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# help right triangles

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I need to find right triangles

Find the number of non-congruent right triangles, where all the sides are positive integers, and one of the legs is $6$.

Aug 24, 2023

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

Aug 24, 2023
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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.

Aug 25, 2023