Of a given perimeter, do there exist infinitely many right-angled triangles?

Question

I'm a high school student (11th grade, CBSE curriculum). My first language isn't English.

So here's my question:

Having a unique perimeter, let's say $x$ units, how many right-angled triangles, if a finite number of them, do exist? Are there infinitely many right-angled triangles having the same perimeter? How do we prove that, in case it's true or false?

(Why is my question put on hold as “off-topic”? What does that even mean)

Answer 1

HINTS

We may assume the perimeter is $1$. Then we want to know the number of solutions, in positive numbers, to $$x^2+y^2=z^2\tag{1}$$ subject to $$x+y+z=1\tag{2}$$ Substitute $(2)$ into $(1)$ to eliminate $z$ and you get $$2x+2y-2xy=1$$

The graph of this equation is a hyperbola, but not all the points on it correspond to positive $x,y,z$.

Take it from here.

Answer 2

A selection of a right triangle comes down to choosing lengths $a$ and $b$ for the legs which will then determine the length $c$ of the hypotenuse by the Pythagorean Theorem $$c=\sqrt{a^2+b^2}.$$ So a natural rephrasing of the question would be: Given a positive number $x$ are there infinitely many pairs $(a,b)$ such that $a+b+\sqrt{a^2+b^2}=x$?

In fact, for each side length $a$ which is between $0$ and $x/2$ there is such a pair. And since there are infinitely many values between $0$ and $x/2$ there are infinitely many right triangles with perimeter $x$.

To actually prove this result you need to use something you will be familiar with if you have taken calculus, the Intermediate Value Theorem or IVT. The IVT states that if you have a continuous (no gaps or breaks in the curve) function $f$ such that $f(a)=c$ and $f(b)=d$, then if some number $q$ is in between $d$ and $e$, there will be a number $p$ that is between $a$ and $b$ such that $f(p)=q$. The IVT is essentially saying that if a curve passes through two values on the vertical ("y") axis, then it must also pass through every value in between.

Solving your problem provides a concrete example of the IVT. Suppose that $a$ is greater than $0$ and less than $x/2$. Think of $a+b+\sqrt{a^2+b^2}$ as a function in which $a$ is fixed and $b$ is variable. Then when $b=0$, $a+b+\sqrt{a^2+b^2}=a+a=2a<x.$ Also, when $b=x$, $a+b+\sqrt{a^2+b^2}=a+x+\sqrt{a^2+x^2)}\geq 2x >x.$ For certain values of $b$ the perimeter will be too small and for certain values of $b$ the perimeter will be too big. Therefore by the IVT there must be a $b$ which is just right, for which $a^2+b^2+\sqrt{a^2+b^2}=x.$ The IVT is our Goldilocks principle.

Since every value of $a$ between $0$ and $x/2$ corresponds to certain $b$ and a certain right triangle with perimeter $x$ there have to be infinitely many right triangles with perimeter $x$.