Proofing a Rhomb - Get the Answer Here

  • Thread starter Thread starter aRTx
  • Start date Start date
  • Tags Tags
    Proof
AI Thread Summary
To prove or disprove the existence of a rhombus in a given geometric configuration, clarity on the properties of the shapes involved is essential. The discussion centers around a regular pentagon, with questions about the presence of parallel lines. Participants express uncertainty regarding the parallel lines but confirm the regularity of the pentagon. The need for further clarification and assistance in understanding the geometric relationships is emphasized. Overall, the conversation highlights the challenges in proving geometric properties without complete information.
aRTx
Messages
3
Reaction score
0
Hi,

Can anyone help me?
How to proof/disproof there is a rhomb?


SyprinaePentagonit.jpg


thnx very very much...
 
Physics news on Phys.org
Is it a regular pentagon? And are any lines given parallel?
 
Norway said:
Is it a regular pentagon? And are any lines given parallel?

Yes, it is regular.

I don't know about the parallel lines (they seems to be parallel...)

I'm looking forward ...

thnx
 
Last edited:
I really need it!
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
Essentially I just have this problem that I'm stuck on, on a sheet about complex numbers: Show that, for ##|r|<1,## $$1+r\cos(x)+r^2\cos(2x)+r^3\cos(3x)...=\frac{1-r\cos(x)}{1-2r\cos(x)+r^2}$$ My first thought was to express it as a geometric series, where the real part of the sum of the series would be the series you see above: $$1+re^{ix}+r^2e^{2ix}+r^3e^{3ix}...$$ The sum of this series is just: $$\frac{(re^{ix})^n-1}{re^{ix} - 1}$$ I'm having some trouble trying to figure out what to...

Similar threads

Back
Top