Through the illustrations in **Figures 2-8** (preceding webpage), we have come to understand Liu Hui's ideas about cube (and rectangular prism) dissection and the three important Chinese terms used in his description, *quiandu, yangma,* and *bie'nao.* But what is the volume of a *yangma* (rectangular pyramid) that results from such a dissection? And how does the *bie'nao* (triangular pyramid) compare with the *yangma* in terms of volume? Although it may be evident in the case of a cube that the *yangma* volume is one third of the cube volume, the *yangma* volume is not as clear in the general case of a rectangular prism. Liu Hui, in his commentary, wanted to give a more rigorous justification for the volumes of a *yangma* and a *bie'nao. *Toward this goal, he started with a cube and divided it into two congruent *qiandu,* then divided one of the *qiandu* into a *yangma* and a *bie’nao*. He then made his crucial argument that a *yangma* is twice as much as a *bie'nao* in volume, and, once he had established that relationship, concluded that each *yangma* is one third of the cube volume and each *bie'nao* is one sixth of the cube volume. Finally, he extended these relationships to a general rectangular prism, but without detailed discussion. How he did so has been an area of research in the history of Chinese mathematics (Wagner, 1979; Shen, Crossley, & Lun, 1999). Another source of confusion is the modern interpretation of ancient Chinese. Ancient Chinese texts can be potentially interpreted in multiple ways. However, there is general agreement today among researchers that, with technical emendations, Liu Hui's approach presented in the case of a cube is still valid in the case of a rectangular prism.

**Figure 9. **A *yangma* and a *bie'nao* are bisected in all three dimensions for a comparison of their volumes. On a smaller scale, the *yangma* (at right) consists of a cube, two *qiandu*, and two *yangma*; the *bie'nao* (at left) has two *qiandu* and two *bie'nao*.

As noted above, Liu Hui sought a general solution on the basis of the special case of a cube. He started with a cube that is two *chi* in all three dimensions. From such a cube, he took a *yangma* and a *bie'nao*. Then, in order to show that the volume of the *yangma* is twice that of the *bie'nao,* he dissected the *yangma* in half along all three dimensions into, on the smaller 1-*chi* scale, a cube, two *qiandu*, and two *yangma*, as illustrated in **Figure 9** (above). The *bie'nao* was dissected in the same way, producing, on the 1-*chi* scale, two *qiandu* and two *bie'nao*. After the bisections, it is clear that the original *yangma* consists of a smaller cube (equivalent to two *qiandu*), two smaller *qiandu*, and two smaller *yangma*; and the original *bie'nao* has two smaller* qiandu* and two smaller *bie'nao*. Now, because two smaller *qiandu* make a smaller cube on a 1-*chi *scale, the original *yangma* is equivalent to two smaller cubes plus two smaller *yangma*, and the original *bie'nao* is equivalent to one smaller cube and two smaller *bie'nao*. The same bisection can be applied to the smaller *yangma* and *bie'nao* iteratively.

Modern mathematicians have tried hard to make sense of Liu Hui's ideas. The following is a synthesis of modern interpretative efforts in the literature (Guo, 2009; Quan & Ke, 2009; Shen, Crossley, & Lun, 1999; Li, 2002; Wagner, 1979). To show the progression of Liu Hui's iterative approach, let us use \(Y_n\) and \(B_n\) for the volumes of the *yangma* and the *bie'nao, *respectively, after the \(n\)th iteration, with \(n=0\) for the initial volumes so that \(Y_0\) is the volume of the original *yangma* and \(B_0\) is the volume of the original *bie'nao*. We further use \(V\) for the volume of a unit cube, substituting \(V=1\) only at the end of the process. Then, we have the following expressions for the volumes of the original *yangma* and *bie'nao*:

When \(n=1,\) we have \[Y_0 = 2V + 2Y_1,\,\,\,\, B_0 = V + 2B_1.\]

When \(n=2,\) we have $$Y_0= 2V + 2\left(2\cdot{\frac{1}{8}}V + 2Y_2\right) = 2V\left(1+\frac{2}{8}\right) + 2^2Y_2$$ and \[B_0 = V + 2\left(\frac{1}{8}V + 2B_2\right) = V\left(1+\frac{2}{8}\right) + 2^2B_2.\]

