help right triangles

6, 8, 10 is the only one that I found.

I think you can mathematically show that the 6-8-10 right triangle that PurpleWasp discovered is the only one. Consider the diagram below, which showcase the two possibilities for a non-congruent right triangle. Either the side length of 6 is the leg of the right triangle or it is the hypotenuse of the triangle.

Let's consider \(\triangle {\text{ABC}}\) first. We can use Pythagorean's Theorem to find the relationship between the lengths of the sides of the triangles, and simplify to a point where we can begin trying solutions.

\(h^2 = s_1^2 + 6^2 \\ h^2 = s_1^2 + 36 \\ 36 = h^2 - s_1^2 \\ 36 = (h + s_1)(h - s_1)\)

We have now changed the problem to factoring the number 36 where one factor is \(h + s_1\) and the other is \(h - s_1\). The number 36 has a lot of factors, but there are a lot of restrictions on these factors. For one, \(h \text{ and } s_1\) are both lengths of a triangle, so we need not waste time considering negative factors. We should also realize that \(h + s_1 > h - s_1\) since adding two positive numbers should always be greater than their corresponding difference. We can now list all the factors that meet this criteria.

\(36 = 36 \times 1 \\ 36 = 18 \times 2 \\ 36 = 12 \times 3 \\ 36 = 9 \times 4\)

Now, we have found a few candidates for possible factors of 36. We can solve the corresponding systems of equations and then find the mystery side lengths. I will start with the first set of factors.

\(\begin{cases} h + s_1 = 36 \\ h - s_1 = 1 \end{cases} \\\)

Probably simplest is to use the elimination method to solve this system of equations. This system would then simplify to \(2h = 37\). We can stop here and not go any further because we recall that \(h \in \mathbb{Z}\), and \(h = \frac{37}{2} \not\in \mathbb{Z}\). With this insight, we recognize that any two factors that have one even and one odd factor cannot have integer solutions. This leaves us with one possibility left, that is \(36 = 18 \times 2\). Let's solve this one with the elimination method!

\(\begin{cases} h + s_1 = 18 \\ h + s_2 = 2 \end{cases} \\ 2h = 20 \\ h = 10 \\ s_1 = 8\)

This answer is recognizable. This is the 6-8-10 right triangle that PurpleWasp discovered. We have no more possibilities for \(\triangle \text{ABC}\), so we move on to \(\triangle \text{EDF}\). 

Once again, we use the Pythagorean's Theorem to find a relationship that may give us insight on possible integer solutions.

\(6^2 = s_1^2 + s_2^2 \\ s_1^2 = 36 - s_2^2 \\ s_1^2 = (6 - s_2)(6 + s_2)\)

Since \(h = 6\) in this case, we know that \(s_2 < 6\) since the hypotenuse is the longest side of a right triangle. The question for this factor becomes when  \(6 - s_2 \text{ and } 6 + s_2\) are both perfect squares. Since \(0 < s_2 < 6\), \(6 - s_2\) will be a perfect square when \(6 - s_2 = 4 \text{ or } 6 - s_2 = 1\) because they are the only two perfect below 6. The only options are \(s_2 = 2 \text{ or } s_2 = 5\). Let's check both of those options.

\(s_1^2 = (6 - s_2)(6 + s_2) \\ \begin{align*} \text{Let } s_2 = 2:& 6 + s_2 = 8 \\ \text{Let } s_2 = 5:& 6 + s_2 = 11 \end{align*}\)

Notice how in both cases the \(6 + s_2\) term is not a perfect square. This means that \(s_1\) will be an irrational length, which is not acceptable given the parameters of the question. Therefore, no options exist for \(\triangle \text{EDF}\).

This leaves the 6-8-10 right triangle as the only option.