This problem appears in the finals of 13 years old chinese students
In a circunference of any radius, two particles start from the same point at the same time and start moving at the same constant speed without stopping. One goes through the diameter of the circunference, turning back once it reaches the end, and the other one follows the line of the circunference.
When will both particles meet?
Never? The arc length that the red bit has to travel to go from the top to the bottom is longer than the diameter of the circle ( the path of the blue bit) if they are both going at the same speed and start at the same time it will just never happen..
It was already solved on /sci
they will meet in the infinite
I'd say never, since there can be no integer ratio between the diameter and circumference.
Let me have my cheap win you fucking sook losers
they will eventually meet when n*diameter = n*pi*diameter
Lets make r=pi.
Hol up
How does this work?
>circunference
Then the circumfrence is 2pi^2 and that has no integer ratios
it doesn't
Well half a circumference would be pi*radius, those are the distances traveled where the outer point is at an intersection.
Each 2r traveled would imply a point where the inner point intersects.
So the solution would be at whatever point distance*pi*r is evenly divisible by distance*2*r into a whole number. More simply pi/2 must be a whole number which is impossible, because pi/2 isnt whole. They never meet.
Am i autism now?
No circumference=pi*diameter
if it's a simple yes or no, the answer is no... the circumfrence of a circle will always be longer than double the diameter... whoever thought this is a 'good' question are retarded
word it better otherwise they instantly meet
When the both started! But I assume that doesn't count so never. Let's say the blue particle touches the circle at times 0, x, 2x, 3x, 4x, etc. Then the red particle touches the possible collision points at 0, π/2*x, π*x, 3π/2 * x, etc.
Let's assume we have an integer a and b such that ax = bπ/2 * x. Dividing both sides by bx gives a/b = π/2. Thus a rational number equal an irrational number which is a contradiction. Therefore they obviously can't collide.
niggers tongue my anus
Its half the circumference because it meets the line twice per full circle, hence pi*r instead of pi*d
why did you make the a and b integers?
they have to be because the particles cannot be at that exact position a fractional amount of times
I mean, what do those letters mean? the time?
a is the number of times the blue particle touched a collision point, and b is how many times the red particle has touched a collision point. Obviously these are whole numbers.
Let x1 be the position of the particle transiting the circumference of the circle and let x2 be the position of the particle transiting the diameter of the circle. Furthermore, let (x,y)=(r,0) be the position at which the two particles are initially located and let z be the angle.
X1 is given by:
X1(t)= r + rsinz*dz/dt*t
X2 (t)= r + (r/pi)sinz*dz/dt*t
Let t' be the time at which the two particles meet. Then:
X1 (t')=X2 (t')
And:
rsinz*dz/dt*t=(r/pi)sinz*dz/dt*t which implies that r=(r/pi) which is impossible unless r=0 which, incidentally, wouldn't correspond to a circle.
>circunference
nice bait
2(pi)r = circumference
(pi)r = 1/2 circumference
d = 2r
assume 1d/sec x = seconds
((pi)r)/2rx = 2rx
r(pi) = (2rx)^2
r(pi) = (2r)^2(x^2)
r(pi)/(2r)^2 = x^2
x = sqrt(r(pi)/2r^2)
what does dz/dt mean?
The derivative of the angle z with respect to time.
isn't there a simpler way to solve it?
then a and b are the same
I'm sure there is, but this is the only proof I thought of.
I suppose you could simply argue that both particles move linearly with respect to velocity and since two linear functions intersect at only one point and the systems began with both functions intersecting, they don't intersect at all in the future.
What does 1d/sec x represent
>circunference