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考题22-23 2015 AMC 10A

2018-08-07 重点归纳

AMC10数学竞赛是美国高中数学竞赛中的一项，是针对高中一年级及初中三年级学生的数学测试，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。不管是对高校申请还是今后在数学领域的发展都极其有利！那么接下来跟随小编来看一下AMC10数学竞赛真题以及官方解答吧：

Problem 22

Eight people are sitting around a circular table, each holding a fair coin. All eight people flip their coins and those who flip heads stand while those who flip tails remain seated. What is the probability that no two adjacent people will stand?

$\textbf{(A)}\dfrac{47}{256}\qquad\textbf{(B)}\dfrac{3}{16}\qquad\textbf{(C) }\dfrac{49}{256}\qquad\textbf{(D) }\dfrac{25}{128}\qquad\textbf{(E) }\dfrac{51}{256}$

Solution

Solution 1

We will count how many valid standing arrangements there are (counting rotations as distinct), and divide by $2^8 = 256$ at the end. We casework on how many people are standing.

Case $1:$ $0$ people are standing. This yields $1$ arrangement.

Case $2:$ $1$ person is standing. This yields $8$ arrangements.

Case $3:$ $2$ people are standing. This yields $AMC10数学竞赛真题$ arrangements, because the two people cannot be next to each other.

Case $4:$ $4$ people are standing. Then the people must be arranged in stand-sit-stand-sit-stand-sit-stand-sit fashion, yielding $2$ possible arrangements.

More difficult is:

Case $5:$ $3$ people are standing. First, choose the location of the first person standing ( $8$ choices). Next, choose $2$ of the remaining people in the remaining $5$ legal seats to stand, amounting to $6$ arrangements considering that these two people cannot stand next to each other. However, we have to divide by $3,$ because there are $3$ ways to choose the first person given any three. This yields $\dfrac{8 \cdot 6}{3} = 16$ arrangements for Case $5.$

Alternate Case $5:$ Use complementary counting. Total number of ways to choose 3 people from 8 which is $\dbinom{8}{3}$ . Sub-case $1:$ three people are next to each other which is $\dbinom{8}{1}$ . Sub-case $2:$ two people are next to each other and the third person is not $\dbinom{8}{1}$ $\dbinom{4}{1}$ . This yields $AMC10数学竞赛真题$

Summing gives $1 + 8 + 20 + 2 + 16 = 47,$ and so our probability is $\boxed{\textbf{(A) } \dfrac{47}{256}}$ .

Solution 2

We will count how many valid standing arrangements there are counting rotations as distinct and divide by $256$ at the end. Line up all $8$ people linearly. In order for no two people standing to be adjacent, we will place a sitting person to the right of each standing person. In effect, each standing person requires $2$ spaces and the standing people are separated by sitting people. We just need to determine the number of combinations of pairs and singles and the problem becomes very similar to pirates and gold aka stars and bars aka sticks and stones aka balls and urns.

If there are $4$ standing, there are ${4 \choose 4}=1$ ways to place them. For $3,$ there are ${3+2 \choose 3}=10$ ways. etc. Summing, we get $AMC10数学竞赛真题$ ways.

Now we consider that the far right person can be standing as well, so we have $AMC10数学竞赛真题$ ways

Together we have $34+13=47$ , and so our probability is $\boxed{\textbf{(A) } \dfrac{47}{256}}$ .

Solution 3

We will count how many valid standing arrangements there are (counting rotations as distinct), and divide by $2^8 = 256$ at the end. If we suppose for the moment that the people are in a line, and decide from left to right whether they sit or stand. If the leftmost person sits, we have the same number of arrangements as if there were only $7$ people. If they stand, we count the arrangements with $6$ instead because the person second from the left must sit. We notice that this is the Fibonacci sequence, where with $1$ person there are two ways and with $2$ people there are three ways. Carrying out the Fibonacci recursion until we get to $8$ people, we find there are $55$ standing arrangements. Some of these were illegal however, since both the first and last people stood. In these cases, both the leftmost and rightmost two people are fixed, leaving us to subtract the number of ways for $4$ people to stand in a line, which is $8$ from our sequence. Therefore our probability is $\frac{55 - 8}{256} = \boxed{\textbf{(A) } \dfrac{47}{256}}$

Solution 4

We will count the number of valid arrangements and then divide by $2^8$ at the end. We proceed with casework on how many people are standing.

