Okay, I've been struggling on this for a couple days now. Given a 45-45-90 triangle with legs of length n, extend three perpendicular rays, one from each segment, such that they intersect within the body of the right triangle, and such that each is of equivalent length. What is that length?
It's empirically possible, obviously, but I can't seem to solve for x to save my life.
The intersection point represents the physical rotation point of the overall triangle, such that every face can rest flat against the X axis and that intersection point will travel only laterally, never vertically.
The closest guess I've arrived at is n-((n√2)/2), but I'll be damned if I know WHY (or even if that's actually correct); it just seems to fit a series of test inputs to an approximate degree, though I cannot prove it right. I cannot seem to find a critical length whatever approach I try.
(Forgive the crudeness of the diagram; I did that on my phone)
You are constructing what is known scholastically as the incenter of the triangle, and $x$ is the radius of the incircle. For circumscribed polygons (which all triangles are) we have the relation $A=\frac12p r$, where $A$ is the area, $p$ is the perimeter and $r$ is the inradius. Therefore here $$\frac12 n^2=\frac{n+n+\sqrt2n}2 x\\ x=\frac{n}{2+\sqrt2}$$