So we all know the infamous Two-Pole Problem, which doesn't rely on the distances between the two poles. I have a situation where there are a n poles. Each pair of consecutive poles is used to create a Two-Pole situation, thus creating n-1 shorter poles. Using these, n-2 shorter poles are created in a similar fashion. This continues until there is one pole, which for reference I will call the "supporting pole". if \(x_{1}, x_{2}, x_{3}, ..., x_{n}\) are the lengths of the starting poles, find a formula for the length of the supporting pole in a n-Pole Problem.

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TopNewest\(\frac{1}{\sum_{i=1}^n\frac{1}{x_{i}}}\) – Mohammad Sarfaraz · 2 years, 11 months ago

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– Tristan Shin · 2 years, 11 months ago

That's close. Try a few other cases that can disprove this(try 5 poles, for instance).Log in to reply

– Mohammad Sarfaraz · 2 years, 11 months ago

I don't get it...so please can you reveal the answer...:)Log in to reply

I think now you should reveal the solution..... – Mohammad Sarfaraz · 2 years, 11 months ago

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I'll reveal this solution within the next week(April 21-April 25). – Tristan Shin · 2 years, 11 months ago

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For clarification, the Two-Pole Problem is where there are two poles of length a and b. Lines are drawn from the top of each to the bottom of the other. From the intersection of these two new lines, a vertical line segment parallel to the original poles is drawn. It turns out that the distance between the two poles is irrelevant. To solve this problem, try playing around with a few two-pole scenarios. – Tristan Shin · 2 years, 11 months ago

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I don't know that problem (you said it's infamous then why not show it) – Amr Saber · 2 years, 11 months ago

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We use the formula \(\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{z}\) where \(x,y\) are the two poles and \(z\) is the height of the middle pole. Solving for the supporting pole is pretty trivial, just plug 'n' chug. – Daniel Liu · 2 years, 11 months ago

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– Tristan Shin · 2 years, 11 months ago

But is there a generic formula for the supporting pole, not just plugging it in?Log in to reply