### 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?

# A Little Light Thinking

##### Stage: 4 Challenge Level:

Zafirah and Folashade suggested some ideas for attacking the problem:

Do not use random numbers when solving this problem go up in a strategic method.  Before trying to work out anything at least get three numbers which light up each light to see if there is any pattern occouring. Look closely at the relationship between the numbers for each colour. Once you find out the relationship between the numbers, start to look if you are able to put the method into a formula.

Isaac from Bryn Offa school used this method and tried numbers in ascending order until he had enough to spot a pattern.

Natalie explained how she worked out the rules that turned on the lights:

The sequences and rules for each light follows:
Red: $7, 15, 23, 31$     The difference between $7$ and $15$ is $8$ so the beginning
of the rule is $8n$ and then $1 \times 8$ is $8$, but the first term of the sequence
is $7$. To change the $8$ to $7$ you subtract $1$. So the rule is $8n - 1$.

Yellow: $5, 15, 25$      The difference between $5$ and $15$ is $10$ so the
beginning of the rule is $10n$. However, the first term of the sequence is $5$
so to change the $5$ into 10, you add $5$. The rule is $10n + 5$.

Green: $10, 20, 30$      The difference between $10$ and $20$ is $10$ so the
beginning of the rule is $10n$. the first term of the sequence is $10$, so the
rule doesn't need to change. The rule is therefore $10n$

Blue: $2, 6, 10, 14, 17, 22, 26, 30, 34$   The difference between $6$ and $10$ is
$4$ so the beginning of the rule is $4n$. The first term of the sequence is $2$,
so to change the $4$ to a $2$ you subtract $2$. The rule is $4n - 2$.

The numbers that light 2 lights are:
10, 15 and 30.

Some of you found sequences which didn't ever light both lights. Josh, Christian, Pavan, Alex and Vishaal found one such scenario:

We found the rules: $3n+1$ and $6n+2$. We have decided that the two are impossible to light up together, as it will never both have the same number. The reason for that is because $3n+1$ will always be one more than a multiple of three whereas $6n+2$ will always two more then a multiple of three.

Rajeev gave some examples when both lights could be lit:

$8n+3$ and $6n+5$... the number which would turn both lights on is $11$ because $8+3$ and $6+5$ both equal $11$.
The other numbers which would work are $35, 83,131$.
The formula is $24n-13$.

$11n+10$ and $10n+1$... the number which would turn both lights on is $21$ because $11+10$ and $20+1$ both equal $21$.
The other numbers are $131,241,351$ and so on.
The formula is $110n-89$.

Jessica and Sam suggested some other combined sequences:

$4n+1$ and $5m-2$ have a combined rule of $20k-7$
$10n+4$ and $3m-1$ have a combine rule of $30k+14$
$11n-5$ and $6m-5$ have a combined rule of $66k-5$

A pattern we can see for the combined rules is that to find the combined rules times the coefficients of $n$ together and then find the common differences in the new sequence.

Finally Herschel gave an insight into the method known as the Chinese Remainder Theorem:

We want to find a number that is part of two Arithmetic Progressions (APs): $an+b$ and $cn+d$. To put it another way, we are searching for a number $x$ that satisfies the congruences $x = b$ (mod$a$) and $x = d$ (mod $c$). (The "mod $a$" part means that you look at the remainder on each side of the equation after dividing by "$a$".)

Finding this value of $x$ that will satisfy both equations and hence turn on both lights can be done using Chinese Remainder Theorem, which deals precisely with this problem - solving congruences.