What is the sum of the angles a+b+c?

Question

For reference:

(figure without scale)

My progress:

Circle centers are collinear. I think that the straight line joining the centers passes through the point of tangency with the side of the triangle. Therefore, the triangles formed will be rectangles and $\measuredangle a$ and $\measuredangle b$ will be $90^\circ$. It remains to demonstrate that $\measuredangle c = 45^\circ$.

Update:

Answer 1

Yes, you are correct, $\measuredangle b$ is $90^\circ$, but not $\measuredangle a$.

Hint: Consider the pentagon (sum of interior angles=$540^\circ$). Can you take it from here?

Answer 2

Note that $PS = PA$ and $O_aA = O_aS$. $PO_a$ is perp bisector of $AS$.

So, $\angle SAO_a = \frac{\angle P}{2}$

Now what is the value of $\angle TCO_c ~$?

As, $\angle MO_aA + \angle MO_bB = 180^\circ$, $\angle MAO_a + \angle MBO_b = 90^\circ$

Now what is $\angle NBO_b + \angle NCO_c ~ $?

Adding them all up, what do you get?

Answer 3

Further to Jean Marie’s comment and my own, this is solvable without assuming that the centres are collinear. Begin as per ACB’s answer, but instead consider the hexagon (not pentagon) bounded by the legs of the triangle and the radii of the circles:

The radii are perpendicular to the tangents, so three of the six angles are 90°. The angles at the circle centres are twice the angles at their circumferences. Since the latter are $a$, $b$ and $c$, the former are $2a$, $2b$, and $2c$. Lastly, the angle sum of a hexagon is 720°. Hence, $2a+2b+2c+270°=720°$, and thus $a+b+c=225°$.

Incidentally, I’ve had a lot of fun playing around with the Desmos Geometry tool, which I’m sad to say I only recently discovered existed! I made a construction of the problem to empirically explore it before I came up with the solution; a variation on that was used for the image above. </plug>