MMM #34 Winner, Sander Land!!!

MMM #32 Winner

The winner for the last MMM at Blinkdagger is Sander Land. A lot of you mistook this problem for the classic bridge/flashlight problem, but it was actually a little more complicated than that one. You can view the problem statement here. As mentioned previously, this is the last MMM to be conducted here at blinkdagger. We appreciate everyone for participating in this contest. It was a lot of fun while it lasted and we hope that there will be many new puzzles to solve over at wildaboutmath.com! Good luck to all!

The Answer by Sander Land

First, simply consider a general strategy for crossing the bridge.
It is clear that
- both Daniel and Quan need to use the bike at some point
- Sol needs to be the one to transfer the bike from one of them to the other.

This leaves a strategy like:
- Daniel takes the bike from the start to point x1, dismounting, and
continues to walk.
- Sol walks to point x1, and takes the bike back to point x2, and
walks to the end point.
- Quan walks to point x2, picks up the bike, and bikes to the end.

Taking into account the fact that any solution which has
- one of them arrive earlier than the others or
- one of them waiting at any time (including t=0)
can be improved, and normalizing the length of the bridge to 1, the
equations for this are:

D: x1 + 15*(1-x1) = T
Q: 8*x2 + (1-x2) = T
S: 3 + 4*(x1-x2) = T (i.e. walking the entire length, and
both walking and biking x1-x2)

Which matlab can help us solve:
>> D=15; Q=8; S=3;
>> R=[-D -1 -S]‘
>> A=[-(D-1) 0 -1 ; 0 (Q-1) -1 ; (S+1) -(S+1) -1]
>> A\R
0.7692 => x1 is at 77% of the bridge length
0.4615 => x2 is at 46% of the bridge length
4.2308 => They can cross the bridge in about 4 minutes and 14 seconds.

The other possible strategy (Quan starts on the bike) can also be
tried by simply switching Daniel and Quan’s velocities. This results
in the same shortest time.
>> D=8; Q=15; S=3;
0.5385
0.2308
4.2308

5 Responses to “MMM #34 Winner, Sander Land!!!”

on 19 Jun 2009 at 12:54 pm 1Andrew Johnson

i think the winner was incorect with his assumption that only sol can deliver the bike to the person in back. this leaves the bike waiting for sol, which causes an inefficiency. i solved the transfer to 3.8 minutes

on 19 Jun 2009 at 12:59 pm 2Andrew Johnson

i think the winner was incorrect because of his assumption that only sol can deliver the bike between the other 2. this leaves the bike waiting for sol, which causes an inefficiency. calculated the time to be 3.8 minutes for all three to cross assuming all three share bike delivery duty and the bike is never left waiting

on 22 Jun 2009 at 8:37 am 3MMM #35: Fibonacci Fun » Fun Math Blog

[...] has announced the winner and the solution for MMM #34. Now, it’s time for MMM [...]

on 22 Jun 2009 at 1:11 pm 4Nate

I also believe the winner is incorrect. I solved the puzzle with the total time of just under 3 minutes and 28 seconds

The reason why the winner is incorrect is because the point x2 is not optimal. The point x2 should be determined by where Sol and Quan meet. If this new x2 position is used a third bike swap takes place at x3 such that all three cross the bridge at the same time.

This processes can be described as:
D bikes to x and starts walking
S walks to x and bikes back until he meets Q
Q bikes to x3 and starts walking
S walks to x3 and bikes rest of the way

x and x3 are set such that D, S, and Q all cross the bridge at the same time

In addition, it should be noted that this process can be repeated any number of times across the bridge (val) so that in the limit all three move as the same speed of 548/1899 assuming unit bridge length.

syms x x3 val
solve('(x3-4*x/9)+(val-x3)*1/8 = (x3-4*x/9)*1/3+(val-x3)',x3)
solve('(val-x)*15+x = 32*x/9 + (64/333*x+21/37*val - 4*x/9) + (val-64/333*x+21/37*val)*1/8',x)
 
Totaltime = ((val-903/1096*val)*15+903/1096*val)/val

If anyone finds a problem with this or finds a more optimal solution please let us know.

on 23 Jun 2009 at 8:14 am 5Clueless

Hi Nate,

Following your methodology, I arrive at an answer of 4.38 minutes, which is longer than the winner’s answer.

In your first equation, should you have (x3-4x/9)+(val -x3)*8 =(x3-4x/9)*3+(val-x3) ?

The speeds are 1/8 and 1/3, and I assume you are trying to equate times, which is distance/speed??

Clueless

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