Babylonian Mathematics/Examples/Division of Triangular Field
Examples of Babylonian Mathematics
A triangular field is to be divided between $6$ brothers by equidistant lines parallel to one of the sides.
Expressed in Babylonian notation:
What is the difference between the brothers' shares?
Solution
The difference between each successive share is:
- $37; 55$ in Babylonian notation
- $37 \frac {11} {12}$ in mixed fractions.
Proof
Let $\triangle ABC$ be the triangular field in question.
Let $d$ be the marked side.
Let $a$ be the side which is parallel to the dividing lines.
Let $\AA$ be the total area of $ABC$.
Let $\AA_1, \AA_2, \ldots, \AA_6$ be the areas of each of the divisions of $ABC$ such that $\AA_1 > \AA_2 > \cdots > \AA_6$.
Let $a_1, a_2, \ldots, a_5$ denote the dividing lines such that $a_1 > a_2 > \cdots > a_5$.
From Area of Triangle, we have that:
- $\AA = \dfrac 1 2 k d a$
where $k = \sin \angle CAB$.
Then:
\(\ds \AA_1\) | \(=\) | \(\ds \dfrac {\paren {a + a_1} k d} {2 \times 6}\) | Area of Trapezium | |||||||||||
\(\ds \AA_j\) | \(=\) | \(\ds \dfrac {\paren {a_j + a_{j - 1} } k d} {2 \times 6}\) | for $j = 2$ to $5$ | |||||||||||
\(\ds \AA_6\) | \(=\) | \(\ds \dfrac {a_5 k d} {2 \times 6}\) | Area of Triangle |
We have:
\(\ds a_j\) | \(=\) | \(\ds a \paren {1 - \dfrac j 6}\) | for $j = 1$ to $5$ |
and so:
\(\ds \AA_{j - 1} - \AA_j\) | \(=\) | \(\ds \dfrac {\paren {a_{j - 1} + a_{j - 2} } k d} {2 \times 6} - \dfrac {\paren {a_j + a_{j - 1} } k d} {2 \times 6}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {\paren {a_{j - 2} - a_j} k d} {2 \times 6}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {\paren {a \paren {1 - \dfrac {j - 2} 6} - a \paren {1 - \dfrac j 6} } k d} {2 \times 6}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {a k d} {36}\) | after tedious algebra | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {\AA} {18}\) | substituting for $\AA$ |
We have that:
\(\ds \AA\) | \(=\) | \(\ds 11, 22; 30\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 11 \times 60 + 22 + \dfrac 1 2\) | expressing in decimal notation | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \dfrac {\AA} {18}\) | \(=\) | \(\ds \dfrac {682} {18} + \dfrac 1 {36}\) | |||||||||||
\(\ds \) | \(=\) | \(\ds 37 + \frac 8 9 + \dfrac 1 {36}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 37 + \frac {32} {36} + \dfrac 1 {36}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 37 + \frac {33} {36}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 37 + \frac {11} {12}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 37 + \frac {55} {60}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 37; 55\) |
Note that it is not necessary to know how long the marked side is.
$\blacksquare$
Sources
- 1945: O. Neugebauer and A. Sachs: Mathematical Cuneiform Texts
- 1992: David Wells: Curious and Interesting Puzzles ... (previous) ... (next): Dividing a Field: $12$