A few weeks ago I remembered a puzzle I once saw known as the crossed ladders theorem. The way I remembered was:
I have two poles sticking out of the ground. One is 4m tall and the other is 6m tall. I take a piece of rope and connect the top of one tree to the bottom of the other. I then take another piece of rope and connect the other top to the other other bottom. How far off the ground is the intersection of the two pieces of rope?
I won't go through the solution here but the answer is 2.4m. More generally if we have two poles of lengths $x$ and $y$ then the intersection will be at $\frac{xy}{x+y}.$ This means that the length $\frac{xy}{x+y}$ is constructable just from knowing the lengths $x$ and $y$. This slightly surprised me. Usually when you multiply lengths together using a compass and straightedge you need to know how "big" 1 is. i.e. you need a line which represents how big the unit is. Its quite obvious this is the case when you think about squaring a given length. If you square something less than one then it decreases but if you square something greater than one then it increases. So without knowing if your length is bigger or smaller than 1 you can't really do anything.
My intuition as to why this is the case is because $\frac{xy}{x+y}$ is "linear". By that I mean it is a degree 2 polynomial over a degree 1 polynomial. This led me to wonder which other rational functions can be constructed. After a little bit of thinking I realised that $\frac{xy+y}{x+y}$ is not constructible. So I need more than just the degree of the numerator being one more than that of the denominator. Maths Jam Stephen suggested that the polynomials be homogenous. So I wrote down a simple homogenous polynomial and gave it a go:
$\frac{x^2+y^2}{x+y}=\frac{(x+y)^{2}}{x+y}=x+y-2\frac{xy}{x+y}$
which is constructible. Just to recap a polynomial is homogeneous if each term in the polynomial has the same degree.
This led me to make the following conjecture:
Let $x_{1},...,x_{n}$ be $n$ length in the plane. Let $P$ be a homogeneous polynomial of degree $d+1$ and let $Q$ be a homogeneous polynomial of degree $d$ both of $n$ variables. Then
$$\frac{P(x_{1},...,x_{n})}{Q(x_{1},...,x_{n})}$$
is constructible without needing a copy of the unit.
The rest of this post will be quite odd. I currently have no proof of this claim and will be updating this post as I try and prove it. Either in one big swoop or in partial steps. Whatever happens.
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Cp has suggested I write down a degree 3 over a degree 2 and try and construct it. I am going with
$$\frac{x^{3}+y^{3}}{x^{2}+y^{2}}.$$
Turns out my conjecture is false!!! I think. Consider the polynomials $P(x,y,z)=\frac{x^{2}+yz}$ and $Q(x,y,z)=zx.$ Then $\frac{P}{Q}$ is not constructible without knowing your unit.
I need to rethink about how to phrase the problem now :(
For now that can be the problem:
What is the set of rational functions which are constructible without knowing your unit?
---------------------------------
Pretty sure that the polynomials need to be symmetric and that that may also be sufficient.
I have two poles sticking out of the ground. One is 4m tall and the other is 6m tall. I take a piece of rope and connect the top of one tree to the bottom of the other. I then take another piece of rope and connect the other top to the other other bottom. How far off the ground is the intersection of the two pieces of rope?
I won't go through the solution here but the answer is 2.4m. More generally if we have two poles of lengths $x$ and $y$ then the intersection will be at $\frac{xy}{x+y}.$ This means that the length $\frac{xy}{x+y}$ is constructable just from knowing the lengths $x$ and $y$. This slightly surprised me. Usually when you multiply lengths together using a compass and straightedge you need to know how "big" 1 is. i.e. you need a line which represents how big the unit is. Its quite obvious this is the case when you think about squaring a given length. If you square something less than one then it decreases but if you square something greater than one then it increases. So without knowing if your length is bigger or smaller than 1 you can't really do anything.
My intuition as to why this is the case is because $\frac{xy}{x+y}$ is "linear". By that I mean it is a degree 2 polynomial over a degree 1 polynomial. This led me to wonder which other rational functions can be constructed. After a little bit of thinking I realised that $\frac{xy+y}{x+y}$ is not constructible. So I need more than just the degree of the numerator being one more than that of the denominator. Maths Jam Stephen suggested that the polynomials be homogenous. So I wrote down a simple homogenous polynomial and gave it a go:
$\frac{x^2+y^2}{x+y}=\frac{(x+y)^{2}}{x+y}=x+y-2\frac{xy}{x+y}$
which is constructible. Just to recap a polynomial is homogeneous if each term in the polynomial has the same degree.
This led me to make the following conjecture:
Let $x_{1},...,x_{n}$ be $n$ length in the plane. Let $P$ be a homogeneous polynomial of degree $d+1$ and let $Q$ be a homogeneous polynomial of degree $d$ both of $n$ variables. Then
$$\frac{P(x_{1},...,x_{n})}{Q(x_{1},...,x_{n})}$$
is constructible without needing a copy of the unit.
The rest of this post will be quite odd. I currently have no proof of this claim and will be updating this post as I try and prove it. Either in one big swoop or in partial steps. Whatever happens.
-------------------------------------------------------------------
Cp has suggested I write down a degree 3 over a degree 2 and try and construct it. I am going with
$$\frac{x^{3}+y^{3}}{x^{2}+y^{2}}.$$
Turns out my conjecture is false!!! I think. Consider the polynomials $P(x,y,z)=\frac{x^{2}+yz}$ and $Q(x,y,z)=zx.$ Then $\frac{P}{Q}$ is not constructible without knowing your unit.
I need to rethink about how to phrase the problem now :(
For now that can be the problem:
What is the set of rational functions which are constructible without knowing your unit?
---------------------------------
Pretty sure that the polynomials need to be symmetric and that that may also be sufficient.
