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考题22-23 2015 AMC 12B

2018-08-06 重点归纳

AMC12是针对高中学生的数学测验，该竞赛开始于2000年，分A赛和B赛，于每年的2月初和2月中举行，学生可任选参加一项即可。其主要目的在于激发学生对数学的兴趣，参予AMC12的学生应该不难发现测验的问题都很具挑战性，但测验的题型都不会超过学生的学习范围。这项测验希望每个考生能从竞赛中享受数学。那么接下来跟随小编来看一下AMC12的官方真题以及官方解答吧：

Problem 22

Six chairs are evenly spaced around a circular table. One person is seated in each chair. Each person gets up and sits down in a chair that is not the same chair and is not adjacent to the chair he or she originally occupied, so that again one person is seated in each chair. In how many ways can this be done?

$考题22-23 2015 AMC 12B$

Solution 1

Consider shifting every person over three seats (left or right) after each person has gotten up and sat back down again. Now, instead of each person being seated not in the same chair and not in an adjacent chair, each person will be seated either in the same chair or in an adjacent chair. The problem now becomes the number of ways in which six people can sit down in a chair that is either the same chair or an adjacent chair in a circle.

Consider the similar problem of $n$ people sitting in a chair that is either the same chair or an adjacent chair in a row. Call the number of possibilities for this $F_n$ . Then if the leftmost person stays put, the problem is reduced to a row of $n-1$ chairs, and if the leftmost person shifts one seat to the right, the new person sitting in the leftmost seat must be the person originally second from the left, reducing the problem to a row of $n-2$ chairs. Thus, $F_n = F_{n-1} + F_{n-2}$ for $n \geq 3$ . Clearly $F_1 = 1$ and $F_2 = 2$ , so $F_3 = 3$ , $F_4 = 5$ , and $F_5 = 8$ .

Now consider the six people in a circle and focus on one person. If that person stays put, the problem is reduced to a row of five chairs, for which there are $F_5 = 8$ possibilities. If that person moves one seat to the left, then the person who replaces him in his original seat will either be the person originally to the right of him, which will force everyone to simply shift over one seat to the left, or the person originally to the left of him, which reduces the problem to a row of four chairs, for which there are $F_4 = 5$ possibilities, giving $1 + 5 = 6$ possibilities in all. By symmetry, if that person moves one seat to the right, there are another $6$ possibilities, so we have a total of $8+6+6 = \boxed{\textbf{(D)}\; 20}$ possibilities.

Solution 2

Label the people sitting at the table $A, B, C, D, E, F,$ and assume that they are initially seated in the order $ABCDEF$ . The possible new positions for $A, B, C, D, E,$ and $F$ are respectively (a dash indicates a non-allowed position):

$考题22-23 2015 AMC 12B$

The permutations we are looking for should use one letter from each column, and there should not be any repeated letters:

$考题22-23 2015 AMC 12B$

There are $\boxed{\textbf{(D)}\; 20}$ such permutations.

Solution 3

We can represent each rearrangement as a permutation of the six elements $\{1,2,3,4,5,6\}$ in cycle notation. Note that any such permutation cannot have a 1-cycle, so the only possible types of permutations are 2,2,2-cycles, 4,2-cycles, 3,3-cycles, and 6-cycles. We deal with each case separately.

For 2,2,2-cycles, suppose that one of the 2-cycles switches the people across from each other, i.e. $(14)$ , $(25)$ , or $(36)$ . WLOG, we may assume it to be $(14)$ . Then we could either have both of the other 2-cycles be across from each other, giving the permutation $(14)(25)(36)$ , or else neither of the other 2-cycles are across from each other, in which case the only possible permutation is $(14)(26)(35)$ . This can happen for $(25)$ and $(36)$ as well. So since the first permutation is not counted twice, we find a total of $1+3=4$ permutations that are 2,2,2-cycles where at least one of the 2-cycles switches people diametrically opposite from each other. Otherwise, since the elements in a 2-cycle cannot differ by 1, 3, or 5 mod 6, they must differ by 2 or 4 mod 6, i.e. they must be of the same parity. But since we have three odd and three even elements, this is impossible. Hence there are exactly 4 such permutations that are 2,2,2-cycles.

For 4,2-cycles, we assume for the moment that 1 is part of the 2-cycle. Then the 2-cycle can be $(13)$ , $(15)$ , or $(14)$ . The first two are essentially the same by symmetry, and we must arrange the elements 2, 4, 5, 6 into a 4-cycle. However, 5 must have two neighbors that are not next to it, which is impossible, hence the first two cases yield no permutations. If the 2-cycle is $(14)$ , then we must arrange the elements 2, 3, 5, 6 into a 4-cycle. Then 2 must have the neighbors 5 and 6. We find that the 4-cycles $(2536)$ and $(2635)$ satisfy the desired properties, yielding the permutations $(14)(2536)$ and $(14)(2635)$ . This can be done for the 2-cycles $(25)$ and $(36)$ as well, so we find a total of 6 such permutations that are 4,2-cycles.

