## Monday, 30 September 2013

### Mathematics Talks 4

Francis Aznaran (S6P) and Josh Robinson (MF6B) will be giving talks on 'What are the Mathematical implications of Gödelian incompleteness?' and 'How should you fight job promotion competitions?' on Monday lunchtime at 1.40 in M23. All are welcome.

## Monday, 23 September 2013

### Mathematics Talks 2

Great to see so many boys at the Maths and the City talk today. Please let Mr Ottewill know if you are keen to attend follow up sessions.

## Monday, 16 September 2013

### Mathematics Talks 1

Louis Smith (S6A) and Anjie Tang (S6C) will be giving talks on Monday lunchtime in M23 at 1.40 entitled 'Can we predict the primes?' and 'Malthusian model to logistic model and its applications'.  All are welcome.

Chris Ottewill

## Sunday, 8 September 2013

### DC Mathematics Extension Class

A relatively well known problem often asked at interviews.

You are told that there is a pack of cards with a letter printed on one side of each card and number on the other.  Four cards are dealt and show E, 3, T, 8.  Which of these four cards do you need to turn over to test the hypothesis that if a card has a vowel on one side then it has an even number on the other?

A worthy mention in dispatches for the first DC boy to provide a well outlined solutions, is on offer. Good luck

Contributed by C J Ottewill, DC Mathematics Department

## Friday, 19 July 2013

### Further Algebraic Problem

By Yussof MF6B (2013)
One Thursday morning, Mr Lord came dashing through my form class and presented us with a maths problem that was given to him by Dr O’Neill. I looked at it and thought it was challenging so I gave it a go.
Given: $$a + b + c = 0$$, prove that: $$\left( {\frac{{b - c}}{a} + \frac{{c - a}}{b} + \frac{{a - b}}{c}} \right)\left( {\frac{a}{{b - c}} + \frac{b}{{c - a}} + \frac{c}{{a - b}}} \right) = 9$$
When I first looked at this problem, I thought it would be wise to expand the two brackets straight on and tried to simplify it but I ended up having something really messy so I tried a different approach.
Let: $$x = \frac{{b - c}}{a},$$ $$y = \frac{{c - a}}{b},$$ $$z = \frac{{a - b}}{c}$$
Therefore we get:
$= \left( {x + y + z} \right)\left( {\frac{1}{x} + \frac{1}{y} + \frac{1}{z}} \right)$
Expanding these two brackets:
$= 3 + \frac{{z + y}}{x} + \frac{{z + x}}{y} + \frac{{x + y}}{z} \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \left( * \right)$
Now let us simplify: $$\begin{array}{c} = \frac{{z + y}}{x}\\ = \frac{1}{x}\left( {z + y} \right)\\ = \left( {\frac{a}{{b - c}}} \right)\left( {\frac{{a - b}}{c} + \frac{{c - a}}{b}} \right)\\ = \left( {\frac{a}{{b - c}}} \right)\left[ {\frac{{b\left( {a - b} \right) + c\left( {c - a} \right)}}{{bc}}} \right]\\ = \left( {\frac{a}{{b - c}}} \right)\left( {\frac{{ab - {b^2} + {c^2} - ac}}{{bc}}} \right)\\ = \left( {\frac{a}{{b - c}}} \right)\left( {\frac{{ab - ac - \left( {{b^2} - {c^2}} \right)}}{{bc}}} \right)\\ = \left( {\frac{a}{{b - c}}} \right)\left( {\frac{{a\left( {b - c} \right) - \left( {b - c} \right)\left( {b + c} \right)}}{{bc}}} \right)\\ = \left( {\frac{a}{{b - c}}} \right)\left( {b - c} \right)\left( {\frac{{a - b - c}}{{bc}}} \right)\\ = \left( a \right)\left( {\frac{{a - b - c}}{{bc}}} \right)\end{array}$$
Using $$a + b + c = 0$$ so, $$a = - b - c$$ : $$\begin{array}{c} = \left( a \right)\left( {\frac{{a + a}}{{bc}}} \right)\\ = \left( a \right)\left( {\frac{{2a}}{{bc}}} \right)\\ = \left( {\frac{{2{a^2}}}{{bc}}} \right)\end{array}$$
By symmetry, we can note that:
$\frac{{z + x}}{y} = \frac{{2{b^2}}}{{ac}}$ & $$\frac{{x + y}}{z} = \frac{{2{c^2}}}{{ab}}$$
Therefore going back to equation (*) from above: $$\begin{array}{c} = 3 + \frac{{2{a^2}}}{{bc}} + \frac{{2{b^2}}}{{ac}} + \frac{{2{c^2}}}{{ab}}\\ = 3 + \left( {\frac{{2{a^3} + 2{b^3} + 2{c^3}}}{{abc}}} \right)\\ = 3 + 2\left( {\frac{{{a^3} + {b^3} + {c^3}}}{{abc}}} \right)\end{array}$$
Note from here onwards I did get stuck for a while because I could not simplify $${a^3} + {b^3} + {c^3}$$ until I found the formula for sum of three cubes# which is:
${a^3} + {b^3} + {c^3} = \left( {a + b + c} \right)\left( {{a^2} + {b^2} + {c^2} - ab - bc - ac} \right) + 3abc$
And since $$a + b + c = 0$$ , $${a^3} + {b^3} + {c^3} = 3abc$$.
Going back to the equation: $$\begin{array}{c} = 3 + 2\left( {\frac{{{a^3} + {b^3} + {c^3}}}{{abc}}} \right)\\ = 3 + 2\left( {\frac{{3abc}}{{abc}}} \right)\\ = 3 + 2\left( 3 \right)\\ = 9\left( {{\rm{QED}}} \right)\end{array}$$

# Source article/book should be credited