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2012 AMC 12B 第 25 题

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所有题目均经美国数学协会（MAA）官方合法授权使用。

25.

设 S={(x,y):x∈{0,1,2,3,4}S = \{(x, y) : x \in \{0, 1, 2, 3, 4\}，y∈{0,1,2,3,4,5}y \in \{0, 1, 2, 3, 4, 5\}，且 (x,y)≠(0,0)}(x, y) \ne (0, 0)\}。设 TT 为所有顶点在 SS 中的直角三角形的集合。对每个直角三角形 t=△ABCt = \triangle ABC，其顶点 AA，BB，CC 按逆时针顺序排列，且直角在 AA，令 f(t)=tan⁡(∠CBA)f(t) = \tan(\angle CBA)。求 ∏t∈Tf(t)？\prod_{t \in T} f(t)\text{？}

Let S={(x,y):x∈{0,1,2,3,4},S = \{(x, y) : x \in \{0, 1, 2, 3, 4\}, y∈{0,1,2,3,4,5},y \in \{0, 1, 2, 3, 4, 5\}, and (x,y)≠(0,0)}.(x, y) \ne (0, 0)\}. Let TT be the set of all right triangles whose vertices are in S.S. For every right triangle t=△ABCt = \triangle ABC with vertices A,A, B,B, and CC in counter-clockwise order and right angle at A,A, let f(t)=tan⁡(∠CBA).f(t) = \tan(\angle CBA). What is ∏t∈Tf(t)?\prod_{t \in T} f(t)?

11

625144\dfrac{625}{144}

12524\dfrac{125}{24}

66

62524\dfrac{625}{24}

答案：B
知识点：三角学裂项相消对称性
难度评级：2650
小提示：

把一个三角形与它的反射配对，会交换 tan⁡(∠CBA)\tan(\angle CBA) 和 tan⁡(∠ACB)\tan(\angle ACB)，二者乘积为 11

Pairing a triangle with its reflection swaps tan⁡(∠CBA)\tan(\angle CBA) with tan⁡(∠ACB),\tan(\angle ACB), whose product is 11

大提示：

反射会把整个乘积化简到少数几个经过原点的三角形，最后只剩六个需要计算

Reflections collapse the whole product down to a handful of triangles through the origin, and only six survive

解答：

等腰直角三角形贡献 f(t)=1f(t)=1。 对于三边互不相等的直角三角形，关于合适的直线反射可把它与三角形 t1t_1 配对，使得 f(t)f(t1)f(t)f(t_1) =tan⁡(∠CBA)tan⁡(∠ACB)=\tan(\angle CBA)\tan(\angle ACB) =1=1。

连续反射（关于 x=2x=2， 再关于 x=yx=y， 再关于 y=52y=\tfrac52）把乘积化简为仅六个形如 OYZOYZ 且 YY 在最上排的三角形对应乘积的倒数。

这六个三角形给出 15⋅25⋅35⋅45⋅322⋅4222=144625， \begin{aligned} &\frac15\cdot\frac25\cdot\frac35\cdot\frac45\cdot\frac{3\sqrt2}{\sqrt2} \\ &\quad {}\cdot\frac{4\sqrt2}{2\sqrt2}=\frac{144}{625} \end{aligned}\text{，} 因此所求乘积是其倒数 625144\dfrac{625}{144}。

因此正确答案是 B。

Isosceles right triangles contribute f(t)=1.f(t)=1. For a scalene right triangle, reflecting across a suitable line pairs it with a triangle t1t_1 so that f(t)f(t1)f(t)f(t_1) =tan⁡(∠CBA)tan⁡(∠ACB)=\tan(\angle CBA)\tan(\angle ACB) =1.=1.

Successive reflections (across x=2,x=2, then x=y,x=y, then y=52y=\tfrac52) reduce the product to the reciprocal of the product over just six triangles of the form OYZOYZ with YY on the top row.

Those six give 15⋅25⋅35⋅45⋅322⋅4222=144625, \begin{aligned} &\frac15\cdot\frac25\cdot\frac35\cdot\frac45\cdot\frac{3\sqrt2}{\sqrt2} \\ &\quad {}\cdot\frac{4\sqrt2}{2\sqrt2}=\frac{144}{625}, \end{aligned} so the required product is its reciprocal, 625144.\dfrac{625}{144}.

Thus, the correct answer is B.

第 24 题#24
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