/sci/ - Science & Math

>>12620492
let [math]r[/math] be the radius of the inscribed circle, [math]r = \sqrt{\frac{1}{s}(s-a)(s-b)(s-c)}[/math], where [math]r = \frac{1}{2}+\frac{i}{2}[/math]
and [math]s[/math] the semi-perimeter [math]s = \frac{a+b+c}{2}[/math], where [math]s = \frac{1}{2}+\frac{i}{2}[/math]

From the law of cotangents we have
[math]\cot \frac{\alpha}{2} = \frac{s - a}{r}[/math]

So we get
[math]\cot \frac{\alpha}{2} = \frac{\frac{1}{2}+\frac{i}{2} - i}{\frac{1}{2}+\frac{i}{2}}[/math]
[math]\cot \frac{\alpha}{2} = -i[/math]

Extending the law further:

[math]\cot \frac{\beta}{2} = \frac{\frac{1}{2}+\frac{i}{2} - 1}{\frac{1}{2}+\frac{i}{2}}[/math]
[math]\cot \frac{\beta}{2} = i[/math]

[math]\cot \frac{\gamma}{2} = \frac{\frac{1}{2}+\frac{i}{2} - 0}{\frac{1}{2}+\frac{i}{2}}[/math]
[math]\cot \frac{\gamma}{2} = 1[/math]

We can then conclude with the last identity which is
[math]\frac{\cot \frac{\alpha}{2}}{s - a} = \frac{\cot \frac{\beta}{2}}{s - b} = \frac{\cot \frac{\gamma}{2}}{s - c} = \frac{1}{r}[/math]
which becomes
[math]\frac{-i}{s - a} = \frac{i}{s - b} = \frac{1}{s - c} = \frac{1}{\frac{1}{2}+\frac{i}{2}}[/math]
which is proven to be true here

https://www.wolframalpha.com/input/?i=%28-i%2F%28s-a%29%29%3D%28i%2F%28s-b%29%29%3D%281%2F%28s-c%29%29%3D%281%2F%281%2F2%2Bi%2F2%29%29+where+a%3Di%2C+b%3D1%2C+c%3D0%2C+s%3D%281%2F2%2Bi%2F2%29