A Guide to Growing Vegetables in Containers

[Albert1]

View attachment 3307

 

[anemone]

I'm thinking perhaps you have another elegant solution at hand but I'm going to post my "weak" solution here nevertheless:

Spoiler

First we let $\angle AED=\alpha$, and $DE=2k,\,AB=k$.

Applying the Sine Rule on both the triangles $AED$ and $ABE$ we see that

$\dfrac{2k}{\sin 90^{\circ}}=\dfrac{AE}{\sin (90-\alpha)^{\circ}}$ and $\dfrac{k}{\sin \alpha}=\dfrac{AE}{\sin (\alpha-18)^{\circ}}$

Dividing the first by the second equation yields

$\begin{align*}\sin (\alpha-18)^{\circ}&=2\sin \alpha \sin (90-\alpha)^{\circ}\\&=2\sin \alpha \cos \alpha\\&=\sin 2\alpha\end{align*}$

Since $0<2\alpha<180^{\circ}$, we must alter the expression inside the sine function of $\sin (\alpha-18)^{\circ}$ such that we have

$\sin (180-(\alpha-18))^{\circ}=\sin 2\alpha$

Solving it for $\alpha$ we obtain $\alpha=66^{\circ}$.

 

[Albert1]

I'm thinking perhaps you have another elegant solution at hand but I'm going to post my "weak" solution here nevertheless:

Spoiler

First we let $\angle AED=\alpha$, and $DE=2k,\,AB=k$.

Applying the Sine Rule on both the triangles $AED$ and $ABE$ we see that

$\dfrac{2k}{\sin 90^{\circ}}=\dfrac{AE}{\sin (90-\alpha)^{\circ}}$ and $\dfrac{k}{\sin \alpha}=\dfrac{AE}{\sin (\alpha-18)^{\circ}}$

Dividing the first by the second equation yields

$\begin{align*}\sin (\alpha-18)^{\circ}&=2\sin \alpha \sin (90-\alpha)^{\circ}\\&=2\sin \alpha \cos \alpha\\&=\sin 2\alpha\end{align*}$

Since $0<2\alpha<180^{\circ}$, we must alter the expression inside the sine function of $\sin (\alpha-18)^{\circ}$ such that we have

$\sin (180-(\alpha-18))^{\circ}=\sin 2\alpha$

Solving it for $\alpha$ we obtain $\alpha=66^{\circ}$.

your solution is also very nice!

 

[Albert1]

View attachment 3307

Spoiler

View attachment 3329

 

[johng1]

Albert.
I found your parallelogram very interesting. You may well have observed this, but here is something. If angle ABC in the parallelogram is "arbitrary" \(\displaystyle \theta\), the parallelogram is constructible with ruler and compass iff \(\displaystyle {\theta\over 3}\) is constructible. Shown is an angle of 60 degrees. Of course, a 20 degree angle is not constructible.View attachment 3336

 

[Albert1]

Albert.
I found your parallelogram very interesting. You may well have observed this, but here is something. If angle ABC in the parallelogram is "arbitrary" \(\displaystyle \theta\), the parallelogram is constructible with ruler and compass iff \(\displaystyle {\theta\over 3}\) is constructible. Shown is an angle of 60 degrees. Of course, a 20 degree angle is not constructible.https://www.physicsforums.com/attachments/3336

yes it is very interesting
given $\angle ABC=72^o$
using compass and ruler only ,
Can we construct an angle=$24^o ?$

 

[johng1]

Albert,
Years ago I taught constructibility of numbers from an algebraic standpoint (degree of an extension field). So I remembered cos(20 degree) is not constructible. For 24 degrees:
First, the regular pentagon is constructible and hence an angle of 72. So 3=(72-(45+30)) is constructible. Thus 24 degrees is constructible by constructing 3 degree angles 8 times.

I surfed a little and found the 2500 year old trisection method (due to Hippocrates?) -- Angle Trisection by Hippocrates This is really just a variation of your parallelogram.

 

[Albert1]

Albert,
Years ago I taught constructibility of numbers from an algebraic standpoint (degree of an extension field). So I remembered cos(20 degree) is not constructible. For 24 degrees:
First, the regular pentagon is constructible and hence an angle of 72. So 3=(72-(45+30)) is constructible. Thus 24 degrees is constructible by constructing 3 degree angles 8 times.

I surfed a little and found the 2500 year old trisection method (due to Hippocrates?) -- Angle Trisection by Hippocrates This is really just a variation of your parallelogram.

johng
well done!