Pythagorean triples in a triangle inside a rectangle

Question

Suppose there's a right-angled triangle inside a rectangle in the following way:

(just take any right-angled rectangle, "rotate" it and draw a rectangle around it)

There are four right-angled triangles in that image.

Does there exist a triangle and rectangle combo so that all the resulting triangles have Pythagorean triples as their sides?

Answer 1

I realize this is rather hastily done but I'm at work so I'll have to fix this later. Fixed, my earlier "solution" was super wrong check it out to find a great logical fallacy.

I checked the answer from the link you posted and this problem is a direct concequence. So let's look at the solution from the other question:

Note that triangles $ABC$ and $CDE$ must be similar if $BCD$ is to be a straight line, so both must be built from integer multiples of the same primitive pythagorean triple $T= {a,b,c}$ with hypotenuse $c$.

Suppose $ABC$ is $pT$ and $CDE$ is $qT$ then $AC=pc$ and $CE=qc$ and $AE^{2}=p^{2}c^{2}+q^{2}c^{2}$ so that $AE$ is divisible by $c$ and equals $rc$.

You can build your trapezium from any pair of pythagorean triples (not necessarily primitive)

$T={a,b,c}$ and $U= {p,q,r}$

with $ABC = \{pa,pb,pc\}, CDE = \{qb,qa,qc\}, ACE = \{pc,qc,rc\}$.

And this is essentially the only way of doing it.

Now lets keep going from where he left off and expand $BA$ to $G$ to make the right angled triangle $AGE$ and we'll see if it also has integer side lengths.

$$ AG = qb - pa \\ GE = pb + qa \\ AE = cr $$

Yup it does.

That's it. But let's verify that it really is a right angled triangle: $$ AG^{2} + GE^{2} = AE^{2} \\ \Leftrightarrow (qb - pa)^{2}+(pb + qa)^{2} = (rc)^{2} \\ \Leftrightarrow (qb)^{2}+(pa)^{2}+(pb)^{2}+(qa)^{2} = (rc)^{2} \\ \Leftrightarrow^{*} (qc)^{2} + (pc)^{2} =^{**} (rc)^{2} $$

$*\: a, b, c$ is a pythagorean triplet
$**\: q, p, r$ is a pythagorean triplet

QED

Answer 2

Yes. One example would be a rectangle as you have drawn, $100$ horizontal and $96$ vertical. The left side is divided $75+21$ and the bottom is $72+28$. The upper line is $100$, the left lower is $75$, and the right lower is $100$. You can make one starting with any Pythagorean triangle you like. See the figure below. $s=\sin \theta, c=\cos \theta$ are both rational based on your favorite triangle. You need $\theta$ to be the small angle so you can double it. The labels on the segments come from trig, including the double angle formulas. As all the sides are rational, scale up and you have integers. In my example the top two triangles are both $75-100-125$, the lower left is $21-72-75$ and the right one is $28-96-100$