Quadrilateral Incircle

The incentre $I$ of the quadrilateral will lie on its angle bisectors, just as it does for a triangle. By symmetry, the angle bisector of $\angle BGD$ is just the line $GD$ . To find $I$ , and the inradius $r$ , we can focus on triangle $\Delta GDC$ :

From known angles in the pentagon, $\angle CGD=72^\circ$ , $\angle GCD=\angle GDC=54^\circ$ .

In triangle $\Delta ICD$ , then, $\angle IDC=27^\circ$ , $\angle DCI=54^\circ$ , $\angle CID=99^\circ$ and $CD=1$ (given).

The inradius $r$ is just the length of the altitude $IH$ ; we can find this by computing the area of $\Delta ICD$ . Using a formula for the area in terms of the angles and one side, this is $[ICD]=\frac{CD^2 \sin \angle ICD \cdot \sin \angle IDC}{2\sin \angle CID} = \frac{\sin 27^\circ \cdot \sin 54^\circ}{2\sin 99^\circ}$

We then have $r=\frac{2[ICD]}{CD}$ which works out to be $r=\frac14 \left(-2 (2 + \sqrt5) + \sqrt{50 + 22 \sqrt5}\right)$

This has minimal polynomial $f(x)=16 x^4 + 64 x^3 - 44 x^2 + 4 x + 1$

and $f(1)=\boxed{41}$ .

---

I had to use a lot of trig simplification here - basically bashing away at Wolfram|Alpha to get the algebraic form. An alternative approach might be to couch this in terms of complex numbers, or coordinate geometry, and try to avoid trig as much as possible.