Spain Mathematical Olympiad Problem 2

The problem is fairly straightforward. Note that a 'shortest' path is comprised of exactly 3+4=7 edges. Now, let's say that both A and B move at a speed of 1m/s and that each edge is of length 1 metre . Then, the possible positions of A at different times (in seconds) will be as shown:

From this, and a similar consideration for B , it should be clear that they will meet at t=3.5 seconds at middle of the specific edges shown in the figure below:

I have labelled certain edges with the same symbol because they are symmetric with respect to each other. That is to say that whatever probability corresponds to the upper edge a also corresponds to the lower edge a . This is basically equivalent to switching the positions of A and B . This will help us simplify the problem. Now, consider the case where A and B meet at the mid-point of edge d . For this, A must reach the point (2,1) (with A as origin) through any of the dotted red lines (the continuous red path for example), from the figure below:

Thus, A has three possible paths to reach (2,1) Then, A must also choose the green path. At each junction, A has 2 choices, to go horizontal or to go vertical. Thus, the probability for this to occur is: ${ P }_{ A,d }={ 3(\frac { 1 }{ 2 } ) }^{ 3 }{ (\frac { 1 }{ 2 } ) }$ Using similar notation for B : ${ P }_{ B,d }={ 3(\frac { 1 }{ 2 } ) }^{ 3 }{ (\frac { 1 }{ 2 } ) }$ Thus, the probability of A and B to meet at d is: ${ P }_{ d }={ P }_{ A,d }.{ P }_{ B,d }={ 9(\frac { 1 }{ 2 } ) }^{ 8 }$ Note that the path that A and B choose after meeting does not affect this in any way. With similar reasoning, the other parts are easy: ${ P }_{ c }={ 9(\frac { 1 }{ 2 } ) }^{ 8 }$ ${ P }_{ b }={ 3(\frac { 1 }{ 2 } ) }^{ 8 }$ ${ P }_{ a }={ (\frac { 1 }{ 2 } ) }^{ 7 }$ Finally, the required probability is: $P={ 2P }_{ a }+{ 2P }_{ b }+{ 2P }_{ c }+{ P }_{ d }$ Substituting the values, we get: $P=\frac { 37 }{ 256 }$ Which gives the answer: $37+256=\boxed{293}$