Question

十个人围成一圈。每个人选取一个数，并将它告诉圆圈中与自己相邻的两个人。然后，每个人计算并公布自己两位邻居所选数字的平均数。图中显示的是每个人公布的平均数（不是此人原来选取的数）。公布平均数 $6$ 的人所选的数是

Ten people form a circle. Each picks a number and tells it to the two neighbors adjacent to him in the circle. Then each person computes and announces the average of the numbers of his two neighbors. The figure shows the average announced by each person (not the original number the person picked). The number picked by the person who announced the average $6$ was

$1$

$5$

$6$

$10$

无法根据已知信息唯一确定

not uniquely determined from the given information

答案：A

知识点：recurrence linear system cyclic arrangement

难度评级：2500

小提示：

若 $x_i$ 是选取的数， $a_i$ 是图中显示的平均数，则 $x_{i-1}+x_{i+1}=2a_i$

If $x_i$ is a picked number and $a_i$ the displayed average, then $x_{i-1}+x_{i+1}=2a_i$

大提示：

从相邻的两个未知原数开始，沿圆圈递推

Start with two unknown adjacent picked numbers and propagate around the circle

解答：

按顺时针方向，将图中显示的平均数 $a_0,a_1,\ldots,a_9$ 依次编号为 $1,2,\ldots,10$ ，并令 $x_i$ 为相应位置所选的数。记 $x_0=t,\ x_1=u$ 。由 $x_{i-1}+x_{i+1}=2a_i$ ，依次得到 $\begin{aligned} x_2&=4-t, &x_3&=6-u,\\ x_4&=4+t, &x_5&=4+u,\\ x_6&=8-t, &x_7&=10-u,\\ x_8&=8+t, &x_9&=8+u \end{aligned}\text{。}$ 最后两个闭合方程给出 $t=6$ 和 $u=-3$ 。因此 $x_5=4+u=1$ 。

所以正确答案是 A。

Index the displayed averages $a_0,a_1,\ldots,a_9$ clockwise as $1,2,\ldots,10,$ and let $x_i$ be the corresponding picked numbers. Write $x_0=t,\ x_1=u.$ From $x_{i-1}+x_{i+1}=2a_i,$ successive values are $\begin{aligned} x_2&=4-t, &x_3&=6-u,\\ x_4&=4+t, &x_5&=4+u,\\ x_6&=8-t, &x_7&=10-u,\\ x_8&=8+t, &x_9&=8+u. \end{aligned}$ The two closing equations give $t=6$ and $u=-3.$ Hence $x_5=4+u=1.$

Thus the correct answer is A.

1990 AMC 12 第 26 题

答案：A

难度评级：2500

小提示：

大提示：

解答：

其他年份的第 26 题