Wednesday 3 February 2010

Triangle Area Formulas, Old and New

Ok, I know Heron's Formula (if your teacher calls it Hero's formula, it's the same)...

Heron of Alexandria, sometimes called Hero, lived around the year 100 AD and is most often remembered for a formula for the area of a triangle. The formula gives a method of computing the area from the lengths of the three sides. If we call the sides a, b, and c; then the area is given by where the "s" stands for the semi-perimeter, . You can find a nice geometric proof of Heron's formula at this link to the Dr. Math site. The proof was done by Dr. Floor, who credits the method to Paul Yiu of Florida Atlantic University.

Documents from the Arabic writers indicate Archimedes may well have known this formula 300 years before Heron. In 1896 a copy of Heron's Metrica was recovered in Constantinople (now Istanbul) that had been copied around 1100 AD. It contains the oldest known demonstration of the formula. Heron is also remembered for his invention of a primitive steam engine and one of the earliest forerunners of the thermometer. The image at right shows a picture of a reconstruction of Heron's steam engine. The image is from the Smith College meuseum of Ancient Inventions where you can find more about Heron's, and many other's, interesting creations.
An extension of Heron's area formula for cyclic quadrilaterals is known as Brahmagupta's Formula

Heron's Metrica also contains one of the earliest examples of a method of finding square roots that is called the divide and average model. To find an approximate square root of a number, N, think of any number smaller than N, which we will call M. Then find a new approximation by letting E = (M + N/M)/2. Another approximation can be found by repeating the method with this new approximation. For example, beginning with N=20 and M= 2, we get E= (2 + 20/2) / 2 or E= (2+10)/2 = 6.
Repeating with M= 6 we get E= (6+ 20/6)/2 = ( 6 + 3 1/3 )/2 = 14/3 or 4 2/3. After only two iterations from a very bad starting guess the approximation is within .2 of the correct value.

Heron is also remembered for a problem he solved in Catoprica; Given two points, A and B, on the same side of a line, find a point X on the line so that the total distance AX+XB is a minimum. The solution may come quickly if you know that the translation of Catoprica is "About Mirrors". The solution given by Heron is to find the mirror reflection of point B in the line, B', and draw a straight line from A to B'. Where it intersects the line is the choice of point X.

Ok, so much for the old news... but recently I was going through some old journals that Dave Refro sends me from time to time to keep me out of mischief, and I came across an article in the 1885 Annals of Mathematics which listed 105 different formulas for the area of a triangle ( things to do on a rainy afternoon, list 110 different formulae for the area of a triangle). One was the well known Heron's formula above and then there was another that looked strikingly similar. If we let MA be the length of the median to vertex A, and similarly for MB and MC . The we can call sigma 1/2 the sum of MA + MBa+MC. Then we can write.

Now that is a new one to me..

1 comment:

Paddy said...

Very nice. I used those formulas to combine them with AGI to give this:

A more generalised result is here: