This apparently simple statement is of relatively recent origins. It's been associated with the names of Lowry, W. Wallace, Farkas Bolyai, and P. Gerwien and usually goes by the name of Wallace-Bolyai-Gerwien Theorem. The accounts, however, differ. According to Greg Frederickson, Lowry (1814) provided a simple explanation in answer to a problem posed by Wallace around 1808. (Wallace presumably had a solution at that time, which he gave in expanded form in 1831.) Frederickson acknowledges the methods of Bolyai (1832) and Gerwien (1833).
According to Ian Stewart, the statement is usually called
According to Andreescu and Gelca, the property was proved independently by F. Bolyai (1833) and Gerwien (1835). A Russian math encyclopedia concurs.
(It should be mentioned that the Bolyai in question was a noted Hungarian mathematician Wolfgang Farkas Bolyai (1775-1856) and, sometimes Farkas Wolfgang Bolyai, father of Janos Bolyai, the co-inventor of the non-Euclidean geometry, and a dear friend of Johann Carl Friedrich Gauss. Like the birthplace of I. Kant, the birthplace of F. Bolyai has also changed hands. It is now a part of Romania.)
More accurately the Wallace-Bolyai-Gerwien Theorem states
Any two simple polygons of equal area can be dissected into a finite number of congruent polygonal pieces.
In other words, one of the polygons can be dissected into polygonal pieces which, after rearrangement, combine into the other polygon.
(The word "dissect" and the mention of the "polygonal" results of the dissection are important, because a sheer unrestricted equidecomposibility admits weirdest decompositions imaginable.)
The proof of the Wallace-Bolyai-Gerwien Theorem proceeds in steps. The idea is that a polygon is equidecomposable with a square of equal area. Two polygons are dissected into pieces that combine into and overlap in the same square, where the pieces can be further cut into common parts.
Any simple polygon admits many types of triangulation, the diagonal triangulation being just one among many.
Each of the triangles is equidecomposable with a rectangle.
- T. Andreescu, R. Gelca, Mathematical Olympiad Challenges, Birkhäuser, 2004, 5th printing, pp. 13-14
- G. Frederickson, Dissections: Plane & Fancy, Cambridge University Press, 1997, p. 222
- I. Stewart, From Here To Infinity, Oxford University Press, 1996, p. 169
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