1987 AMC 12 第 17 题

Question

在一次数学竞赛中，比尔与迪克的分数之和等于安与卡萝尔的分数之和。如果交换比尔与卡萝尔的分数，安与卡萝尔的分数之和将超过另外两人的分数之和。此外，迪克的分数超过比尔与卡萝尔的分数之和。假设所有分数均为非负数，求四名选手从高到低的名次顺序。

In a mathematics competition, the sum of the scores of Bill and Dick equalled the sum of the scores of Ann and Carol. If the scores of Bill and Carol had been interchanged, then the sum of the scores of Ann and Carol would have exceeded the sum of the scores of the other two. Also, Dick’s score exceeded the sum of the scores of Bill and Carol. Determine the order in which the four contestants finished, from highest to lowest. Assume all scores were nonnegative.

迪克、安、卡萝尔、比尔

Dick, Ann, Carol, Bill

迪克、安、比尔、卡萝尔

Dick, Ann, Bill, Carol

迪克、卡萝尔、比尔、安

Dick, Carol, Bill, Ann

安、迪克、卡萝尔、比尔

Ann, Dick, Carol, Bill

安、迪克、比尔、卡萝尔

Ann, Dick, Bill, Carol

答案：E

知识点：inequalities ordering algebraic comparison

难度评级：1830

小提示：

按姓名顺序用

A,B,C,D

表示四人的分数

Represent the four scores by

A,B,C,D

in name order

大提示：

将等式

A+C=B+D

与不等式

A+B\gt C+D

相加和相减

Add and subtract the equality

A+C=B+D

and the inequality

A+B\gt C+D

解答：

设

A,B,C,D

分别表示安、比尔、卡萝尔、迪克的分数。条件为

\begin{aligned} A+C&=B+D,\\ A+B&\gt C+D,\\ D&\gt B+C \end{aligned}\text{。}

将前两个比较关系相加，得到

A\gt D

。用不等式减去等式，得到

B\gt C

。最后，

D\gt B+C\ge B

。因此

A\gt D\gt B\gt C\text{。}

所以正确答案是 E。

Let

A,B,C,D

be the scores of Ann, Bill, Carol, and Dick. The conditions are

\begin{aligned} A+C&=B+D,\\ A+B&\gt C+D,\\ D&\gt B+C. \end{aligned}

Adding the first two comparisons gives

A\gt D.

Subtracting the equality from the inequality gives

B\gt C.

Finally,

D\gt B+C\ge B.

Hence

A\gt D\gt B\gt C.

Thus the correct answer is E.

1987 AMC 12 第 17 题

答案：E

难度评级：1830

小提示：

大提示：

解答：

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