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int64 1
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3,701
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_19
| 2
|
Suppose that $13$ cards numbered $1, 2, 3, \ldots, 13$ are arranged in a row. The task is to pick them up in numerically increasing order, working repeatedly from left to right. In the example below, cards $1, 2, 3$ are picked up on the first pass, $4$ and $5$ on the second pass, $6$ on the third pass, $7, 8, 9, 10$ on the fourth pass, and $11, 12, 13$ on the fifth pass. For how many of the $13!$ possible orderings of the cards will the $13$ cards be picked up in exactly two passes?
[asy] size(11cm); draw((0,0)--(2,0)--(2,3)--(0,3)--cycle); label("7", (1,1.5)); draw((3,0)--(5,0)--(5,3)--(3,3)--cycle); label("11", (4,1.5)); draw((6,0)--(8,0)--(8,3)--(6,3)--cycle); label("8", (7,1.5)); draw((9,0)--(11,0)--(11,3)--(9,3)--cycle); label("6", (10,1.5)); draw((12,0)--(14,0)--(14,3)--(12,3)--cycle); label("4", (13,1.5)); draw((15,0)--(17,0)--(17,3)--(15,3)--cycle); label("5", (16,1.5)); draw((18,0)--(20,0)--(20,3)--(18,3)--cycle); label("9", (19,1.5)); draw((21,0)--(23,0)--(23,3)--(21,3)--cycle); label("12", (22,1.5)); draw((24,0)--(26,0)--(26,3)--(24,3)--cycle); label("1", (25,1.5)); draw((27,0)--(29,0)--(29,3)--(27,3)--cycle); label("13", (28,1.5)); draw((30,0)--(32,0)--(32,3)--(30,3)--cycle); label("10", (31,1.5)); draw((33,0)--(35,0)--(35,3)--(33,3)--cycle); label("2", (34,1.5)); draw((36,0)--(38,0)--(38,3)--(36,3)--cycle); label("3", (37,1.5)); [/asy] $\textbf{(A) } 4082 \qquad \textbf{(B) } 4095 \qquad \textbf{(C) } 4096 \qquad \textbf{(D) } 8178 \qquad \textbf{(E) } 8191$
|
Since the $13$ cards are picked up in two passes, the first pass must pick up the first $n$ cards and the second pass must pick up the remaining cards $m$ through $13$ .
Also note that if $m$ , which is the card that is numbered one more than $n$ , is placed before $n$ , then $m$ will not be picked up on the first pass since cards are picked up in order. Therefore we desire $m$ to be placed before $n$ to create a second pass, and that after the first pass, the numbers $m$ through $13$ are lined up in order from least to greatest.
To construct this, $n$ cannot go in the $n$ th position because all cards $1$ to $n-1$ will have to precede it and there will be no room for $m$ . Therefore $n$ must be in slots $n+1$ to $13$ .
Let's do casework on which slot $n$ goes into to get a general idea for how the problem works.
Case 1: With $n$ in spot $n+1$ , there are $n$ available slots before $n$ , and there are $n-1$ cards preceding $n$ . Therefore the number of ways to reserve these slots for the $n-1$ cards is $\binom{n}{n-1}$ . Then there is only $1$ way to order these cards (since we want them in increasing order). Then card $m$ goes into whatever slot is remaining, and the $13-m$ cards are ordered in increasing order after slot $n+1$ , giving only $1$ way. Therefore in this case there are $\binom{n}{n-1}$ possibilities.
Case 2: With $n$ in spot $n+2$ , there are $n+1$ available slots before $n$ , and there are $n-1$ cards preceding $n$ . Therefore the number of ways to reserve slots for these cards are $\binom{n+1}{n-1}$ . Then there is one way to order these cards. Then cards $m$ and $m+1$ must go in the remaining two slots, and there is only one way to order them since they must be in increasing order. Finally, cards $m+2$ to $13$ will be ordered in increasing order after slot $n+1$ , which yields $1$ way. Therefore, this case has $\binom{n+1}{n-1}$ possibilities.
I think we can see a general pattern now. With $n$ in slot $x$ , there are $x-1$ slots to distribute to the previous $n-1$ cards, which can be done in $\binom{x-1}{n-1}$ ways. Then the remaining cards fill in in just $1$ way. Since the cases of $n$ start in slot $n+1$ and end in slot $13$ , this sum amounts to: \[\binom{n}{n-1}+\binom{n+1}{n-1}+\binom{n+2}{n-1} + \cdots + \binom{12}{n-1}\] for any $n$
Hmmm ... where have we seen this before?
We use wishful thinking to add a term of $\binom{n-1}{n-1}$ \[\binom{n-1}{n-1}+\binom{n}{n-1}+\binom{n+1}{n-1}+\binom{n+2}{n-1} + \cdots + \binom{12}{n-1}\]
This is just the hockey stick identity! Applying it, this expression is equal to $\binom{13}{n}$ . However, we added an extra term, so subtracting it off, the total number of ways to order the $13$ cards for any $n$ is \[\binom{13}{n}-1\]
Finally, to calculate the total for all $n$ , we sum from $n=0$ to $13$ . This yields us:
\[\sum_{n=0}^{13} \binom{13}{n}-1 \implies \sum_{n=0}^{13} \binom{13}{n} - \sum_{n=0}^{13} 1\] \[\implies 2^{13} - 14 = 8192 - 14 = 8178 = \boxed{8178}.\]
| 178
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3,702
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_19
| 3
|
Suppose that $13$ cards numbered $1, 2, 3, \ldots, 13$ are arranged in a row. The task is to pick them up in numerically increasing order, working repeatedly from left to right. In the example below, cards $1, 2, 3$ are picked up on the first pass, $4$ and $5$ on the second pass, $6$ on the third pass, $7, 8, 9, 10$ on the fourth pass, and $11, 12, 13$ on the fifth pass. For how many of the $13!$ possible orderings of the cards will the $13$ cards be picked up in exactly two passes?
[asy] size(11cm); draw((0,0)--(2,0)--(2,3)--(0,3)--cycle); label("7", (1,1.5)); draw((3,0)--(5,0)--(5,3)--(3,3)--cycle); label("11", (4,1.5)); draw((6,0)--(8,0)--(8,3)--(6,3)--cycle); label("8", (7,1.5)); draw((9,0)--(11,0)--(11,3)--(9,3)--cycle); label("6", (10,1.5)); draw((12,0)--(14,0)--(14,3)--(12,3)--cycle); label("4", (13,1.5)); draw((15,0)--(17,0)--(17,3)--(15,3)--cycle); label("5", (16,1.5)); draw((18,0)--(20,0)--(20,3)--(18,3)--cycle); label("9", (19,1.5)); draw((21,0)--(23,0)--(23,3)--(21,3)--cycle); label("12", (22,1.5)); draw((24,0)--(26,0)--(26,3)--(24,3)--cycle); label("1", (25,1.5)); draw((27,0)--(29,0)--(29,3)--(27,3)--cycle); label("13", (28,1.5)); draw((30,0)--(32,0)--(32,3)--(30,3)--cycle); label("10", (31,1.5)); draw((33,0)--(35,0)--(35,3)--(33,3)--cycle); label("2", (34,1.5)); draw((36,0)--(38,0)--(38,3)--(36,3)--cycle); label("3", (37,1.5)); [/asy] $\textbf{(A) } 4082 \qquad \textbf{(B) } 4095 \qquad \textbf{(C) } 4096 \qquad \textbf{(D) } 8178 \qquad \textbf{(E) } 8191$
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To solve this problem, we can use recursion on $n$ . Let $A_n$ be the number of arrangements for $n$ numbers. Now, let's look at how these arrangements are formed by case work on the first number $a_1$
If $a_1 = 1$ , the remaining $n-1$ numbers from $2$ to $n$ are arranged in the same way just like number 1 to $n-1$ in the case of $n-1$ numbers. So there are $A_{n-1}$ arrangements.
If $a_1 = 2$ , then we need to choose 1 position from position 2 to $n-1$ to put 1, and all remaining numbers must be arranged in increasing order, so there are $\binom{n-1}{1}$ such arrangements.
If $a_1 = k$ , then we need to choose $k-1$ positions from position 2 to $n-1$ to put $1, 2,\cdots k-1$ , and all remaining numbers must be arranged in increasing order, so there are $\binom{n-1}{k-1}$ such arrangements.
So we can write \[A_n = A_{n-1} + \binom{n-1}{1} + \binom{n-1}{2} + \cdots + \binom{n-1}{n-1}\] which can be simplified to \[A_n = A_{n-1} + 2^{n-1} - 1\] We can solve this recursive sequence by summing up $n-1$ lines of the recursive formula \[A_n - A_{n-1} = 2^{n-1} - 1\] \[A_{n-1} - A_{n-2} = 2^{n-2} - 1\] \[\cdots\] \[A_2 - A_{1} = 2^{1} - 1\] to get \[A_n - A_1 = \sum_{k=1}^{n-1} (2^k - 1) = 2^n - 2 - (n-1) = 2^n - n - 1\] since $A_1=0$ , we have \[A_n = 2^n - n - 1\] and $A_{13} = 2^{13} - 14 = \boxed{8178}$
| 178
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3,703
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_19
| 4
|
Suppose that $13$ cards numbered $1, 2, 3, \ldots, 13$ are arranged in a row. The task is to pick them up in numerically increasing order, working repeatedly from left to right. In the example below, cards $1, 2, 3$ are picked up on the first pass, $4$ and $5$ on the second pass, $6$ on the third pass, $7, 8, 9, 10$ on the fourth pass, and $11, 12, 13$ on the fifth pass. For how many of the $13!$ possible orderings of the cards will the $13$ cards be picked up in exactly two passes?
[asy] size(11cm); draw((0,0)--(2,0)--(2,3)--(0,3)--cycle); label("7", (1,1.5)); draw((3,0)--(5,0)--(5,3)--(3,3)--cycle); label("11", (4,1.5)); draw((6,0)--(8,0)--(8,3)--(6,3)--cycle); label("8", (7,1.5)); draw((9,0)--(11,0)--(11,3)--(9,3)--cycle); label("6", (10,1.5)); draw((12,0)--(14,0)--(14,3)--(12,3)--cycle); label("4", (13,1.5)); draw((15,0)--(17,0)--(17,3)--(15,3)--cycle); label("5", (16,1.5)); draw((18,0)--(20,0)--(20,3)--(18,3)--cycle); label("9", (19,1.5)); draw((21,0)--(23,0)--(23,3)--(21,3)--cycle); label("12", (22,1.5)); draw((24,0)--(26,0)--(26,3)--(24,3)--cycle); label("1", (25,1.5)); draw((27,0)--(29,0)--(29,3)--(27,3)--cycle); label("13", (28,1.5)); draw((30,0)--(32,0)--(32,3)--(30,3)--cycle); label("10", (31,1.5)); draw((33,0)--(35,0)--(35,3)--(33,3)--cycle); label("2", (34,1.5)); draw((36,0)--(38,0)--(38,3)--(36,3)--cycle); label("3", (37,1.5)); [/asy] $\textbf{(A) } 4082 \qquad \textbf{(B) } 4095 \qquad \textbf{(C) } 4096 \qquad \textbf{(D) } 8178 \qquad \textbf{(E) } 8191$
|
When we have $3$ cards arranged in a row, after listing out all possible arrangements, we see that we have $4$ ones: $(1, 3, 2), (2, 1, 3), (2, 3, 1),$ and $(3, 1, 2)$ . When we have $4$ cards, we find $11$ possible arrangements: $(1, 2, 4, 3), (1, 3, 2, 4), (1, 3, 4, 2), (1, 4, 2, 3), (2, 1, 3, 4), (2, 3, 1, 4), (2, 3, 4, 1), (3, 1, 2, 4), (3, 1, 4, 2), (3, 4, 1, 2),$ and $(4, 1, 2, 3).$ Hence, we recognize the pattern that for $n$ cards, we have $2^n - n - 1$ valid arrangements, so our answer is $2^{13} - 13 - 1 = \boxed{8178}.$ ~eibc
| 178
|
3,704
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_19
| 5
|
Suppose that $13$ cards numbered $1, 2, 3, \ldots, 13$ are arranged in a row. The task is to pick them up in numerically increasing order, working repeatedly from left to right. In the example below, cards $1, 2, 3$ are picked up on the first pass, $4$ and $5$ on the second pass, $6$ on the third pass, $7, 8, 9, 10$ on the fourth pass, and $11, 12, 13$ on the fifth pass. For how many of the $13!$ possible orderings of the cards will the $13$ cards be picked up in exactly two passes?
[asy] size(11cm); draw((0,0)--(2,0)--(2,3)--(0,3)--cycle); label("7", (1,1.5)); draw((3,0)--(5,0)--(5,3)--(3,3)--cycle); label("11", (4,1.5)); draw((6,0)--(8,0)--(8,3)--(6,3)--cycle); label("8", (7,1.5)); draw((9,0)--(11,0)--(11,3)--(9,3)--cycle); label("6", (10,1.5)); draw((12,0)--(14,0)--(14,3)--(12,3)--cycle); label("4", (13,1.5)); draw((15,0)--(17,0)--(17,3)--(15,3)--cycle); label("5", (16,1.5)); draw((18,0)--(20,0)--(20,3)--(18,3)--cycle); label("9", (19,1.5)); draw((21,0)--(23,0)--(23,3)--(21,3)--cycle); label("12", (22,1.5)); draw((24,0)--(26,0)--(26,3)--(24,3)--cycle); label("1", (25,1.5)); draw((27,0)--(29,0)--(29,3)--(27,3)--cycle); label("13", (28,1.5)); draw((30,0)--(32,0)--(32,3)--(30,3)--cycle); label("10", (31,1.5)); draw((33,0)--(35,0)--(35,3)--(33,3)--cycle); label("2", (34,1.5)); draw((36,0)--(38,0)--(38,3)--(36,3)--cycle); label("3", (37,1.5)); [/asy] $\textbf{(A) } 4082 \qquad \textbf{(B) } 4095 \qquad \textbf{(C) } 4096 \qquad \textbf{(D) } 8178 \qquad \textbf{(E) } 8191$
|
Notice that for each card "position", we can choose for it to be picked up on the first or second pass, for a total of $2^{13}$ options. However, if all of the cards selected to be picked up first are before all of the cards to be picked up second, then this means that the list is in consecutive ascending order (and thus all cards will be picked up on the first pass instead). This can happen in 14 ways, so our answer is $2^{13}-14=\boxed{8178}$
| 178
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3,705
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_24
| 1
|
How many strings of length $5$ formed from the digits $0$ $1$ $2$ $3$ $4$ are there such that for each $j \in \{1,2,3,4\}$ , at least $j$ of the digits are less than $j$ ? (For example, $02214$ satisfies this condition
because it contains at least $1$ digit less than $1$ , at least $2$ digits less than $2$ , at least $3$ digits less
than $3$ , and at least $4$ digits less than $4$ . The string $23404$ does not satisfy the condition because it
does not contain at least $2$ digits less than $2$ .)
$\textbf{(A) }500\qquad\textbf{(B) }625\qquad\textbf{(C) }1089\qquad\textbf{(D) }1199\qquad\textbf{(E) }1296$
|
For some $n$ , let there be $n+1$ parking spaces counterclockwise in a circle. Consider a string of $n$ integers $c_1c_2 \ldots c_n$ each between $0$ and $n$ , and let $n$ cars come into this circle so that the $i$ th car tries to park at spot $c_i$ , but if it is already taken then it instead keeps going counterclockwise and takes the next available spot. After this process, exactly one spot will remain empty.
Then the strings of $n$ numbers between $0$ and $n-1$ that contain at least $k$ integers $<k$ for $1 \leq k \leq n$ are exactly the set of strings that leave spot $n$ empty. Also note for any string $c_1c_2 \ldots c_n$ , we can add $1$ to each $c_i$ (mod $n+1$ ) to shift the empty spot counterclockwise, meaning for each string there exists exactly one $j$ with $0 \leq j \leq n$ so that $(c_1+j)(c_2+j) \ldots (c_n+j)$ leaves spot $n$ empty. This gives there are $\frac{(n+1)^{n}}{n+1} = (n+1)^{n-1}$ such strings.
Plugging in $n = 5$ gives $\boxed{1296}$ such strings.
| 296
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3,706
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_24
| 2
|
How many strings of length $5$ formed from the digits $0$ $1$ $2$ $3$ $4$ are there such that for each $j \in \{1,2,3,4\}$ , at least $j$ of the digits are less than $j$ ? (For example, $02214$ satisfies this condition
because it contains at least $1$ digit less than $1$ , at least $2$ digits less than $2$ , at least $3$ digits less
than $3$ , and at least $4$ digits less than $4$ . The string $23404$ does not satisfy the condition because it
does not contain at least $2$ digits less than $2$ .)
$\textbf{(A) }500\qquad\textbf{(B) }625\qquad\textbf{(C) }1089\qquad\textbf{(D) }1199\qquad\textbf{(E) }1296$
|
Note that a valid string must have at least one $0.$
We perform casework on the number of different digits such strings can have. For each string, we list the digits in ascending order, then consider permutations:
Together, the answer is $1+75+500+600+120=\boxed{1296}.$
| 296
|
3,707
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_24
| 3
|
How many strings of length $5$ formed from the digits $0$ $1$ $2$ $3$ $4$ are there such that for each $j \in \{1,2,3,4\}$ , at least $j$ of the digits are less than $j$ ? (For example, $02214$ satisfies this condition
because it contains at least $1$ digit less than $1$ , at least $2$ digits less than $2$ , at least $3$ digits less
than $3$ , and at least $4$ digits less than $4$ . The string $23404$ does not satisfy the condition because it
does not contain at least $2$ digits less than $2$ .)
$\textbf{(A) }500\qquad\textbf{(B) }625\qquad\textbf{(C) }1089\qquad\textbf{(D) }1199\qquad\textbf{(E) }1296$
|
Denote by $N \left( p, q \right)$ the number of $p$ -digit strings formed by using numbers $0, 1, \cdots, q$ , where for each $j \in \{1,2, \cdots , q\}$ , at least $j$ of the digits are less than $j$
We have the following recursive equation: \[N \left( p, q \right) = \sum_{i = 0}^{p - q} \binom{p}{i} N \left( p - i, q - 1 \right) , \ \forall \ p \geq q \mbox{ and } q \geq 1\] and the boundary condition $N \left( p, 0 \right) = 1$ for any $p \geq 0$
By solving this recursive equation, for $q = 1$ and $p \geq q$ , we get \begin{align*} N \left( p , 1 \right) & = \sum_{i = 0}^{p - 1} \binom{p}{i} N \left( p - i, 0 \right) \\ & = \sum_{i = 0}^{p - 1} \binom{p}{i} \\ & = \sum_{i = 0}^p \binom{p}{i} - \binom{p}{p} \\ & = 2^p - 1 . \end{align*}
For $q = 2$ and $p \geq q$ , we get \begin{align*} N \left( p , 2 \right) & = \sum_{i = 0}^{p - 2} \binom{p}{i} N \left( p - i, 1 \right) \\ & = \sum_{i = 0}^{p - 2} \binom{p}{i} \left( 2^{p - i} - 1 \right) \\ & = \sum_{i = 0}^p \binom{p}{i} \left( 2^{p - i} - 1 \right) - \sum_{i = p - 1}^p \binom{p}{i} \left( 2^{p - i} - 1 \right) \\ & = \sum_{i = 0}^p \left( \binom{p}{i} 1^i 2^{p - i} - \binom{p}{i} 1^i 1^{p - i} \right) - p \\ & = \left( 1 + 2 \right)^p - \left( 1 + 1 \right)^p - p \\ & = 3^p - 2^p - p . \end{align*}
For $q = 3$ and $p \geq q$ , we get \begin{align*} N \left( p , 3 \right) & = \sum_{i = 0}^{p - 3} \binom{p}{i} N \left( p - i, 2 \right) \\ & = \sum_{i = 0}^{p - 3} \binom{p}{i} \left( 3^{p - i} - 2^{p - i} - \left( p - i \right) \right) \\ & = \sum_{i = 0}^p \binom{p}{i} \left( 3^{p - i} - 2^{p - i} - \left( p - i \right) \right) - \sum_{i = p - 2}^p \binom{p}{i} \left( 3^{p - i} - 2^{p - i} - \left( p - i \right) \right) \\ & = \sum_{i = 0}^p \left( \binom{p}{i} 1^i 3^{p - i} - \binom{p}{i} 1^i 2^{p - i} - \binom{p}{i} \left( p - i \right) \right) - \frac{3}{2} p \left( p - 1 \right) \\ & = \left( 1 + 3 \right)^p - \left( 1 + 2 \right)^p - \frac{d \left( 1 + x \right)^p}{dx} \bigg|_{x = 1} - \frac{3}{2} p \left( p - 1 \right) \\ & = 4^p - 3^p - 2^{p-1} p - \frac{3}{2} p \left( p - 1 \right) . \end{align*}
For $q = 4$ and $p = 5$ , we get \begin{align*} N \left( 5 , 4 \right) & = \sum_{i = 0}^1 \binom{5}{i} N \left( 5 - i , 3 \right) \\ & = N \left( 5 , 3 \right) + 5 N \left( 4 , 3 \right) \\ & = \boxed{1296}
| 296
|
3,708
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_24
| 4
|
How many strings of length $5$ formed from the digits $0$ $1$ $2$ $3$ $4$ are there such that for each $j \in \{1,2,3,4\}$ , at least $j$ of the digits are less than $j$ ? (For example, $02214$ satisfies this condition
because it contains at least $1$ digit less than $1$ , at least $2$ digits less than $2$ , at least $3$ digits less
than $3$ , and at least $4$ digits less than $4$ . The string $23404$ does not satisfy the condition because it
does not contain at least $2$ digits less than $2$ .)
$\textbf{(A) }500\qquad\textbf{(B) }625\qquad\textbf{(C) }1089\qquad\textbf{(D) }1199\qquad\textbf{(E) }1296$
|
The number of strings is $(n+1)^{(n-1)}$ as shown by Solution 1 (Parking Function), which is always equivalent to 1 (mod n). Thus you can choose $\boxed{1296}$
| 296
|
3,709
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_24
| 5
|
How many strings of length $5$ formed from the digits $0$ $1$ $2$ $3$ $4$ are there such that for each $j \in \{1,2,3,4\}$ , at least $j$ of the digits are less than $j$ ? (For example, $02214$ satisfies this condition
because it contains at least $1$ digit less than $1$ , at least $2$ digits less than $2$ , at least $3$ digits less
than $3$ , and at least $4$ digits less than $4$ . The string $23404$ does not satisfy the condition because it
does not contain at least $2$ digits less than $2$ .)
$\textbf{(A) }500\qquad\textbf{(B) }625\qquad\textbf{(C) }1089\qquad\textbf{(D) }1199\qquad\textbf{(E) }1296$
|
Solution 4 tried to observe the answer modulo $5$ to easily solve the problem, but apparently had faulty logic. This solution is still completely viable though:
Notice that for any valid set $\{a_1, a_2, a_3, a_4, a_5\}$ , if there is at least one element in the set that is unique (i.e. there is at least one digit in the set that is found nowhere else in the set), then the number of distinct permutations of the set is clearly divisible by $5$ . Therefore to evaluate the answer modulo $5$ , we only need to look at sets where each element has a multiplicity of at least $2$ (i.e. appears twice or more in the set).
These sets are of the form $\{a,a,b,b,b\}$ and $\{a,a,a,a,a\}$ . The first set can be permuted in $\binom{5}{2,3}=10 \equiv 0\pmod{5}$ , and the second set can be permuted one way, and the only set of the form $\{a,a,a,a,a\}$ is $\{0,0,0,0,0\}$ . Therefore the answer is congruent to $1\pmod{5}$ and you CAN choose $\boxed{1296}$
| 296
|
3,710
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_25
| 1
|
A circle with integer radius $r$ is centered at $(r, r)$ . Distinct line segments of length $c_i$ connect points $(0, a_i)$ to $(b_i, 0)$ for $1 \le i \le 14$ and are tangent to the circle, where $a_i$ $b_i$ , and $c_i$ are all positive integers and $c_1 \le c_2 \le \cdots \le c_{14}$ . What is the ratio $\frac{c_{14}}{c_1}$ for the least possible value of $r$
$\textbf{(A)} ~\frac{21}{5} \qquad\textbf{(B)} ~\frac{85}{13} \qquad\textbf{(C)} ~7 \qquad\textbf{(D)} ~\frac{39}{5} \qquad\textbf{(E)} ~17$
|
Suppose that with a pair $(a_i,b_i)$ the circle is an excircle. Then notice that the hypotenuse must be $(r-x)+(r-y)$ , so it must be the case that \[a_i^2+b_i^2=(2r-a_i-b_i)^2.\] Similarly, if with a pair $(a_i,b_i)$ the circle is an incircle, the hypotenuse must be $(x-r)+(y-r)$ , leading to the same equation.
Notice that this equation can be simplified through SFFT to \[(a_i-2r)(b_i-2r)=2r^2.\] Thus, we want the smallest $r$ such that this equation has at least $14$ distinct pairs $(a_i,b_i)$ for which this holds. The obvious choice to check is $r=6$ . In this case, since $2r^2=2^3\cdot 3^2$ has $12$ positive factors, we get $12$ pairs, and we get another two if the factors are $-8,-9$ or vice versa. One can check that for smaller values of $r$ , we do not even get close to $14$ possible pairs.
When $r=6$ , the smallest possible $c$ -value is clearly when the factors are negative. When this occurs, $a_i=4, b_i=3$ (or vice versa), so the mimimal $c$ is $5$ . The largest possible $c$ -value occurs when the largest of $a_i$ and $b_i$ are maximized. This occurs when the factors are $72$ and $1$ , leading to $a_i=84, b_i=13$ (or vice-versa), leading to a maximal $c$ of $85$
Hence the answer is $\frac{85}5=\boxed{17}$
| 17
|
3,711
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_25
| 2
|
A circle with integer radius $r$ is centered at $(r, r)$ . Distinct line segments of length $c_i$ connect points $(0, a_i)$ to $(b_i, 0)$ for $1 \le i \le 14$ and are tangent to the circle, where $a_i$ $b_i$ , and $c_i$ are all positive integers and $c_1 \le c_2 \le \cdots \le c_{14}$ . What is the ratio $\frac{c_{14}}{c_1}$ for the least possible value of $r$
$\textbf{(A)} ~\frac{21}{5} \qquad\textbf{(B)} ~\frac{85}{13} \qquad\textbf{(C)} ~7 \qquad\textbf{(D)} ~\frac{39}{5} \qquad\textbf{(E)} ~17$
|
Case 1: The tangent and the origin are on the opposite sides of the circle.
