2012 AMC 12B 第 25 题

Question

设

S = \{(x, y) : x \in \{0, 1, 2, 3, 4\}

，

y \in \{0, 1, 2, 3, 4, 5\}

，且

(x, y) \ne (0, 0)\}

。设

T

为所有顶点在

S

中的直角三角形的集合。对每个直角三角形

t = \triangle ABC

，其顶点

A

，

B

，

C

按逆时针顺序排列，且直角在

A

，令

f(t) = \tan(\angle CBA)

。求

\prod_{t \in T} f(t)\text{？}

Let

S = \{(x, y) : x \in \{0, 1, 2, 3, 4\},

y \in \{0, 1, 2, 3, 4, 5\},

and

(x, y) \ne (0, 0)\}.

Let

T

be the set of all right triangles whose vertices are in

S.

For every right triangle

t = \triangle ABC

with vertices

A,

B,

and

C

in counter-clockwise order and right angle at

A,

let

f(t) = \tan(\angle CBA).

What is

\prod_{t \in T} f(t)?

1

\dfrac{625}{144}

\dfrac{125}{24}

6

\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(\angle CBA)

和

\tan(\angle ACB)

，二者乘积为

1

Pairing a triangle with its reflection swaps

\tan(\angle CBA)

with

\tan(\angle ACB),

whose product is

1

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)=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

t_1

配对，使得

f(t)f(t_1)

=\tan(\angle CBA)\tan(\angle ACB)

=1

。

连续反射（关于

x=2

，再关于

x=y

，再关于

y=\tfrac52

）把乘积化简为仅六个形如

OYZ

且

Y

在最上排的三角形对应乘积的倒数。

这六个三角形给出

\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

\dfrac{625}{144}

。

因此正确答案是 B。

Isosceles right triangles contribute

f(t)=1.

For a scalene right triangle, reflecting across a suitable line pairs it with a triangle

t_1

so that

f(t)f(t_1)

=\tan(\angle CBA)\tan(\angle ACB)

=1.

Successive reflections (across

x=2,

then

x=y,

then

y=\tfrac52

) reduce the product to the reciprocal of the product over just six triangles of the form

OYZ

with

Y

on the top row.

Those six give

\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,

\dfrac{625}{144}.

Thus, the correct answer is B.

2012 AMC 12B 第 25 题

答案：B

难度评级：2650

小提示：

大提示：

解答：

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