2012 AMC 12B Problems/Problem 25

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 tanx × tan(90-x) = tan x × cot x = 1

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