Such triangles are interesting in that both their side lengths and areas are integer. It should be fairly easy to see that all pythagorean triangles fulfil this property (which, by the way, shows that my heron picture is in fact not made of heronian triangles at all – as all the right angle triangles are isoceles which means they cannot have all integer sides…)
I’ve been having a little look at a problem asking for all triangles with integer sides which have an integer ratio of area to perimeter. The problem is delimited by giving a maximum value of that ratio – for the original see Problem 283 at projecteuler.net. A little thought shows that to have an integral ratio, the ratio must first be rational. For a triangle to have integer sides and a rational area:perimeter ratio, the area must be integral – therefore the triangles in question must be Heronian.
What makes the puzzle a bit tricky is that not all Heronian triangles are right angled (pythagorean). If they were – it is relatively easy to generate all pythagorean triples with the required property. However, the non-pythagorean Heronian triangles are harder to generate and it is rather harder to convert the limit on the ratio to the largest Heronian triangle which might fall within the constraint.
This post simply records my line of thinking on a first inspection of the problem:
Consider pythagorean triples (3,4,5) and (5,12,13). The triangle with sides 3,4,5 has area 6 (=(3*4)/2) and perimeter 12. Thus, it has an area:perimeter ratio 0.5. For ease of notation – let’s call the area:perimeter ratio r. If we double the side lengths (i.e. triangle with sides 6,8,10) the triangle has area 24 and perimeter 24 => r=1. So, we have a triangle fulfilling our criterion. Similarly with 5,12,13: area=30, perimeter=30 => r=1.
Now consider right angled triangles 12,16,20 (3,4,5 * 4) and 5,12,13. The former has r=2 and the latter r=1 – both within the set we are looking for. If we then sit them together with the edges of length 12 touching (adjoin them), we form another triangle with side lengths 13,20 and 21. As the right angled triangles have integral ares, the triangle formed by their adjunction must also have integral area and therefore this must be Heronian. In this case, the triangle so formed has area = 126 and perimeter = 54. r = 7/3. It doesn’t fit our criterion, but all is not lost. If we multiply the whole thing by 3, it does! Thus – triangle with sides 39,60,63 is a Heronian triangle with r=1 again. This is the adjunction of right angled triangles 36,48,60 and 15,36,39 with r=6 and r=3 respectively (non-primitive pythagorean triangles = (3,4,5)*12 and (5,12,13)*3) to give a Heronian with r=1.
So, we can see that adjoining two right angled triangles with (relatively) large r gives a Heronian with fairly large side lengths and a perimeter larger than either right angled triangle, but with a much smaller r (=1 in the example case). This means we might have to find some very large Heronian triangles to ensure I’ve caught all of them with ratio <= 1000. So, thoughts continue. Parameterisations should help with both generation of the triangles and turning the limits on r into limits on the size of the triangle (either in terms of perimeter or longest side). In fact turning the limit on r into a limit on size of right-angled triangles is fairly easy – we need to find triangles with a maximum non-hypotenuse side length of 2*maximum_r + 1. I’ll sign off now – proof of that in a future post, maybe.