 Ostomachion

Ostomachion, also known as loculus Archimedius (Archimedes' box in Latin) and also as syntomachion, is a mathematical treatise attributed to Archimedes. This work has survived fragmentarily in an Arabic version and in a copy of the original ancient Greek text made in Byzantine times.^{[1]} The word Ostomachion has as its roots in the Greek "Ὀστομάχιον",^{[2]} which means "bonefight", from "ὀστέον" (osteon), "bone"^{[3]} + "μάχη" (mache), "fight, battle, combat".^{[4]} Note that the manuscripts refer to the word as "Stomachion", an apparent corruption of the original Greek. Ausonius gives us the correct name "Ostomachion" (quod Graeci ostomachion vocavere). The Ostomachion which he describes was a puzzle similar to tangrams and was played perhaps by several persons with pieces made of bone. It is not known which is older, Archimedes' geometrical investigation of the figure, or the game.
Contents
Game
The game is a 14piece dissection puzzle forming a square. One form of play to which classical texts attest is the creation of different objects, animals, plants etc. by rearranging the pieces: an elephant, a tree, a barking dog, a ship, a sword, a tower etc. Another suggestion is that it exercised and developed memory skills in the young. James Gow, in his Short History of Greek Mathematics (1884), footnotes that the purpose was to put the pieces back in their box, and this was also a view expressed by W. W. Rouse Ball in some intermediate editions of Mathematical Essays and Recreations, but edited out from 1939.
Mathematical problem
In the fragment in Arabic translated by Heinrich Suter it is shown that each of the pieces has an area that is a rational fraction of the total area of the dissected parallelogram. In the Greek version of the treatise, some investigation is made as to the sizes of the angles of the pieces to see which could go together to make a straight line. This might have been preparatory to consideration of the number of ways the pieces might reform some prescribed shape, for example the box in which the pieces were contained, although there is not enough of the Greek text remaining to be sure. If this is the case, then Archimedes anticipated aspects of combinatorics. Combinatorics often involves finding the number of ways a given problem can be solved, subject to welldefined constraints. For example, the number of ways of reforming a square using the pieces as proposed by Suter is 536 without distinguishing the result up to rotations and reflections of the square, but allowing the pieces to be turned over.^{[1]} One such solution is shown in colour to the right. However, if pieces are not allowed to be turned over, for example, if obverse and reverse can be distinguished or, in the case of Suter's pieces, on account of the sharpness of some angles, the corresponding number is 4. If rotations and reflections are treated as distinct, these numbers rise to 17,152 and 64, respectively. The counts of 4 and 64 may be verified easily as lower bounds by elementary group theory, but were confirmed as exact by Bill Cutler shortly after his determination of the count 536 and by the same computer program. So, there are at least four different answers that we might give just considering Suter's proposal. Clearly, to count, you have to know what counts. When, as here, the number of outcomes is so sensitive to the assumptions made, it helps to state them explicitly. Put another way, combinatorics can help sharpen our awareness of tacit assumptions. If, say, answers like 4 or 64 are unacceptable for some reason, we have to reexamine our presumptions, possibly questioning whether Suter's pieces can be turned over in reforming their square. As emerges below, there is also some objection to Suter's proposal which would render this combinatorial discussion of the Suter board academic.
The Greek text of the fragment can also be found in the Bibliotheca Augustana website.
This is an English translation, with added disambiguation in brackets, of the text of the fragmentary Arabic manuscript (translated from Heinrich Suter's German translation in: Archimedis opera omnia, vol. 2, p. 420 sqq., ed. J. L. Heiberg, Leipzig 1881, as published in the Bibliotheca Augustana website):
" We draw a [rectangular] parallelogram ABGD, we bisect BG in E and draw EZ perpendicular to BG [to intersect AD], we draw the diagonals AG, BZ [intersecting AG at L], and ZG, we also bisect BE in H, and draw HT perpendicular to BE [to intersect BZ], then we put the ruler at point H and  looking to point A  we draw HK [to intersect BZ], then bisect AL in M, and draw BM. So the AE rectangle is divided into seven parts. Now we bisect DG in N, ZG in C, we draw EC and attaching the ruler to the points B and C we draw CO [to intersect DG], furthermore CN. Thus the rectangle ZG is also divided in seven parts, but in another way than the first one. Therefore, the whole square has fourteen parts.
We now demonstrate that each of the fourteen parts is in rational relationship to the whole square.
Because ZG is the diagonal of the rectangle ZG, the triangle DZG is half of this rectangle, that means 1/4 of the square. But the triangle GNC is 1/4 of triangle DZG, because, if we extend the line EC, it comes to point D, and that means triangle GDC has half area of the triangle DZG and is equal to the two triangles GNC and DNC taken together; that means triangle GNC is 1/16 of the square. If we presume that line OC is orientated to point B, as we have drawn it before, so the line NC is parallel to BG, which is the side of the square and of the triangle OBG, so we get the proportion
BG : NC = GO : NO.
