of ways is 5.
Step 3: For each unselected group, we determine which vertex is red.
The number of ways is $2^4$
Following from the rule of product, the total number of configurations in this case is $6 \cdot 5 \cdot 2^4 = 30 \cdot 2^4$
Putting all cases together, the total number of feasible configurations is \[ 2^6 + 6 \cdot 2^6 + 30 \cdot 2^4 = \boxed{928}. \] | null | 928 |
d59714d615a1262fb813437141ed2c62 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_II_Problems/Problem_7 | Each vertex of a regular dodecagon ( $12$ -gon) is to be colored either red or blue, and thus there are $2^{12}$ possible colorings. Find the number of these colorings with the property that no four vertices colored the same color are the four vertices of a rectangle. | Note that a rectangle is formed if and only if it's diagonals pass through the center of the dodecagon and the diagonal's endpoints are the same color.
Consider the $6$ diagonals that pass through the center. A rectangle is formed if the endpoints of some pair of them are all the same color. We can now perform casework on the diagonals:
Case $1:$ The endpoints of all the diagonals are all different (e.g. for all diagonals, the endpoints are red and blue in some order).
The amount of ways to do this is $2^6=64,$ since there are $2$ ways to order which vertice is red and which is blue for each diagonal.
Case $2:$ There exists one diagonal such that the endpoints are the same color.
There are $6$ ways to choose this diagonal, $2$ ways to choose which color it is, and $2^5$ ways to color the rest of the diagonals. Therefore, the total for this case is $6 \cdot 2 \cdot 2^5 = 384.$
Case $3:$ There exists two diagonals such that the endpoints are the same color, but they are different colors
Then, there are $6 \cdot 5=30$ ways to choose these two diagonals, and $2^4=16$ ways to color the rest, so the total for this case is $30 \cdot 16 = 480.$
Summing all of the cases, we have $64+384+480=\boxed{928}.$ | null | 928 |
d59714d615a1262fb813437141ed2c62 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_II_Problems/Problem_7 | Each vertex of a regular dodecagon ( $12$ -gon) is to be colored either red or blue, and thus there are $2^{12}$ possible colorings. Find the number of these colorings with the property that no four vertices colored the same color are the four vertices of a rectangle. | \[\text{First, we notice that a rectangle is made from two pairs of vertices 1/2 turn away from each other.}\]
\[\textit{Note: The image is }\frac{\textit{280}}{\textit{841}}\approx\frac{\textit{1}}{\textit{3}}\textit{ size.}\]
\[\text{For there to be no rectangles, there can be at most one same-colored pair for each color and the rest need to have one red and blue.}\]
\[\textit{\underline{Case 1: \textbf{No pairs}}}\]
\[\text{Each pair has two ways to color: One red or the other red. Thus, there are }2^6=\boxed{64}\] | null | 64 |
6ad2336d5795e491643e8b55a6463bf2 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_II_Problems/Problem_8 | Let $\omega = \cos\frac{2\pi}{7} + i \cdot \sin\frac{2\pi}{7},$ where $i = \sqrt{-1}.$ Find the value of the product \[\prod_{k=0}^6 \left(\omega^{3k} + \omega^k + 1\right).\] | For any $k\in Z$ , we have, \begin{align*} & \left( \omega^{3k} + \omega^k + 1 \right) \left( \omega^{3\left( 7 - k \right)} + \omega^{\left( 7 - k \right)} + 1 \right) \\ & = \omega^{3 \cdot 7} + \omega^{2k + 7} + \omega^{3k} + \omega^{-2k + 3 \cdot 7} + \omega^7 + \omega^k + \omega^{3\left( 7 - k \right)} + \omega^{\left( 7 - k \right)} + 1 \\ & = 1 + \omega^{2k} + \omega^{3k} + \omega^{-2k} + 1 + \omega^k + \omega^{-3k} + \omega^{-k} + 1 \\ & = 2 + \omega^{-3k} \sum_{j=0}^6 \omega^{j k} \\ & = 2 + \omega^{-3k} \frac{1 - \omega^{7 k}}{1 - \omega^k} \\ & = 2 . \end{align*} The second and the fifth equalities follow from the property that $\omega^7 = 1$
Therefore, \begin{align*} \Pi_{k=0}^6 \left( \omega^{3k} + \omega^k + 1 \right) & = \left( \omega^{3 \cdot 0} + \omega^0 + 1 \right) \Pi_{k=1}^3 \left( \omega^{3k} + \omega^k + 1 \right) \left( \omega^{3\left( 7 - k \right)} + \omega^{\left( 7 - k \right)} + 1 \right) \\ & = 3 \cdot 2^3 \\ & = \boxed{024} | null | 024 |
6ad2336d5795e491643e8b55a6463bf2 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_II_Problems/Problem_8 | Let $\omega = \cos\frac{2\pi}{7} + i \cdot \sin\frac{2\pi}{7},$ where $i = \sqrt{-1}.$ Find the value of the product \[\prod_{k=0}^6 \left(\omega^{3k} + \omega^k + 1\right).\] | Because the answer must be a positive integer, it is just equal to the modulus of the product. Define $z_n = \left(\textrm{cis }\frac{2n\pi}{7}\right)^3 + \textrm{cis }\frac{2n\pi}{7} + 1$
