Question

一支大蜡烛高 $119$ 厘米。它的设计是刚点燃时烧得较快，越接近底部烧得越慢。具体地，从顶端烧掉第一厘米需要 $10$ 秒，烧掉第二厘米需要 $20$ 秒，烧掉第 $k$ 厘米需要 $10k$ 秒。设蜡烛完全烧完需要 $T$ 秒。那么点燃后 $\frac{T}{2}$ 秒时，蜡烛高度为 $h$ 厘米。求 $10h$ 。

A large candle is $119$ centimeters tall. It is designed to burn down more quickly when it is first lit and more slowly as it approaches its bottom. Specifically, the candle takes $10$ seconds to burn down the first centimeter from the top, $20$ seconds to burn down the second centimeter, and $10k$ seconds to burn down the $k$ -th centimeter. Suppose it takes $T$ seconds for the candle to burn down completely. Then $\frac{T}{2}$ seconds after it is lit, the candle’s height in centimeters will be $h.$ Find $10h.$

答案

答案：350

知识点：等差数列求和因式分解

难度评级：1970

小提示：

烧掉前 $x$ 厘米需要 $10(1 + 2 + \cdots + x) = 5x(x+1)$ 秒。

Burning the first $x$ centimeters takes $10(1 + 2 + \cdots + x) = 5x(x+1)$ seconds

大提示：

令 $5x(x+1)$ 等于 $\frac{T}{2} = 5 \cdot 60 \cdot 119$ ，并把 $7140$ 写成两个连续整数的乘积。

Set $5x(x+1)$ equal to $\frac{T}{2} = 5 \cdot 60 \cdot 119$ and write $7140$ as a product of two consecutive integers

解答：

烧掉前 $x$ 厘米需要 $10(1 + 2 + \cdots + x) = 5x(x+1)$ 秒，所以 $T = 5 \cdot 119 \cdot 120 = 71400$ ，且 $\frac{T}{2} = 35700$ 。

令 $5x(x+1) = 35700$ ，得到 $x(x+1) = 7140 = 84 \cdot 85$ ，所以在 $\frac{T}{2}$ 时刻蜡烛恰好烧掉 $84$ 厘米。它的高度为 $h = 119 - 84 = 35$ ，因此 $10h = 350$ 。

Burning the first $x$ centimeters takes $10(1 + 2 + \cdots + x) = 5x(x+1)$ seconds, so $T = 5 \cdot 119 \cdot 120 = 71400$ and $\frac{T}{2} = 35700.$

Setting $5x(x+1) = 35700$ gives $x(x+1) = 7140 = 84 \cdot 85,$ so at time $\frac{T}{2}$ the candle has burned down exactly $84$ centimeters. Its height is $h = 119 - 84 = 35,$ and $10h = 350.$

2013 AIME II 第 3 题

答案：350

难度评级：1970

小提示：

大提示：

解答：

其他年份的第 3 题