Here is an excellent solution from Andrei Lazanu, age 14, School
No. 205, Bucharest, Romania.
I divided the solution into two parts: the demonstration of the
congruence of angles, and the demonstration that the three angles
add up to 180o.
1. Congruence of angles in a lune
Let O1 and O2 be the centres of two circles, A and B the
intersection points and M and N the intersections of the
tangents to these two circles. I observed that triangles O1AO2
and O1BO2 are congruent, because they have a common side
O1O2, and the other two sides are radii in the two circles.
So, angle O1AO2 and O1BO2 are equal to say θ.
A tangent to a circle and the radius to the point of contact are
perpendicular so ∠MAO2 = ∠NAO1 = ∠MBO2 = ∠NBO1 = 90o. Hence ∠O1AM = ∠O2AN = ∠O1BM = ∠ = ∠O2BAN = θ− 90o. So, ∠MAN = ∠MBN.
2. Let O1, O2 and O3 be the centres of the three circles
respectively, D the point common to all three circles, and
T1T1 ' , T2T2 ' , T3T3 ' the three tangents.
In my figure the radius and the tangent to circle 1 are in red, to
circle 2 in blue and to circle 3 in green respectively.
From the previous point, I see that angle α has the same
measure with angle T2DT1 ' and to the angle T1DT2 ' (the
last two being vertically opposite angles).
In the same manner I found: β = ∠T1 ' DT3 ' = ∠T1DT3 and γ = ∠T3DT2 = ∠T3 ' DT2 ' .
This way, it is easy to see that the three angles add up to
180o: α+ β+ γ = T2DT1 ' + T1 ' DT3 ' +T3 ' DT2 ' = T2DT2 ' = 180o.
For the second case, I follow exactly the same steps.
I find α = ∠T2DT1 ' , β = ∠T2 ' DT3 ' = ∠T2DT3 and γ∠T3DT1 = ∠T3 ' DT1 ' .
The result is the same, i.e. that in this case the three angles
add up to 180o again.