Problem:
Let S={(x,y):xβ{0,1,2,3,4},yβ{0,1,2,3,4,5}, and (x,y)ξ =(0,0)}. Let T be the set of all right triangles whose vertices are in S. For every right triangle t=β³ABC with vertices A,B, and C in counter-clockwise order and right angle at A, let f(t)=tan(β CBA). What is
tβTββf(t)?
Answer Choices:
A. 1
B. 144625β
C. 24125β
D. 6
E. 24625β 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=β³ABC with vertices in Sβͺ{O}. Let R1β be the reflection with respect to the line with equation x=2. Let A1β=R1β(A),B1β=R1β(B),C1β=R1β(C), and t1β=β³A1βB1βC1β. Note that β³ABCβ β³A1βB1βC1β with right angles at A and A1β, but the counterclockwise order of the vertices of t1β is A1β,C1β, and B1β. Thus f(t1β)=tan(β A1βC1βB1β)=tan(β ACB) and
The reflection R1β is a bijection of Sβͺ{O} and it induces a partition of the triangles in pairs (t,t1β) such that f(t)f(t1β)=1. Thus the product over all triangles in Sβͺ{O} is equal to 1 , and thus the required product is equal to the reciprocal of βtβT1ββf(t), where T1β is the set of triangles with vertices in Sβͺ{O} having O as one vertex.
Let S1β={(x,y):xβ{0,1,2,3,4}, and yβ{0,1,2,3,4}} and let R2β be the reflection with respect to the line with equation x=y. For every right triangle t=β³OBC with vertices B and C in S1β, let B2β=R2β(B),C2β=R2β(C), and t2β=β³OB2βC2β. Similarly as before, R2β is a bijection of S1β and it induces a partition of the triangles in pairs (t,t2β) such that f(t)f(t2β)=1. Thus βtβT1ββf(t)=βtβT2ββf(t), where T2β is the set of triangles with vertices in Sβͺ{O} with O as one vertex, and another vertex with y coordinate equal to 5 .
Next, consider the reflection R3β with respect to the line with equation y=25β. Let X=(0,5). For every right triangle t=β³OXC with C in S, let C3β=R3β(C), and t3β=β³OXC3β. As before R3β induces a partition of these triangles in pairs (t,t3β) such that f(t)f(t3β)=1. Therefore to calculate βtβT2ββf(t), the only triangles left to consider are the triangles of the form t=β³OYZ where Yβ{(x,5):xβ{1,2,3,4}} and ZβS\{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 x5β with 1β€xβ€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 1β€xβ€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
