# Incommensurability at an equilateral triangle

#### mathmari

##### Well-known member
MHB Site Helper
Hey!! I want to prove the incommensurability at an equilateral triangle.

The way of such a proof is the following:

1) We are looking for a "common measure" of two lines $a$, $b$, i.e. a line $e$ that measures both $a$ and $b$ integer (i.e. there are natural numbers $m$, $n$with $a = me, b = ne$).

2) We take the shorter of the two parts (e.g., $b$) away from the longer (e.g., $a$) until the remaining piece $r_1$ is shorter than $b$. We're taking away $r_1$ from $b$ as often as ... and so on.

3) If the process breaks down, there is a common measure, the stretches are then commensurate; otherwise incommensurable.

I found the proof (in german) here (pages 941-942) but I haven't really understood the method.

What I understand, is the following:

We consider an equilateral triangle ABC with side length $a$. Let CD be the height of the triangle from $C$.

We assume that $AD$ and $CD$ have a common $e$ measure.

Then, for appropriate $m$ and $n$: $AD = ne$ and $CD = me$.

Now we take the shorter of the two stretches, $AD$ off the longer, $CD$, until the remaining $r_1$ is shorter than $AD$.

To do this, we take a circle around the point $D$ with radius $AD$, so this intersects the height $CD$ at point $E$.

We have that $CE=CD-ED=CD-AD=a-b$

We draw then a circle with center $E$ and radius $CE=a-b$. This intersects the segment $CA$ at the point $F$.

We have that $EF=CE=a-b$.

We draw a parallel line to $BC$ from $F$. This intersects the segment $AD$ at $A_1$.

The triangle $AFA_1$ is equilateral, since $FA_1$ is parallel to $BC$ and so all the angles of $AFA_1$ are equal to $60^{\circ}$.

We construct an equilateral triangle $A_1B_1C_1$ with side length $2\cdot A_1D$.

Is everything correct so far? Or have I understood the construction wrong? What I don't understand how we can determine the length of the sides of the new triangle. Could you explain to me how we can find these values? #### Klaas van Aarsen

##### MHB Seeker
Staff member
What I understand, is the following:

We consider an equilateral triangle ABC with side length $a$. Let CD be the height of the triangle from $C$.
Can it be that it should be side length $2b$ and height $a$? We assume that $AD$ and $CD$ have a common $e$ measure.

Then, for appropriate $m$ and $n$: $AD = ne$ and $CD = me$.

Now we take the shorter of the two stretches, $AD$ off the longer, $CD$, until the remaining $r_1$ is shorter than $AD$.

To do this, we take a circle around the point $D$ with radius $AD$, so this intersects the height $CD$ at point $E$.

We have that $CE=CD-ED=CD-AD=a-b$

We draw then a circle with center $E$ and radius $CE=a-b$. This intersects the segment $CA$ at the point $F$.

We have that $EF=CE=a-b$.

We draw a parallel line to $BC$ from $F$. This intersects the segment $AD$ at $A_1$.

The triangle $AFA_1$ is equilateral, since $FA_1$ is parallel to $BC$ and so all the angles of $AFA_1$ are equal to $60^{\circ}$.
I don't think $AFA_1$ is equilateral.
And $A_1$ is not in the segment $AD$ either. Can it be that that we should draw a line parallel to $AC$ from $F$ that intersects the segment $AD$ at $A_1$?
And a line parallel to $BC$ from $F$ that intersects the segment $DB$ at $B_1$?
If so then it does follow that $A_1FB_1$ is equilateral. We construct an equilateral triangle $A_1B_1C_1$ with side length $2\cdot A_1D$.

Is everything correct so far? Or have I understood the construction wrong?
If we additionally let $C_1$ be the same point as $F$, then we have an equilateral triangle $A_1B_1C_1$ with side length $2\cdot A_1D$. What I don't understand how we can determine the length of the sides of the new triangle. Could you explain to me how we can find these values? I think we don't need to.
It suffices that after 2 iterations we have a strictly smaller equilateral triangle that has a positive side length.
We can now repeat the argument ad infinitum can't we?
And we'll never reach zero will we? #### mathmari

##### Well-known member
MHB Site Helper
I don't think $AFA_1$ is equilateral.
And $A_1$ is not in the segment $AD$ either. Why is $A_1$ not in $AD$ ?

Can it be that that we should draw a line parallel to $AC$ from $F$ that intersects the segment $AD$ at $A_1$?
And a line parallel to $BC$ from $F$ that intersects the segment $DB$ at $B_1$?
How can we draw a line parallel to $AC$ from $F$ ? Isn't $F$ on $AC$ ? Or do you mean from $E$ instead of $F$ ?

