I put the title in quotes because this problem is totally optional I did it for fun because I love physics problems. Here is the problem I was given:
If a dog and a rabbit both travel at the speed of light divided by the number of hairs on their respective bodies, but the dog sheds 23% of its fur whereas the rabbit only sheds 19%, yet the rabbit normally only has 0.394 times the number of hairs of the dog, and the dog swims twice as fast as the rabbit unless the water is below 40 deg F, in which case the dog travels 10 mph slower for every degree under 40, and they begin a race at 7:53 AM (Pacific Standard Time) on April 3 from Omaha to Tokyo following the Great Circle Route, who would win? Extra Credit: The visionary tetrametric runes of which melancholy diminutive Lithuanian poet/mechanic of the mid19th century anticipated the abovementioned contest?

As best as I can gather, these are the facts:
Number of hairs on the dog = $\displaystyle N$
Number of hairs on the rabbit = $\displaystyle 0.394(N)$
(in both cases, I attempted to get a concrete number by visiting a veterinary hospital and asking the receptionist if they had data regarding average dog weight with/without hair, from which I could weigh a single hair and get a "rough number" to use in calculations. Aside from a very strange look, I did not get much farther and thus decided to assign a value for N, discussed in the calculations)
Speed of light = $\displaystyle 3*10^8 m/s ~= 6.7*10^8 mi/h$
Distance from Omaha, Nebraska to Tokyo, Japan along the Great Circle Route (approximate): 14,500 mi (source:
WikiAnswers  What is the distance from the Tokyo Japan to Omaha NE )
Average surface temperature of the Pacific Ocean (approximate): 55 deg F (source:
Marine Weather : Weather Underground )
Because the average surface temperature is higher than 40 deg F, we can therefore assume that the dog will travel faster than the rabbit and, therefore, as long as the dog beats the rabbit to the shoreline, it will win the race. Otherwise, we will need to do calculations for the entire trip.
Now, finally, for some MATH!
I solved for the "average velocity" of the dog by integrating over its number of hairs (N) as follows, given that it's traveling at the speed of light divided by N:
$\displaystyle V(avg) = integral(C / N)dN$, which turns out to be $\displaystyle C(ln(N))$, evaluated from N to 0.77N, as it loses 23% of its fur.
With the same logic being applied to the rabbit, we find $\displaystyle C(ln(0.394N))$, evaluated from N to 0.81N (as it loses 19% of its fur).
To simplify calculations, I made $\displaystyle N = e$, or 2.71828182818 because it makes calculations simple and, logically, it shouldn't matter what value I use for N, so long as I use the same value in all instances where N appears.
From this, we can conclude that the dog's average speed is $\displaystyle [C  0.77C]$, or $\displaystyle 0.23C$, and the rabbit's average speed is $\displaystyle [C(0.394)  C(0.81(0.394))]$, or $\displaystyle ~0.075C$. From this, we can clearly see that the dog is traveling at a far greater rate of speed than the rabbit (and the speed of sound!) over land where N = e. Since we've already established that the dog will swim faster than the rabbit in the water, it would seem that the dog is the victor of this race. Provided he can swim that far.
As for the extra credit, I'll let the literary majors figure that out.
For those of you who are wondering "who made this problem?", it was printed on a graduation card that I received from a friend as a joke two years ago. Obviously, he underestimated my resolve to solve it as well as the card company. I had my professor doublecheck my math before the semester was over, and he suggested I doublecheck it with others as well. He also gave me a strange look.
Would someone be willing to doublecheck my logic and reasoning? I'm planning on sending this to the card company as an "I beat your impossible problem" kind of thing, but at the same time, I don't want one of their guys to come back and say, "You missed this little thing in your math, genius."
Thanks in advance!
P.S. I'm not a fan of working with English units, but they're what the card gave, so short of converting everything vs. working with the numbers they had, I chose the latter.