2013 AIME II 第 3 题

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3.

一支大蜡烛高 119119 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 1010 秒,烧掉第二厘米需要 2020 秒,烧掉第 kk 厘米需要 10k10k 秒。设蜡烛完全烧完需要 TT 秒。那么点燃后 T2\frac{T}{2} 秒时,蜡烛高度为 hh 厘米。求 10h10h

A large candle is 119119 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 1010 seconds to burn down the first centimeter from the top, 2020 seconds to burn down the second centimeter, and 10k10k seconds to burn down the kk-th centimeter. Suppose it takes TT seconds for the candle to burn down completely. Then T2\frac{T}{2} seconds after it is lit, the candle's height in centimeters will be h.h. Find 10h.10h.

答案:350
知识点:等差数列求和因式分解
难度评级:1970
解答:

烧掉前 xx 厘米需要 10(1+2++x)=5x(x+1)10(1 + 2 + \cdots + x) = 5x(x+1) 秒,所以 T=5119120=71400T = 5 \cdot 119 \cdot 120 = 71400,且 T2=35700\frac{T}{2} = 35700

5x(x+1)=357005x(x+1) = 35700,得到 x(x+1)=7140=8485x(x+1) = 7140 = 84 \cdot 85,所以在 T2\frac{T}{2} 时刻蜡烛恰好烧掉 8484 厘米。它的高度为 h=11984=35h = 119 - 84 = 35,因此 10h=35010h = 350

Burning the first xx centimeters takes 10(1+2++x)=5x(x+1)10(1 + 2 + \cdots + x) = 5x(x+1) seconds, so T=5119120=71400T = 5 \cdot 119 \cdot 120 = 71400 and T2=35700.\frac{T}{2} = 35700.

Setting 5x(x+1)=357005x(x+1) = 35700 gives x(x+1)=7140=8485,x(x+1) = 7140 = 84 \cdot 85, so at time T2\frac{T}{2} the candle has burned down exactly 8484 centimeters. Its height is h=11984=35,h = 119 - 84 = 35, and 10h=350.10h = 350.

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