Using Ptolemy's Theorem to prove simple (yet unique) cases?

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The discussion revolves around proving that for point P on arc AB of a regular hexagon's circumcircle, the equation PD + PE = PA + PB + PC + PF holds true, with an emphasis on using Ptolemy's theorem. The original poster expresses concern about the complexity of the proof involving various quadrilaterals and considers a simpler analogy from triangles. Additionally, there is a secondary question regarding a scenario where point P forms an obtuse triangle with segments PA, PB, and PC, leading to confusion about the figure's construction and the problem's wording. Ultimately, the poster resolves their confusion regarding the problems after reviewing solutions, indicating a shift in focus to other pressing questions. The discussion highlights the challenges of geometric proofs and the importance of clear problem statements.
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Problem: Point P is on arc AB of the circumcircle of regular hexagon ABCDEF. Prove that PD + PE = PA + PB + PC + PF.

I'm aware that I'm supposed to use Ptolemy's theorem, which states that
if a quadrilateral ABCD is cyclic, then AC x BD = AB x CD + AD x BC.
I've drawn the hexagon and it seems like an unreasonably long proof is required if I wanted to prove it based on various quadrilaterals inherent within the circle, but I've got a hunch that I can use the analog for a triangle:
If point P is on arc AB of the circumcircle of equilateral triangle ABC, then PC = PA + PB,
which I've already proved in a separate question.

Any tips for getting the ball rolling for this proof?

Also, there's another question that I have based on using Ptolemy's theorem:
Let P be a point in the plane of triangle ABC such that the segments PA, PB, and PC are the sides of an obtuse triangle. Assume that in this triangle the obtuse angle opposes the side congruent to PA. Prove that angle BAC is acute.
For this problem, primarily, I'm having a difficult time determining how to draw the figure. I'm sure that once I am able to, I'll be able to work through to a solution using Ptolemy's inequality, but how should I go about making a figure for this? I find the wording to be a bit abstract compared to how it usually is in these types of problems.
 
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theJorge551 said:
I'm aware that I'm supposed to use Ptolemy's theorem, which states that

AC ?? BD = AB ?? CD + AD ?? BC
Whatever operators you have there, they don't show up for me. (Other than as 0x95).
 
Edited out the weird symbols and replaced them. Should be all clear now. Sorry for the confusion.
 
theJorge551 said:
Also, there's another question that I have based on using Ptolemy's theorem:For this problem, primarily, I'm having a difficult time determining how to draw the figure.

There must be an error somewhere. How can 3 lines radiating from a point form the sides of a triangle? And what is meant by "the side congruent to PA"? Are there meant to be two triangles?

Too many errors.
 
NascentOxygen said:
There must be an error somewhere. How can 3 lines radiating from a point form the sides of a triangle? And what is meant by "the side congruent to PA"? Are there meant to be two triangles?

Too many errors.

I went over the problem today, and it turns out that the wording was fallacious. I've seen the solutions for both of these problems now (the way they were meant to be interpreted) and I no longer need help with them...but the ones in my other thread are much more pressing. :P
 
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Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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