《孫子算經卷下》 "Sun Tzŭ's Computational Classic: Volume III"
§25. Utilising similar triangles

This section gives a word problem where similar triangles are used to determine an unknown length.

The relevant unit conversions for length are

\begin{aligned} 1 \unit{rod~(丈)} &= 10 \unit{rules~(尺)} \\ 1 \unit{rule~(尺)} &= 10 \unit{inches~(寸)}. \end{aligned}

See Vol. I §1 (Units of length).

Translation

Chinese source text: Version A, Version B, Version C, Version D.
Unless noted otherwise, I follow the text from Version D, 《知不足齋叢書》本.

Source text Target text Notes
今有竿不知長短。度其影、得一丈五尺。別立一表、長一尺五寸、影得五寸。 Suppose there be [a] pole [we] know not [the] length of. Measuring its shadow, resulteth in one rod [and] five rules. Separately erecting one post, of length one rule [and] five inches, [its] shadow resulteth in five inches.
  • 度： measuring

    度、入聲、 Cantonese: tok9, Mandarin: duò (Government-regulated 統讀: duó)

  • 表： post; or pillar
問竿長幾何。 [We] ask, how much [be the] pole's length?
答曰、四丈五尺。 Answer saith: four rods [and] five rules.
術曰、置竿影一丈五尺、以表長一尺五寸乘之、上十之、得二十二丈五尺。 Method saith: put [down the] pole's shadow one rod [and] five rules; multiplying it by [the] post's length one rule [and] five inches, [and] decupling it upward, resulteth in twenty-two rods [and] five rules.
  • 二十二丈五尺： twenty-two rods [and] five rules

    This quantity is actually an area. Writing s for shadow and l for length, this area is

    \begin{aligned} s(\text{pole}) \cdot l(\text{post}) &= 1.5 \unit{rods} \times 1.5 \unit{rules} \times \frac{10 \unit{inches}}{\unit{rule}} \\[\tallspace] &= 22.5 \unit{rod} \unit{inches}. \end{aligned}

    上十之 "decupling it upward" is, in modern notation, the multiplication by 10 \unit{inches} / {\unit{rule}}, which converts rules to inches.

    The area needs to be in \unit{rod} \unit{inches} because it is to be divided by the post's shadow, which is given in inches. But since Chinese mathematics had no notion of dimensional analysis, \unit{rod} \unit{inches} were simply referred to as \unit{rods}.

以表影五寸除之、即得。 Dividing it by [the] post's shadow five inches, [we] are done.
  • In modern notation,
    \begin{aligned} l(\text{pole}) &= \frac{ s(\text{pole}) \cdot l(\text{post}) }{ s(\text{post}) } \\[\tallspace] &= \frac{22.5 \unit{rod} \unit{inches}}{5 \unit{inches}} \\[\tallspace] &= 4.5 \unit{rods}. \end{aligned}

Cite this page

Conway (2023). "Sun Tzŭ's Computational Classic: Volume III §25". <https://yawnoc.github.io/sun-tzu/iii/25> Accessed yyyy-mm-dd.