Rescue The Navigator

A downed navigator departs from a site marked by in a lake region shown in Figure 1and proceeds to the local town of Topologee Figure 2. Can he successfully cross each bridge in Topologee without repeating his path? He is rescued and brought to a site somewhere on Earth depicted in Figure 3 where he proceeds to walk one mile south, one mile west and one mile north ending back at the site. As a navigator, he is not astonished. All distance references are in nautical miles abbreviated nmi.

Figure 1. Lake region On land or water?

Figure 2. The town of Topologee.
Can the bridges of Topologee be crossed only once to complete a circuit?

Figure 3. Where on Earth is the navigator?
Where was the navigator when he left his site marked by in the lake region (keep the right side of the figure covered); could he cross each bridge only once without repeating the circuit; where was the final site?
a. on land; no; at the North Pole.
b. on land; yes; at the intersections of the Tropic of Cancer or Tropic of Capricorn and any meridian.
c. on a lake; yes; at the intersection of the equator and the Greenwich meridian
d. on a lake; no; at the North Pole and near the South Pole
The answer is d.
The reader becomes acquainted with the theories of topology in solving these three scenarios. In the first scenario the navigator is on a lake (a closed curve). One is in or on a closed curve if it takes an odd number of crossings to be out of the closed curve. In this case it takes three crossings.
In the second scenario the navigator must cross nine bridges without repeating a crossing. Use your finger or pencil to trace your path. It is not possible to complete this circuit without repeating a crossing. We will demonstrate this impossibility by examining the diagram that is depicted in Figure 4. We will use an example with seven bridges (equivalent for path logic). The key to solving this problem is attributed to Euler the father of topology. He created a network diagram to analyze the path logic. He used points (vertices) to convey land regions and arcs to convey bridges connecting the land regions. He discovered that if a vertex had an odd number of joining arcs (referred to as an odd vertex) it could be a starting or ending point of a path once reached that could permit only one departure and one return without repeating the path. This was first noted in the famous "Bridges of Konigsberg " paradox that Euler solved.
bridge configuration and its simplified equivalent to illustrate the path logic. We find that each of the four vertices have a convergence of three arcs, According to Euler's theorem, when a network has more than two vertices each with an odd number of arcs there is no continuous path from start to the completion of the circuit. A network that can be crossed by a singular path through each arc without repeating is an Euler path. When a network has two or less vertices with an odd number of arcs, it contains at least one Euler path.
Figure 4. A network with four vertices with odd number of joining arcs demonstrating no continuous path without repeating.
There are two cases where the navigator returns to his site by traveling the path of one mi south, one mi. west and one nmi. north. One site is at the North Pole shown in Figure 5 and the other very close to the South Pole is shown Figure 6.

In Figure 5 both AB and CA are arcs of meridians (or great circles) where an arc minute is equal to 1 nmi. Arc BC is along a small circle (rhumb line). The property of the Earth, which we assume as a perfect sphere (for simplicity) , that allows the return of the navigator to his origin is its sphericity. On a sphere the meridians converge at the poles. The small circles of latitude decrease in circumference as a function of the cosine of latitude as we approach the poles.

Figure 5. The site at the North Pole

In Figure 6 let point A be the other site. The navigator proceeds one nmi. south from A to point B along the arc of a meridian. He then proceeds one nmi. west along the small circle BCB returning to the arc of the meridian. He then proceeds one nmi. north along BA to return to his site. We know the circumference of the small circle BCB to be one nmi. We also know that the the small circle contains 360 degrees with each degree for this unknown latitude equal to 1 nmi./ 360 degrees. Therefore a degree at this unknown latitude is equal to 0.0027777 nmi. We now can determine the latitude for this small circle. The length of a degree of longitude difference is equal to the cos latitude X 60 nmi. Thus arc cos lat. = 0.0027777(1/60) = 0.0000462 with latitude = 89.997348 degrees S. Therefore the latitude of the site at A is one nmi north of point B. One nmi. is 1 arc min which is 1/ 60 of a degree or 0.0166666 degrees places the latitude of point A at 89.980681 degrees S. The site is then 0.019319 (90.000000 - 89.980681) degrees north of the South Pole which is 1.15914 nmi above the South Pole. Note that the site can be located anywhere along the upper small circle AD of latitude 89.980681degrees S.

Figure 6. The other site close to the South Pole (not to scale).

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