For 3,3-cycles, note that if 1 neighbors 4, then the third element in the cycle will neighbor one of 1 and 4, so this is impossible. Therefore, the 3-cycle containing 1 must consist of the elements 1, 3, and 5. Therefore, we obtain the four 3,3-cycles $(135)(246)$ , $(153)(246)$ , $(135)(264)$ , and $(153)(264)$ .

For 6-cycles, note that the neighbors of 1 can be 3 and 4, 3 and 5, or 4 and 5. In the first case, we may assume that it looks like $(314\dots)$ -- the form $(413\dots)$ is also possible, but equivalent to this case. Then we must place the elements 2, 5, and 6. Note that 5 and 6 cannot go together, so 2 must go in between them. Also, 5 cannot neighbor 4, so we are left with one possibility, namely $(314625)$ , which has an analogous possibility $(413526)$ . In the second case, we assume that it looks like $(315\dots)$ . Clearly, the 2 must go next to the 5, and the 6 must go last (to neighbor the 3), so the only possibility here is $(315246)$ , with the analogous possibility $(513642)$ . In the final case, we may assume that it looks like $(415\dots)$ . Then the 2 and 3 cannot go together, so the 6 must go in between them. Therefore, the only possibility is $(415362)$ , with the analogous possibility $(514263)$ . We have covered all possibilities for 6-cycles, and we have found 6 of them.

Therefore, there are $考题22-23 2015 AMC 12B$ such permutations.

Solution 4 (Based off of AoPS MathJam)

Note that each person can't end up in his/her original seat, or either of the two adjacent seats. This means everyone must essentially "move across the table" to the other three seats. Notice we can recast the problem here into two steps:

1) First, everyone move to the seat diametrically across their original seat

2) Second, everyone either stays put, or moves to the chair to their immediate left or right (such that no two people attempt to sit together).

Note that the first step really does nothing, so we only have to consider the second step. Now we can simply consider cases (Do this as an exercise; these cases are much simpler than the other solutions):

The answer is $4+6+9+1=20$ so $D$ is the answer.

- Idea came from Jeremy Copeland

Problem 23

A rectangular box measures $a \times b \times c$ , where $a$ , $b$ , and $c$ are integers and $1\leq a \leq b \leq c$ . The volume and the surface area of the box are numerically equal. How many ordered triples $(a,b,c)$ are possible?

$\textbf{(A)}\; 4 \qquad\textbf{(B)}\; 10 \qquad\textbf{(C)}\; 12 \qquad\textbf{(D)}\; 21 \qquad\textbf{(E)}\; 26$

Solution

The surface area is $2(ab+bc+ca)$ , and the volume is $abc$ , so equating the two yields

$2(ab+bc+ca)=abc.$

Divide both sides by $2abc$ to obtain

$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{2}.$

First consider the bound of the variable $a$ . Since $\frac{1}{a}<\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{2},$ we have $a>2$ , or $a\geqslant3$ .

Also note that $c \geq b \geq a > 0$ , hence $\frac{1}{a} \geq \frac{1}{b} \geq \frac{1}{c}$ . Thus, $考题22-23 2015 AMC 12B$ , so $a \leq 6$ .

So we have $a=3, 4, 5$ or $6$ .

Before the casework, let's consider the possible range for $b$ if $\frac{1}{b}+\frac{1}{c}=k>0$ . From $\frac{1}{b}<k$ , we have $b>\frac{1}{k}$ . From $\frac{2}{b} \geq \frac{1}{b}+\frac{1}{c}=k$ , we have $b \leq \frac{2}{k}$ . Thus $\frac{1}{k}<b \leq \frac{2}{k}$ .

When $a=3$ , we get $\frac{1}{b}+\frac{1}{c}=\frac{1}{6}$ , so $b=7, 8, 9, 10, 11, 12$ . We find the solutions $(a, b, c)=(3, 7, 42)$ , $(3, 8, 24)$ , $(3, 9, 18)$ , $(3, 10, 15)$ , $(3, 12, 12)$ , for a total of $5$ solutions.

When $a=4$ , we get $考题22-23 2015 AMC 12B$ , so $b=5, 6, 7, 8$ . We find the solutions $(a, b, c)=(4, 5, 20)$ , $(4, 6, 12)$ , $(4, 8, 8)$ , for a total of $3$ solutions.