In Plate 2 Freese claimed that there are only two isosceles triangles for each of which the pieces of a 6-part dissection will also form either of two different pairs of equiareal isosceles triangles. This is clearly wrong: Take the four pieces in the dissection on the right in the third row of Plate 27. They can be rearranged to form two isosceles triangles of equal area but different shape, as I discuss in A square to one isosceles triangle and also to another, regarding Plate 28. If we take an isosceles right triangle of area equal to the total of the four pieces, we get a set of five pieces (not even six!) that form a large isosceles right triangle. Clearly the five pieces form two different pairs of isosceles triangles of equal area. Perhaps Freese meant to write "congruent" rather than "equiareal". But we would still need a proof.
Copyright 2018, Greg N. Frederickson.
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Last updated February 18, 2018.