Then, our product is equal to
\[|z_0||z_1||z_2||z_3||z_4||z_5||z_6|.\]
$z_0 = 0$ , and we may observe that $z_x$ and $z_{7-x}$ are conjugates for any $x$ , meaning that their magnitudes are the same. Thus, our product is
\[3|z_1|^2|z_2|^2|z_3|^2\] \[= 3\left((\cos \frac{6\pi}{7} + \cos \frac{2\pi}{7} + 1)^2 + (\sin \frac{6\pi}{7} + \sin \frac{2\pi}{7})^2\right) \left((\cos \frac{12\pi}{7} + \cos \frac{4\pi}{7} + 1)^2 + (\sin \frac{12\pi}{7} + \sin \frac{4\pi}{7})^2\right) \left((\cos \frac{4\pi}{7} + \cos \frac{6\pi}{7} + 1)^2 + (\sin \frac{4\pi}{7} + \sin \frac{6\pi}{7})^2\right)\]
Let us simplify the first term. Expanding, we obtain
\[\cos^2 \frac{6\pi}{7} + \cos^2 \frac{2\pi}{7} + 1 + 2\cos \frac{6\pi}{7}\cos \frac{2\pi}{7} + 2\cos \frac{6\pi}{7} + 2\cos \frac{2\pi}{7} + \sin^2 \frac{6\pi}{7} + \sin^2 \frac{2\pi}{7} + 2\sin \frac{6\pi}{7}\sin \frac{2\pi}{7}.\]
Rearranging and cancelling, we obtain
\[3 + 2\cos \frac{6\pi}{7} + 2\cos \frac{2\pi}{7} + 2\cos \frac{6\pi}{7}\cos \frac{2\pi}{7} + 2\sin \frac{6\pi}{7}\sin \frac{2\pi}{7}.\]
By the cosine subtraction formula, we have $2\cos \frac{6\pi}{7}\cos \frac{2\pi}{7} + 2\sin \frac{6\pi}{7}\sin \frac{2\pi}{7} = \cos \frac{6\pi - 2\pi}{7} = \cos \frac{4\pi}{7}$
Thus, the first term is equivalent to
\[3 + 2(\cos \frac{2\pi}{7} + \cos \frac{4\pi}{7} + \cos \frac{6\pi}{7}).\]
Similarly, the second and third terms are, respectively,
\[3 + 2(\cos \frac{4\pi}{7} + \cos \frac{8\pi}{7} + \cos \frac{12\pi}{7}),\textrm{ and}\] \[3 + 2(\cos \frac{6\pi}{7} + \cos \frac{12\pi}{7} + \cos \frac{4\pi}{7}).\]
Next, we have $\cos \frac{2\pi}{7} + \cos \frac{4\pi}{7} + \cos \frac{6\pi}{7} = -\frac{1}{2}$ . This is because
\[\cos \frac{2\pi}{7} + \cos \frac{4\pi}{7} + \cos \frac{6\pi}{7} = \frac{1}{2}(\textrm{cis }\frac{2\pi}{7} + \textrm{cis }\frac{4\pi}{7} + \textrm{cis }\frac{6\pi}{7} + \textrm{cis }\frac{8\pi}{7} + \textrm{cis }\frac{10\pi}{7} + \textrm{cis }\frac{12\pi}{7})\]
\[= \frac{1}{2}(-1)\] \[= -\frac{1}{2}.\]
Therefore, the first term is simply $2$ . We have $\cos x = \cos 2\pi - x$ , so therefore the second and third terms can both also be simplified to $3 + 2(\cos \frac{2\pi}{7} + \cos \frac{4\pi}{7} + \cos \frac{6\pi}{7}) = 2$ . Thus, our answer is simply
\[3 \cdot 2 \cdot 2 \cdot 2\] \[= \boxed{024}.\] | null | 024 |
6ad2336d5795e491643e8b55a6463bf2 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_II_Problems/Problem_8 | Let $\omega = \cos\frac{2\pi}{7} + i \cdot \sin\frac{2\pi}{7},$ where $i = \sqrt{-1}.$ Find the value of the product \[\prod_{k=0}^6 \left(\omega^{3k} + \omega^k + 1\right).\] | We write out the product in terms of $\omega$ \[\prod_{k=0}^6 \left(\omega^{3k} + \omega^k + 1\right)=3(\omega^3+\omega+1)(\omega^6+\omega^2+1)(\omega^9+\omega^3+1)(\omega^{12}+\omega^4+1)(\omega^{15}+\omega^5+1)(\omega^{18}+\omega^6+1).\]
Grouping the terms in the following way exploits the fact that $\omega^{7k}=1$ for an integer $k$ , when multiplying out two adjacent products from left to right:
\[\frac{1}{3} \prod_{k=0}^6 \left(\omega^{3k} + \omega^k + 1\right)=(\omega^3+\omega+1)(\omega^{18}+\omega^6+1)(\omega^6+\omega^2+1)(\omega^{15}+\omega^5+1)(\omega^9+\omega^3+1)(\omega^{12}+\omega^4+1).\]
When multiplying two numbers with like bases, we add the exponents. We can now rewrite the exponents of each product (two at a time, where $1$ is treated as the identity) as a series of arrays:
\[\textbf{(A)}\begin{bmatrix} 3&1 &0 \\ 18&6&0\\ \end{bmatrix}\]
\[\textbf{(B)}\begin{bmatrix} 6&2 &0 \\ 15&5&0\\ \end{bmatrix}\]
\[\textbf{(C)}\begin{bmatrix} 9&3 &0 \\ 12&4&0\\ \end{bmatrix}.\]
Note that $\omega=e^{\frac{2\pi i}{7}}$ . When raising $\omega$ to a power, the numerator of the fraction is $2$ times whatever power $\omega$ is raised to, multiplied by $\pi i$ . Since the period of $\omega$ is $2\pi,$ we multiply each array by $2$ then reduce each entry $\mod{14},$ as each entry in an array represents an exponent which $\omega$ is raised to.
\[\textbf{(A)}\begin{bmatrix} 6&2 &0 \\ 8&12&0\\ \end{bmatrix}\]
\[\textbf{(B)}\begin{bmatrix} 12&4 &0 \\ 2&10&0\\ \end{bmatrix}\]
\[\textbf{(C)}\begin{bmatrix} 4&6 &0 \\ 10&8&0\\ \end{bmatrix}.\]
To obtain the correct exponents, we seperately add each element of the lower row to one element of the top row.
