Points on a sphere (part 2)
Suppose someone selects three points at random from the surface of a sphere. Then what is the the probability, in percent, that you can draw a great circle--the largest circle you can possibly draw on the sphere--such that all 3 of the points lie on the same hemisphere (divided by the great circle)?
Details and Assumptions:
-
Points on the boundary of a hemisphere are considered to lie on the hemisphere.
-
Enter, for example, percent like 50% as 50.
The answer is 100.
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
2 solutions
My answer to this question was based on the assumption (which I admit is not written in the text) that the great circle is already given, then you have the three random points. In this case if you name a hemisphere '0' and the other '1', you have 2 out of 8 cases when the three dots belong to the same hemisphere ('000' and '111'), which is 25 percent.
Log in to reply
I said 25% because their is a 50% chance two will land on the same hemisphere and adding another point means a 50% chance of it landing on the others hemisphere, which makes the probability 25%
Another way to word it is: Draw a great circle through any two of the three randomly placed points on the sphere. As the two points lie on the great circle dividing the sphere into two hemispheres, by the assumption above they belong to to both hemispheres concurrently. Therefore wherever on the sphere the third point is it will always belong to a hemisphere shared with any other two points on the sphere through which the hemisphere's bounding great circle runs.
for me it is not clear what it means points on the boundary lie on the hemisphere...which hemisphere? so i did neglect on purpose these extreme case ( laying on the boundary) and i got the same result as Ricardo. 25 percent. you can even try doing a tree diagram something like - consider naming the points A B and C. you can get: ABC all on the same hemis. A in one B on the other C on the other A in one B on the same C on the other A in one B on the other C on the same so possibility is 1/4 which is 25 % Patricia L
Log in to reply
It want you to choose the points first, then find the hemisphere.
Aren't hemispheres the two parts that result when the sphere is divided into 2 equal parts? Or can we place "the circle" that you talked about, which cuts the sphere, anywhere and not just in exact center of the sphere?
Log in to reply
A great circle on a sphere is a circle that exactly divides the sphere into two equal hemispheres. For example, on Earth any of the circles that go through both the North and South Geographic Poles (and thus form the lines of longitude) are great circles.
Where does question says we have to take points on surface only?
"Suppose someone selects three points at random from surface of a sphere"
Log in to reply
I don't see the distinction here. Can you elaborate on this?
Log in to reply
- its a sphere (3d object), it has volume (can be hollow or solid)
- it is not a circle (2d object)
Log in to reply
@Akshay Gupta – You're confused. The term for the surface of a perfectly round 3-dimensional object is a sphere .
The mathematical term for the entire object is a ball .
And for those unaware of that fact, the question even uses the term "surface."
Log in to reply
@Denton Young – yes u r ri8, should have focussed on term "surface" rather than sphere.
Because the condition "Points on the boundary of a hemisphere are considered to lie on the hemisphere." is given, and there are infinite different places you can draw the hemisphere boundary on the sphere; then there is always a hemisphere where all three points can reside no matter where the points are placed.
I had just assumed that "hemisphere" was fixed, like northern / southern hemisphere on the earth. (Probably the prior problem about the way the half-moon appears biased me to thinking about planets). When "25" didn't work, I was just baffled. If the problem were rephrased as "...what is the probability that there exists a hemisphere that includes all three points?", then I might have interpreted it correctly.
Log in to reply
Yeah, the previous format was a little bit ambiguous. I see that it's been fixed.
Assuming one defines the hemipheres after choosing the points.
Poorly worded question, I should've figured it out when the 25 was too easy.
Not sure where one would get 25%. Of any set of 3 points, with a pre-defined hemisphere, 2 of the points will lie in the same hemisphere. The 3rd point has a 50% chance of being in the same hemisphere as the other 2.
What am I missing?
Yes, I realize this is not the assumption that "Brilliant" wanted us to use.
This answer doesn't really explain anything. Because A and B then there is always C. But why? Why does C follow from A and B? There's a big gap in the middle of the explanation.
@Scott Kettler, I for one was going at it from the perspective of whether or not each of the two other points were on the same hemisphere as the first point, whichever one that happened to be. Or, to go the route of mapping out all possibilities with a fixed hemisphere, there are 2 out of 8 possibilities that satisfy the condition.
There is no assumption in which the three points could only be always in the same hemisphere. Because let say you traced a circle dividing the sphere in 2 hemispheres: then you can put one point in one hemisphere the other in the opposed and the third on the circle, which is considered in either one the hemispheres. So, 66% are in one hemisphere, never less than that. Also you can put all three i the same hemisphere meaning 100%. Now it doesn't matter how many circle you are tracing after you have chose the 3 point location: in one hemisphere you have 66.66666666% or 100% in the opposed one you have also 66.6666666% or 0.
Log in to reply
The question wants to select the points first, then draw a great circle.
Draw a great circle through two of the points. This divides the sphere into two hemispheres.
Case 1: The third point lies on this great circle. In this case, the three points lie on the same hemisphere by the assumption above.
Case 2: The third point does not lie on this great circle. In that case, it lies in the interior of one of the two hemispheres. Select this hemisphere.
So either way, we have found a hemisphere to which all 3 points belong.