Problem:
Let S={(x,y):x∈{0,1,2,3,4},y∈{0,1,2,3,4,5}, and (x,y)=(0,0)}. Let T be the set of all right triangles whose vertices are in S. For every right triangle t=△ABC with vertices A,B, and C in counter-clockwise order and right angle at A, let f(t)=tan(∠CBA). What is
t∈T∏f(t)?
Answer Choices:
A. 1
B. 144625
C. 24125
D. 6
E. 24625 Solution:
First note that the isosceles right triangles t can be excluded from the product because f(t)=1 for these triangles. All triangles mentioned from now on are scalene right triangles. Let O=(0,0). First consider all triangles t=△ABC with vertices in S∪{O}. Let R1 be the reflection with respect to the line with equation x=2. Let A1=R1(A),B1=R1(B),C1=R1(C), and t1=△A1B1C1. Note that △ABC≅△A1B1C1 with right angles at A and A1, but the counterclockwise order of the vertices of t1 is A1,C1, and B1. Thus f(t1)=tan(∠A1C1B1)=tan(∠ACB) and
f(t)f(t1)=tan(∠CBA)tan(∠ACB)=ABAC⋅ACAB=1.
The reflection R1 is a bijection of S∪{O} and it induces a partition of the triangles in pairs (t,t1) such that f(t)f(t1)=1. Thus the product over all triangles in S∪{O} is equal to 1 , and thus the required product is equal to the reciprocal of ∏t∈T1f(t), where T1 is the set of triangles with vertices in S∪{O} having O as one vertex.
Let S1={(x,y):x∈{0,1,2,3,4}, and y∈{0,1,2,3,4}} and let R2 be the reflection with respect to the line with equation x=y. For every right triangle t=△OBC with vertices B and C in S1, let B2=R2(B),C2=R2(C), and t2=△OB2C2. Similarly as before, R2 is a bijection of S1 and it induces a partition of the triangles in pairs (t,t2) such that f(t)f(t2)=1. Thus ∏t∈T1f(t)=∏t∈T2f(t), where T2 is the set of triangles with vertices in S∪{O} with O as one vertex, and another vertex with y coordinate equal to 5 .
Next, consider the reflection R3 with respect to the line with equation y=25. Let X=(0,5). For every right triangle t=△OXC with C in S, let C3=R3(C), and t3=△OXC3. As before R3 induces a partition of these triangles in pairs (t,t3) such that f(t)f(t3)=1. Therefore to calculate ∏t∈T2f(t), the only triangles left to consider are the triangles of the form t=△OYZ where Y∈{(x,5):x∈{1,2,3,4}} and Z∈S\{X}.
The following argument shows that there are six such triangles. Because the y coordinate of Y is greater than zero, the right angle of t is not at O. The slope of the line OY has the form x5 with 1≤x≤4, so if the right angle were at Y, then the vertex Z would need to be at least 5 horizontal units away from Y, which is impossible. Therefore the right angle is at Z. There are 4 such triangles with Z on the x-axis, with vertices O,Z=(x,0), and Y=(x,5) for 1≤x≤4. There are two more triangles: with vertices O,Z=(3,3), and Y=(1,5), and with vertices O,Z=(4,4), and Y=(3,5). The product of the values f(t) over these six triangles is equal to
