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60 degree triangle preserving matrices.


The 120 degree triangle preserving matrix

$\displaystyle \mathbf{M}=\begin{pmatrix}
4 & 3 & 6\\
3 & 4 & 6\\
4 & 4 & 7
\end{pmatrix}$

we found in the previous section, can be transformed into four 60 degree triangle preserving matrices. To do so, we will need the following identity,

$\displaystyle x^2+y^2+xy=(x+y)^2+y^2-(x+y)y=x^2+(x+y)^2-x(x+y).$ (40)


We have $ a^2+b^2-ab=(-a)^2+b^2+(-a)b=c^2$ . Then using $ \mathbf{M}$ ,

$\displaystyle \begin{pmatrix}
-a&b&c
\end{pmatrix}\begin{pmatrix}
4 & 3 & 6\\
3 & 4 & 6\\
4 & 4 & 7
\end{pmatrix} =
\begin{pmatrix}
u &v & w
\end{pmatrix}$

where $ u^2+v^2+uv=w^2$ . However, from equation (40), $ (u+v)^2+v^2-(u+v)v=w^2$ . Hence

$\displaystyle \begin{pmatrix}
-a&b&c
\end{pmatrix}\begin{pmatrix}
4+3 & 3 & 6\\...
...nd{pmatrix}\begin{pmatrix}
-7 & -3 & -6\\
7 & 4 & 6\\
8 & 4 & 7
\end{pmatrix}$

where $ a^2+b^2-ab=c^2$ is a 60 degree triangle. Similarly, if $ a_1^2+(-b_1)^2+a_1(-b_1)=c_1^2$ , then

$\displaystyle \begin{pmatrix}
a_1&b_1&c_1
\end{pmatrix}\begin{pmatrix}
7 & 3 &6...
...-6\\
8 & 4 & 7
\end{pmatrix}=
\begin{pmatrix}
u_1+v_1 &v_1 & w_1
\end{pmatrix}$

where $ a_1^2+b_1^2-a_1 b_1=c_1^2$ and $ (u_1+v_1)^2+v_1^2-(u_1+v_1)v_1=w_1^2$ .


Let

$\displaystyle \boxed{\mathbf{S}_1=\begin{pmatrix}
7 & 3 &6\\
-7 & -4 & -6\\
8...
...f{T}_1=
\begin{pmatrix}
-7 & -3 & -6\\
7 & 4 & 6\\
8 & 4 & 7
\end{pmatrix}.}
$

Then, clearly,

$\displaystyle \boxed{\mathbf{S}_2=\begin{pmatrix}
4 & 7 &6\\
-3 & -7 & -6\\
4...
...bf{T}_2=
\begin{pmatrix}
-4 & -7 & -6\\
3 & 7 & 6\\
4 & 8 & 7
\end{pmatrix}}
$

are also 60 degree triangle preserving matrices.



For example, since $ 8^2+3^2-(8)(3)=7^2$ ,

$\displaystyle \begin{pmatrix}
8 & 3 & 7
\end{pmatrix}\mathbf{S}_1=
\begin{pmatr...
...& 3 & 7
\end{pmatrix}\mathbf{S}_2=
\begin{pmatrix}
51 & 91 & 79
\end{pmatrix},
$

$\displaystyle \begin{pmatrix}
8 & 3 & 7
\end{pmatrix}\mathbf{T}_1=
\begin{pmatr...
...
8 & 3 & 7
\end{pmatrix}\mathbf{T}_2=
\begin{pmatrix}
5 & 21 & 19
\end{pmatrix}$

where

$\displaystyle 91^2+40^2-(91)(40)=79^2,\qquad 51^2+91^2-(51)(91)=79^2,
$

$\displaystyle 21^2+16^2-(21)(16)=19^2,\qquad 5^2+21^2-(5)(21)=19^2.
$

Download these 3 sections in pdf form. triples8.pdf


next up previous contents
Next: 120 degree triples and Up: 120 degree primitive triples,(a,b,c), Previous: A 120 degree triangle   Contents
f. barnes 2008-04-29
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