Puzzle time - Beetles edition

[jon-nyc]

Four tiny beetles, are on each of the vertices of a square of side length s. At the same moment each beetle starts walking (continuously) toward the beetle to its left. All beetles walk at the same speed. What distance does each beetle travel before they all meet in the middle?

Now, generalize it for n beetles on an n-sides polygon.

(This is what I was doing from at 4am this morning. Friend gave me the first problem, generalizing it was my idea)

[Klaus]

I don't understand. If they walk to the beetle to the left, how would they ever meet in the middle?

[jon-nyc]

They don’t stay on the square. They continuously head toward the next beetle in a straight line, who is moving toward the next in a straight line, etc

[Klaus]

Oh, now I see. You mean they constantly change direction to take the shortest path to the next beetle, whereever that beetle is at that time, right? But they do not "foresee" the future movement of that other beetle to adapt their "shortest path" calculation, right?

That sounds like a hard puzzle.

[jon-nyc]

Exactly.

There are several ways to solve it, my friend had two, I came up with a third.

[AndyD]

@Klaus said in Puzzle time - Beetles edition:

That sounds like a hard puzzle.

Good enough answer for me. Advanced maths or what ! Trying to imagine an increasingly spiral path?

[Klaus]

I also wonder about the limit case: infinite number of beetles on a circle. I'd say the solution is that no traveling occurs and they don't meet in the middle. But for finite n it seems to be the case that they would meet in the middle. That's a strange discontinuity.

[jon-nyc]

It’s a limit, not a discontinuity. The distance approaches infinity as n does.

[Klaus]

Oh I see. So in the limit case the beetles just all walk simultaneously on the perimeter of the circle.

[jon-nyc]

I would start with the square case then generalize, not start with the general case.

[Klaus]

I'll start with the 2-sided polygon.

If the length of the line is s, then every beetle travels s/2.

And as a bonus, for a 1-sided polygon, the distance is 0.

Do I get the prize?

[jon-nyc]

What do you think this is, everybody gets a trophy?

[Klaus]

:::

OK, slightly more seriously:

I'd say the beetles will always form a square at each point in time, but that square rotates and shrinks.

At every time, the movement vectors will hence be orthogonal to each other. Which means that every beetle has to move distance s before the size of the square becomes 0.

:::

[jon-nyc]

That’s right!

Now generalize to N sides.

[Klaus]

But then the "orthogonality" thing doesn't hold anymore and things get complicated. Please don't tell me you want me to write down some complicated trigonometric functions and differential equations!

[Klaus]

Also, bonus question for Jon:

How many times do the beetles spin around each other before they meet?

[jon-nyc]

@Klaus said in Puzzle time - Beetles edition:

But then the "orthogonality" thing doesn't hold anymore and things get complicated. Please don't tell me you want me to write down some complicated trigonometric functions and differential equations!

I didn't have to.

[jon-nyc]

@Klaus said in Puzzle time - Beetles edition:

Also, bonus question for Jon:

How many times do the beetles spin around each other before they meet?

Define 'spin around each other'. You mean that the square formed by the four of them rotates 360 degrees?

[Klaus]

yes

[jon-nyc]

@Klaus said in Puzzle time - Beetles edition:

But then the "orthogonality" thing doesn't hold anymore and things get complicated. Please don't tell me you want me to write down some complicated trigonometric functions and differential equations!

I could give you a hint. Really a way to reframe the orthogonality that will generalize to other n