2012 AMC 12B Problem 25

Question

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}

Pairing a triangle with its reflection swaps

\tan(\angle CBA)

with

\tan(\angle ACB),

whose product is

1

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

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.

Problem 25 in other years