But BG is four times NC, and in the same way GO four times NO; therefore is GN three times NO, and triangle GNC = 3 ONC. However, as we have shown, triangle GNC is 1/16 of the square, that means triangle ONC = 1/48 of the square. Furthermore, as triangle GDZ = 1/4 of the square, and therefore GNC = 1/16 of that triangle and NCO = 1/48 of that, it remains for the quadrilateral DOCZ = 1/6 of the square’s area. According to the proposition that line NC [extended] intersects [ZE at] point F, and GE is parallel to CF, [and labelling the intersection of AG and CE as Q,] we get the proportion
EC : CF = EQ : CQ = GQ : FQ.
Because EQ = 2 CQ and GQ = 2 FQ, triangle EQG is double to the two triangles GCQ and EFQ. It is clear, that triangle EGZ = 2 times triangle EFG, because ZE = 2 FE. As the triangle EGZ = 1/4 of the square, that means triangle EFG = 1/8 of the square. This triangle is three times as big as each of the two triangles EFQ and GCQ, so each of these two triangles = 1/24 of the square AG. And the triangle EGQ is double to each of the two triangles EFQ and GCQ, so it is = 1/12 of the square. Furthermore because ZF = EF, triangle ZFG = triangle EFG. If we now take away triangle GCQ (= triangle EFQ), it leaves quadrilateral FQCZ (= triangle EGQ), therefore quadrilateral FQCZ = 1/12 of the square AG.
We have now divided the rectangle ZG in 7 parts, and go on to divide the other rectangle.
Because BZ and EC are two parallel diagonals, and ZF = EF, therefore triangle ZLF = EFQ, and also triangle ZLF = 1/24 of the square AG. Because BH = HE, triangle BEZ is four times the triangle BHT, because each of them is rectangular. As triangle BEZ = 1/4 of the square ABGD, triangle BHT = 1/16 of that. According to our proposition the line HK [extended[ intersects point A, so we get the proportion
AB : HT = BK : KT.
Because AB = 2 HT, and BK = 2 KT and BT = 3 KT, triangle BHT is three times the triangle KHT. However, because triangle BHT = 1/16 of the whole square, triangle KHT = 1/48 of that. Triangle BKH is double the triangle KHT, so = 1/24 of the square. Further, as BL = 2 ZL, and AL = 2 LF, triangle ABL is twice the triangle ALZ, and ALZ double the triangle ZLF. However, because triangle ZLF = 1/24 of the whole square, triangle ALZ = 1/12 of that, so triangle ABL = 1/6. But triangle ABM = triangle BML, so each of these two triangles = 1/12 of the square. It leaves the pentagon LFEHT = 7/48 of the entire square.
We have now also divided the square AE into 7 sections, therefore, the whole figure ABGD in 14 parts. Each of these fourteen parts is in rational relationship to the whole, and that is what we wanted."
However, Suter was translating unpointed Arabic, in which, as he concedes, equals and twice are easily confused. At a crucial point in his translation he ignores this in making his figure a square; instead he makes a typographical error, equating, not the sides, but a side and a diagonal, in which case the figure cannot be a rectangle. Suter does know that, for the areal results discussed in the text, it suffices that the figure be a parallelogram. Suter may have been under the impression that the figure had to be a square, so made it one.
The dissection lines given by Archimedes' Stomachion from the Archimedes Codex , as agreed by Heiberg and Dijksterhuis, are a subset of those of Suter's square board when the latter is subjected to a lateral stretch by a factor of two. This was recognised by Richard Dixon Oldham, FRS, in a letter to Nature in March, 1926. The letter triggered a wave of interest in the dissection puzzle, with kits for sale and a feature article in The New York Times that August; Popular Science Monthly had related items in its issues for November, 1926 and February, March, May and June, 1927, including Stomachion competitions with cash prizes. As each of the two unit squares is cut diagonally, the pieces can be arranged in a single square of side the square root of two, after the manner deployed by Socrates in Plato's Meno. Thus, the 14 pieces still form a square, even when the dissected board is not a square. Naturally, the count of solutions also changes, but here we now have three possible boxes: the two unit squares side by side; the two unit squares one on top of the other; and this single square of side the square root of two. The single square can be formed from four congruent right isosceles triangles, but this can also be accomplished with three right isosceles triangles, two congruent and the third double their area, as in Tangram, as well as in a third way where only one right isosceles triangle is formed. (Bill Cutler has made a comprehensive study of the associated counts comparable to those mentioned for the Suter board.) With two exceptions, the pieces can be formed from right triangles with legs in the ratios 1:1, 1:2 and 1:3, with some pieces similar to others, possibly at different scales; this observation provides a comparatively easy means to determine the proportional areas of the pieces.
