2011 AIME I 第 3 题

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3.

LL 是斜率为 512\frac{5}{12} 且经过点 A=(24,1)A = (24, -1) 的直线,设 MM 是垂直于直线 LL 且经过点 B=(5,6)B = (5, 6) 的直线。原来的坐标轴被擦去,并把直线 LL 作为 xx-轴,直线 MM 作为 yy-轴。在新的坐标系中,点 AA 在正 xx-轴上,点 BB 在正 yy-轴上。原坐标系中坐标为 (14,27)(-14, 27) 的点 PP,在新坐标系中的坐标为 (α,β)(\alpha, \beta)。求 α+β\alpha + \beta

Let LL be the line with slope 512\frac{5}{12} that contains the point A=(24,1),A = (24, -1), and let MM be the line perpendicular to line LL that contains the point B=(5,6).B = (5, 6). The original coordinate axes are erased, and line LL is made the xx-axis and line MM the yy-axis. In the new coordinate system, point AA is on the positive xx-axis, and point BB is on the positive yy-axis. The point PP with coordinates (14,27)(-14, 27) in the original system has coordinates (α,β)(\alpha, \beta) in the new coordinate system. Find α+β.\alpha + \beta.

答案:31
知识点:坐标几何距离公式变换
难度评级:2390
解答:

直线 LL5x12y132=05x - 12y - 132 = 0,直线 MM12x+5y90=012x + 5y - 90 = 0。一个点的新 xx-坐标是它到直线 MM 的有向距离,以含有 AA 的一侧为正;新 yy-坐标是它到直线 LL 的有向距离,以含有 BB 的一侧为正。

P=(14,27)P = (-14, 27) 代入 12x+5y9012x + 5y - 90,得到 168+13590=123-168 + 135 - 90 = -123,而代入 AA 得到 193>0193 \gt 0;除以 122+52=13\sqrt{12^2 + 5^2} = 13,得 α=12313\alpha = -\frac{123}{13}。将 PP 代入 5x12y1325x - 12y - 132,得到 70324132=526-70 - 324 - 132 = -526,代入 BB 得到 179-179,所以 PPBBLL 的同一侧, β=52613\beta = \frac{526}{13}

因此 α+β=123+52613=40313=31\alpha + \beta = \frac{-123 + 526}{13} = \frac{403}{13} = 31

Line LL is 5x12y132=05x - 12y - 132 = 0 and line MM is 12x+5y90=0.12x + 5y - 90 = 0. The new xx-coordinate of a point is its signed distance to line M,M, counted positive on the side containing A,A, and the new yy-coordinate is its signed distance to line L,L, positive on the side containing B.B.

Substituting P=(14,27)P = (-14, 27) into 12x+5y9012x + 5y - 90 gives 168+13590=123,-168 + 135 - 90 = -123, while AA gives 193>0;193 \gt 0; dividing by 122+52=13,\sqrt{12^2 + 5^2} = 13, we get α=12313.\alpha = -\frac{123}{13}. Substituting PP into 5x12y1325x - 12y - 132 gives 70324132=526,-70 - 324 - 132 = -526, and BB gives 179,-179, so PP lies on the same side of LL as BB and β=52613.\beta = \frac{526}{13}.

Therefore α+β=123+52613=40313=31.\alpha + \beta = \frac{-123 + 526}{13} = \frac{403}{13} = 31.

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