When \(n=3,\) we have $$Y_0= 2V\left(1+\frac{2}{8}\right) + 2^2\left(2\cdot\frac{1}{8^2}V + 2Y_3\right) =2V\left(1+\frac{2}{8}+\frac{2^2}{8^2}\right) + 2^3Y_3$$ and \[B_0 = V\left(1+\frac{2}{8}\right) + 2^2\left(\frac{1}{8^2}V + 2B_3\right) =V\left(1+\frac{2}{8}+\frac{2^2}{8^2}\right) + 2^3B_3.\]

For the general case, after the \(n\)th iteration, with \(V=1,\) we have \[Y_0 = 2\,{\sum_{i=0}^{n-1} \frac{1}{4^i}} + 2^nY_n,\] and \[B_0 = \sum_{i=0}^{n-1} \frac{1}{4^i} + 2^nB_n.\]

In the equations above, the last two components, \(2^nY_n\) and \(2^nB_n,\) are particularly interesting. Recall that at each step of Liu Hui's dissection of the *yangma *and the *bie'nao, *there are two smaller *yangma* and two smaller *bie'nao* left over, which together form a smaller cube with one eighth of the previous cube's volume (see **Figures** **10** and **11**). Therefore, remembering that \(V=1,\)

\[2^nY_n+2^nB_n=2^{n-1}(2Y_n+2B_n)= 2^{n-1}\cdot\frac{1}{8^{n-1}}V=\frac{1}{4^{n-1}},\]

which means the sum \(2^nY_n + 2^nB_n\) goes to zero as the process is iterated indefinitely. Indeed, Liu Hui seems to have argued that this “remainder,” the leftover *yangma* and *bie'nao* at each step of the iteration, disappears (Shen, Crossley, & Lun, 1999, p. 270):

To exhaust the calculation, halve the remaining breadth, length and altitude respectively, … . The smaller the halves, the finer the remainder. Extreme fineness means infinitesimal, which is formless. In that case, how can one have a remainder?

Since, with each iteration, dissection of the *yangma* produces two smaller cubes while dissection of the *bie’nao* produces just one smaller cube, Liu Hui concluded that, in a *qiandu* formed from a *yangma* and a *bie'nao*, there is a 2:1 ratio between the volumes of the *yangma* and the *bie'nao*. Thus, the original *yangma* is twice as much as the original *bie'nao* in terms of volume. In other words, two *bie'nao* make one *yangma*, which is often called “Liu Hui's Principle” (Shen, Crossley, & Lun, 1999, p. 275). Since the *qiandu* has half the volume of the original cube, it follows that a *yangma* is one third and a *bie'nao* is one sixth of the original cube volume. A little algebraic notation can sometimes help convince prospective teachers (and others) of these conclusions. Thus, let us assume that the *bie'nao *has a volume of *B*. Then, the *yangma *has a volume of *2B*, and the *qiandu* has a volume of *3B*. The whole cube then has a volume of *6B*. Therefore, the *yangma* is one third and the *bie'nao* is one sixth of the cube volume.

Although Liu Hui presented his approach in the case of a cube, the iterative process can be extended to the general case of a rectangular prism, using the same analysis shown above. Now, since an arbitrary rectangular prism consists of two *qiandu* and a *qiandu* can be dissected into a *yangma* and a *bie'nao* in a 2:1 ratio, then a *yangma* is one third and a *bie'nao* is one sixth of the original rectangular prism in volume. That is essentially Liu Hui's argument for the volume of a *yangma*, which involves an iterative process of bisection of edges and, at least in its modern interpretation, the idea of a limit, instead of using an experimental approach that allows three *yangma* to fit in a cube or a rectangular prism. It seems that Liu Hui had the general case in mind when he was using a cube to illustrate his dissecting strategy.

