One stick and Pythagoras

Question

I am strugling with a question: How can you design a right-angled triangle if you want a certain perimeter? let's say I have a 1 meter stick and I want to turn it into a right-angled triangle. Is there only 1 solution?

Thanks!

Answer 1

Given a 1 metre sick and calling the lengths of the sides of the triangle we want $a$, $b$ and $c$ (with $c$ the hypotenuse), we have by Pythagoras: $$ 1= a + b + c = a + b + \sqrt{a^2 + b^2},$$ leading to: $$ 1-(a+b) = \sqrt{a^2 + b^2}\\ 1 -2a-2b+a^2+2ab+b^2=a^2+b^2 \\ 1=2a+2b - 2ab=2(a+b(1-a)).$$ Given $a \in (0,1)$, we can solve for $b$: $$ b = \frac{\tfrac{1}{2} - a}{1-a}.$$ This means the solution to the problem will definitely not be unique. It also shows we need $a \in (0, \frac{1}{2})$.

Answer 2

There are an infinite number of solutions, but they can be parametrised in the length of one leg; we'll call this $a$. If the other leg is $b$ and we start with a unit stick, Pythagoras's theorem gives $$a^2+b^2=(1-a-b)^2=1+a^2+b^2-2a-2b+2ab$$ $$1=2-2a-2b+2ab=2(1-a)(1-b)$$ $$b=1-\frac1{2(1-a)}=\frac{1-2a}{2(1-a)}$$ $a$ has to be in $(0,\frac12)$ for a non-degenerate right triangle.