Ok so, obviously you do moments around A.
To find the length from A to C (the two points touching the bowl). You make an isocelese triangle AOB (AO and BO are radius') and then split that in half to find length AC, comes out as AC = 2acos(theta).
Now you need to find the reaction force at C. So because you know that the reaction force at A works through the centre, you can resolve it into horizontal and vertical parts.
So R(vertical) = RSin(2Theta)
R(horizontal)=RCos(2theta)
Then do the same with reaction force at C:
S(vertical) = Scos(theta)S
(horizontal) = SSin(theta).
You know that the vertical parts add up to the weight:
RSin(2theta)+SCos(theta) = mg
and you know that the two horizontal forces must be equal:
SSin(theta) = RCos(2theta)
Rearrange for R: R=(SSin(theta)/cos(2theta)).
Sub that into the vertical equation and you get:
S*Sin(theta)*Sin(2theta)/cos(2theta) + Scos(theta) = mg.
Therefore S=(mg)/(Sin(theta)Tan(2theta)+cos(theta))
Times top and bottom by cos(2theta):
S=(mgcos(2theta))/(sin(theta)*sin(2theta)+cos(theta)*cos(2theta))
Then do moments:
Weight/half-length = reaction force at C*AC
0.5L*cos(theta)*mg = 2aCos(theta)*(mgcos(2theta))/Cos(theta)
(note that Sin(theta)*Sin(2theta)+cos(theta)*cos(2theta) = cos(theta))
mg cancels out. Times by 2. Cos(thetas) cancel out.
L=(4acos(2theta))/cos(theta)).
Done! I think this is probably unreadable, but hopefully you can make some sense of it.