Liu Hui's original comment, however, is slightly different than the modern interpretation given above and has proven “very difficult to understand” (Shen, Crossley, & Lun, 1999, p. 277). After bisecting the *yangma* and *bie'nao* along all three dimensions, Liu Hui proposed that the two blocks be colored black and red, respectively, and combined into a *qiandu*, as shown in **Figures 10** and **11** (below). Then, he halved the breadth and the height and focused on the two black-and-red *qiandu* (**Figure 11**), each of which is made of a smaller *yangma* and a smaller *bie'nao* (Wagner, 1979, pp. 179-182). Liu Hui thus stated that (Shen, Crossley, & Lun, 1999, p. 270):

[T]he red and black right triangular prisms constitute, in such a case, a cube with altitude of 1 *chi* and a square base of 1 [square] *chi*. Every two *bie'nao* make one *yangma.*

Liu Hui further indicated an iterative process and the applicability of his approach to an arbitrary rectangular prism without giving detailed justification. Wagner*'*s (1979) interpretation of Liu Hui's original comment has been generally accepted as appropriate since the 1980s by Chinese mathematicians and historians of mathematics (Shen, Crossley, & Lun, 1999). According to Wagner, Liu Hui may have just used the colors to explore the relationships among the various pieces. In the case of a cube, the two red *qiandu* in **Figure 10** make a cube, and the two black *qiandu* also make a cube. By contrast, in the general case, the two red *qiandu* in **Figure 10 **do not fit in a rectangular prism; nor do the two black *qiandu*. However, two *qiandu* in the same color have the same volume as a corresponding black-and-red rectangular prism (Guo, 2009). To justify this claim, note that in the general case a red *qiandu* and a corresponding black *qiandu* make a black-and-red rectangular prism. Thus, a red *qiandu* is one half of the black-and-red rectangular prism. Therefore, the two red *qiandu* in the *bie'nao* have the same volume as the black-and-red rectangular prism, although they do not fit in such a rectangular prism. The two black *qiandu* in the *yangma*, by the same argument, also have the same volume as the black-and-red rectangular prism. In short, the analytic process used for the cube applies to the general case of a rectangular prism. Liu Hui's ingenious dissection techniques and his use of coloring are of pedagogical significance today, especially so when we consider the use of 3-D modeling technologies.

**Figure 10.** A *yangma* and a *bie'nao* are bisected in all three dimensions and colored black and red, respectively.

**Figure 11.** A *yangma* and a *bie'nao*, after their bisections and coloring, are recombined to form a *qiandu*. The smaller black-and-red *qiandu*, each made of a *yangma* and *bie'nao*, constitute a smaller cube. Note that some of the colors from **Figure 10** are removed to highlight the two black-and-red *qiandu*.

In summary, Liu Hui elaborated on a systematic approach to the dissection of various polyhedral solids in his commentaries on the *Nine Chapters on the Mathematical Art*. His purpose was obviously to calculate the volume of such solids for practical needs in farming and earthworks. An arbitrary rectangular prism, using his approach, can be decomposed into a *qiandu* (triangular prism), a *yangma* (rectangular pyramid), and a *bie'nao* (triangular pyramid). While a *qiandu* is one-half of the original volume of the rectangular prism, a *yangma* is one third, and a *bie'nao* is one-sixth. Every *qiandu *can be further dissected into a *yangma* and a *bie'nao,* with the latter two pieces in a 2-to-1 ratio in volume. Toward the end of his comment on Problem Fifteen of Chapter Five, Liu Hui summarized his systematic method of cube dissection—his overarching goal and the interconnections among the constituent solids (Shen, Crossley, & Lun, 1999, p. 270):

Exhaustive calculation flows from the situation without involving counting rods. The form of the *bie'nao* is different from a practical object, and that of a *yangma* also shows variations in dimensions. Nevertheless, without the definite volume of the *bie'nao*, there is no way to survey the *yangm*a, and without the definite volume of the *yangma*, there is no way to know the volume of solids such as cones [*zhui*, 锥] and frusta [*ting*, 亭], which are fundamental to the calculations of the tasks of labourers and volumes.

Although Liu Hui used a cube in his original text to illustrate the process of dissection, his method applies to arbitrary rectangular prisms and the various relationships remain constant. In the following sections, however, we focus on the case of cube dissection for its rich symmetry and pedagogical implications. For the purpose of this article, we refer to Liu Hui's cube dissection as "Liu Hui's Cube Puzzle," to suggest how it might be approached in the classroom. Indeed, a puzzle designation may well be appropriate for Liu Hui's conception of cube dissection in light of the way he sought to understand the ratio between a *yangma* and a *bie'nao* in the context of a cube (Martzloff, 1987/1999, p. 283). With a working knowledge of Liu Hui's approach to cube dissection, we now move on to discuss its classroom applications, starting with paper folding and GeoGebra-based modeling.