Two objects move in circular paths of radius 1 so that each intersects

>>8537373
Set them equal and solve for t.

They might not collide ever jsyk.

>>8537373
>>8537375
To elaborate on this: there's formula for the difference of two sines and cosines. Your problem is solvable analytically.

>>8537388
Just say you don't know, bro.

>>8537373
t = pi(2a+1/3) or pi(2a-1/3)

t^2 = pi(2b+2/3) or pi(2b-2/3)

Where a and b are integers.

So there is no solution otherwise pi would be rational

>>8537477
What's the lowest integer power greater than 1 that t would have to be raised to in the second function in order for a collision to take place?

If two objects more in a circular path like stated in the OP, and they are not moving at the same velcoity, then no matter what will they collide/intersect/whatsvsr you want to say given enough time?

>>8537540
None, because that would imply that an integer power of pi is rational.

>>8537541
They can probably get arbitrarily close

>>8537541
No, already proved it can't here >>8537477

For you brainlets, first we see that the objects can only collide at the two intersections of the circle. Since the intersections and centers form equilateral triangles, the top intersection can conveniently be described as:

p1 = < cos(2piA+pi/3), sin(2piA+pi/3) >
p2 = < cos(2piB+2pi/3) + 1, sin(2piB+2pi/3) >

The bottom intersection is

p1 = < cos(2piA-pi/3), sin(2piA-pi/3) >
p2 = < cos(2piB-2pi/3) + 1, sin(2piB-2pi/3) >

So we have

t = pi(2A+1/3) or pi(2A-1/3)
t^2 = pi^2(2A+1/3)^2 or pi^2(2A-1/3)^2

and

t^2 = pi(2B+2/3) or pi(2B-2/3)

pi^2(2A+1/3)^2 = pi(2B+2/3)
pi = (2B+2/3) / (2A+1/3)^2

Which would make pi rational.

For p2 = < cos(t^n) + 1, sin(t^n) >

We get pi^(n-1) = (2B+2/3) / (2A+1/3)^n

Which would mean p^(n-1) is rational. This cannot be since pi is transcendental.

>>8537373

In five minutes.

File: 1481671344833.jpg (9KB, 200x151px)

9KB, 200x151px

>>8537373
ive marked it for you on the diagram

>>8538113
>five
No.

>minutes
No.