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Incommensurability at an equilateral triangle

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
3,964
Hey!! :eek:

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? (Wondering)

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? (Wondering)
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,625
Leiden
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 $a$? (Wondering)

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. (Worried)

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. (Thinking)

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 $2\cdot A_1D$. (Thinking)

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? (Wondering)
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? (Thinking)
 

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
3,964
I don't think $AFA_1$ is equilateral.
And $A_1$ is not in the segment $AD$ either. (Worried)
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? (Thinking)
Why will we not reach zero?


(Wondering)
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,625
Leiden
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? (Wondering)

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? (Wondering)
 

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
3,964
I thought that we get the following construction:

incommensurability.JPG

Isn't this correct? (Wondering)
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,625
Leiden
I thought that we get the following construction:

Isn't this correct? (Wondering)
Ah okay. My mistake. (Blush)

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? (Wondering)
 

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
3,964
Ah okay. My mistake. (Blush)

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? (Wondering)
Yes! (Nod)


The triangle $AFA_1$ is equilateral, isn't it?

Can we calculate the length of the sides of the new triangle $A_1B_1C_1$ ? (Wondering)
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,625
Leiden
Yes! (Nod)

The triangle $AFA_1$ is equilateral, isn't it?

Can we calculate the length of the sides of the new triangle $A_1B_1C_1$ ? (Wondering)
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? (Wondering)