+69 If 4 random points are placed on a sphere, what is the probability that the triangular prism made crosses over the center of the sphere?
I've been working on this for at least 20 mins and I'm stuck. Halp
+2446 This puzzle has a very elegant solution to it!
Actually, I think the best way to think about this one is to think two-dimensionally first! We have 3 points that we place on a circle!
I have picked arbitrary points for A, B, and C. I want to determine the probability when the three points form a triangle wherein the center is included. No matter what three points you pick, there are four options
Option 1: Point A, Point B and Point C are where they are in the diagram.
Option 2: Point A is moved to the other side of line, Point B stays and Point C stays
Option 3: Point A stays, Point B moves to the other side of line, and Point C stays
Option 4: Point A moves to the other side of the line, Point B moves to the other side of the line, and Point C stays
In these 4 cases, only one of the situations has a polygon that includes the center, so for a circle, the probability is \(\frac{1}{4}\).
Actually, let's go down to one dimension! Let's just worry about a segment.
There are two possibilities.
1) Point B is on the left of O
2) Point B is on the right of O
In this case, there is 1 case that results in the center being included, so the probability is \(\frac{1}{2}\).
If a one-dimensional case yielded \(\frac{1}{2^1}\) and the two-dimensional case yielded \(\frac{1}{2^2}\), then it would seem logical to me to assume that the three-dimensional case would be \(\frac{1}{2^3}=\frac{1}{8}\). In fact, if you try the cases again, you will see that this is the case.
+2446 This puzzle has a very elegant solution to it!
Actually, I think the best way to think about this one is to think two-dimensionally first! We have 3 points that we place on a circle!
I have picked arbitrary points for A, B, and C. I want to determine the probability when the three points form a triangle wherein the center is included. No matter what three points you pick, there are four options
Option 1: Point A, Point B and Point C are where they are in the diagram.
Option 2: Point A is moved to the other side of line, Point B stays and Point C stays
Option 3: Point A stays, Point B moves to the other side of line, and Point C stays
Option 4: Point A moves to the other side of the line, Point B moves to the other side of the line, and Point C stays
In these 4 cases, only one of the situations has a polygon that includes the center, so for a circle, the probability is \(\frac{1}{4}\).
Actually, let's go down to one dimension! Let's just worry about a segment.
There are two possibilities.
1) Point B is on the left of O
2) Point B is on the right of O
In this case, there is 1 case that results in the center being included, so the probability is \(\frac{1}{2}\).
If a one-dimensional case yielded \(\frac{1}{2^1}\) and the two-dimensional case yielded \(\frac{1}{2^2}\), then it would seem logical to me to assume that the three-dimensional case would be \(\frac{1}{2^3}=\frac{1}{8}\). In fact, if you try the cases again, you will see that this is the case.
