### Diophantine N-tuples

Can you explain why a sequence of operations always gives you perfect squares?

### DOTS Division

Take any pair of two digit numbers x=ab and y=cd where, without loss of generality, ab > cd . Form two 4 digit numbers r=abcd and s=cdab and calculate: {r^2 - s^2} /{x^2 - y^2}.

### Sixational

The nth term of a sequence is given by the formula n^3 + 11n . Find the first four terms of the sequence given by this formula and the first term of the sequence which is bigger than one million. Prove that all terms of the sequence are divisible by 6.

# Pythagoras Perimeters

##### Age 14 to 16Challenge Level

Siddhant from Indus International School of Bangalore in India, Noraisha, Venus, Aditya, Ahn, Jemma and Minjun from West Island School in Hong Kong, Guruvignesh from England, Agathiyan from Hymers College in the UK, Vo from ISM in Vietnam and Sophie and Milly from Trinity Catholic High School in the UK wrote correct proofs which used the steps in the correct order. This is Jemma's proof and explanation:

g) $a + b + c =12$ What we know is that the perimeter of the triangle is twelve and the perimeter is all the sides added up. So that is $a$ plus $b$ plus $c$

c) $a + b=12âˆ’c$ In order to get the correct order for Pythagoras theorem ($a^2 + b^2 =c^2$) you must put the $c$ onto the other side of the equal sign. The positive $c$ will then turn into a negative number.

a) $a^2+ 2ab +b^2=144-24c+c^2$ We now can square both sides of the equation so that we are able to simplify when we reach the next step.
So $(a+b)^2 = a^2+ 2ab +b^2$ and for the other side $12\times12=144$ (The twelve is from the perimeter) and the $24 = 12\times2$

f) $a^2 + b^2 =c^2$ Now we can use the Pythagoras theorem

d) $2ab=144-24c$ When we get rid of the $a^2,b^2$ and $c^2$ it will just equal $2ab=144-24c$

h) $ab=72-12c$ We can then simplify the equation by dividing everything by two so then $2ab\div2=ab,$ $144\div2=72$ and $24c\div2=12c$

e) Area of triangle$= \frac{ab}2$ To find the area of a triangle the equation is $\frac{a\times b}2$

b) $36-6c$ Therefore we know $a$ and we know $b$ so the equation must be $36-6c$

The last step comes from dividing both sides of $ab=72-12c$ by $2$.

Adithya from Hymers College in the UK and Piyush, Anna, Vaibhavi, Nicholas, Ahan, Sahiti, Caitlin, Winston, Natalia, Stephanie and Wing Tung from West Island School in Hong Kong sent in proofs that included the steps in a slightly different order. These proofs didn't flow as neatly, but are still convincing. For example, this is Natalia's proof. Note that Natalia lists all of the information available near the beginning, which is sensible, but less easy to follow than using the information only when it is needed:

g) $a+b+c=12$

c) $a+b=12âˆ’c$

e) Area of the triangle $=\frac{ab}2$

f) By Pythagoras' Theorem, $a^2+b^2=c^2$

a) Squaring both sides: $a^2+2ab+b^2=144âˆ’24c+c^2$

d) So $2ab=144âˆ’24c$

h) Dividing by $2$: $ab=72âˆ’12c$

b) So Area of the triangle $=36âˆ’6c$

Minhaj from St Ivo School in England wrote a proof that didn't use the steps suggested. This is Minhaj's proof:

$a + b + c = 12$ (as perimeter is $12$ units)
Using Pythagoras' theorem,
$c = (a^2+b^2)^\frac12\\ \Rightarrow a + b + (a^2+b^2)^\frac12 = 12\\ \Rightarrow (a^2+b^2)^\frac12 = 12 - a - b\\ \Rightarrow a^2+b^2 = (12 - a - b)^2 = 144 - 24a - 24b + 2ab + a^2 + b^2\\ \Rightarrow 144 - 24a - 24b + 2ab = 0$
Dividing by $4$ gives $36 - 6a - 6b + \frac12 ab = 0\\ \Rightarrow\frac12 ab = 6 (a + b) - 36$
The area of the triangle is $\frac12 ab$ and $a + b = 12 - c$
$\Rightarrow\frac12 ab = 6 ( 12 - c ) -36 = 72 - 6c - 36 = 36 - 6c\\ \therefore\text{Area}= 36 - 6c$ square units

Adithya, Siddhant, Minhaj, Anna, Caitlin, Winston, Wing Tung, Minjun, Guruvignesh, Agathiyan, Sophie and Milly proved that a right-angled triangle whose perimeter is $30$ units has an area of $225-15c$ units. This is Milly's proof:

Amrit from Hymers College in the UK and Vo both proved this by first finding and proving the area in the general case (in the extension), and then substituting in $p=12$. This is Amrit's proof of the general case (extension) and also the second part.

A right angled triangle has sides $a,b$ (the base and height) and $c$ (the hypotenuse) which make up its perimeter $p$. Therefore,

$a+b+c=p$

Subtracting $c$ from both sides,

$a+b=p-c$

Squaring both sides,

$a^2+2ab+b^2=p^2-2pc+c^2$

By Pythagoras' theorem, $a^2+b^2=c^2$. Plugging $c^2$ in for $a^2+b^2$ on the LHS,

$2ab+c^2=p^2-2pc+c^2$

Subtracting $c^2$ from both sides,

$2ab=p^2-2pc$

Factorising,

$2ab=p(p-2c)$

Dividing both sides by $4$, we get that the Area of the triangle, $A$, is

$A=\dfrac{ab}2=\dfrac{p(p-2c)}4$

When the perimeter is $30$, the area of the triangle in terms of $c$ is

$\dfrac{30(30-2c)}4$

Factorising,

$\dfrac{60(15-c)}4$

Multiplying by $\dfrac{1/4}{1/4}$,
$15(15-c)$

Expanding the brackets, the area of the triangle is

$225-15c$

Well done to Shreya, Apurva and Aryan from Dhirubhai Ambani international School in India, and to Caitlin, Winston, Wing Tung, Minjun, Guruvignesh, Agathiyan, Matt, Sophie and Milly who also correctly found and proved the general expression for the area of a right-angled triangle. This is Guruvignesh's proof, which is slightly different:

$P = a+b+c$ (Equation for the perimeter of a right-angled triangle)

$P-c = a+b$ (Subtract $a$ from both sides)

$P^2-2Pc+c^2 = a^2+2ab+b^2$ (Square both sides)

$P^2-2Pc = a^2+2ab+b^2-c^2$ (Subtract $c^2$ from both sides)

$A = \frac{ab}2$ (Equation for the area of a triangle)

$4A = 2ab$ (Multiply both sides by $4$)

$P^2-2Pc = a^2+b^2+4A -c^2$ (Since, $2ab=4A$, we can substitute that into the previous
equation)

$a^2+b^2 = c^2$ (Pythagorean Theorem)

$P^2-2Pc = c^2-c^2+4A$ (Substitute the Pythagorean Theorem into the equation)

$P^2-2Pc = 4A$

$\dfrac{P^2-2Pc}4 = A$ (Divide both sides by $4$ to express $A$ in terms of $P$ and $c$)