The Astute Ant

First we observe that to move from $A$ to $B$ , along the surface, the ant has to cross the plane shaded in the following figure.

Now, to cross the plane shown, it has to cross the line $POQ$ at some point in between $P$ and $Q$ on $POQ$ .

So we consider the following cases where the ant can cross the line $POQ$ at $X$ between $O$ and $Q$ , or at $O$ , or at $X'$ between $O$ and $P$ .

Case 1

The ant crosses the line $POQ$ at a point between $O$ and $Q$ , say $X$ .

Now we unfold the part of the block given by $CROKFJIEHQC$ so that we can find the straight line distance from $X$ to $B$ , and compare the distance from $A$ to $B$ along $X$ to that along $O$ .

( Note

When we fold the piece above back to it's original form, the ant can actually walk along the path $XB$ though it will not be a straight line, it will undergo a change in it's slope at the two points where it meets the edges. )

Now we see, that in triangle $AXB$ , $O$ is an internal point, and we know that $AX+XB>AO+OB$ , from Dogleg's Rule , so for any point $X$ between $O$ and $Q$ , the total path length is greater than if the ant goes along $O$ .

Case 2

The ant crosses the line $POQ$ at a point between $O$ and $P$ , say $X'$ .

Now we unfold the part of the block given by $CRGPFJIKOQC$ so that we can find the straight line distance from $X$ to $B$ , and compare the distance from $A$ to $B$ along $X'$ to that along $O$ .

Again we see in triangle $AX'B$ , $O$ being an internal point, $AX'+X'B>AO+OB$ by Dogleg's rule again.

So we have to cross the said plane at point $O$ only.

Calculation

We can check that the distance $AO+OB$ is same in both cases. $AO=\sqrt{1^2+1^2}=\sqrt{2}$ and $OB=\sqrt{3^2+1^2}=\sqrt{10}$ , so that we get $m=2$ and $n=10$ .

Note

The unfolding of the whole block could not be done at the very beginning because in that case the faces $OQHE$ and $OKIE$ would overlap and the path chalked out for the ant would not be along the surface of the block, it would be a path suspended in air.

$\text{As shown in the sketch, Shortest path is brown+yellow+brown. But yellow is in air. So to clear shortest path}\\ \text{on the surface, the ant has to go diagonally to the corner. This distance is } \sqrt2.\\ \text{Now it is in a position to go on the shortest surface paths. It has to travel one unit down and 2+1=3 }\\ \text{units so to say right. But these are legs of the right triangle. So the shortest is the hypotunuse } \sqrt{10}.\\ \therefore m+n=12.$