Case $1:$ $0$ people are standing. This yields $1$ arrangement.

Case $2:$ $1$ person is standing. This yields $8$ arrangements.

Case $3:$ $2$ people are standing. To do this, we imagine having 6 people with tails in a line first. Notate "tails" with $T$ . Thus, we have $TTTTTT$ . Now, we look to distribute the 2 $H$ 's into the 7 gaps made by the $T$ 's. We can do this in ${7 \choose 2}$ ways. However, note one way does not work, because we have two H's at the end, and the problem states we have a table, not a line. So, we have $AMC10数学竞赛真题$ arrangements.

Case $4:$ $3$ people are standing. Similarly, we imagine 5 $T$ 's. Thus, we have $TTTTT$ . We distribute 3 $H$ 's into the gaps, which can be done ${6 \choose 3}$ ways. However, 4 arrangements will not work. (See this by putting the H's at the ends, and then choosing one of the remaining 4 gaps: ${4 \choose 1}$ =4) Thus, we have ${6 \choose 3}-4=16$ arrangements.

Case $5:$ $4$ people are standing. This can clearly be done in 2 ways: $HTHTHTHT$ or $THTHTHTH$ . This yields $2$ arrangements.

Summing the cases, we get $1+8+20+16+2=47$ arrangements. Thus, the probability is $\boxed{\textbf{(A) } \dfrac{47}{256}}$

Problem 23

The zeroes of the function $f(x)=x^2-ax+2a$ are integers. What is the sum of the possible values of $a?$

$\textbf{(A)}\ 7\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 16\qquad\textbf{(D)}\ 17\qquad\textbf{(E)}\ 18$

Solution 1

By Vieta's Formula, $a$ is the sum of the integral zeros of the function, and so $a$ is integral.

Because the zeros are integral, the discriminant of the function, $a^2 - 8a$ , is a perfect square, say $k^2$ . Then adding 16 to both sides and completing the square yields $(a - 4)^2 = k^2 + 16.$ Therefore $(a-4)^2 - k^2 = 16$ and $((a-4) - k)((a-4) + k) = 16.$ Let $(a-4) - k = u$ and $(a-4) + k = v$ ; then, $a-4 = \dfrac{u+v}{2}$ and so $a = \dfrac{u+v}{2} + 4$ . Listing all possible $(u, v)$ pairs (not counting transpositions because this does not affect ( $u + v$ ), $AMC10数学竞赛真题$ , yields $a = 9, 8, -1, 0$ . These $a$ sum to $16$ , so our answer is $\boxed{\textbf{(C) }16}$ .

Solution 2

Let $r_1$ and $r_2$ be the integer zeroes of the quadratic. Since the coefficient of the $x^2$ term is $1$ , the quadratic can be written as $(x - r_1)(x - r_2)=x^2 - (r_1 + r_2)x + r_1r_2$

By comparing this with $x^2 - ax + 2a$ , $r_1 + r_2 = a\text{ and }r_1r_2 = 2a.$

Plugging the first equation in the second, $r_1r_2 = 2 (r_1 + r_2).$ Rearranging gives $AMC10数学竞赛真题$ These factors can be $(1, 4), (-1, -4), (4, 1), (-4, -1), (2, 2),$ or $(-2, -2).$

We want the number of distinct $a = r_1 + r_2$ , and these factors gives $a = -1, 0, 8, 9$ . So the answer is $-1 + 0 + 8 + 9 = \boxed{\textbf{(C) }16}$ .