Written by Colin+ in geometry, ninja maths, quadratics.

Dear Uncle Colin,

I've got a funny square and I can't find $x$. Can you help?

– Oughta Be Simple, Can't Unravel Resulting Equations

Hi, OBSCURE, and thanks for your message! You're right, it ought to be simple… but it turns out not to be.

It *is* simple enough to set up some equations.

The sine of the 'big' triangle's bottom-right angle is clearly $\frac{1}{x+1}$.

The cosine of the top-left angle of the medium triangle (the one with the right-angle in the top-right, and which is similar to the big triangle) is $\frac{1}{x}$.

Those two angles are the same, which means $\left(\frac{1}{x+1}\right)^2 + \left( \frac{1}{x} \right)^2 = 1$.

Multiplying everything by the denominators gives:

$x^2 + (x^2 + 2x + 1) = x^2(x^2 + 2x+ 1)$

This simplifies to

$0 = x^4 + 2x^3 – x^2 – 2x – 1$.

Ah, a quartic. My old nemesis! We meet again.

I don't know (apart from numerically and via Wolfram) how to solve quartics, although I ought to. However, I have an idea: how about I try to get rid of the $x^3$ term by using a substitution?

If I let $x = y – \frac{1}{2}$, what happens? Well, after a horrible bit of algebra, I get $y^4 – \frac{5}{2}y^2 – \frac{7}{16} = 0$. What do you know? It turned into a quadratic!

Multiplying by 16 for aesthetic reasons gives $16y^4 – 40y^2 – 7 = 0$. Completing the square tells me $(4y^2 – a)^2 + b \equiv 16y^2 – 40y – 7$, so $16y^2 – 8a + a^2 + b \equiv 16y^4 – 40y^2 – 7$. In the end, $a=5$ and $b=-32$.

So, $(4y^2-5)^2 – 32 = 0$, which gives $(4y^2-5) = \pm 4\sqrt{2}$, or $y = \pm \sqrt{ \frac{5}{4} \pm \sqrt{2}}$. For the second $\pm$, only the positive root makes sense, since $\frac{5}{4} < \sqrt{2}$.

Taking away the half again gives $x = -\frac{1}{2} \pm

\sqrt{ \frac{5}{4} + \sqrt{2}}$. For the remaining $\pm$, only the positive root makes sense (or else we'd have a negative value for a distance.)

We end up with $x = -\frac{1}{2} + \sqrt{ \frac{5}{4} + \sqrt{2}}$ as the only positive real root of the equation – about 1.13, which looks plausible.

“If you write that as $2x = -1 + \sqrt{5 + 4\sqrt{2}}$, it’s rather neater. $4\sqrt{2}$ is about 5.65, so we need the square root of $10.65$. That’s obviously about 3.25…”

“Obviously… ow.”

“More accurately, you say? 10.24 is $3.2^2$, so $3.25^2$ is 10.5625. Error is about 0.09, divided by 6.5, 0.18 over 13 is 0.01385 or so. So 3.264 seems better. Can I carry on now?”

“Yes, sensei.”

“Take away 1 to get 2.264 and halve the result. Hey presto: 1.132, as prophesied.”

Whoosh.

Phew! I hope that helps!

– Uncle Colin