Problem:
Let x,y, and z be positive real numbers satisfying the system of equations:
2x−xy+2y−xy=12y−yz+2z−yz=22z−zx+2x−zx=3
Then [(1−x)(1−y)(1−z)]2 can be written as nm, where m and n are relatively prime positive integers. Find m+n.
Solution:
Create a triangle with sides 2x,2z and an altitude from that vertex to the last side of length xz.
Let the angle that "points" at the 2x side be θ.
By we have that
sinθ=2yxy=2x
Notice this can be rewritten as x=2sin2θ making the 2x side be written as 2⋅2sin2θ=2sinθ.
This is particularly useful using the main part of law of sines, particularly that
sinαA=2R
In this case, we'd have that
sinθ2sinθ=2R→R=1
This is particularly useful because this same idea can be extended to the other equations, and shows that the three triangles we'd create can all be circumscribed by a circle with radius 1.
Let's assign more angles. We assigned the inscribed angle to a chord with length 2x in a circle with radius 1 as θ.
Now, let's assign the inscribed angle to a chord with length 2z as α and do similarly for 2y, assigning that angle as β.
In similar fashion to finding that x=2sin2θ, we can similarly assign z=2sin2α, and y=2sin2β, and keep this in mind for later.
The triangle with sides 2x,2z and altitude xz, has its third side of length 3, which means the inscribed angle to it would be 2120=60∘, or consequently yield that \theta + \alpha = 180 - 60 = 120^
With the first equation, notice that the inscribed angle that would have a chord length of 1 is just an equilateral triangle, and so we'd have that
θ+β+260=180→θ+β=150
Lastly, for the second equation, the inscribed angle with a chord length of 2 in this circle of radius 1 would be 90∘, so we'd have