### Rotating Triangle

What happens to the perimeter of triangle ABC as the two smaller circles change size and roll around inside the bigger 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?

### Polycircles

Show that for any triangle it is always possible to construct 3 touching circles with centres at the vertices. Is it possible to construct touching circles centred at the vertices of any polygon?

##### Stage: 4 Challenge Level:

Like Quad in Quad (June 1998) the key to this problem is using the fact that, when the midpoints of the sides of a triangle are joined, the join is parallel to the third side and half the length of the third side.

This solution came from Murat from Turkey.

Due to the Thales relationship, SP and QR are both parallel to BD, thus to each other. Moreover they are half as long as BD. A similar relationship holds among PQ, RS and AC. Therefore PQRS is a parallelogram.

Let T be the midpoint of AC.

In the above figure, the areas of the triangles marked by the same symbol are obviously equal (look for triangles on equal bases which have the same height). Moreover Area (CST) = S 1 , Area (CPT)= S 2 and Area (TPS)= S 3 . The following equations are obviously true:

 Area (ABCD) = 4( S 1 + S 2 ) Area (PQRS) = ( S 1 + S 2 + S 3 ) + ( S 1 + S 2 - S 3 ) = 2( S 1 + S 2 ) = $\frac {1}{2}$ Area (ABCD)