In the figure, $ABCD$ is a rectangle, $M$ and $N$ are points of tangency . Knowing that $MB= \sqrt2, OC =3 , ND=2$ calculate the radius of the circle. (Answer: $\sqrt3$)
My progress:
Let $MB = a,~ OC = b$ and $ND = c$
$A$ is the point of tangency?
Therefore $O,~A,~B$ are collinear
Using Pythagoras in $\triangle OBM$ and $\triangle BCO$
$OB^2=r^2+a^2\tag I$ $b^2=c^2+OB^2\tag{II}$
From $(I)$ and $(II)$: $r^2=b^2-a^2-c^2$
Therefore $r=\sqrt{9-4-2}=\sqrt3$
In this solution, point $A$ was considered as the point of tangency. But this was not mentioned in the statement. Is it possible to demonstrate this or is there another kind of solution?
The solution is indeed quite simple. We just need a preliminary result (easy to prove): if $S$ is the midpoint of a segment $PQ$ and $H$ is the projection of point $R$ on line $PQ$, then the signed distance of $H$ from $S$ is: $$ HS={PR^2-QR^2\over 2PQ}. $$
Let then $H$, $K$ be the projections of $O$ on lines $AD$, $BC$. Their signed distances from the midpoints of $AD$ and $BC$ must be the same, thus from the above lemma we get: $$ {OA^2-OD^2\over2AD}={OB^2-OC^2\over2BC}. $$ But $AD=BC$, hence it must be $OA^2-OD^2=OB^2-OC^2$, that is: $$ (r^2+2)-9=r^2-(r^2+4), \quad\text{i.e.}\quad r^2=3. $$