2011 AIME I 第 3 题

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3.

设 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
小提示:

PP 的新坐标是它到新坐标轴的有向距离:α\alpha 是到直线 MM 的距离,β\beta 是到直线 LL 的距离

The new coordinates of PP are its signed distances to the new axes: α\alpha is the distance to line MM and β\beta the distance to line LL

大提示:

使用点到直线的距离公式,并通过检查 PP 是否与 AA(对于 α\alpha)或 BB(对于 β\beta)在同一侧来确定符号

Use the point-to-line distance formula, fixing each sign by checking whether PP is on the same side as AA (for α\alpha) or as BB (for β\beta)

解答:

直线 LL 为 5x−12y−132=05x - 12y - 132 = 0,直线 MM 为 12x+5y−90=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。

将 P=(−14,27)P = (-14, 27) 代入 12x+5y−9012x + 5y - 90,得到 −168+135−90=−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 5x−12y−1325x - 12y - 132,得到 −70−324−132=−526-70 - 324 - 132 = -526,代入 BB 得到 −179-179,所以 PP 与 BB 在 LL 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 5x−12y−132=05x - 12y - 132 = 0 and line MM is 12x+5y−90=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+5y−9012x + 5y - 90 gives −168+135−90=−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 5x−12y−1325x - 12y - 132 gives −70−324−132=−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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