Cyclic Quadrilateral Ain't Enough

Note that the area of the triangle with the given values is2.5sqrt(3), so the other part of the quadrilateral must have area 1.5sqrt(3). Since the middle angle must be 120 degrees, the product of the sides must be 6 by the sine area formula. Using the law of cosines, the shared side of these two triangles has length sqrt(19), so if the two unknown sides are a and b:

a^2 + b^2 +ab = 19 -> a^2 + 2ab + b^2 = 25 -> a + b = 5.

Rogers has given an excellent solution, but just for sake of variety, here is a different approach.

The area $K$ of a cyclic quadrilateral with successive sides $a,b,c,d$ and angle $B$ between sides $a$ and $b$ is

$K = \frac{1}{2}(ab + cd)\sin(B)$ .

Plugging in $a = 2, b = 5, B = \frac{\pi}{3}$ and $K = 4\sqrt{3}$ yields

$4\sqrt{3} = \frac{1}{2}(10 + cd)*\frac{\sqrt{3}}{2} \Longrightarrow cd = 6$ .

Now we also have the area formula

$K = \sqrt{(s - a)(s - b)(s - c)(s - d)}$ where $s = \dfrac{a + b + c + d}{2}$ .

In this case, with $s = \dfrac{7 + c + \frac{6}{c}}{2}$ , we end up with

$K = \frac{1}{4}\sqrt{(3 + (c + \frac{6}{c}))(-3 + (c + \frac{6}{c}))(7 - (c - \frac{6}{c}))(7 + (c - \frac{6}{c}))} =$

$\Longrightarrow 16\sqrt{3} = \sqrt{(3 + R)(61 - R)}$ where $R = c^{2} + \frac{36}{c^{2}}$ .

Squaring and rearranging gives us

$R^{2} - 58*R + 585 = 0 \Longrightarrow (R - 45)(R - 13) = 0$ .

Now $R = 13$ yields side lengths $c,d$ of $2,3$ , (or $3,2$ ). Now the diagonal opposite sides $a,b$ has length $\sqrt{4 + 25 - 20\cos(\frac{\pi}{3})} = \sqrt{19}$ , which is the same length as we would obtain using sides length $2,3$ and angle $\frac{2\pi}{3}$ .

For $R = 45$ we would have side lengths $c,d$ of about $0.9026, 6.6472$ , which with angle $\frac{2\pi}{3}$ would give a diagonal of length $\sqrt{51}$ . Hence this is an extraneous solution.

The final (unique) answer is then $2 + 3 = \boxed{5}$ .