Ostomachion 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. The word Ostomachion has as its roots the Greek word Ὀστομάχιον, which means "bone-fight". 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 - called in Latin also "loculus Archimedius" - 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.
The game is a 14-piece dissection puzzle originally forming a square. The object of the game was to form different objects, animals, plants etc. by rearranging the pieces: an elephant, a tree, a barking dog, a ship, a sword, a tower etc.
Archimedes demonstrates in the treatise that each of the pieces has an area that is an integer fraction of the total area of the square. In the Greek version of the treatise, he also investigates the sizes of the angles of the pieces to see which could go together to make a straight line. The implication is that he was attempting to show how many ways in which the pieces could be assembled to form a square, although there is not enough of the Greek text remaining to be sure. If this is the case, then Archimedes anticipated aspects of combinatorics. The goal of combinatorics is to determine how many ways a given problem can be solved.
The Greek text of the fragment can also be found in the Bibliotheca Augustana website.
This is an English translation of the text of the fragmentary Arabic manuscript (translated from Hermann 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, we draw the diagonals AG, BZ, and ZG, we also bisect BE in H, and draw HT perpendicular to BE, then we put the ruler at point H and - looking to point A - we draw HK, then bisect AL in M, and draw BM. So the A-E 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, 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 Z-G, 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 rectangle DOCZ = 1/6 of the square’s area. According to the proposition that line NC intersects point F, and GE is parallel to CF, 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. Otherwise, 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 A-G. 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 remains rectangle FQCZ (= triangle EGQ), therefore rectangle FQCZ = 1/12 of the square A-G.
We have now divided the rectangle Z-G 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 A-G. 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 intersects point A, so we get the proportion
AB: HT = BK: KT.
Because AB = 2 HT, including 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. As further 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's left 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."