Is there a sequence here? (no)

Let $x$ be the diameter’s length. $\triangle ABD$ and $\triangle ACD$ are right triangles. Using Pythagorean theorem on them we have

$BD=\sqrt{A{{D}^{2}}-A{{B}^{2}}}\Rightarrow BD=\sqrt{{{x}^{2}}-1}$ and

$AC=\sqrt{A{{D}^{2}}-C{{D}^{2}}}\Rightarrow AC=\sqrt{{{x}^{2}}-81}$

By Ptolemy’s theorem on cyclic quadrilateral $ABCD$ , $\begin{aligned} AC\cdot BD=AB\cdot CD+BC\cdot AD & \Leftrightarrow \sqrt{{{x}^{2}}-81}\cdot \sqrt{{{x}^{2}}-1}=1\times 9+4x \\ & \Leftrightarrow \left( {{x}^{2}}-81 \right) \left( {{x}^{2}}-1 \right)={{\left( 9+4x \right)}^{2}} \\ & \Leftrightarrow {{x}^{4}}-98{{x}^{2}}-72x=0 \\ & \overset{x>0}{\mathop{\Leftrightarrow }}\,{{x}^{3}}-98x-72=0 \\ \end{aligned}$ This cubic has one positive solution, $x\approx 10.2482009806474$ .

For the answer, $\left\lfloor AD\cdot {{10}^{4}} \right\rfloor =\boxed{102482}$ .

Method 1

Let $AD = x$ and draw $BD$ . By Thales's Theorem , $\angle ABD = 90°$ , so that $BD = \sqrt{x^2 - 1}$ , $\sin A = \frac{\sqrt{x^2 - 1}}{x}$ , and $\tan A = \sqrt{x^2 - 1}$ .

As a cyclic quadrilateral , opposite angles are supplementary, so $\sin A = \sin D$ .

The area of the quadrilateral is then $A_{ABCD} = A_{\triangle ABD} + A_{\triangle BCD} = \frac{1}{2} x \sin A + \frac{1}{2} 36 \sin D = \frac{1}{2x}(x + 36) \sqrt{x^2 - 1}$ .

The area of the quadrilateral is also given by $A_{ABCD} = \frac{1}{4}(a^2 - b^2 - c^2 + d^2)\tan A = \frac{1}{4}(1^2 - 4^2 - 9^2 + x^2) \sqrt{x^2 - 1} = \frac{1}{4}(x^2 - 96) \sqrt{x^2 - 1}$ .

Therefore, $A_{ABCD} = \frac{1}{2x}(x + 36) \sqrt{x^2 - 1} = \frac{1}{4}(x^2 - 96) \sqrt{x^2 - 1}$ , which rearranges to $x^3 -98x - 72 = 0$ and solves to $x \approx 10.2482$ for $x > 0$ and $x \neq 1$ .

Therefore, $\lfloor AD \cdot 10^4 \rfloor = \boxed{102482}$ .

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Method 2

Let $a = AB = 1$ , $b = BC = 4$ , $c = CD = 9$ , and $d = AB = x$ .

By Parameshvara's Theorem , $R = \cfrac{1}{4}\sqrt{\cfrac{(ab + cd)(ac + bd)(ad + bc)}{(s - a)(s - b)(s - c)(s - d)}}$ , or $\cfrac{x}{2} = \cfrac{1}{4}\sqrt{\cfrac{(4 + 9x)(9 + 4x)(x + 36)}{\frac{1}{16}(-x + 14)(x + 12)(x + 6)(x - 4)}}$ .

This expands to $x^6 - 196 x^4 - 144 x^3 + 9604 x^2 + 14112 x + 5184 = 0$ .

Taking the square root of both sides gives $x^3 -98x - 72 = 0$ which solves to $x \approx 10.2482$ for $x > 0$ .

Therefore, $\lfloor AD \cdot 10^4 \rfloor = \boxed{102482}$ .

Let $\color{#20A900}{\theta}$ denote the measure of the angle $\color{#20A900}{\angle ABC}$ .

Since $ABCD$ is a cyclic quadrilateral, we have ${ \color{#69047E}{\angle ADC}} = {\color{#69047E}{180^\circ - \theta}}$ .
${ \color{#3D99F6}{\angle AOC} }= 2\times \angle ACD = { \color{#3D99F6}{360^\circ- 2\theta}}$ .
Thus, ${\color{#EC7300}{ \angle COD}} = 180^\circ - \angle AOC ={\color{#EC7300}{ 2\theta - 180^\circ}} > 0$ .
And so, $90^\circ < \theta < 180^\circ$ .

And let $r$ denote the radius of the circle. We want to find the diameter $AD = 2r$ .

Apply cosine rule on $\triangle COD$ , $2r^2 - 2r^2 \cos(2\theta - 180^\circ) = 9^2 \quad \Leftrightarrow \quad 2r^2( \underbrace{1 + \cos(2\theta)}_{=2\cos^2\theta}) = 81 \quad \Leftrightarrow \quad r \cos \theta = -\frac92 \quad (\because \cos\theta < 0)$

Similarly, apply cosine rule on $\triangle ABC$ , $(AC)^2 = 1^2 + 4^2 - 2\cdot 1 \cdot 4 \cos \theta = 17 - 8\cos\theta$

Likewise, apply cosine rule on $\triangle AOC$ , $(AC)^2 = 2r^2 \Big(\underbrace{1 - \cos(360^\circ - 2\theta)}_{=1-\cos(2\theta)=2\sin^2\theta} \Big) =4r^2 \sin^2\theta = 4r^2 - 4r^2 \cos^2 \theta = 4r^2 - 4 \left( -\frac92 \right) ^2 = 4r^2 - 81$

Comparing the last two equations above, $\begin{array} { r c l } 17 - 8\cos \theta &=& 4r^2 - 81 \\ 17r - 8r \cos \theta &=& 4r^3 - 81r \\ 17r - 8(-\tfrac 92) &=& 4r^3 - 81r \\ 2r^3 - 49r - 18 &=& 0 \end{array}$

Thus, $r \approx 5.12141$ . The answer is $\lfloor AD \cdot 10^4 \rfloor = \boxed{102482}$ .