**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.

Department of Mathematics, Dulwich College

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.

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

Chris Ottewill

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

Contributed by C J Ottewill, DC Mathematics Department

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: \(\begin{array}{c} = 3 + 2\left( {\frac{{{a^3} + {b^3} +
{c^3}}}{{abc}}} \right)\\ = 3 + 2\left( {\frac{{3abc}}{{abc}}} \right)\\ = 3 +
2\left( 3 \right)\\ = 9\left( {{\rm{QED}}} \right)\end{array}\)

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