In this case, $a, b > 2r$
We can easily prove that \[a + b - 2 r = c . \hspace{1cm} (1)\]
Recall that $c = \sqrt{a^2 + b^2}$
Taking square of (1) and reorganizing all terms, (1) is converted as \[\left( a - 2 r \right) \left( b - 2 r \right) = 2 r^2 .\]
Case 2: The tangent and the origin are on the same sides of the circle.
In this case, $0 < a, b < r$
We can easily prove that \[2 r - a - b = c . \hspace{1cm} (2)\]
Recall that $c = \sqrt{a^2 + b^2}$
Taking square of (2) and reorganizing all terms, (2) is converted as \[\left( a - 2 r \right) \left( b - 2 r \right) = 2 r^2 .\]
Putting both cases together, for given $r$ , we look for solutions of $a$ and $b$ satisfying \[\left( a - 2 r \right) \left( b - 2 r \right) = 2 r^2 ,\] with either $a, b > 2r$ or $0 < a, b < r$
Now, we need to find the smallest $r$ , such that the number of feasible solutions of $(a, b)$ is at least 14.
For equation \[uv = 2 r^2 ,\] we observe that the R.H.S. is a not a perfect square. Thus, the number of positive $(u, v)$ is equal to the number of positive divisors of $2 r^2$
Second, for each feasible positive solution $(u, v)$ , its opposite $(-u, -v)$ is also a solution. However, $(u,v)$ corresponds to a feasible solution if $(a, b)$ with $a = u + 2r$ and $b = v + 2r$ , but $(-u, -v)$ may not lead to a feasible solution if $(a, b)$ with $a = 2 r - u$ and $b = 2 r - v$
Recall that we are looking for $r$ that leads to at least 14 solutions.
Therefore, the above observations imply that we must have $r$ , such that $2 r^2$ has least 7 positive divisors.
Following this guidance, we find the smallest $r$ is 6. This leads to the following solutions:
$\left( a_1, b_1, c_1 \right) = \left( 3, 4, 5 \right)$ $\left( a_2, b_2, c_2 \right) = \left( 4, 3, 5 \right)$
$\left( a_3, b_3, c_3 \right) = \left( 20, 21, 29 \right)$ $\left( a_4, b_4, c_4 \right) = \left( 21, 20, 29 \right)$
$\left( a_5, b_5, c_5 \right) = \left( 18, 24, 30 \right)$ $\left( a_6, b_6, c_6 \right) = \left( 24, 18, 30 \right)$
$\left( a_7, b_7, c_7 \right) = \left( 16, 30, 34 \right)$ $\left( a_8, b_8, c_8 \right) = \left( 30, 16, 34 \right)$
$\left( a_9, b_9, c_9 \right) = \left( 15, 36, 39 \right)$ $\left( a_{10}, b_{10}, c_{10} \right) = \left( 36, 15, 39 \right)$
$\left( a_{11}, b_{11}, c_{11} \right) = \left( 14, 48, 50 \right)$ $\left( a_{12}, b_{12}, c_{12} \right) = \left( 48, 14, 50 \right)$
$\left( a_{13}, b_{13}, c_{13} \right) = \left( 13, 84, 85 \right)$ $\left( a_{14}, b_{14}, c_{14} \right) = \left( 84, 13, 85 \right)$
Therefore, $\frac{c_{14}}{c_1} = \boxed{17}$
| 17
|
3,712
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_25
| 3
|
A circle with integer radius $r$ is centered at $(r, r)$ . Distinct line segments of length $c_i$ connect points $(0, a_i)$ to $(b_i, 0)$ for $1 \le i \le 14$ and are tangent to the circle, where $a_i$ $b_i$ , and $c_i$ are all positive integers and $c_1 \le c_2 \le \cdots \le c_{14}$ . What is the ratio $\frac{c_{14}}{c_1}$ for the least possible value of $r$
$\textbf{(A)} ~\frac{21}{5} \qquad\textbf{(B)} ~\frac{85}{13} \qquad\textbf{(C)} ~7 \qquad\textbf{(D)} ~\frac{39}{5} \qquad\textbf{(E)} ~17$
|
As $c_i$ is the length of the segment $(0,a_i)$ and $(b_i,0)$ $a_i^2+b_i^2=c_i^2$ . The equation for the line that passes through $(0,a_i)$ and $(b_i,0)$ is $a_ix+b_iy-a_ib_i=0$
By the point-line distance formula from point $(r,r)$ to line $a_ix+b_iy-a_ib_i=0$
\[r = \frac{ |a_ir+b_ir-a_ib_i| }{ \sqrt{a_i^2+b_i^2} }, \quad c_i = \frac{ |a_ir+b_ir-a_ib_i| }{r} = |a_i+b_i- \frac{a_ib_i}{r}|\]
\[\because c_i \quad \text{is an integer}, \quad \therefore r|a_ib_i\]
To find $c_1$ and $c_{14}$ for the smallest $r$ , we will list Pythagorean triples $(a_i,b_i,c_i)$ with relatively prime elements from the smallest. Notice that we only need to find $7$ triples with $a_i<b_i$ , the $7$ other triples will simply be $(b_i,a_i,c_i)$ $a_i$ will not equal $b_i$ because then $c_i = a_i \cdot \sqrt{2}$ , meaning that $a_i$ $b_i$ , and $c_i$ cannot all be integers.
The $7$ smallest triples are: $(3,4,5)$ $(5,12,13)$ $(7,24,25)$ $(8,15,17)$ $(11,60,61)$ $(12,35,37)$ $(13,84,85)$
Therefore, $\frac{c_{14}}{c_1} = \frac{85}{5} = \boxed{17}$
| 17
|
3,713
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_1
| 1
|
Define $x\diamond y$ to be $|x-y|$ for all real numbers $x$ and $y.$ What is the value of \[(1\diamond(2\diamond3))-((1\diamond2)\diamond3)?\]
$\textbf{(A)}\ {-}2 \qquad \textbf{(B)}\ {-}1 \qquad \textbf{(C)}\ 0 \qquad \textbf{(D)}\ 1 \qquad \textbf{(E)}\ 2$
|
We have \begin{align*} (1\diamond(2\diamond3))-((1\diamond2)\diamond3) &= |1-|2-3|| - ||1-2|-3| \\ &= |1-1| - |1-3| \\ &= 0-2 \\ &= \boxed{2} ~MRENTHUSIASM
| 2
|
3,714
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_1
| 2
|
Define $x\diamond y$ to be $|x-y|$ for all real numbers $x$ and $y.$ What is the value of \[(1\diamond(2\diamond3))-((1\diamond2)\diamond3)?\]
$\textbf{(A)}\ {-}2 \qquad \textbf{(B)}\ {-}1 \qquad \textbf{(C)}\ 0 \qquad \textbf{(D)}\ 1 \qquad \textbf{(E)}\ 2$
|
Observe that the $\diamond$ function is simply the positive difference between two numbers. Thus, we evaluate: the difference between $2$ and $3$ is $1;$ the difference between $1$ and $1$ is $0;$ the difference between $1$ and $2$ is $1;$ the difference between $1$ and $3$ is $2;$ and finally, $0-2=\boxed{2}.$
| 2
|
3,715
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_2
| 1
|
In rhombus $ABCD$ , point $P$ lies on segment $\overline{AD}$ so that $\overline{BP}$ $\perp$ $\overline{AD}$ $AP = 3$ , and $PD = 2$ . What is the area of $ABCD$ ? (Note: The figure is not drawn to scale.)
[asy] import olympiad; size(180); real r = 3, s = 5, t = sqrt(r*r+s*s); defaultpen(linewidth(0.6) + fontsize(10)); pair A = (0,0), B = (r,s), C = (r+t,s), D = (t,0), P = (r,0); draw(A--B--C--D--A^^B--P^^rightanglemark(B,P,D)); label("$A$",A,SW); label("$B$", B, NW); label("$C$",C,NE); label("$D$",D,SE); label("$P$",P,S); [/asy]
$\textbf{(A) }3\sqrt 5 \qquad \textbf{(B) }10 \qquad \textbf{(C) }6\sqrt 5 \qquad \textbf{(D) }20\qquad \textbf{(E) }25$
|
[asy] pair A = (0,0); label("$A$", A, SW); pair B = (2.25,3); label("$B$", B, NW); pair C = (6,3); label("$C$", C, NE); pair D = (3.75,0); label("$D$", D, SE); pair P = (2.25,0); label("$P$", P, S); draw(A--B--C--D--cycle); draw(P--B); draw(rightanglemark(B,P,D)); [/asy]
\[\textbf{Figure redrawn to scale.}\]
$AD = AP + PD = 3 + 2 = 5$
$ABCD$ is a rhombus, so $AB = AD = 5$
$\bigtriangleup APB$ is a $3-4-5$ right triangle, hence $BP = 4$
The area of the rhombus is base times height: $bh = (AD)(BP) = 5 \cdot 4 = \boxed{20}$
| 20
|
3,716
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_2
| 2
|
In rhombus $ABCD$ , point $P$ lies on segment $\overline{AD}$ so that $\overline{BP}$ $\perp$ $\overline{AD}$ $AP = 3$ , and $PD = 2$ . What is the area of $ABCD$ ? (Note: The figure is not drawn to scale.)
[asy] import olympiad; size(180); real r = 3, s = 5, t = sqrt(r*r+s*s); defaultpen(linewidth(0.6) + fontsize(10)); pair A = (0,0), B = (r,s), C = (r+t,s), D = (t,0), P = (r,0); draw(A--B--C--D--A^^B--P^^rightanglemark(B,P,D)); label("$A$",A,SW); label("$B$", B, NW); label("$C$",C,NE); label("$D$",D,SE); label("$P$",P,S); [/asy]
$\textbf{(A) }3\sqrt 5 \qquad \textbf{(B) }10 \qquad \textbf{(C) }6\sqrt 5 \qquad \textbf{(D) }20\qquad \textbf{(E) }25$
|
[asy] pair A = (0,0); label("$A$", A, SW); pair B = (2.25,3); label("$B$", B, NW); pair C = (6,3); label("$C$", C, NE); pair D = (3.75,0); label("$D$", D, SE); pair P = (2.25,0); label("$P$", P, S); draw(A--B--C--D--cycle); draw(D--B); draw(B--P); draw(rightanglemark(B,P,D)); [/asy]
The diagram is from as Solution 1, but a line is constructed at $BD$
When it comes to the sides of a rhombus, their opposite sides are congruent and parallel. This means that $\angle ABD \cong \angle BDC$ , by the Alternate Interior Angles Theorem.
By SAS Congruence, we get $\triangle ABD \cong \triangle BDC$
Since $AP=3$ and $AB=5$ , we know that $BP=4$ because $\triangle APB$ is a 3-4-5 right triangle, as stated in Solution 1.
The area of $\triangle ABD$ would be $10$ , since the area of the triangle is $\frac{bh}{2}$
Since we know that $\triangle ABD \cong \triangle BDC$ and that $ABCD=\triangle ABD + \triangle BDC$ , so we can double the area of $\triangle ADB$ to get $10 \cdot 2 = \boxed{20}$
| 20
|
3,717
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_3
| 1
|
How many of the first ten numbers of the sequence $121, 11211, 1112111, \ldots$ are prime numbers?
$\textbf{(A) } 0 \qquad \textbf{(B) }1 \qquad \textbf{(C) }2 \qquad \textbf{(D) }3 \qquad \textbf{(E) }4$
|
The $n$ th term of this sequence is \[\sum_{k=n}^{2n}10^k + \sum_{k=0}^{n}10^k = 10^n\sum_{k=0}^{n}10^k + \sum_{k=0}^{n}10^k = \left(10^n+1\right)\sum_{k=0}^{n}10^k.\] It follows that the terms are \begin{align*} 121 &= 11\cdot11, \\ 11211 &= 101\cdot111, \\ 1112111 &= 1001\cdot1111, \\ & \ \vdots \end{align*} Therefore, there are $\boxed{0}$ prime numbers in this sequence.
| 0
|
3,718
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_3
| 2
|
How many of the first ten numbers of the sequence $121, 11211, 1112111, \ldots$ are prime numbers?
$\textbf{(A) } 0 \qquad \textbf{(B) }1 \qquad \textbf{(C) }2 \qquad \textbf{(D) }3 \qquad \textbf{(E) }4$
|
Denote this sequence as $a_{n}$ , then we can find that \begin{align*} a_{1} &= 121 = 10^2 + 2\cdot10 + 1 = (10^2 + 10) + (10 + 1), \\ a_{2} &= 11211 = (10^4 + 10^3 + 10^2) + (10^2 + 10 + 1), \\ a_{3} &= 1112111 = (10^6 + 10^5 + 10^4 + 10^3) + (10^3 + 10^2 + 10 + 1), \\ & \ \vdots \end{align*} So, we can induct that the general term is \begin{align*} a_n &= (10^{2n} + 10^{2n-1} + \ldots + 10^{n+1} + 10^n) + (10^n + 10^{n-1} + \ldots +10 + 1) \\ &= 10^n\cdot(10^n + 10^{n-1} + \ldots +10 + 1) + (10^n + 10^{n-1} + \ldots +10 + 1) \\ &= \left(10^n+1\right)\sum_{k=0}^{n}10^k. \end{align*} Therefore, there are $\boxed{0}$ prime numbers in this sequence.
| 0
|
3,719
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_3
| 3
|
How many of the first ten numbers of the sequence $121, 11211, 1112111, \ldots$ are prime numbers?
$\textbf{(A) } 0 \qquad \textbf{(B) }1 \qquad \textbf{(C) }2 \qquad \textbf{(D) }3 \qquad \textbf{(E) }4$
|
Observe how \begin{align*} 121 &= 110 + 11, \\ 11211 &= 11100 + 111, \\ 1112111 &= 1111000 + 1111, \\ & \ \vdots \end{align*} all take the form of \[\underbrace{111\ldots}_{n+1}\underbrace{00\ldots}_{n} + \underbrace{111\ldots}_{n+1} = \underbrace{111\ldots}_{n+1}(10^{n} + 1).\] Factoring each of the sums, we have \[11(10+1), 111(100+1), 1111(1000+1), \ldots\] respectively. With each number factored, there are $\boxed{0}$ primes in the set.
| 0
|
3,720
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_3
| 4
|
How many of the first ten numbers of the sequence $121, 11211, 1112111, \ldots$ are prime numbers?
$\textbf{(A) } 0 \qquad \textbf{(B) }1 \qquad \textbf{(C) }2 \qquad \textbf{(D) }3 \qquad \textbf{(E) }4$
|
Note that $121$ is divisible by $11$ and $11211$ is divisible by $3$ . Because this is Problem 6 of the AMC 10, we assume we do not need to check two-digit prime divisibility or use obscure theorems. Therefore, the answer is $\boxed{0}.$
| 0
|
3,721
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_4
| 2
|
For how many values of the constant $k$ will the polynomial $x^{2}+kx+36$ have two distinct integer roots?
$\textbf{(A)}\ 6 \qquad\textbf{(B)}\ 8 \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ 14 \qquad\textbf{(E)}\ 16$
|
Note that $k$ must be an integer. Using the quadratic formula $x=\frac{-k \pm \sqrt{k^2-144}}{2}.$ Since $4$ divides $144$ evenly, $k$ and $k^2-144$ have the same parity, so $x$ is an integer if and only if $k^2-144$ is a perfect square.
Let $k^2-144=n^2.$ Then, $(k+n)(k-n)=144.$ Since $k$ is an integer and $144$ is even, $k+n$ and $k-n$ must both be even. Assuming that $k$ is positive, we get $5$ possible values of $k+n$ , namely $2, 4, 8, 6, 12$ , which will give distinct positive values of $k$ , but $k+n=12$ gives $k+n=k-n$ and $n=0$ , giving $2$ identical integer roots. Therefore, there are $4$ distinct positive values of $k.$ Multiplying that by $2$ to take the negative values into account, we get $4\cdot2=\boxed{8}$ values of $k$
| 8
|
3,722
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_4
| 3
|
For how many values of the constant $k$ will the polynomial $x^{2}+kx+36$ have two distinct integer roots?
$\textbf{(A)}\ 6 \qquad\textbf{(B)}\ 8 \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ 14 \qquad\textbf{(E)}\ 16$
|
Proceed similar to Solution 2 and deduce that the discriminant of $x^{2}+kx+36$ must be a perfect square greater than $0$ to satisfy all given conditions. Seeing something like $k^2-144$ might remind us of a right triangle, where $k$ is the hypotenuse, and $12$ is a leg. There are four ways we could have this: a $9$ $12$ $15$ triangle, a $12$ $16$ $20$ triangle, a $5$ $12$ $13$ triangle, and a $12$ $35$ $37$ triangle.
Multiply by $2$ to account for negative $k$ values (since $k$ is being squared), and our answer is $\boxed{8}$
| 8
|
3,723
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_4
| 4
|
For how many values of the constant $k$ will the polynomial $x^{2}+kx+36$ have two distinct integer roots?
$\textbf{(A)}\ 6 \qquad\textbf{(B)}\ 8 \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ 14 \qquad\textbf{(E)}\ 16$
|
Since $36 = 2^2\cdot3^2$ , that means there are $(2+1)(2+1) = 9$ possible factors of $36$ . Since $6 \cdot 6$ violates the distinct root condition, subtract $1$ from $9$ to get $8$ . Each sum is counted twice, and we count of those twice for negatives. This cancels out, so we get $\boxed{8}$
| 8
|
3,724
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_6
| 1
|
Consider the following $100$ sets of $10$ elements each: \begin{align*} &\{1,2,3,\ldots,10\}, \\ &\{11,12,13,\ldots,20\},\\ &\{21,22,23,\ldots,30\},\\ &\vdots\\ &\{991,992,993,\ldots,1000\}. \end{align*} How many of these sets contain exactly two multiples of $7$
$\textbf{(A)}\ 40\qquad\textbf{(B)}\ 42\qquad\textbf{(C)}\ 43\qquad\textbf{(D)}\ 49\qquad\textbf{(E)}\ 50$
|
We apply casework to this problem. The only sets that contain two multiples of seven are those for which:
Each case has $\left\lfloor\frac{100}{7}\right\rfloor=14$ sets. Therefore, the answer is $14\cdot3=\boxed{42}.$
| 42
|
3,725
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_6
| 2
|
Consider the following $100$ sets of $10$ elements each: \begin{align*} &\{1,2,3,\ldots,10\}, \\ &\{11,12,13,\ldots,20\},\\ &\{21,22,23,\ldots,30\},\\ &\vdots\\ &\{991,992,993,\ldots,1000\}. \end{align*} How many of these sets contain exactly two multiples of $7$
$\textbf{(A)}\ 40\qquad\textbf{(B)}\ 42\qquad\textbf{(C)}\ 43\qquad\textbf{(D)}\ 49\qquad\textbf{(E)}\ 50$
|
We find a pattern. \begin{align*} &\{1,2,3,\ldots,10\}, \\ &\{11,12,13,\ldots,20\},\\ &\{21,22,23,\ldots,30\},\\ &\vdots\\ &\{991,992,993,\ldots,1000\}. \end{align*} We can figure out that the first set has $1$ multiple of $7$ . The second set also has $1$ multiple of $7$ . The third set has $2$ multiples of $7$ . The fourth set has $1$ multiple of $7$ . The fifth set has $2$ multiples of $7$ . The sixth set has $1$ multiple of $7$ . The seventh set has $2$ multiples of $7$ . Calculating this pattern further, we can see (reasonably) that it repeats for each $7$ sets.
We see that the pattern for the number of multiples per $7$ sets goes: $1,1,2,1,2,1,2.$ So, for every $7$ sets, there are three sets with $2$ multiples of $7$ . We calculate $\left\lfloor\frac{100}{7}\right\rfloor$ and multiply that by $3$ . (We also disregard the remainder of $2$ since it doesn't add any extra sets with $2$ multiples of $7$ .). We get $14\cdot3= \boxed{42}$
| 42
|
3,726
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_7
| 1
|
Camila writes down five positive integers. The unique mode of these integers is $2$ greater than their median, and the median is $2$ greater than their arithmetic mean. What is the least possible value for the mode?
$\textbf{(A)}\ 5 \qquad\textbf{(B)}\ 7 \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ 11 \qquad\textbf{(E)}\ 13$
|
Let $M$ be the median. It follows that the two largest integers are both $M+2.$
Let $a$ and $b$ be the two smallest integers such that $a<b.$ The sorted list is \[a,b,M,M+2,M+2.\] Since the median is $2$ greater than their arithmetic mean, we have $\frac{a+b+M+(M+2)+(M+2)}{5}+2=M,$ or \[a+b+14=2M.\] Note that $a+b$ must be even. We minimize this sum so that the arithmetic mean, the median, and the unique mode are minimized. Let $a=1$ and $b=3,$ from which $M=9$ and $M+2=\boxed{11}.$
| 11
|
3,727
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_7
| 2
|
Camila writes down five positive integers. The unique mode of these integers is $2$ greater than their median, and the median is $2$ greater than their arithmetic mean. What is the least possible value for the mode?
$\textbf{(A)}\ 5 \qquad\textbf{(B)}\ 7 \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ 11 \qquad\textbf{(E)}\ 13$
|
We can also easily test all the answer choices. (This strategy is generally good to use for multiple-choice questions if you don't have a concrete method to proceed with!)
For answer choice $\textbf{(A)},$ the mode is $5,$ the median is $3,$ and the arithmetic mean is $1.$ However, we can quickly see this doesn't work, as there are five integers, and they can't have an arithmetic mean of $1$ while having a mode of $5.$
Trying answer choice $\textbf{(B)},$ the mode is $7,$ the median is $5,$ and the arithmetic mean is $3.$ From the arithmetic mean, we know that all the numbers have to sum to $15.$ We know three of the numbers: $\underline{\hspace{3mm}},\underline{\hspace{3mm}},5,7,7.$ This exceeds the sum of $15.$
Now we try answer choice $\textbf{(C)}.$ The mode is $9,$ the median is $7,$ and the arithmetic mean is $5.$ From the arithmetic mean, we know that the list sums to $25.$ Three of the numbers are $\underline{\hspace{3mm}},\underline{\hspace{3mm}},7,9,9,$ which is exactly $25.$ However, our list needs positive integers, so this won't work.
Since we were really close on answer choice $\textbf{(C)},$ we can intuitively feel that the answer is probably going to be $\textbf{(D)}.$ We can confirm this by creating a list that satisfies the problem and choose $\textbf{(D)}: 1,3,9,11,11.$
So, our answer is $\boxed{11}.$
| 11
|
3,728
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_9
| 1
|
The sequence $a_0,a_1,a_2,\cdots$ is a strictly increasing arithmetic sequence of positive integers such that \[2^{a_7}=2^{27} \cdot a_7.\] What is the minimum possible value of $a_2$
$\textbf{(A)}\ 8 \qquad \textbf{(B)}\ 12 \qquad \textbf{(C)}\ 16 \qquad \textbf{(D)}\ 17 \qquad \textbf{(E)}\ 22$
|
We can rewrite the given equation as $2^{a_7-27}=a_7$ . Hence, $a_7$ must be a power of $2$ and larger than $27$ . The first power of 2 that is larger than $27$ , namely $32$ , does satisfy the equation: $2^{32 - 27} = 2^5 = 32$ . In fact, this is the only solution; $2^{a_7-27}$ is exponential whereas $a_7$ is linear, so their graphs will not intersect again.
Now, let the common difference in the sequence be $d$ . Hence, $a_0 = 32 - 7d$ and $a_2 = 32 - 5d$ . To minimize $a_2$ , we maxmimize $d$ . Since the sequence contains only positive integers, $32 - 7d > 0$ and hence $d \leq 4$ . When $d = 4$ $a_2 = \boxed{12}$
| 12
|
3,729
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_11
| 3
|
Let $f(n) = \left( \frac{-1+i\sqrt{3}}{2} \right)^n + \left( \frac{-1-i\sqrt{3}}{2} \right)^n$ , where $i = \sqrt{-1}$ . What is $f(2022)$
$\textbf{(A)}\ -2 \qquad \textbf{(B)}\ -1 \qquad \textbf{(C)}\ 0 \qquad \textbf{(D)}\ \sqrt{3} \qquad \textbf{(E)}\ 2$
|
We begin by recognizing this form looks similar to the definition of cosine: \[\cos(x)=\frac{e^{ix}+e^{-ix}}{2}.\] We can convert our two terms into exponential form to find \[f(n) = \left( e^{\frac{2\pi i}{3}} \right ) ^n + \left ( e^{-\frac{2\pi i}{3}} \right ) ^n=e^{\frac{2 \pi i n}{3}} + e^{-\frac{2\pi i n}{3}}.\] This simplifies nicely: \[f(n)=2\cos\left( \frac{2\pi n}{3} \right).\] Thus, \[f(2022)=2\cos \left ( \frac{2\pi (2022) }{3} \right) = 2\cos(1348 \pi) = \boxed{2}.\]
| 2
|
3,730
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_11
| 4
|
Let $f(n) = \left( \frac{-1+i\sqrt{3}}{2} \right)^n + \left( \frac{-1-i\sqrt{3}}{2} \right)^n$ , where $i = \sqrt{-1}$ . What is $f(2022)$
$\textbf{(A)}\ -2 \qquad \textbf{(B)}\ -1 \qquad \textbf{(C)}\ 0 \qquad \textbf{(D)}\ \sqrt{3} \qquad \textbf{(E)}\ 2$
|
Notice how this looks like the closed form of the Fibonacci sequence except different roots. This is motivation to turn this closed formula into a recurrence relation. The base of the exponents are the roots of the characteristic equation $r^3-1=0$ . So we have \begin{align*} a_0&=2\\ a_1&=-1\\ a_2&=-1\\ a_n&=a_{n-3} \end{align*} Every time $n$ is multiple of $3$ as is true when $n=2022$ $a_n= \boxed{2}$ ~lopkiloinm
| 2
|
3,731
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_11
| 5
|
Let $f(n) = \left( \frac{-1+i\sqrt{3}}{2} \right)^n + \left( \frac{-1-i\sqrt{3}}{2} \right)^n$ , where $i = \sqrt{-1}$ . What is $f(2022)$
$\textbf{(A)}\ -2 \qquad \textbf{(B)}\ -1 \qquad \textbf{(C)}\ 0 \qquad \textbf{(D)}\ \sqrt{3} \qquad \textbf{(E)}\ 2$
|
Converting the two terms into rectangular form,
\[f(2022)=\left(\cos{\frac{2\pi}{3}}+i\sin{\frac{2\pi}{3}}\right)^{2022}+\left(\cos{\frac{4\pi}{3}}+i\sin{\frac{4\pi}{3}}\right)^{2022}.\]
By DeMoivre's Theorem,
\[f(2022)=\left(\cos{\left(\frac{2\pi}{3}\cdot{2022}\right)}+i\sin{\left(\frac{2\pi}{3}\cdot{2022}\right)}\right)+\left(\cos{\left(\frac{4\pi}{3}\cdot{2022}\right)}+i\sin{\left(\frac{4\pi}{3}\cdot{2022}\right)}\right).\]
Note that $\cos{\pi\cdot{k}}=1$ if $k$ is even and $-1$ if $k$ is odd, and that $\sin{\pi\cdot{k}}=0$ for all integers $k$
All arguments are even in the second equation for $f(2022)$ , so the two $\cos$ terms are equal to $1$ , and the two $\sin$ terms are equal to $0$
Therefore the answer is $1+1=\boxed{2}.$
| 2
|
3,732
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_17
| 1
|
How many $4 \times 4$ arrays whose entries are $0$ s and $1$ s are there such that the row sums (the sum of the entries in each row) are $1, 2, 3,$ and $4,$ in some order, and the column sums (the sum of the entries in each column) are also $1, 2, 3,$ and $4,$ in some order? For example, the array \[\left[ \begin{array}{cccc} 1 & 1 & 1 & 0 \\ 0 & 1 & 1 & 0 \\ 1 & 1 & 1 & 1 \\ 0 & 1 & 0 & 0 \\ \end{array} \right]\] satisfies the condition.
$\textbf{(A) }144 \qquad \textbf{(B) }240 \qquad \textbf{(C) }336 \qquad \textbf{(D) }576 \qquad \textbf{(E) }624$
|
Note that the arrays and the sum configurations have one-to-one correspondence. Furthermore, the row sum configuration and the column sum configuration are independent of each other. Therefore, the answer is $(4!)^2=\boxed{576}.$
| 576
|
3,733
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_17
| 2
|
How many $4 \times 4$ arrays whose entries are $0$ s and $1$ s are there such that the row sums (the sum of the entries in each row) are $1, 2, 3,$ and $4,$ in some order, and the column sums (the sum of the entries in each column) are also $1, 2, 3,$ and $4,$ in some order? For example, the array \[\left[ \begin{array}{cccc} 1 & 1 & 1 & 0 \\ 0 & 1 & 1 & 0 \\ 1 & 1 & 1 & 1 \\ 0 & 1 & 0 & 0 \\ \end{array} \right]\] satisfies the condition.
$\textbf{(A) }144 \qquad \textbf{(B) }240 \qquad \textbf{(C) }336 \qquad \textbf{(D) }576 \qquad \textbf{(E) }624$
|
In this problem, we call a matrix that satisfies all constraints given in the problem a feasible matrix.
First, we observe that if a matrix is feasible, and we swap two rows or two columns to get a new matrix, then this new matrix is still feasible.
Therefore, any feasible matrix can be obtained through a sequence of such swapping operations from a feasible matrix where for all $i \in \left\{ 1, 2, 3 ,4 \right\}$ , the sum of entries in row $i$ is $i$ and the sum of entries in column $i$ is $i$ , hereafter called as a benchmark matrix.
Second, we observe that there is a unique benchmark matrix, as shown below: \[\left[ \begin{array}{cccc} 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 1 \\ 0 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 \\ \end{array} \right]\] With above observations, we now count the number of feasible matrixes.
We construct a feasible matrix in the following steps.
Step 1: We make a permutation of rows of the benchmark matrix.
The number of ways is $4!$
Step 2: We make a permutation of columns of the matrix obtained after Step 1.
The number of ways is $4!$
Following from the rule of product, the total number of feasible matrixes is $4! \cdot 4! = \boxed{576}$
| 576
|
3,734
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_17
| 3
|
How many $4 \times 4$ arrays whose entries are $0$ s and $1$ s are there such that the row sums (the sum of the entries in each row) are $1, 2, 3,$ and $4,$ in some order, and the column sums (the sum of the entries in each column) are also $1, 2, 3,$ and $4,$ in some order? For example, the array \[\left[ \begin{array}{cccc} 1 & 1 & 1 & 0 \\ 0 & 1 & 1 & 0 \\ 1 & 1 & 1 & 1 \\ 0 & 1 & 0 & 0 \\ \end{array} \right]\] satisfies the condition.
$\textbf{(A) }144 \qquad \textbf{(B) }240 \qquad \textbf{(C) }336 \qquad \textbf{(D) }576 \qquad \textbf{(E) }624$
|
Of the sixteen entries in one such array, there are six $0$ s and ten $1$ s. The rows and the columns each have zero, one, two, and three $0$ s, in some order. Once we decide the positions of the $0$ s, we form one such array.
We perform the following steps:
Together, the answer is $4\cdot4\cdot3\cdot3\cdot4=\boxed{576}.$
| 576
|
3,735
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_17
| 4
|
How many $4 \times 4$ arrays whose entries are $0$ s and $1$ s are there such that the row sums (the sum of the entries in each row) are $1, 2, 3,$ and $4,$ in some order, and the column sums (the sum of the entries in each column) are also $1, 2, 3,$ and $4,$ in some order? For example, the array \[\left[ \begin{array}{cccc} 1 & 1 & 1 & 0 \\ 0 & 1 & 1 & 0 \\ 1 & 1 & 1 & 1 \\ 0 & 1 & 0 & 0 \\ \end{array} \right]\] satisfies the condition.
$\textbf{(A) }144 \qquad \textbf{(B) }240 \qquad \textbf{(C) }336 \qquad \textbf{(D) }576 \qquad \textbf{(E) }624$
|
Since exactly $1$ row sum is $4$ and exactly $1$ column sum is $4$ , there is a unique entry in the array such that it, and every other entry in the same row or column, is a $1.$ Since there are $16$ total entries in the array, there are $16$ ways to choose the entry with only $1$ s in its row and column.
Without loss of generality, let that entry be in the top-left corner of the square. Note that there is already $1$ entry numbered $1$ in each unfilled row, and $1$ entry numbered $1$ in each unfilled column. Since exactly $1$ row sum is $1$ and exactly $1$ column sum is $1$ , there is a unique entry in the $3\times3$ array of the empty squares such that it, and every other entry in the same row or column in the $3\times3$ array is a $0.$ Using a process similar to what we used in the first paragraph, we can see that there are $9$ ways to choose the entry with only $0$ in its row and column (in the $3\times3$ array).
Without loss of generality, let that entry be in the bottom-right corner of the square. Then, the remaining empty squares are the $4$ center squares. Of these, one of the columns of the empty $2\times2$ array will have two $1$ s and the other column will have one $1.$ That happens if and only if exactly $1$ of the remaining squares is filled with a $0$ , and there are $4$ ways to choose that square. Filling that square with a $0$ and the other $3$ squares with $1$ s completes the grid.
All in all, there are $4\cdot9\cdot16=\boxed{576}$ ways to complete the grid.
| 576
|
3,736
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_17
| 5
|
How many $4 \times 4$ arrays whose entries are $0$ s and $1$ s are there such that the row sums (the sum of the entries in each row) are $1, 2, 3,$ and $4,$ in some order, and the column sums (the sum of the entries in each column) are also $1, 2, 3,$ and $4,$ in some order? For example, the array \[\left[ \begin{array}{cccc} 1 & 1 & 1 & 0 \\ 0 & 1 & 1 & 0 \\ 1 & 1 & 1 & 1 \\ 0 & 1 & 0 & 0 \\ \end{array} \right]\] satisfies the condition.
$\textbf{(A) }144 \qquad \textbf{(B) }240 \qquad \textbf{(C) }336 \qquad \textbf{(D) }576 \qquad \textbf{(E) }624$
|
Note that swapping any two rows or any two columns or both from the given example array, leads to a new array that satisfies the condition. There are $4$ rows, and you choose $2$ to swap, so $\frac{4!}{2!2!} = 6$ ; likewise for columns. Therefore, $6\cdot6 = 36$ . Clearly, the final answer must be divisible by $36$ , so we can eliminate $\textbf{(B)}$ $\textbf{(C)}$ and $\textbf{(E)}$
Now, you are in exam mode with not much time left. You see that $144 = 4\cdot36$ while $576 = 16\cdot36$ . There are $16$ elements in the array ( $4$ rows and $4$ columns), so you go with $\boxed{576}$ , and this is indeed the correct answer!
| 576
|
3,737
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_18
| 1
|
Each square in a $5 \times 5$ grid is either filled or empty, and has up to eight adjacent neighboring squares, where neighboring squares share either a side or a corner. The grid is transformed by the following rules:
A sample transformation is shown in the figure below. [asy] import geometry; unitsize(0.6cm); void ds(pair x) { filldraw(x -- (1,0) + x -- (1,1) + x -- (0,1)+x -- cycle,mediumgray,invisible); } ds((1,1)); ds((2,1)); ds((3,1)); ds((1,3)); for (int i = 0; i <= 5; ++i) { draw((0,i)--(5,i)); draw((i,0)--(i,5)); } label("Initial", (2.5,-1)); draw((6,2.5)--(8,2.5),Arrow); ds((10,2)); ds((11,1)); ds((11,0)); for (int i = 0; i <= 5; ++i) { draw((9,i)--(14,i)); draw((i+9,0)--(i+9,5)); } label("Transformed", (11.5,-1)); [/asy] Suppose the $5 \times 5$ grid has a border of empty squares surrounding a $3 \times 3$ subgrid. How many initial configurations will lead to a transformed grid consisting of a single filled square in the center after a single transformation? (Rotations and reflections of the same configuration are considered different.) [asy] import geometry; unitsize(0.6cm); void ds(pair x) { filldraw(x -- (1,0) + x -- (1,1) + x -- (0,1)+x -- cycle,mediumgray,invisible); } for (int i = 1; i < 4; ++ i) { for (int j = 1; j < 4; ++j) { label("?",(i + 0.5, j + 0.5)); } } for (int i = 0; i <= 5; ++i) { draw((0,i)--(5,i)); draw((i,0)--(i,5)); } label("Initial", (2.5,-1)); draw((6,2.5)--(8,2.5),Arrow); ds((11,2)); for (int i = 0; i <= 5; ++i) { draw((9,i)--(14,i)); draw((i+9,0)--(i+9,5)); } label("Transformed", (11.5,-1)); [/asy] $\textbf{(A)}\ 14 \qquad\textbf{(B)}\ 18 \qquad\textbf{(C)}\ 22 \qquad\textbf{(D)}\ 26 \qquad\textbf{(E)}\ 30$
|
There are two cases for the initial configuration:
Together, the answer is $2+20=\boxed{22}.$
| 22
|
3,738
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_19
| 1
|
In $\triangle{ABC}$ medians $\overline{AD}$ and $\overline{BE}$ intersect at $G$ and $\triangle{AGE}$ is equilateral. Then $\cos(C)$ can be written as $\frac{m\sqrt p}n$ , where $m$ and $n$ are relatively prime positive integers and $p$ is a positive integer not divisible by the square of any prime. What is $m+n+p?$
$\textbf{(A) }44 \qquad \textbf{(B) }48 \qquad \textbf{(C) }52 \qquad \textbf{(D) }56 \qquad \textbf{(E) }60$
|
Let $AG = AE = GE = CE = 1$ . Since $G$ is the centroid, $DG = \frac12$ $BG = 2$
\[\angle BGD = \angle AGE = 60^{\circ}\]
By the Law of Cosine in $\triangle BGD$
\[BD^2 = BG^2 + DG^2 - 2 \cdot BG \cdot DG \cdot \cos \angle BGD\]
\[BD = \sqrt {2^2 + \left( \frac{1}{2} \right)^2 - 2 \cdot 2 \cdot \frac12 \cdot \cos \angle BGD} = \frac{\sqrt{13}}{2}, \quad CD = \frac{\sqrt{13}}{2}\]
By the Law of Cosine in $\triangle ACD$
\[AD^2 = AC^2 + CD^2 - 2 \cdot AC \cdot CD \cdot \cos \angle C\]
\[\cos \angle C = \frac{ AC^2 + CD^2 - AD^2 }{ 2 \cdot AC \cdot CD } = \frac{ 2^2 + \left( \frac{\sqrt{13}}{2} \right)^2 - \left( \frac{3}{2} \right)^2 }{ 2 \cdot 2 \cdot \frac{\sqrt{13}}{2} } = \frac{ 5 \sqrt{13} }{26}\]
\[5 + 13 + 26 = \boxed{44}\]
| 44
|
3,739
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_19
| 2
|
In $\triangle{ABC}$ medians $\overline{AD}$ and $\overline{BE}$ intersect at $G$ and $\triangle{AGE}$ is equilateral. Then $\cos(C)$ can be written as $\frac{m\sqrt p}n$ , where $m$ and $n$ are relatively prime positive integers and $p$ is a positive integer not divisible by the square of any prime. What is $m+n+p?$
$\textbf{(A) }44 \qquad \textbf{(B) }48 \qquad \textbf{(C) }52 \qquad \textbf{(D) }56 \qquad \textbf{(E) }60$
|
Using reference triangle $\triangle AGE$ , we can let \[A=(1,0,0),G=(0,1,0),E=(0,0,1),C=(-1,0,2),D=(-\tfrac{1}{2},\tfrac{3}{2},0),B=(0,3,-2).\] If we move $A,B,C$ each over by $(1,0,-2)$ , leaving $\angle C$ unchanged, we have \[A=(2,0,-2),B=(1,3,-4),C=(0,0,0).\] The angle $\theta$ between vectors $\overrightarrow{CA}$ and $\overrightarrow{CB}$ satisfies \[\cos\theta=\frac{(2)(1)+(0)(3)+(-2)(-4)}{\sqrt{\left[2^{2}+0^{2}+(-2)^{2}\right]\left[1^{2}+3^{2}+(-4)^{2}\right]}}=\frac{10}{\sqrt{8\cdot 26}}=\frac{10}{4\sqrt{13}}=\frac{5\sqrt{13}}{26},\] giving the answer, $5+13+26=\boxed{44}$
| 44
|
3,740
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_20
| 1
|
Let $P(x)$ be a polynomial with rational coefficients such that when $P(x)$ is divided by the polynomial $x^2 + x + 1$ , the remainder is $x+2$ , and when $P(x)$ is divided by the polynomial $x^2+1$ , the remainder
is $2x+1$ . There is a unique polynomial of least degree with these two properties. What is the sum of
the squares of the coefficients of that polynomial?
$\textbf{(A)}\ 10 \qquad\textbf{(B)}\ 13 \qquad\textbf{(C)}\ 19 \qquad\textbf{(D)}\ 20 \qquad\textbf{(E)}\ 23$
|
Given that all the answer choices and coefficients are integers, we hope that $P(x)$ has positive integer coefficients.
Throughout this solution, we will express all polynomials in base $x$ . E.g. $x^2 + x + 1 = 111_{x}$
We are given: \[111a + 12 = 101b + 21 = P(x).\] We add $111$ and $101$ to each side and balance respectively: \[111(a - 1) + 123 = 101(b - 1) + 122 = P(x).\] We make the unit's digits equal: \[111(a - 1) + 123 = 101(b - 2) + 223 = P(x).\] We now notice that: \[111(a - 11) + 1233 = 101(b - 12) + 1233 = P(x).\] Therefore $a = 11_{x} = x + 1$ $b = 12_{x} = x + 2$ , and $P(x) = 1233_{x} = x^3 + 2x^2 + 3x + 3$ $3$ is the minimal degree of $P(x)$ since there is no way to influence the $x$ ‘s digit in $101b + 21$ when $b$ is an integer. The desired sum is $1^2 + 2^2 +3^2+ 3^2 = \boxed{23}$
| 23
|
3,741
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_20
| 2
|
Let $P(x)$ be a polynomial with rational coefficients such that when $P(x)$ is divided by the polynomial $x^2 + x + 1$ , the remainder is $x+2$ , and when $P(x)$ is divided by the polynomial $x^2+1$ , the remainder
is $2x+1$ . There is a unique polynomial of least degree with these two properties. What is the sum of
the squares of the coefficients of that polynomial?
$\textbf{(A)}\ 10 \qquad\textbf{(B)}\ 13 \qquad\textbf{(C)}\ 19 \qquad\textbf{(D)}\ 20 \qquad\textbf{(E)}\ 23$
|
Let $P(x) = Q(x)(x^2+x+1) + x + 2$ , then $P(x) = Q(x)(x^2+1) + xQ(x) + x + 2$ , therefore $xQ(x) + x + 2 \equiv 2x + 1 \pmod{x^2+1}$ , or $xQ(x) \equiv x-1 \pmod{x^2+1}$ . Clearly the minimum is when $Q(x) = x+1$ , and expanding gives $P(x) = x^3+2x^2+3x+3$ . Summing the squares of coefficients gives $\boxed{23}$
| 23
|
3,742
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_20
| 3
|
Let $P(x)$ be a polynomial with rational coefficients such that when $P(x)$ is divided by the polynomial $x^2 + x + 1$ , the remainder is $x+2$ , and when $P(x)$ is divided by the polynomial $x^2+1$ , the remainder
is $2x+1$ . There is a unique polynomial of least degree with these two properties. What is the sum of
the squares of the coefficients of that polynomial?
$\textbf{(A)}\ 10 \qquad\textbf{(B)}\ 13 \qquad\textbf{(C)}\ 19 \qquad\textbf{(D)}\ 20 \qquad\textbf{(E)}\ 23$
|
Let $P(x) = (x^2+x+1)Q_1(x) + x + 2$ ,
then $P(x) = (x^2+1)Q_1(x) + xQ_1(x) + x + 2$
Also $P(x) = (x^2+1)Q_2(x) + 2x + 1$
We infer that $Q_1(x)$ and $Q_2(x)$ have same degree, we can assume $Q_1(x) = x + a$ , and $Q_2(x) = x + b$ , since $P(x)$ has least degree. If this cannot work, we will try quadratic, etc.
Then we get: $(x^2+1)(Q_1(x) - Q_2(x)) + xQ_1(x) - x + 1 = 0$
The constant term gives us: $(Q_1(x) - Q_2(x)) + 1 = 0$
So $Q_1(x) - Q_2(x) = -1$
Substituting this in gives: $-(x^2+1) + xQ_1(x) - x + 1 = 0$
Solving this equation, we get $Q_1(x) = x + 1$
Plugging this into our original equation we get $P(x) = x^3 + 2x^2 + 3x + 3$
Verify this works with $P(x) = (x^2+1)Q_2(x) + 2x + 1$
Therefore the answer is $1^2 + 2^2 + 3^2 + 3^2 = \boxed{23}$
| 23
|
3,743
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_20
| 4
|
Let $P(x)$ be a polynomial with rational coefficients such that when $P(x)$ is divided by the polynomial $x^2 + x + 1$ , the remainder is $x+2$ , and when $P(x)$ is divided by the polynomial $x^2+1$ , the remainder
is $2x+1$ . There is a unique polynomial of least degree with these two properties. What is the sum of
the squares of the coefficients of that polynomial?
$\textbf{(A)}\ 10 \qquad\textbf{(B)}\ 13 \qquad\textbf{(C)}\ 19 \qquad\textbf{(D)}\ 20 \qquad\textbf{(E)}\ 23$
|
Notice that we cannot have the quotients equal to some constants, since the same constant will yield different constant terms for $P(x)$ (which is bad) and different constants will yield different first coefficients (also bad). Thus, we try setting the quotients equal to linear terms (for minimizing degree).
Let $P(x)=(x^2+x+1)(ax+b)+(x+2)$ and $P(x)=(x^2+1)(ax+c)+(2x+1)$ . The quotients have the same $x$ coefficient, since $P(x)$ must have the same $x^3$ coefficient in both cases. Expanding, we get \[P(x)=ax^3+(a+b)x^2+(a+b+1)x+(b+2)\] and \[P(x)=ax^3+cx^2+(a+2)x+(c+1).\]
Equating coefficients, we get $b+2=c+1$ $a+b+1=a+2$ , and $a+b=c$ . From the second equation, we get $b=1$ , then substituting into the first, $c=2$ . Finally, from $a+b=c$ , we have $a=1$ . Now, $P(x)=(x^2+x+1)(ax+b)+(x+2)=(x^2+x+1)(x+1)+(x+2)=x^3+2x^2+3x+3$ and our answer is \[1^2+2^2+3^2+3^2=\boxed{23}.\]
| 23
|
3,744
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_20
| 5
|
Let $P(x)$ be a polynomial with rational coefficients such that when $P(x)$ is divided by the polynomial $x^2 + x + 1$ , the remainder is $x+2$ , and when $P(x)$ is divided by the polynomial $x^2+1$ , the remainder
is $2x+1$ . There is a unique polynomial of least degree with these two properties. What is the sum of
the squares of the coefficients of that polynomial?
$\textbf{(A)}\ 10 \qquad\textbf{(B)}\ 13 \qquad\textbf{(C)}\ 19 \qquad\textbf{(D)}\ 20 \qquad\textbf{(E)}\ 23$
|
We construct the following equations in terms of $P(x)$ and the information given by the problem: \[\textbf{(1) } P(x)=(x^2+x+1)\cdot Q(x)+x+2\] \[\textbf{(2) } P(x)=(x^2+1)\cdot R(x)+2x+1\] Upon inspection, $Q(x)$ and $R(x)$ cannot be constant, so the smallest possible degree of $P(x)$ is $3,$ and both $Q(x)$ and $R(x)$ are linear.
Let $Q(x)=x-q$ and $R(x)=x-r.$ We know there will be values for $q$ and $r$ that make the below equation hold, so we can assume that $P(x)$ has a leading coefficient of $1$
Substituting these values in, and setting $\textbf{(1)}$ and $\textbf{(2)}$ equal to each other, \[(x^2+x+1)(x-q)+x+2=(x^2+1)(x-r)+2x+1.\] We plug in $x=0$ , yielding $r+1=q.$ Substituting this value into the above equation, \[(x^2+x+1)(x-r-1)+x+2=(x^2+1)(x-r)+2x+1.\] Letting $x=1,$ we conclude that $r=-2,$ so $R(x)=x+2.$ Therefore, \[P(x)=(x^2+1)(x+2)+2x+1 = x^3+2x^3+3x+3.\] The requested sum is \[1^2+2^2+3^2+3^2=\boxed{23}\]
| 23
|
3,745
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_20
| 6
|
Let $P(x)$ be a polynomial with rational coefficients such that when $P(x)$ is divided by the polynomial $x^2 + x + 1$ , the remainder is $x+2$ , and when $P(x)$ is divided by the polynomial $x^2+1$ , the remainder
is $2x+1$ . There is a unique polynomial of least degree with these two properties. What is the sum of
the squares of the coefficients of that polynomial?
$\textbf{(A)}\ 10 \qquad\textbf{(B)}\ 13 \qquad\textbf{(C)}\ 19 \qquad\textbf{(D)}\ 20 \qquad\textbf{(E)}\ 23$
|
By remainder theorem, the polynomial can be written as follows.
\[P(x) = (x^2+x+1)Q_{1}(x)+x+2 = (x^2+1)Q_{2}(x)+2x+1\] This is a timed exam, we can use the information given by answer choices. The answer choices tell us this is the polynomial with integer coefficients, and we need to find the polynomial with the least degree so we can assume both $Q_{1}(x)$ and $Q_{2}(x)$ are linear (the coefficient of x should be same).
Then we can write $P(x)$ as a cubic polynomial.
\[P(x) = (x^2+x+1)(ax+b)+x+2 = (x^2+1)(ax+c)+2x+1\] Substituting $x=0,1,-1$ to determine the value of $a$ and $b$
We have: \[P(0) = b+2 = c+1\] \[P(1) = 3a+3b+3 = 2a+2c+3\] \[P(-1) = -a+b+1 = -2a+2c-1\]
We can solve the simultaneous equations: $a=1,b=1,c=2$
Hence, $P(x)=(x^2+x+1)(x+1)+x+2=x^3+2x^2+3x+3$ . The answer is $1^2+2^2+3^2+3^2=\boxed{23}$
| 23
|
3,746
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_21
| 1
|
Let $S$ be the set of circles in the coordinate plane that are tangent to each of the three circles with equations $x^{2}+y^{2}=4$ $x^{2}+y^{2}=64$ , and $(x-5)^{2}+y^{2}=3$ . What is the sum of the areas of all circles in $S$
$\textbf{(A)}~48\pi\qquad\textbf{(B)}~68\pi\qquad\textbf{(C)}~96\pi\qquad\textbf{(D)}~102\pi\qquad\textbf{(E)}~136\pi\qquad$
|
We denote by $C_1$ the circle that has the equation $x^2 + y^2 = 4$ .
We denote by $C_2$ the circle that has the equation $x^2 + y^2 = 64$ .
We denote by $C_3$ the circle that has the equation $(x-5)^2 + y^2 = 3$
We denote by $C_0$ a circle that is tangent to $C_1$ $C_2$ and $C_3$ .
We denote by $\left( u, v \right)$ the coordinates of circle $C_0$ , and $r$ the radius of this circle.
From the graphs of circles $C_1$ $C_2$ $C_3$ , we observe that if $C_0$ is tangent to all of them, then $C_0$ must be internally tangent to $C_2$ .
We have \[ u^2 + v^2 = \left( 8 - r \right)^2 . \hspace{1cm} (1) \]
We do the following casework analysis in terms of the whether $C_0$ is externally tangent to $C_1$ and $C_3$
Case 1: $C_0$ is externally tangent to $C_1$ and $C_3$
We have \[ u^2 + v^2 = \left( r + 2 \right)^2 \hspace{1cm} (2) \] and \[ (u-5)^2 + v^2 = \left( r + \sqrt{3} \right)^2 . \hspace{1cm} (3) \]
Taking $(2) - (1)$ , we get $r + 2 = 8 - r$ . Thus, $r = 3$ .
