### Baby Circle

A small circle fits between two touching circles so that all three circles touch each other and have a common tangent? What is the exact radius of the smallest circle?

### Pericut

Two semicircle sit on the diameter of a semicircle centre O of twice their radius. Lines through O divide the perimeter into two parts. What can you say about the lengths of these two parts?

### Kissing

Two perpendicular lines are tangential to two identical circles that touch. What is the largest circle that can be placed in between the two lines and the two circles and how would you construct it?

# Lunar Angles

##### Stage: 5 Challenge Level:
Here is an excellent solution from Andrei from 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 $180^o$.

1. Congruence of angles in a lune

Let $O_1$ and $O_2$ 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 $O_1AO_2$ and $O_1BO_2$ are congruent, because they have a common side $O_1O_2$, and the other two sides are radii in the two circles. So, angle $O_1AO_2$ and $O_1BO_2$ are equal to say $\theta$. A tangent to a circle and the radius to the point of contact are perpendicular so $\angle MAO_2 = \angle NAO_1 =\angle MBO_2 = \angle NBO_1 = 90^o$. Hence $\angle O_1AM = \angle O_2AN =\angle O_1BM =\angle = \angle O_2BAN = \theta - 90^o$. So, $\angle MAN = \angle MBN$.

2. Let $O_1$, $O_2$ and $O_3$ be the centres of the three circles respectively, D the point common to all three circles, and $T_1T_1'$, $T_2T_2'$, $T_3T_3'$ 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 $\alpha$ has the same measure with angle $T_2DT_1'$ and to the angle $T_1DT_2'$ (the last two being vertically opposite angles).

In the same manner I found: $\beta = \angle T_1'DT_3' =\angle T_1DT_3$ and $\gamma = \angle T_3DT_2 = \angle T_3'DT2'$. \par This way, it is easy to see that the three angles add up to $180^o$: $\alpha + \beta + \gamma = T_2DT_1' + T_1'DT_3' + T_3'DT_2' = T_2DT_2' = 180^o$. \par For the second case, I follow exactly the same steps.

I find $\alpha = \angle T_2DT_1'$, $\beta = \angle T_2'DT_3' = \angle T_2DT_3$ and $\gamma \angle T_3DT_1 = \angle T_3'DT_1'$.

The result is the same, i.e. that in this case the three angles add up to $180^o$ again.