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# Equilateral Triangle Tangency

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.

Note by Daniel Liu
2 years, 7 months ago

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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 · 2 years, 7 months ago

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. Staff · 2 years, 7 months ago

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. · 2 years, 7 months ago

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! Staff · 2 years, 7 months ago

Oooh that is tough. But won't this note help people with the problem? :P · 2 years, 7 months 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. · 2 years, 7 months ago

How? · 2 years, 7 months ago

Sorry, edited comment to be more clear $$\ddot\smile$$ · 2 years, 7 months ago