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:
 * the length of the marked side is $6; 30$
 * the area of the triangle is $11; 22, 30$.

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

 * Division-of-Triangular-Field.png

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 $k = \sin \angle CAB$.

From Area of Trapezoid:
 * $\AA_1 = \dfrac {\paren {a + a_1} k d} {2 \times 6}$