A Passing Fancy

[plainglazed]

Six (really tiny) racecars all set out in the same direction starting at the intersection of a figure eight shaped track. Each travels at a unique constant integer speed (greater than 0). The intersection of the figure eight is the only spot on the track where the racecars can pass one another. If the six cars can race in this manner for an indefinitely long time, what is the minimum speed of the fastest of the six cars?

[bhramarraj]

I'm glad you hold our feet to the fire, but try a 1:3 ratio, as bhramarraj suggests. I think it works and is the max ratio. I agree with you that more multiplication is doomed to failure

I disagree with your last statement, bhramarraj, although I agree with the max limit on their ratio being 3. But, if cars A, B, and C have speeds in ratio 1:2:3, it appears that each pair can drive around without crashing, even the 2:3 pair. Somehow we have to squeeze in the other cars between them (multiplied up to integer speeds, of course).

I am still confused, although I agree to your version that three cars can move on the 8 shaped track with the speeds in ratio 1:2:3, when it is allowed to cross other cars at X, irrespective of their directions at X.

But how will it be possible to insert fourth car on the track...? The speed should fall between the ratio 1:2:3. But fourth car should also reach X when slowest car reaches X, and that too with the speed lesser than the second faster or third fastest car. Then what will happen...?

I think the fastest car is definitely going to overtake fourth car, in between the track at a place other than X.

[CaptainEd]

Bhramarraj, look at the 4-car case 2:3:4:6.

Clearly the 2, 4, and 6 represent the 1:2:3, and we've inserted one between the first two.

When looking at a pair of cars, I like to divide their speeds by the smaller.

So, looking at the 3:4 case, I think of that as 1:1.33. This means, when the slower one has reached the intersection for the first time (1), the faster has gone one third of the way around the loop beyond. When the slower one has reached for the second time, the faster is now 2.66, and when the slower reaches for the third time, the faster is reaching it for the fourth time.

The other cases are familiar from the 1:2:3 situation.

As it turns out, to get more than 4 cars with compatible speeds, we have to abandon the 1:3 ratio.

Is this a useful explanation?

[CaptainEd]

Finally, I wrote a Python program, benefiting from the iterators over combinations.

(180, 200, 210, 216, 220, 225, 240)

there are none with the lowest speed < 1000 for eight cars

Edited September 9, 2011 by CaptainEd