Question

起初 Alex、Betty 和 Charlie 一共有 $444$ 颗花生。Charlie 的花生最多，Alex 的花生最少。三人各自拥有的花生数成等比数列。Alex 吃掉了 $5$ 颗，Betty 吃掉了 $9$ 颗，Charlie 吃掉了 $25$ 颗。现在三人各自剩下的花生数成等差数列。求 Alex 起初有多少颗花生。

Initially Alex, Betty, and Charlie had a total of $444$ peanuts. Charlie had the most peanuts, and Alex had the least. The three numbers of peanuts that each person had form a geometric progression. Alex eats $5$ of his peanuts, Betty eats $9$ of her peanuts, and Charlie eats $25$ of his peanuts. Now the three numbers of peanuts that each person has form an arithmetic progression. Find the number of peanuts Alex had initially.

答案

答案：108

知识点：等比数列等差数列方程组

难度评级：2050

解答：

吃掉花生后，还剩 $444 - 5 - 9 - 25 = 405$ 颗，并且三个数量成等差数列，所以中间项，也就是 Betty 的数量，是 $\frac{405}{3} = 135$ 。因此 Betty 起初有 $135 + 9 = 144$ 颗花生。

起始数量成等比数列，所以可写成 $\frac{144}{r}$ 、 $144$ 和 $144r$ ，其中 $r \gt 1$ （因为 Charlie 最多而 Alex 最少）。于是 $\frac{144}{r} + 144 + 144r = 444,$ 化简得 $12r^2 - 25r + 12 = 0$ ，根为 $r = \frac{4}{3}$ 与 $r = \frac{3}{4}$ 。因为 $r \gt 1$ ，取 $r = \frac{4}{3}$ 。

所以 Alex 起初有 $144 \cdot \frac{3}{4} = 108$ 颗花生。（检验：吃完后数量为 $103$ 、 $135$ 、 $167$ ，公差为 $32$ 。）

After the eating, $444 - 5 - 9 - 25 = 405$ peanuts remain, and the three amounts form an arithmetic progression, so the middle amount, Betty's, is $\frac{405}{3} = 135.$ Hence Betty started with $135 + 9 = 144$ peanuts.

The starting amounts form a geometric progression, so they are $\frac{144}{r},$ $144,$ and $144r$ with $r \gt 1$ (Charlie had the most and Alex the least). Then $\frac{144}{r} + 144 + 144r = 444,$ which simplifies to $12r^2 - 25r + 12 = 0,$ with roots $r = \frac{4}{3}$ and $r = \frac{3}{4};$ since $r \gt 1,$ we take $r = \frac{4}{3}.$

So Alex initially had $144 \cdot \frac{3}{4} = 108$ peanuts. (Check: after eating, the amounts $103,$ $135,$ $167$ increase by $32$ each.)

2016 AIME II 第 1 题

答案：108

难度评级：2050

解答：

其他年份的第 1 题