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Be done the triangle ABC, it is known the method to finding the point Q that minimises the sum QA+QB+QC among all points Q in the plane (The Fermat point). I want a hint for solving this problem using the construction of the tangent to ellipse. (Hadamard, Lesson de Geometrie Elementaire, II, problem no. 745).

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Please tag as "exercise". –  Scott Morrison Oct 24 '09 at 18:37
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I have the vague idea that Hadamard is referring to the construction where you erect equilateral triangles BCA', CAB' and ABC' on the sides of the triangle, as described here. The Fermat point is the intersection of the cevians AA', BB' and CC'. It can also be constructed using the various angles of 60 resp. 120 degrees.

In the construction of the tangent from a point P to an ellipse with foci F and F' (in the book you cite), they consider an additional point f. The correspondence should be

F ↔ C,

F' ↔ A,

P ↔ B,

f ↔ P'.

The general philosophy behind both, I think, is to convert a sum of segments into a single segment.

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