We can further compute (omitted here) that there exist feasible $(u,v)$ with this given $r$
Case 2: $C_1$ is internally tangent to $C_0$ and $C_3$ is externally tangent to $C_0$
We have \[ u^2 + v^2 = \left( r - 2 \right)^2 \hspace{1cm} (2) \] and \[ (u-5)^2 + v^2 = \left( r + \sqrt{3} \right)^2 . \hspace{1cm} (3) \]
Taking $(2) - (1)$ , we get $r - 2 = 8 - r$ . Thus, $r = 5$ .
We can further compute (omitted here) that there exist feasible $(u,v)$ with this given $r$
Case 3: $C_1$ is externally tangent to $C_0$ and $C_3$ is internally tangent to $C_0$
We have \[ u^2 + v^2 = \left( r + 2 \right)^2 \hspace{1cm} (2) \] and \[ (u-5)^2 + v^2 = \left( r - \sqrt{3} \right)^2 . \hspace{1cm} (3) \]
Taking $(2) - (1)$ , we get $r + 2 = 8 - r$ . Thus, $r = 3$ .
We can further compute (omitted here) that there exist feasible $(u,v)$ with this given $r$
Case 4: $C_1$ is internally tangent to $C_0$ and $C_3$ is internally tangent to $C_0$
We have \[ u^2 + v^2 = \left( r - 2 \right)^2 \hspace{1cm} (2) \] and \[ (u-5)^2 + v^2 = \left( r - \sqrt{3} \right)^2 . \hspace{1cm} (3) \]
Taking $(2) - (1)$ , we get $r - 2 = 8 - r$ . Thus, $r = 5$ .
We can further compute (omitted here) that there exist feasible $(u,v)$ with this given $r$
Because the graph is symmetric with the $x$ -axis, and for each case above, the solution of $v$ is not 0. Hence, in each case, there are two congruent circles whose centers are symmetric through the $x$ -axis.
Therefore, the sum of the areas of all the circles in $S$ is $2\left( 3^2 \pi +5^2 \pi +3^2 \pi +5^2 \pi \right) = \boxed{136}$
| 136
|
3,747
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_23
| 1
|
Let $x_0,x_1,x_2,\dotsc$ be a sequence of numbers, where each $x_k$ is either $0$ or $1$ . For each positive integer $n$ , define \[S_n = \sum_{k=0}^{n-1} x_k 2^k\] Suppose $7S_n \equiv 1 \pmod{2^n}$ for all $n \geq 1$ . What is the value of the sum \[x_{2019} + 2x_{2020} + 4x_{2021} + 8x_{2022}?\] $\textbf{(A) } 6 \qquad \textbf{(B) } 7 \qquad \textbf{(C) }12\qquad \textbf{(D) } 14\qquad \textbf{(E) }15$
|
In binary numbers, we have \[S_n = (x_{n-1} x_{n-2} x_{n-3} x_{n-4} \ldots x_{2} x_{1} x_{0})_2.\] It follows that \[8S_n = (x_{n-1} x_{n-2} x_{n-3} x_{n-4} \ldots x_{2} x_{1} x_{0}000)_2.\] We obtain $7S_n$ by subtracting the equations: \[\begin{array}{clccrccccccr} & (x_{n-1} & x_{n-2} & x_{n-3} & x_{n-4} & \ldots & x_2 & x_1 & x_0 & 0 & 0 & 0 \ )_2 \\ -\quad & & & & (x_{n-1} & \ldots & x_5 & x_4 & x_3 & x_2 & x_1 & x_0)_2 \\ \hline & & & & & & & & & & & \\ [-2.5ex] & ( \ \ ?& ? & ? & 0 \ \ \ & \ldots & 0 & 0 & 0 & 0 & 0 & 1 \ )_2 \\ \end{array}\] We work from right to left: \begin{alignat*}{6} x_0=x_1=x_2=1 \quad &\implies \quad &x_3 &= 0& \\ \quad &\implies \quad &x_4 &= 1& \\ \quad &\implies \quad &x_5 &= 1& \\ \quad &\implies \quad &x_6 &= 0& \\ \quad &\implies \quad &x_7 &= 1& \\ \quad &\implies \quad &x_8 &= 1& \\ \quad &\quad \vdots & & & \end{alignat*} For all $n\geq3,$ we conclude that
Finally, we get $(x_{2019},x_{2020},x_{2021},x_{2022})=(0,1,1,0),$ from which \[x_{2019} + 2x_{2020} + 4x_{2021} + 8x_{2022} = \boxed{6}.\] ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
| 6
|
3,748
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_23
| 2
|
Let $x_0,x_1,x_2,\dotsc$ be a sequence of numbers, where each $x_k$ is either $0$ or $1$ . For each positive integer $n$ , define \[S_n = \sum_{k=0}^{n-1} x_k 2^k\] Suppose $7S_n \equiv 1 \pmod{2^n}$ for all $n \geq 1$ . What is the value of the sum \[x_{2019} + 2x_{2020} + 4x_{2021} + 8x_{2022}?\] $\textbf{(A) } 6 \qquad \textbf{(B) } 7 \qquad \textbf{(C) }12\qquad \textbf{(D) } 14\qquad \textbf{(E) }15$
|
First, notice that \[x_{2019} + 2x_{2020} + 4x_{2021} + 8x_{2022} = \frac{S_{2023} - S_{2019}}{2^{2019}}.\] Then since $S_n$ is the modular inverse of $7$ in $\mathbb{Z}_{2^n}$ , we can perform the Euclidean Algorithm to find it for $n = 2019,2023$
Starting with $2019$ \begin{align*} 7S_{2019} &\equiv 1 \pmod{2^{2019}} \\ 7S_{2019} &= 2^{2019}k + 1. \end{align*} Now, take both sides $\operatorname{mod} \ 7$ \[0 \equiv 2^{2019}k + 1 \pmod{7}.\] Using Fermat's Little Theorem, \[2^{2019} = (2^{336})^6 \cdot 2^3 \equiv 2^3 \equiv 1 \pmod{7}.\] Thus, \[0 \equiv k + 1 \pmod{7} \implies k \equiv 6 \pmod{7} \implies k = 7j + 6.\] Therefore, \[7S_{2019} = 2^{2019} (7j + 6) + 1 \implies S_{2019} = \frac{2^{2019} (7j + 6) + 1}{7}.\]
We may repeat this same calculation with $S_{2023}$ to yield \[S_{2023} = \frac{2^{2023} (7h + 3) + 1}{7}.\] Now, we notice that $S_n$ is basically an integer expressed in binary form with $n$ bits.
This gives rise to a simple inequality, \[0 \leqslant S_n \leqslant 2^n.\] Since the maximum possible number that can be generated with $n$ bits is \[\underbrace{{11111\dotsc1}_2}_{n} = \sum_{k=0}^{n-1} 2^k = 2^n - 1 \leqslant 2^n.\] Looking at our calculations for $S_{2019}$ and $S_{2023}$ , we see that the only valid integers that satisfy that constraint are $j = h = 0$ \[\frac{S_{2023} - S_{2019}}{2^{2019}} = \frac{\tfrac{2^{2023} \cdot 3 + 1}{7} - \tfrac{2^{2019} \cdot 6 + 1}{7}}{2^{2019}} = \frac{2^4 \cdot 3 - 6}{7} = \boxed{6}.\] ~zoomanTV
| 6
|
3,749
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_23
| 4
|
Let $x_0,x_1,x_2,\dotsc$ be a sequence of numbers, where each $x_k$ is either $0$ or $1$ . For each positive integer $n$ , define \[S_n = \sum_{k=0}^{n-1} x_k 2^k\] Suppose $7S_n \equiv 1 \pmod{2^n}$ for all $n \geq 1$ . What is the value of the sum \[x_{2019} + 2x_{2020} + 4x_{2021} + 8x_{2022}?\] $\textbf{(A) } 6 \qquad \textbf{(B) } 7 \qquad \textbf{(C) }12\qquad \textbf{(D) } 14\qquad \textbf{(E) }15$
|
Note that, as in Solution 2, we have \[x_{2019} + 2x_{2020} + 4x_{2021} + 8x_{2022} = \frac{S_{2023} - S_{2019}}{2^{2019}}.\] This is because \[S_{2023} = x_{0}2^{0} + x_{1}2^{1} + \cdots + x_{2019}2^{2019} + \cdots + x_{2022}2^{2022}\] and \[S_{2019} = x_{0}2^{0} + x_{1}2^{1} + \cdots + x_{2018}2^{2018}.\] Note that \[S_{2023} - S_{2019} = x_{2019}2^{2019} + \cdots + x_{2022}2^{2022} = 2^{2019}(x_{2019} + 2x_{2020} + 4x_{2021} + 8x_{2022}).\] Therefore, \[x_{2019} + 2x_{2020} + 4x_{2021} + 8x_{2022} = \frac{S_{2023} - S_{2019}}{2^{2019}}.\] Multiplying both sides by 7 gives us \[7(x_{2019} + 2x_{2020} + 4x_{2021} + 8x_{2022}) = \frac{7S_{2023} - 7S_{2019}}{2^{2019}}.\] We can write \[7S_{2023} = 1\pmod{2^{2023}} = 1 + 2^{2023}a = 1 + 2^{2019}*16a\] and \[7S_{2019} = 1\pmod{2^{2019}} = 1 + 2^{2019}b\] for some a and b. Substituting, we get \[7(x_{2019} + 2x_{2020} + 4x_{2021} + 8x_{2022}) = \frac{(1 + 2^{2019} * 16a) - (1 + 2^{2019}b)}{2^{2019}} = 16a - b.\] Therefore, our answer can be written as \[\frac{16a - b}{7}.\] Another thing to notice is that a and b are integers between 0 and 6. This is because \[7(1 + 2 + 4 + 8 + \cdots + 2^{2022}) \geqslant 7S_{2023} = 1 + 2^{2023}a\] which is \[7(2^{2023}) - 7 \geqslant 1 + 2^{2023}a\] \[(7-a) \geqslant \frac{8}{2^{2023}}\] which only holds when a is less than 7 because the right is very small positive number, so the left must be positive, too. Clearly, a is also non-negative, because otherwise, \[7S_{2023} = 1 + 2^{2023}a < 0\] which would mean \[S_{2023} < 0\] which cannot happen, so a is greater than 0. A similar explanation for b shows that b is an integer between 0 and 6 inclusive.
Going back to the solution, if our answer to the problem is n, then \[16a - b = 7n\] and \[16a = 7n + b,\] so we can try the five option choices and see which one, when multiplied by 7 and added to some whole number between 0 and 6 results in a multiple of 16. Trying all the option choices, we see that you need to add 7n to something more than 6 to equal a multiple of 16 other than for option A. Therefore, the answer is $\boxed{6}.$
| 6
|
3,750
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_24
| 1
|
The figure below depicts a regular $7$ -gon inscribed in a unit circle. [asy] import geometry; unitsize(3cm); draw(circle((0,0),1),linewidth(1.5)); for (int i = 0; i < 7; ++i) { for (int j = 0; j < i; ++j) { draw(dir(i * 360/7) -- dir(j * 360/7),linewidth(1.5)); } } for(int i = 0; i < 7; ++i) { dot(dir(i * 360/7),5+black); } [/asy] What is the sum of the $4$ th powers of the lengths of all $21$ of its edges and diagonals?
$\textbf{(A) }49 \qquad \textbf{(B) }98 \qquad \textbf{(C) }147 \qquad \textbf{(D) }168 \qquad \textbf{(E) }196$
|
There are $7$ segments whose lengths are $2 \sin \frac{\pi}{7}$ $7$ segments whose lengths are $2 \sin \frac{2 \pi}{7}$ $7$ segments whose lengths are $2 \sin \frac{3\pi}{7}$
Therefore, the sum of the $4$ th powers of these lengths is \begin{align*} 7 \cdot 2^4 \sin^4 \frac{\pi}{7} + 7 \cdot 2^4 \sin^4 \frac{2 \pi}{7} + 7 \cdot 2^4 \sin^4 \frac{3 \pi}{7} & = \frac{7 \cdot 2^4}{(2i)^4} \left( e^{i \frac{\pi}{7}} - e^{i \frac{\pi}{7}} \right)^4 + \frac{7 \cdot 2^4}{(2i)^4} \left( e^{i \frac{2 \pi}{7}} - e^{i \frac{2 \pi}{7}} \right)^4 + \frac{7 \cdot 2^4}{(2i)^4} \left( e^{i \frac{3 \pi}{7}} - e^{i \frac{4 \pi}{7}} \right)^4 \\ & = 7 \left( e^{i \frac{4 \pi}{7}} - 4 e^{i \frac{2 \pi}{7}} + 6 - 4 e^{- i \frac{2 \pi}{7}} + e^{- i \frac{4 \pi}{7}} \right) \\ & \quad + 7 \left( e^{i \frac{8 \pi}{7}} - 4 e^{i \frac{4 \pi}{7}} + 6 - 4 e^{- i \frac{4 \pi}{7}} + e^{- i \frac{8 \pi}{7}} \right) \\ & \quad + 7 \left( e^{i \frac{12 \pi}{7}} - 4 e^{i \frac{6 \pi}{7}} + 6 - 4 e^{- i \frac{6 \pi}{7}} + e^{- i \frac{12 \pi}{7}} \right) \\ & = 7 \left( e^{i \frac{4 \pi}{7}} + e^{i \frac{8 \pi}{7}} + e^{i \frac{12 \pi}{7}} + e^{-i \frac{4 \pi}{7}} + e^{-i \frac{8 \pi}{7}} + e^{-i \frac{12 \pi}{7}} \right) \\ & \quad - 7 \cdot 4 \left( e^{i \frac{2 \pi}{7}} + e^{i \frac{4 \pi}{7}} + e^{i \frac{6 \pi}{7}} + e^{-i \frac{2 \pi}{7}} + e^{-i \frac{4 \pi}{7}} + e^{-i \frac{6 \pi}{7}} \right) \\ & \quad + 7 \cdot 6 \cdot 3 \\ & = 7 \left( e^{i \frac{4 \pi}{7}} + e^{-i \frac{6 \pi}{7}} + e^{-i \frac{2 \pi}{7}} + e^{-i \frac{4 \pi}{7}} + e^{i \frac{6 \pi}{7}} + e^{i \frac{2 \pi}{7}} \right) \\ & \quad - 7 \cdot 4 \left( e^{i \frac{2 \pi}{7}} + e^{i \frac{4 \pi}{7}} + e^{i \frac{6 \pi}{7}} + e^{-i \frac{2 \pi}{7}} + e^{-i \frac{4 \pi}{7}} + e^{-i \frac{6 \pi}{7}} \right) \\ & \quad + 7 \cdot 6 \cdot 3 \\ & = -7 + 7 \cdot 4 + 7 \cdot 6 \cdot 3 \\ & = \boxed{147} ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
| 147
|
3,751
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_24
| 2
|
The figure below depicts a regular $7$ -gon inscribed in a unit circle. [asy] import geometry; unitsize(3cm); draw(circle((0,0),1),linewidth(1.5)); for (int i = 0; i < 7; ++i) { for (int j = 0; j < i; ++j) { draw(dir(i * 360/7) -- dir(j * 360/7),linewidth(1.5)); } } for(int i = 0; i < 7; ++i) { dot(dir(i * 360/7),5+black); } [/asy] What is the sum of the $4$ th powers of the lengths of all $21$ of its edges and diagonals?
$\textbf{(A) }49 \qquad \textbf{(B) }98 \qquad \textbf{(C) }147 \qquad \textbf{(D) }168 \qquad \textbf{(E) }196$
|
There are $7$ segments whose lengths are $2 \sin \frac{\pi}{7}$ $7$ segments whose lengths are $2 \sin \frac{2 \pi}{7}$ $7$ segments whose lengths are $2 \sin \frac{3\pi}{7}$
Therefore, the sum of the $4$ th powers of these lengths is \begin{align*} & 7 \cdot 2^4 \sin^4 \frac{\pi}{7} + 7 \cdot 2^4 \sin^4 \frac{2 \pi}{7} + 7 \cdot 2^4 \sin^4 \frac{3 \pi}{7} \\ & = 7 \cdot 2^4 \left( \frac{1 - \cos \frac{2 \pi}{7}}{2} \right)^2 + 7 \cdot 2^4 \left( \frac{1 - \cos \frac{4 \pi}{7}}{2} \right)^2 + 7 \cdot 2^4 \left( \frac{1 - \cos \frac{6 \pi}{7}}{2} \right)^2 \\ & = 7 \cdot 2^2 \left( 1 - 2 \cos \frac{2 \pi}{7} + \cos^2 \frac{2 \pi}{7} \right) + 7 \cdot 2^2 \left( 1 - 2 \cos \frac{4 \pi}{7} + \cos^2 \frac{4 \pi}{7} \right) + 7 \cdot 2^2 \left( 1 - 2 \cos \frac{6 \pi}{7} + \cos^2 \frac{6 \pi}{7} \right) \\ & = 7 \cdot 2^2 \cdot 3 - 7 \cdot 2^3 \left( \cos \frac{2 \pi}{7} + \cos \frac{4 \pi}{7} + \cos \frac{6 \pi}{7} \right) + 7 \cdot 2^2 \left( \cos^2 \frac{2 \pi}{7} + \cos^2 \frac{4 \pi}{7} + \cos^2 \frac{6 \pi}{7} \right) \\ & = 7 \cdot 2^2 \cdot 3 - 7 \cdot 2^3 \left( \cos \frac{2 \pi}{7} + \cos \frac{4 \pi}{7} + \cos \frac{6 \pi}{7} \right) + 7 \cdot 2^2 \left( \frac{1 + \cos \frac{4 \pi}{7} }{2} + \frac{1 + \cos \frac{8 \pi}{7} }{2} + \frac{1 + \cos \frac{12 \pi}{7} }{2} \right) \\ & = 7 \cdot 2^2 \cdot 3 - 7 \cdot 2^3 \left( \cos \frac{2 \pi}{7} + \cos \frac{4 \pi}{7} + \cos \frac{6 \pi}{7} \right) + 7 \cdot 2 \cdot 3 + 7 \cdot 2 \left( \cos \frac{4 \pi}{7} + \cos \frac{8 \pi}{7} + \cos \frac{12 \pi}{7} \right) \\ & = 7 \cdot 2^2 \cdot 3 - 7 \cdot 2^3 \left( \cos \frac{2 \pi}{7} + \cos \frac{4 \pi}{7} + \cos \frac{6 \pi}{7} \right) + 7 \cdot 2 \cdot 3 + 7 \cdot 2 \left( \cos \frac{4 \pi}{7} + \cos \frac{6 \pi}{7} + \cos \frac{2 \pi}{7} \right) \\ & = 7 \cdot 2 \cdot 3 \left( 2 + 1 \right) - 7 \cdot 2 \left( 4 - 1 \right) \left( \cos \frac{4 \pi}{7} + \cos \frac{6 \pi}{7} + \cos \frac{2 \pi}{7} \right) \\ & = 7 \cdot 2 \cdot 3 \left( 2 + 1 \right) - 7 \cdot 2 \left( 4 - 1 \right) \cdot \left( - \frac{1}{2} \right) \\ & = \boxed{147} . \]
| 147
|
3,752
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_24
| 3
|
The figure below depicts a regular $7$ -gon inscribed in a unit circle. [asy] import geometry; unitsize(3cm); draw(circle((0,0),1),linewidth(1.5)); for (int i = 0; i < 7; ++i) { for (int j = 0; j < i; ++j) { draw(dir(i * 360/7) -- dir(j * 360/7),linewidth(1.5)); } } for(int i = 0; i < 7; ++i) { dot(dir(i * 360/7),5+black); } [/asy] What is the sum of the $4$ th powers of the lengths of all $21$ of its edges and diagonals?
$\textbf{(A) }49 \qquad \textbf{(B) }98 \qquad \textbf{(C) }147 \qquad \textbf{(D) }168 \qquad \textbf{(E) }196$
|
As explained in Solutions 1 and 2, what we are trying to find is $7 \cdot 2^4 \sin^4 \frac{\pi}{7} + 7 \cdot 2^4 \sin^4 \frac{2 \pi}{7} + 7 \cdot 2^4 \sin^4 \frac{3 \pi}{7}$ . Using trig we get \begin{align*} & \sin^4 \frac{\pi}{7} + \sin^4 \frac{2 \pi}{7} + \sin^4 \frac{3 \pi}{7} \\ = & \sin^2 \frac{\pi}{7} \left(1 - \cos^2 \frac{\pi}{7} \right) + \sin^2 \frac{2\pi}{7} \left(1 - \cos^2 \frac{2\pi}{7} \right) + \sin^2 \frac{3\pi}{7} \left(1 - \cos^2 \frac{3\pi}{7} \right) \\ = & \sin^2 \frac{\pi}{7} - \left(\frac{1}{2} \sin \frac{2\pi}{7}\right)^2 + \sin^2 \frac{2\pi}{7} - \left(\frac{1}{2} \sin \frac{4\pi}{7}\right)^2 + \sin^2 \frac{3\pi}{7} - \left(\frac{1}{2} \sin \frac{6\pi}{7}\right)^2\\ = & \sin^2 \frac{\pi}{7} - \frac{1}{4} \sin^2 \frac{2\pi}{7} + \sin^2 \frac{2\pi}{7} - \frac{1}{4} \sin^2 \frac{4\pi}{7} + \sin^2 \frac{3\pi}{7} - \frac{1}{4} \sin^2 \frac{6\pi}{7} \\ = & \frac{3}{4} \left(\sin^2 \frac{\pi}{7} + \sin^2 \frac{2\pi}{7} + \sin^2 \frac{3\pi}{7}\right) \\ = & \frac{3}{4} \cdot \frac{1}{2} \left(1 - \cos \frac{2\pi}{7} + 1 - \cos \frac{4\pi}{7} + 1 - \cos \frac{6\pi}{7} \right)\\ = & \frac{3}{4} \cdot \frac{1}{2} \left(3 - \left(-\frac{1}{2}\right)\right) \\ = & \frac{21}{16}. \end{align*} Like in the second solution, we also use the fact that $\cos \frac{2\pi}{7} + \cos \frac{4\pi}{7} + \cos \frac{6\pi}{7} = -\frac{1}{2}$ , which admittedly might need some explanation.
For explanation see supplement
Notice that \begin{align*} \cos \frac{2\pi}{7} + \cos \frac{4\pi}{7} + \cos \frac{6\pi}{7} & = \frac{1}{2}\left(e^\frac{2i\pi}{7}+ e^{-\frac{2i\pi}{7}} \right) + \frac{1}{2}\left( e^\frac{4i\pi}{7}+ e^{-\frac{4i\pi}{7}} \right) + \frac{1}{2}\left( e^\frac{6i\pi}{7}+ e^{-\frac{6i\pi}{7}}\right) \\ & = \frac{1}{2}\left(e^\frac{2i\pi}{7}+ e^{-\frac{2i\pi}{7}}+ e^\frac{4i\pi}{7}+ e^{-\frac{4i\pi}{7}} +e^\frac{6i\pi}{7}+ e^{-\frac{6i\pi}{7}} + 1\right) - \frac{1}{2} \end{align*} In the brackets we have the sum of the roots of the polynomial $x^7 - 1 = 0$ . These sum to $0$ by Vieta’s formulas , and the desired identity follows. See Roots of unity if you have not seen this technique.
Going back to the question: \[7 \cdot 2^4 \sin^4 \frac{\pi}{7} + 7 \cdot 2^4 \sin^4 \frac{2 \pi}{7} + 7 \cdot 2^4 \sin^4 \frac{3 \pi}{7} = 7 \cdot 2^4 \left(\sin^4 \frac{\pi}{7} + \sin^4 \frac{2 \pi}{7} + \sin^4 \frac{3 \pi}{7}\right) = 7 \cdot 2^4 \cdot \frac{21}{16} = \boxed{147}.\] ~obscene_kangaroo
| 147
|
3,753
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_24
| 4
|
The figure below depicts a regular $7$ -gon inscribed in a unit circle. [asy] import geometry; unitsize(3cm); draw(circle((0,0),1),linewidth(1.5)); for (int i = 0; i < 7; ++i) { for (int j = 0; j < i; ++j) { draw(dir(i * 360/7) -- dir(j * 360/7),linewidth(1.5)); } } for(int i = 0; i < 7; ++i) { dot(dir(i * 360/7),5+black); } [/asy] What is the sum of the $4$ th powers of the lengths of all $21$ of its edges and diagonals?
$\textbf{(A) }49 \qquad \textbf{(B) }98 \qquad \textbf{(C) }147 \qquad \textbf{(D) }168 \qquad \textbf{(E) }196$
|
This solution follows the same steps as the trigonometry solutions (Solutions 2 and 3), except it gives an alternate way to prove the statement below true without complex numbers:
\[\cos \frac{2\pi}{7} + \cos \frac{4\pi}{7} + \cos \frac{6\pi}{7} = -\frac{1}{2}\]
\begin{align*} S &= \cos \frac{2\pi}{7} + \cos \frac{4\pi}{7} + \cos \frac{6\pi}{7}, \\ S^2 &= \cos^2 \frac{2\pi}{7} + \cos^2 \frac{4\pi}{7} + \cos^2 \frac{6\pi}{7} + 2\cos \frac{2\pi}{7}\cos \frac{4\pi}{7} + 2\cos \frac{2\pi}{7}\cos \frac{6\pi}{7} + 2\cos \frac{4\pi}{7}\cos \frac{6\pi}{7} \\ &= \left(\frac{1+ \cos \frac{4\pi}{7}}{2}\right) + \left(\frac{1+ \cos \frac{8\pi}{7}}{2}\right) + \left(\frac{1+ \cos \frac{12\pi}{7}}{2}\right) + 2\cos \frac{2\pi}{7}\cos \frac{4\pi}{7} + 2\cos \frac{2\pi}{7}\cos \frac{6\pi}{7} + 2\cos \frac{4\pi}{7}\cos \frac{6\pi}{7} \\ &= \frac{1}{2}(3 + S) + \left(\cos \frac{6\pi}{7} + \cos \frac{2\pi}{7}\right) + \left(\cos \frac{8\pi}{7} + \cos \frac{4\pi}{7}\right) + \left(\cos \frac{10\pi}{7} + \cos \frac{2\pi}{7}\right) \\ &= \frac{1}{2}(3 + S) + 2\cos \frac{2\pi}{7} + 2\cos \frac{4\pi}{7} + 2\cos \frac{6\pi}{7}\\ &= \frac{1}{2}(3 + S) + 2S. \\ \end{align*} We end up with \[2S^2 - 5S - 3 = 0.\] Using the quadratic formula, we find the solutions for $S$ to be $-\frac{1}{2}$ and $3$ . Because $3$ is impossible, $S = -\frac{1}{2}$ .
