2012 AIME II 第 4 题

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4.

安娜、鲍勃和曹分别以每秒 8.68.6 米、每秒 6.26.2 米和每秒 55 米的恒定速度骑自行车。他们同时从一个长方形场地的东北角出发,该场地较长的一边正好沿东西方向,向西延伸。安娜沿场地边缘骑行,起初向西;鲍勃沿场地边缘骑行,起初向南;曹沿直线穿过场地骑到南边上的一点 DD。曹到达点 DD 的时间,正好等于安娜和鲍勃第一次到达 DD 的时间。场地的长、宽以及点 DD 到场地东南角的距离之比可表示为 p:q:rp : q : r,其中 ppqqrr 是正整数,且 ppqq 互质。求 p+q+rp + q + r

Ana, Bob, and Cao bike at constant rates of 8.68.6 meters per second, 6.26.2 meters per second, and 55 meters per second, respectively. They all begin biking at the same time from the northeast corner of a rectangular field whose longer side runs due west. Ana starts biking along the edge of the field, initially heading west, Bob starts biking along the edge of the field, initially heading south, and Cao bikes in a straight line across the field to a point DD on the south edge of the field. Cao arrives at point DD at the same time that Ana and Bob arrive at DD for the first time. The ratio of the field’s length to the field’s width to the distance from point DD to the southeast corner of the field can be represented as p:q:r,p : q : r, where p,p, q,q, and rr are positive integers with pp and qq relatively prime. Find p+q+r.p + q + r.

答案:61
知识点:路程、速度与时间方程组因式分解
难度评级:2390
小提示:

设场地长为 LL、宽为 WW,并设 xxDD 到东南角的距离;令三名骑车者的行进时间相等。

Let the field be LL by WW and let xx be the distance from DD to the southeast corner; set the three riders’ travel times equal

大提示:

令鲍勃和曹的时间相等可得 168W2625Wx+168x2=0168W^2 - 625Wx + 168x^2 = 0,它可以分解成两个一次因式。

Equating Bob’s and Cao’s times gives 168W2625Wx+168x2=0,168W^2 - 625Wx + 168x^2 = 0, which factors into two linear factors

解答:

设场地长为 LL(向西),宽为 WW(向南),且 L>WL \gt W,并设 xxDD 到东南角的距离。安娜绕边缘骑行距离 2L+Wx2L + W - x,鲍勃骑行距离 W+xW + x,曹骑行距离 W2+x2\sqrt{W^2 + x^2},且三人用时相同:2L+Wx8.6=W+x6.2=W2+x25 \begin{aligned} \frac{2L + W - x}{8.6} &= \frac{W + x}{6.2} \\ &= \frac{\sqrt{W^2 + x^2}}{5} \end{aligned}\text{。}

第一个等式给出 L=6W+37x31L = \frac{6W + 37x}{31}。将第二个等式平方,得到 25(W+x)2=38.44(W2+x2)25(W + x)^2 = 38.44\,(W^2 + x^2),化简为 168W2625Wx+168x2=0168W^2 - 625Wx + 168x^2 = 0,并可分解为 (24W7x)(7W24x)=0(24W - 7x)(7W - 24x) = 0

x=7W24x = \frac{7W}{24} 给出 L=13W24<WL = \frac{13W}{24} \lt W,不可能,所以 x=24W7x = \frac{24W}{7},进而 L=30W7L = \frac{30W}{7}。比例为 L:W:x=30:7:24L : W : x = 30 : 7 : 24,所以 p+q+r=30+7+24=61p + q + r = 30 + 7 + 24 = 61

Let the field have length LL (west) and width WW (south) with L>W,L \gt W, and let xx be the distance from DD to the southeast corner. Ana rides around the perimeter a distance 2L+Wx,2L + W - x, Bob rides W+x,W + x, and Cao rides W2+x2,\sqrt{W^2 + x^2}, all in the same time: 2L+Wx8.6=W+x6.2=W2+x25. \begin{aligned} \frac{2L + W - x}{8.6} &= \frac{W + x}{6.2} \\ &= \frac{\sqrt{W^2 + x^2}}{5}. \end{aligned}

The first equality gives L=6W+37x31.L = \frac{6W + 37x}{31}. Squaring the second, 25(W+x)2=38.44(W2+x2),25(W + x)^2 = 38.44\,(W^2 + x^2), which simplifies to 168W2625Wx+168x2=0,168W^2 - 625Wx + 168x^2 = 0, factoring as (24W7x)(7W24x)=0.(24W - 7x)(7W - 24x) = 0.

The root x=7W24x = \frac{7W}{24} gives L=13W24<W,L = \frac{13W}{24} \lt W, which is impossible, so x=24W7x = \frac{24W}{7} and then L=30W7.L = \frac{30W}{7}. The ratio is L:W:x=30:7:24,L : W : x = 30 : 7 : 24, and p+q+r=30+7+24=61.p + q + r = 30 + 7 + 24 = 61.

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