2020 AIME II 第 4 题

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4.

三角形 △ABC\triangle ABC 与 △A′B′C′\triangle A'B'C' 位于坐标平面内,顶点分别为 A(0,0)A(0, 0)、B(0,12)B(0, 12)、C(16,0)C(16, 0)、A′(24,18)A'(24, 18)、B′(36,18)B'(36, 18)、C′(24,2)C'(24, 2)。绕点 (x,y)(x, y) 顺时针旋转 mm 度,其中 0<m<1800 \lt m \lt 180,会把 △ABC\triangle ABC 变换为 △A′B′C′\triangle A'B'C'。求 m+x+ym + x + y。

Triangles △ABC\triangle ABC and △A′B′C′\triangle A'B'C' lie in the coordinate plane with vertices A(0,0),A(0, 0), B(0,12),B(0, 12), C(16,0),C(16, 0), A′(24,18),A'(24, 18), B′(36,18),B'(36, 18), C′(24,2).C'(24, 2). A rotation of mm degrees clockwise around the point (x,y),(x, y), where 0<m<180,0 \lt m \lt 180, will transform △ABC\triangle ABC to △A′B′C′.\triangle A'B'C'. Find m+x+y.m + x + y.

答案:108
知识点:变换坐标几何
难度评级:2300
小提示:

线段 ABAB 是竖直的,而它的像 A′B′A'B' 是水平的,所以旋转角必须是 9090 度

Segment ABAB is vertical while its image A′B′A'B' is horizontal, so the rotation must be by 9090 degrees

大提示:

绕 (a,b)(a, b) 顺时针旋转 90∘90^\circ 会把 (p,q)(p, q) 送到 (a+q−b, b−p+a)(a + q - b,\ b - p + a);匹配 A↦A′A \mapsto A'

A 90∘90^\circ clockwise rotation about (a,b)(a, b) sends (p,q)(p, q) to (a+q−b, b−p+a);(a + q - b,\ b - p + a); match A↦A′A \mapsto A'

解答:

向量 AB→=(0,12)\overrightarrow{AB} = (0, 12) 是竖直的,而 A′B′→=(12,0)\overrightarrow{A'B'} = (12, 0) 是水平的且长度相同,所以旋转把方向顺时针转了 90∘90^\circ,因此 m=90m = 90。

绕 (a,b)(a, b) 顺时针旋转 90∘90^\circ 会把 (p,q)(p, q) 送到 (a+q−b, b−p+a)(a + q - b,\ b - p + a)。将其用于 A=(0,0)A = (0, 0),并令像等于 A′=(24,18)A' = (24, 18),得到 a−b=24a - b = 24 且 a+b=18a + b = 18,所以 a=21a = 21,b=−3b = -3。检查另外两个顶点:B=(0,12)B = (0, 12) 映到 (21+12+3, −3+21)(21 + 12 + 3,\ -3 + 21) =(36,18)=B′= (36, 18) = B',且 C=(16,0)C = (16, 0) 映到 (21+3, −3−16+21)(21 + 3,\ -3 - 16 + 21) =(24,2)=C′= (24, 2) = C'。

因此 m+x+y=90m + x + y = 90 +21+ 21 +(−3)=108+ (-3) = 108。

The vector AB→=(0,12)\overrightarrow{AB} = (0, 12) is vertical, while A′B′→=(12,0)\overrightarrow{A'B'} = (12, 0) is horizontal and of the same length, so the rotation turns directions by 90∘90^\circ clockwise, and m=90.m = 90.

A 90∘90^\circ clockwise rotation about (a,b)(a, b) sends (p,q)(p, q) to (a+q−b, b−p+a).(a + q - b,\ b - p + a). Applying this to A=(0,0)A = (0, 0) and setting the image equal to A′=(24,18)A' = (24, 18) gives a−b=24a - b = 24 and a+b=18,a + b = 18, so a=21a = 21 and b=−3.b = -3. Checking the other vertices: B=(0,12)B = (0, 12) maps to (21+12+3, −3+21)(21 + 12 + 3,\ -3 + 21) =(36,18)=B′,= (36, 18) = B', and C=(16,0)C = (16, 0) maps to (21+3, −3−16+21)(21 + 3,\ -3 - 16 + 21) =(24,2)=C′.= (24, 2) = C'.

Therefore m+x+y=90m + x + y = 90 +21+ 21 +(−3)=108.+ (-3) = 108.

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