Therefore (after reducing $\mod 14$ again), we get the following sets:
\[\textbf{(A)}\ \{0, 4, 6, 10, 0, 2, 8, 12, 0\}\] \[\textbf{(B)}\ \{0, 8, 12, 6, 0, 4, 2, 10, 0\}\] \[\textbf{(C)}\ \{0, 12, 4, 2, 0, 8, 10, 8, 0\}.\]
Raising $\omega$ to the power of each element in every set then multiplying over $\textbf{(A)}, \textbf{(B)},$ and $\textbf{(C)}$ yields
\[\frac{1}{3} \prod_{k=0}^6 \left(\omega^{3k} + \omega^k + 1\right)=\left(\sum_{a\in \textbf{(A)}} \omega^a\right)\left(\sum_{b\in \textbf{(B)}} \omega^b\right)\left(\sum_{c\in \textbf{(C)}} \omega^c\right)\]
\[=\left(\sum_{a\in \textbf{(A)}} \omega^a\right)^3\]
\[=\left(\omega^0+\omega^4+\omega^6+\omega^{10}+\omega^0+\omega^2+\omega^8+\omega^{12}+\omega^0\right)^3\]
\[=\left(3+\omega^2+\omega^4+\omega^6+\omega^8+\omega^{10}+\omega^{12}\right)^3,\] as these sets are all identical.
Summing as a geometric series,
\[\frac{1}{3} \prod_{k=0}^6 \left(\omega^{3k} + \omega^k + 1\right)=\left(3+\frac{\omega^2(\omega^{12}-1)}{\omega^2-1}\right)^3\]
\[=\left(3+\frac{\omega^{14}-\omega^2}{\omega^2-1}\right)^3\]
\[=\left(3+\frac{1-\omega^2}{\omega^2-1}\right)^3\]
\[=(3-1)^3=8.\]
Therefore,
\[\frac{1}{3} \prod_{k=0}^6 \left(\omega^{3k} + \omega^k + 1\right)=8,\] and \[\prod_{k=0}^6 \left(\omega^{3k} + \omega^k + 1\right)=3\cdot8=\boxed{024}.\] | null | 024 |
6ad2336d5795e491643e8b55a6463bf2 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_II_Problems/Problem_8 | Let $\omega = \cos\frac{2\pi}{7} + i \cdot \sin\frac{2\pi}{7},$ where $i = \sqrt{-1}.$ Find the value of the product \[\prod_{k=0}^6 \left(\omega^{3k} + \omega^k + 1\right).\] | The product can be factored into $-(r-1)(s-1)(t-1)(r-w)(s-w)(t-w)(r-w^2)(s-w^2)(t-w^2)....(r-w^6)(s-w^6)(t-w^6)$
where $r,s,t$ are the roots of the polynomial $x^3+x+1=0$
This is then $-(r^7-1)(s^7-1)(t^7-1)$ because $(r^7-1)$ and $(r-1)(r-w)(r-w^2)...(r-w^6)$ share the same roots.
To find $-(r^7-1)(s^7-1)(t^7-1)$
Notice that $(r^7-1)=(r-1)(r^6+r^5+r^4+r^3+r^2+r+1)$ . Since r satisfies $x^3+x+1=0$ $r^6+r^4+r^3=0$
Substituting, you are left with $r^5+r^2+r+1$ . This is $r^2(r^3+1)+r+1$ , and after repeatedly substituting $r^3+x+1=0$ you are left with $-2r^3$
So now the problem is reduced to finding $-(r-1)(s-1)(t-1)(-2r^3)(-2s^3)(-2t^3)=8(rst)^3(r-1)(s-1)(t-1)$ , and vietas gives you the result of $\boxed{24}$ -resources | null | 24 |
5170cc9bf060c5226bcdc641080ec3ec | https://artofproblemsolving.com/wiki/index.php/2023_AIME_II_Problems/Problem_10 | Let $N$ be the number of ways to place the integers $1$ through $12$ in the $12$ cells of a $2 \times 6$ grid so that for any two cells sharing a side, the difference between the numbers in those cells is not divisible by $3.$ One way to do this is shown below. Find the number of positive integer divisors of $N.$ \[\begin{array}{|c|c|c|c|c|c|} \hline \,1\, & \,3\, & \,5\, & \,7\, & \,9\, & 11 \\ \hline \,2\, & \,4\, & \,6\, & \,8\, & 10 & 12 \\ \hline \end{array}\] | We replace the numbers which are 0 mod(3) to 0, 1 mod(3) to 1, and 2 mod(3) to 2.
Then, the problem is equivalent to arranging 4 0's,4 1's, and 4 2's into the grid (and then multiplying by $4!^3$ to account for replacing the remainder numbers with actual numbers) such that no 2 of the same numbers are adjacent.
Then, the numbers which are vertically connected must be different. 2 1s must be connected with 2 2s, and 2 1s must be connected with 2 2s (vertically), as if there are less 1s connected with either, then there will be 2s or 3s which must be connected within the same number.
For instance if we did did:
- 12 x 3
- 13 x 1
Then we would be left with 23 x1, and 2 remaining 3s which aren't supposed to be connected, so it is impossible. Similar logic works for why all 1s can't be connected with all 2s.
Thus, we are left with the problem of re-arranging 2x 12 pairs, 2x 13 pairs, 2x 23 pairs. Notice that the pairs can be re-arranged horizontally in any configuration, but when 2 pairs are placed adjacent there is only 1 configuration for the rightmost pair to be set after the leftmost one has been placed.
We have $\frac{6!}{2!2!2!}=90$ ways to horizontally re-arrange the pairs, with 2 ways to set the initial leftmost column. Thus, there are 180 ways to arrange the pairs. Accounting for the initial simplification of the problem with 1-12 -> 0-3, we obtain the answer is:
\[2488320=2^{11}\cdot3^5\cdot5^1\]
The number of divisors is: $12\cdot6\cdot2=\boxed{144}.$ ~SAHANWIJETUNGA | null | 144 |
5170cc9bf060c5226bcdc641080ec3ec | https://artofproblemsolving.com/wiki/index.php/2023_AIME_II_Problems/Problem_10 | Let $N$ be the number of ways to place the integers $1$ through $12$ in the $12$ cells of a $2 \times 6$ grid so that for any two cells sharing a side, the difference between the numbers in those cells is not divisible by $3.$ One way to do this is shown below. Find the number of positive integer divisors of $N.$ \[\begin{array}{|c|c|c|c|c|c|} \hline \,1\, & \,3\, & \,5\, & \,7\, & \,9\, & 11 \\ \hline \,2\, & \,4\, & \,6\, & \,8\, & 10 & 12 \\ \hline \end{array}\] | Let's carry out an archaeological study, that is, we will find the bones (the base) and "build up the meat."
