Irrational Constructions
What questions does the spiral construction prompt you to ask?
Problem
Take a look at the image below.
Image
What do you make of it?
What questions might you ask yourself about this image?
Take a look at the hint to see the questions we thought of.
This is an Underground Mathematics resource.
Underground Mathematics is hosted by Cambridge Mathematics. The project was originally funded by a grant from the UK Department for Education to provide free web-based resources that support the teaching and learning of post-16 mathematics.
Visit the site at undergroundmathematics.org to find more resources, which also offer suggestions, solutions and teacher notes to help with their use in the classroom.
Underground Mathematics is hosted by Cambridge Mathematics. The project was originally funded by a grant from the UK Department for Education to provide free web-based resources that support the teaching and learning of post-16 mathematics.
Visit the site at undergroundmathematics.org to find more resources, which also offer suggestions, solutions and teacher notes to help with their use in the classroom.
Getting Started
Here are some questions we asked about this image that you might like to think about, if you have not already considered them.
- Can I recreate the image using only straight-edge and compasses?
- Can I find line segments of lengths $2$, $3$, $4$, $\dots$?
- Are there any lengths that do not occur in the diagram?
- How many steps would I need to construct $\sqrt{5}$ in the diagram?
- Can I construct $\sqrt{13}$ in any other way (using straight-edge and compasses)?
- Which other lengths can I construct in other ways (using straight-edge and compasses)?
- How will the spiral change if I start with a $2\times 2$ right-angled triangle instead?
- How will the spiral change if I start with a right-angled triangle that isn't isosceles?
Student Solutions
Demirhan from Dubai British School has made a good start on this investigation by finding formulae for the side lengths of the Nth triangle in the spiral:
In the irrational construction problem I took into fact that the bottom side of the triangle will always be 1cm as shown in the picture. While the other side which isn't the hypotenuse on the first triangle was 1cm. Using the Pythagoras theorem ($A^2+B^2=C^2$) I figured out the first hypotenuse of the triangle as $\sqrt{1+1}$, which is $\sqrt{2}$. The hypotenuse of one triangle is the same as the non-hypotenuse side of the triangle after it. So the hypotenuse of the second triangle is going to be $\sqrt{2+1} = \sqrt{3}$. So the hypotenuse of the third triangle is going to be $\sqrt{3+1}=\sqrt{4}$.
In general, If $N$ is the number for the consecutive triangle then:
The Hypotenuse of the triangle will be $\sqrt{N+1}$
So this shows we can find any length of the form $\sqrt{N}$ in the diagram!
Sergio from Kings College of Alicante has suggested another way to construct $\sqrt{13}$
If 13 is the hypotenuse of the triangle and a,b are the two sides, following pythagoras theory then $a$ squared plus $b$ square must add up to to 13. I looked at sqaure numbers and I simply figured which number a and b had to be in seconds: $a=2$ and $b=3$.
Pablo from Kings College of Alicante has found a way to construct this diagram exactly using only a straight-edge and compass, which is explained here.
Image