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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TopNewestI 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 – Hitesh Sai · 3 years, 1 month ago

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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. – Calvin Lin Staff · 3 years, 1 month ago

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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. – Daniel Liu · 3 years, 1 month ago

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Haha, I assumed that \( r_1 \neq r_2 \), so that's another assumption that needed to be corrected. Thanks! – Calvin Lin Staff · 3 years, 1 month ago

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Oooh that is tough. But won't this note help people with the problem? :P – Finn Hulse · 3 years, 1 month ago

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– Daniel Liu · 3 years, 1 month ago

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.Log in to reply

– Finn Hulse · 3 years, 1 month ago

How?Log in to reply

– Daniel Liu · 3 years, 1 month ago

Sorry, edited comment to be more clear \(\ddot\smile\)Log in to reply