## A circle inscribing a pentagon

This is also from The Irrationals – though I had to ask for assistance over at Stack Exchange to get the right answer (as is so often the case the solution is reasonably obvious when you are presented with it).

Anyway, the question is: given a regular pentagon (of sides with length of 10 units) which is inscribed by a circle, what is the diameter of the circle?

This figure helps illustrate the problem: We are trying to find $2R$ and we know that $x=10$.

If we knew $r$ then we could answer as $R^2 = r^2 + (\frac{x}{2})^2$. We do not know that but we do know that $\sin(\frac{\pi}{5})=\frac{x}{2R}$ and hence, from the $\cos^2 + \sin^2 = 1$ identity: $\sin\frac{\pi}{5} = \frac{10}{2R} = \sqrt{1-\cos^2(\frac{\pi}{5})}$.

From our knowledge of the pentagon with sides of unit length (you’ll have to trust me on this or look it up – it’s too much extra to fit in here) we also know that $\cos(\frac{\pi}{5}) = \frac{\phi}{2} = \frac{1}{4}(1+\sqrt{5})$, where $\phi$ is the golden ratio.

Hence $2R =$ … well, the rest is left as an exercise for the reader 🙂

Posted on

## The proof that frightened Pythagoras

I have been meaning to do this for a while …

Pythagoras was said to be shocked by the proof that $\sqrt 2$ is an irrational number (i.e., that it cannot be represented as a ratio of two whole numbers). The proof (by contradiction) is a simple one.

If $\sqrt 2$ is rational then $\sqrt 2 = \frac{a}{b}$ and $2 = \frac{a^2}{b^2}$ where $a$ and $b$ are the smallest nominator and denominator possible, i.e, in the lowest terms.

Hence $a^2 = 2b^2$ and so $a$ must be even (as two odd numbers multiplied always give an odd number – let $l = 2x$ and $m = 2y$ be even numbers then $(l + 1)(m + 1) = lm + l + m + 1$). Hence $a = 2k$.

But we also have $4k^2 = a^2$ and thus $4k^2 = 2b^2$ and $b^2 = 2k^2$ and so $b$ is even.

So we have $a$ and $b$ sharing a common factor of 2, so they cannot be the nominator and denominator of the fraction in the lowest terms.

But if $a$ and $b$ are both even then they share a common factor of 2 and $\frac{a}{b} = \frac{2k}{2p} = \frac{k}{p}$: implying that $a = k$ and $a = 2k$, an obvious contradiction: hence $\sqrt 2$ cannot be a rational.

Update: I have made the final step of this shorter and clearer.

Further update: I have been told (see comments below) I would have been better sticking with a clearer version of the original ending ie., we state that $\frac{a}{b}$ are the lowest terms. Then, plainly as they have a factor in common (2) they cannot be the lowest terms and so we have a contradiction. Would be great if someone could explain why we cannot use the $a = 2k = k$ contradiction.

And another update: Should have stuck with the original explanation – which I have now restored in a hopefully clearer way. The comments below are really interesting and from serious mathematicians, so please have a look!