Suter's translation can now be checked against the fragment of Archimedes' Stomachion in the Archimedes Codex, as was presumably not open to Suter himself when he made the translation. Heiberg, followed by Dijksterhius, seem to have thought that the two texts, the Greek discussing angles, the Arabic areas, were so different there was no relation between the two, so may have felt there was nothing to check. But, in Suter's square, the diagonals cross at right angles, whence the nature the angles examined in the first proposition of Archimedes' Stomachion, whether acute or obtuse, is immediate, making it puzzling why Archimedes would prove this as a proposition. Instead, the first proposition, if it is seen to relate to the underlying figure of the dissected Stomachion board, sets up two squares side by side, in which setting the nature of the angles is more sensitive, but still tractable. The two squares separately suggest two frames of an iterative process. If this process is continued, rational approximations of the square root of two are obtained; these are the sidediameter numbers familiar to the Greeks. At each step of the process, the nature of the angles switches back and forth, highlighting the significance of the first proposition. Whether or not this has any bearing on Archimedes' Stomachion is another matter, but it does provide a geometrical means to handle recurrence relations if you do not have algebraic notation, such as subscripts, and the geometry is fully within Archimedes' capabilities. The same technique readily produces the bounds for the square root of three that Archimedes assumes without comment in Measurement of a Circle; if it were applied to the dissection of a square proposed by Suter, analogous bounds for the square root of five are obtained. The geometry can be worked independently of knowledge of Pell's equation or of the properties of convergents of a continued fraction, but to similar effect.
The dissection of the two squares side by side can also be seen as a layered composite or collage of instances of the diagrams for Elements II.9, 10. Viewed numerically, these propositions provide the link between the legs of right triangles in the unit square grid and those of right triangles in the overlaid diagonal square grid where the right triangles share their hypotenuse, although Euclid proves them in terms of geometrical lines. Thus, pairs of successive side numbers in the overlaid grid are associated with pairs of successive diameter numbers in the unit square grid, as remarked by Proclus, rather than side and diameter numbers being paired. The ratios within these pairs are then rational approximations for the tangent of π/8, in agreement with the iterative process of the previous paragraph, which is, in effect, an angle bisection algorithm. This reminds us that a natural setting for the sidediameter numbers is a nest of regular octagons. The analogue for the square root of three is not an angle bisection algorithm, but it does produce rational approximations to the tangent of π/12, with the regular dodecagon as a setting. Instead of simply approximating the square roots in each case, we can approximate the corresponding regular polygons. But, again, whether this was actually Archimedes' agenda is another matter.
Irrespective of the parallelogram adopted for the Stomachion board, the presence of several centroids of triangles on the board as further points of it might seem another Archimedean theme, not least as they facilitate the computation of the relative areas of the pieces. Most obviously, Q is the centroid of the right triangle GZE, so the pieces in that triangle are either one sixth or one third of it. But similarly K is the centroid of right triangle EAB, so triangle HKB has area one sixth of it. In Equilibrium of Planes, Archimedes makes use of the concurrence of the three medians of a triangle in the centroid, but does not prove it. As it happens, for both K and Q, we see two medians intersecting in them, but the pairings are not the same, although the triangles GZE and EAB are translations of one another.
Notes
 ^ ^{a} ^{b} Darling, David (2004). The universal book of mathematics: from Abracadabra to Zeno's paradoxes. John Wiley and Sons, p. 188. ISBN 0471270474
 ^ ὀστομάχιον, Henry George Liddell, Robert Scott, A GreekEnglish Lexicon, on Perseus Digital Library
 ^ ὀστέον, Henry George Liddell, Robert Scott, A GreekEnglish Lexicon, on Perseus Digital Library
 ^ μάχη, Henry George Liddell, Robert Scott, A GreekEnglish Lexicon, on Perseus Digital Library
Selected bibliography
 J. L. Heiberg, Archimedis opera omnia, vol. 2, S. 420 ff., Leipzig: Teubner 1881
 Reviel Netz & William Noel, The Archimedes Codex (Weidenfeld & Nicholson, 2007)
 J. Väterlein, Roma ludens (Heuremata  Studien zu Literatur, Sprachen und Kultur der Antike, Bd. 5), Amsterdam: Verlag B. R. Grüner bv 1976
External links
 Heinrich Suter, Loculus
 James Gow, Short History
 W. W. R. Ball, Recreations and Essays
 Ostomachion, a GraecoRoman puzzle
 Professor Chris Rorres
 Kolata, Gina. "In Archimedes' Puzzle, a New Eureka Moment." The New York Times. December 14, 2003
 A tour of Archimedes' Stomachion, by Fan Chung and Ronald Graham.
Categories: Ancient Greek mathematical works
 Puzzles
 Tiling puzzles
 Works by Archimedes
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