Question

A triangle has vertices A, B, and C. Vertex A has an angle of `pi/8`, vertex B has an angle of `( pi)/4`, and the triangle's area is `42`. What is the area of the triangle's incircle?

Answer 1

`text{incircle_area} = {2 pi mathcal{A} tan^2(A/2) tan^2(C/2)}/{ tan A(tan(A/2) + tan(C/2))^2} approx 19.6399`

Explanation:

The incircle is tangent to all the sides. The incenter, its center, is the meet of the angle bisectors.

When I have trouble getting started I pin my triangle to the Cartesian plane. `C(0,0), A(a,0), B(b,c)`.

Let's express our facts;

`A=\pi/8`

`B=pi/4`

`C = \pi-pi/8-pi/4={5pi}/8`

`tan A = c/a`

`tan C = c/b`

Let's call the area `mathcal{A=42}`.

`mathcal{A} = 1/2 a c`

That's three equations in three unknowns. We mostly care about `a`.

`2 mathcal{A} = ac = a (a tan A)=a^2 tan A`

`a = \sqrt{ {2 mathcal{A} }/ tan A}`

Tangents are slopes. The perpendicular bisector of C has equation

`y = x tan (C/2)`

The perpendicular bisector of A has equation

`y = -(x-a) tan (A/2)`

Those meet at the incenter, when

`x tan(C/2) = -x tan(A/2) + a tan (A/2)`

`x = {a tan(A/2)}/{tan(A/2) + tan(C/2)}`

`y = {a tan(A/2) tan(C/2)}/{tan(A/2) + tan(C/2)}`

The incenter is `(x,y)` and the incircle is tangent to AC, the x axis, so the radius is `y` and the incircle's area is `pi y^2`.

Let's put everything together.

`text{incircle_area} = {2 pi mathcal{A} tan^2(A/2) tan^2(C/2)}/{ tan A(tan(A/2) + tan(C/2))^2}`

There's probably some simplification to be done here, but let's plug in the numbers to see if I've gone off the deep end. I'd guess from my rough sketch an area a little less than half, around 20.

Feeding it to Alpha gets an approxmate incircle area of `19.6399`, closer to my guess than I would have guessed. Hard to believe I didn't make a mistake getting this far, but let's proceed.

That's motivation to work out the exact answer but I'm getting length warnings, so let's stop here.