Are you sure that’s a right angle?

What's that, @pickover? Shiver in ecstasy, you say? Just for a change.

shapes

That's neat. But why?

Let's suppose the circles all have radius 1, without loss of generality. Then the triangle's side lengths are (in decreasing order) $2\sin\left( \piby 5 \right)$, $2\sin\left( \piby 6 \right)$ and $2\sin\left( \piby {10} \right)$.

Those will form a right-angled triangle if $\sin^2\left( \piby 5 \right) = \sin^2\left( \piby 6 \right) + \sin^2\left( \piby {10} \right)$. And… well, is it?

Let's consider $Z = \sin^2\left( \piby 5 \right) – \sin^2\left( \piby 6 \right) – \sin^2\left( \piby {10} \right)$, which ought to end up as 0 to give us a right angle.

My first thought is, I don't much like those $\sin^2$s. They'd be much nicer if they were $\cos$s. Luckily, there's an identity: $2\sin^2(x) \equiv 1- \cos(2x)$.

If we double everything, that makes it a bit easier:

$Z = 1 – \cos\left(\frac{2}{5}\pi\right) – 2 + \cos\left(\piby 3\right) + \cos\left(\piby 5 \right)$

And of course, we know that $\cos\left(\piby 3\right) = \frac {1}{2}$, so we can tidy up:

$Z = -\frac{1}{2} + \cos\left(\piby 5 \right) – \cos\left(\frac{2}{5}\pi\right)$. (*)

Now for the cosines

Now let's have a look at $\cos\left(\piby 5 \right) – \cos\left(\frac{2}{5}\pi\right)$. I shall call that $H$.

Using the cosine difference formula, $H = 2 \cos\left(\frac{2}{5}\pi\right)\cos\left(\piby 5\right)$.

Squaring the original $H$ gives $H^2 = \cos^2\left(\piby 5\right) – H + \cos^2\left(\frac{2}{5}\pi\right)$.

Doubling everything, as is normal procedure when $\cos^2$s show up: $2H^2 + 2H = 2\cos^2\left(\piby 5\right) + 2\cos^2\left(\frac{2}{5}\pi\right)$.

However, we know that $2\cos^2(x) = \cos(2x)+1$, which gives:

$2H^2 + 2H = \left(1+ \cos\left(\frac{2}{5}\pi\right)\right) + \left(1 + \cos\left(\frac{4}{5}\pi\right)\right)$.

Meanwhile, $\cos\left(\frac{4}{5}\pi\right) = – \cos\left(\piby 5\right)$, so the right hand side becomes $2 – H$.

We end up with $2H^2 + 3H – 2 = 0$, which factorises nicely as $(2H-1)(H+2)=0$, so $H = \frac{1}{2}$ or $H= -2$. Since the component parts of $H$ are both smaller in magnitude than 1, $H = \frac{1}{2}$ is the only sensible answer.

Argh! A Ninja!

“Roots of unity!” hissed the Mathematical Ninja. “$2\cos\left(\frac{2}{5}\pi\right) + 2\cos\left(\frac{4}{5}\pi\right) + 1 = 0$ so $2\left( \cos\left(\frac{2}{5}\pi\right) – \cos\left(\frac{1}{5}\pi\right)\right) = -1$. That’s $-2H = -1$. Would you like a tombstone?”

“Uh…”

“$\blacksquare$.”

“Thank you.”

So, where were we?

Oh yeah! Going back to (*), we've got $Z = -\frac{1}{2} + H$, so $Z = 0$ and our triangle is right-angled. Phew!

That might make one shiver in ecstasy, I suppose, if one were so inclined.

* Many thanks to @dragon_dodo who got the hard bit of the proof and then invited the Ninja in.

Colin

Colin is a Weymouth maths tutor, author of several Maths For Dummies books and A-level maths guides. He started Flying Colours Maths in 2008.

He lives with an espresso pot and nothing to prove.

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