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

$1$

$\dfrac{625}{144}$

$\dfrac{125}{24}$

$6$

$\dfrac{625}{24}$

Answer: B

Concepts:trigonometry telescoping symmetry

Difficulty rating: 2650

Solution:

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 Problem 25

Answer: B

Difficulty rating: 2650

Solution:

Problem 25 in Other Years