Show the constructibility of $\sqrt{\sqrt{2}+5}$ with a diagram

Question

I am trying to show that $\sqrt{\sqrt{2}+5}$ is constructible through a diagram.

I know how to show something of the form $\sqrt[n]{a}$ is constructible through a diagram, but I am really having a difficult time with this one.

Any tips? Thanks.

Answer 1

The key here is that if you can construct two numbers, you can construct their sum by laying two line segments end to end, and if you can construct some length, then you can construct its square root as follows:

Consider a segment $AB$ of length $a$. On the line through $AB$, mark a point $C$ such that $B$ is between $A$ and $C$ and $BC$ has length $1$. Draw a circle with diameter $AC$. Then the perpendicular to the line $AC$ through $B$ intersects the circle in two points. Call one of these points $D$. Then by playing with similar triangles, you can see that $BD$ has length $\sqrt{a}$.

With this in mind, construct a segment of length $\sqrt 2$ (either by using a right triangle or using this construction), add to it 5 unit length segments, and then use this construction to take the square root.

I should note that you shouldn't be able to take $n$th roots unless $n$ is a power of $2$, which you can do by taking square roots repeatedly.

Answer 2

Well, consider rewriting the expression:

$$\sqrt{\sqrt{2}+5} = \sqrt{(2^{\frac{1}{4}})^2 + (\sqrt{5})^2}$$

You can construct $2^{\frac{1}{4}}$ and you can construct $\sqrt{5}$. Then, turn them into the sides of a right-angled triangle and you get the hypotenuse as your construction for this expression.

Answer 3

Visualization of Aaron's answer.

If $AB$ is of length $1$, then $IQ$ is of length $\sqrt{\sqrt2+5}$.

Coloured lines represent constructions of perpendicular bisectors.

We see that $KI = 1$, $IC = 5$, $JC = CF = \sqrt2$, and $KN = NJ$.

Finally $\triangle KIQ \sim \triangle QIJ$.

I believe Abhi's answer gives a more compact diagram but is harder to unravel.