1. Let "bones" of number $X$ be $X \pmod 3.$ Then the “skeleton” of the original table is
\[1 0 2 1 0 2\] \[2 1 0 2 1 0\] By condition, the table cannot have a column of two identical numbers (the difference of such numbers is a multiple of $3).$
There are $4$ zeros, $4$ ones and $4$ twos in the table (the order of the numbers in the columns is not important).
Therefore, there cannot be more than two identical columns (otherwise there will be four identical numbers in the remaining three, that is, the numbers in one column are the same).
Any such table has exactly $2$ columns $(0,1), 2$ columns $(0,2)$ and $2$ columns $(1,2).$
The number of possible tables of six elements of three types is \[m = \frac {6!}{2! \cdot 2! \cdot 2!} = 2 \cdot 3^2 \cdot 5.\] The number of possible tables of six elements, if the order of the digits is important, is \[M = 2^6 \cdot m = 2^7 \cdot 3^2 \cdot 5.\] 2. We are looking for the total number of options.
The column $(0,1)$ can contain the following $4^2 = 16$ pairs of numbers (the order is not important, it is already taken into account) \[(1,3), (1,6), (1,9), (1, 12), (4.3), (4.6), (4.9) (4.12), (7.3), (7.6), (7.9), (7.12), (11,3), (11,6), (11,9), (11,12).\] The second column $(0,1)$ can contain $3^2 = 9$ pairs of numbers (excluding the two that are in the first column).
Similarly with the other two columns, i.e. the total number of choices \[M \cdot 16 \cdot 9 \cdot 3 = 2^{11} \cdot 3^5 \cdot 5.\] Number of divisors \[(11+1) \cdot (5+1) \cdot (1 + 1) = \boxed{144}.\] [email protected], vvsss | null | 144 |
c59d6c093f8590f5f74c7201541cbfa5 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_II_Problems/Problem_11 | Find the number of collections of $16$ distinct subsets of $\{1,2,3,4,5\}$ with the property that for any two subsets $X$ and $Y$ in the collection, $X \cap Y \not= \emptyset.$ | Denote by $\mathcal C$ a collection of 16 distinct subsets of $\left\{ 1, 2, 3, 4, 5 \right\}$ .
Denote $N = \min \left\{ |S|: S \in \mathcal C \right\}$
Case 1: $N = 0$
This entails $\emptyset \in \mathcal C$ .
Hence, for any other set $A \in \mathcal C$ , we have $\emptyset \cap A = \emptyset$ . This is infeasible.
Case 2: $N = 1$
Let $\{a_1\} \in \mathcal C$ .
To get $\{a_1\} \cap A \neq \emptyset$ for all $A \in \mathcal C$ .
We must have $a_1 \in \mathcal A$
The total number of subsets of $\left\{ 1, 2, 3, 4, 5 \right\}$ that contain $a_1$ is $2^4 = 16$ .
Because $\mathcal C$ contains 16 subsets.
We must have $\mathcal C = \left\{ \{a_1\} \cup A : \forall \ A \subseteq \left\{ 1, 2, 3, 4, 5 \right\} \backslash \left\{a_1 \right\} \right\}$ .
Therefore, for any $X, Y \in \mathcal C$ , we must have $X \cap Y \supseteq \{a_1\}$ .
So this is feasible.
Now, we count the number of $\mathcal C$ in this case.
We only need to determine $a_1$ .
Therefore, the number of solutions is 5.
Case 3: $N = 2$
Case 3.1: There is exactly one subset in $\mathcal C$ that contains 2 elements.
Denote this subset as $\left\{ a_1, a_2 \right\}$ .
We then put all subsets of $\left\{ 1, 2, 3, 4, 5 \right\}$ that contain at least three elements into $\mathcal C$ , except $\left\{ a_3, a_4, a_5 \right\}$ .
This satisfies $X \cap Y \neq \emptyset$ for any $X, Y \in \mathcal C$
Now, we count the number of $\mathcal C$ in this case.
We only need to determine $\left\{ a_1, a_2 \right\}$ .
Therefore, the number of solutions is $\binom{5}{2} = 10$
Case 3.2: There are exactly two subsets in $\mathcal C$ that contain 2 elements.
They must take the form $\left\{ a_1, a_2 \right\}$ and $\left\{ a_1, a_3 \right\}$
We then put all subsets of $\left\{ 1, 2, 3, 4, 5 \right\}$ that contain at least three elements into $\mathcal C$ , except $\left\{ a_3, a_4, a_5 \right\}$ and $\left\{ a_2, a_4, a_5 \right\}$ .
This satisfies $X \cap Y \neq \emptyset$ for any $X, Y \in \mathcal C$
Now, we count the number of $\mathcal C$ in this case.
We only need to determine $\left\{ a_1, a_2 \right\}$ and $\left\{ a_1, a_3 \right\}$ .
Therefore, the number of solutions is $5 \cdot \binom{4}{2} = 30$
Case 3.3: There are exactly three subsets in $\mathcal C$ that contain 2 elements.
They take the form $\left\{ a_1, a_2 \right\}$ $\left\{ a_1, a_3 \right\}$ $\left\{ a_1, a_4 \right\}$
We then put all subsets of $\left\{ 1, 2, 3, 4, 5 \right\}$ that contain at least three elements into $\mathcal C$ , except $\left\{ a_3, a_4, a_5 \right\}$ $\left\{ a_2, a_4, a_5 \right\}$ $\left\{ a_2, a_3, a_5 \right\}$ .
This satisfies $X \cap Y \neq \emptyset$ for any $X, Y \in \mathcal C$
Now, we count the number of $\mathcal C$ in this case.
We only need to determine $\left\{ a_1, a_2 \right\}$ $\left\{ a_1, a_3 \right\}$ $\left\{ a_1, a_4 \right\}$ .
