Another “method” that gives some of the solutions given above is to start with your favourite triangle, and apply the matrix I displayed where x, y, and z are carefully chosen integers. For instance, starting from sides of length 3 and 4 you get the family of triangles (0,0,0), (3(x^2-y^2-z^2), 6xy, 6xz), (8xy,4(y^2-x^2-z^2),8yz). Substituting x=y=z=1 gives (-3,6,6), (8,-4,8), while substituting (x,y,z)=(3,1,2) yields 2 times (6,9,18),(12,-24,8)

Further to my last post, whenever 1/A=1/B+1/C, it follows that A,B,C,B+C-A is a Pythagorean quadruple. For

(B+C-A)^2=B^2+C^2+A^2+2BC-2AC-2BC

=A^2+B^2+C^2+2ABC(1/A-1/B-1/C)=A^2+B^2+C^2

So the crucial equation is 1/A=1/B+1/C. Reciprocals of integers are called unit fractions and the cases of the sum of unit fractions being a unit fraction must be known. Another example is 1/5+1/20 =1/4. More generally, 1/n+1/(n^2-n)=1/(n-1). I suspect that if you insist on there being no common divisor for A,B,C this is the only example, but I have not proved this and have not found it by googling.

There’s also \frac{1}{2} + \frac{1}{-3} = \frac{1}{6}, assuming that you meant it when you said “integers”.

EDIT: Or, d’oh, \frac{1}{6} +\frac{1}{3} = \frac{1}{2}

\frac{1}{10}+\frac{1}{15}=\frac{1}{6}, etc

Excellent. So when can a pair of unit fractions sum to a unit fraction?

I did not know there was a quiz … just pointing out that your example is just a case of: if m and k are relatively prime, then consider

I don’t think that you even need m and k to be relatively prime, there: Considering k=2, m=4, then you get 1/12 + 1/24 = 1/8, which is still true.

It does obviously raise the question, though, of whether there are any such unit-fraction triples not of that form.

They are not all of that form; e.g., \frac{1}{8}+\frac{1}{24} = \frac{1}{6}. However, in that example multiplying by 2 yields \frac{1}{4}+\frac{1}{12}=\frac{1}{3}, which is of that form.

The relatively prime restriction was about there being no common divisor to all the denominators. As for *all* examples, first of all we can cast out those common divisors and reduce to that case, and then you should be able to put it in that form

I’m aneued by my error.

I feel silly not seeing that when a=b+c, then divide by abc to get 1/bc=1/ac+1/ab. For example, when a=37,\ b=10,\ c=27, this yields 1/270=1/370+1/999. The converse is nearly true, every example of 1/A=1/B+1/C comes from this kind of equation, up to a constant factor. Putting this together with @brossa’s construction, we conclude that if b,\ c are arbitrary integers (they can be negative or even 0, but that latter would not lead to an interesting solution), and (e.f.g is a Pythagorean triple, then

(0,0,0),(bce,(b+c)ce,(b+c)be),((b+c)bf,bcf,(b+c)cf)

will be the vertices of a lattice triangle with all integer lengths.

Nice one.

I can follow that (no way in hell I could have constructed it!). My contribution, in its entirety, is this:

(B+C−A)2=B2+C2+A2+2BC−2AC−2BC should be:

(B+C−A)2=B2+C2+A2+2BC−2AC−2AB

See? I can’t even get the formatting right. Jeezus. Everything else checks out. Really incredible work.

ETA: Oh! and the last vertex needs to be ((b+c)bf, bcf, **-**(b+c)cf) - a negative z component.

You are correct of course. Brain cramp. It was an interesting exercise.

Incidentally, the triangles constructed using @brossa’s construction are all right triangles. For a non-right triangle, rotate the one with vertices (0,0,0), (5,12,0), (-9,12,0) (with side lengths 13, 14, 15) through some angle, say using 1/3i+2/3j+2/3k and then multiplying by 9 to turn the rational vertices into integers.

The right-triangle effect was a goal of my method. The idea was to create a triangle that could be hand drawn, skewed, in orthographic projection, then manipulated through various rotations and scalings and finally redrawn to show its true size and shape. The right angle and integer sides serve as a way to check the accuracy of the manipulations. It’s all in service of taking a class in stereotomy, so as to be able to hand-draft plans for stuff like this:

It is true that any triangle with integer side lengths would work for this, but I like seeing the right angle pop up for the first time after eight or ten views in which no angles are ‘true’.

Let me apologize for obsessing over this. I tried to reconcile @brossa’s construction with my example in post 12 above. Here is what I found.

Start with an isosceles right triangle. Take the origin as the right angle and let (A,B,C),(A',B',C') be the other two vertices and assume they have common length D. Then we must have A^2+B^2+C^2=D^2=A'^2+B'^2+C'^2 and AA'+BB'+CC"=0, the latter equation to make the two edges meet at a right angle. Then, whenever e^2+f^2=g^2, the triangles with vertices (0,0,0),(Ae,Be,Ce),(A'f,B'f,C'f) will have edges of length De,Df,Dg. This explains all of @brossa’s examples since the two conditions are obviously satisfied by (A,B.C),(C,A,-B) whenever 1/A=1/B+1/C. Incidentally, (A,B,C),(B,-C,A) will also work. And my example comes from (-7,4,4), (4,-1,8) with e=4,f=3,g=5. And any two rows of the matrix in @DPRK’s post 19 will always work, once you clear the fractions