With this result, following similar to steps to Solutions 2 and 3 will get $\boxed{147}$
| 147
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3,754
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_24
| 5
|
The figure below depicts a regular $7$ -gon inscribed in a unit circle. [asy] import geometry; unitsize(3cm); draw(circle((0,0),1),linewidth(1.5)); for (int i = 0; i < 7; ++i) { for (int j = 0; j < i; ++j) { draw(dir(i * 360/7) -- dir(j * 360/7),linewidth(1.5)); } } for(int i = 0; i < 7; ++i) { dot(dir(i * 360/7),5+black); } [/asy] What is the sum of the $4$ th powers of the lengths of all $21$ of its edges and diagonals?
$\textbf{(A) }49 \qquad \textbf{(B) }98 \qquad \textbf{(C) }147 \qquad \textbf{(D) }168 \qquad \textbf{(E) }196$
|
Let $x$ $y$ , and $z$ be the lengths of the chords with arcs $\frac{2\pi}{7}$ $\frac{4\pi}{7}$ and $\frac{6\pi}{7}$ respectively.
Then by the law of cosines we get: \begin{align*} x^2 &= 2\left(1-\cos\frac{2\pi}{7}\right), \\ y^2 &= 2\left(1-\cos\frac{4\pi}{7}\right), \\ z^2 &= 2\left(1-\cos\frac{6\pi}{7}\right). \end{align*} The answer is then just $7(x^4+y^4+z^4)$ (since there's $7$ of each diagonal/side), obtained by summing the squares of the above equations and then multiplying by $7$ \begin{align*} & \quad7\cdot2^2\left( \left(1-\cos\frac{2\pi}{7}\right)^2 + \left(1-\cos\frac{4\pi}{7}\right)^2 + \left(1-\cos\frac{6\pi}{7}\right)^2 \right) \\ &= 7\cdot4\left( \left(1-2\cos\frac{2\pi}{7}+\cos^2\frac{2\pi}{7}\right) + \left(1-2\cos\frac{4\pi}{7}+\cos^2\frac{4\pi}{7}\right) + \left(1-2\cos\frac{6\pi}{7}+\cos^2\frac{6\pi}{7}\right) \right) \\ &= 7\cdot4\left(3-2\left(\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}\right) + \left(\cos^2\frac{2\pi}{7}+\cos^2\frac{4\pi}{7}+\cos^2\frac{6\pi}{7}\right) \right) \\ &= 7\cdot4\left(3-2\left(\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}\right) + \frac12 \left(1+\cos\frac{4\pi}{7}+1+\cos\frac{8\pi}{7}+1+\cos\frac{12\pi}{7}\right) \right) \\ &= 7\cdot4\left(\frac{9}{2}-2\left(\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}\right) + \frac12 \left(\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}+\cos\frac{2\pi}{7}\right) \right) \\ &= 7\cdot4\left(\frac{9}{2}-\frac{3}{2}\left(\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}\right) \right) \\ &= 7\cdot4\left(\frac{9}{2}-\frac{3}{2}\left(-\frac{1}{2}\right)\right) \\ &= \boxed{147}
| 147
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3,755
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_24
| 6
|
The figure below depicts a regular $7$ -gon inscribed in a unit circle. [asy] import geometry; unitsize(3cm); draw(circle((0,0),1),linewidth(1.5)); for (int i = 0; i < 7; ++i) { for (int j = 0; j < i; ++j) { draw(dir(i * 360/7) -- dir(j * 360/7),linewidth(1.5)); } } for(int i = 0; i < 7; ++i) { dot(dir(i * 360/7),5+black); } [/asy] What is the sum of the $4$ th powers of the lengths of all $21$ of its edges and diagonals?
$\textbf{(A) }49 \qquad \textbf{(B) }98 \qquad \textbf{(C) }147 \qquad \textbf{(D) }168 \qquad \textbf{(E) }196$
|
Hope you had a ruler handy! This problem can be done with a ruler and basic estimation.
First, measuring the radius of the circle obtains $2.9$ cm (when done on the paper version). Thus, any other measurement we get for the sides/diagonals should be divided by $2.9$
Measuring the sides of the circle gets $2.5$ cm. The shorter diagonals are $4.5$ cm, and the longest diagonals measure $5.6$ cm. Thus, we'd like to estimate \[7\left(\frac{2.5}{2.9}\right)^4 + 7\left(\frac{4.5}{2.9}\right)^4 + 7\left(\frac{5.6}{2.9}\right)^4.\]
We know $\left(\frac{2.5}{2.9}\right)^4$ is slightly less than $1.$ Let's approximate it as 1 for now. Thus, $7\left(\frac{2.5}{2.9}\right)^4 \approx 7.$
Next, $\left(\frac{4.5}{2.9}\right)^4$ is slightly more than $\left(\frac{4.5}{3}\right)^4.$ We know $\left(\frac{4.5}{3}\right)^4 = 1.5^4 = \frac{81}{16},$ slightly more than $5,$ so we can approximate $\left(\frac{4.5}{2.9}\right)^4$ as $5.5.$ Thus, $7\left(\frac{4.5}{2.9}\right)^4 \approx 38.5.$
Finally, $\left(\frac{5.6}{2.9}\right)^4$ is slightly less than $\left(\frac{5.6}{2.8}\right)^4 = 2^4 = 16.$ We say it's around $15,$ so then $7\left(\frac{5.6}{2.9}\right)^4 \approx 105.$
Adding what we have, we get $105 + 38.5 + 1 = 144.5$ as our estimate. We see $\boxed{147}$ is very close to our estimate, so we have successfully finished the problem.
| 147
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3,756
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_24
| 7
|
The figure below depicts a regular $7$ -gon inscribed in a unit circle. [asy] import geometry; unitsize(3cm); draw(circle((0,0),1),linewidth(1.5)); for (int i = 0; i < 7; ++i) { for (int j = 0; j < i; ++j) { draw(dir(i * 360/7) -- dir(j * 360/7),linewidth(1.5)); } } for(int i = 0; i < 7; ++i) { dot(dir(i * 360/7),5+black); } [/asy] What is the sum of the $4$ th powers of the lengths of all $21$ of its edges and diagonals?
$\textbf{(A) }49 \qquad \textbf{(B) }98 \qquad \textbf{(C) }147 \qquad \textbf{(D) }168 \qquad \textbf{(E) }196$
|
First, we put the figure in the coordinate plane with the center of the circle at the origin and a vertex on the positive x-axis. Thus, the coordinates of the vertices will be the terminal points of integer multiples of the angle $\frac{2\pi}{7},$ which are \[\left(\cos\dfrac{2\pi n}{7}, \sin\dfrac{2\pi n}{7}\right)\] for integer $n.$ Then, we notice there are three types of diagonals: the ones with chords of arcs $\dfrac{2\pi}{7}, \dfrac{4\pi}{7},$ and $\dfrac{6\pi}{7}.$ We notive there are here are $7$ of each type of diagonal. Then, we use the pythagorean theorem to find the distance from $\left(\cos\frac{2\pi n}{7}, \sin\frac{2\pi n}{7}\right)$ to $(1, 0)$ \[\left(\sqrt{\left(\cos\frac{2\pi n}{7}-1\right)^2+\sin^2\frac{2\pi n}{7}}\right)^4=\left(\sqrt{\cos^2\frac{2\pi n}{7}+\sin^2\frac{2\pi n}{7}-2\cos\frac{2\pi n}{7}+1}\right)^4\] \[=\left(\sqrt{2-2\cos\frac{2\pi n}{7}}\right)^4\] \[=4\cos^2\frac{2\pi n}{7}-8\cos\frac{2\pi n}{7}+4.\] By the cosine double angle identity, $\cos{2\theta}=2\cos^2\theta-1.$ This means that $2\cos^2\theta=\cos{2\theta}+1.$ Substituting this in, \[4\cos^2\frac{2\pi n}{7}-8\cos\frac{2\pi n}{7}+4=2\left(\cos\frac{4\pi n}{7}+1\right)-8\cos\frac{2\pi n}{7}+4=2\cos\frac{4\pi n}{7}-8\cos\frac{2\pi n}{7}+6.\] Summing this up for $n=1,2,3,$ \[2\cos\frac{4\pi}{7}-8\cos\frac{2\pi}{7}+6+2\cos\frac{8\pi}{7}-8\cos\frac{4\pi}{7}+6+2\cos\frac{12\pi}{7}-8\cos\frac{6\pi}{7}+6\] \[=2\left(\cos\frac{4\pi}{7}+\cos\frac{8\pi}{7}+\cos\frac{12\pi}{7}\right)-8\left(\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}\right)+18\] \[=2\left(\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}+\cos\frac{2\pi}{7}\right)-8\left(\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}\right)+18\] \[=2(-0.5)-8(-0.5)+18=21.\] (These equalities are based on $\cos\theta=\cos(2\pi-\theta)$ and $\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}=-\frac{1}{2}.$ ) (For explanation of this see supplement .)
Finally, because there are $7$ of each type of diagonal, the answer is $7\cdot 21=\boxed{147}.$
| 147
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3,757
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_24
| 8
|
The figure below depicts a regular $7$ -gon inscribed in a unit circle. [asy] import geometry; unitsize(3cm); draw(circle((0,0),1),linewidth(1.5)); for (int i = 0; i < 7; ++i) { for (int j = 0; j < i; ++j) { draw(dir(i * 360/7) -- dir(j * 360/7),linewidth(1.5)); } } for(int i = 0; i < 7; ++i) { dot(dir(i * 360/7),5+black); } [/asy] What is the sum of the $4$ th powers of the lengths of all $21$ of its edges and diagonals?
$\textbf{(A) }49 \qquad \textbf{(B) }98 \qquad \textbf{(C) }147 \qquad \textbf{(D) }168 \qquad \textbf{(E) }196$
|
Place the figure on the complex plane and let $w = e^{\frac{2\pi}{7}}$ (a $7$ th root of unity). The vertices of the $7$ -gon are $w^0,w^1,w^2,\dots,w^6$ . We wish to find \[\sum_{i=0}^5\sum_{j=i+1}^6\lvert w^i-w^j\rvert^4 = \frac{1}{2}\sum_{i=0}^6\sum_{j=0}^6\lvert w^i-w^j\rvert^4.\] The second expression is more convenient to work with. The factor of $\tfrac{1}{2}$ is because it double-counts each edge (and includes when $i=j$ , but the length is $0$ so it doesn't matter).
Recall the identity $|z|^2 = z\overline{z}$ . Since $|w| = 1$ $\overline{w} = w^{-1}$ , and \begin{align*} \lvert w^i-w^j\rvert^4 &= ((w^i-w^j)(w^{-i}-w^{-j}))^2 \\ &= (2-w^{i-j}-w^{j-i})^2 \\ &= 4+w^{2i-2j}+w^{2j-2i}+2(-2w^{i-j}-2w^{j-i}+1) \\ &= 6-4w^{i-j}-4w^{j-i}+w^{2i-2j}+w^{2j-2i}. \end{align*}
But notice that, for a fixed value of $i$ , over all values of $j$ from $0$ to $6$ , by properties of modular arithmetic with $w^7 = w^0$ (or expanding),
each of the $w^\bullet$ -terms takes on every value in $w^0,w^1,w^2,\dots,w^6$ exactly once. Since $\sum_{j=0}^6 w^j = 0$ \begin{align*} \frac{1}{2}\sum_{i=0}^6\sum_{j=0}^6\lvert w^i-w^j\rvert^4 &= \frac{1}{2}\sum_{i=0}^6\sum_{j=0}^6(6-4w^{i-j}-4w^{j-i}+w^{2i-2j}+w^{2j-2i}) \\ &= \frac{1}{2}\sum_{i=0}^6\left(\sum_{j=0}^6 6-4\sum_{j=0}^6 w^{i-j}-4\sum_{j=0}^6 w^{j-i}+\sum_{j=0}^6 w^{2i-2j}+\sum_{j=0}^6 w^{2j-2i}\right) \\ &= \frac{1}{2}\sum_{i=0}^6\left(\sum_{j=0}^6 6-4\sum_{j=0}^6 w^j-4\sum_{j=0}^6 w^j+\sum_{j=0}^6 w^j+\sum_{j=0}^6 w^j\right) \\ &= \frac{1}{2}\sum_{i=0}^6\sum_{j=0}^6 6 \\ &= \frac{1}{2}\cdot7\cdot7\cdot6 \\ &= \boxed{147}
| 147
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3,758
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_24
| 9
|
The figure below depicts a regular $7$ -gon inscribed in a unit circle. [asy] import geometry; unitsize(3cm); draw(circle((0,0),1),linewidth(1.5)); for (int i = 0; i < 7; ++i) { for (int j = 0; j < i; ++j) { draw(dir(i * 360/7) -- dir(j * 360/7),linewidth(1.5)); } } for(int i = 0; i < 7; ++i) { dot(dir(i * 360/7),5+black); } [/asy] What is the sum of the $4$ th powers of the lengths of all $21$ of its edges and diagonals?
$\textbf{(A) }49 \qquad \textbf{(B) }98 \qquad \textbf{(C) }147 \qquad \textbf{(D) }168 \qquad \textbf{(E) }196$
|
This is how I solve this problem:
It's easy to solve for $3$ -gon, $4$ -gon, and $6$ -gon inscribed in a unit circle. (Okay, it's just the weird names for triangle, square, and hexagon)
For $3$ -gon, the sum is equal to $3$ times the $4$ th power of an edge. Thus, \[S_3=3\,\cdot\,\left(\sqrt{3}\right)^4=27.\]
For $4$ -gon, the sum is equal to $4$ times the $4$ th power of an edge, and $2$ times the $4$ th power of the diagonal. Thus, \[S_4=4\,\cdot\,\left(\sqrt{2}\right)^4+2\,\cdot\,\left(2\right)^4=48.\]
For $6$ -gon, the sum is equal to $6$ times the $4$ th power of an edge, $6$ times the $4$ th power of the short diagonal, and $3$ times the $4$ th power of the long diagonal. Thus, \[S_6=6\,\cdot\,\left(1\right)^4+6\,\cdot\,\left(\sqrt{3}\right)^4+3\,\cdot\,\left(2\right)^4=108.\]
Then, I quickly noticed that $27=3\,\cdot\,3^2$ $48=3\,\cdot\,4^2$ , and $108=3\,\cdot\,6^2$ . So reasonably, it will work out this formula, $S_n=3n^2$ . (This step is purely out of guessing, maybe have a look at Solution 8 for more info...)
By inductive reasoning, we got $S_7=3\,\cdot\,7^2=\boxed{147}$
| 147
|
3,759
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_25
| 1
|
Four regular hexagons surround a square with side length 1, each one sharing an edge with the square,
as shown in the figure below. The area of the resulting 12-sided outer nonconvex polygon can be
written as $m \sqrt{n} + p$ , where $m$ $n$ , and $p$ are integers and $n$ is not divisible by the square of any prime.
What is $m+n+p$ [asy] import geometry; unitsize(3cm); draw((0,0) -- (1,0) -- (1,1) -- (0,1) -- cycle); draw(shift((1/2,1-sqrt(3)/2))*polygon(6)); draw(shift((1/2,sqrt(3)/2))*polygon(6)); draw(shift((sqrt(3)/2,1/2))*rotate(90)*polygon(6)); draw(shift((1-sqrt(3)/2,1/2))*rotate(90)*polygon(6)); draw((0,1-sqrt(3))--(1,1-sqrt(3))--(3-sqrt(3),sqrt(3)-2)--(sqrt(3),0)--(sqrt(3),1)--(3-sqrt(3),3-sqrt(3))--(1,sqrt(3))--(0,sqrt(3))--(sqrt(3)-2,3-sqrt(3))--(1-sqrt(3),1)--(1-sqrt(3),0)--(sqrt(3)-2,sqrt(3)-2)--cycle,linewidth(2)); [/asy] $\textbf{(A) } -12 \qquad \textbf{(B) }-4 \qquad \textbf{(C) } 4 \qquad \textbf{(D) }24 \qquad \textbf{(E) }32$
|
[asy] import geometry; unitsize(3cm); draw((0,0) -- (1,0) -- (1,1) -- (0,1) -- cycle); draw(shift((1/2,1-sqrt(3)/2))*polygon(6)); draw(shift((1/2,sqrt(3)/2))*polygon(6)); draw(shift((sqrt(3)/2,1/2))*rotate(90)*polygon(6)); draw(shift((1-sqrt(3)/2,1/2))*rotate(90)*polygon(6)); draw((0,1-sqrt(3))--(1,1-sqrt(3))--(3-sqrt(3),sqrt(3)-2)--(sqrt(3),0)--(sqrt(3),1)--(3-sqrt(3),3-sqrt(3))--(1,sqrt(3))--(0,sqrt(3))--(sqrt(3)-2,3-sqrt(3))--(1-sqrt(3),1)--(1-sqrt(3),0)--(sqrt(3)-2,sqrt(3)-2)--cycle,linewidth(1.5)); draw((3-sqrt(3),3-sqrt(3)) -- (3-sqrt(3),sqrt(3)-2) -- (sqrt(3)-2,sqrt(3)-2) -- (sqrt(3)-2,3-sqrt(3)) -- cycle,linewidth(1.5)); label("$O (0, 0)$",(0.5,0.5),S); dot((0.5,0.5)); label("$A$", (3-sqrt(3), 3-sqrt(3)), NE); label("$B$", (sqrt(3) - 2, 3-sqrt(3)), NW); label("$M$", (0, sqrt(3)), NW); label("$N$", (1, sqrt(3)), NE); [/asy]
Refer to the diagram above.
Let the origin be at the center of the square, $A$ be the intersection of the top and right hexagons, $B$ be the intersection of the top and left hexagons, and $M$ and $N$ be the top points in the diagram.
By symmetry, $A$ lies on the line $y = x$ . The equation of line $AN$ is $y = -x\sqrt{3} + \frac{3}{2}\sqrt{3} - \frac{1}{2}$ (due to it being one of the sides of the top hexagon). Thus, we can solve for the coordinates of $A$ by finding the intersection of the two lines: \[x = -x\sqrt{3} + \frac{3\sqrt{3} - 1}{2}\] \[x(\sqrt{3} + 1) = \frac{3\sqrt{3} - 1}{2}\] \[x = \frac{3\sqrt{3}-1}{2} \cdot \frac{1}{\sqrt{3} + 1}\] \[= \frac{3\sqrt{3}-1}{2(\sqrt{3} + 1)} \cdot \frac{\sqrt{3} - 1}{\sqrt{3} - 1}\] \[= \frac{10 - 4\sqrt{3}}{4}\] \[= \frac{5}{2} - \sqrt{3}\] \[\therefore A = \left(\frac{5}{2} - \sqrt{3}, \frac{5}{2} - \sqrt{3}\right).\]
This means that we can find the length $AB$ , which is equal to $2(\frac{5}{2} - \sqrt{3}) = (5 - 2\sqrt{3}$ . We will next find the area of trapezoid $ABMN$ . The lengths of the bases are $1$ and $5 - 2\sqrt{3}$ , and the height is equal to the $y$ -coordinate of $M$ minus the $y$ -coordinate of $A$ . The height of the hexagon is $\sqrt{3}$ and the bottom of the hexagon lies on the line $y = \frac{1}{2}$ . Thus, the $y$ -coordinate of $M$ is $\sqrt{3} - \frac{1}{2}$ , and the height is $2\sqrt{3} - 3$ . We can now find the area of the trapezoid: \[[ABMN] = (2\sqrt{3} - 3)\left(\frac{1 + 5 - 2\sqrt{3}}{2}\right)\] \[= (2\sqrt{3} - 3)(3 - \sqrt{3})\] \[= 6\sqrt{3} + 3\sqrt{3} - 9 - 6\] \[= 9\sqrt{3} - 15.\]
The total area of the figure is the area of a square with side length $AB$ plus four times the area of this trapezoid: \[\textrm{Area} = (5 - 2\sqrt{3})^2 + 4(9\sqrt{3} - 15)\] \[= 37 - 20\sqrt{3} + 36\sqrt{3} - 60\] \[= 16\sqrt{3} - 23.\]
Our answer is $16 + 3 - 23 = \boxed{4}$
| 4
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3,760
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_25
| 2
|
Four regular hexagons surround a square with side length 1, each one sharing an edge with the square,
as shown in the figure below. The area of the resulting 12-sided outer nonconvex polygon can be
written as $m \sqrt{n} + p$ , where $m$ $n$ , and $p$ are integers and $n$ is not divisible by the square of any prime.
What is $m+n+p$ [asy] import geometry; unitsize(3cm); draw((0,0) -- (1,0) -- (1,1) -- (0,1) -- cycle); draw(shift((1/2,1-sqrt(3)/2))*polygon(6)); draw(shift((1/2,sqrt(3)/2))*polygon(6)); draw(shift((sqrt(3)/2,1/2))*rotate(90)*polygon(6)); draw(shift((1-sqrt(3)/2,1/2))*rotate(90)*polygon(6)); draw((0,1-sqrt(3))--(1,1-sqrt(3))--(3-sqrt(3),sqrt(3)-2)--(sqrt(3),0)--(sqrt(3),1)--(3-sqrt(3),3-sqrt(3))--(1,sqrt(3))--(0,sqrt(3))--(sqrt(3)-2,3-sqrt(3))--(1-sqrt(3),1)--(1-sqrt(3),0)--(sqrt(3)-2,sqrt(3)-2)--cycle,linewidth(2)); [/asy] $\textbf{(A) } -12 \qquad \textbf{(B) }-4 \qquad \textbf{(C) } 4 \qquad \textbf{(D) }24 \qquad \textbf{(E) }32$
|
[asy] import geometry; unitsize(3cm); draw((0,0) -- (1,0) -- (1,1) -- (0,1) -- cycle); draw(shift((1/2,1-sqrt(3)/2))*polygon(6)); draw(shift((1/2,sqrt(3)/2))*polygon(6)); draw(shift((sqrt(3)/2,1/2))*rotate(90)*polygon(6)); draw(shift((1-sqrt(3)/2,1/2))*rotate(90)*polygon(6)); draw((0,1-sqrt(3))--(1,1-sqrt(3))--(3-sqrt(3),sqrt(3)-2)--(sqrt(3),0)--(sqrt(3),1)--(3-sqrt(3),3-sqrt(3))--(1,sqrt(3))--(0,sqrt(3))--(sqrt(3)-2,3-sqrt(3))--(1-sqrt(3),1)--(1-sqrt(3),0)--(sqrt(3)-2,sqrt(3)-2)--cycle,linewidth(2)); draw((0.5,sqrt(3))--(0.5,1-sqrt(3)),linewidth(2)); draw((1-sqrt(3),0.5)--(sqrt(3),0.5),linewidth(2)); draw((-2+sqrt(3),-2+sqrt(3))--(3-sqrt(3),3-sqrt(3)),linewidth(2)); draw((1, sqrt(3))--(1,1),linewidth(2)); label("$O$",(0.5,0.5),SE); dot((0.5,0.5)); label("$A$", (3-sqrt(3), 3-sqrt(3)), NE); label("$B$", (1, sqrt(3)), NE); label("$C$", (1,1), E); label("$D$", (1/2, sqrt(3)), N); [/asy]
Begin by dividing the figure as shown above. Clearly, the entire figure has 8-fold symmetry. Therefore, we can calculate the area of $ODBA$ and multiply it by 8. We split $[ODBA]$ into $[ODBC]+[ABC]$
Knowing the side length of the hexagon is $1$ , we can use 30-60-90 triangles within the hexagon to find the total distance between opposite edges is $2\cdot \frac{\sqrt{3}}{2}=\sqrt{3}.$ Thus, $OD=\sqrt{3}-\frac{1}{2}$ and $BC=\sqrt{3}-1.$ Recognizing $DB=\frac{1}{2}$ and $ODBC$ is a trapezoid, \[8\cdot[ODBC]=2\left(\sqrt{3}-\frac{1}{2}+\sqrt{3}-1\right)=4\sqrt{3}-3.\]
Next, we aim to find $[ABC]$ . By angle chasing, we find $\angle A=105^\circ,$ $\angle B=30^\circ,$ and $\angle C = 45^\circ.$ We can use the law of sines to find $AB$
\[\frac{\sin(105^\circ)}{\sqrt{3}-1}=\frac{\sin(45^\circ)}{AB}\implies AB=\frac{\sqrt{6}-\sqrt{2}}{2\sin(105^\circ)}.\]
We may not know what $\sin(105^\circ)$ is by memory, but we can cleverly calculate it using a common trig identity:
\begin{align*} \sin(105^\circ)&=\sin(60^\circ+45^\circ),\\ &=\sin(60^\circ)\cos(45^\circ)+\cos(60^\circ)\sin(45^\circ),\\ &=\frac{\sqrt{3}}{2}\cdot\frac{\sqrt{2}}{2}+\frac{1}{2}\cdot\frac{\sqrt{2}}{2},\\ &=\frac{\sqrt{6}+\sqrt{2}}{4}. \end{align*}
With some simplification, we'll find $AB=4-2\sqrt{3}$ . Now, we can easily calculate $8\cdot [ABC]$ as \[8\cdot\frac{1}{2}\cdot(\sqrt{3}-1)(4-2\sqrt{3})\sin(30^\circ)=12\sqrt{3}-20.\]
Thus, the area of the dodecagon is $8\cdot [ODBA] = 8\cdot [ODBC] + 8 \cdot [ABC] =4\sqrt{3}-3+12\sqrt{3}-20=16\sqrt{3}-23.$
Finally, we find \[16+3-23=\boxed{4}.\]
| 4
|
3,761
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_25
| 3
|
Four regular hexagons surround a square with side length 1, each one sharing an edge with the square,
as shown in the figure below. The area of the resulting 12-sided outer nonconvex polygon can be
written as $m \sqrt{n} + p$ , where $m$ $n$ , and $p$ are integers and $n$ is not divisible by the square of any prime.