Therefore, the number of solutions is $5 \cdot \binom{4}{3} = 20$
Case 3.4: There are exactly three subsets in $\mathcal C$ that contain 2 elements.
They take the form $\left\{ a_1, a_2 \right\}$ $\left\{ a_1, a_3 \right\}$ $\left\{ a_2, a_3 \right\}$
We then put all subsets of $\left\{ 1, 2, 3, 4, 5 \right\}$ that contain at least three elements into $\mathcal C$ , except $\left\{ a_3, a_4, a_5 \right\}$ $\left\{ a_2, a_4, a_5 \right\}$ $\left\{ a_1, a_4, a_5 \right\}$ .
This satisfies $X \cap Y \neq \emptyset$ for any $X, Y \in \mathcal C$
Now, we count the number of $\mathcal C$ in this case.
We only need to determine $\left\{ a_1, a_2 \right\}$ $\left\{ a_1, a_3 \right\}$ $\left\{ a_2, a_3 \right\}$ .
Therefore, the number of solutions is $\binom{5}{3} = 10$
Case 3.5: There are exactly four subsets in $\mathcal C$ that contain 2 elements.
They take the form $\left\{ a_1, a_2 \right\}$ $\left\{ a_1, a_3 \right\}$ $\left\{ a_1, a_4 \right\}$ $\left\{ a_1, a_5 \right\}$
We then put all subsets of $\left\{ 1, 2, 3, 4, 5 \right\}$ that contain at least three elements into $\mathcal C$ , except $\left\{ a_3, a_4, a_5 \right\}$ $\left\{ a_2, a_4, a_5 \right\}$ $\left\{ a_1, a_4, a_5 \right\}$ $\left\{ a_2, a_3, a_4 \right\}$ .
This satisfies $X \cap Y \neq \emptyset$ for any $X, Y \in \mathcal C$
Now, we count the number of $\mathcal C$ in this case.
We only need to determine $\left\{ a_1, a_2 \right\}$ $\left\{ a_1, a_3 \right\}$ $\left\{ a_1, a_4 \right\}$ $\left\{ a_1, a_5 \right\}$ .
Therefore, the number of solutions is 5.
Putting all subcases together, the number of solutions is this case is $10 + 30 + 20 + 10 + 5 = 75$
Case 4: $N \geq 3$
The number of subsets of $\left\{ 1, 2, 3, 4, 5 \right\}$ that contain at least three elements is $\sum_{i=3}^5 \binom{5}{i} = 16$ .
Because $\mathcal C$ has 16 elements, we must select all such subsets into $\mathcal C$ .
Therefore, the number of solutions in this case is 1.
Putting all cases together, the total number of $\mathcal C$ is $5 + 75 + 1 = \boxed{081}$ | null | 081 |
c59d6c093f8590f5f74c7201541cbfa5 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_II_Problems/Problem_11 | Find the number of collections of $16$ distinct subsets of $\{1,2,3,4,5\}$ with the property that for any two subsets $X$ and $Y$ in the collection, $X \cap Y \not= \emptyset.$ | Denote the $A$ as $\{ 1,2,3,4,5 \}$ and the collection of subsets as $S$
Case 1: There are only sets of size $3$ or higher in $S$ Any two sets in $S$ must have at least one element common to both of them (since $3+3>5$ ). Since there are $16$ subsets of $A$ that have size $3$ or higher, there is only one possibility for this case.
Case 2: There are only sets of size $2$ or higher in $S$ Firstly, there cannot be a no size $2$ set $S$ , or it will be overcounting the first case.
If there is only one such size $2$ set, there are $10$ ways to choose it. That size $2$ set, say $X$ , cannot be in $S$ with $Y = A/X$ (a three element set). Thus, there are only $15$ possible size $3$ subsets that can be in $S$ , giving us $10$ for this case.
If there are two sets with size $2$ in $S$ , we can choose the common elements of these two subsets in $5$ ways, giving us a total of $5\cdot 6 = 30$
If there are three sets with size $2$ , they can either share one common element, which can be done in $5 \cdot 4 = 20$ ways, or they can share pairwise common elements (sort of like a cycle), which can be done $\binom{5}{2} = 10$ ways. In total, we have $30$ possibilities.
If there are four sets with size $2$ , they all have to share one common element, which can be done in $5\cdot 1$ ways.
Thus, summing everything up, this will give us $75$ possible sets $S$
Case 3: There is a set with size $1$ in $S$ Notice that be at most one size $1$ subset. There are $5$ ways to choose that single element set. Say it's $\{ 1\}$ . All other subsets in $S$ must have a $1$ in them, but there are only $2^4-1 = 15$ of them. Thus this case yields $5\cdot 1 = 5$ possibilities.
Thus, the total number of sets $S$ would be $1+75+5 = \boxed{081}$ | null | 081 |
c59d6c093f8590f5f74c7201541cbfa5 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_II_Problems/Problem_11 | Find the number of collections of $16$ distinct subsets of $\{1,2,3,4,5\}$ with the property that for any two subsets $X$ and $Y$ in the collection, $X \cap Y \not= \emptyset.$ | Firstly, there cannot be two subsets with cardinality 1, or they will not intersect.
If there is one subset $A$ with cardinality $1$ ; let the element in $A$ be $a$ , then there are $2^4=16$ subsets that do not include $a$ so they do not work. Every remaining subsets $S$ will have $a$ as an element so $S\cap{A}\geq1$ , since we just excluded all that do not. Since there are 15 subsets left, all are forced into the group of 16 subsets, so we just choose the number of $a$ to determine the set $A$ $A\in\{1,2,3,4,5\}$ so there are 5 ways.
For the rest of the cases, we assume there are no sets $A$ with cardinality 1. Notice that the only way to "violate" the condition is to have subsets $X$ and $Y$ with cardinalities 2 and 3 in some order. Otherwise, by the Pigeonhole Principle, if two sets both have cardinalities more than 3, they are bound to have one element of intersection. Say a set $S$ has $|S|=2$ , then there is clearly only one set $s$ that will make $S\cap{s}=0$ . By our previous claim, all other subsets that have cardinality $c\geq{3}$ will work.
