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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?


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.

Equilateral Areas

ABC and DEF are equilateral triangles of side 3 and 4 respectively. Construct an equilateral triangle whose area is the sum of the area of ABC and DEF.


Age 14 to 16 Challenge Level:

Sunil sent us his work on this problem:

When I tried this out, no matter what numbers I started with, I always seemed to find that the $6$th term was the same as the $1$st term, and the $7$th the same as the $2$nd, so it kept repeating. I used a spreadsheet to help me do the calculations! Then I used algebra to try to explain why it worked.

The first term is $a_1$.

The second term is $a_2$.

The third term is $\frac{1+a_2}{a_1}$.

The fourth term is $\frac{1+a_1+a_2}{a_1 a_2}$.

The fifth term is $\left(\frac{1+a_1+a_2+a_1 a_2}{a_1 a_2}\right)\times\left(\frac {a_1}{1+a_2}\right)=\left(\frac{(1+a_1)(1+a_2)}{a_1 a_2}\right)\times\left( \frac{a_1}{1+a_2}\right)=\frac{1+a_1}{a_2}$.

The sixth term is $\left(\frac{1+a_1+a_2}{a_2}\right)\times\left(\frac{a_1 a_2} {1+a_1+a_2}\right)=a_1$.

The seventh term is $\left(a_1+1\right)\times\left(\frac{a_2}{1+a_1}\right)=a_2$.

This explains why the pattern always works, no matter what you start with! (As long as the bottom is never $0$, so we can't have $a_1=0$, or $a_2=0$, or $a_1=-1$, or $a_2=-1$, or $a_1+a_2=-1$.)