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We received three good solutions to this
problem. First of all, Josh from Lyng Hall wrote in
The area of the square you make can either be worked out by:
or,
- area of triangle x number of triangles
Dealing with the triangles first, they are right angled with 30 and
60 degrees in the other two corners and unit lenghth hypotenuse.
This means, by trigonometry, the other two sides will be
$\frac{1}{2}$ and $\frac{\sqrt{3}}{2}$. From this we can deduce the
area of the triangle will be: $$\frac{1}{2} \times
\frac{\sqrt{3}}{2} \times \frac{1}{2} = \frac{\sqrt{3}}{8}$$ The
square must be a multiple of this so its area is:
$$n\frac{\sqrt{3}}{8}$$ The edges of the square must be made up of
a combination of sides of the triangle so can be written as
follows:
- Base: $a\times \frac{1}{2} + b\times \frac{\sqrt{3}}{2} +
c\times 1$
- Height: $d\times \frac{1}{2} + e\times \frac{\sqrt{3}}{2}\times
\frac{\sqrt{3}}{2} + f\times 1$
So its area is: $$\left(\frac{a}{2} + b\frac{\sqrt{3}}{2} +
c\right)\times\left(\frac{d}{2} + e\frac{\sqrt{3}}{2} + f\right) =
\frac{ad}{4} + ae\frac{\sqrt{3}}{4}+ a\frac{f}{2} +
bd\frac{\sqrt{3}}{4} + be\frac{3}{4} + bf\frac{\sqrt{3}}{2} +
c\frac{d}{2} + ce\frac{\sqrt{3}}{2} + cf$$ If these two areas are
equal then the parts without $\sqrt{3}$ must cancel out. All of a,
b, c, d, e and f cannot be negative as they are the number of edges
so ad, af, be, cd, cf must be zero. If we take a to be zero then
the first two are zero, if c is zero then the last two are zero,
this leaves be. If b is zero then the base has no length so it must
be e that is zero. If you apply this to the expansion then you are
left with: $bd\frac{\sqrt{3}}{4}+ bf\frac{\sqrt{3}}{2}$ which could
equal $n\frac{\sqrt{3}}{8}$. But a square also has to have both
side lengths equal this means (with a,c and e zero) that:
$$b\frac{\sqrt{3}}{2} =\frac{d}{2} + f$$ This cannot be true as the
right hand side cannot become a multiple of root3. This can be done
with any selection being zero, but there is only one other possible
combination. This is b, f and d = zero. You end up with; $$a/2 + c
= e\frac{\sqrt{3}}{2}$$ which is again impossible.
Steven from City of Sunderland College sent
in a similar excellent solution. You can read his write-up in this
attached word
document.
Robert also gave a solution, and realised
that a very advanced (university-level) proof would require us to
justify the following results, for which he gave three good proofs.
This section would make good reading for any students considering
taking a mathematics degree.
In order to solve this problem I will first state and prove some
theorems concerning irrational numbers.
Theorem: $\sqrt{3}$ is
irrational :
Proof: Assume $\sqrt{3} =
a/b$ where $a$ and $b$ are integers and $a/b$ is a fraction in its
simplest terms. Then $$ 3 = a^2/b^2 \Rightarrow 3b^2 = a^2
\Rightarrow 3|a^2 \Rightarrow 3|a \Rightarrow 9|a^2 $$ So let $a^2
= 9m$ for some integer m. $\Rightarrow 3b^2 = 9m \Rightarrow b^2 =
3m \Rightarrow 3|b^2 \Rightarrow 3|b$ Therefore if 3 divides both a
and b, the fraction $a/b$ is not in its simplest form, which is a
contradiction. Thus $\sqrt{3}$ does not equal $a/b$ (for any
integers a or b) and so $\sqrt{3}$ is irrational.
Now I shall state and prove some theorems:
Theorem: An irrational number
multiplied by a rational number is irrational.
Proof: Let m be an
irrational number and let n be a rational number. Thus $n = a/b$,
for integers $a$ and $b$. So let us assume that $mn$ is rational
thus: $mn = c/d$ for some integers $c$ and $d$. Therefore, $$
(am)/b = c/d \Rightarrow am = (bc)/d \Rightarrow m = (bc)/(ad) $$
Because $bc$ and $ad$ will both be integers, say $e$ and $f$
respectively, then $(bc)/(ad) = e/f$, which is rational, however
$e/f$ also equals $m$ which is known to be irrational, which is a
contradiction. Therefore an irrational multiplied by a rational
must be irrational.
Theorem: A rational number divided
by an irrational number is irrational.
Proof: Let m be irrational and n be rational and let us assume that
$n/m = a/b$ where $a$ and $b$ are both integers. Now because $n$ is
rational we can denote it by $c/d$ where $c$ and $d$ are both
integers. Therefore, $$ c/(dm) = a/b \Rightarrow m = bc/ad $$ and
because $ad$ and $bc$ will both be integers say $e$ and $f$
respectively, then: $m = f/e$, which is a contradiction because m
is irrational and f/e is rational thus $n/m$ must be irrational.
Theorem: A rational number plus an
irrational number is irrational.
Proof: Let m be irrational
and n be rational, hence $n$ can be expressed as $a/b$ for integers
$a$ and $b$. Let us assume that $m$ plus $n$ is rational thus, $m +
n = c/d$, for integers $c$ and $d$ $\Rightarrow m + a/b = c/d
\Rightarrow m = (bc - ad)/(bd)$, which will be rational, but we
have said that $m$ is irrational so this a contradiction and thus
$(m + n)$ must be irrational.