Now if we generalize a bit: If a subset has $N$ 2-element subsets which belong to set $M$ , then there are exactly $N$ subsets with cardinality 3 that don't work. Therefore, the number of "violating subsets" are all subsets with cardinality $c\leq{1}$ , all 2-element subsets that are not in $M$ , and all corresponding cardinality 3 subsets. Subtracting from the total 32 subsets, we get that $32-(1+5+(10-N)+N)=16$ subsets that do work. This includes all subsets in $M$ , so the remaining non-violating subsets are forced. This is equivalent now to choosing $N$ 2 element subsets.
Following casework on the number of 2-element subsets:
If $N=1$ : There are $\binom{5}{2}=10$ ways.
If $N=2$ : There are $5$ ways to choose the intersection between the 2 sets (remember they have to have at least one element of intersection) and $\binom{4}{2}=6$ ways to choose the distinct elements in the subsets, so there are $5*6=30$ ways.
If $N=3$ : It can be a cycle, where WLOG let the elements be $a,b,c$ so the sets are $\{a,b\}$ $\{b,c\}$ , and $\{c,a\}$ . This is just $\binom{5}{3}=10$ . Alternatively, it can also be the case where all sets share one element. There are 5 ways to choose this element and $\binom{4}{2}=6$ ways to choose the remaining elements to assign to each set. There are $30$ ways.
If $N=4$ : By the Pigeonhole Principle, the only way all pairwise sets have at one common intersection is if all share one element in common. There are 5 ways to choose this element and the remaining numbers are forced. There are 5 ways.
$N\geq{5}$ does not provide any valid cases since to have all pairwise elements to intersect one element, they must be the same element by the Pigeonhole Principle, but there are not enough subsets.
If $N=0$ , then there is only one way since $\binom{5}{3}+\binom{5}{4}+\binom{5}{5}=16$
Adding all cases yields $5+10+30+30+5+1=\boxed{81}$ ways! | null | 81 |
99fc444e2f3ebf7cc9b02b19481099f6 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_II_Problems/Problem_12 | In $\triangle ABC$ with side lengths $AB = 13,$ $BC = 14,$ and $CA = 15,$ let $M$ be the midpoint of $\overline{BC}.$ Let $P$ be the point on the circumcircle of $\triangle ABC$ such that $M$ is on $\overline{AP}.$ There exists a unique point $Q$ on segment $\overline{AM}$ such that $\angle PBQ = \angle PCQ.$ Then $AQ$ can be written as $\frac{m}{\sqrt{n}},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$ | Because $M$ is the midpoint of $BC$ , following from the Stewart's theorem, $AM = 2 \sqrt{37}$
Because $A$ $B$ $C$ , and $P$ are concyclic, $\angle BPA = \angle C$ $\angle CPA = \angle B$
Denote $\theta = \angle PBQ$
In $\triangle BPQ$ , following from the law of sines, \[ \frac{BQ}{\sin \angle BPA} = \frac{PQ}{\angle PBQ} \]
Thus, \[ \frac{BQ}{\sin C} = \frac{PQ}{\sin \theta} . \hspace{1cm} (1) \]
In $\triangle CPQ$ , following from the law of sines, \[ \frac{CQ}{\sin \angle CPA} = \frac{PQ}{\angle PCQ} \]
Thus, \[ \frac{CQ}{\sin B} = \frac{PQ}{\sin \theta} . \hspace{1cm} (2) \]
Taking $\frac{(1)}{(2)}$ , we get \[ \frac{BQ}{\sin C} = \frac{CQ}{\sin B} \]
In $\triangle ABC$ , following from the law of sines, \[ \frac{AB}{\sin C} = \frac{AC}{\sin B} . \hspace{1cm} (3) \]
Thus, Equations (2) and (3) imply \begin{align*} \frac{BQ}{CQ} & = \frac{AB}{AC} \\ & = \frac{13}{15} . \hspace{1cm} (4) \end{align*}
Next, we compute $BQ$ and $CQ$
We have \begin{align*} BQ^2 & = AB^2 + AQ^2 - 2 AB\cdot AQ \cos \angle BAQ \\ & = AB^2 + AQ^2 - 2 AB\cdot AQ \cos \angle BAM \\ & = AB^2 + AQ^2 - 2 AB\cdot AQ \cdot \frac{AB^2 + AM^2 - BM^2}{2 AB \cdot AM} \\ & = AB^2 + AQ^2 - AQ \cdot \frac{AB^2 + AM^2 - BM^2}{AM} \\ & = 169 + AQ^2 - \frac{268}{2 \sqrt{37}} AQ . \hspace{1cm} (5) \end{align*}
We have \begin{align*} CQ^2 & = AC^2 + AQ^2 - 2 AC\cdot AQ \cos \angle CAQ \\ & = AC^2 + AQ^2 - 2 AC\cdot AQ \cos \angle CAM \\ & = AC^2 + AQ^2 - 2 AC\cdot AQ \cdot \frac{AC^2 + AM^2 - CM^2}{2 AC \cdot AM} \\ & = AC^2 + AQ^2 - AQ \cdot \frac{AC^2 + AM^2 - CM^2}{AM} \\ & = 225 + AQ^2 - \frac{324}{2 \sqrt{37}} AQ . \hspace{1cm} (6) \end{align*}
Taking (5) and (6) into (4), we get $AQ = \frac{99}{\sqrt{148}}$
Therefore, the answer is $99 + 148 = \boxed{247}$ | null | 247 |