What is $m+n+p$ [asy] import geometry; unitsize(3cm); draw((0,0) -- (1,0) -- (1,1) -- (0,1) -- cycle); draw(shift((1/2,1-sqrt(3)/2))*polygon(6)); draw(shift((1/2,sqrt(3)/2))*polygon(6)); draw(shift((sqrt(3)/2,1/2))*rotate(90)*polygon(6)); draw(shift((1-sqrt(3)/2,1/2))*rotate(90)*polygon(6)); draw((0,1-sqrt(3))--(1,1-sqrt(3))--(3-sqrt(3),sqrt(3)-2)--(sqrt(3),0)--(sqrt(3),1)--(3-sqrt(3),3-sqrt(3))--(1,sqrt(3))--(0,sqrt(3))--(sqrt(3)-2,3-sqrt(3))--(1-sqrt(3),1)--(1-sqrt(3),0)--(sqrt(3)-2,sqrt(3)-2)--cycle,linewidth(2)); [/asy] $\textbf{(A) } -12 \qquad \textbf{(B) }-4 \qquad \textbf{(C) } 4 \qquad \textbf{(D) }24 \qquad \textbf{(E) }32$
|
[asy] import geometry; unitsize(3cm); draw((0,0) -- (1,0) -- (1,1) -- (0,1) -- cycle); draw(shift((1/2,1-sqrt(3)/2))*polygon(6)); draw(shift((1/2,sqrt(3)/2))*polygon(6)); draw(shift((sqrt(3)/2,1/2))*rotate(90)*polygon(6)); draw(shift((1-sqrt(3)/2,1/2))*rotate(90)*polygon(6)); draw((0,1-sqrt(3))--(1,1-sqrt(3))--(3-sqrt(3),sqrt(3)-2)--(sqrt(3),0)--(sqrt(3),1)--(3-sqrt(3),3-sqrt(3))--(1,sqrt(3))--(0,sqrt(3))--(sqrt(3)-2,3-sqrt(3))--(1-sqrt(3),1)--(1-sqrt(3),0)--(sqrt(3)-2,sqrt(3)-2)--cycle,linewidth(2)); draw((1,1)--(3-sqrt(3),3-sqrt(3)),linewidth(2)); draw((0,1)--(sqrt(3)-2,3-sqrt(3)),linewidth(2)); draw((0,0)--(sqrt(3)-2,sqrt(3)-2),linewidth(2)); draw((1,0)--(3-sqrt(3),sqrt(3)-2),linewidth(2)); draw((0, sqrt(3))--(0, -sqrt(3)+1),linewidth(2)); draw((1, sqrt(3))--(1, -sqrt(3)+1),linewidth(2)); draw((1-sqrt(3),0)--(sqrt(3),0),linewidth(2)); draw((1-sqrt(3),1)--(sqrt(3),1),linewidth(2)); label("$A$", (3-sqrt(3), 3-sqrt(3)), NE); label("$B$", (1, sqrt(3)), NE); label("$C$", (1,1), SW); label("$D$", (1,0), NW); label("$E$", (1, -sqrt(3)+1), SE); label("$F$", (0, -sqrt(3)+1), SW); label("$G$", (0, sqrt(3)), NW); [/asy]
Divide this polygon into a 'cross' and $8$ triangles like $\triangle ABC$ $BD = 2 \cdot \frac{ \sqrt{3} }{2} = \sqrt{3}$ $BE = 2BD - CD = 2\sqrt{3} - 1$ $EF = 1$ $[BEFG] = 1 \cdot (2\sqrt{3} - 1) = 2\sqrt{3} - 1$ $[\text{cross}] = 2\cdot [BEFG] - 1 = 2(2\sqrt{3} - 1)- 1 = 4\sqrt{3} - 3$
$\angle ABC = 120^{\circ} - 90^{\circ} = 30^{\circ}$ $\angle ACB = \frac{90^{\circ}}{2} = 45^{\circ}$ $\angle BAC = 180^{\circ} - 30^{\circ} - 45^{\circ} = 105^{\circ}$ $BC = BD - CD = \sqrt{3} - 1$ . Draw a perpendicular line $AG$ to side $BC$
$BG + CG = BC$ $AG \cdot \sqrt{3} + AG = \sqrt{3} - 1$ $AG = \frac{\sqrt{3} - 1}{\sqrt{3} + 1} = \frac{(\sqrt{3} - 1)^2}{2} = \frac{4 - 2\sqrt{3}}{2} = 2 - \sqrt{3}$ $[ABC] = \frac12 \cdot AG \cdot BC = \frac12 (2 - \sqrt{3})(\sqrt{3} - 1)$
$[\text{12-sided outer polygon}] = [\text{cross}] + 8 \cdot [ABC] = 4\sqrt{3} - 3 + 8 \cdot \frac12 (2 - \sqrt{3})(\sqrt{3} - 1) = 4\sqrt{3} - 3 + 4 (3\sqrt{3} - 5) = 16\sqrt{3}-23$
\[16+3-23=\boxed{4}\]
| 4
|
3,762
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_25
| 4
|
Four regular hexagons surround a square with side length 1, each one sharing an edge with the square,
as shown in the figure below. The area of the resulting 12-sided outer nonconvex polygon can be
written as $m \sqrt{n} + p$ , where $m$ $n$ , and $p$ are integers and $n$ is not divisible by the square of any prime.
What is $m+n+p$ [asy] import geometry; unitsize(3cm); draw((0,0) -- (1,0) -- (1,1) -- (0,1) -- cycle); draw(shift((1/2,1-sqrt(3)/2))*polygon(6)); draw(shift((1/2,sqrt(3)/2))*polygon(6)); draw(shift((sqrt(3)/2,1/2))*rotate(90)*polygon(6)); draw(shift((1-sqrt(3)/2,1/2))*rotate(90)*polygon(6)); draw((0,1-sqrt(3))--(1,1-sqrt(3))--(3-sqrt(3),sqrt(3)-2)--(sqrt(3),0)--(sqrt(3),1)--(3-sqrt(3),3-sqrt(3))--(1,sqrt(3))--(0,sqrt(3))--(sqrt(3)-2,3-sqrt(3))--(1-sqrt(3),1)--(1-sqrt(3),0)--(sqrt(3)-2,sqrt(3)-2)--cycle,linewidth(2)); [/asy] $\textbf{(A) } -12 \qquad \textbf{(B) }-4 \qquad \textbf{(C) } 4 \qquad \textbf{(D) }24 \qquad \textbf{(E) }32$
|
We calculate the area as the area of the red octagon minus the four purple congruent triangles: [asy] import geometry; unitsize(3cm); draw((1-sqrt(3),1-sqrt(3))--(1-sqrt(3),sqrt(3))--(sqrt(3),sqrt(3))--(sqrt(3),1-sqrt(3))--cycle,dashed); filldraw((0,1-sqrt(3))--(1,1-sqrt(3))--(sqrt(3),0)--(sqrt(3),1)--(1,sqrt(3))--(0,sqrt(3))--(1-sqrt(3),1)--(1-sqrt(3),0)--cycle,red*0.2+white,red); filldraw((1,1-sqrt(3))--(3-sqrt(3),sqrt(3)-2)--(sqrt(3),0)--cycle,purple*0.2+white,blue); filldraw((sqrt(3),1)--(3-sqrt(3),3-sqrt(3))--(1,sqrt(3))--cycle,purple*0.2+white,blue); filldraw((0,sqrt(3))--(sqrt(3)-2,3-sqrt(3))--(1-sqrt(3),1)--cycle,purple*0.2+white,blue); filldraw((0,1-sqrt(3))--(1-sqrt(3),0)--(sqrt(3)-2,sqrt(3)-2)--cycle,purple*0.2+white,blue); draw((0,0) -- (1,0) -- (1,1) -- (0,1) -- cycle); draw(shift((1/2,1-sqrt(3)/2))*polygon(6)); draw(shift((1/2,sqrt(3)/2))*polygon(6)); draw(shift((sqrt(3)/2,1/2))*rotate(90)*polygon(6)); draw(shift((1-sqrt(3)/2,1/2))*rotate(90)*polygon(6)); [/asy] We first find the important angles in the figure. We note that 2 adjacent hexagons are rotated $90^\circ$ with respect to the other, so the angles between any sides is $150^\circ$ . In particular, as the purple triangles are isosceles, they have angles $150^\circ,15^\circ$ , and $15^\circ$ , and the octagon is equiangular (all its angles are $135^\circ$ ). Thus, we can draw a square around the octagon, and we note that the ``cut out" triangles are all isosceles right triangles.
Now, we calculate the side length of the square. Note that the hexagon has a height of $\sqrt 3$ , so the length of a side of the square is $2\sqrt 3-1$ . In particular, the horizontal/vertical sides of the octagon have length $1$ , so the legs of the isosceles triangles are \[\frac{2\sqrt3-1-1}2=\sqrt3-1\] Thus, the area of the octagon is \[(2\sqrt3-1)^2-4\cdot\frac 12(\sqrt3-1)^2=5\] Now, we calculate the area of one of the four isosceles triangles. The base of the triangle is $(\sqrt 3-1)\sqrt 2$ , so the area is \[\frac 14\mathrm{base length}^2\cdot\tan(\mathrm{base angle})=\frac 14((\sqrt3-1)\sqrt2)^2\cdot\tan15^\circ=\frac 14(8-4\sqrt3)(2-\sqrt3)=7-4\sqrt 3\] Thus, the area of the dodecagon is \[5-4(7-4\sqrt3)=16\sqrt3-23\] Thus the answer is $16+3-23=-4$ , or $\boxed{4}$
| 4
|
3,763
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_25
| 5
|
Four regular hexagons surround a square with side length 1, each one sharing an edge with the square,
as shown in the figure below. The area of the resulting 12-sided outer nonconvex polygon can be
written as $m \sqrt{n} + p$ , where $m$ $n$ , and $p$ are integers and $n$ is not divisible by the square of any prime.
What is $m+n+p$ [asy] import geometry; unitsize(3cm); draw((0,0) -- (1,0) -- (1,1) -- (0,1) -- cycle); draw(shift((1/2,1-sqrt(3)/2))*polygon(6)); draw(shift((1/2,sqrt(3)/2))*polygon(6)); draw(shift((sqrt(3)/2,1/2))*rotate(90)*polygon(6)); draw(shift((1-sqrt(3)/2,1/2))*rotate(90)*polygon(6)); draw((0,1-sqrt(3))--(1,1-sqrt(3))--(3-sqrt(3),sqrt(3)-2)--(sqrt(3),0)--(sqrt(3),1)--(3-sqrt(3),3-sqrt(3))--(1,sqrt(3))--(0,sqrt(3))--(sqrt(3)-2,3-sqrt(3))--(1-sqrt(3),1)--(1-sqrt(3),0)--(sqrt(3)-2,sqrt(3)-2)--cycle,linewidth(2)); [/asy] $\textbf{(A) } -12 \qquad \textbf{(B) }-4 \qquad \textbf{(C) } 4 \qquad \textbf{(D) }24 \qquad \textbf{(E) }32$
|
Note that each of the green sections is a rectangle, so its interior angles are all $90^{\circ}.$ Since $\angle{ABC}=120^{\circ}$ , every one of the orange sections is a $30-60-90$ right triangle.
Define $x$ to be the distance from the corner of the square with side length $1$ to the corner of the larger blue square. Due to the sides of the two squares being parallel to each other, the large blue triangle is a $45-45-90$ right triangle. By $AAA$ similarity, the smaller blue triangles are also $45-45-90,$ and have side lengths of $\frac{x\sqrt{2}}{2}, \frac{x\sqrt{2}}{2},$ and $x$ . By $30-60-90$ triangle relations, the largest altitude of the orange triangle is $\frac{x\sqrt{6}}{2}.$
Now, we can find the height of the hexagon to obtain an equation in terms of $x$ . Consider a hexagon with side length $1$ , where point $P$ is the foot of the perpendicular dropped from $B$ , bisecting $\angle{ABC}$ [asy] import geometry; import fontsize; defaultpen(fontsize(9pt)); unitsize(3cm); draw(shift((1/2,1-sqrt(3)/2))*polygon(6)); draw((1,1-sqrt(3))--(1, 1)--cycle,linewidth(1)); draw((3/2,1-sqrt(3)/2)--(1, 1-sqrt(3)/2)--cycle,linewidth(1)); draw((1, 1-sqrt(3)/2)--(1.1, 1-sqrt(3)/2)--(1.1, 1-sqrt(3)/2 +0.1)--(1, 1-sqrt(3)/2+0.1)--cycle,linewidth(1)); draw((1, 1-sqrt(3)/2)--(1.1, 1-sqrt(3)/2)--(1.1, 1-sqrt(3)/2 -0.1)--(1, 1-sqrt(3)/2-0.1)--cycle,linewidth(1)); label("$A$", (1, 1), N); label("$P$", (1, 1-sqrt(3)/2), W); label("$C$", (1, 1-sqrt(3)), S); label("$B$", (3/2, 1-sqrt(3)/2), E); label("$60^{\circ}$", (3/2-0.025, 1-sqrt(3)/2), NW); label("$60^{\circ}$", (3/2-0.025, 1-sqrt(3)/2), SW); [/asy] Note that triangles $\triangle{APB}$ and $\triangle{CPB}$ are congruent $30-60-90$ triangles, by SAS congruence. Since the side length of this hexagon is $1$ , the length of $AP$ is $\frac{1}{2}\cdot \sqrt{3} =\frac{\sqrt{3}}{2}$ , by $30-60-90$ triangle relations. The height of the hexagon is twice this value, or $\sqrt{3}.$
The height is also equal to the sum of the values along the long blue line, in the first diagram. Therefore, $1+\frac{x\sqrt{2}}{2}+\frac{x\sqrt{6}}{2}=\sqrt{3}.$ Solving and rationalizing, $x=2\sqrt{2}-\sqrt{6}.$
The area of the dodecagon is equal to the sum of the areas of the four rectangles, eight orange triangles, and purple square. In terms of $x$ , this is \[8\cdot \frac{1}{2}\cdot \frac{x\sqrt{2}}{2}\cdot \frac{x\sqrt{6}}{{2}}+4\cdot 1\cdot\frac{x\sqrt{6}}{2}+(x\sqrt{2}+1)^2.\]
Plugging in $x=2\sqrt{2}-\sqrt{6}$ , the area of the dodecagon is $16\sqrt{3}-23$ . Therefore, the answer is $16+3-23=$ $\boxed{4}.$
| 4
|
3,764
|
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12A_Problems/Problem_1
| 1
|
What is the value of $\frac{(2112-2021)^2}{169}$
$\textbf{(A) } 7 \qquad\textbf{(B) } 21 \qquad\textbf{(C) } 49 \qquad\textbf{(D) } 64 \qquad\textbf{(E) } 91$
|
We have \[\frac{(2112-2021)^2}{169}=\frac{91^2}{169}=\frac{91^2}{13^2}=\left(\frac{91}{13}\right)^2=7^2=\boxed{49}.\] ~MRENTHUSIASM
| 49
|
3,765
|
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12A_Problems/Problem_1
| 2
|
What is the value of $\frac{(2112-2021)^2}{169}$
$\textbf{(A) } 7 \qquad\textbf{(B) } 21 \qquad\textbf{(C) } 49 \qquad\textbf{(D) } 64 \qquad\textbf{(E) } 91$
|
We have \[\frac{(2112-2021)^2}{169}=\frac{91^2}{169}=\frac{(10^2-3^2)^2}{13^2}=\frac{((10+3)(10-3))^2}{13^2}=\frac{(13\cdot7)^2}{13^2}=\frac{13^2 \cdot 7^2}{13^2}=7^2=\boxed{49}.\]
| 49
|
3,766
|
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12A_Problems/Problem_1
| 3
|
What is the value of $\frac{(2112-2021)^2}{169}$
$\textbf{(A) } 7 \qquad\textbf{(B) } 21 \qquad\textbf{(C) } 49 \qquad\textbf{(D) } 64 \qquad\textbf{(E) } 91$
|
We know that $2112-2021 = 91$ . Approximate this as $100$ as it is pretty close to it. Also, approximate $169$ to $170$ . We then have \[\frac{(2112 - 2021)^2}{169} \approx \frac{100^2}{170} \approx \frac{1000}{17} \approx 58.\] Now check the answer choices. The two closest answers are $49$ and $64$ . As the numerator is actually bigger than it should be, it should be the smaller answer, or $\boxed{49}$
| 49
|
3,767
|
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12A_Problems/Problem_2
| 1
|
Menkara has a $4 \times 6$ index card. If she shortens the length of one side of this card by $1$ inch, the card would have area $18$ square inches. What would the area of the card be in square inches if instead she shortens the length of the other side by $1$ inch?
$\textbf{(A) } 16 \qquad\textbf{(B) } 17 \qquad\textbf{(C) } 18 \qquad\textbf{(D) } 19 \qquad\textbf{(E) } 20$
|
We construct the following table: \[\begin{array}{c||c|c||c} & & & \\ [-2.5ex] \textbf{Scenario} & \textbf{Length} & \textbf{Width} & \textbf{Area} \\ [0.5ex] \hline & & & \\ [-2ex] \text{Initial} & 4 & 6 & 24 \\ \text{Menkara shortens one side.} & 3 & 6 & 18 \\ \text{Menkara shortens other side instead.} & 4 & 5 & 20 \end{array}\] Therefore, the answer is $\boxed{20}.$
| 20
|
3,768
|
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12A_Problems/Problem_4
| 1
|
The six-digit number $\underline{2}\,\underline{0}\,\underline{2}\,\underline{1}\,\underline{0}\,\underline{A}$ is prime for only one digit $A.$ What is $A?$
$\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 3 \qquad\textbf{(C)}\ 5 \qquad\textbf{(D)}\ 7 \qquad\textbf{(E)}\ 9$
|
First, modulo $2$ or $5$ $\underline{20210A} \equiv A$ .
Hence, $A \neq 0, 2, 4, 5, 6, 8$
Second modulo $3$ $\underline{20210A} \equiv 2 + 0 + 2 + 1 + 0 + A \equiv 5 + A$ .
Hence, $A \neq 1, 4, 7$
Third, modulo $11$ $\underline{20210A} \equiv A + 1 + 0 - 0 - 2 - 2 \equiv A - 3$ .
Hence, $A \neq 3$
Therefore, the answer is $\boxed{9}$
| 9
|
3,769
|
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12A_Problems/Problem_4
| 2
|
The six-digit number $\underline{2}\,\underline{0}\,\underline{2}\,\underline{1}\,\underline{0}\,\underline{A}$ is prime for only one digit $A.$ What is $A?$
$\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 3 \qquad\textbf{(C)}\ 5 \qquad\textbf{(D)}\ 7 \qquad\textbf{(E)}\ 9$
|
Any number ending in $5$ is divisible by $5$ . So we can eliminate option $\textbf{(C)}$
If the sum of the digits of a number is divisible by $3$ , the number is divisible by $3$ . The sum of the digits of this number is $2 + 0 + 2 + 1 + 0 + A = 5 + A$ . If $5 + A$ is divisible by $3$ , the number is divisible by $3$ . Thus we can eliminate options $\textbf{(A)}$ and $\textbf{(D)}$
So the correct option is either $\textbf{(B)}$ or $\textbf{(E)}$ . Let's try dividing the number with some integers.
$20210A/7 = 2887x$ , where $x$ is $1A/7$ . Since $13$ and $19$ are both indivisible by $7$ , this does not help us narrow the choices down.
$20210A/11 = 1837x$ , where $x$ is $3A/11$ . Since $33/11 = 3$ , option $\textbf{(B)}$ would make $20210A$ divisible by $11$ . Thus, by elimination, the correct choice must be option $\boxed{9}$
| 9
|
3,770
|
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12A_Problems/Problem_4
| 3
|
The six-digit number $\underline{2}\,\underline{0}\,\underline{2}\,\underline{1}\,\underline{0}\,\underline{A}$ is prime for only one digit $A.$ What is $A?$
$\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 3 \qquad\textbf{(C)}\ 5 \qquad\textbf{(D)}\ 7 \qquad\textbf{(E)}\ 9$
|
$202100 \implies$ divisible by $2$
$202101 \implies$ divisible by $3$
$202102 \implies$ divisible by $2$
$202103 \implies$ divisible by $11$
$202104 \implies$ divisible by $2$
$202105 \implies$ divisible by $5$
$202106 \implies$ divisible by $2$
$202107 \implies$ divisible by $3$
$202108 \implies$ divisible by $2$
This leaves only $A=\boxed{9}$
| 9
|
3,771
|
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12A_Problems/Problem_5
| 1
|
Elmer the emu takes $44$ equal strides to walk between consecutive telephone poles on a rural road. Oscar the ostrich can cover the same distance in $12$ equal leaps. The telephone poles are evenly spaced, and the $41$ st pole along this road is exactly one mile ( $5280$ feet) from the first pole. How much longer, in feet, is Oscar's leap than Elmer's stride?
$\textbf{(A) }6\qquad\textbf{(B) }8\qquad\textbf{(C) }10\qquad\textbf{(D) }11\qquad\textbf{(E) }15$
|
There are $41-1=40$ gaps between the $41$ telephone poles, so the distance of each gap is $5280\div40=132$ feet.
Each of Oscar's leaps covers $132\div12=11$ feet, and each of Elmer's strides covers $132\div44=3$ feet.
Therefore, Oscar's leap is $11-3=\boxed{8}$ feet longer than Elmer's stride.
| 8
|
3,772
|
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12A_Problems/Problem_5
| 2
|
Elmer the emu takes $44$ equal strides to walk between consecutive telephone poles on a rural road. Oscar the ostrich can cover the same distance in $12$ equal leaps. The telephone poles are evenly spaced, and the $41$ st pole along this road is exactly one mile ( $5280$ feet) from the first pole. How much longer, in feet, is Oscar's leap than Elmer's stride?
$\textbf{(A) }6\qquad\textbf{(B) }8\qquad\textbf{(C) }10\qquad\textbf{(D) }11\qquad\textbf{(E) }15$
|
There are $41-1=40$ gaps between the $41$ telephone poles, so Elmer takes $44 \cdot 40 = 1760$ strides in total, and Oscar takes $12 \cdot 40 = 480$ leaps in total. Therefore, the answer is $(5280 \div 480) - (5280 \div 1760) = 11-3=\boxed{8}$
| 8
|
3,773
|
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12A_Problems/Problem_6
| 1
|
As shown in the figure below, point $E$ lies on the opposite half-plane determined by line $CD$ from point $A$ so that $\angle CDE = 110^\circ$ . Point $F$ lies on $\overline{AD}$ so that $DE=DF$ , and $ABCD$ is a square. What is the degree measure of $\angle AFE$
[asy] size(6cm); pair A = (0,10); label("$A$", A, N); pair B = (0,0); label("$B$", B, S); pair C = (10,0); label("$C$", C, S); pair D = (10,10); label("$D$", D, SW); pair EE = (15,11.8); label("$E$", EE, N); pair F = (3,10); label("$F$", F, N); filldraw(D--arc(D,2.5,270,380)--cycle,lightgray); dot(A^^B^^C^^D^^EE^^F); draw(A--B--C--D--cycle); draw(D--EE--F--cycle); label("$110^\circ$", (15,9), SW); [/asy]
$\textbf{(A) }160\qquad\textbf{(B) }164\qquad\textbf{(C) }166\qquad\textbf{(D) }170\qquad\textbf{(E) }174$
|
By angle subtraction, we have $\angle ADE = 360^\circ - \angle ADC - \angle CDE = 160^\circ.$ Note that $\triangle DEF$ is isosceles, so $\angle DFE = \frac{180^\circ - \angle ADE}{2}=10^\circ.$ Finally, we get $\angle AFE = 180^\circ - \angle DFE = \boxed{170}$ degrees.
| 170
|
3,774
|
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12A_Problems/Problem_6
| 2
|
As shown in the figure below, point $E$ lies on the opposite half-plane determined by line $CD$ from point $A$ so that $\angle CDE = 110^\circ$ . Point $F$ lies on $\overline{AD}$ so that $DE=DF$ , and $ABCD$ is a square. What is the degree measure of $\angle AFE$
[asy] size(6cm); pair A = (0,10); label("$A$", A, N); pair B = (0,0); label("$B$", B, S); pair C = (10,0); label("$C$", C, S); pair D = (10,10); label("$D$", D, SW); pair EE = (15,11.8); label("$E$", EE, N); pair F = (3,10); label("$F$", F, N); filldraw(D--arc(D,2.5,270,380)--cycle,lightgray); dot(A^^B^^C^^D^^EE^^F); draw(A--B--C--D--cycle); draw(D--EE--F--cycle); label("$110^\circ$", (15,9), SW); [/asy]
$\textbf{(A) }160\qquad\textbf{(B) }164\qquad\textbf{(C) }166\qquad\textbf{(D) }170\qquad\textbf{(E) }174$
|
We can extend $\overline{AD}$ to $G$ , making $\angle CDG$ a right angle. It follows that $\angle GDE$ is $110^\circ - 90^\circ = 20^\circ$ , as shown below. [asy] size(6cm); pair A = (0,10); label("$A$", A, N); pair B = (0,0); label("$B$", B, S); pair C = (10,0); label("$C$", C, S); pair D = (10,10); label("$D$", D, SW); pair EE = (15,11.8); label("$E$", EE, N); pair F = (3,10); label("$F$", F, N); pair G = (15,10); label("$G$", G, E); filldraw(D--arc(D,2.5,270,380)--cycle,lightgray); dot(A^^B^^C^^D^^EE^^F^^G); draw(A--B--C--D--G--cycle); draw(D--EE--F--cycle); [/asy] Since $\angle DFE = \angle DEF$ , we see that $\angle DFE = \angle DEF = \frac{20}{2} = 10^\circ$ . Thus, $\angle AFE = 180^\circ - 10^\circ = \boxed{170}$ degrees.
