### 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?

### Nim

Start with any number of counters in any number of piles. 2 players take it in turns to remove any number of counters from a single pile. The loser is the player who takes the last counter.

### Loopy

Investigate sequences given by $a_n = \frac{1+a_{n-1}}{a_{n-2}}$ for different choices of the first two terms. Make a conjecture about the behaviour of these sequences. Can you prove your conjecture?

# In a Spin

##### Age 14 to 16 Challenge Level:

Correct solutions were received from Joshua T. and Andrew I. Congratulations to both of you. I have used a mixture of the solutions to form the basis of what follows.

The first thing to observe is that by rotating the triangle about the hypotenuse I obtain two cones having the same base and opposite vertices.

To find the volume of a cone you need its height and the radius of its base. So you need to calculate the radius and the height of each cone.

$$V = \frac{1}{3}\pi r^2 h$$

The radius of the base of the cones is the altitude of the right-angled triangle ($r$), and the corresponding heights are $AX$ and $CX$.

Working directly on the general case - as it seems simpler. Let the sides of the triangle be $a$, $c$ and $b$ ($b$- the hypotenuse).

\begin{eqnarray}V &=& \frac{1}{3}\pi r^2 h \\ &=& \frac{1}{3}\pi r^2 h_{1}+\frac{1}{3}\pi r^2 h_{2}\\ &=&\frac{1}{3}\pi r^2 (h_{1}+h_{2}) \\ &=& \frac{1}{3}\pi r^2 b \end{eqnarray}

Finding the area of a right-angled triangle in two ways:

$$\frac{ac}{2}=\frac{rb}{2}$$

Therefore $$r = \frac{ac}{b}$$

Substituting for $r$ in the equation for the volume you get:

\begin{eqnarray} V &=& \frac{1}{3}\pi r^2 b \\ &=& \frac{1}{3}\pi \left( {\frac{{ac}}{b}} \right)^2 b \\ &=& \frac{1}{3}\pi \frac{{(ac)}^2}{b}\end{eqnarray}

From here you can substitute for the particular case given at the start of the problem.