The semicircle with centre $O$ and the diameter $AC$ is divided in two arcs $AB$ and $BC$ with ratio $1: 3$. $M$ is the midpoint of the radium $OC$. Let $T$ be the point of arc $BC$ such that the area of the cuadrylateral $OBTM$ is maximum. Find such area in fuction of the radium.
$[OBTM]=[OBM]+[MBT]$
$[OBM]$ is always fixed it is not dependent on point $T$.
Draw altitude $TH$ of $\triangle MBT$
$\displaystyle [MBT]=\frac{1}{2} (BM)(TH)$
If we choose $T$ such that $TH$ is a maximum, $O$, $H$, and $T$ are collinear, meaning that the tangent line containing $T$ is perpendicular to $TH$ and parallel to $BM$. This is because if we extend $BM$ past $M$ to meet circle $O$ at point $D$, the $T$ that maximizes the height of $\triangle MBT$ bisects arc $BD$.
Now $[OBTM]$ can be expressed as $[OBTM]=\frac{1}{2}[BM][OT]$, since the diagonals are perpendicular.
Using Pythagoras,
$[OBTM]=\frac{1}{2}\left(r\frac{\sqrt{5+2\sqrt{2}}}{2}\right)\left(r\right)=\frac{\sqrt{5+2\sqrt{2}}}{4}r^2$