A different kind of pursuit .....
Consider two concentric circles, radii 1 and 2 respectively, centered at the origin. Particles are situated on each of the circles, at ( 1 , 0 ) and ( 2 , 0 ) respectively, and then simultaneously begin to move counterclockwise around their respective circles, both at a rate of 1 unit per second.
The time that has elapsed when the line joining the two particles is tangent to the smaller circle for the first time is b a π seconds, where a and b are positive coprime integers. Find a + b .
The answer is 5.
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2 solutions
An interesting connection from Astronomy:
If the two 'particles' are two planets (of course they will never move at same linear speed), then observed from planet A, planet B seems to go forth, hesitate and stop and go little backward before resuming its onward journey. This is called the retrograde motion. The start and stop of the retrograde phase is marked by this 'tangency' configuration.
Here is a video showing this mechanism:
https://www.youtube.com/watch?v=j-iZmuqexEQ
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Great video, Ujjwal; thanks for posting the link. This was in fact what I was visualizing when I posted this question. :)
(P.S.. There is one small typo in your solution. Near the end you have "Put OB = 1" when it should be "Put OB = 2".)
The equations of motion for the particle on the radius 1 circle are
x = cos ( t ) , y = sin ( t ) ,
and for the particle on the radius 2 circle they are
x = 2 ∗ cos ( 2 t ) , y = 2 ∗ sin ( 2 t ) .
The tangent line to the smaller circle at time t has slope
d x d y = d t d x d t d y = − sin ( t ) cos ( t ) .
The line joining the two particles at time t has slope
2 ∗ cos ( 2 t ) − cos ( t ) 2 ∗ sin ( 2 t ) − sin ( t ) .
Thus we are looking for the least positive t such that these two slopes are equal. This is the case when
2 ∗ sin ( t ) ∗ sin ( 2 t ) ∗ ( 1 − cos ( 2 t ) ) = cos ( t ) ∗ ( cos ( t ) − 2 ∗ cos ( 2 t ) ) .
Using the identities
sin ( t ) = 2 ∗ s i n ( 2 t ) ∗ cos ( 2 t ) and
cos ( t ) = 2 ∗ cos 2 ( 2 t ) − 1
we can simplify to find that cos ( 2 t ) = 2 1 , and so the desired time is t = 3 2 π .
Thus a = 2 , b = 3 and a + b = 5 .
I did it by setting the distance between the two points equal to 3 .
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Yes, that does work out nicely; no calculus necessary. When I was writing my solution I had a sense I was over-thinking it. However, it's never hurts to have a variety of approaches presented. :)
A, B = particles on circles of radius 1 and 2 respectively,
Since the radii are 1 and 2, but the linear speeds are the same, the angular speeds (angles traversed in unit time) will be 2:1
Hence angle covered by A will be 2 x angle covered by B
To meet the tangency condition, angle OAB = 90°
Applying sine rule to triangle OAB:
OB/sin90° = AB/sin(alfa)
Put OB = 2, AB = sqrt(3)
Alfa = arcsin(sqrt(3)/2) = 60° giving traversed distance = 2 x pi/3 on the outer circle. Since velocity is unity this also gives the time.