It Golden!

By cosine rule , $\begin{cases} BD^2 = a^2x^2 + 1 - 2ax\cos 60^\circ = a^2x^2 - ax + 1 \\ DC^2 = x^2 + 1 - 2x \cos 60^\circ = x^2 - x + 1 \end{cases}$

By angle bisector theorem , $\dfrac {BD}{AB} = \dfrac {DC}{CA} \implies \dfrac {BD}{ax} = \dfrac {DC}x \implies BD = a \cdot DC$ . Therefore,

$\begin{aligned} BD^2 & = a^2 \cdot DC^2 \\ a^2x^2 - ax + 1 & = a^2x^2 - a^2x + a^2 \\ a^2x - ax & = a^2 - 1 \\ \implies x & = \frac {a^2-1}{a(a-1)} = \frac {a+1}a \end{aligned}$

By cosine rule again,

$\begin{aligned} BC^2 & = AB^2 + CA^2 - 2AB \cdot CA \cos \angle BAC \\ \left(\frac {(a+1)\sqrt{2(a+1)}}a\right)^2 & = a^2x^2 + x^2 - 2ax^2 \cos 120^\circ \\ \frac {2(a+1)^3}{a^2} & = a^2x^2 + x^2 + ax^2 = (a^2+a+1)x^2 \\ \frac {2(a+1)^3}{a^2} & = (a^2+a+1)\left(\frac {a+1}a\right)^2 \\ 2(a+1) & = a^2 + a + 1 \\ a^2 - a - 1 & = 0 \\ \implies a & = \varphi & \small \blue{\text{where }\varphi \text{ is golden ratio.}} \end{aligned}$

Then $x = \dfrac {\varphi+1}\varphi = \dfrac {\varphi^2}\varphi = \varphi$ and $A_{\triangle ABC} = \dfrac {ax \cdot x \sin 120^\circ}2 = \dfrac {\sqrt 3 \varphi^3}4$ . Therefore $k+q = 3+4 = \boxed 7$ .

$\alpha + \lambda = 60^{\circ} \implies \alpha = 60^{\circ} - \lambda$

Using the law of cosines on $\triangle{ABC}$ with included $\angle{BAC} \implies y = \sqrt{a^2 + a + 1}x = px$ , where $p = \sqrt{a^2 + a + 1}$ .

Using the law of sines on $\triangle{ABC} \implies \dfrac{ax}{\sin(\lambda)} = \dfrac{px}{\sin(120^{\circ})} = \dfrac{2px}{\sqrt{3}} \implies$

$\sin(\lambda) = \dfrac{\sqrt{3}a}{2p}$

Using the law of sines on $\triangle{DAC} \implies \dfrac{1}{\sin(\lambda)} = \dfrac{n}{\sin(60^{\circ})} = \dfrac{2n}{\sqrt{3}} \implies$

$n = \dfrac{\sqrt{3}}{2\sin(\lambda)} = \boxed{\dfrac{p}{a}}$

Similarly using the law of sines on $\triangle{BAD} \implies m = \dfrac{\sqrt{3}}{2\sin(\alpha)} =$ $\dfrac{\sqrt{3}}{2\sin(60^{\circ} - \lambda)} =$

$\dfrac{\sqrt{3}}{\sqrt{3}\cos(\lambda) - \sin(\lambda)} = \dfrac{2p}{\sqrt{4p^2 - 3a^2} - a} =$ $\dfrac{2p}{2} = \boxed{p}$

where $\cos(\lambda) = \dfrac{\sqrt{4p^2 - 3a^2}}{2p} = \dfrac{a + 2}{2p}$

$\implies y = m + n = \dfrac{p(a + 1)}{a} = \dfrac{a + 1}{a}\sqrt{a^2 + a + 1} =$ $\dfrac{a + 1}{a}\sqrt{2(a + 1)}$

$\implies a^2 - a - 1 = 0 \implies a = \dfrac{1 + \sqrt{5}}{2} = \phi$ , for $a > 0$

$\implies y = \dfrac{a + 1}{a}\sqrt{a^2 + a + 1} = \dfrac{3 + \sqrt{5}}{1 + \sqrt{5}}\sqrt{3 + \sqrt{5}} =$

$(\dfrac{1 + \sqrt{5}}{2})\sqrt{3 + \sqrt{5}} = \sqrt{3 + \sqrt{5}}\phi$

and $y = px = p\dfrac{a + 1}{a} \implies x = \dfrac{a + 1}{a} = \phi = a$

and $\sin(\lambda) = \dfrac{\sqrt{3}}{2}\dfrac{a}{p} = \dfrac{\sqrt{3}}{2}\dfrac{1}{\sqrt{3 + \sqrt{5}}}\phi$

$\implies$ the height $h = x\sin(\lambda) = \dfrac{\sqrt{3}}{2}\dfrac{1}{\sqrt{3 + \sqrt{5}}}\phi^2$

$\implies A_{\triangle{ABC}} = \dfrac{1}{2}yh = \dfrac{\sqrt{3}}{4}\phi^3 =$ $\dfrac{\sqrt{k}}{q}\phi^3 \implies k + q = \boxed{7}$ .

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Below is the triangle with the assigned values:

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