It suffices that after 2 iterations we have a strictly smaller equilateral triangle that has a positive side length.
We can now repeat the argument ad infinitum can't we?
And we'll never reach zero will we? Why will we not reach zero? #### Klaas van Aarsen

##### MHB Seeker
Staff member
Why is $A_1$ not in $AD$ ?
Because:
\begin{tikzpicture}[scale=3]
\coordinate[label=left:A] (A) at (-1,0);
\coordinate[label=right:B] (B) at (1,0);
\coordinate[label=C] (C) at (0,{sqrt(3)});
\coordinate[label=below: D] (D) at (0,0);
\coordinate[label=left:E] (E) at (0,1);
\coordinate[label=left:F] (F) at (0,{2-sqrt(3)});
\coordinate[label=below:$A_1$] (A1) at ({2/sqrt(3)-1},0);
\draw (A) -- (B) -- (C) -- cycle;
\draw (D) -- (C);
\draw[fill] (E) circle (.01);
\draw (F) -- (A1);
\end{tikzpicture}
Or did I misunderstand? How can we draw a line parallel to $AC$ from $F$ ? Isn't $F$ on $AC$ ? Or do you mean from $E$ instead of $F$ ?
Because:
\begin{tikzpicture}[scale=3]
\coordinate[label=left:A] (A) at (-1,0);
\coordinate[label=right:B] (B) at (1,0);
\coordinate[label=C] (C) at (0,{sqrt(3)});
\coordinate[label=below: D] (D) at (0,0);
\coordinate[label=left:E] (E) at (0,1);
\coordinate[label=left:F] (F) at (0,{2-sqrt(3)});
\coordinate[label=below:$A_1$] (A1) at ({1-2/sqrt(3)},0);
\coordinate[label=below:$B_1$] (B1) at ({2/sqrt(3)-1},0);
\draw (A) -- (B) -- (C) -- cycle;
\draw (D) -- (C);
\draw[fill] (E) circle (.01);
\draw (F) -- (A1);
\draw (F) -- (B1);
\end{tikzpicture}

Why will we not reach zero?
We have a new equilateral triangle that has non-zero size.
After 2 more iterations, we will have yet again a new equilateral triangle that has non-zero size, won't we? MHB Site Helper

#### Klaas van Aarsen

##### MHB Seeker
Staff member
I thought that we get the following construction:

Isn't this correct? Ah okay. My mistake. So we have:
\begin{tikzpicture}[scale=3]
\coordinate[label=left:A] (A) at (-1,0);
\coordinate[label=right:B] (B) at (1,0);
\coordinate[label=C] (C) at (0,{sqrt(3)});
\coordinate[label=below: D] (D) at (0,0);
\coordinate[label=right:E] (E) at (0,1);
\coordinate[label=left:F] (F) at ({-1/sqrt(3)},{sqrt(3)-1});
\coordinate[label=right:$C_1$] (C1) at (0,{2-sqrt(3)});
\coordinate[label=below:$A_1$] (A1) at ({1-2/sqrt(3)},0);
\coordinate[label=below:$B_1$] (B1) at ({2/sqrt(3)-1},0);
\draw (A) -- (B) -- (C) -- cycle;
\draw (D) -- (C);
\draw[fill] (E) circle (.01);
\draw (F) -- (A1);
\draw (C1) -- (A1);
\draw (C1) -- (B1);
\end{tikzpicture}
don't we? #### mathmari

##### Well-known member
MHB Site Helper
Ah okay. My mistake. So we have:
\begin{tikzpicture}[scale=3]
\coordinate[label=left:A] (A) at (-1,0);
\coordinate[label=right:B] (B) at (1,0);
\coordinate[label=C] (C) at (0,{sqrt(3)});
\coordinate[label=below: D] (D) at (0,0);
\coordinate[label=right:E] (E) at (0,1);
\coordinate[label=left:F] (F) at ({-1/sqrt(3)},{sqrt(3)-1});
\coordinate[label=right:$C_1$] (C1) at (0,{2-sqrt(3)});
\coordinate[label=below:$A_1$] (A1) at ({1-2/sqrt(3)},0);
\coordinate[label=below:$B_1$] (B1) at ({2/sqrt(3)-1},0);
\draw (A) -- (B) -- (C) -- cycle;
\draw (D) -- (C);
\draw[fill] (E) circle (.01);
\draw (F) -- (A1);
\draw (C1) -- (A1);
\draw (C1) -- (B1);
\end{tikzpicture}
don't we? Yes! The triangle $AFA_1$ is equilateral, isn't it?

Can we calculate the length of the sides of the new triangle $A_1B_1C_1$ ? #### Klaas van Aarsen

##### MHB Seeker
Staff member
Yes! The triangle $AFA_1$ is equilateral, isn't it?

Can we calculate the length of the sides of the new triangle $A_1B_1C_1$ ? Suppose we divide CEF into 2 right triangles.
Then we get:
\begin{tikzpicture}[scale=3]
\coordinate[label=left:A] (A) at (-1,0);
\coordinate[label=right:B] (B) at (1,0);
\coordinate[label=C] (C) at (0,{sqrt(3)});
\coordinate[label=below: D] (D) at (0,0);
\coordinate[label=right:E] (E) at (0,1);
\coordinate[label=left:F] (F) at ({-(3-sqrt(3))/2},{(3-sqrt(3))/2});
\coordinate[label=left:G] (G) at ({-(3-sqrt(3))/4},{(3+sqrt(3))/4});
%\coordinate[label=right:$C_1$] (C1) at (0,{2-sqrt(3)});
\coordinate[label=below:$A_1$] (A1) at ({-(2-sqrt(3))},0);
%\coordinate[label=below:$B_1$] (B1) at ({2/sqrt(3)-1},0);
\draw (A) -- (B) -- (C) -- cycle;
\draw (D) -- (E) -- node
{$\sqrt 3 - 1$} (C);
\draw (E) -- node[below,xshift=4] {$\sqrt 3 - 1$} (F) -- (A1);
\draw (E) -- (G);
\path (C) -- node
{$\frac 12(3-\sqrt 3)$} (G) -- node
{$\frac 12(3-\sqrt 3)$} (F) -- (A);
\end{tikzpicture}

Now we can calculate $AF$ and $A_1D$ can't we? 