# Find the missing angles in the picture containing squares and circles

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.

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.
• Good solution but we should first show $PQ$ is parallel to $BC$, which is easy but at least that needs to be said. Commented Aug 26, 2021 at 17:30