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March 28th, 2019, 12:32 PM   #1
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Question How to find the angle when a hexagon is rotated along one of its corners?

I've been confused with this problem but I've found some solution from looking it many times:

The problem is as follows:
A certain protein is under investigation in a laboratory in Taichung. The atoms are arranged along the corners of an hexagon and to examine its optic properties, the crystal is rotated counterclockwise $30^{\circ}$ so that the opposing side forms a $90^{\circ}$ as shown in the figure. If it is known that the light passing through the crystal bends exactly the angle $\angle ABC$. Find the bending angle of the light labeled as $\phi$.


The alternatives shown in my book are as follows:

$\begin{array}{ll}
1.&60^{\circ}\\
2.&37^{\circ}\\
3.&53^{\circ}\\
4.&75^{\circ}\\
5.&45^{\circ}\\
\end{array}$

What I did in my attempt to solve the problem is summarized in the sketch from [url=https://i.imgur.com/Uy0JiS4.png]below[url].



In other words, I did spotted that there is an hexagon in $EAHDGF$ so that the total sum of its interior angles would be equal to $6\times 120^{\circ}$. Although some of its corners have different angles. Since what it is being asked is $\angle ABC$. It's already known that $\angle ABH = 30^{\circ}$ as $\angle AEB= 60^{\circ}$.

From this I inferred the following:

$6\left(120^{\circ}\right)=90^{\circ}+4\left(120^{ \circ}\right)+180^{\circ}-2\omega$

Therefore:

$2\omega=180^{\circ}-2\left(120^{\circ}\right)+90^{\circ}$

$2\omega=270^{\circ}-240^{\circ}=30^{\circ}$

$\omega=15^{\circ}$

Now all that is left to do is to sum $\omega +30^{\circ}=\phi$

therefore:

$15^{\circ}+30^{\circ}=45^{\circ}$

However to establish this answer I had to take for granted that $\triangle BDC$ is isosceles. This part is where I'm still stuck as I couldn't find a way to prove that. Can somebody help me with this matter?. I'd like to know if there are other ways to get this answer. By looking in my book the answer I got is correct. But still I feel dubious if what I did was the right thing to do.
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March 28th, 2019, 02:59 PM   #2
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Originally Posted by Chemist116 View Post
A certain protein is under investigation in a laboratory in Taichung. The atoms are arranged along the corners of an hexagon and to examine its optic properties, the crystal is rotated counterclockwise $30^{\circ}$ so that the opposing side forms a $90^{\circ}$ as shown in the figure. If it is known that the light passing through the crystal bends exactly the angle $\angle ABC$. Find the bending angle of the light labeled as $\phi$.
.................................................. .........

I did spotted that there is an hexagon in $EAHDGF$ so that the total sum of its interior angles would be equal to $6\times 120^{\circ}$. Although some of its corners have different angles.
Nice Diagrams!!

Why complicate the problem by introducing proteins and atoms?!
Problem can be stated very simply:
2 congruent hexagons are drawn as shown. Find angleABC.

And why mention hexagon EAHDGF? NOT necessary.

Can't tell what you're asking.
You calculated angleABC = 45, which is correct.
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March 28th, 2019, 03:20 PM   #3
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Originally Posted by Denis View Post
Nice Diagrams!!

Why complicate the problem by introducing proteins and atoms?!
Problem can be stated very simply:
2 congruent hexagons are drawn as shown. Find angleABC.

And why mention hexagon EAHDGF? NOT necessary.

Can't tell what you're asking.
You calculated angleABC = 45, which is correct.
Thanks!. I had to draw it using Inkscape (not sure if you're familiar with). Although what you say is reasonable. I preferred to put the problem as it was posed. It looks to me that you find a bit annoying at reading the details given to the question. I believe the reason why this problem was stated this way was an effort to link the mathematical aspect with some application. This can be a matter of taste an opinions so I'll leave it there.

Going straight to the maths, I feel that there's an unattended question and that is, how to solve the problem in another way and Is the triangle mentioned above is isosceles or not? and does it exist a way to simplify its solution? or perhaps another way to solve it without referring to polygons?. This is my question.
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March 28th, 2019, 04:21 PM   #4
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March 30th, 2019, 10:03 PM   #5
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Originally Posted by skeeter View Post
Brilliant!. It seems that I have to perfect my figure spotting abilities. Although I am still wondering if there is a way to answer my unattended question which was if there is a way to prove $DC \cong DB$ so that $\triangle DBC$ is isosceles?.
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