How can construct three circles in a given triangle such that three internal tangent form an equilateral triangle?
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2$\begingroup$ Heuristically, there ought to be a one-parameter family of such circles by counting degrees of freedom, Since we have two degrees of freedom (radii of three circles) and only two conditions (two angles are $60^\circ$, the third is automatic) $\endgroup$– Sam ZbarskyJun 28, 2020 at 12:57
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2$\begingroup$ Why is this tagged as an open problem? Is it stated somewhere that it is unsolved? It is not an "open problem" merely because the answer is not known to the OP. $\endgroup$– Michael RenardyJun 28, 2020 at 13:39
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1$\begingroup$ There seems to be even a one-parameter family of such circles with these three tangents intersecting at a single point. $\endgroup$– მამუკა ჯიბლაძეJun 29, 2020 at 5:45
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1$\begingroup$ Say you decrease the radius of each circle by the same amount, and move all lines (including triangle edges) to maintain tangency with the circles. As this operation would merely replace each line with a parallel line, the new triangle will be similar to the original one, and the "equilateral triangle" property will be preserved. By taking the radius decrease to be the radius of the smallest circle, it seems that if there's a solution, then there's always a solution in which one of the circles has radius 0. Trying to find such a solution may be simpler than the original problem. $\endgroup$– user1020406Jun 30, 2020 at 1:03
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1$\begingroup$ No no please don't be sorry. You are asking question, you want to know the answer, this is nothing to be sorry about - on the contrary, you contribute to this site and you are doing good thing, I believe. $\endgroup$– მამუკა ჯიბლაძეJun 30, 2020 at 9:28
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1 Answer
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Draw three congruent equilateral triangles of an arbitrary size, with bases on the three sides centered at the incircle tangency points. Then the desired circles are inscribed in the kites delimited by the extended sides of these equilateral triangles as shown. Their common tangents form an equilateral triangle congruent to the other three.
This construction is most easily justified by working backwards. Find a solution among the one-parameter family with an equilateral triangle of the given size. Reflecting it over the common diameter of any two of the circles produces equilateral triangles with bases on the three sides. The quadrilaterals delimited by their extended sides are circumscribed about the circles, and so are kites by the equal opposite angles. That means any two of them are equidistant from the vertex between them, which means their bases are centered at the incircle tangency points.
