Babylonian math problem - Solution

Thursday, 21 November, 2019

This is the solution to the monthly maths problem from RTSdécouverte Babylonian math problem. 




The length of the left side is 1,45 (or 105 in decimal) and the length of the right side is 15.

Trapezium [UNIGE - Mathscope]


The scribe reasons as follows:

Multiply 1,45 with 1,45: this gives 3,3,45

Multiply 15 with 15: this gives 3,45

Add to 3,3,45: this gives 3,7,30

Divide 3,7,30 in two: this gives 1,33,45

What is the square root of 1,33,45? It is 1,15.


In decimal this is 75. What was the scribes reasoning?

If the left side has length L and the right side length l, then the sought after length x is

This is the formula the scribe is using. Where does it come from?

Recall that the area of a trapezium is obtained as half the result of multiplying the height by the sum of the short base and the long base. With the notation above:

Let M and N denote the two ends of the segment x dividing the trapezium into two equal bands. The area of the ABCD trapezium is twice the area of the BCMN trapezium.

This can be re-arranged to give an expression for y as follows.

Let us now consider a parallel to segment AB which passes through point C. This line defines two similar triangles CMP and CDQ (see sketch).  We can apply Thales' theorem to these two triangles - although note the anachronism in doing this. We then obtain:



From this we get another expression for y as:


Setting these to expressions for y equal we obtain:


Simplifying and solving for x yields:



The only solution which is a positive number is the one given:


As to whether the Babylonians knew the Theorem of Thales, that is another story…


You can read the original problem here: Babylonian math problem


Original RTS Problèmes du mois (FR)