This Problem Is Gold!

Since the circle is the incircle of $\triangle ABC$ , $BD=BE = a$ , $CD=CF = b$ and let $AE=AF = c$ . Then

$\begin{aligned} c & = r \cot \frac A2 = r \cot \left(\frac {180^\circ - B - C}2\right) = r \tan \left(\frac B2 + \frac C2\right) \\ & = r \left(\frac {\tan \frac B2+\tan \frac C2}{1-\tan \frac B2 \cdot \tan \frac C2} \right) = r \left(\frac {\frac ra + \frac rb}{1-\frac {r^2}{ab}} \right) = \frac {(a+b)r^2}{ab-r^2} \end{aligned}$

Note that the area of $\triangle ABC$ is given by:

$\begin{aligned} A_\triangle & = (a+b+c)r \\ \frac {A_\triangle}{a+b} & = \frac {(a+b+c)r}{a+b} \\ \sqrt{ab} & = \frac {(a+b+c)r}{a+b} \\ \implies \frac r{\sqrt{ab}} & = \frac {a+b}{a+b+c} = \frac 1{1+\frac {r^2}{ab-r^2}} = \frac {ab-r^2}{ab} = 1 - \frac {r^2}{ab} \end{aligned}$

$\begin{aligned} \implies \left(\frac r{\sqrt{ab}} \right)^2 + \frac r{\sqrt{ab}} & = 1 \\ \left(\frac r{\sqrt{ab}} + \frac 12 \right)^2 & = 1 + \frac 14 \\ \implies \frac r{\sqrt{ab}} & = \frac {\sqrt 5}2 - \frac 12 \approx \boxed{0.618} \end{aligned}$

In the diagram above the lengths are filled in by noting that two tangents to a circle from an outside point are equal in length.

$A_{\triangle{ABC}} = \dfrac{r}{2}(2(a + b) + 2x) = r(a + b + x)$

Using Heron's formula for $\triangle{ABC}$ we have:

$s = a + b + x \implies A_{\triangle{ABC}} = \sqrt{(ab)(a + b + x) x} = r(a + b + x) \implies$

$(a + b)abx + abx^2 = (a + b)^2r^2 + 2(a + b)r^2x + r^2x^2 \implies$

$(ab - r^2)x^2 + (a + b)(ab - 2r^2)x - (a + b)^2r^2 = 0 \implies$

$x = \dfrac{(a + b)r^2}{ab - r^2} > 0$ dropping the root $-(a + b) < 0$ $\implies A_{\triangle{ABC}} = \dfrac{ab(a + b)r}{ab - r^2} \implies$

$\dfrac{A_{\triangle{ABC}}}{a + b} = \dfrac{abr}{ab - r^2} = \sqrt{ab} \implies$

$(ab)^{\frac{1}{2}}r^2 + abr - (ab)^{\frac{3}{2}} = 0 \implies r = \sqrt{ab}(\dfrac{\pm\sqrt{5} - 1}{2})$

$r > 0 \implies r = \sqrt{ab}(\dfrac{\sqrt{5} - 1}{2}) = \sqrt{ab}(\phi - 1) \implies$

$\dfrac{r}{\sqrt{ab}} = \phi - 1 \approx \boxed{0.6180339887498948}$ .