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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

CoopTheDupe  Mar 5, 2018

Best Answer 

 #1
avatar+2264 
+2

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.

TheXSquaredFactor  Mar 5, 2018
 #1
avatar+2264 
+2
Best Answer

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.

TheXSquaredFactor  Mar 5, 2018
 #2
avatar+91141 
0

Very nice, X2   !!!

 

 

 

cool cool cool

CPhill  Mar 6, 2018

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