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[ed boy]

Does infinity/infinity have a solution? The only possible answers I can see are infinity or one. Am I wrong, or is one of these correct?

You must remember that there is no such number as infinity.

Mathematicians use the word infinite a lot, but that is merely as a shorthand to avoid needless repetition of lengthy phrases.

It helps to replace "infinite" with "unlimited", and "finite" with "limited".

For example, when we say that a set in infinite, we mean that there is no natural number x such that the set has x elements.

When you approach maths at a university level, you can learn about the extended complex numbers, which is a system that contains a concept called infinity that is treated similarly to a number. However, there are multiple distinct x such that x multiplied by infinity equals infinity, so we cannot have an inverse for multiplying by infinity, so dividing by infinity is impossible.

[Dsarker]

Okay, thanks!

[RedWarrior0]

Okay, physics problem. Show the flight time of an object, launched from y-value 0 at angle θ with initial velocity v₀ to a landing point at y-value h, to be (.5)(T₀)(1+sqrt[1-(h/H)]), where T₀ is the flight time of h=0, and H is the maximum height ([v₀²*sin²(θ)]/2g, by the way)

It's really annoying me. I feel like I should be able to get it, but I can't for some reason.

[Darvi]

*nevermind*

I could get the formula for dynamic flight, but it's 2:30 and I don't feel like it.

[ed boy]

It's a quadratic equation. (I'm going to assume that air resitance is not a factor)

The vertical acceleration is a constant of -g, so the vertical velocity is linear A-gt for some A, so the position is quatratic B+At-(1/2)gt²

We know that when t=0, the vertical position is 0, so we know that B is 0.
So we solve -h-At+(1/2)gt²=0.

We also know that initially the vertical velocity is vsin(θ), so A=vsin(θ).

Simply solve the quadratic equation, substitute in 2gH for v², and you will get your answer. Keep in mind that your answer will be in a slightly different form depending on whether you are taking g as a negative number in a positive direction or a positive number in a negative direction (I went for the latter).

[MonkeyHead]

Used to have a script I wrote in VBasic that did projecticle calculations like that one. Cant for the life of me find it now tho, it was years ago...

[Bauglir]

Random geometry question. Is it possible to have a polygon enclosing a circle such that no line perpendicular to any edge of the polygon passes through the circle? My gut says no, but it would be really conveniently awesome if my gut were wrong.

EDIT: Oh, and the internal (on the inside of the polygon) angle of every vertex needs to be less than 180 degrees. That's kind of important.

[Virex]

No, since for every edge of the polygon, the set of lines perpendicular to that edge covers the entire polygon. You'd have nowhere to put the circle.

[Bauglir]

No, since for every edge of the polygon, the set of lines perpendicular to that edge covers the entire polygon. You'd have nowhere to put the circle.

Wait, that can't be true. Regular pentagon. Doesn't meet all my requirements, obviously, but the lines perpendicular to each edge leave some area unfilled.

[eerr]

No, since for every edge of the polygon, the set of lines perpendicular to that edge covers the entire polygon. You'd have nowhere to put the circle.

Wait, that can't be true. Regular pentagon. Doesn't meet all my requirements, obviously, but the lines perpendicular to each edge leave some area unfilled.

There are an infinite number of lines perpendicular to another line. Is there another constraint?

[Bauglir]

No, since for every edge of the polygon, the set of lines perpendicular to that edge covers the entire polygon. You'd have nowhere to put the circle.

Wait, that can't be true. Regular pentagon. Doesn't meet all my requirements, obviously, but the lines perpendicular to each edge leave some area unfilled.

There are an infinite number of lines perpendicular to another line. Is there another constraint?

Oh, right, it has to cross that edge too. I'm dumb, I should've specified that and that is why Virex's answer works for what I'd laid out. Sorry, Virex. Heck, I wrote up a whole post about how yours was irrelevant, eerr, since there are infinitely many points on any finite line, but then I realized what you actually meant.

To be clearer, the requirements for the polygon would be:

A) No internal angles greater than 180 degrees
B) A finite area within the polygon is not crossed by any line that
   a) is perpendicular to any edge
   b) crosses the edge to which it is perpendicular

[Jim Groovester]

What are the requirements for the polygon to enclose the circle?

[Bauglir]

Well, the circle's area just has to be enclosed entirely within the polygon's is what I meant by the phrase, but then I realized that I could just inscribe a circle on any finite area within the polygon so the circle isn't really a necessity anymore with the requirements listed in my last post.

It can just be tacked on later, and has more to do with the thing I was thinking about when the question came to me (which, for what it's worth, was considering how to maximize the extent to which a given piece of armor deflects incoming attacks; I figured it'd do a better job if some component of any attack that can actually strike the wearer must have some component parallel to the surface of the armor, wasting that portion of the incoming energy).

[Christes]

As far as I understand it, here is how I would restate your problem (in a slightly stronger form):

For any convex polygon P and and point p inside of it, there exists an edge E and line L, such that L passes through p, intersects E, and is perpendicular to E.

I claim this is true.  (Note that it is false if you omit convex.  An easy counterexample is a star)

Proof:  Draw segments from p to each vertex of P, breaking the interior of P into triangles.  For each triangle, call the angle at P the "top" angle and the others the "base" angles.  It suffices to show that one of these triangles has base angles that are both acute.  One of these line segments from p to the vertices must be the shortest.  Look at the two triangles containing it.  Call them T_1 and T_2

The next part is a lot more delicate, and hard to explain.  Consider the base angles of T_1 and T_2.  Assume that each has a base angle greater than 90 degrees.  Since we assumed the shared side is the shortest, the obtuse angle cannot be opposite it in either triangle.  So the obtuse angles must be adjacent.  Thus the polygon has an interior angle greater than 180 degrees, which is a contradiction.

I could make a picture if you really want to understand the proof.

[Soadreqm]

Does infinity/infinity have a solution? The only possible answers I can see are infinity or one. Am I wrong, or is one of these correct?

The limit of x/x when x approaches infinity is one. This is not actually the same as ∞/∞, just as dividing by zero is not the same as dividing by a variable that approaches zero, but depending on what you're doing, it can be close enough.

Just don't try it if any actual mathematicians can see.