Triangle Triplets

Let the lengths of the various segments be as marked. If the three incircles all have radius $r$ , then $\begin{aligned} \tfrac34\sqrt{3}x & = \; \tfrac12r(3+x+w) \\ \tfrac14\sqrt{3}(3-x)y &=\; \tfrac12r(3-x+y+z) \\ \tfrac94\sqrt{3} - \tfrac34\sqrt{3}x - \tfrac14\sqrt{3}(3-x)y & = \; \tfrac12r(3-y + w + z) \end{aligned}$ and these equations can be solved to write $r,y,z$ in terms of $x,w$ : $\begin{aligned} r & = \; \frac{3\sqrt{3}x}{2(3 + x +w)} \\ y & = \; \frac{3(2x^2 + 2wx - 3w - 9)}{2(x^2 + wx + 3x - 3w - 9)} \\ z & = \; \frac{(3+w)(3-2x)}{2x} \end{aligned}$ Since $z>0$ , $0 < x < \tfrac32$ . Since $w^2 = x^2 - 3x + 9$ and $z^2 = (3-x)^2 + y^2 - (3-x)y$ we deduce that

$0 \; = \; \frac{(3-x)\Big((2x^2-9) + (2x-3)\sqrt{x^2 - 3x + 9}\Big)\Big((6x^4 - 126x^3 + 729x^2 - 1539x + 972) + 3(3-2x)(x^2 + 27x - 36)\sqrt{x^2 - 3x + 9}\Big)}{4x^2\Big((x^2 + 3x - 9) - (3 - x)\sqrt{x^2 - 3x + 9}\Big)}$ If $(2x^2-9) + (2x-3)\sqrt{x^2 - 3x + 9} = 0$ then $0 \; =\; (2x^2 - 9)^2 - (x^2 - 3x + 9)(2x-3)^2 \; = \; 3x(x-3)(8x-15)$ which has no solution with $0 < x < \tfrac32$ . Thus we deduce that $\begin{aligned} (6x^4 - 126x^3 + 729x^2 - 1539x + 972) + 3(3-2x)(x^2 + 27x - 36)\sqrt{x^2 - 3x + 9} & = \; 0 \\ (6x^4 - 126x^3 + 729x^2 - 1539x + 972)^2 - 9(3-2x)^2(x^2 + 27x - 36)^2(x^2 - 3x + 9) & = \; 0 \\ 81x^2(15-8x)(5x^3 - 9x^2 + 27x - 27) & = \; 0 \end{aligned}$ and so, since $0 < x <\tfrac32$ , we deduce that $f(x) \; = \; 5x^3 - 9x^2 + 27x - 27 \; = \; 0$ which makes the answer $f(2) = \boxed{31}$ . Note that the value of $x$ is approximately $1.15946$ .