Find triangles with areas that are integers

[Mikestone]

TL;DR Summary

Can one find which triangles (other than right-angled ones) have areas wheich are integers?

It is pretty obvious that all right-angled triangles whose sides are integers will have areas which are also integers. Since either the base or height will be an even number, half base x height will always come out exactly.

However, I have only found one non-right-angled triangle where this is the case. If the sides are 13, 14 and 15 then (if I've done it right) this gives an area of 84.

Are there any other such triangles with exact integer areas as well as sides, and if so is there any rule for finding them?

 

[Stephen Tashi]

It is pretty obvious that all right-angled triangles whose sides are integers will have areas which are also integers.

Can't we have a right triangle with two sides that are both of length 1? How are you defining "right-angled" triangle?

Edit: I see what you're asking. Some use the convention that a right triangle has 2 "sides" and a "hypotenuse".

 

[fresh_42]

Summary:: Can one find which triangles (other than right-angled ones) have areas wheich are integers?

Sure. You can choose two side lengths as any numbers you wish.

It is pretty obvious that all right-angled triangles whose sides are integers will have areas which are also integers.

It is obvious that this is not the case. Height and base can be the two perpendicular legs, hence the area is a half integer.

Since either the base or height will be an even number, half base x height will always come out exactly.

Why not choose $g=h=1$?

However, I have only found one non-right-angled triangle where this is the case. If the sides are 13, 14 and 15 then (if I've done it right) this gives an area of 84.

Are there any other such triangles with exact integer areas as well as sides, and if so is there any rule for finding them?

Why do you want to know? Which other regular triangles have you tried?

 

[undefined314]

Heron's formula may be helpful to you. A triangle of side lengths a, b, and c has area given by the following equation:

$$Area = \sqrt{s(s-a)(s-b)(s-c)},$$

Where s is the semiperimeter of the triangle. Restrict areas to positive integers, and see what falls out in combination with other conditions (such as the triangle inequality).

 

[mathman]

Take any right triangle with integer sides and join it with its mirror image along one of its sides to form an isosceles triangle. Area is obviously double that of the right triangle.

 

[Mikestone]

Can't we have a right triangle with two sides that are both of length 1? How are you defining "right-angled" triangle?

Edit: I see what you're asking. Some use the convention that a right triangle has 2 "sides" and a "hypotenuse".

Whoops, my mistake. I should probably have said "Pythagorean" rather than "right-angled'. On the other hand, if a right triangle had sides 1 and 1 its hypotenuse would of course *not* be an integer.

 

[fresh_42]

On the other hand, if a right triangle had sides 1 and 1 its hypotenuse would of course *not* be an integer.

... which you never explicitly required! We all had to guess what you meant.

 

[arydberg]

... which you never explicitly required! We all had to guess what you meant.

How about 5 5 6

 

[Merlin3189]

It surprises me that there are many. Actually I was surprised there were any.
Apart from 1-2-3, 2-3-5, 3-4-7, 2-13-15, etc. which are a bit trivial, they seem to be at least as common as Pythagorean triples.
4-13-15, 9-10-17, 11-13-20, 7-15-20, 10-17-21, 13-20-21, 12-17-25, 3-25-26, ... , 26-35-51, (34 so far)

 

[phyzguy]

How about 5 5 6

This is an example of @mathman's post #5, starting with a 3-4-5 right triangle.

 

[pbuk]

... which you never explicitly required! We all had to guess what you meant.

I think that's a bit unfair, the meaning was clear (to me at least, and also to @Stephen Tashi on second reading) from the context.

 

[pbuk]

These are called Heronian triangles, entry A072294 in the On-Line Encyclopedia of Integer Sequences is a good start for further enquiry.

 

[phyzguy]

These are called Heronian triangles, entry A072294 in the On-Line Encyclopedia of Integer Sequences is a good start for further enquiry.

The Wikipedia entry also has a lot of information.

 

[Mikestone]

Heron's formula may be helpful to you. A triangle of side lengths a, b, and c has area given by the following equation:

$$Area = \sqrt{s(s-a)(s-b)(s-c)},$$

Where s is the semiperimeter of the triangle. Restrict areas to positive integers, and see what falls out in combination with other conditions (such as the triangle inequality).

Thanks. I do know that formula, and indeed it was while toying with it that I happened upon the 13,14,15 triangle. I tried various other combinations but was evidently unlucky in my choices as they all produced irrational numbers. As I was already familiar with a set of formulae to produce Pythagorean Triples, I wondered if there was anysuch way of deriving non-Pythagorean ones.

 

[Mikestone]

The Wikipedia entry also has a lot of information.

Thanks a lot. The Wiki entry seems to give me what I was looking for.

Also thanks to everyone else who has tried to help. In my schooldays I was good at maths, but there's clearly still a lot of terminology with which I am unfamiliar.

 

[fresh_42]

I think that's a bit unfair, the meaning was clear (to me at least, and also to @Stephen Tashi on second reading) from the context.

Which was what I have said: implicit, not explicit.

 

[mathman]

It surprises me that there are many. Actually I was surprised there were any.
Apart from 1-2-3, 2-3-5, 3-4-7, 2-13-15, etc. which are a bit trivial, they seem to be at least as common as Pythagorean triples.
4-13-15, 9-10-17, 11-13-20, 7-15-20, 10-17-21, 13-20-21, 12-17-25, 3-25-26, ... , 26-35-51, (34 so far)

Actually there are an infinite number, which can be easily generated. Take any two different integers $m$ and $n$, let $a=2mn$, $b=n^2-m^2$ and $c=n^2+m^2$. To get primitives, $m$ and $n$ should be relatively prime and one of them should be even, while the other is odd.

 

[Mikestone]

Actually there are an infinite number, which can be easily generated. Take any two different integers $m$ and $n$, let $a=2mn$, $b=n^2-m^2$ and $c=n^2+m^2$. To get primitives, $m$ and $n$ should be relatively prime and one of them should be even, while the other is odd.

Aren't those specifically the formulae for generating *Pythagorean* triangles (ie right-angled ones) rather than heronian ones in general?

 

[mathman]

Aren't those specifically the formulae for generating *Pythagorean* triangles (ie right-angled ones) rather than heronian ones in general?

Yes.

https://en.wikipedia.org/wiki/Heronian_triangle
has formula to get all Heronian triangles

 

[Mikestone]

Yes.

https://en.wikipedia.org/wiki/Heronian_triangle
has formula to get all Heronian triangles

Thanks.

I did find the formulae, though you rather need to "read the fine print", as the results which they produce often have to be divided by their highest common factor in order to get the triangle in its "primitive" form.

 

[pbuk]

For a really comprehensive investigation of this and related topics you could start with this relatively recent paper and follow the references back.