| 170
|
3,775
|
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12A_Problems/Problem_8
| 1
|
Let $M$ be the least common multiple of all the integers $10$ through $30,$ inclusive. Let $N$ be the least common multiple of $M,32,33,34,35,36,37,38,39,$ and $40.$ What is the value of $\frac{N}{M}?$
$\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 2 \qquad\textbf{(C)}\ 37 \qquad\textbf{(D)}\ 74 \qquad\textbf{(E)}\ 2886$
|
By the definition of least common mutiple, we take the greatest powers of the prime numbers of the prime factorization of all the numbers, that we are taking the $\text{lcm}$ of. In this case, \[M = 2^4 \cdot 3^3 \cdot 5^2 \cdot 7 \cdot 11 \cdot 13 \cdot 17 \cdot 19 \cdot 23 \cdot 29.\] Now, using the same logic, we find that \[N = M \cdot 2 \cdot 37,\] because we have an extra power of $2$ and an extra power of $37.$ Thus, $\frac{N}{M} = 2\cdot 37 = \boxed{74}.$
| 74
|
3,776
|
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12A_Problems/Problem_9
| 1
|
A right rectangular prism whose surface area and volume are numerically equal has edge lengths $\log_{2}x, \log_{3}x,$ and $\log_{4}x.$ What is $x?$
$\textbf{(A)}\ 2\sqrt{6} \qquad\textbf{(B)}\ 6\sqrt{6} \qquad\textbf{(C)}\ 24 \qquad\textbf{(D)}\ 48 \qquad\textbf{(E)}\ 576$
|
The surface area of this right rectangular prism is $2(\log_{2}x\log_{3}x+\log_{2}x\log_{4}x+\log_{3}x\log_{4}x).$
The volume of this right rectangular prism is $\log_{2}x\log_{3}x\log_{4}x.$
Equating the numerical values of the surface area and the volume, we have \[2(\log_{2}x\log_{3}x+\log_{2}x\log_{4}x+\log_{3}x\log_{4}x)=\log_{2}x\log_{3}x\log_{4}x.\] Dividing both sides by $\log_{2}x\log_{3}x\log_{4}x,$ we get \[2\left(\frac{1}{\log_{4}x}+\frac{1}{\log_{3}x}+\frac{1}{\log_{2}x}\right)=1. \hspace{15mm} (\bigstar)\] Recall that $\log_{b}a=\frac{1}{\log_{a}b}$ and $\log_{b}\left(a^n\right)=n\log_{b}a,$ so we rewrite $(\bigstar)$ as \begin{align*} 2(\log_{x}4+\log_{x}3+\log_{x}2)&=1 \\ 2\log_{x}24&=1 \\ \log_{x}576&=1 \\ x&=\boxed{576} ~MRENTHUSIASM
| 576
|
3,777
|
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12A_Problems/Problem_15
| 1
|
Recall that the conjugate of the complex number $w = a + bi$ , where $a$ and $b$ are real numbers and $i = \sqrt{-1}$ , is the complex number $\overline{w} = a - bi$ . For any complex number $z$ , let $f(z) = 4i\hspace{1pt}\overline{z}$ . The polynomial \[P(z) = z^4 + 4z^3 + 3z^2 + 2z + 1\] has four complex roots: $z_1$ $z_2$ $z_3$ , and $z_4$ . Let \[Q(z) = z^4 + Az^3 + Bz^2 + Cz + D\] be the polynomial whose roots are $f(z_1)$ $f(z_2)$ $f(z_3)$ , and $f(z_4)$ , where the coefficients $A,$ $B,$ $C,$ and $D$ are complex numbers. What is $B + D?$
$(\textbf{A})\: {-}304\qquad(\textbf{B}) \: {-}208\qquad(\textbf{C}) \: 12i\qquad(\textbf{D}) \: 208\qquad(\textbf{E}) \: 304$
|
By Vieta's formulas, $z_1z_2+z_1z_3+\dots+z_3z_4=3$ , and $B=(4i)^2\left(\overline{z}_1\,\overline{z}_2+\overline{z}_1\,\overline{z}_3+\dots+\overline{z}_3\,\overline{z}_4\right).$
Since $\overline{a}\cdot\overline{b}=\overline{ab},$ \[B=(4i)^2\left(\overline{z_1z_2}+\overline{z_1z_3}+\overline{z_1z_4}+\overline{z_2z_3}+\overline{z_2z_4}+\overline{z_3z_4}\right).\] Since $\overline{a}+\overline{b}=\overline{a+b},$ \[B=(4i)^2\left(\overline{z_1z_2+z_1z_3+\dots+z_3z_4}\right)=-16(\overline{3})=-48\]
Also, $z_1z_2z_3z_4=1,$ and \[D=(4i)^4\left(\overline{z}_1\,\overline{z}_2\,\overline{z}_3\,\overline{z}_4\right)=256\left(\overline{z_1z_2z_3z_4}\right)=256(\overline{1})=256.\]
Our answer is $B+D=256-48=\boxed{208}.$
| 208
|
3,778
|
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12A_Problems/Problem_16
| 1
|
An organization has $30$ employees, $20$ of whom have a brand A computer while the other $10$ have a brand B computer. For security, the computers can only be connected to each other and only by cables. The cables can only connect a brand A computer to a brand B computer. Employees can communicate with each other if their computers are directly connected by a cable or by relaying messages through a series of connected computers. Initially, no computer is connected to any other. A technician arbitrarily selects one computer of each brand and installs a cable between them, provided there is not already a cable between that pair. The technician stops once every employee can communicate with each other. What is the maximum possible number of cables used?
$\textbf{(A)}\ 190 \qquad\textbf{(B)}\ 191 \qquad\textbf{(C)}\ 192 \qquad\textbf{(D)}\ 195 \qquad\textbf{(E)}\ 196$
|
We claim that to maximize the number of cables used, we isolate one computer and connect all cables for the remaining $29$ computers, then connect one more cable for the isolated computer.
If a brand A computer is isolated, then the technician can use at most $19\cdot10+1=191$ cables. If a brand B computer is isolated instead, then the technician can use at most $20\cdot9+1=181$ cables. Therefore, the answer is $\boxed{191}.$
| 191
|
3,779
|
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12A_Problems/Problem_17
| 2
|
For how many ordered pairs $(b,c)$ of positive integers does neither $x^2+bx+c=0$ nor $x^2+cx+b=0$ have two distinct real solutions?
$\textbf{(A) } 4 \qquad \textbf{(B) } 6 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 12 \qquad \textbf{(E) } 16 \qquad$
|
Similar to Solution 1, use the discriminant to get $b^2\leq 4c$ and $c^2\leq 4b$ . These can be rearranged to $c\geq \frac{1}{4}b^2$ and $b\geq \frac{1}{4}c^2$ . Now, we can roughly graph these two inequalities, letting one of them be the $x$ axis and the other be $y$ .
The graph of solutions should be above the parabola and under its inverse, meaning we want points on the graph or in the first area enclosed by the two graphs: [asy] unitsize(2); Label f; f.p=fontsize(6); xaxis("$x$",0,5,Ticks(f, 1.0)); yaxis("$y$",0,5,Ticks(f, 1.0)); real f(real x) { return 0.25x^2; } real g(real x) { return 2*sqrt(x); } dot((1,1)); dot((2,1)); dot((1,2)); dot((2,2)); dot((3,3)); dot((4,4)); draw(graph(f,0,sqrt(20))); draw(graph(g,0,5)); [/asy] We are looking for lattice points (since $b$ and $c$ are positive integers), of which we can count $\boxed{6}$
| 6
|
3,780
|
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12A_Problems/Problem_17
| 3
|
For how many ordered pairs $(b,c)$ of positive integers does neither $x^2+bx+c=0$ nor $x^2+cx+b=0$ have two distinct real solutions?
$\textbf{(A) } 4 \qquad \textbf{(B) } 6 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 12 \qquad \textbf{(E) } 16 \qquad$
|
We need to solve the following system of inequalities: \[ \left\{ \begin{array}{ll} b^2 - 4 c \leq 0 \\ c^2 - 4 b \leq 0 \end{array} \right.. \] Feasible solutions are in the region formed between two parabolas $b^2 - 4 c = 0$ and $c^2 - 4 b = 0$
Define $f \left( b \right) = \frac{b^2}{4}$ and $g \left( b \right) = 2 \sqrt{b}$ .
Therefore, all feasible solutions are in the region formed between the graphs of these two functions.
For $b = 1$ , we have $f(b) = \frac{1}{4}$ and $g(b) = 2$ .
Hence, the feasible $c$ are $1, 2$
For $b = 2$ , we have $f(b) = 1$ and $g(b) = 2 \sqrt{2}$ .
Hence, the feasible $c$ are $1, 2$
For $b = 3$ , we have $f(b) = \frac{9}{4}$ and $g(b) = 2 \sqrt{3}$ .
Hence, the feasible $c$ is $3$
For $b = 4$ , we have $f(b) = 4$ and $g(b) = 4$ .
Hence, the feasible $c$ is $4$
For $b > 4$ , we have $f(b) > g(b)$ . Hence, there is no feasible $c$
Putting all cases together, the correct answer is $\boxed{6}$
| 6
|
3,781
|
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12A_Problems/Problem_17
| 4
|
For how many ordered pairs $(b,c)$ of positive integers does neither $x^2+bx+c=0$ nor $x^2+cx+b=0$ have two distinct real solutions?
$\textbf{(A) } 4 \qquad \textbf{(B) } 6 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 12 \qquad \textbf{(E) } 16 \qquad$
|
A quadratic equation $Ax^2+Bx+C=0$ has one real solution if and only if $\sqrt{B^2-4AC}=0.$ Similarly, it has imaginary solutions if and only if $\sqrt{B^2-4AC}<0.$ We proceed as following:
We want both $x^2+bx+c$ to be $1$ value or imaginary and $x^2+cx+b$ to be $1$ value or imaginary. $x^2+4x+4$ is one such case since $\sqrt {b^2-4ac}$ is $0.$ Also, $x^2+3x+3, x^2+2x+2, x^2+x+1$ are always imaginary for both $b$ and $c.$ We also have $x^2+x+2$ along with $x^2+2x+1$ since the latter has one solution, while the first one is imaginary. Therefore, we have $\boxed{6}$ total ordered pairs of integers.
| 6
|
3,782
|
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12A_Problems/Problem_17
| 5
|
For how many ordered pairs $(b,c)$ of positive integers does neither $x^2+bx+c=0$ nor $x^2+cx+b=0$ have two distinct real solutions?
$\textbf{(A) } 4 \qquad \textbf{(B) } 6 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 12 \qquad \textbf{(E) } 16 \qquad$
|
We see that $b^2 \leq 4c$ and $c^2 \leq 4b.$ WLOG, assume that $b \geq c.$ Then we have that $b^2 \leq 4c \leq 4b$ , so $b^2 \leq 4b$ and therefore $b \leq 4$ , also meaning that $c \leq 4.$ This means that we only need to try 16 cases. Now we can get rid of the assumption that $b \geq c$ , because we want ordered pairs. For $b = 1$ and $b = 2$ $c = 1$ and $c = 2$ work. When $b = 3$ $c$ can only be $3$ , and when $b = 4$ , only $c = 4$ works, for a total of $\boxed{6}$ ordered pairs of integers.
| 6
|
3,783
|
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12A_Problems/Problem_17
| 6
|
For how many ordered pairs $(b,c)$ of positive integers does neither $x^2+bx+c=0$ nor $x^2+cx+b=0$ have two distinct real solutions?
$\textbf{(A) } 4 \qquad \textbf{(B) } 6 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 12 \qquad \textbf{(E) } 16 \qquad$
|
We need both $b^2\leq 4c$ and $c^2\leq 4b$
If $b=c$ then the above become $b^2\leq 4b\iff b\leq 4$ , so we have four solutions $(k,k)$ , where $k=1$ $2$ $3$ $4$
If $b<c$ then we only need $c^2\leq 4b$ since it implies $b^2< 4c$ . Now $c^2\leq 4b\leq 4(c-1) \implies (c-2)^2\leq 0 \implies c=2$ , so $b=1$ . We plug $b=1$ $c=2$ back into $c^2\leq 4b$ and it works. So there is another solution $(1,2)$
By symmetry, if $b>c$ then $(b,c)=(2,1)$
Therefore the total number of solutions is $\boxed{6}$
| 6
|
3,784
|
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12A_Problems/Problem_17
| 7
|
For how many ordered pairs $(b,c)$ of positive integers does neither $x^2+bx+c=0$ nor $x^2+cx+b=0$ have two distinct real solutions?
$\textbf{(A) } 4 \qquad \textbf{(B) } 6 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 12 \qquad \textbf{(E) } 16 \qquad$
|
Since $b^{2} - 4c \le 0$ and $c^{2} - 4b \le 0$ , adding the two together yields $b^{2} + c^{2} \le 4(c+b)$ . Obviously, this is not true if either $b$ or $c$ get too large, and they are equal when $b = c = 4$ , so the greatest pair is $(4,4)$ and both numbers must be lesser for further pairs. For there to be two distinct real solutions, we can test all these pairs where $(b,c)$ are less than 4 (except for the already valid solution) on the original quadratics, and we find the working pairs are $(1,1)$ $(2,1)$ $(2,1)$ $(2,2)$ $(3,3)$ $(4,4)$ meaning there are $\boxed{6}$ pairs.
| 6
|
3,785
|
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12A_Problems/Problem_18
| 1
|
Each of the $20$ balls is tossed independently and at random into one of the $5$ bins. Let $p$ be the probability that some bin ends up with $3$ balls, another with $5$ balls, and the other three with $4$ balls each. Let $q$ be the probability that every bin ends up with $4$ balls. What is $\frac{p}{q}$
$\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 4 \qquad\textbf{(C)}\ 8 \qquad\textbf{(D)}\ 12 \qquad\textbf{(E)}\ 16$
|
For simplicity purposes, we assume that the balls and the bins are both distinguishable.
Recall that there are $5^{20}$ ways to distribute $20$ balls into $5$ bins. We have \[p=\frac{5\cdot4\cdot\binom{20}{3,5,4,4,4}}{5^{20}} \text{ and } q=\frac{\binom{20}{4,4,4,4,4}}{5^{20}}.\] Therefore, the answer is \[\frac pq=\frac{5\cdot4\cdot\binom{20}{3,5,4,4,4}}{\binom{20}{4,4,4,4,4}}=\frac{5\cdot4\cdot\frac{20!}{3!\cdot5!\cdot4!\cdot4!\cdot4!}}{\frac{20!}{4!\cdot4!\cdot4!\cdot4!\cdot4!}}=\frac{5\cdot4\cdot(4!\cdot4!\cdot4!\cdot4!\cdot4!)}{3!\cdot5!\cdot4!\cdot4!\cdot4!}=\frac{5\cdot4\cdot4}{5}=\boxed{16}.\] ~MRENTHUSIASM ~Jesshuang
| 16
|
3,786
|
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12A_Problems/Problem_18
| 2
|
Each of the $20$ balls is tossed independently and at random into one of the $5$ bins. Let $p$ be the probability that some bin ends up with $3$ balls, another with $5$ balls, and the other three with $4$ balls each. Let $q$ be the probability that every bin ends up with $4$ balls. What is $\frac{p}{q}$
$\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 4 \qquad\textbf{(C)}\ 8 \qquad\textbf{(D)}\ 12 \qquad\textbf{(E)}\ 16$
|
For simplicity purposes, we assume that the balls and the bins are both distinguishable.
Let $q=\frac{x}{a},$ where $a$ is the total number of combinations and $x$ is the number of cases where every bin ends up with $4$ balls.
We can take $1$ ball from one bin and place it in another bin so that some bin ends up with $3$ balls, another with $5$ balls, and the other three with $4$ balls each. Note that one configuration of $4{-}4{-}4{-}4{-}4$ corresponds to $5\cdot4\cdot4=80$ configurations of $3{-}5{-}4{-}4{-}4.$ On the other hand, one configuration of $3{-}5{-}4{-}4{-}4$ corresponds to $5$ configurations of $4{-}4{-}4{-}4{-}4.$
Therefore, we have \[p = \frac{80}{5}\cdot\frac{x}{a} = 16\cdot\frac{x}{a},\] from which $\frac{p}{q} = \boxed{16}.$
| 16
|
3,787
|
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12A_Problems/Problem_18
| 3
|
Each of the $20$ balls is tossed independently and at random into one of the $5$ bins. Let $p$ be the probability that some bin ends up with $3$ balls, another with $5$ balls, and the other three with $4$ balls each. Let $q$ be the probability that every bin ends up with $4$ balls. What is $\frac{p}{q}$
$\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 4 \qquad\textbf{(C)}\ 8 \qquad\textbf{(D)}\ 12 \qquad\textbf{(E)}\ 16$
|
Since both of the cases will have $3$ bins with $4$ balls in them, we can leave those out. There are $2 \cdot \binom {5}{2} = 20$ ways to choose where to place the $3$ and the $5$ . After that, there are $\binom {8}{3} = 56$ ways to put the $3$ and $5$ balls being put into the bins. For the $4,4,4,4,4$ case, after we canceled the $4,4,4$ out, we have $\binom {8}{4} = 70$ ways to put the $4$ balls inside the bins. Therefore, we have $\frac {56\cdot 20}{70}$ which is equal to $8 \cdot 2 = \boxed{16}$
| 16
|
3,788
|
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12A_Problems/Problem_18
| 4
|
Each of the $20$ balls is tossed independently and at random into one of the $5$ bins. Let $p$ be the probability that some bin ends up with $3$ balls, another with $5$ balls, and the other three with $4$ balls each. Let $q$ be the probability that every bin ends up with $4$ balls. What is $\frac{p}{q}$
$\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 4 \qquad\textbf{(C)}\ 8 \qquad\textbf{(D)}\ 12 \qquad\textbf{(E)}\ 16$
|
Construct the set $A$ consisting of all possible $3{-}5{-}4{-}4{-}4$ bin configurations, and construct set $B$ consisting of all possible $4{-}4{-}4{-}4{-}4$ configurations. If we let $N$ be the total number of configurations possible, it's clear we want to solve for $\frac{p}{q} = \frac{\frac{|A|}{N}}{\frac{|B|}{N}} = \frac{|A|}{|B|}$
Consider drawing an edge between an element in $A$ and an element in $B$ if it is possible to reach one configuration from the other by moving a single ball (Note this process is reversible.). Let us consider the total number of edges drawn.
For any element in $A$ , we may choose one of the $5$ balls in the $5$ -bin and move it to the $3$ -bin to get a valid element in $B$ . This implies the number of edges is $5|A|$
On the other hand, for any element in $B$ , we may choose one of the $20$ balls and move it to one of the other $4$ -bins to get a valid element in $A$ . This implies the number of edges is $80|B|$
We equate the expressions to get $5|A| = 80|B|$ , from which $\frac{|A|}{|B|} = \frac{80}{5} = \boxed{16}$
| 16
|
3,789
|
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12A_Problems/Problem_19
| 1
|
Let $x$ be the least real number greater than $1$ such that $\sin(x)= \sin(x^2)$ , where the arguments are in degrees. What is $x$ rounded up to the closest integer?
$\textbf{(A) } 10 \qquad \textbf{(B) } 13 \qquad \textbf{(C) } 14 \qquad \textbf{(D) } 19 \qquad \textbf{(E) } 20$
|
The smallest $x$ to make $\sin(x) = \sin(x^2)$ would require $x=x^2$ , but since $x$ needs to be greater than $1$ , these solutions are not valid.
The next smallest $x$ would require $x=180-x^2$ , or $x^2+x=180$
After a bit of guessing and checking, we find that $12^2+12=156$ , and $13^2+13=182$ , so the solution lies between $12{ }$ and $13$ , making our answer $\boxed{13}.$
| 13
|
3,790
|
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12A_Problems/Problem_20
| 2
|
For each positive integer $n$ , let $f_1(n)$ be twice the number of positive integer divisors of $n$ , and for $j \ge 2$ , let $f_j(n) = f_1(f_{j-1}(n))$ . For how many values of $n \le 50$ is $f_{50}(n) = 12?$
$\textbf{(A) }7\qquad\textbf{(B) }8\qquad\textbf{(C) }9\qquad\textbf{(D) }10\qquad\textbf{(E) }11$
|
First, take note that the maximum possible value of $f_1(n)$ for $1 \le n \le k$ increases as $k$ increases (it is a step function), i.e. it is increasing. Likewise, as $k$ decreases, the maximum possible value of $f_1(n)$ decreases as well. Also, let $f_1(n) = 2d(n)$ where $d(n)$ is the number of divisors of n.
Since $n \le 50$ $f_1(n) <= 20$ . This maximum occurs when $d(n) = 10 \implies n = 2^4 \cdot 3 = 48$ . Next, since $f_1(n) <=20$ $f_1(f_1(n)) \le 12 \implies f_2(n) \le 12$ . This maximum occurs when $d(f_1(n)) = 6 \implies n = 2 \cdot 3^2 = 18, n = 2^2 \cdot 3 = 12$ . Since $f_2(n) \le 12$ $f_1(f_2(n)) \le 12 \implies f_3(n) \le 12$ , once again. This maximum again occurs when $d(f_2(n)) = 6 \implies f_2(n) = 2^2 \cdot 3 = 12$ . Now, suppose for the sake of contradiction that $f_2(n) < 12$ . Then, $f_3(n) < 12$ (since $f_2(n) = 12$ was the only number that would maximize $f_3(n))$ for $f_2(n) \le 12$ ). As a result, since $f_1(n)$ is increasing, and because $12$ is where $f_1$ steps down from a maximum of $6 \cdot 2 = 12$ , we must have that $f_1(f_3(n)) < f_1(12) = 12 \implies f_4(n) < 12$ . We continue applying $f_1$ on both sides (which is possible since $f_1$ is increasing) until we reach $f_50$ , giving us that $f_50(n) < 12$ . However, $f_50(n) = 12$ , which is a contradiction. Thus, $f_2(n) = 12$
Now, let us finally solve for the solutions. $f_2(n) = 12 \implies f_1(f_1(n)) = 12 \implies d(f_1(n)) = 6$ $d(f_1(n)) = 6 \implies f_1(n) = p^2 \cdot q$ where $p$ and $q$ are primes. Since $f_1(n) \le 20$ $f_1(n)$ can only be $12$ $18$ , or $20$ . If $f_1(n) = 12$ , then $d(n) = 6 \implies n = p^5, p^2 \cdot q \implies n \in \{ 12, 18, 20, 28, 32, 44, 45, 50 \}$ , resulting in 8 solutions. If $f_1(n) = 18$ , then $d(n) = 9 \implies n = p^8, p^2 \cdot q^2 \implies n = 36$ , giving us one more solution. Finally, $f_1(n) = 20 \implies d(n) = 10 \implies n = p^9, p^4 \cdot q \implies n = 48$ . Thus, in total, we have $\boxed{10}$ solutions.
| 10
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3,791
|
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12A_Problems/Problem_20
| 3
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For each positive integer $n$ , let $f_1(n)$ be twice the number of positive integer divisors of $n$ , and for $j \ge 2$ , let $f_j(n) = f_1(f_{j-1}(n))$ . For how many values of $n \le 50$ is $f_{50}(n) = 12?$
$\textbf{(A) }7\qquad\textbf{(B) }8\qquad\textbf{(C) }9\qquad\textbf{(D) }10\qquad\textbf{(E) }11$
|
$\textbf{Observation 1}$ $f_1 \left( 12 \right) = 12$
Hence, if $n$ has the property that $f_j \left( n \right) = 12$ for some $j$ , then $f_k \left( n \right) = 12$ for all $k > j$
$\textbf{Observation 2}$ $f_1 \left( 8 \right) = 8$
Hence, if $n$ has the property that $f_j \left( n \right) = 8$ for some $j$ , then $f_k \left( n \right) = 8$ for all $k > j$
$\textbf{Case 1}$ $n = 1$
We have $f_1 \left( n \right) = 2$ $f_2 \left( n \right) = f_1 \left( 2 \right) = 4$ $f_3 \left( n \right) = f_1 \left( 4 \right) = 6$ $f_4 \left( n \right) = f_1 \left( 6 \right) = 8$ . Hence, Observation 2 implies $f_{50} \left( n \right) = 8$
$\textbf{Case 2}$ $n$ is prime.
We have $f_1 \left( n \right) = 4$ $f_2 \left( n \right) = f_1 \left( 4 \right) = 6$ $f_3 \left( n \right) = f_1 \left( 6 \right) = 8$ . Hence, Observation 2 implies $f_{50} \left( n \right) = 8$
$\textbf{Case 3}$ : The prime factorization of $n$ takes the form $p_1^2$
We have $f_1 \left( n \right) = 6$ $f_2 \left( n \right) = f_1 \left( 6 \right) = 8$ . Hence, Observation 2 implies $f_{50} \left( n \right) = 8$
$\textbf{Case 4}$ : The prime factorization of $n$ takes the form $p_1^3$
We have $f_1 \left( n \right) = 8$ . Hence, Observation 2 implies $f_{50} \left( n \right) = 8$
$\textbf{Case 5}$ : The prime factorization of $n$ takes the form $p_1^4$
We have $f_1 \left( n \right) = 10$ $f_2 \left( n \right) = f_1 \left( 10 \right) = 8$ . Hence, Observation 2 implies $f_{50} \left( n \right) = 8$
$\textbf{Case 6}$ : The prime factorization of $n$ takes the form $p_1^5$
We have $f_1 \left( n \right) = 12$ . Hence, Observation 1 implies $f_{50} \left( n \right) = 12$
In this case the only $n$ is $2^5 = 32$
$\textbf{Case 7}$ : The prime factorization of $n$ takes the form $p_1 p_2$
We have $f_1 \left( n \right) = 8$ . Hence, Observation 2 implies $f_{50} \left( n \right) = 8$
$\textbf{Case 8}$ : The prime factorization of $n$ takes the form $p_1 p_2^2$
We have $f_1 \left( n \right) = 12$ . Hence, Observation 1 implies $f_{50} \left( n \right) = 12$
In this case, all $n$ are $12, 18, 20, 28, 44, 45,$ and $50$
$\textbf{Case 9}$ : The prime factorization of $n$ takes the form $p_1 p_2^3$
We have $f_1 \left( n \right) = 16$ $f_2 \left( n \right) = f_1 \left( 16 \right) = 10$ $f_3 \left( n \right) = f_1 \left( 10 \right) = 8$ . Hence, Observation 2 implies $f_{50} \left( n \right) = 8$
$\textbf{Case 10}$ : The prime factorization of $n$ takes the form $p_1 p_2^4$
We have $f_1 \left( n \right) = 20$ $f_2 \left( n \right) = f_1 \left( 20 \right) = 12$ . Hence, Observation 1 implies $f_{50} \left( n \right) = 12$
In this case, the only $n$ is $48$
$\textbf{Case 11}$ : The prime factorization of $n$ takes the form $p_1^2 p_2^2$
We have $f_1 \left( n \right) = 18$ $f_2 \left( n \right) = f_1 \left( 18 \right) = 12$ . Hence, Observation 1 implies $f_{50} \left( n \right) = 12$
In this case, the only $n$ is $36$
$\textbf{Case 12}$ : The prime factorization of $n$ takes the form $p_1 p_2 p_3$
We have $f_1 \left( n \right) = 16$ $f_2 \left( n \right) = f_1 \left( 16 \right) = 10$ $f_3 \left( n \right) = f_2 \left( 10 \right) = 8$ . Hence, Observation 2 implies $f_{50} \left( n \right) = 8$
Putting all cases together, the number of feasible $n \leq 50$ is $\boxed{10}$
| 10
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3,792
|
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12A_Problems/Problem_22
| 1
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Azar and Carl play a game of tic-tac-toe. Azar places an in $X$ one of the boxes in a $3$ -by- $3$ array of boxes, then Carl places an $O$ in one of the remaining boxes. After that, Azar places an $X$ in one of the remaining boxes, and so on until all boxes are filled or one of the players has of their symbols in a row—horizontal, vertical, or diagonal—whichever comes first, in which case that player wins the game. Suppose the players make their moves at random, rather than trying to follow a rational strategy, and that Carl wins the game when he places his third $O$ . How many ways can the board look after the game is over?
