In the following figure, $ABCD$ is a square. $\stackrel{\frown} {AB}$ and $\stackrel{\frown}{AC}$ are semicircles and quarter circles respectively. The blue circle touches $\stackrel{\frown} {AB}$ at $P$, $\stackrel{\frown} {AC}$ at $Q$ and line $BC$. $MN$ is a line such that $MN\parallel PQ$ and touches the blue circle at $R$. If $\angle MPR=\alpha$ and $\angle NQR=\beta$, find $\alpha+\beta$.
Here is my progress in solving the problem:
I could just find the radius of the blue circle. Let $OG=r$ and $AB=2$. So, $BE=1$, $BO=2-r$, $OE=1+r$, $EH=1-r$. Then, $BG^2=HO^2=4-4r$ and $4-4r+(1-r)^2=(1+r)^2$. This gives $r=\frac 12$. I can't proceed from here.
Also while creating the figures, I found the answer:
$\alpha=45^\circ$ and $\beta=30^\circ$
So, I need a solution to the problem. All solutions are welcome, but as I'm always interested in synthetic solutions I'll most likely accept an answer with synthetic solution.
Best Answer
As @MathLover pointed out in the comments, we should first prove that $PQ \parallel BC$. To show this we observe that the distance of $P$ from $BC$ is the same as the distance of $Q$ from $BC$. Indeed, the former is equal to $\frac 23 OG + \frac 13 EB = \frac 23$ and the latter equals $\frac 43 OG = \frac 23$.
Next, since $r=\frac 12$, we see that the points $M$ and $N$ lie on the perpendicular bisector of $AB$.
Considering a homothety centered at $P$, we see that $B, P, R$ are collinear. Therefore $$\angle RPM = 180^\circ - \angle MPB = \angle BAM = 45^\circ.$$
Similarly you can see that $Q,R,A$ are collinear, so $$\angle NQR = \angle NQA = \frac 12 \angle NBA = \frac 12 \cdot 60^\circ = 30^\circ$$ where the equality $\angle NBA = 60^\circ$ follows from the observation that the triangle $ABN$ is equilateral.