**Simple But Tricky Math Questions - 1 October**

Three cars are driving on a track that forms a perfect circle and is wide enough that multiple cars can pass anytime. The car that is leading in the race right now is driving at 55 MPH and the car that is trailing at the last is going at 45 MPH. The car that is in the middle is somewhere between these two speeds.

Right now, you can assume that there is a distance of x miles between the leading car and the middle car and x miles between the middle car and the last car and also, x is not equal to 0 or 1.

The cars maintain their speed till the leading car catches up with the last car and then every car stops. In this scenario, do you think of any point when the distance between any two pairs will again be x miles i.e. the pairs will be x distance apart at the same time ?

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If I understand the problem correctly (it's not defined very strictly) it's obvious that this can not happen for the very same distance x between 1st and 2nd , and 2nd and 3rd respectively, since x(i) is monotonically increasing ,since the speeds are constants and SpeedA>speedB>speedC. If the question is about the same random distance x, it is obvious that there will be a time moment when the order (along the direction of cars' motion) will be B C A and distance BC=distanceCA =x'

ReplyDeleteUnless the second car came up behind the third car with the first car in the rear. This would be the only way I can picture it - reverse order but don't know how to prove without going to trig and radians. Depending upon the track length this should be possible, but before the race shuts down? Anyway, I see it as C, B, A.

ReplyDeleteNot Possible

ReplyDeleteAs per the question, the middle car is running right now approximately at 50 MPH. Therefore, with time, the distance between them will keep increasing by x miles every hour for a certain time. After that, it will start decreasing between the leading car and the last car till they meet.

Therefore, you will find no point where the distance between the cars is x miles again.

this is impossible with tangential distance, but not with linear distance! remember the cars are on a circular track! It's trivial to show that there is necessarily one point at which the distance x will occur again, since all distances between x up to some y>x and then back to 0 must occur as the lead car circumnavigates the track. To make this problem simpler, view the rear car as standing still. Have fun with this problem as it's interesting when you discover the simple proof.

ReplyDelete