$\textbf{(A) } 36 \qquad\textbf{(B) } 112 \qquad\textbf{(C) } 120 \qquad\textbf{(D) } 148 \qquad\textbf{(E) } 160$
|
We need to find out the number of configurations with 3 $O$ and 3 $X$ with 3 $O$ in a row, and 3 $X$ not in a row.
$\textbf{Case 1}$ : 3 $O$ are in a horizontal row or a vertical row.
Step 1: We determine the row that 3 $O$ occupy.
The number of ways is 6.
Step 2: We determine the configuration of 3 $X$
The number of ways is $\binom{6}{3} - 2 = 18$
In this case, following from the rule of product, the number of ways is $6 \cdot 18 = 108$
$\textbf{Case 2}$ : 3 $O$ are in a diagonal row.
Step 1: We determine the row that 3 $O$ occupy.
The number of ways is 2.
Step 2: We determine the configuration of 3 $X$
The number of ways is $\binom{6}{3} = 20$
In this case, following from the rule of product, the number of ways is $2 \cdot 20 = 40$
Putting all cases together, the total number of ways is $108 + 40 = 148$
Therefore, the answer is $\boxed{148}$
| 148
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3,793
|
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12A_Problems/Problem_24
| 1
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Convex quadrilateral $ABCD$ has $AB = 18, \angle{A} = 60^\circ,$ and $\overline{AB} \parallel \overline{CD}.$ In some order, the lengths of the four sides form an arithmetic progression, and side $\overline{AB}$ is a side of maximum length. The length of another side is $a.$ What is the sum of all possible values of $a$
$\textbf{(A) } 24 \qquad \textbf{(B) } 42 \qquad \textbf{(C) } 60 \qquad \textbf{(D) } 66 \qquad \textbf{(E) } 84$
|
Let $E$ be a point on $\overline{AB}$ such that $BCDE$ is a parallelogram. Suppose that $BC=ED=b, CD=BE=c,$ and $DA=d,$ so $AE=18-c,$ as shown below. [asy] /* Made by MRENTHUSIASM */ size(250); pair A, B, C, D, E; A = (0,0); B = (18,0); D = A+9*dir(60); C = D+(7,0); E = D+(B-C); dot("$A$",A,1.5*SW,linewidth(4)); dot("$B$",B,1.5*SE,linewidth(4)); dot("$C$",C,1.5*NE,linewidth(4)); dot("$D$",D,1.5*NW,linewidth(4)); dot("$E$",E,1.5*S,linewidth(4)); draw(A--B--C--D--cycle); draw(D--E,dashed); label("$60^\circ$",A,2.5*dir(30),fontsize(10)); label("$18-c$",midpoint(A--E),1.5*S,red); label("$c$",midpoint(E--B),2.25*S,red); label("$b$",midpoint(B--C),scale(1.5)*rotate(90)*dir(midpoint(B--C)--B),red); label("$b$",midpoint(D--E),scale(1.5)*rotate(90)*dir(midpoint(E--D)--E),red); label("$c$",midpoint(C--D),1.5*N,red); label("$d$",midpoint(D--A),scale(1.5)*rotate(90)*dir(midpoint(D--A)--D),red); [/asy] We apply the Law of Cosines to $\triangle ADE:$ \begin{align*} AD^2 + AE^2 - 2\cdot AD\cdot AE\cdot\cos 60^\circ &= DE^2 \\ d^2 + (18-c)^2 - d(18-c) &= b^2 \\ (18-c)^2 - d(18-c) &= b^2 - d^2 \\ (18-c)(18-c-d) &= (b+d)(b-d). \hspace{15mm}(\bigstar) \end{align*} Let $k$ be the common difference of the arithmetic progression of the side-lengths. It follows that $b,c,$ and $d$ are $18-k, 18-2k,$ and $18-3k,$ in some order. It is clear that $0\leq k<6.$
If $k=0,$ then $ABCD$ is a rhombus with side-length $18,$ which is valid.
If $k\neq0,$ then we have six cases:
Together, the sum of all possible values of $a$ is $18+(13+3+8)+(14+12+16)=\boxed{84}.$
| 84
|
3,794
|
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12A_Problems/Problem_24
| 2
|
Convex quadrilateral $ABCD$ has $AB = 18, \angle{A} = 60^\circ,$ and $\overline{AB} \parallel \overline{CD}.$ In some order, the lengths of the four sides form an arithmetic progression, and side $\overline{AB}$ is a side of maximum length. The length of another side is $a.$ What is the sum of all possible values of $a$
$\textbf{(A) } 24 \qquad \textbf{(B) } 42 \qquad \textbf{(C) } 60 \qquad \textbf{(D) } 66 \qquad \textbf{(E) } 84$
|
Let $b, c$ , and $d$ denote the sides $BC, CD$ , and $AD$ respectively. [asy] size(250); pair A, B, C, D, E; A = (0,0); B = (18,0); D = A+9*dir(60); C = D+(7,0); E = D+(B-C); pen pdot=linewidth(3)+fontsize(12); dot("$A$",A,SW,pdot); dot("$B$",B,SE,pdot); dot("$C$",C,NE,pdot); dot("$D$",D,NW,pdot); draw(A--B--C--D--cycle); label("$60^\circ$",A,5*dir(30),fontsize(10)); label("$\theta$", B, 5*dir(155),fontsize(10)); pen plabel=red+fontsize(12); label("$18$",midpoint(A--B),1.5*S,plabel); label("$b$", midpoint(B--C), scale(1.5)*rotate(90)* dir((B+C)/2--B), plabel); label("$c$", (C+D)/2,1.5*N, plabel); label("$d$",(D+A)/2, scale(1.5)*rotate(90)*dir((D+A)/2--D), plabel); [/asy] Since $AB\parallel CD$ , we get \[\tfrac{\sqrt 3}{2}\ d = b\sin\theta \quad \textrm{and}\quad \tfrac 12 d + c + b\cos\theta = 18.\] Using $b^2\sin^2\theta + b^2\cos^2\theta = b^2$ , we eliminate $\theta$ from above to get $(36-2c-d)^2+3d^2=4b^2$ , which rearranges to $(36-2c-d)^2-d^2=4(b^2-d^2)$ , and, upon factoring, yields \begin{align} (18-c)(18-c-d)=(b+d)(b-d). \end{align} We divide into two cases, depending on whether $c$ is the smallest side.
If $c$ is not the smallest side then $18-c=\pm (b-d)$ . If $c=18$ , we get a rhombus of side $18$ , so one possible value is $a=18$ . Otherwise, we can cancel the common factor from $(1)$ . After rearranging we get \[18-c=-b \quad \textrm{or}\quad 18-c=b+2d.\] The first condition is false because $-b< 0 <18-c$ ; the second condition is false because $b+2d > |b-d| = 18-c$
If $c$ is the smallest side, then $18-c = \pm 3(b-d)$ . Assuming $c<18$ we can cancel common factors in $(1)$ to get \[8b=13d \quad \textrm{or}\quad 8b=7d.\] The first condition yields the solution $(c,d,b)=(3,8,13)$ and the second condition yields the solution $(c,b,d)=(12,14,16)$
Together, the sum of all possible values of $a$ is $18+(3+8+13)+(12+14+16)=\boxed{84}.$
| 84
|
3,795
|
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12A_Problems/Problem_24
| 3
|
Convex quadrilateral $ABCD$ has $AB = 18, \angle{A} = 60^\circ,$ and $\overline{AB} \parallel \overline{CD}.$ In some order, the lengths of the four sides form an arithmetic progression, and side $\overline{AB}$ is a side of maximum length. The length of another side is $a.$ What is the sum of all possible values of $a$
$\textbf{(A) } 24 \qquad \textbf{(B) } 42 \qquad \textbf{(C) } 60 \qquad \textbf{(D) } 66 \qquad \textbf{(E) } 84$
|
Denote $x = AD$ $\theta = \angle B$ .
Hence, $BC = \frac{\sqrt{3}}{2} \cdot \frac{x}{\sin \theta}$ $DC = 18 - \frac{x}{2} - \frac{\sqrt{3}}{2} x \cot \theta$
$\textbf{Case 1}$ $DC = AD = BC = AB$
This is a rhombus. So each side has length $18$
For the following cases, we consider four sides that have distinct lengths.
To make their lengths an arithmetic sequence, we must have $\theta \neq 120^\circ$
Therefore, in the subsequent analysis, we exclude the solution $\theta = 120^\circ$
$\textbf{Case 2}$ $DC < AD < BC < AB$
Because the lengths of these sides form an arithmetic sequence, we have the following system of equations: \[ AB - BC = BC - AD = AD - DC . \]
Hence, \begin{eqnarray*} & 18 - \frac{\sqrt{3}}{2}\cdot\frac{x}{\sin \theta} = \frac{\sqrt{3}}{2}\cdot\frac{x}{\sin \theta} - x = x - \left( 18 - \frac{x}{2} - \frac{\sqrt{3}}{2} x \cot \theta \right) . & \end{eqnarray*}
By solving this system of equations, we get $\left( \cos \theta , \sin \theta , x\right) = \left( \frac{11}{13} , \frac{4 \sqrt{3}}{13} , 8 \right)$
Thus, in this case, $DC = 3$ $AD = 8$ $BC = 13$
$\textbf{Case 3}$ $DC < BC < AD < AB$
Because the lengths of these sides form an arithmetic sequence, we have the following system of equations: \[ AB - AD = AD - BC = BC - DC . \]
Hence, \begin{eqnarray*} & 18 - x = x - \frac{\sqrt{3}}{2}\cdot\frac{x}{\sin \theta} = \frac{\sqrt{3}}{2}\cdot\frac{x}{\sin \theta} - \left( 18 - \frac{x}{2} - \frac{\sqrt{3}}{2} x \cot \theta \right) . & \end{eqnarray*}
By solving this system of equations, we get $\left( \cos \theta , \sin \theta , x\right) = \left( - \frac{1}{7} , \frac{4 \sqrt{3}}{7} , 16 \right)$
Thus, in this case, $DC = 12$ $AD = 16$ $BC = 14$
$\textbf{Case 4}$ $BC < CD < AD < AB$
By doing the similar analysis, we can show there is no solution in this case.
$\textbf{Case 5}$ $BC < AD < CD < AB$
By doing the similar analysis, we can show there is no solution in this case.
$\textbf{Case 6}$ $AD < CD < BC < AB$
By doing the similar analysis, we can show there is no solution in this case.
$\textbf{Case 7}$ $AD < BC < CD < AB$
By doing the similar analysis, we can show there is no solution in this case.
Therefore, the sum of all possible values of $a$ is \begin{align*} 18 + \left( 3 + 8 + 13 \right) + \left( 12 + 14 + 16 \right) & = 84 . \end{align*}
Therefore, the answer is $\boxed{84}$
| 84
|
3,796
|
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12A_Problems/Problem_25
| 1
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Let $m\ge 5$ be an odd integer, and let $D(m)$ denote the number of quadruples $(a_1, a_2, a_3, a_4)$ of distinct integers with $1\le a_i \le m$ for all $i$ such that $m$ divides $a_1+a_2+a_3+a_4$ . There is a polynomial \[q(x) = c_3x^3+c_2x^2+c_1x+c_0\] such that $D(m) = q(m)$ for all odd integers $m\ge 5$ . What is $c_1?$
$\textbf{(A)}\ {-}6\qquad\textbf{(B)}\ {-}1\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 11$
|
For a fixed value of $m,$ there is a total of $m(m-1)(m-2)(m-3)$ possible ordered quadruples $(a_1, a_2, a_3, a_4).$
Let $S=a_1+a_2+a_3+a_4.$ We claim that exactly $\frac1m$ of these $m(m-1)(m-2)(m-3)$ ordered quadruples satisfy that $m$ divides $S:$
Since $\gcd(m,4)=1,$ we conclude that \[\{k+4(0),k+4(1),k+4(2),\ldots,k+4(m-1)\}\] is the complete residue system modulo $m$ for all integers $k.$
Given any ordered quadruple $(a'_1, a'_2, a'_3, a'_4)$ in modulo $m,$ it follows that exactly one of these $m$ ordered quadruples has sum $0$ modulo $m:$ \[\begin{array}{c|c} & \\ [-2.5ex] \textbf{Ordered Quadruple} & \textbf{Sum Modulo }\boldsymbol{m} \\ [0.5ex] \hline & \\ [-2ex] (a'_1, a'_2, a'_3, a'_4) & S'+4(0) \\ [0.5ex] (a'_1+1, a'_2+1, a'_3+1, a'_4+1) & S'+4(1) \\ [0.5ex] (a'_1+2, a'_2+2, a'_3+2, a'_4+2) & S'+4(2) \\ [0.5ex] \cdots & \cdots \\ [0.5ex] (a'_1+m-1, a'_2+m-1, a'_3+m-1, a'_4+m-1) & S'+4(m-1) \\ [0.5ex] \end{array}\] We conclude that $q(m)=\frac1m\cdot[m(m-1)(m-2)(m-3)]=(m-1)(m-2)(m-3),$ so \[q(x)=(x-1)(x-2)(x-3)=c_3x^3+c_2x^2+c_1x+c_0.\] By Vieta's Formulas, we get $c_1=1\cdot2+1\cdot3+2\cdot3=\boxed{11}.$
| 11
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3,797
|
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12A_Problems/Problem_25
| 2
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Let $m\ge 5$ be an odd integer, and let $D(m)$ denote the number of quadruples $(a_1, a_2, a_3, a_4)$ of distinct integers with $1\le a_i \le m$ for all $i$ such that $m$ divides $a_1+a_2+a_3+a_4$ . There is a polynomial \[q(x) = c_3x^3+c_2x^2+c_1x+c_0\] such that $D(m) = q(m)$ for all odd integers $m\ge 5$ . What is $c_1?$
$\textbf{(A)}\ {-}6\qquad\textbf{(B)}\ {-}1\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 11$
|
Define \[ b_i = \left\{ \begin{array}{ll} a_i & \mbox{ if } 1 \leq a_i \leq \frac{m-1}{2} \\ a_i - m & \mbox{ if } \frac{m-1}{2} + 1 \leq a_i \leq m - 1 \\ 0 & \mbox{ if } a_i = m \end{array} \right.. \]
Hence, $b_i$ is a one-to-one and onto function of $a_i$ , and the range of $b_i$ is $\left\{- \frac{m-1}{2} , \cdots , \frac{m-1}{2} \right\}$
Therefore, to solve this problem, it is equivalent for us to count the number of tuples $\left( b_1 , b_2 , b_3 , b_4 \right)$ that are all distinct and satisfy $m | b_1 + b_2 + b_3 + b_4$
Denote by $d \left( m \right)$ the number of such tuples that are also subject to the constraint $b_1 < b_2 < b_3 < b_4$
Hence, $D \left( m \right) = 4! d \left( m \right) = 24 d \left( m \right)$
We do the following casework analysis to compute $d \left( m \right)$
$\textbf{Case 1}$ : There is one $0$ in $\left( b_1 , b_2 , b_3 , b_4 \right)$
Denote by $d_{1i} \left( m \right)$ the number of tuples with $b_i = 0$
By symmetry, $d_{11} \left( m \right)= d_{14} \left( m \right)$ and $d_{12} \left( m \right)= d_{13} \left( m \right)$
$\textbf{Case 2}$ : There is no $0$ in $\left( b_1 , b_2 , b_3 , b_4 \right)$
Denote by $d_{2i} \left( m \right)$ the number of tuples with $i$ positive entries.
By symmetry, $d_{20} \left( m \right) = d_{24} \left( m \right)$ and $d_{21} \left( m \right) = d_{23} \left( m \right)$
Therefore, \begin{align*} D \left( m \right) & = 24 d \left( m \right) \\ & = 24 \left( \sum_{i=1}^4 d_{1i} \left( m \right) + \sum_{i=0}^4 d_{2i} \left( m \right) \right) \\ & = 24 \left( 2 d_{11} \left( m \right) + 2 d_{12} \left( m \right) + 2 d_{24} \left( m \right) + 2 d_{23} \left( m \right) + d_{22} \left( m \right) \right) . \end{align*}
Now, we compute $D \left( m \right)$ for $m = 5 , 7 , 9 , 11$
$\underline{\textbf{SCENARIO}}$ $m = 5$
We have $b_i \in \left\{ - 2 , \cdots , 2 \right\}$
$\textbf{Case 1}$
$\textbf{Case 1.1}$ $b_1 = 0$
We cannot have $3$ distinct positive integers. So $d_{11} \left( 5 \right) = 0$
$\textbf{Case 1.2}$ $b_2 = 0$
Because there are $2$ positive integers, we must have $b_3 = 1$ $b_4 = 2$ .
Hence, $b_1 = - 3$ . However, this is out of the range of $b_i$ .
Thus, $d_{12} \left( 5 \right) = 0$
$\textbf{Case 2}$
$\textbf{Case 2.1}$ $b_1 > 0$
We cannot have $4$ distinct positive integers. So $d_{24} \left( 5 \right) = 0$
$\textbf{Case 2.2}$ $b_1 < 0 < b_2$
We cannot have $3$ distinct positive integers. So $d_{23} \left( 5 \right) = 0$
$\textbf{Case 2.3}$ $b_2 < 0 < b_3$
The only solution is $\left( b_1 , b_2 , b_3 , b_4 \right) = \left( - 2 , - 1 , 1 , 2 \right)$ . So $d_{22} \left( 5 \right) = 1$
Therefore, $D \left( 5 \right) = 24$
$\underline{\textbf{SCENARIO}}$ $m = 7$
We have $b_i \in \left\{ - 3 , \cdots , 3 \right\}$
$\textbf{Case 1}$
$\textbf{Case 1.1}$ $b_1 = 0$
We have no feasible solution. Thus, $d_{11} \left( 7 \right) = 0$
$\textbf{Case 1.2}$ $b_2 = 0$
The only solution is $\left( b_1 , b_3 , b_4 \right) = \left( - 3 , 1 , 2 \right)$ .
Thus, $d_{12} \left( 7 \right) = 1$
$\textbf{Case 2}$
$\textbf{Case 2.1}$ $b_1 > 0$
We cannot have $4$ distinct positive integers. So $d_{24} \left( 7 \right) = 0$
$\textbf{Case 2.2}$ $b_1 < 0 < b_2$
To get $3$ distinct positive integers, we have $\left( b_2 , b_3 , b_4 \right) = \left( 1 , 2 , 3 \right)$ . This implies $b_1 = - 6$ . However, this is out of the range of $b_1$ . So $d_{23} \left( 6 \right) = 0$
$\textbf{Case 2.3}$ $b_2 < 0 < b_3$
We have $d_{22} \left( 7 \right) = \binom{3}{2} = 3$
Therefore, $D \left( 7 \right) = 24 \cdot 5$
$\underline{\textbf{SCENARIO}}$ $m = 9$
We have $b_i \in \left\{ - 4 , \cdots , 4 \right\}$
$\textbf{Case 1}$
$\textbf{Case 1.1}$ $b_1 = 0$
The only solution is $\left( b_2 , b_3 , b_4 \right) = \left( 2, 3, 4 \right)$ .
Thus, $d_{11} \left( 9 \right) = 1$
$\textbf{Case 1.2}$ $b_2 = 0$
The feasible solutions are $\left( b_1 , b_3 , b_4 \right) = \left( - 3 , 1 , 2 \right)$ $\left( - 4 , 1 , 3 \right)$ .
Thus, $d_{12} \left( 9 \right) = 2$
$\textbf{Case 2}$
$\textbf{Case 2.1}$ $b_1 > 0$
There is no feasible solution. So $d_{24} \left( 9 \right) = 0$
$\textbf{Case 2.2}$ $b_1 < 0 < b_2$
To get $3$ distinct positive integers, we have $b_2 + b_3 + b_4 \geq 1 + 2 + 3 = 6$ . This implies $b_1 = - 6$ . However, this is out of the range of $b_1$ . So $d_{23} \left( 9 \right) = 0$
$\textbf{Case 2.3}$ $b_2 < 0 < b_3$
We have $d_{22} \left( 9 \right) = 8$
Therefore, $D \left( 9 \right) = 24 \cdot 14$
$\underline{\textbf{SCENARIO}}$ $m = 11$
We have $b_i \in \left\{ - 5 , \cdots , 5 \right\}$
$\textbf{Case 1}$
$\textbf{Case 1.1}$ $b_1 = 0$
The only solution is $\left( b_2 , b_3 , b_4 \right) = \left( 2, 4, 5 \right)$ .
Thus, $d_{11} \left( 11 \right) = 1$
$\textbf{Case 1.2}$ $b_2 = 0$
The feasible solutions are $\left( b_1 , b_3 , b_4 \right) = \left( - 3 , 1 , 2 \right)$ $\left( - 4 , 1 , 3 \right)$ $\left( - 5 , 1 , 4 \right)$ $\left( - 5 , 2 , 3 \right)$ .
Thus, $d_{12} \left( 11 \right) = 4$
$\textbf{Case 2}$
$\textbf{Case 2.1}$ $b_1 > 0$
The only feasible solution is $\left( b_1 , b_2 , b_3 , b_4 \right) = \left( 1 , 2, 3, 5 \right)$ . So $d_{24} \left( 11 \right) = 1$
$\textbf{Case 2.2}$ $b_1 < 0 < b_2$
The only feasible solution is $\left( b_1 , b_2 , b_3 , b_4 \right) = \left( -1 , 3, 4, 5 \right)$ . So $d_{23} \left( 11 \right) = 1$
$\textbf{Case 2.3}$ $b_2 < 0 < b_3$
We have $d_{22} \left( 11 \right) = 16$
Therefore, $D \left( 11 \right) = 24 \cdot 30$
We know that $q \left( m \right) = D \left( m \right)$ for odd $m \geq 5$
Plugging $m = 5, 7, 9, 11$ into this equation, we get \begin{align*} c_3 5^3 + c_2 5^2 + c_1 5 + c_0 & = 24 \cdot 1 && (1.1) \\ c_3 7^3 + c_2 7^2 + c_1 7 + c_0 & = 24 \cdot 5 && (1.2) \\ c_3 9^3 + c_2 9^2 + c_1 9 + c_0 & = 24 \cdot 14 && (1.3) \\ c_3 11^3 + c_2 11^2 + c_1 11 + c_0 & = 24 \cdot 30 && (1.4) \end{align*}
Now, we solve this system of equations.
Taking $\frac{(1.2)-(1.1)}{2}$ $\frac{(1.3)-(1.2)}{2}$ $\frac{(1.4)-(1.2)}{2}$ , we get \begin{align*} c_3 109 + c_2 12 + c_1 & = 48 && (2.1) \\ c_3 193 + c_2 16 + c_1 & = 108 && (2.2) \\ c_3 301 + c_2 20 + c_1 & = 192 && (2.3) \end{align*}
Taking $\frac{(2.2)-(2.1)}{4}$ $\frac{(2.3)-(2.2)}{4}$ , we get \begin{align*} c_3 21 + c_2 & = 15 && (3.1) \\ c_3 27 + c_2 & = 21 && (3.2) \end{align*}
Taking $\frac{(3.2)-(3.1)}{6}$ , we get $c_3 = 1$
Plugging $c_3$ into Equation (3.1), we get $c_2 = - 6$
Plugging $c_2$ and $c_3$ into Equation (2.1), we get $c_1 = 11$
Therefore, the answer is $\boxed{11}$
| 11
|
3,798
|
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12A_Problems/Problem_25
| 3
|
Let $m\ge 5$ be an odd integer, and let $D(m)$ denote the number of quadruples $(a_1, a_2, a_3, a_4)$ of distinct integers with $1\le a_i \le m$ for all $i$ such that $m$ divides $a_1+a_2+a_3+a_4$ . There is a polynomial \[q(x) = c_3x^3+c_2x^2+c_1x+c_0\] such that $D(m) = q(m)$ for all odd integers $m\ge 5$ . What is $c_1?$
$\textbf{(A)}\ {-}6\qquad\textbf{(B)}\ {-}1\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 11$
|
As before, note that we have $m(m-1)(m-2)$ numbers we can choose as $a,b,c.$ From here, there is exactly one possible value of $1 \leq d \leq m$ that could make $S=a+b+c+d$ divisible by $m.$ However, there is a $\frac{3}{m}$ chance that this value of $d$ has already been chosen as $a,b$ or $c$ . Thus our polynomial is $m(m-1)(m-2)\left(1-\frac{3}{m}\right)=m(m-1)(m-2)\left(\frac{m-3}{m}\right)=(m-1)(m-2)(m-3)$ . By Vieta's, $c_1 = 2+3+6=\boxed{11}$
| 11
|
3,799
|
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12B_Problems/Problem_1
| 1
|
What is the value of $1234 + 2341 + 3412 + 4123$
$\textbf{(A)}\: 10000\qquad\textbf{(B)} \: 10010\qquad\textbf{(C)} \: 10110\qquad\textbf{(D)} \: 11000\qquad\textbf{(E)} \: 11110$
|
We see that $1, 2, 3,$ and $4$ each appear in the ones, tens, hundreds, and thousands digit exactly once. Since $1+2+3+4=10$ , we find that the sum is equal to \[10\cdot(1+10+100+1000)=\boxed{11110}.\] Note that it is equally valid to manually add all four numbers together to get the answer.
| 110
|
3,800
|
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12B_Problems/Problem_1
| 2
|
What is the value of $1234 + 2341 + 3412 + 4123$
$\textbf{(A)}\: 10000\qquad\textbf{(B)} \: 10010\qquad\textbf{(C)} \: 10110\qquad\textbf{(D)} \: 11000\qquad\textbf{(E)} \: 11110$
|
We have \[1234 + 2341 + 3412 + 4123 = 1111 \left( 1 + 2 + 3 + 4 \right) = \boxed{11110}.\] ~Steven Chen (www.professorchenedu.com)
| 110
|
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