Analytic geometry

[Albert Chan]

Trivia: If triangle inscribed unit circle, Δarea = Δhalf-perimeter

Prove:
Again, assume t1=0, and normalized t2, t3, such that 2Pi > t3 > t2 > 0
To have unit circle inside triangle require these conditions:

0 < t2 < Pi             ; made triangle angle Pi - t2
Pi < t3 < Pi + t2     ; made triangle angle t3 - Pi

-> tan(t2/2) > 0, tan(t3/2) < 0, tan((t3-t2)/2) > 0

a = | tan(t3/2) - tan(t2/2) |         = tan(t2/2) - tan(t3/2)
b = | tan(-t2/2) - tan(½(t3-t2)) | = tan(t2/2) + tan(½(t3-t2))
c = | tan(-t3/2) - tan(½(t2-t3)) | = -tan(t3/2) + tan(½(t3-t2))

s = ½(a + b + c)
= tan(t2/2) - tan(t3/2) + tan(½(t3-t2))
= tan(½(t3-t2)) * (1 - (1 + tan(t2/2) tan(t3/2)))
= - tan(t2/2) tan(t3/2) tan(½(t3-t2))

Add back absolute function to remove sign, and remove t1=0 restriction:

s = |tan(½(t2-t1))| + |tan(½(t3-t1))| + |tan(½(t3-t2))|
   = |tan(½(t2-t1)) tan(½(t3-t1)) tan(½(t3-t2))|

Match previously derived Δarea formula. QED

Comment: |tan(...)| peices are length of circle tangents to triangle vertice.