Hint: IMO 2008, Madrid, Spain

This is a simpler(much simpler!) case of the first problem of $IMO$ $2008$ , Spain. The solution is long only because it is explained in complete detail!

Solution:

You may reach this conclusion from the solution of the fore-stated problem or simple observation : The centre of the circumscribing circle of the quadrilateral is the circumcentre of the $\Delta ABC$ i.e. The circumcircles of the triangle $ABC$ and the quadrilateral $PQRS$ are concentric .

( This proof : Because $O$ is the circumcentre of the triangle, $OP = OQ$ (can be proved by congruence of $\Delta OMQ$ and $\Delta OMP$ ) and by symmetry $OQ = OR$ while $OP = OS$ (because $O$ lies on the symmetry line $AD$ ) So, we have $OP=OQ=OR=OS$ .) So, we basically need to find, say, $OP$ .

An elementary (and so often used in Olympiads!) result is that the diametrically opposite point to $C$ is the same as the reflection of orthocentre $H$ in $M$ . i.e. $H,M,C"$ are collinear and $HM = MC"$ i.e. $HAC"B$ is a $parallelogram$ .

( This proof : I hope it is known that $\angle AHB$ can be found by simple angle chasing as $180 - \angle C$ which of course is the same as $\angle AC"B$ (cyclic quadrilateral $CAC"B$ ). Now if $\angle HBC"$ comes out as $\angle C$ , our job is done as then $HAC"B$ will be a $parallelogram$ .

$How?:$ See $Figure 1.$ $\angle ABL = 90 - \angle A$ (in $\Delta ABL$ ).

$\angle AC"C = \angle B$ (both subtended by chord $AC$ ) $\implies \angle ACC"$ is $90 - \angle B$ (because $\Delta ACC"$ has $\angle CAC" =90$ because $CC"$ is $diameter$ ),

But then $\angle ABC" = \angle ACC" = 90-\angle B$ (angles subtended by chord $AC"$ ). So, $\angle HBC"= \angle ABH + \angle ABC" = 90-\angle A + 90-\angle B = \angle C$ . $Q.E.D.$

So, $MP = MC" = MH$ .

So, our answer is $OP^2 = OM^2 + MP^2 = OM^2 + MH^2$ (from above)..(*)

Now, the track is straight forward: Use:

In a parallelogram $ABCD$ , $AB^2 + BC^2 = \frac{1}{2}\times (AC^2 + BD^2)$

This proof : Basically sum of two expressions for the diagonals using cosine rules; i.e. Apply cosine rule to find $AC^2$ in $\Delta ABC$ and then cosine rule in $\Delta BCD$ to find $BD^2$ and add the two equations. Use $cos(180-\theta)=-cos\theta$ )

Mark $K$ as the midpoint of $CH$ . Now since $CH = 2\times OM$ ,

(It can be proved by similiarity of $\Delta MOG$ and $\Delta CHG$ where $G$ is the centroid, and noting that $CG = 2\times GM$ . We in herently use that $H, G, O$ are collinear (Euler's line). So, now $KH = OM$ .

Thus, $KOMH$ is a parallelogram. Apply the result (1.). continuing the statement in (*)

So, $OP^2 = OM^2 + MH^2 = \frac{1}{2}\times (OH^2 + MK^2)$

Note immediately that $OCKM$ is also a parallelogram, so $MK = OC = \large{R}$ , the circumradius. So, we update the above statement to:

$OP^2 =\frac{1}{2}\times (OH^2 + \large{R}^2)$ .....(**)

Finally, the "bomb" part of the problem!

Till now the acuteness or obtuseness of angle $A$ did not matter. To find $OH$ , it $does$ matter. Suppose on $AD$ , the order of $O,G,H$ be as in $OPTION 1$ in figure. However this is wrong, why? Because of the simple result:

In an isosceles triangle iff $AO:OD =2$ (where $AO, OD$ are as in this problem), the triangle is equilateral, iff $AO > 2\times OD$ , the triangle is obtuse at $A$ and iff $AO < 2\times OD$ , the triangle is acute at $A$ .

Here, we say now, that the triangle is obtuse: So, $H$ is closer to $A$ than $O$ . Now take $OH$ as $x$ . By the division in Euler's line segment : $HG = 2\times GO$ , so $HG = \frac{2x}{3}$ and $GO = \frac{x}{3}$ . So, $AG = \frac{2x}{3} + 8-x = \frac{24-x}{3}$ , $GM = \frac{x}{3} + 1 = \frac{x+3}{3}$

Now, $AG = 2\times GM \implies : 24-x = 2x + 6 \implies x=6$ .

So, $OH =6, \large{R} =8$ ,

Using (**), $OP^2 = \rho^2 = 50$ .

$\text{Let O be the center of the circumcircle } C_T~of ~ \triangle ~ABC . \\Let~C_Q\text{ be the circumcircle of quadrilateral PQRS.}\\In ~~rt \angle d ~\Delta~ DOB, \angle DOB= Cos^{-1}\dfrac{OD}{OB}=\dfrac{1}{8}=2*41.41=2*\alpha. \\\implies~ in~ isosceles ~\triangle ~OAB, ~~\angle~OAB=\alpha.\\\therefore~AB=AD*Sin\alpha = 9*Sin41.41 =12, ~~\implies~AM=MB=6..(1)\\In ~\Delta~C^"CB, ~OD=1, also~ O~and~D~are ~midpionts,\\\therefore~BC^"~=~2~and~ \parallel OD \implies \parallel AD. \\So ~\angle ~MBC^"=\alpha......Applying ~Cos~Law~to~\Delta~MBC^",\\ MC^{"2}=22.~~~But ~MC^{"2}=MQ^2=22...\text{MQ is the radius of } C_Q ...(2) \\ \text{AD is the line of symmetry for both } \Delta~ABC~and~quadrilateral ~PQRS. \\ \text{M is the midpoint of the chord AB of } C_T~ and ~also~ of~chord~QR~of ~C_Q.\\So \perp \text{ to the chords from M intersects AD, the line of symmetry at the }\\\text{common center O. So both cord distances from their common center}\\are~ same.~~\implies~Radius^2 -(half~chord)^2 ~is~same~for~both. \\ \implies~OA^2-AM^2 = \rho ^2- MQ^2~~\therefore~ \rho ^2=8^2-6^2+22=\boxed{50}$

I solved it almost the same way Aditya (but I did use the hint!LOL)

Apart from that: To find $OH$ , an easier way is this: $AH = 2\times OD$

which in a way you have mentioned in your proof as well. ( $CH = 2\times OM$ )

So, $AH = 2$

So, $OH =6$ . Bingo!