# 2012 AMC 12B Problems/Problem 25

## Problem 25

Let . Let be the set of all right triangles whose vertices are in . For every right triangle with vertices , , and in counter-clockwise order and right angle at , let . What is

## Solution 1

Four points in a rectangular arrangement allow 4 possible right angle triangles. These four congruent triangles will form two pairs of different vertice labelling - two ABC and two ACB. These pairs will multiple to equal 1 due to the fact that

Due to the missing point (0,0) in the grid then in the rectangular arrangement (0,0), (x,0), (x,y), (0,y) there is only one triangle which will not cancel out to zero with a reflected version.

So we need to consider all triangles of the form A(x,y), B(0,y), C(x,0). For these triangle tanB = y/x.
Multiplying them all together gives: 1/1 * 1/2 * 1/3 * 1/4 * 2/1 * 2/2 * 2/3 * 2/4 * 3/1 * 3/2 * 3/3 * 3/4 * 4/1 * 4/2 * 4/3 * 4/4 * 5/1 * 5/2 * 5/3 * 5/4 = 625/24

==Solution 2==y Consider reflection. For any right triangle with the right labeling described in the problem, any reflection labeled that way will give us . First we consider the reflection about the line . Then only those triangle that has one vertex at does not . Within those triangles, consider a reflection about the line . Then only those triangle that has one vertex on the line . There are three cases:

Case 1: . Then is impossible.

Case 2: . Then we look for such that and that . They are: , and . The product of their tangents of is .

Case 3: . Then is impossible.

Therefore is the answer.