Let ABC be the triangle and A the starting vertex. Let B* and C* be circles
of radius h/2 and centers B and C respectively. Clearly, to detect mines at
B and C, his path must contain points on circles B* and C*. WOLOG let his path
be the line segments AM and MN (where M and N are points on circles C* and B*
respectively). Let A', B', and C' be the midpoints of sides BC, CA, and AB
respectively. Let D be the intersection of CC' and A'B', E the intersection
of DB with circle B*, and P the intersection of MN with A'B'.
AD + DE + EB = AD + DB <= AP + PB <= AM + MP + PN + NB = AM + MN + NB
therefore,
AD + DE <= AM + MN.
Since M and N are arbitrary points on C* and B* respectively and D and E
are points on C* and B* respectively, path AD + DE is our shortest path.
Disks of radius h/2 centered at points A, D, and E cover triangle ABC.
AD + DE = AD + DB  EB = 2 AD  EB
= 2 sqrt(AC'^{2} + C'D^{2})  EB
= 2 sqrt(h^{2}/3 + h^{2}/4)  h/2
= h (2 sqrt(21)  3)/6
~= 1.02752523 h
