After looking at the diagram of Extremes with Equilaterals, a question came to mind:

Given two concentric circles of radii \(r_1\) and \(r_2\), find a function \(f(r_1,r_2)\) that outputs the side length of the smallest equilateral triangle such that one side is tangent to one of the circle and another side is tangent to the other circle. The sides themselves must be touching the circle, not just the extensions of the sides.

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## Comments

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TopNewestOooh that is tough. But won't this note help people with the problem? :P

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No, it won't. In fact, I am pretty sure it is impossible in my original problem for this double-tangency situation to occur. This new problem was just inspired by the previous problem.

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How?

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My first thought is that such a configuration is not possible. I found an assumption, corrected for it, and only have solutions where \( r_1 = \frac{1}{2} r_2 \). I'd be interesting in seeing the general case.

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The vertices of the equilateral triangle not necessarily have to lie on the circles.

Also you missed the case where \(r_1=r_2\) I think. (using your interpretation) or I'm missing something on what you are interpreting it as.

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Ah yes. I was misled by the reference to the other question.

Haha, I assumed that \( r_1 \neq r_2 \), so that's another assumption that needed to be corrected. Thanks!

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I think that r2 = 0 and then the second circle is a point circle then the r1 will be altitude of the equilateral triangle then side of the equilateral triangle 2r1÷√3

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