8
$\begingroup$

The cubic formula and modern math is not allowed, only algebra, geometry, and the like. Supposedly this problem was given to Fibonacci. Here is the whole paragraph I read:

In Flos Fibonacci gives an accurate approximation to a root of $10x + 2x^2 + x^3 = 20$, one of the problems that he was challenged to solve by Johannes of Palermo. This problem was not made up by Johannes of Palermo, rather he took it from Omar Khayyam's algebra book where it is solved by means of the intersection of a circle and a hyperbola. Fibonacci proves that the root of the equation is neither an integer nor a fraction, nor the square root of a fraction. He then continues:-

And because it was not possible to solve this equation in any other of the above ways, I worked to reduce the solution to an approximation. Without explaining his methods, Fibonacci then gives the approximate solution in sexagesimal notation as $1.22.7.42.33.4.40$ (this is written to base $60$, so it is $1 + 22/60 + 7/60^2 + 42/60^3 + \dots$). This converts to the decimal $1.3688081075$ which is correct to nine decimal places, a remarkable achievement.

Source: : Leonardo Pisano Fibonacci (short biography), School of Mathematics and Statistics, University of St Andrews, Scotland

This is just out of curiosity, I have no idea how this problem could be solved in terms of a circle and a hyperbola.

$\endgroup$
  • 1
    $\begingroup$ Try this automated GeoGebra applet. It helps to show how he approached it, very clever! Regards $\endgroup$ – Amzoti Sep 7 '13 at 22:25
  • $\begingroup$ @Amzoti Thanks! $\endgroup$ – Ovi Sep 7 '13 at 22:50
6
$\begingroup$

Well, it's actually pretty complicated... Because it's not just one circle and one hyperbola...

If you are really motivated, I think you may find this reading interesting : http://www.math.cornell.edu/~dwh/papers/geomsolu/geomsolu.html

Good luck ! I had once to learn how this work and it's pretty logical and impressive at the end, but not always as easy to understand as it seems at first sight. ^^'

PS : note that they don't discuss this equation in particular.

$\endgroup$
  • $\begingroup$ Thanks! I will upvote your answer when my voting limit is reset $\endgroup$ – Ovi Sep 7 '13 at 22:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.