MMM #32 Winner

mmmwin.jpg

The winner for this edition of MMM is Bhaskar Bhattacharya. Although the problem seemed quite simple, it was in fact a pretty complex problem! As you can already probably guess, the answer was NOT that the two plots of land are equal. You can view the problem statement here.

The Answer by Mark Eichenlaub

Land 2, with long N-S fences and short E-W fences, is larger. I’ll try to give intuitive and computational justification.

For most places on Earth the described plots are impossible. If the two E-W fences are different distances from the equator, they will subtend different azimuthal angles. But because they are connected by perfect N-S fences, they must have their endpoints on the same meridians, and so must subtend the same azimuthal angles. They can’t be different distances from the equator, so they must be the same distance. This means the equator runs dead center through the plot.

The qualitative question is then between wide, short plots and tall, thin ones. Imagine a wide, short plot (long E-W fences and short N-S fences). Exaggerate the situation, imagining is wraps a quarter around the Earth. If we grab the N-S fences and pull them even further apart still, we make the enclosure bigger because those N-S fences sweep out additional area as we pull them. The E-W fences are basically unaffected, though. They get longer, but only wind up running a little further along a line of latitude.

On the other hand, imagine a tall, thin plot (long N-S fences and short E-W fences). Exaggerate again, imagining the plot wraps a quarter around the globe. As we grab the short E-W and pull them further apart, they sweep out area as they’re pulled, exactly as before. But there’s a new effect. They also subtend a larger azimuthal angle as they go higher. In order to run perfectly N-S, the long fences develop a wider and wider bulge at the equator. In this case the N-S fences also sweep out new area. By the time we draw the E-W fences out near the poles, the plot covers most of the Earth. Compare this to the counter situation. When we draw the short wide plot most way around the equator, it still just covers one small band that’s a tiny fraction of the globe.

This argument shows that if we start with a 10mile*10mile plot, pulling the E-W fences apart towards the poles adds more area than pulling the N-S fences around the equator, so land 2 with long N-S fences is bigger.

Now let’s calculate the area of each plot. For land 1 (short wide plot), the N-S fences are ten miles long. The zenith angle they subtend is then ten miles divided by the radius of the Earth, or 1/395 radians. Half is above the equator, and half below, so the plot goes from pi/2-1/790 to pi/2+1/790 radians in the zenith angle. In the azimuthal angle, the E-W fences extend 20 miles. The angle they subtend is their length divided by the distance around the Earth 1/790 radians above the equator. This is 20miles/[cos(1/790)*3950miles]. We can then take the azimuthal angle to go from zero to this value.

The differential element of area on the surface of a sphere is r^2*sin(theta)*d(theta)d(phi), with ‘theta’ the zenith angle and ‘phi’ the azimuthal angle. Integrating this expression over the limits of the angles found above yields the area. Doing the computation, I found an area of 158,000*tan(1/790)mi^2. This is 200.000107 mi^2, which is pretty much what you’d expect for a rectangle 10mi*20mi.

Doing it all over for the tall, thin plot is mostly the same. Now the N-S fences subtend a zenith angle of 2/395 radians, and the E-W fences an angle of 1/[cos(1/395)*395] radians. Putting in the new limits of integration, I get an area of 79000*tan(1/395)mi^2. This is 200.000427 mi^2. Land 2 is larger by 3.2*10^-4 mi^2, or about 9000 square feet. They could flood the extra space to make a nice big pond and swim to cool off from the tropical sun.

Bhaskar’s MATLAB Code

Bhaskar also provided some MATLAB code to get the area of each plot:

function calcrect()
    S = 10; L = 20;
    calcarea(S,L)
    S = 20; L = 10;
    calcarea(S,L)
 
    function calcarea(S,L)
        % S is north-south and L is east-west
        R = 3950;
        Aring = 4*pi*(R^2)*sin(S/(2*R));
        Langle = L/(R*cos(S/(2*R)));
        Arect = Aring*Langle/(2*pi)
    end
 
end

Running it produces –

>> calcrect

Arect =

2.000001068205567e+002

Arect =

2.000004272830481e+002