Dear colleagues, I have a new original geometry problem. Please try to solve it and evaluate the author's idea:
Reconstruct triangle ABC from points A, E, F, and H1 (see the picture).)
One side of the triangle is the line through H1 perpendicular to AH1. Now find the midpoint G of EF. Then the length GH1 is the inradius. Construct circles centred at E and F with radius GH1. The intersection of the two circles is the incentre I. Construct the incircle centred at I with radius IE. The tangents to the incircle through A are the remaining two sides of the triangle.
Since there may be two intersections of the constructed circles, there may be up to two distinct solutions (one of which is a reflection of the other in the line AH1).
Edward Doolittle
Wow, thank you, colleague!
Your response is even better than I could have imagined!
If the midpoint of the segment EF coincides with the midpoint of the segment AH1, then vertex B or vertex C go to points at infinity.
Sergey E. Gladun
If the midpoint of the altitude coincides with the midpoint of chord EF of the inscribed circle, then the altitude is equal to two radii of the inscribed circle, which is impossible, as such a triangle does not exist.
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