Heron and his Formula(s)

 Ok, you probably 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...no angles required. If we call the sides a, b, and c; then the area is given by \(A= \sqrt {s(s-a)(s-b)(s-c)}\)    where the "s" stands for the semi-perimeter, \(s=\frac{a + b + c}{2}\). 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.  Wikipedia also has several nice proofs of the theorem, including one derived from the Pythagorean Thm. 

I recently (2024) found a very similar formula for the square of the volume of a Tetrahedron in a paper in Letters to the editor of The Mathematical Intellingencer by Martin Lukarevski.  The Formula applies to a tetrahedron with all four faces congruent with edges a, b, c.    I did a little simple algebra to enhance the similarity to Heron's formula.

\(V(T)= \sqrt(\frac{( -a^2 + b^2 + c^2)( +a^2 - b^2 + c^2)( +a^2 + b^2 - c^2)}{72})\)

 Heron is also remembered for his invention of a primitive steam engine and many early automatons, and a coin operated vending machine,  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 museum 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 

  

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 M_A be the length of the median to vertex A, and similarly for M_B and M_C . The we can call sigma 1/2 the sum of M_A + M_B + M_C. Then we can write.  

\(A= \frac{4}{3}\sqrt {\sigma(\sigma - M_A)(\sigma - M_B)(\sigma - M_C)}\)Now that is a new one to me..

Here is one more I only learned recently.  The triangle at right has the lengths s-a, etc shown, and you realize that they are the radii of Soddy "kissing circles".(below) If we think of the side a, as opposite angle A (as we like to do in geometry) the a= (s-b )+(s-c), and b=(s-a)+(s-c), and you've figured out already that c=(s-a)+(s-b). Now if we use a' for s-a, and b' for s-b... then we have  a= b' + c',  and b=a'+c' and c=a' + b'.  

"But Why?", you ask.  Because the the formula can be written

as \( \sqrt{a' b' c' (a'+b'+c'}\), which looks much simpler.