99fc444e2f3ebf7cc9b02b19481099f6 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_II_Problems/Problem_12 | In $\triangle ABC$ with side lengths $AB = 13,$ $BC = 14,$ and $CA = 15,$ let $M$ be the midpoint of $\overline{BC}.$ Let $P$ be the point on the circumcircle of $\triangle ABC$ such that $M$ is on $\overline{AP}.$ There exists a unique point $Q$ on segment $\overline{AM}$ such that $\angle PBQ = \angle PCQ.$ Then $AQ$ can be written as $\frac{m}{\sqrt{n}},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$ | Define $L_1$ to be the foot of the altitude from $A$ to $BC$ . Furthermore, define $L_2$ to be the foot of the altitude from $Q$ to $BC$ . From here, one can find $AL_1=12$ , either using the 13-14-15 triangle or by calculating the area of $ABC$ two ways. Then, we find $BL_1=5$ and $L_1C = 9$ using Pythagorean theorem. Let $QL_2=x$ . By AA similarity, $\triangle{AL_1M}$ and $\triangle{QL_2M}$ are similar. By similarity ratios, \[\frac{AL_1}{L_1M}=\frac{QL_2}{L_2M}\] \[\frac{12}{2}=\frac{x}{L_2M}\] \[L_2M = \frac{x}{6}\] Thus, $BL_2=BM-L_2M=7-\frac{x}{6}$ . Similarly, $CL_2=7+\frac{x}{6}$ . Now, we angle chase from our requirement to obtain new information. \[\angle{PBQ}=\angle{PCQ}\] \[\angle{QCM}+\angle{PCM}=\angle{QBM}+\angle{PBM}\] \[\angle{QCL_2}+\angle{PCM}=\angle{QBL_2}+\angle{PBM}\] \[\angle{PCM}-\angle{PBM}=\angle{QBL_2}-\angle{QCL_2}\] \[\angle{MAB}-\angle{MAC}=\angle{QBL_2}-\angle{QCL_2}\text{(By inscribed angle theorem)}\] \[(\angle{MAL_1}+\angle{L_1AB})-(\angle{CAL_1}-\angle{MAL_1})=\angle{QBL_2}-\angle{QCL_2}\] \[2\angle{MAL_1}+\angle{L_1AB}-\angle{CAL_1}=\angle{QBL_2}-\angle{QCL_2}\] Take the tangent of both sides to obtain \[\tan(2\angle{MAL_1}+\angle{L_1AB}-\angle{CAL_1})=\tan(\angle{QBL_2}-\angle{QCL_2})\] By the definition of the tangent function on right triangles, we have $\tan{MAL_1}=\frac{7-5}{12}=\frac{1}{6}$ $\tan{CAL_1}=\frac{9}{12}=\frac{3}{4}$ , and $\tan{L_1AB}=\frac{5}{12}$ . By abusing the tangent angle addition formula, we can find that \[\tan(2\angle{MAL_1}+\angle{L_1AB}-\angle{CAL_1})=\frac{196}{2397}\] By substituting $\tan{\angle{QBL_2}}=\frac{6x}{42-x}$ $\tan{\angle{QCL_2}}=\frac{6x}{42+x}$ and using tangent angle subtraction formula we find that \[x=\frac{147}{37}\] Finally, using similarity formulas, we can find \[\frac{AQ}{AM}=\frac{12-x}{x}\] . Plugging in $x=\frac{147}{37}$ and $AM=\sqrt{148}$ , we find that \[AQ=\frac{99}{\sqrt{148}}\] Thus, our final answer is $99+148=\boxed{247}$ .
~sigma | null | 247 |
99fc444e2f3ebf7cc9b02b19481099f6 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_II_Problems/Problem_12 | In $\triangle ABC$ with side lengths $AB = 13,$ $BC = 14,$ and $CA = 15,$ let $M$ be the midpoint of $\overline{BC}.$ Let $P$ be the point on the circumcircle of $\triangle ABC$ such that $M$ is on $\overline{AP}.$ There exists a unique point $Q$ on segment $\overline{AM}$ such that $\angle PBQ = \angle PCQ.$ Then $AQ$ can be written as $\frac{m}{\sqrt{n}},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$ | It is clear that $BQCP$ is a parallelogram. By Stewart's Theorem, $AM=\sqrt{148}$ , POP on $M$ tells $PM=\frac{49}{\sqrt{148}}$
As $QM=PM, AQ=AM-PM=\frac{99}{\sqrt{148}}$ leads to $\boxed{247}$ | null | 247 |
99fc444e2f3ebf7cc9b02b19481099f6 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_II_Problems/Problem_12 | In $\triangle ABC$ with side lengths $AB = 13,$ $BC = 14,$ and $CA = 15,$ let $M$ be the midpoint of $\overline{BC}.$ Let $P$ be the point on the circumcircle of $\triangle ABC$ such that $M$ is on $\overline{AP}.$ There exists a unique point $Q$ on segment $\overline{AM}$ such that $\angle PBQ = \angle PCQ.$ Then $AQ$ can be written as $\frac{m}{\sqrt{n}},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$ | First, note that by Law of Sines, $\frac{\sin(\angle{QBP})}{QP}=\frac{\sin(\angle{BPQ})}{BQ}$ and that $\frac{\sin(\angle{QCP})}{QP}=\frac{\sin(\angle{QPC})}{QP}$ . Equating the 2 expressions, you get that $\frac{\sin(\angle{BPQ})}{BQ}=\frac{\sin(\angle{QPC})}{QP}$ . Now drop the altitude from $A$ to $BC$ . As it is commonly known that the dropped altitude forms a $5-12-13$ and a $9-12-15$ triangle, you get the measures of $\angle{ABC}$ and $\angle{ACB}$ respectively, which are $\arcsin(\frac{12}{13})$ and $\arcsin(\frac{4}{5})$ . However, by the inscribed angle theorem, you get that $\angle{BPQ}=\arcsin(\frac{4}{5})$ and that $\angle{QPC}=\arcsin(\frac{12}{13})$ , respectively. Therefore, by Law of Sines (as previously stated) $\frac{BQ}{CQ}=\frac{13}{15}$
Now commence coordbashing. Let $B$ be the origin, and $A$ be the point $(5,12)$ . As $AP$ passes through $A$ , which is $(5,12)$ , and $M$ , which is $(7,0)$ , it has the equation $-6x+42$ , so therefore a point on this line can be written as $(x,42-6x)$ . As we have the ratio of the lengths, which prompts us to write the lengths in terms of the distance formula, we can just plug and chug it in to get the ratio $\frac{\sqrt{37x^2-504x+1764}}{\sqrt{37x^2-532x+1960}}=\frac{13}{15}$ . This can be squared to get $\frac{37x^2-504x+1764}{37x^2-532x+1960}=\frac{169}{225}$ . This can be solved to get a solution of $x=\frac{469}{74}$ , and an extraneous solution of $5$ which obviously doesn’t work.
Plugging $x$ into the line equation gets you $y=\frac{147}{37}$ . The distance between this point and $A$ , which is $(5,12)$ is $\sqrt{\frac{9801}{148}}$ , or simplified to $\frac{99}{\sqrt{148}}\Longrightarrow99+148=\boxed{247}$ | null | 247 |
99fc444e2f3ebf7cc9b02b19481099f6 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_II_Problems/Problem_12 | In $\triangle ABC$ with side lengths $AB = 13,$ $BC = 14,$ and $CA = 15,$ let $M$ be the midpoint of $\overline{BC}.$ Let $P$ be the point on the circumcircle of $\triangle ABC$ such that $M$ is on $\overline{AP}.$ There exists a unique point $Q$ on segment $\overline{AM}$ such that $\angle PBQ = \angle PCQ.$ Then $AQ$ can be written as $\frac{m}{\sqrt{n}},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$ | We use the law of Cosine and get \[AB^2 = AM^2 + BM^2 - 2 AM \cdot BM \cos \angle AMB,\] \[AC^2 = AM^2 + CM^2 + 2 AM \cdot CM \cos \angle AMB \implies\] \[AM^2 = \frac {AB^2 + AC^2}{2}- BM^2 = \sqrt{148} \approx 12.\] We use the power of point $M$ with respect circumcircle $\triangle ABC$ and get \[AM \cdot MP = BM \cdot CM = BM^2 \implies\] \[PM = \frac {49}{\sqrt {148}} \approx \frac {48}{12} \approx 4 < AM.\] It is clear that if $Q = P,$ then $\angle PBQ = \angle PCQ = 0 \implies$
if $Q$ is symmetric to $P$ with respect $M$ then $\angle PBQ = \angle PCQ.$
There exists a unique point $Q$ on segment $\overline{AM}, PM < AM \implies$ \[PQ = AM - PM = \frac{99}{\sqrt{148}} \implies \boxed{247}.\] [email protected], vvsss | null | 247 |
4c13d067238106648d920d3957d2437e | https://artofproblemsolving.com/wiki/index.php/2023_AIME_II_Problems/Problem_13 | Let $A$ be an acute angle such that $\tan A = 2 \cos A.$ Find the number of positive integers $n$ less than or equal to $1000$ such that $\sec^n A + \tan^n A$ is a positive integer whose units digit is $9.$ | Denote $a_n = \sec^n A + \tan^n A$ .
For any $k$ , we have \begin{align*} a_n & = \sec^n A + \tan^n A \\ & = \left( \sec^{n-k} A + \tan^{n-k} A \right) \left( \sec^k A + \tan^k A \right) - \sec^{n-k} A \tan^k A - \tan^{n-k} A \sec^k A \\ & = a_{n-k} a_k - 2^k \sec^{n-k} A \cos^k A - 2^k \tan^{n-k} A \cot^k A \\ & = a_{n-k} a_k - 2^k a_{n-2k} . \end{align*}
Next, we compute the first several terms of $a_n$
By solving equation $\tan A = 2 \cos A$ , we get $\cos A = \frac{\sqrt{2 \sqrt{17} - 2}}{4}$ .
Thus, $a_0 = 2$ $a_1 = \sqrt{\sqrt{17} + 4}$ $a_2 = \sqrt{17}$ $a_3 = \sqrt{\sqrt{17} + 4} \left( \sqrt{17} - 2 \right)$ $a_4 = 9$
In the rest of analysis, we set $k = 4$ .
Thus, \begin{align*} a_n & = a_{n-4} a_4 - 2^4 a_{n-8} \\ & = 9 a_{n-4} - 16 a_{n-8} . \end{align*}
Thus, to get $a_n$ an integer, we have $4 | n$ .
In the rest of analysis, we only consider such $n$ . Denote $n = 4 m$ and $b_m = a_{4n}$ .
Thus, \begin{align*} b_m & = 9 b_{m-1} - 16 b_{m-2} \end{align*} with initial conditions $b_0 = 2$ $b_1 = 9$
To get the units digit of $b_m$ to be 9, we have \begin{align*} b_m \equiv -1 & \pmod{2} \\ b_m \equiv -1 & \pmod{5} \end{align*}
Modulo 2, for $m \geq 2$ , we have \begin{align*} b_m & \equiv 9 b_{m-1} - 16 b_{m-2} \\ & \equiv b_{m-1} . \end{align*}
Because $b_1 \equiv -1 \pmod{2}$ , we always have $b_m \equiv -1 \pmod{2}$ for all $m \geq 2$
Modulo 5, for $m \geq 5$ , we have \begin{align*} b_m & \equiv 9 b_{m-1} - 16 b_{m-2} \\ & \equiv - b_{m-1} - b_{m-2} . \end{align*}
We have $b_0 \equiv 2 \pmod{5}$ $b_1 \equiv -1 \pmod{5}$ $b_2 \equiv -1 \pmod{5}$ $b_3 \equiv 2 \pmod{5}$ $b_4 \equiv -1 \pmod{5}$ $b_5 \equiv -1 \pmod{5}$ $b_6 \equiv 2 \pmod{5}$ .
Therefore, the congruent values modulo 5 is cyclic with period 3.
To get $b_m \equiv -1 \pmod{5}$ , we have $3 \nmid m \pmod{3}$
From the above analysis with modulus 2 and modulus 5, we require $3 \nmid m \pmod{3}$
For $n \leq 1000$ , because $n = 4m$ , we only need to count feasible $m$ with $m \leq 250$ .
The number of feasible $m$ is \begin{align*} 250 - \left\lfloor \frac{250}{3} \right\rfloor & = 250 - 83 \\ & = \boxed{167} | null | 167 |
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