Another Cyclic Quad

soln

The area $A$ of a cyclic quadrilateral with successive sides $a,b,c,d$ and $\angle B$ between sides $a$ and $b$ is given by the formula

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

So with $A = 6\sqrt{3}, a = 3, b = 6$ and $\angle B = \frac{\pi}{3}$ we have that

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

Now by the Cosine rule, the diagonal $L$ opposite $\angle B$ will be such that

$L^{2} = 3^{2} + 6^{2} - 2(3)(6)\cos(\frac{\pi}{3}) \Longrightarrow L^{2} = 45 - 18 \Longrightarrow L = 3\sqrt{3}.$

But $L$ is also the diagonal opposite $\angle D = (\pi - \angle B)$ between sides $c$ and $d$ , and so again by the Cosine rule we have that

$L^{2} = c^{2} + d^{2} - 2cd\cos(\pi - \frac{\pi}{3}) \Longrightarrow 27 = c^{2} + d^{2} + cd.$

But from before we have that $cd = 6$ , so $c^{2} + d^{2} = 21.$ So with $d = \dfrac{6}{c}$ we now have the equation

$c^{2} + \dfrac{36}{c^{2}} = 21 \Longrightarrow c^{4} - 21c^{2} + 36 = 0$ ,

which is a quadratic in $c^{2}$ , and thus

$c^{2} = \dfrac{21 \pm \sqrt{(-21)^{2} - 4*36}}{2} \Longrightarrow c = \sqrt{\dfrac{21 \pm \sqrt{297}}{2}}.$

The two resulting roots are correspond to the lengths of the remaining two sides, so the length of the longer of these is

$\sqrt{\dfrac{21 + \sqrt{297}}{2}} = \boxed{4.372}$ to $3$ decimal places.

6 Sqrt (3) = (1/ 2)(3)(6) Sin (Pi/ 3) + x

x = (3/ 2) Sqrt (3) = (1/ 2) a b Sin (2 Pi/ 3)

a b = 6

c^2 = 3^2 + 6^2 - 2 (3)(6) Cos (Pi/ 3) = a^2 + b^2 - 2 a b Cos (2 Pi/ 3)

27 = a^2 + b^2 + a b

a^2 + b^2 = 21

Substitutes b = 6/ a into it,

a^4 - 21 a^2 + 36 = 0

a^2 = (21 + Sqrt (21^2 - 4 x 36))/ 2 = (21 + Sqrt (297))/ 2

a = 4.3722813232690143299253057341095

b = 1.3722813232690143299253057341095

The answer for longer side is a = 4.372 (correct to 3 decimal places)