Twofer, Part Two

Let's plot this on a Cartesian plane with coordinates $A(0,0), B(1,0), C(1,1), D(0,1), E(t,1), F(\tfrac t2, \tfrac12)$

By Pythagorean theorem , $AE = \sqrt{1 + t^2}$ .

The radius on an incircle of a triangle is given to be $r = \dfrac{\mathcal A}{s}$ , where $\mathcal A$ and $s$ are the area and semiperimeter of the triangle.

In this case, the radius of incircle of triangle $ADE$ is $r := \left(\frac12 \cdot 1 \cdot t \right) \div \left(\dfrac{1 + t + \sqrt{1+t^2}}2 \right) = \dfrac t{1 + t + \sqrt{1+t^2}} \tag 1$

Likewise, we can compute the radius of the incircle of triangle $EFB$ , $r = \dfrac{\text{Area of triangle }EFB}{\text{semiperimeter of triangle }EFB} \tag 2$

We can compute the area of the triangle $EFB$ by simply finding the complementary areas. That is, $1 - [EFB] = [DEA] + [FAB] + [EBC]$

Simplify all these, we get $\text{Area of triangle }EFB = \dfrac14$ .

Using the distance formula, we can find all the lengths of the triangle $FBE$ . And so (after some tedious algebra) we can obtain its semiperimeter to be $\dfrac14 \left(\sqrt{1+t^2} + \sqrt{t^2-4t+5} + 2 \sqrt{t^2 - 2t + 2} \right)$

Substitute these expressions into $(2)$ , we get $r = \dfrac1{\sqrt{1+t^2} + \sqrt{t^2 - 4t + 5} + 2 \sqrt{t^2 - 2t + 2} } \tag 3$

Because $(1) = (3)$ ,

$t \left( \sqrt{1+t^2} + \sqrt{t^2 - 4t + 5} + 2 \sqrt{t^2 - 2t + 2} \right) = 1 + t + \sqrt{t^2 + 1}$

Through repeated squaring and plenty of oh-so-tedious algebraic manipulation, we (magically) arrive at $32 t^9 - 208 t^8 + 656 t^7 - 1236 t^6 + 1568 t^5 - 1436 t^4 + 948 t^3 - 344 t^2 - 108 t + 79 = 0.$

Hence, $f(x) = \pm \bigg( 32 x^9 - 208 x^8 + 656 x^7 - 1236 x^6 + 1568 x^5 - 1436 x^4 + 948 x^3 - 344 x^2 - 108 x + 79 \bigg) \Rightarrow |f(2)| = \boxed{1271} .$