There are two large circular continents on a planet, each occupies 10% of total planet surface?

Question

The continents are 'randomly' drifitng due to planet tectonics.
('randomly' is in quotes because I do not know what it means.
Perhaps independently and not interacting, except when collide and bounce off)

What is probability that at given moment of time there is a pair of diametrally opposite points both on land?

Answer 1

Okay, let's first find out the size of round continent that covers 10% of the planet. Let planet radius = 1. Then the area of the spherical cap is 2πh, where h is the height of the cap. Thus we know that (1/10)4π = 2πh, or h = 1/5. Now, using some trigonometry, we find the area of the cap where the angular width is exactly twice that of the continents, by first figuring the height of the other:

2ArcCos(1-(1/5)) = ArcCos(1-H)

or H = 18/25, so that the area that we seek is:

A = 36π/25

Now the area that the center of one of the continents is free to roam without colliding the other is 4π - A, but if it enters the "antipodal continent" of area A on the other side, there exists at least a pair of antipodal points. So, the probability of this occuring is

A / (4π - A)

which comes out to 9/16, or about 56.25%

Answer 2

OK I'll bite.

The area of the "spherical circle" (in quotes because I don't know what the correct terminology is) is given by:
A = 2πR^2 * (1 - cosθ)
where 2θ = angle formed by the conical section at the center of the planet.

In this case, A = 4πR^2 / 10 = 2πR^2 / 5
So 1 - cosθ = 1/5
cosθ = 4/5
cos(2θ) = 7/25

Now let's fix one of the continents and allow the other one to drift randomly. The fixed planet occupies a section with an angle 2θ at the center, and if any part of the other continent is touching the corresponding region on the other side, there will be diametrically opposite points. So essentially the circumference of the drifting continent can roll along the circumference of the fixed region (really loose terminology here).

This means the center of the drifting region has an angle 4θ to play with. So the allowable area
= 2πR^2 * (1 - cos2θ)
= 2πR^2 * 18/25

The total possible area for the center of the drifting planet to occupy is the total surface - the same area around the fixed continent
= 4πR^2 - 2πR^2 * 18/25
= 2πR^2 ( 2 - 18/25)

So the required probability is
(18/25) / (2 - 18/25)
= 9/16

Answer 3

Random means that it is impossible to tell given a continent's current or past position what it's position will be at any point in the future.

There is a 100% probability that some point will be on land at a given moment. The question then becomes, "What is the probability that the diametrically opposite point is also on land." Luckily that is a simple answer. Since the land masses each occupy 10% of the planet's surface, the probability is 10%.

Answer 4

planet's radius = R
surface area = 4πR^2

area of 1 continent
= 0.4πR^2

possible area for the other continent
= 3^2 * 0.4πR^2
= 3.6πR^2

P(both end on land)
= (possible area for the other continent) / (planet's area - 1 continent's area)
= 3.6πR^2 / (4πR^2 - 0.4πR^2)
= 1

not so sure because i don't really know about spherical circle, not even their triangles.

edit
i think the 2 continents will always have a pair of diametrally opposite points both on land even when the 2 are touching/colliding with each other.
the diameter, or rather the "surface" distance of 2 farthest point on a continent (like when a car is to travel from 1 point on the island to the opposite side) is actually 2πR/4 = 0.5πR.
so they are, at some point on each island/continent, oppositing each other at all time.